Research ArticleStationary Gas Networks with Compressor Control andRandom Loads Optimization with Probabilistic Constraints
Martin Gugat and Michael Schuster
Lehrstuhl Angewandte Mathematik 2 Department Mathematik Friedrich-Alexander Universitat Erlangen-Nurnberg (FAU)Cauerstr 11 91058 Erlangen Germany
Correspondence should be addressed to Michael Schuster michischusterfaude
Received 27 April 2018 Revised 29 July 2018 Accepted 4 September 2018 Published 26 September 2018
Guest Editor Gerhard-WilhelmWeber
Copyright copy 2018 Martin Gugat and Michael Schuster This is an open access article distributed under the Creative CommonsAttribution License which permits unrestricted use distribution and reproduction in any medium provided the original work isproperly cited
We introduce a stationary model for gas flow based on simplified isothermal Euler equations in a non-cycled pipeline networkEspecially the problem of the feasibility of a random load vector is analyzed Feasibility in this context means the existence of aflow vector meeting these loads which satisfies the physical conservation laws with box constraints for the pressure An importantaspect of the model is the support of compressor stations which counteract the pressure loss caused by friction in the pipes Thenetwork is assumed to have only one influx node all other nodes are efflux nodes With these assumptions the set of feasible loadscan be characterized analytically In addition we show the existence of optimal solutions for some optimization problems withprobabilistic constraints A numerical example based on real data completes this paper
1 Introduction
In this paper we present a way to model stationary gasnetworks with random loads Natural gas as energy sourcehas been popular for decades Because of the nuclear powerphase-out and aim to leave energy gained by coal behindnatural gas as energy source is more current than ever So inthis paper we study the problem of gas transport in a pipelinenetworkmathematically The aim of this paper is to get resultsin optimization problems with probabilistic constraints usingthe example of gas networks
There are several studies about the mathematical problemof gas transport A gas transport network in general can bemodeled as a system of hyperbolic balance laws like egthe isothermal Euler equations In [1] one can find a greatoverview about existing models and their application areas Anetwork system is modeled as a graph with nodes end edgesIt can be extended by different elements like compressorstations valves and resistors In [2] the authors describes thefunctionality of these elements and how they can be includedto the model
We assume an active stationary state for our model thatmeans we use compressor stations as controllable elements in
a state of time-independent flows and pressures Ourmodel isbased on the Weymouth-Equation (see [3]) a simplificationof the isothermal Euler equation (see eg [4 5]) Further[3] also studies the isothermal Euler equations for real gasWe assume our gas to be ideal The optimal control problemin pipeline networks has been studied in eg [4 6] Inour work we assume the loads to be random so this leadsto optimization problems with probabilistic constraints (see[7])
In the next section we introduce our gas network modelwhich is based on themodel in [8]This is a passive onewhichwe extend by compressor stations So our model containsinner control variables which makes our model an activeone These controls will be one goal of the optimization inSection 4
In Section 3 we characterize the set of feasible loadsInstead of searching a flow and a pressure vector (based ona given load vector) which fulfills the physical balance lawswe look for these vectors fulfilling a system of inequalitieswhich depend on bounds for the pressure In addition weassume the load vector to be random So we adapt the systemto the spheric-radial decomposition which is our central toolto handle the uncertainty in the loads
HindawiMathematical Problems in EngineeringVolume 2018 Article ID 7984079 17 pageshttpsdoiorg10115520187984079
2 Mathematical Problems in Engineering
In Section 4 we consider some optimization problemswith and without probabilistic constraints The existence ofoptimal solutions is based on the analysis we did in Section 3This should built a base for further works about optimalcontrol problems with probabilistic constraints using theexample of gas networks
In Section 5 we present three numerical examples Thefirst is a minimal graph without inner control which showsthe idea of the spheric-radial decomposition The second oneis a minimal graph with inner control and the third one isbased on real data for which we present real values for anetwork
In the last section we present a few ideas of extending thispaper One main aspect is the turnpike-theory
2 Mathematical Modeling
We start with the introduction of the model that is also usedin [8] This model does not include compressor stations Weextend this model to include compressor stations and wediscuss the network analysisThen themain tool the so calledspheric-radial decomposition is introduced to handle thestochastic uncertainty
21 Model Description We consider a connected directedgraph 119866 = (V+E) which represents a pipeline gas transportnetwork We assume that there are no cycles in the graphso the network is a tree We set |V+| fl 119899 + 1 and |E| fl119898 = 119899 (119899119898 isin N) In our model every node is either ainflux node (gas enters the network) or an efflux node (gasleaves the network) An edge can either be a flux edge so thepressure decreases along the edge or a compressor edge sothe pressure increases along the edge We define E119865 as theset of all flux edges and E119862 as the set of compressor edges itfollows E = E119865 cupE119862 with E119865 cap E119862 = 0
Let 119887+ isin R119899+1 with 1119879119899+1119887+ = 0 denote the balanced load
vector and assume 119887119894 le 0 for nodes with gas influx and 119887119894 ge 0for nodes with gas efflux (119894 isin 0 119899) The vector 1119899+1
denotes the vector of all ones in the dimension 119899+1 Further-more a vector 119902 isin R119898 describes the flows through the edgesresp through the pipesThe pressures at the nodes are definedin a vector 119901+ isin R119899+1 Let pressure bounds 119901+119898119894119899 119901+119898119886119909 isinR119899+1 be given For the pressure we consider the constraints119901+ isin [119901+119898119894119899 119901+119898119886119909] For further modeling we need the fol-lowing definition
Definition 1 Consider the graph 119866 = (V+E)(i) ℎ(119890) denotes the head node of an edge 119890 and 119891(119890)
denotes the feet of an edge 119890 for all 119890 isin E(ii) 1198640(V) fl 119890 isin E | ℎ(119890) = V or 119891(119890) = V denotes all
edges which are connected to node V(iii) The matrix 119860+
119894119895 isin R119899+1times119898 119860+119894119895 fl 120590(V119894 119890119895) with
120590 (V 119890) fl minus1 if 119891 (119890) = V and 119890 isin 1198640 (V)1 if ℎ (119890) = V and 119890 isin 1198640 (V)0 if 119890 notin 1198640 (V) (1)
is called the incidence matrix of the graph 119866
(iv) For V isin V+ is Π(V) fl V cup⋃119896isinV+|exist119890isinE119891(119890)=119896andℎ(119890)=VΠ(119896) the (unique) directedpath from the root to V
(v) For V 119908 isin V+ is Π119908(V) fl Π(V) Π(119908) the (unique)directed path from119908 to V
Note that definition (iv) only makes sense since weassumed the graph to be tree-structured so there are nocycles and the union is finite The model is based upon theconservation equations of mass and momentum The massequation for the nodes is formulated for every node by
sum119890isin1198640(V)
119902119890 = 119887+V forallV isin V+ (2)
This is equivalent to Kirchhoff rsquos first law (see [9]) and byusing the definition of the incidence matrix the equation formass conservation for the whole graph is
119860+119902 = 119887+ resp 119860119902 = 119887 with
119860+ = [1198600119860 ] isin R1times119898
isin R119899times119898and 119887+ = [1198870119887 ] isin Risin R119899
(3)
As balance laws we use the isothermal Euler equations(see [3 4 10]) for a horizontal pipe so the flow through everyflux edge is modeled by
120588119890119905 + 119902119890119909 = 0
119902119890119905 + ((119902119890)2120588119890+ 1198862120588119890)
119909
= minus 1205821198902119863119890
10038161003816100381610038161199021198901003816100381610038161003816 119902119890120588119890
forall119890 isin E119865
(4)
Here 120588 is the gas density 119886 is the speed of sound 120582 is a(constant) friction coefficient and 119863 is the diameter of thepipe
We assume first that the network is in a steady state sothe time derivatives are equal to zero And second we assumethe gas flow to be slow so the coefficient (119902119890119886)2 is negligibleThese assumptions simplify the equations enormously Withthe ideal gas equation 119901 = 120588119877119878119879 (see [9]) we get adirect dependency between pressure and density Solving thissimplified equation leads us to the so called Weymouth-Equation (see [3])
1199012119891(119890) minus 1199012
ℎ(119890) = 120601119890 10038161003816100381610038161199021198901003816100381610038161003816 119902119890 with 120601119890 = (119877119878119879)2 120582119890119863119890
1198711198901198862forall119890 isin E119865 (5)
Here 119877119878 is the specific gas constant 119879 is the temperature ofthe gas and 119871119890 is the length of the pipe 119890 This equation showsthe pressure drop along the pipes (see Figure 1) Some articlesconsider the isothermal Euler equations in gas transportwithout these simplifications eg [5 10]
Compressor stations counteract the pressure drop alongthe pipes These stations are modeled as pipes without
Mathematical Problems in Engineering 3
times104
times106
15
2
25
3
35
4
45
5
55
6
pres
sure
in P
a
1 2 3 4 5 6 7 80x in m
Figure 1 Pressure drop along a pipe for 119863 = 0762119898 119879 = 28915119870119877119878 = 50027119869119896119892119870 120582 = 0005 119886 = 42917119898119904 119902 = 7951198961198921198982119904 and119901(0) = 60 sdot 105119875119886frictionWe use the followingmodel equation (see [1] section747 [11] section 23)
119888120574 ((119901119900119906119905119901119894119899
)120574 minus 1) = (6)
Here 119901119900119906119905 and 119901119894119899 are the pressures at the end resp at thebeginning of one pipe 119888 and 120574 are constants and whichis our control is the specific change in adiabatic enthalpyMore information about the parameters can be found in[11] In addition [2] contains more information about thefunctionality of compressor stations and a deduction of theequation A short transformation of the equation leads toour representation of the pressure rise along the compressoredges
(119901ℎ(119890)119901119891(119890)
)2 = 119906119890 with 119906119890 = (120574119888 + 1)2120574 forall119890 isin E119862 (7)
It holds 119906119890 ge 1 where 119906119890 = 1 means that the compressorstation is switched off (see Figure 2)
Now we define the set of feasible loads We say that avector 119887+ isin R119899+1 is feasible if (3) (5) and (7) are fulfilledSo the feasible set (here called ) is defined as follows
= 119887+ isin R119899+1 | 1119879
119899+1119887+ = 0 and exist (119901+ 119902) isin R119899+1
times R119898 119901+
isin [119901+119898119894119899 119901+119898119886119909] and (3) (5) (7) are fulfilled (8)
22 Stochastic Uncertainty in the Loads In reality the futuredemand for gas can never be known exactly a priori Itdepends on various physical effects like eg the temperatureof the environment Also personal reasons can be responsiblefor a higher or lower use of gas Thus the load vector is
times105
times10755
05
1
15
2
25
3
35
4
45
5
pres
sure
in P
a
15 2 25 3 35 4 45 51u in Jkg
Figure 2 Compressor property for 119888 = 1446526119869119896119892 120574 = 0225and 119901119894119899 = 40 sdot 105119875119886considerd to be a random vector Many papers study howthe distribution functions of the load vector can be identifiedeg [12] This article also describes why a multivariate Gaus-sian distribution is a good choice for random gas demandSo we use some random vector 120585(120596) with 120585 sim N(120583 Σ) fora mean vector 120583 and a positive definite covariance matrix Σon a suitable probability space (ΩAP) and our aim is tocalculate the following probability
P 120596 isin Ω | 120585 (120596) isin 119872 (9)
for a suitable set119872In our case we assume that the tree-structured graph
has only one influx node which gets the number zeroall remaining nodes are efflux nodes Then the graph isnumerated with breadth-first search or depth-first search andevery edge 119890 isin E gets the number maxℎ(119890) 119891(119890) Our set119872is defined as follows
119872 fl 119887 isin R119899ge0 | (minus1119879
119899119887 119887) isin (10)
There are many studies for special sets 119872 eg for poly-hedral sets (see [13]) but our set is very general So we need avery general method for calculating the probability (9) Ourmethod of choice is the so called spheric-radial decompositionIt became popular for dealing with probabilistic constraints(eg see [8 14ndash19])
Theorem 2 (spheric-radial decomposition) Let 120585 sim N(0 119877)be the 119899-dimensional standard Gaussian distribution with zeromean and positive definite correlation matrix 119877 Then for anyBorel measurable subset119872 sube R119899 it holds that
P (120585 isin 119872) = intS119899minus1
120583120594 119903 ge 0 | 119903119871V isin 119872119889120583120578 (V) (11)
where S119899minus1 is the (119899 minus 1)-dimensional sphere in R119899 120583120578 is theuniform distribution on S119899minus1 120583120594 denotes the 120594-distribution
4 Mathematical Problems in Engineering
with 119899 degrees of freedom and 119871 is such that 119877 = 119871119871119879 (egCholesky decomposition)
Thedifference with our problem is that our load vector isnormal distributed with mean value 120583 and positive definitecovariance Σ We want to use the result of Theorem 2 togeneral Gaussian distributions (see [8]) Therefore we set120585lowast fl 119863minus1 (120585 minus 120583) sim N (0 119877) with
119863 fl diag (radicΣ119894119894)119894=1119899
and 119877 fl 119863minus1Σ119863minus1 (12)
and observe that P(120585 isin 119872) = P(120585lowast isin 119863minus1(119872 minus 120583)) So ouraim is to compute the following probability
P (120585lowast isin 119863minus1 (119872 minus 120583)) (13)
So the calculation of the probability (9) for a very generalset 119872 is equivalent to compute the integral in Theorem 2which is not easy to compute analytically too But the integralcan be computed numerically in a easy and nearly exactwayTherefore we formulate the following algorithm (see [8]Algorithm 4)
Algorithm 3 Let 120585 sim N(120583 Σ) and 119871 such that 119871119871119879 = Σ(1) Sample 119873 points V1 V119873 uniformly distributed
on the sphere S119899minus1(2) Compute the one-dimensional sets 119872119894 fl 119903 ge 0 |119903119871V119894 + 120583 isin 119872 for 119894 = 1 119873(3) Set P(120585 isin 119872) asymp (1119873)sum119873
119894=1 120583120594(119872119894)The first step is quite easy There are many possibilities
to sample a set of uniformly distributed points on the sphereS119899minus1 eg Monte-Carlo sampling or the randn-generator inMATLAB Normalizing these points leads us to the samplingwe are looking for Alternatively quasi Monte-Carlo methodscan be used In [8] the authors use a quasi Monte-Carlomethod for their computations and they show its advantagesin Figure 6
Themain challenge is to handle the one-dimensional setsof the second step If the set119872 is convex these sets are closedintervals but 119872 need not to be convex The aim is to get arepresentation of 119872119894 (119894 = 1 119873) as a union of disjointintervals With this the third step can be done efficiently Thenext section deals with this representation in detail
The third step is the approximation of the integral Thevalues 120583120594(119872119894) can be calculated easily with the distributionfunction F120594 of the distribution (see [20] for details of thatdistribution) With the representation 119872119894 = ⋃119905
119895=1[119886119895 119886119895] the120594-distribution can be calculated the following
120583120594 (119872119894) = 119905sum119895=1
F120594 (119886119895) minusF120594 (119886119895) (14)
This leads to the wanted probability The 120594-distributionfunction is given by the following (see [20] p 132)
F120594 (119909) = 121198732Γ (1198732) int119909
0V1198732minus1119890minusV2119889V (15)
2 4 6 8 10 12 14 16 18 200x
0010203040506070809
1
F(x)
Figure 3 Distribution function of the Chi-square distribution for 3(blue) 5 (red) 8 (green) and 12 (purple) degrees of freedom
with Γ(12) = 120587 Γ(1) = 1 and Γ(119903 + 1) = 119903Γ(119903) for all 119903 isin R+Note that there are very precise numerical approximationsfor the distribution function eg the chi2cdf -function inMATLAB (see Figure 3)
3 Explicit Characterization of the Gas Flow
In this section we want to characterize the feasible set to useAlgorithm 3 The aim is to get a representation of the one-dimensional sets of step (2) as a union of disjoint intervalsTherefore we need another representation of the feasible setas a system of inequalitiesThis is an idea from [8]Theorem 1In the proof the authors use a representation of the pressureloss for the full network graph
(119860+)119879 (119901+)2 = minusΦ 10038161003816100381610038161199021003816100381610038161003816 119902 (16)
which is equivalent to (5) if and only if E119862 = 0 In this paperwe also consider the case E119862 = 0 The matrix Φ is diagonalwith the values 120601119890 as defined in (5) at its diagonal Thisrepresentation is not possible yet because we extended ourmodel by compressor edges so this equation does not holdfor all edges Sowe have to find away to use a similar equationlike (16) to state a suitable theorem The idea that allowsto include compressor stations is to remove compressoredges from the graph and replace them by suitable algebraicconditions First we consider the equation of masssum
119890isin1198640(V)119902119890 = 119887+V lArrrArr
sum119890isin1198640(V)capE119865
119902119890 = 119887+V minus sum119890isin1198640(V)capE119862
119902119890 (17)
We want to treat the flows through the compressor edges asloads sowehave to adapt the load vectorThenew load vector+ is defined as follows+V
fl
119887+V + 119902119890 forall119890 isin 1198640 (V) with 119890 isin E119862 and 119891 (119890) = V119887+V minus 119902119890 forall119890 isin 1198640 (V) with 119890 isin E119862 and ℎ (119890) = V119887+V else
(18)
The resulting graph 119866 = (V+E119865) is no longer connected(scheme is shown in Figure 4)
Mathematical Problems in Engineering 5
p0
p1
p2
p3
p4
p5
p6
p7
p8
p9
q1
q2
q4
q6
q7
q8
q9
b0
b1
b2
b3
b4
b5
b6
b7
b8
b9
p0
p1
p2
p3
p4
p5
p6
p7
p8
p9
q1
q2
q3 q3
q4
q5 q5
q6
q7
q8
q9
b0
b1
b2
b3
b4
b5
b6
b7
b8
b9q5 isin ℰC
q3 isin ℰC
Figure 4 Network graph with two compressor edges (left) and new not connected network graph with three connected subgraphs afterremoving the compressor edges (right)
We set |E119865| fl 1198981 and |E119862| fl 1198982 with 1198981 + 1198982 = 119898so 119866 consists of 1198982 + 1 subgraphs which are all connectedNote that 119902 is a vector in1198771198981 nowWe determine that119866119894 is the
119894-th self-connected subgraph of119866 (depends on numbering ofthe graph) Analogously 119901+
119894119895 is the 119895-th component of the 119894-thsubgraph The terms 119902119894119895 and 119860 119894119895119896 are defined similarly Thefeasible set for 119866 is then
=+ isin R119899+1 |
1119879119899+1+ = 0 andexist (119901+ 119902) isin R119899+1
ge0 times R119898 with 119901+ isin [119901+119898119894119899 119901+119898119886119909] 119860 119894119902119894 = 119894 forall119894 = 1 1198982 + 1(119860+119894 )119879 (119901+
119894 )2 = minusΦ11989410038161003816100381610038161199021198941003816100381610038161003816 ∘ 119902119894 forall119894 = 1 1198982 + 1
(119901ℎ(119890)119901119891(119890)
)2 = 119906119890 forall119890 isin E119862
(19)
The last equation in is the same as in (8) that modelsthe pressure rise caused by the compressor With this we canpresent the equation for pressure loss in the form of (16) forevery subgraph Before we state a theorem for characterizingthe feasible set we show the equivalence of and
Lemma4 Let + be as defined in (18)The feasible sets definedin (8) and (19) are equivalent in the following sense
(i) For all 119887+ isin there is + isin (ii) For all + isin there is 119887+ isin (20)
Proof (i) Consider a vector 119887+ isin It holds 1119879119899+1119887+ = 0 and
because of (18) it follows directly 1119879119899+1+ = 0 After removing
the compressor edges from119860+119902 = 119887+ it follows with (17) that119860+119894 119902119894 = +119894 which is equivalent to 119860 119894119902119894 = 119894 The pressure loss
is fulfilled for all flux edges in and because the flow vector119902 doesnrsquot change in the relevant coefficients while removingthe compressor edges the pressure loss is fulfilled in tooThe compressor property is independent of 119887+ and 119902 so it alsoholds From this it follows + isin 119872
(ii) The other way around is quite similar We consider avector + isin From 1119879
119899+1+ = 0 it follows with (18) that1119879119899+1119887+ = 0 The equation for mass conservation is fulfilled
in every subgraph So with (17) it follows 119860+119902 = 119887+ whichimplies the mass conservation in119872 The pressure loss holds
in all self-connected parts of the graph too and because therelevant coefficients of 119902 doesnrsquot change while adding thecompressor edges the pressure loss is fulfilled in119872 too Lastas said before the compressor property is independent of +and 119902 it is also fulfilled in119872 This completes the proof
Because our network graph is a tree the mass conserva-tion is fulfilledwith a full rank incidencematrix119860of size 119899times119899so 119902 can be computed explicit by the following
119902 = 119860minus1119887 (21)
which makes us able to use Lemma 4With this result we canwrite the equation formomentum
conservation depending on the incidence matrices for theparts of the tree This makes us able to formulate the theoremfor characterizing the feasible set Therefore we will use thefollowing function
119892 R119899 997888rarr R
119909 997891997888rarr (119860119879)minus1Φ 10038161003816100381610038161003816119860minus111990910038161003816100381610038161003816 (119860minus1119909) (22)
which describes the pressure loss between the nodes We alsointroduce a new notation
(i) 119906(119894119895)119896 is the 119896-th control on the path from V1198940 to V1198950(ii) 119892(119894119895)119896 is the function 119892 for the 119896-th subgraph on the
path from V1198940 to V1198950 whereas 119892(119894119895)1 is the function 119892
6 Mathematical Problems in Engineering
of the graph 119866119894 According to this 119892(119894119895)119896ℓ is the ℓ-thcomponent of the function 119892(119894119895)119896
(iii) (119894119895)119896 is the load vector of the 119896-th subgraph in thepath from V1198940 to V1198950 whereas (119894119895)1 is the load vectorof the graph 119866119894 and (119894119895)119896ℓ is the ℓ-th component of(119894119896)119896
Analogously we define 119860 (119894119895)119896 119901(119894119895)119896 and 119902(119894119895)119896 Furthermorewe define the following values
(i) 119896lowast119894119895 fl max119896 isin 1 1198982 + 1 | V1198960 isin Π(V1198940) and V1198960 isinΠ(V1198950) as the largest index of all subgraph the pathsto V1198940 and V1198950 pass through
(ii) 119899lowast119894119895 |119896 isin 1 1198982 + 1 | V1198960 isin ΠV1198940(V1198950)| as thenumber of subgraph which are on the path from V1198940from V1198950 containing 119866119894 and 119866119895
(iii) 119898lowast119894119895 fl |119890 isin E119862 | ℎ(119890) isin ΠV1198940(V1198950)and119891(119890) isin ΠV1198940(V1198950)|
as the number of controls which are between V1198940 andV1198950
With this notation and these values we define a sum for 119894 119895 isin11198982 + 1 If V1198940 isin Π(V1198950) we set sum119894119894 fl 0 and else we set
Σ119896lowast119894 fl119899lowastminus2sum119896=1
1prod119898lowastminus119896ℓ=1 119906(119896lowast119894)ℓsdot 119892(119896lowast119894)119899lowastminus119896119891(119890119906
(119896lowast119894)119899lowastminus119896) ((119896lowast119894)119899lowastminus119896)
+ 119892(119896lowast119894)1119891(119890119906(119896lowast119894)1
) ((119896lowast119894)1) (23)
where we write only 119899lowast and 119898lowast instead of 119899lowast119896lowast119894 and 119898lowast119896lowast119894 for
better reading Last we define a productΠ119894119894 fl 1 andΠ119896lowast119894 fl
119898lowastprod119896=1
119906(119896lowast119894)119896 (24)
From a first point of view the notation seems to be confusingbut using this notation makes sense since we want toguarantee the feasibility of a load vector by comparing thepressure bounds at every node with each other We explainthis in a short example Figure 5 shows a graph after removingthe compressor edges
If we want to compare the pressure bounds of node 11 andnode 12 we need to know how the pressures change on thepath from 2 to 11 resp from 2 to 12 Node 11 is part of thesubgraph 1198665 and node 12 is part of the subgraph 1198666 so itfollows 119896lowast56 = 2 because node 2 is part of subgraph 2 Thenotation can be understood as a numbering along the pathIt is 119906(211)1 = 1199062 the first control on the path from node 2to node 11 and 119906(211)2 = 1199064 the second control on this pathFurther it follows 119899lowast211 = 3 (1198662 1198663 1198665) and119898lowast
211 = 2 (1199062 1199064)Theproduct defined before is the product of all controls alongthis path and the sum is the pressure loss between node 2 andnode 7 Using this notation makes us able to formulate thetheorem for characterizing the feasible set in a readable way
Theorem 5 For given pressure bounds 119901+119898119894119899 119901+119898119886119909 isin R119899+1
and controls 119906119894 isin R (119894 = 1 1198982) the following equivalenceholds
A vector + with 1119879119899+1+ = 0 is feasible ie + isin if and
only if the following inequalities hold For all 119894 = 1 1198982 + 1holds (feasibility inside the subgraphs)
(1199011198981198941198991198940 )2 le min
119896=1119899119894[(119901119898119886119909
119894119896 )2 + 119892119894119896 (119894)] (25)
(1199011198981198861199091198940 )2 ge max
119896=1119899119894[(119901119898119894119899
119894119896 )2 + 119892119894119896 (119894)] (26)
max119896=1119899119894
[(119901119898119894119899119894119896 )2 + 119892119894119896 (119894)]
le min119896=1119899119894
[(119901119898119886119909119894119896 )2 + 119892119894119896 (119894)] (27)
For all 119894 119895 = 1 1198982 + 1 with 119894 lt 119895 holds (feasibility betweenthe subgraphs)1Π119896lowast 119894
(1199011198981198941198991198940 )2 + Σ119896lowast119894 le 1Π119896lowast119895
(1199011198981198861199091198950 )2 + Σ119896lowast119895 (28)
1Π119896lowast 119894
(1199011198981198861199091198940 )2 + Σ119896lowast119894 ge 1Π119896lowast 119895
(1199011198981198941198991198950 )2 + Σ119896lowast119895 (29)
1Π119896lowast 119894
(1199011198981198941198991198940 )2 + Σ119896lowast119894
le 1Π119896lowast119895
min119896=1119899119895
[(119901119898119886119909119895119896 )2 + 119892119895119896 (119895)] + Σ119896lowast119895 (30)
1Π119896lowast 119894
(1199011198981198861199091198940 )2 + Σ119896lowast119895
ge 1Π119896lowast119895
max119896=1119899119895
[(119901119898119894119899119895119896 )2 + 119892119895119896 (119895)] + Σ119896lowast119895 (31)
1Π119896lowast 119894
max119896=1119899119894
[(119901119898119894119899119894119896 )2 + 119892119894119896 (119894)] + Σ119896lowast119894
le 1Π119896lowast119895
(1199011198981198861199091198950 )2 + Σ119896lowast119895 (32)
1Π119896lowast 119894
min119896=1119899119894
[(119901119898119886119909119894119896 )2 + 119892119894119896 (119894)] + Σ119896lowast119894
ge 1Π119896lowast119895
(1199011198981198941198991198950 )2 + Σ119896lowast119895 (33)
1Π119896lowast 119894
max119896=1119899119894
[(119901119898119894119899119894119896 )2 + 119892119894119896 (119894)] + Σ119896lowast119894
le 1Π119896lowast119895
min119896=1119899119895
[(119901119898119886119909119895119896 )2 + 119892119895119896 (119895)] + Σ119896lowast119895 (34)
1Π119896lowast 119894
min119896=1119899119894
[(119901119898119886119909119894119896 )2 + 119892119894119896 (119894)] + Σ119896lowast119894
ge 1Π119896lowast119895
max119896=1119899119895
[(119901119898119894119899119895119896 )2 + 119892119895119896 (119895)] + Σ119896lowast119895 (35)
Mathematical Problems in Engineering 7
0 1 2
3
4
5
6
7
8
9
10
11
12
u1
u2
u3
u4
u5
G1 G2 G3
G4
G5
G6
Figure 5 Example graph for illustration of the used notation
0 1 2q1 q2
b0 b1 b2
Figure 6 Graph of Example 1
Let us have a look at the inequalities Inequalities(25)ndash(27) are well-known from [8] and they describe thefeasibility inside a connected tree The other inequalities(28)ndash(35) guarantee the feasibility between the subgraphsTherefore we compare the pressures of the entries of two parts((28) (29)) of the entry of one part and the exit nodes of theother parts ((30)ndash(33)) and we compare the pressures of allexits ((34) (35)) Because we cannot compare the pressuresof119866119894 and 119866119895 directly (there may be no direct connection) wehave to compare them using the subgraph 119866119896lowast which is thelast part the paths to V1198940 and V1198950 pass Therefore we definedthe sumsΣ119896lowast 119894 andΣ119896lowast119895 which trace back the pressures to119866119896lowast
Remark 6 For an implementation it is wise to define a set119880(119894119895) containing all controls on the path from V1198940 to V1198950 anduse an index running over this set to define the sums and theproducts
Now we want to proof the theorem
Proof of Theorem 5 Because the columns of 119860+ sum up tozero it holdsminus1119879
119899119894119860 119894 = 119860+
1198940 (119894 = 1 1198982 + 1) (36)
ldquo997904rArrrdquo We consider a feasible vector 119887+ isin From theequation of mass conservation of every subgraph it follows
119902119894 = 119860minus1119894 119894 (37)
We take the equation of momentum conservation for the 119894-thsubgraph (119894 isin 1 1198982 +1) and separate 119860+
119894 to119860+1198940 and 119860 119894
(119860+1198940)119879 1199012
1198940 + 119860119879119894 1199012
119894 = minusΦ 10038161003816100381610038161199021198941003816100381610038161003816 119902119894 (38)
Next we insert (36) into (38) and get
minus119860119879119894 1119899119894
11990121198940 + 119860119879
119894 1199012119894 = minusΦ 10038161003816100381610038161199021198941003816100381610038161003816 119902119894 (39)
We multiply this equation from left with (119860119879119894 )minus1 and insert
(37) so it follows
111989911989411990121198940 = 1199012
119894 + (119860119879119894 )minus1Φ 10038161003816100381610038161003816119860minus1
119894 11989410038161003816100381610038161003816 119860minus1119894 119894 (40)
and using the definition of the function 119892 we have
111989911989411990121198940 = 1199012
119894 + 119892119894 (119894) (41)
Note here that 119901119894 is the vector of pressures of the subgraph119866119894
and 119892119894(119894) is the vector-valued function which describes thepressure loss in 119866119894 Equation (41) holds for all subgraphs 119866119894
with 119894 isin 1 1198982 + 1 From this the inequalities (25)ndash(27)follow directly with 119901+ isin [119901+119898119894119899 119901+119898119886119909] Next we use (41)component-by-component For fix 119894 119895 isin 1 1198982 + 1 with119894 lt 119895 it follows
1199012119896lowast0 = 1199012
119896lowast119896 + 119892119896lowast119896 (119896lowast) (119896 = 1 119899119896lowast) (42)
Remember that 119896lowast119894119895 is the largest index of all subgraphs thepaths to V1198940 and V1198950 pass through For better reading we onlywrite 119896lowast instead of 119896lowast119894119895 We use the equation of the compressorproperty for the first compressor on the path from V119896lowast0 to V1198940
1199012ℎ(119890119906(119896lowast119894)1
)1199012119891(119890119906(119896lowast119894)1
)
= 119906(119896lowast119894)1 (43)
and insert it into (42) for 119896 = 119891(119890119906(119896lowast119894)1 )1199012(119896lowast 119894)10 = 1119906(119896lowast119894)1
1199012(119896lowast119894)20
+ 119892(119896lowast 119894)1119891(119890119906(119896lowast119894)1
) ((119896lowast119894)1) (44)
Note that 1199012119896lowast 0 = 1199012
(119896lowast119894)10 119892119896lowast119896(119896lowast) = 119892(119896lowast119894)1119896((119896lowast119894)1) and1199012ℎ(119890119906(119896lowast119894)1
) = 1199012(119896lowast 119894)20 We use the compressor equation for
the second compressor on the path from V119896lowast0 to V1198940 and putit together with (42) into (44)
1199012(119896lowast119894)10 = 1119906(119896lowast119894)1
[ 1119906(119896lowast119894)2
1199012(119896lowast 119894)30
+ 119892(119896lowast119894)2119891(119890119906(119896lowast119894)2
) ((119896lowast119894)2)]+ 119892(119896lowast119894)1119891(119890119906
(119896lowast119894)1) ((119896lowast119894)1)
(45)
8 Mathematical Problems in Engineering
We repeat this procedure for all controls on the path fromV119896lowast0 to V1198940 Although we write 119899lowast and119898lowast instead of 119899lowast119896lowast119894 and119898lowast
119896lowast119894 Thus we get
1199012(119896lowast119894)10 = 1119906(119896lowast119894)1
[ 1119906(119896lowast119894)2
[sdot sdot sdot [ 1119906(119896lowast119894)119898lowast1199012(119896lowast119894)119899lowast0
+ 119892(119896lowast119894)119899lowastminus1119891(119890119906(119896lowast119894)119899lowastminus1
) ((119896lowast119894)119899lowastminus1)] sdot sdot sdot]+ 119892(119896lowast119894)2119891(119890119906
(119896lowast119894)2) ((119896lowast119894)2)]
+ 119892(119896lowast119894)1119891(119890119906(119896lowast119894)1
) ((119896lowast119894)1)
(46)
We take again (42) for the node V119896lowast0 and do the sameprocedure just like before for the path from V119896lowast0 to V1198950 With119899lowast = 119899lowast119896lowast119895 and119898lowast = 119898lowast
119896lowast119895 we get
1199012(119896lowast 119895)10
= 1119906(119896lowast119895)1
[ 1119906(119896lowast119895)2
[sdot sdot sdot [ 1119906(119896lowast119895)119898lowast1199012(119896lowast 119895)119899lowast0
+ 119892(119896lowast119895)119899lowastminus1119891(119890119906(119896lowast119895)119899lowastminus1
) ((119896lowast119895)119899lowastminus1)] sdot sdot sdot]+ 119892(119896lowast119895)2119891(119890119906
(119896lowast119895)2) ((119896lowast119895)2)]
+ 119892(119896lowast119895)1119891(119890119906(119896lowast119895)1
) ((119896lowast119895)1)
(47)
Expanding (46) and (47) and equalizing them (this is possiblebecause 119901(119896lowast 119894)10 = 119901(119896lowast119895)10) leads to1Π119896lowast 119894
11990121198940 + Σ119896lowast119894 = 1Π119896lowast 119895
11990121198950 + Σ119896lowast119895 (48)
with Σ119896lowast119894 resp Σ119896lowast119895 defined before (48) directly implies (28)and (29) Using (48) together with (42) for V1198940 first timefor V1198950 second time and for V1198940 and V1198950 a third time theinequalities (30)ndash(35) follow directly So this part of the proofis complete
ldquolArr997904rdquo For this part we consider a vector 119887+ isin R119899+1
with 1119879119899+1119887+ = 0 which fulfills the inequalities (25)ndash(35) We
define the following sets (119894 = 2 1198982 + 1)11987511 fl [(119901119898119894119899
10 )2 (11990111989811988611990910 )2]
11987521 fl [ max119896=11198991
[(1199011198981198941198991119896 )2 + 1198921119896 (1)]
min119896=11198991
[(1199011198981198861199091119896 )2 + 1198921119896 (1)]]
1198751119894 fl [ 1Π1119894
(1199011198981198941198991198940 )2 + Σ1119894 1Π1198941
(1199011198981198861199091198940 )2 + Σ1119894]
1198752119894 fl [ 1Π1119894
max119896=1119899119894
[(119901119898119894119899119894119896 )2 + 119892119894119896 (119894)] + Σ1119894 1Π1198941
sdot min119896=1119899119894
[(119901119898119886119909119894119896 )2 + 119892119894119896 (119894)] + Σ1119894]
(49)
We set
1198750 fl 1198982+1⋂119896=1
1198751119896 cap 1198982+1⋂119896=1
1198752119896 (50)
and we will define a value 119901210 isin 1198750 later So to guarantee
such a value 119901210 exists we have to show first that 1198750 is not
empty Because 1198750 is a finite intersection of convex intervalswe have to make sure that all intervals are not empty andevery intersection between two intervals is not empty In onedimension this is sufficient that 1198750 is not empty For 119894 =1 1198982 + 11198751119894 = 0 according to requirement1198752119894 = 0 because of (26) (51)
Furthermore its 1198751119894 cap 1198752119894 = 0 because of (25) and (26) Themissing intersections are nonempty too for 119894 119895 isin 1 1198982+1 with 119894 lt 119895 as follows
1198751119894 cap 1198751119895 = 0 because of (28) and (29) 1198751119894 cap 1198752119895 = 0 because of (30) and (31) 1198752119894 cap 1198751119895 = 0 because of (32) and (33) 1198752119894 cap 1198752119895 = 0 because of (34) and (35)
(52)
So the set 1198750 is not empty We define the following values
119901210 isin 119875011990121198940 fl Π1119894 (1199012
10 minus Σ1119894) 119894 = 2 1198982 + 11199012119894119896 fl 1199012
1198940 minus 119892119894119896 (119894) 119894 = 1 1198982 + 1 119896 = 1 119899119894119902119894 fl 119860minus1119894 119894 119894 = 1 1198982 + 1
(53)
Now we show that our choice of pressures and flows isfeasible It is11990110 isin [119901119898119894119899
10 11990111989811988611990910 ] because of1199012
10 isin 11987511 Furtherbecause of 1199012
10 isin 1198751119894 for all 119894 = 2 1198982 + 1 it holds1Π1119894
11990121198940 + Σ1119894 le 1Π1119894
(1199011198981198861199091198940 )2 + Σ1119894 (54)
and 1Π1119894
11990121198940 + Σ1119894 ge 1Π1119894
(1199011198981198941198991198940 )2 + Σ1119894 (55)
Mathematical Problems in Engineering 9
Now we replace 11990121198940 in its definition with (1199012
119894119896 + 119892119894119896(119894)) andbecause of 1199012
1198940 isin 1198752119894 for all 119894 = 1 1198982 + 1 it follows1Π1119894
(1199012119894119896 + 119892119894119896 (119894)) + Σ1119894
le 1Π1119894
min119896=1119899119894
[(119901119898119886119909119894119896 )2 + 119892119894119896 (119894)] + Σ1119894
(56)
and 1Π1119894
(1199012119894119896 + 119892119894119896 (119894)) + Σ1119894
ge 1Π1119894
max119896=1119899119894
[(119901119898119894119899119894119896 )2 + 119892119894119896 (119894)] + Σ1119894 (57)
So for all 119894 = 1 1198982 + 1 and for all 119896 = 1 119899119894 it holds119901119894119896 isin [119901119898119894119899119894119896 119901119898119886119909
119894119896 ] which is equivalent to
119901+ isin [119901+119898119894119899 119901+119898119886119909] (58)
The equation for mass conservation follows directly from thedefinition of 119902119894 (119894 = 1 1198982 + 1) For the pressure loss weuse
1199012119894 = 1119899119894
11990121198940 minus 119892119894 (119894) (59)
And multiply this equation from left with 119860119879119894 and use (36)
With this it follows the equation for momentum conserva-tion
(119860+119894 )119879 (119901+
119894 )2 = minusΦ 10038161003816100381610038161199021198941003816100381610038161003816 119902119894 (60)
The last equation missing is the compressor property For acontrol 119895 isin 1 1198982 we fix a vertex V1198940 (119894 isin 1 1198982 +1)with ℎ(119890119906119895) = V1198940 We use 1199012
1198940 = Π1119894(119901210 minus Σ1119894) and solve the
formula for 119901210 in the following way
119901210 = 1119906(1119894)1
[sdot sdot [ 1119906(1119894)119898lowast1199012(1119894)119899lowast0
+ 119892(1119894)119899lowastminus1119891(119890119906(1119894)119899lowastminus1
) ((1119894)119899lowastminus1)] sdot sdot]+ 119892(1119894)1119891(119890119906(1119894)1 )
((1119894)1) (61)
Consider wlog that the subgraphs along the path from V10to V1198940 are numbered chronological st if ℎ(119890119906119895) isin 119866119894 then119891(119890119906119895) isin 119866119894minus1 We use 1199012
119894minus10 = Π1119894minus1(119901210 minus Σ1119894minus1) to get the
same representation for 119901210 depending on 1199012
(1119894minus1)119899lowast0 whichis equivalent to 119901(1119894)119899lowastminus10 We equalize these representationsdepending on 1199012
1198940 and 1199012119894minus10 and we get
1119906(1119894)119898lowast1199012(1119894)119899lowast0 + 119892(1119894)119899lowastminus1119891(119890119906
(1119894)119899lowastminus1) ((1119894)119899lowastminus1)
= 1199012(1119894)119899lowastminus10 (62)
We chose 119894 isin 1 1198982 + 1 st 119906119895 is the last control on thepath from V10 to V1198940 so it is 119906(1119894)119898lowast = 119906119895 Further we have
1199012(1119894)119899lowast 0 = 1199012
11989401199012(1119894)119899lowastminus10 = 1199012
119894minus10119892(1119894)119899lowastminus1119891(119890119906(1119894)119899lowastminus1
) (119887(1119894)119899lowastminus1) = 119892119894minus1119891(119890119906119895 )(119887119894minus1)
(63)
and we know that 11990121198940 resp 1199012
119894minus10 minus 119892119894minus1119891(119890119906119895 )(119887119894minus1) are the
pressures at the head resp the foot of the compressor edge119890119906119895 So it follows 1119906119895
1199012ℎ(119890119906119895 )
= 1199012119891(119890119906119895 )
(64)
which is equivalent to the compressor property This com-pletes the proof
With this theorem we have another characterization ofthe feasible set The last step in this intersection is theadaption of the results to the spheric-radial decompositionAs mentioned before the main step is to get a representationof the sets119872119894 in step (2) of Algorithm 3
Because we assumed that the graph has only one entry wedefine the following set
119872 fl 119887 isin R119899ge0 | (minus1119879
119899119887 119887) isin (65)
Consider a sampled point V119894 isin S119899minus1 of step (1) of Algorithm 3(119894 = 1 119873) We identify the load vector 119887 with the affinelinear function
119887119894 (119903) fl 119903119871V119894 + 120583 = 119903120596119894 + 120583 with 120596119894 = 119871V119894 (66)
where 120583 is the mean value of the Gaussian distribution and 119871is such that 119871119871119879 = Σ for positive definite covariance matrixΣ Note that the index 119894 here is for the sampled vector V119894 notfor the 119894-th subgraph Because 119887119894 is defined only for the exitnodes it must hold 119887 ge 0 which leads to the definition of theregular range
119877119903119890119892119894 fl 119903 ge 0 | 119887119894 (119903) ge 0 supe 119872119894 (67)
So the feasible set is
119872119894 = 119903 isin 119877119903119890119892119894 | 119887119894 (119903) fulfills (3) (5) and (7) (68)
and with Theorem 5 it holds
119872119894 = 119903 isin 119877119903119890119892119894 | 119894 (119903) fulfills (25) minus (34) (69)
where 119894(119903) is from (18) Because 119894 depends on 119903 now thefunction 119892 depends on 119903 too More exactly 119892 is quadratic in119903 so 119892119896(119887119894(119903)) can be represented as follows
1199032[[119899sum
119895=1
Ψ119896119895( 119899sumℓ=1
119860minus1119895ℓ120596119894ℓ)2]]⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
119886119894
10 Mathematical Problems in Engineering
+ 119903[[2119899sum
119895=1
Ψ119896119895( 119899sumℓ=1
119860minus1119895ℓ120596119894ℓ)( 119899sum
ℓ=1
119860minus1119895ℓ120583ℓ)]]⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
119894
+ [[119899sum
119895=1
Ψ119896119895( 119899sumℓ=1
119860minus1119895ℓ120583ℓ)2]]⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
119888119894
(70)
HereΨ is the product of119860 andΦ So the inequations whichcharacterize119872119894 are also quadratic in 119903 because the minimaland maximal pressures in the inequalities only count to 119888119894Now we can write the feasible sets as follows
119872119894 = 119903 isin 119877119903119890119892119894 | 0 le 1198861198941199051199032 + 119887119894119905119903 + 119888119894119905 (71)
where 119905 is an index for the total number of inequalities in119872119894Now the regular range can be cut with the positive intervals ofthe inequalities and the result is a union of disjoint intervals
119872119894 = 119904119894⋃119896=1
[119886119894119896 119886119894119896] 119904119894 isin N (72)
Here 119904119894 is the number of disjoint intervals in which allinequalities are fulfilled and 119886119894119896 and 119886119894119896 are the intervalbounds they depend on the pressure bounds and the con-trols So the third step of Algorithm 3 can be executed With
P (119887 isin 119872) asymp 1119873 119873sum119894=1
119904119894sum119896=1
F120594 (119886119894119896) minusF120594 (119886119894119896) (73)
the probability for a random load vector to be feasible can becomputed
4 Existence of Optimal Solutions
In this section we want to have a look at some optimizationproblems We distinguish between problems with constantloads (without uncertainty on the demand) and problemswith random loads The latter leads us to optimizationproblems with so called probabilistic constraints or chanceconstraints (see [21]) There are two aims in optimizationMinimize the maximal pressure bounds and minimize thecontrols For minimizing the maximal pressure bounds wedefine the following set
119875 fl119899⋃
119894=0
[119901+119898119894119899119894 infin) (74)
in which we look for them Because the maximal pressuresare in proportion to the cost of pipes an objective functionfor a optimization problem could be 119888119879119901+119898119886119909 for a cost vector119888 isin R119899+1
gt0 By minimizing the controls we use the Euclideannorm of the controls as an objective function
Optimization with Constant Loads Consider the two opti-mization problems
min119901+119898119886119909isin119875
119888119879119901+119898119886119909
subject to 119887+ (119901+119898119886119909) isin 119872 (75)
andmin119906isinR1198982ge1
1199062subject to 119887+ (119906) isin (76)
Lemma 7 There exists a unique solution for (75)
Proof We have a look at the side constraint 119887+(119901+119898119886119909) isin Using Theorem 5 we can write (75) as
min119901+119898119886119909isin119875
119888119879119901+119898119886119909
subject to (25) minus (35) are fulfilled (77)
For 119894 isin 1 1198982 + 1 we write inequality (25) (and allremaining analogous) in the following way
0 le 119892119894119896 (119894) minus (1199011198981198941198991198940 )2 + (119901119898119886119909
119894119896 )2 forall119896 = 1 119899119894 (78)
So every inequality depends only on one upper pressurebound (but more inequalities may depend on the samebound) Because the inequalities are not strict the values119901119896 fl min 119901119898119886119909
119896 isin [0infin) | (25) minus (35) are fulfilled (79)
exist for all 119896 = 1 119899 + 1 and we can define a value119901+119898119886119909119900119901119905119896 fl max 119901119896 119901119898119894119899
119896 (80)
which is obviously a solution of (75) Moreover this solutionis unique because reducing one component of the upperbounds would either hurt an inequality or it would not fulfill119901+119898119886119909 ge 119901+119898119894119899 anymore
For the optimization problem (76) we can formulate anecessary and a sufficient condition for optimal solutions
Lemma8 Consider pressure bounds119901+119898119894119899 119901+119898119886119909 isin R119899+1 st119901+119898119894119899 lt 119901+119898119886119909 and a suitable vector 119887+ isin R119899+1 (ie + fulfills(25)ndash(27)) For all 119896 = 1 1198982 we set 119894(119896) and 119895(119896) such that119891(119890119896) isin V+
119894(119896) and ℎ(119890119896) isin V+119895(119896)
(a) If 119906 isin R1198982 is a solution of (76) it holds for all 119896 =1 1198982 (119901119898119894119899119894(119896)0)2 minus 119892119894(119896)119891(119890119896)
(119894(119896)) le (119901119898119886119909119895(119896)0)2 (81)
(b) If the inequalities
(119901119898119886119909119894(119896)0)2 minus 119892119894(119896)119891(119890119896)
(119894(119896)) ge (119901119898119894119899119895(119896)0)2
and (119901119898119894119899119894(119896)0)2 minus 119892119894(119896)119891(119890119896)
(119894(119896)) le (119901119898119886119909119895(119896)0)2 (82)
are fulfilled for all 119896 = 1 1198982 an optimal solution 119906 isin R1198982ge1
of (76) exists
Mathematical Problems in Engineering 11
Note that a vector 119887+ isin R119899+1 is suitable here if the vector+ is feasible for the subgraphs ie the inequalities (25)-(27)are fulfilled for +Proof (a) For this part we use a proof by contradictionAssume that
(119901119898119894119899119894(119896)0)2 minus 119892119894(119896)119891(119890119896)
(119894(119896)) gt (119901119898119886119909119895(119896)0)2 (83)
Because 119906119896 ge 1 for all 119896 = 1 1198982 it follows
(119901119898119894119899119894(119896)0)2 minus 119892119894(119896)119891(119890119896)
(119894(119896)) gt 1119906119896
(119901119898119886119909119895(119896)0)2 (84)
It it Σ119896lowast119894(119896) = 0 and Π119896lowast 119894(119896) = 1 because the subgraph withindex 119896lowast is equal to the subgraphwith index 119894(119896) And becausethe subgraphs 119894(119896) and 119895(119896) are neighbours (ie directlyconnected) it follows Π119896lowast 119895(119896) = 119906119896 and
Σ119896lowast119895(119896) = 119892(119896lowast119895(119896))1119891(119890119906(119896lowast119895(119896))1
) (119887(119896lowast119895(119896))1)= 119892119894(119896)119891(119890119896)
(119894(119896)) (85)
Thus we have a contradiction to (28) and 119906119896 cannot beoptimal
(b) As mentioned before the vector 119887+ is such that isfeasible for the subgraphs So because of the conditions for119896 isin 1 1198982 there exists a 119901119894(119896)0 isin [119901119898119894119899
119894(119896)0 119901119898119886119909119894(119896)0] and a119901119895(119896) isin [119901119898119894119899
119895(119896) 119901119898119886119909119895(119896) ] so that1199012119894(119896)0 minus 119892119894(119896)119891(119890119896)
(119894(119896)) ge 1199012119895(119896)0 (86)
With 1199012119894(119896)0 minus 119892119894(119896)119891(119890119896)
(119894(119896)) = 1199012119891(119890119896)
and 1199012119895(119896)0 = 1199012
ℎ(119890119896)it
follows
(119901ℎ(119890119896)119901119891(119890119896)
) ge 1 (87)
Therefore we can find a 119906 isin R1198982ge1 that the inequalities (28)-
(35) are fulfilled Now we can use theWeierstraszlig theorem andget the existence of an optimal solution
Note that the sufficient condition in Lemma 8 is com-paratively strong there may exist optimal solutions with lessstrong conditions eg the following
Corollary 9 If the graph has only one compressor edge ie|E119862| = 1 then it holds If 119906 isin Rge1 is an optimal solution of(76) 119906 is unique
This conclusion follows directly from the strict mono-tonicity of the objective function
Optimization with Random Loads In this section we con-sider the load vector to be random as we mentioned inSection 22 In the optimization problems we want to makesure that the feasibility of a random load vector is guaranteedfor a probability 120572 isin (0 1) This leads us to optimizationproblems with so called chance or probabilistic constraints
In general problems with probabilistic constraints have theform
min 119891 (119909 120585)subject to P (ℎ (119909 120585) le 0) ge 120572 (88)
with an objective function 119891 and a vector of uncertainty 120585There exist many works about probabilistic constraints respstochastic programming eg [7] gives an excellent overviewabout chance constraints in theory and application In ourcase we consider the following problem
min119901+119898119886119909isin119875
119888119879119901+119898119886119909
subject to P (119887 (119901+119898119886119909) isin 119872) ge 120572 (89)
To handle this problem resp the probabilistic constraint thespheric-radial decomposition gives us an explicit representa-tion of the constraint for a Gaussian vector 119887 sim N(120583 Σ) withmean value 120583 and positive definite covariance Σ
P (119887 (119901+119898119886119909) isin 119872)= int
S119899minus1120583120594 119903 ge 0 | 119903119871V + 120583 isin 119872119889120583120578 (V) (90)
Further Algorithm 3 gives a way to approximate this integralin an efficient way In Section 3 we showed a way to character-ize the feasible set119872 and how to adapt this characterization tothe spheric-radial decomposition This changes our problemwith probabilistic constraints (89) to
min119901+119898119886119909isin119875
119888119879119901+119898119886119909
subject to 1119873 119873sum119894=1
119904119905sum119905=1
F120594 (119886119894119905) minusF120594 (119886119894119905) ge 120572 (91)
Of course the interval bounds depend on the pressure boundsand the control For this problem we formulate the nextlemma
Lemma 10 Let a sampling V1 V119873 (from step (1) ofAlgorithm 3) and 120572 isin (0 1) be given For a suitable 120583 isin R119899
problem (91) has at least one solution 119901+119898119886119909 isin 119875Proof For the proof we fix a V isin V1 V119873 from the givensampling From (25) and (26) it follows for all 119896 = 1 119899119894with 119894 isin 1 1198982 + 10 le 119892119894119896 (119894 (119903)) + (119901119898119886119909
119894119896 )2 minus (1199011198981198941198991198940 )2
0 le minus119892119894119896 (119894 (119903)) + (1199011198981198861199091198940 )2 minus (119901119898119894119899
119894119896 )2 (92)
With a random vector 119887 sim N(120583 Σ)119892119894119896(119894(119903)) are quadraticfunctions in 119903 and the upper pressure bounds only influencethe constant part with a positive sign
0 le 1198861198941198961199032 + 119887119894119896119903 + 119888119894119896 + (119901119898119886119909119894119896 )2 minus (119901119898119894119899
1198940 )2 0 le minus1198861198941198961199032 minus 119887119894119896119903 minus 119888119894119896 + (119901119898119886119909
1198940 )2 minus (119901119898119894119899119894119896 )2 (93)
12 Mathematical Problems in Engineering
It follows that a rise in 119901119898119886119909119894119896 for all 119896 = 0 119899119894 enlarges
the intervals in which the inequalities hold The same holdsfor all inequalities inTheorem 5The cumulative distributionfunction of theChi-Square-Distribution is strictlymonotonicincreasing and convergent to 1 Thus upper pressure bounds119901119898119886119909119894119896 (for 119894 = 1 119899) can be found so that
1119873 119873sum119895=1
119904119895sum119905=1
F120594 (119886119895119905) minusF120594 (119886119895119905) (94)
is higher that 120572 isin (0 1) The index119873 is for all sampled pointson the sphere and the index 119904119895 is for the union of intervals inwhich all inequalities hold The interval boundaries 119886119895119905 and119886119895119905 depend continuously on the pressure bounds and becausethe inequalities are not strict 119901119898119886119909 can be found st the sideconstraint is fulfilled with equality So there exists at least onesolution of (91)
5 Numerical Results
In this section we show a few results of implementation Atfirst we show the idea of the spheric-radial decomposition byusing an easy example Next we show a easy example with onecompressor edge and at last we use real data of the Greek gasnetworkThe focus of the implementation is on the theoremsproofed in Section 4
Example 1 Our first example is an easy graph without innercontrol (see Figure 6)
Assume that node 1 and 2 are gas consumers with meandemand 120583 = [05 05]119879 The pressure bounds are given by119901+119898119894119899 = [2 1 1]119879 and 119901+119898119886119909 = [3 2 2]119879 and Φ = E2 Thenthe incidence matrix is
119860+ = [[[minus1 01 minus10 1
]]] resp 119860 = [1 minus10 1 ] (95)
Consider the inequalities of Theorem 5 to characterize thefeasible set Because the network has no compressor edgesthe system of inequalities only contains the information (25)-(27) the other inequalities do not occur The function 119892 isgiven by
119892 (119887) = [1 01 1] [1 00 1] sdot1003816100381610038161003816100381610038161003816100381610038161003816[1 10 1](11988711198872)
1003816100381610038161003816100381610038161003816100381610038161003816∘ ([1 10 1](11988711198872)) = ( 11988721 + 211988711198872 + 1198872211988721 + 211988711198872 + 211988722) (96)
By inserting the pressure bounds one can see that inequality(25) is always fulfilled and inequality (27) implies1198872 le radic3 (97)
With this inequality (26) follows for 0 le 1198872 le radic30 le 1198871 le minus1198872 + radic8 minus 11988722 (98)
So the feasible set119872 is the following (see Figure 7)
119872 = 119887 isin R2ge0 | 1198872 le radic3 and 1198871 le minus1198872 + radic8 minus 11988722 (99)
Now consider the optimization problem
min119901+119898119886119909isin119875
11198793119901+119898119886119909
subject to 119887 isin 119872 (119901+119898119886119909) (100)
with 119887 = 120583 = [05 05]119879 Lemma 7 guarantees the existenceof a unique solution of this problem We use the fmincon-function inMATLAB which minimizes a constrained non-linear multivariable function The fmincon default settinguses a interior-point method For more information we referto the official MathWorks homepage (httpswwwmath-workscomhelpoptimugfminconhtml)The solver returnsthe following result
119901+119898119886119909119900119901119905 = [2 radic3radic275]119879 (101)
One can easily see that this is the correct solution The gasdemand at node 2 is 05 which is equal to the flow 1199022 sousing the equation of momentum implies a difference in thequadratic pressures between node 1 and node 2 of 025 Theflow through edge 1 is given by the two exit nodes and it is1199021 = 1 This implies a difference in the quadratic pressuresbetweennode 0 andnode 1 of 1 So the optimal solution119901+119898119886119909
119900119901119905
directly followsNext we use the spheric-radial decomposition especially
Algorithm 3 to calculate the probability P(119887 isin 119872) for arandom vector 119887 sim N(E2 [05 05]119879) Therefore we viewthe results of 8 tests with 1000 sampled points in each one(Table 1)
The probability in all eight tests is nearly the same Themean value is 03479 and the variance is 27723 sdot 10minus6 whichis very small This shows that the implementation is workingnearly exact The efficiency of the implementation of courseis not perfect but that is not part of this work
Last wewant to solve the following optimization problem
min119901+119898119886119909isin119875
11198793119901+119898119886119909
subject to 1119873 119873sum119894=1
119904sum119905=1
F120594 (119886119894119905) minusF120594 (119886119894119905) ge 120572 (102)
We use again the fmincon-function inMATLAB to solve theproblem eight times with 1000 sampled points in every caseBecause Lemma 8 needs a suitable 120583 isin R2
ge0 we assume 119887 simN(01sdot1198642 [15 15])We set120572 fl 09 and the results are shownin Table 2
Example 2 Now we add a compressor edge to the graph ofExample 1 (see Figure 8)
Assume that node 1 and 3 are consumers node 2 is ainner nodeThemeandemand120583 = [05 0 05]119879Thepressure
Mathematical Problems in Engineering 13
minus2 minus1 1 2 3
minus2
minus1
1
2
3
b2
b1
minus1 1 2
minus1
1
2
b2
b1
rLv1
rLv2
rLv3
Scheme of the spheric-radial decomposition with
sphere (yellow)and three points 1 2 3 (red) at the unit
the feasible set Mminus for Σ = E2 = [05 05]TFeasible set M for p+min = [2 1 1]T and
p+max = [3 2 2]T
Figure 7 Feasible set of Example 1
Table 1 Test results of easy example
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 8P(119887 isin 119872) 03486 03481 03450 03501 03484 03499 03459 03475
Table 2 Test results of Example 1
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 81199011198981198861199091199001199011199050 54514 54714 54758 54028 54767 53921 54888 551671199011198981198861199091199001199011199051 29802 30126 30225 29725 30300 29485 30336 300291199011198981198861199091199001199011199052 10084 10004 10776 10001 11221 11390 10285 10558
0 1 2 3q1 q2 q3
b0 b1 b2 b3
e1 isin ℰF e2 isin ℰC e3 isin ℰF
Figure 8 Graph of Example 2
0 1 2 3q1 q2 q2 q3
b0 b1 b2 b3
e1 isin ℰF e3 isin ℰF
Figure 9 Separated graph of Example 2
bounds are given by 119901+119898119894119899 = [2 1 1 1]119879 and 119901+119898119886119909 = [3 22 2]119879 and 1206011198901= 1 and 1206011198903
= 1 The incidence matrix is givenby
119860+ = [[[[[[
minus1 0 01 minus1 00 1 minus10 0 1]]]]]]
resp 119860 = [[[1 minus1 00 1 minus10 0 1
]]] (103)
Before we have a look at the inequalities of Theorem 5we have to remove the compressor edge from the graphThe flows can be computed by using the equation of massconservation and the new load vector is then
= (1198871 + 11990221198872 minus 11990221198873 ) with
119902 = (1198871 + 1198872 + 11988731198872 + 11988731198873)(104)
We assume the control to be switched off so it holds119906 = 1With this the control edge 1198902 is treated as a flow edge withoutfriction The resulting graph is shown in Figure 9 from theinequalities (25)ndash(35) it follows
3 le radic3 and 1 le radic8 minus 23 (105)
14 Mathematical Problems in Engineering
minus1 1 2 3
1
2
3
minus1 1 2 3
1
2
3
b3
b1
Hyperspace of the feasible set MHyperspace of the feasible set M
b1
b3
Figure 10 Feasible set of the Example 2
Table 3 Test results of control example
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 8P(119887 isin 119872) 01531 01500 01540 01536 01530 01550 01567 01528so the feasible set is (see Figure 10)
= + isin R119899+1 | 1119879
4 + = 0 and 0 le 3 le radic3 and 2 = 0and 0 le 1 le radic8 minus 22 (106)
This implies
119872 = 119887 isin R3ge0 | 0 le 1198873 le radic3 and 1198872 = 0 and 0 le 1198871 le minus1198873
+ radic8 minus 11988723 (107)
Solving the problem
min119901+119898119886119909isin119875
11198794119901+119898119886119909
subject to isin (108)
for 119887 = 120583 = [05 0 05]119879 leads us to the optimal solution
119901+119898119886119909119900119901119905 = [2 radic3radic3radic275]119879 (109)
The argumentation for the optimality of this vector is exactlylike before in the easy example Becausewe have a compressoredge in this graph we can minimize the control Thereforewe fix the pressure bounds as we did at the beginning of thisexample and we set 119887 = [1 0 15]119879 So inequality (31) is notfulfilled anymore for 119906 = 1 We use again the fmincon-solverinMATLAB to solve the following problem
min119906ge1
1199062subject to 119887 isin 119872(119906) (110)
The solver returns the optimal solution 119906119900119901119905 = 11818
Here one can see that the sufficient condition in Lemma 8is really strong Itrsquos not fulfilled with the given values but thereexists a solution anyway Of course the necessary conditionis fulfilled
Now we use again Algorithm 3 to compute the proba-bility P(119887 isin 119872) for a random load vector 119887 sim N(E3 [0505 05]119879) Table 3 shows us the results
Again we can see that the probability is nearly the samein all eight cases The mean probability is 01535 and thevariance is 32369 sdot 10minus6 For the optimization we reducethe covariance matrix and assume 119887 sim N(001 sdot E3 [05 0505]119879) For a probability 120572 = 09 Table 4 states the resultsExample 3 In this example we use a part of a real networkThe focus here is on realistic values for our model Weget these values from the GasLib (see [22]) a collection oftechnical gas network data
ldquoThegoal ofGasLib is to promote research on gasnetworks by providing a set of large and realisticbenchmark instancesrdquo (httpgaslibzibde)
We use a part of the Greek gas network (GasLib-134) (seeFigure 11)
For using real values we first need to know the constant120601119890 from (5)
120601119890 = (119877119878119879)2 120582119863 1198711198862 (111)
The value 119877119878 is the specific gas constant which can becalculated by 119877119878 = 119877119872 with the ideal gas constant 119877 =831451119896119892 sdot 11989821199042 sdot 119898119900119897 sdot 119870 (see [23]) and the molar mass119872 = 1662119896119892119896119898119900119897 (from real data) of the used natural gasAlso the temperature 119879 = 28915119870 is given from real dataFor the (constant) friction factor 120582 we use the law of Chen(see [24])
Mathematical Problems in Engineering 15
Table 4 Test results of control example
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 81199011198981198861199091199001199011199050 39013 39223 39039 38901 38879 39039 38901 390981199011198981198861199091199001199011199051 16568 16565 16365 16497 16496 16365 16497 164921199011198981198861199091199001199011199052 16568 16565 16365 16497 16496 16365 16497 164981199011198981198861199091199001199011199053 10000 10000 10000 10000 10002 10000 10000 10000
Table 5 Values based on real data
variable 119877119878 119879 120582 119863 119871 119886 120601value 50027 28915 00093 9144 178693 43862 198 sdot 107unit 11989821199042 sdot 119870 119870 minus 119898119898 119896119898 119898119904 11989821199042
Figure 11 Greek gas network from GasLib [22] red edges are usedin our example
1radic120582 = minus2 log10 ( 11989611986337065minus 50425
Relog10 [(119896119863)1109828257 + 58506
Re08981]) (112)
Diameter 119863 = 9144119898119898 and roughness value 119896 = 8 sdot 10minus6119898for the first pipe (see Figure 11) are known from real data TheReynolds-Number Re can be calculated by Re = (V119863120588)120578with the dynamic viscosity 120578 = 1 sdot 10minus5119875119886 sdot 119904 (see [1])the mean velocity V = 9548119898119904 (calculated with the meanvolume flow rate which is given by the real data) of the gasand the density 120588 = 074331198961198921198983 (from real data) So we getRe = 648959074 asymp 649 sdot 106 and with this the pipe frictioncoefficient after Chen is 120582 = 00093
The length of the first pipe is given by 119871 = 178693119896119898from the real data Last value missing is the sound speed innatural gas which we assume to be ideal The sound speed
119886 depends on the modulus of compressibility (see [9]) Forideal gases the modulus of compressibility only depends onthe adiabatic exponent and the pressure Using the ideal gasequation simplifies the equation for the sound speed to
119886 = radic120581119877119879119872 (113)
The adiabatic exponent for methan which is the main partof natural gas (gt 90) is about 133 (see [25]) Together allvalues are shown in Table 5 (example for the pipe next to thesource in Figure 11)
Nowwewant to optimize the inner control in the networkfor a fix load vector Therefor we choose two nominations2011-11-01 and 2016-02-17 In both cases the fmincon-functioninMATLAB returns
119906119900119901119905 = 1 (114)
as a local minimum which satisfies the constraint 119887 isin 119872This means for these nominations a compressor stationis not needed so it is switched off With this as a lastcalculation in this paper we calculate the probability for thegiven nominations to be feasible We use the values from thenominations as mean value 120583 and set 119887 sim N(120583 sdot 10minus3E 120583)The results are shown in Tables 6 and 7
Again one can see that the results are quite similarThe mean probability in Table 6 is 09860 and the meanprobability in Table 7 is 09760 Both values are nearly 1 sothere must happen something really unforeseeable that aGreek does not get his demanded gas
6 Conclusion
In this paper we have used an existing stationary modelfor gas transport based on the isothermal Euler equationsWe have extended this model by compressor stations andpresented a characterization of the set of feasible loadsEspecially Theorem 5 contains the central idea to get anaccess to the optimization in Section 4 and the spheric-radialdecomposition is our main tool to handle the uncertainty inthe loadsWewere able to show some results for optimizationwith probabilistic constraints in gas transport In a next stepthese results should be generalized
16 Mathematical Problems in Engineering
Table 6 Test results for nomination 2011-11-01
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 8P(119887 isin 119872) 09849 09849 09878 09885 09863 09845 09854 09853
Table 7 Test results for nomination 2016-02-17
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 8P(119887 isin 119872) 09744 09756 09788 09788 09754 09775 09734 09739
With the characterization of the feasible set we haveshown that the feasibility of a load depends significantly onthe pressure bounds These pressure bounds play a mainrole in our optimization problems The other main partsare the inner controls modeled by compressor edges whichshould lead in optimal control problems with probabilisticconstraints Here we have built a base for further works withthis topic
The idea of separating the graph and removing thecompressor edges can be used to extend themodel with othercomponents like eg control valves and resistorsThiswouldchange the inequalities inTheorem 5 but the approach worksalso for thesemodifications Further [8] showed a way to dealwith a networkwhich is a cycle Combining bothmethods weget an access to model gas networks in high detail
Finally the expansion to a dynamic model might bealso possible The idea is to use a space-mapping approachbecause one can use a lot of the analysis we did hereTherefore upcoming papers should deal with these problemsand present results of the turnpike-theory which describesa relation between stationary and instationary solutions Infact turnpike-theory implies that the optimal dynamic stateis close to the optimal static state in the interior of the timeintervals for sufficiently long time horizons AJ Zaslavskiexplains the turnpike-phenomenon in [26] This book alsogives an overview about the theory
Data Availability
The data used to support the findings of this study areavailable from the corresponding author upon request
Conflicts of Interest
The authors declare that there are no conflicts of interestregarding the publication of this paper
Acknowledgments
This work was supported by DFG in the framework of theCollaborative Research Centre CRCTransregio 154 Math-ematical Modeling Simulation and Optimization Using theExample of Gas Networks project C03
References
[1] P Domschke B Hiller J Lang and C Tischendorf ldquoModel-lierung von gasnetzwerken Eine Ubersichtrdquo Tech Rep 2717Technische Universitat Darmstadt 2017
[2] T Koch B Hiller M E Pfetsch and L Schewe Evaluating GasNetwork Capacities MOS-SIAM 2015
[3] M Gugat R Schultz and D Wintergerst ldquoNetworks of pipe-lines for gas with nonconstant compressibility factor stationarystatesrdquoComputational amp AppliedMathematics vol 37 no 2 pp1066ndash1097 2018
[4] M Gugat and M Herty ldquoExistence of classical solutions andfeedback stabilization for the flow in gas networksrdquo ESAIMControl Optimisation and Calculus of Variations vol 17 no 1pp 28ndash51 2011
[5] M K Banda M Herty and A Klar ldquoGas flow in pipeline net-worksrdquo Networks and Heterogeneous Media vol 1 no 1 pp 41ndash56 2006
[6] RM Colombo G GuerraM Herty and V Schleper ldquoOptimalcontrol in networks of pipes and canalsrdquo SIAM Journal onControl and Optimization vol 48 no 3 pp 2032ndash2050 2009
[7] A Shapiro D Dentcheva and A Ruszczynski ectures onStochastic Programming Modeling and Theory MPS-SIAM2009
[8] C Gotzes H Heitsch R Henrion and R Schultz ldquoOn thequantification of nomination feasibility in stationary gas net-works with random loadrdquoMathematical Methods of OperationsResearch vol 84 no 2 pp 427ndash457 2016
[9] D Halliday R Resnick and J Walker Physik Wiley-VCHBachelor edition edition 2007
[10] M Gugat F M Hante M Hirsch-Dick and G LeugeringldquoStationary states in gas networksrdquoNetworks andHeterogeneousMedia vol 10 no 2 pp 295ndash320 2015
[11] D RoseM SchmidtM C Steinbach and BMWillert ldquoCom-putational optimization of gas compressor stations MINLPmodels versus continuous reformulationsrdquoMathematical Meth-ods of Operations Research vol 83 no 3 pp 409ndash444 2016
[12] H Heitsch R Henrion H Leovey et al ldquoEmpirical observa-tions and statistical analysis of gas demandrdquo in Evaluating GasNetwork Capacities T data B KochM E Hiller et al Eds pp273ndash290 MOS-SIAM 2015
[13] A Genz and F Bretz Computation of Multivariate Normal andt Probabilities vol 195 of Lecture Notes in Statistics Springer2009
[14] M H Farshbaf-Shaker R Henrion and D Homberg ldquoProp-erties of Chance Constraints in Infinite Dimensions with anApplication to PDE Constrained Optimizationrdquo Set-Valued andVariational Analysis pp 1ndash21 2017
[15] T Gonzalez Grandon H Heitsch and R Henrion ldquoA jointmodel of probabilisticrobust constraints for gas transportman-agement in stationary networksrdquo Computational ManagementScience vol 14 no 3 pp 443ndash460 2017
[16] W van Ackooij I Aleksovska and M Munoz-Zuniga ldquo(Sub-)Differentiability of Probability Functionswith Elliptical Distri-butionsrdquo Set-Valued and Variational Analysis
Mathematical Problems in Engineering 17
[17] W van Ackooij and R Henrion ldquoGradient formulae for nonlin-ear probabilistic constraints with GAUssian and GAUssian-likedistributionsrdquo SIAM Journal on Optimization vol 24 no 4 pp1864ndash1889 2014
[18] W van Ackooij and R Henrion ldquo(Sub-)gradient formulaefor probability functions of random inequality systems underGaussian distributionrdquo SIAMASA Journal on UncertaintyQuantification vol 5 no 1 pp 63ndash87 2017
[19] W van Ackooij and J Malick ldquoSecond-order differentiabilityof probability functionsrdquo Optimization Letters vol 11 no 1 pp179ndash194 2017
[20] H Eckey R Kosfeld and M Turck Wahrscheinlichkeitsrech-nung und Induktive Statistik - Grundlagen -Methoden - BeispieleGabler Wiesbaden Germany 2005
[21] A Prekopa Stochastic Programming Springer 1st edition 1995[22] I Joormann M Pfetsch R Schwarz et al ldquoGasLibmdashA Library
of Gas Network Instancesrdquo 2017[23] G Wedler Lehrbuch der Physikalischen Chemie Wiley-VCH
5th edition 2004[24] N H Chen ldquoAn explicit equation for friction factor in piperdquo
Industrial amp Engineering Chemistry Fundamentals vol 18 no3 pp 296-297 1979
[25] W Schenk and F Kremer Physikalisches Praktikum Springer14th edition 2014
[26] A J Zaslavski ldquoTurnpike Properties in the Calculus of Varia-tions and Optimal Controlrdquo inNonconvex Optimization and ItsApplications vol 80 Springer 2006
Hindawiwwwhindawicom Volume 2018
MathematicsJournal of
Hindawiwwwhindawicom Volume 2018
Mathematical Problems in Engineering
Applied MathematicsJournal of
Hindawiwwwhindawicom Volume 2018
Probability and StatisticsHindawiwwwhindawicom Volume 2018
Journal of
Hindawiwwwhindawicom Volume 2018
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawiwwwhindawicom Volume 2018
OptimizationJournal of
Hindawiwwwhindawicom Volume 2018
Hindawiwwwhindawicom Volume 2018
Engineering Mathematics
International Journal of
Hindawiwwwhindawicom Volume 2018
Operations ResearchAdvances in
Journal of
Hindawiwwwhindawicom Volume 2018
Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018
International Journal of Mathematics and Mathematical Sciences
Hindawiwwwhindawicom Volume 2018
Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom
The Scientific World Journal
Volume 2018
Hindawiwwwhindawicom Volume 2018Volume 2018
Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in
Nature and SocietyHindawiwwwhindawicom Volume 2018
Hindawiwwwhindawicom
Dierential EquationsInternational Journal of
Volume 2018
Hindawiwwwhindawicom Volume 2018
Decision SciencesAdvances in
Hindawiwwwhindawicom Volume 2018
AnalysisInternational Journal of
Hindawiwwwhindawicom Volume 2018
Stochastic AnalysisInternational Journal of
Submit your manuscripts atwwwhindawicom
2 Mathematical Problems in Engineering
In Section 4 we consider some optimization problemswith and without probabilistic constraints The existence ofoptimal solutions is based on the analysis we did in Section 3This should built a base for further works about optimalcontrol problems with probabilistic constraints using theexample of gas networks
In Section 5 we present three numerical examples Thefirst is a minimal graph without inner control which showsthe idea of the spheric-radial decomposition The second oneis a minimal graph with inner control and the third one isbased on real data for which we present real values for anetwork
In the last section we present a few ideas of extending thispaper One main aspect is the turnpike-theory
2 Mathematical Modeling
We start with the introduction of the model that is also usedin [8] This model does not include compressor stations Weextend this model to include compressor stations and wediscuss the network analysisThen themain tool the so calledspheric-radial decomposition is introduced to handle thestochastic uncertainty
21 Model Description We consider a connected directedgraph 119866 = (V+E) which represents a pipeline gas transportnetwork We assume that there are no cycles in the graphso the network is a tree We set |V+| fl 119899 + 1 and |E| fl119898 = 119899 (119899119898 isin N) In our model every node is either ainflux node (gas enters the network) or an efflux node (gasleaves the network) An edge can either be a flux edge so thepressure decreases along the edge or a compressor edge sothe pressure increases along the edge We define E119865 as theset of all flux edges and E119862 as the set of compressor edges itfollows E = E119865 cupE119862 with E119865 cap E119862 = 0
Let 119887+ isin R119899+1 with 1119879119899+1119887+ = 0 denote the balanced load
vector and assume 119887119894 le 0 for nodes with gas influx and 119887119894 ge 0for nodes with gas efflux (119894 isin 0 119899) The vector 1119899+1
denotes the vector of all ones in the dimension 119899+1 Further-more a vector 119902 isin R119898 describes the flows through the edgesresp through the pipesThe pressures at the nodes are definedin a vector 119901+ isin R119899+1 Let pressure bounds 119901+119898119894119899 119901+119898119886119909 isinR119899+1 be given For the pressure we consider the constraints119901+ isin [119901+119898119894119899 119901+119898119886119909] For further modeling we need the fol-lowing definition
Definition 1 Consider the graph 119866 = (V+E)(i) ℎ(119890) denotes the head node of an edge 119890 and 119891(119890)
denotes the feet of an edge 119890 for all 119890 isin E(ii) 1198640(V) fl 119890 isin E | ℎ(119890) = V or 119891(119890) = V denotes all
edges which are connected to node V(iii) The matrix 119860+
119894119895 isin R119899+1times119898 119860+119894119895 fl 120590(V119894 119890119895) with
120590 (V 119890) fl minus1 if 119891 (119890) = V and 119890 isin 1198640 (V)1 if ℎ (119890) = V and 119890 isin 1198640 (V)0 if 119890 notin 1198640 (V) (1)
is called the incidence matrix of the graph 119866
(iv) For V isin V+ is Π(V) fl V cup⋃119896isinV+|exist119890isinE119891(119890)=119896andℎ(119890)=VΠ(119896) the (unique) directedpath from the root to V
(v) For V 119908 isin V+ is Π119908(V) fl Π(V) Π(119908) the (unique)directed path from119908 to V
Note that definition (iv) only makes sense since weassumed the graph to be tree-structured so there are nocycles and the union is finite The model is based upon theconservation equations of mass and momentum The massequation for the nodes is formulated for every node by
sum119890isin1198640(V)
119902119890 = 119887+V forallV isin V+ (2)
This is equivalent to Kirchhoff rsquos first law (see [9]) and byusing the definition of the incidence matrix the equation formass conservation for the whole graph is
119860+119902 = 119887+ resp 119860119902 = 119887 with
119860+ = [1198600119860 ] isin R1times119898
isin R119899times119898and 119887+ = [1198870119887 ] isin Risin R119899
(3)
As balance laws we use the isothermal Euler equations(see [3 4 10]) for a horizontal pipe so the flow through everyflux edge is modeled by
120588119890119905 + 119902119890119909 = 0
119902119890119905 + ((119902119890)2120588119890+ 1198862120588119890)
119909
= minus 1205821198902119863119890
10038161003816100381610038161199021198901003816100381610038161003816 119902119890120588119890
forall119890 isin E119865
(4)
Here 120588 is the gas density 119886 is the speed of sound 120582 is a(constant) friction coefficient and 119863 is the diameter of thepipe
We assume first that the network is in a steady state sothe time derivatives are equal to zero And second we assumethe gas flow to be slow so the coefficient (119902119890119886)2 is negligibleThese assumptions simplify the equations enormously Withthe ideal gas equation 119901 = 120588119877119878119879 (see [9]) we get adirect dependency between pressure and density Solving thissimplified equation leads us to the so called Weymouth-Equation (see [3])
1199012119891(119890) minus 1199012
ℎ(119890) = 120601119890 10038161003816100381610038161199021198901003816100381610038161003816 119902119890 with 120601119890 = (119877119878119879)2 120582119890119863119890
1198711198901198862forall119890 isin E119865 (5)
Here 119877119878 is the specific gas constant 119879 is the temperature ofthe gas and 119871119890 is the length of the pipe 119890 This equation showsthe pressure drop along the pipes (see Figure 1) Some articlesconsider the isothermal Euler equations in gas transportwithout these simplifications eg [5 10]
Compressor stations counteract the pressure drop alongthe pipes These stations are modeled as pipes without
Mathematical Problems in Engineering 3
times104
times106
15
2
25
3
35
4
45
5
55
6
pres
sure
in P
a
1 2 3 4 5 6 7 80x in m
Figure 1 Pressure drop along a pipe for 119863 = 0762119898 119879 = 28915119870119877119878 = 50027119869119896119892119870 120582 = 0005 119886 = 42917119898119904 119902 = 7951198961198921198982119904 and119901(0) = 60 sdot 105119875119886frictionWe use the followingmodel equation (see [1] section747 [11] section 23)
119888120574 ((119901119900119906119905119901119894119899
)120574 minus 1) = (6)
Here 119901119900119906119905 and 119901119894119899 are the pressures at the end resp at thebeginning of one pipe 119888 and 120574 are constants and whichis our control is the specific change in adiabatic enthalpyMore information about the parameters can be found in[11] In addition [2] contains more information about thefunctionality of compressor stations and a deduction of theequation A short transformation of the equation leads toour representation of the pressure rise along the compressoredges
(119901ℎ(119890)119901119891(119890)
)2 = 119906119890 with 119906119890 = (120574119888 + 1)2120574 forall119890 isin E119862 (7)
It holds 119906119890 ge 1 where 119906119890 = 1 means that the compressorstation is switched off (see Figure 2)
Now we define the set of feasible loads We say that avector 119887+ isin R119899+1 is feasible if (3) (5) and (7) are fulfilledSo the feasible set (here called ) is defined as follows
= 119887+ isin R119899+1 | 1119879
119899+1119887+ = 0 and exist (119901+ 119902) isin R119899+1
times R119898 119901+
isin [119901+119898119894119899 119901+119898119886119909] and (3) (5) (7) are fulfilled (8)
22 Stochastic Uncertainty in the Loads In reality the futuredemand for gas can never be known exactly a priori Itdepends on various physical effects like eg the temperatureof the environment Also personal reasons can be responsiblefor a higher or lower use of gas Thus the load vector is
times105
times10755
05
1
15
2
25
3
35
4
45
5
pres
sure
in P
a
15 2 25 3 35 4 45 51u in Jkg
Figure 2 Compressor property for 119888 = 1446526119869119896119892 120574 = 0225and 119901119894119899 = 40 sdot 105119875119886considerd to be a random vector Many papers study howthe distribution functions of the load vector can be identifiedeg [12] This article also describes why a multivariate Gaus-sian distribution is a good choice for random gas demandSo we use some random vector 120585(120596) with 120585 sim N(120583 Σ) fora mean vector 120583 and a positive definite covariance matrix Σon a suitable probability space (ΩAP) and our aim is tocalculate the following probability
P 120596 isin Ω | 120585 (120596) isin 119872 (9)
for a suitable set119872In our case we assume that the tree-structured graph
has only one influx node which gets the number zeroall remaining nodes are efflux nodes Then the graph isnumerated with breadth-first search or depth-first search andevery edge 119890 isin E gets the number maxℎ(119890) 119891(119890) Our set119872is defined as follows
119872 fl 119887 isin R119899ge0 | (minus1119879
119899119887 119887) isin (10)
There are many studies for special sets 119872 eg for poly-hedral sets (see [13]) but our set is very general So we need avery general method for calculating the probability (9) Ourmethod of choice is the so called spheric-radial decompositionIt became popular for dealing with probabilistic constraints(eg see [8 14ndash19])
Theorem 2 (spheric-radial decomposition) Let 120585 sim N(0 119877)be the 119899-dimensional standard Gaussian distribution with zeromean and positive definite correlation matrix 119877 Then for anyBorel measurable subset119872 sube R119899 it holds that
P (120585 isin 119872) = intS119899minus1
120583120594 119903 ge 0 | 119903119871V isin 119872119889120583120578 (V) (11)
where S119899minus1 is the (119899 minus 1)-dimensional sphere in R119899 120583120578 is theuniform distribution on S119899minus1 120583120594 denotes the 120594-distribution
4 Mathematical Problems in Engineering
with 119899 degrees of freedom and 119871 is such that 119877 = 119871119871119879 (egCholesky decomposition)
Thedifference with our problem is that our load vector isnormal distributed with mean value 120583 and positive definitecovariance Σ We want to use the result of Theorem 2 togeneral Gaussian distributions (see [8]) Therefore we set120585lowast fl 119863minus1 (120585 minus 120583) sim N (0 119877) with
119863 fl diag (radicΣ119894119894)119894=1119899
and 119877 fl 119863minus1Σ119863minus1 (12)
and observe that P(120585 isin 119872) = P(120585lowast isin 119863minus1(119872 minus 120583)) So ouraim is to compute the following probability
P (120585lowast isin 119863minus1 (119872 minus 120583)) (13)
So the calculation of the probability (9) for a very generalset 119872 is equivalent to compute the integral in Theorem 2which is not easy to compute analytically too But the integralcan be computed numerically in a easy and nearly exactwayTherefore we formulate the following algorithm (see [8]Algorithm 4)
Algorithm 3 Let 120585 sim N(120583 Σ) and 119871 such that 119871119871119879 = Σ(1) Sample 119873 points V1 V119873 uniformly distributed
on the sphere S119899minus1(2) Compute the one-dimensional sets 119872119894 fl 119903 ge 0 |119903119871V119894 + 120583 isin 119872 for 119894 = 1 119873(3) Set P(120585 isin 119872) asymp (1119873)sum119873
119894=1 120583120594(119872119894)The first step is quite easy There are many possibilities
to sample a set of uniformly distributed points on the sphereS119899minus1 eg Monte-Carlo sampling or the randn-generator inMATLAB Normalizing these points leads us to the samplingwe are looking for Alternatively quasi Monte-Carlo methodscan be used In [8] the authors use a quasi Monte-Carlomethod for their computations and they show its advantagesin Figure 6
Themain challenge is to handle the one-dimensional setsof the second step If the set119872 is convex these sets are closedintervals but 119872 need not to be convex The aim is to get arepresentation of 119872119894 (119894 = 1 119873) as a union of disjointintervals With this the third step can be done efficiently Thenext section deals with this representation in detail
The third step is the approximation of the integral Thevalues 120583120594(119872119894) can be calculated easily with the distributionfunction F120594 of the distribution (see [20] for details of thatdistribution) With the representation 119872119894 = ⋃119905
119895=1[119886119895 119886119895] the120594-distribution can be calculated the following
120583120594 (119872119894) = 119905sum119895=1
F120594 (119886119895) minusF120594 (119886119895) (14)
This leads to the wanted probability The 120594-distributionfunction is given by the following (see [20] p 132)
F120594 (119909) = 121198732Γ (1198732) int119909
0V1198732minus1119890minusV2119889V (15)
2 4 6 8 10 12 14 16 18 200x
0010203040506070809
1
F(x)
Figure 3 Distribution function of the Chi-square distribution for 3(blue) 5 (red) 8 (green) and 12 (purple) degrees of freedom
with Γ(12) = 120587 Γ(1) = 1 and Γ(119903 + 1) = 119903Γ(119903) for all 119903 isin R+Note that there are very precise numerical approximationsfor the distribution function eg the chi2cdf -function inMATLAB (see Figure 3)
3 Explicit Characterization of the Gas Flow
In this section we want to characterize the feasible set to useAlgorithm 3 The aim is to get a representation of the one-dimensional sets of step (2) as a union of disjoint intervalsTherefore we need another representation of the feasible setas a system of inequalitiesThis is an idea from [8]Theorem 1In the proof the authors use a representation of the pressureloss for the full network graph
(119860+)119879 (119901+)2 = minusΦ 10038161003816100381610038161199021003816100381610038161003816 119902 (16)
which is equivalent to (5) if and only if E119862 = 0 In this paperwe also consider the case E119862 = 0 The matrix Φ is diagonalwith the values 120601119890 as defined in (5) at its diagonal Thisrepresentation is not possible yet because we extended ourmodel by compressor edges so this equation does not holdfor all edges Sowe have to find away to use a similar equationlike (16) to state a suitable theorem The idea that allowsto include compressor stations is to remove compressoredges from the graph and replace them by suitable algebraicconditions First we consider the equation of masssum
119890isin1198640(V)119902119890 = 119887+V lArrrArr
sum119890isin1198640(V)capE119865
119902119890 = 119887+V minus sum119890isin1198640(V)capE119862
119902119890 (17)
We want to treat the flows through the compressor edges asloads sowehave to adapt the load vectorThenew load vector+ is defined as follows+V
fl
119887+V + 119902119890 forall119890 isin 1198640 (V) with 119890 isin E119862 and 119891 (119890) = V119887+V minus 119902119890 forall119890 isin 1198640 (V) with 119890 isin E119862 and ℎ (119890) = V119887+V else
(18)
The resulting graph 119866 = (V+E119865) is no longer connected(scheme is shown in Figure 4)
Mathematical Problems in Engineering 5
p0
p1
p2
p3
p4
p5
p6
p7
p8
p9
q1
q2
q4
q6
q7
q8
q9
b0
b1
b2
b3
b4
b5
b6
b7
b8
b9
p0
p1
p2
p3
p4
p5
p6
p7
p8
p9
q1
q2
q3 q3
q4
q5 q5
q6
q7
q8
q9
b0
b1
b2
b3
b4
b5
b6
b7
b8
b9q5 isin ℰC
q3 isin ℰC
Figure 4 Network graph with two compressor edges (left) and new not connected network graph with three connected subgraphs afterremoving the compressor edges (right)
We set |E119865| fl 1198981 and |E119862| fl 1198982 with 1198981 + 1198982 = 119898so 119866 consists of 1198982 + 1 subgraphs which are all connectedNote that 119902 is a vector in1198771198981 nowWe determine that119866119894 is the
119894-th self-connected subgraph of119866 (depends on numbering ofthe graph) Analogously 119901+
119894119895 is the 119895-th component of the 119894-thsubgraph The terms 119902119894119895 and 119860 119894119895119896 are defined similarly Thefeasible set for 119866 is then
=+ isin R119899+1 |
1119879119899+1+ = 0 andexist (119901+ 119902) isin R119899+1
ge0 times R119898 with 119901+ isin [119901+119898119894119899 119901+119898119886119909] 119860 119894119902119894 = 119894 forall119894 = 1 1198982 + 1(119860+119894 )119879 (119901+
119894 )2 = minusΦ11989410038161003816100381610038161199021198941003816100381610038161003816 ∘ 119902119894 forall119894 = 1 1198982 + 1
(119901ℎ(119890)119901119891(119890)
)2 = 119906119890 forall119890 isin E119862
(19)
The last equation in is the same as in (8) that modelsthe pressure rise caused by the compressor With this we canpresent the equation for pressure loss in the form of (16) forevery subgraph Before we state a theorem for characterizingthe feasible set we show the equivalence of and
Lemma4 Let + be as defined in (18)The feasible sets definedin (8) and (19) are equivalent in the following sense
(i) For all 119887+ isin there is + isin (ii) For all + isin there is 119887+ isin (20)
Proof (i) Consider a vector 119887+ isin It holds 1119879119899+1119887+ = 0 and
because of (18) it follows directly 1119879119899+1+ = 0 After removing
the compressor edges from119860+119902 = 119887+ it follows with (17) that119860+119894 119902119894 = +119894 which is equivalent to 119860 119894119902119894 = 119894 The pressure loss
is fulfilled for all flux edges in and because the flow vector119902 doesnrsquot change in the relevant coefficients while removingthe compressor edges the pressure loss is fulfilled in tooThe compressor property is independent of 119887+ and 119902 so it alsoholds From this it follows + isin 119872
(ii) The other way around is quite similar We consider avector + isin From 1119879
119899+1+ = 0 it follows with (18) that1119879119899+1119887+ = 0 The equation for mass conservation is fulfilled
in every subgraph So with (17) it follows 119860+119902 = 119887+ whichimplies the mass conservation in119872 The pressure loss holds
in all self-connected parts of the graph too and because therelevant coefficients of 119902 doesnrsquot change while adding thecompressor edges the pressure loss is fulfilled in119872 too Lastas said before the compressor property is independent of +and 119902 it is also fulfilled in119872 This completes the proof
Because our network graph is a tree the mass conserva-tion is fulfilledwith a full rank incidencematrix119860of size 119899times119899so 119902 can be computed explicit by the following
119902 = 119860minus1119887 (21)
which makes us able to use Lemma 4With this result we canwrite the equation formomentum
conservation depending on the incidence matrices for theparts of the tree This makes us able to formulate the theoremfor characterizing the feasible set Therefore we will use thefollowing function
119892 R119899 997888rarr R
119909 997891997888rarr (119860119879)minus1Φ 10038161003816100381610038161003816119860minus111990910038161003816100381610038161003816 (119860minus1119909) (22)
which describes the pressure loss between the nodes We alsointroduce a new notation
(i) 119906(119894119895)119896 is the 119896-th control on the path from V1198940 to V1198950(ii) 119892(119894119895)119896 is the function 119892 for the 119896-th subgraph on the
path from V1198940 to V1198950 whereas 119892(119894119895)1 is the function 119892
6 Mathematical Problems in Engineering
of the graph 119866119894 According to this 119892(119894119895)119896ℓ is the ℓ-thcomponent of the function 119892(119894119895)119896
(iii) (119894119895)119896 is the load vector of the 119896-th subgraph in thepath from V1198940 to V1198950 whereas (119894119895)1 is the load vectorof the graph 119866119894 and (119894119895)119896ℓ is the ℓ-th component of(119894119896)119896
Analogously we define 119860 (119894119895)119896 119901(119894119895)119896 and 119902(119894119895)119896 Furthermorewe define the following values
(i) 119896lowast119894119895 fl max119896 isin 1 1198982 + 1 | V1198960 isin Π(V1198940) and V1198960 isinΠ(V1198950) as the largest index of all subgraph the pathsto V1198940 and V1198950 pass through
(ii) 119899lowast119894119895 |119896 isin 1 1198982 + 1 | V1198960 isin ΠV1198940(V1198950)| as thenumber of subgraph which are on the path from V1198940from V1198950 containing 119866119894 and 119866119895
(iii) 119898lowast119894119895 fl |119890 isin E119862 | ℎ(119890) isin ΠV1198940(V1198950)and119891(119890) isin ΠV1198940(V1198950)|
as the number of controls which are between V1198940 andV1198950
With this notation and these values we define a sum for 119894 119895 isin11198982 + 1 If V1198940 isin Π(V1198950) we set sum119894119894 fl 0 and else we set
Σ119896lowast119894 fl119899lowastminus2sum119896=1
1prod119898lowastminus119896ℓ=1 119906(119896lowast119894)ℓsdot 119892(119896lowast119894)119899lowastminus119896119891(119890119906
(119896lowast119894)119899lowastminus119896) ((119896lowast119894)119899lowastminus119896)
+ 119892(119896lowast119894)1119891(119890119906(119896lowast119894)1
) ((119896lowast119894)1) (23)
where we write only 119899lowast and 119898lowast instead of 119899lowast119896lowast119894 and 119898lowast119896lowast119894 for
better reading Last we define a productΠ119894119894 fl 1 andΠ119896lowast119894 fl
119898lowastprod119896=1
119906(119896lowast119894)119896 (24)
From a first point of view the notation seems to be confusingbut using this notation makes sense since we want toguarantee the feasibility of a load vector by comparing thepressure bounds at every node with each other We explainthis in a short example Figure 5 shows a graph after removingthe compressor edges
If we want to compare the pressure bounds of node 11 andnode 12 we need to know how the pressures change on thepath from 2 to 11 resp from 2 to 12 Node 11 is part of thesubgraph 1198665 and node 12 is part of the subgraph 1198666 so itfollows 119896lowast56 = 2 because node 2 is part of subgraph 2 Thenotation can be understood as a numbering along the pathIt is 119906(211)1 = 1199062 the first control on the path from node 2to node 11 and 119906(211)2 = 1199064 the second control on this pathFurther it follows 119899lowast211 = 3 (1198662 1198663 1198665) and119898lowast
211 = 2 (1199062 1199064)Theproduct defined before is the product of all controls alongthis path and the sum is the pressure loss between node 2 andnode 7 Using this notation makes us able to formulate thetheorem for characterizing the feasible set in a readable way
Theorem 5 For given pressure bounds 119901+119898119894119899 119901+119898119886119909 isin R119899+1
and controls 119906119894 isin R (119894 = 1 1198982) the following equivalenceholds
A vector + with 1119879119899+1+ = 0 is feasible ie + isin if and
only if the following inequalities hold For all 119894 = 1 1198982 + 1holds (feasibility inside the subgraphs)
(1199011198981198941198991198940 )2 le min
119896=1119899119894[(119901119898119886119909
119894119896 )2 + 119892119894119896 (119894)] (25)
(1199011198981198861199091198940 )2 ge max
119896=1119899119894[(119901119898119894119899
119894119896 )2 + 119892119894119896 (119894)] (26)
max119896=1119899119894
[(119901119898119894119899119894119896 )2 + 119892119894119896 (119894)]
le min119896=1119899119894
[(119901119898119886119909119894119896 )2 + 119892119894119896 (119894)] (27)
For all 119894 119895 = 1 1198982 + 1 with 119894 lt 119895 holds (feasibility betweenthe subgraphs)1Π119896lowast 119894
(1199011198981198941198991198940 )2 + Σ119896lowast119894 le 1Π119896lowast119895
(1199011198981198861199091198950 )2 + Σ119896lowast119895 (28)
1Π119896lowast 119894
(1199011198981198861199091198940 )2 + Σ119896lowast119894 ge 1Π119896lowast 119895
(1199011198981198941198991198950 )2 + Σ119896lowast119895 (29)
1Π119896lowast 119894
(1199011198981198941198991198940 )2 + Σ119896lowast119894
le 1Π119896lowast119895
min119896=1119899119895
[(119901119898119886119909119895119896 )2 + 119892119895119896 (119895)] + Σ119896lowast119895 (30)
1Π119896lowast 119894
(1199011198981198861199091198940 )2 + Σ119896lowast119895
ge 1Π119896lowast119895
max119896=1119899119895
[(119901119898119894119899119895119896 )2 + 119892119895119896 (119895)] + Σ119896lowast119895 (31)
1Π119896lowast 119894
max119896=1119899119894
[(119901119898119894119899119894119896 )2 + 119892119894119896 (119894)] + Σ119896lowast119894
le 1Π119896lowast119895
(1199011198981198861199091198950 )2 + Σ119896lowast119895 (32)
1Π119896lowast 119894
min119896=1119899119894
[(119901119898119886119909119894119896 )2 + 119892119894119896 (119894)] + Σ119896lowast119894
ge 1Π119896lowast119895
(1199011198981198941198991198950 )2 + Σ119896lowast119895 (33)
1Π119896lowast 119894
max119896=1119899119894
[(119901119898119894119899119894119896 )2 + 119892119894119896 (119894)] + Σ119896lowast119894
le 1Π119896lowast119895
min119896=1119899119895
[(119901119898119886119909119895119896 )2 + 119892119895119896 (119895)] + Σ119896lowast119895 (34)
1Π119896lowast 119894
min119896=1119899119894
[(119901119898119886119909119894119896 )2 + 119892119894119896 (119894)] + Σ119896lowast119894
ge 1Π119896lowast119895
max119896=1119899119895
[(119901119898119894119899119895119896 )2 + 119892119895119896 (119895)] + Σ119896lowast119895 (35)
Mathematical Problems in Engineering 7
0 1 2
3
4
5
6
7
8
9
10
11
12
u1
u2
u3
u4
u5
G1 G2 G3
G4
G5
G6
Figure 5 Example graph for illustration of the used notation
0 1 2q1 q2
b0 b1 b2
Figure 6 Graph of Example 1
Let us have a look at the inequalities Inequalities(25)ndash(27) are well-known from [8] and they describe thefeasibility inside a connected tree The other inequalities(28)ndash(35) guarantee the feasibility between the subgraphsTherefore we compare the pressures of the entries of two parts((28) (29)) of the entry of one part and the exit nodes of theother parts ((30)ndash(33)) and we compare the pressures of allexits ((34) (35)) Because we cannot compare the pressuresof119866119894 and 119866119895 directly (there may be no direct connection) wehave to compare them using the subgraph 119866119896lowast which is thelast part the paths to V1198940 and V1198950 pass Therefore we definedthe sumsΣ119896lowast 119894 andΣ119896lowast119895 which trace back the pressures to119866119896lowast
Remark 6 For an implementation it is wise to define a set119880(119894119895) containing all controls on the path from V1198940 to V1198950 anduse an index running over this set to define the sums and theproducts
Now we want to proof the theorem
Proof of Theorem 5 Because the columns of 119860+ sum up tozero it holdsminus1119879
119899119894119860 119894 = 119860+
1198940 (119894 = 1 1198982 + 1) (36)
ldquo997904rArrrdquo We consider a feasible vector 119887+ isin From theequation of mass conservation of every subgraph it follows
119902119894 = 119860minus1119894 119894 (37)
We take the equation of momentum conservation for the 119894-thsubgraph (119894 isin 1 1198982 +1) and separate 119860+
119894 to119860+1198940 and 119860 119894
(119860+1198940)119879 1199012
1198940 + 119860119879119894 1199012
119894 = minusΦ 10038161003816100381610038161199021198941003816100381610038161003816 119902119894 (38)
Next we insert (36) into (38) and get
minus119860119879119894 1119899119894
11990121198940 + 119860119879
119894 1199012119894 = minusΦ 10038161003816100381610038161199021198941003816100381610038161003816 119902119894 (39)
We multiply this equation from left with (119860119879119894 )minus1 and insert
(37) so it follows
111989911989411990121198940 = 1199012
119894 + (119860119879119894 )minus1Φ 10038161003816100381610038161003816119860minus1
119894 11989410038161003816100381610038161003816 119860minus1119894 119894 (40)
and using the definition of the function 119892 we have
111989911989411990121198940 = 1199012
119894 + 119892119894 (119894) (41)
Note here that 119901119894 is the vector of pressures of the subgraph119866119894
and 119892119894(119894) is the vector-valued function which describes thepressure loss in 119866119894 Equation (41) holds for all subgraphs 119866119894
with 119894 isin 1 1198982 + 1 From this the inequalities (25)ndash(27)follow directly with 119901+ isin [119901+119898119894119899 119901+119898119886119909] Next we use (41)component-by-component For fix 119894 119895 isin 1 1198982 + 1 with119894 lt 119895 it follows
1199012119896lowast0 = 1199012
119896lowast119896 + 119892119896lowast119896 (119896lowast) (119896 = 1 119899119896lowast) (42)
Remember that 119896lowast119894119895 is the largest index of all subgraphs thepaths to V1198940 and V1198950 pass through For better reading we onlywrite 119896lowast instead of 119896lowast119894119895 We use the equation of the compressorproperty for the first compressor on the path from V119896lowast0 to V1198940
1199012ℎ(119890119906(119896lowast119894)1
)1199012119891(119890119906(119896lowast119894)1
)
= 119906(119896lowast119894)1 (43)
and insert it into (42) for 119896 = 119891(119890119906(119896lowast119894)1 )1199012(119896lowast 119894)10 = 1119906(119896lowast119894)1
1199012(119896lowast119894)20
+ 119892(119896lowast 119894)1119891(119890119906(119896lowast119894)1
) ((119896lowast119894)1) (44)
Note that 1199012119896lowast 0 = 1199012
(119896lowast119894)10 119892119896lowast119896(119896lowast) = 119892(119896lowast119894)1119896((119896lowast119894)1) and1199012ℎ(119890119906(119896lowast119894)1
) = 1199012(119896lowast 119894)20 We use the compressor equation for
the second compressor on the path from V119896lowast0 to V1198940 and putit together with (42) into (44)
1199012(119896lowast119894)10 = 1119906(119896lowast119894)1
[ 1119906(119896lowast119894)2
1199012(119896lowast 119894)30
+ 119892(119896lowast119894)2119891(119890119906(119896lowast119894)2
) ((119896lowast119894)2)]+ 119892(119896lowast119894)1119891(119890119906
(119896lowast119894)1) ((119896lowast119894)1)
(45)
8 Mathematical Problems in Engineering
We repeat this procedure for all controls on the path fromV119896lowast0 to V1198940 Although we write 119899lowast and119898lowast instead of 119899lowast119896lowast119894 and119898lowast
119896lowast119894 Thus we get
1199012(119896lowast119894)10 = 1119906(119896lowast119894)1
[ 1119906(119896lowast119894)2
[sdot sdot sdot [ 1119906(119896lowast119894)119898lowast1199012(119896lowast119894)119899lowast0
+ 119892(119896lowast119894)119899lowastminus1119891(119890119906(119896lowast119894)119899lowastminus1
) ((119896lowast119894)119899lowastminus1)] sdot sdot sdot]+ 119892(119896lowast119894)2119891(119890119906
(119896lowast119894)2) ((119896lowast119894)2)]
+ 119892(119896lowast119894)1119891(119890119906(119896lowast119894)1
) ((119896lowast119894)1)
(46)
We take again (42) for the node V119896lowast0 and do the sameprocedure just like before for the path from V119896lowast0 to V1198950 With119899lowast = 119899lowast119896lowast119895 and119898lowast = 119898lowast
119896lowast119895 we get
1199012(119896lowast 119895)10
= 1119906(119896lowast119895)1
[ 1119906(119896lowast119895)2
[sdot sdot sdot [ 1119906(119896lowast119895)119898lowast1199012(119896lowast 119895)119899lowast0
+ 119892(119896lowast119895)119899lowastminus1119891(119890119906(119896lowast119895)119899lowastminus1
) ((119896lowast119895)119899lowastminus1)] sdot sdot sdot]+ 119892(119896lowast119895)2119891(119890119906
(119896lowast119895)2) ((119896lowast119895)2)]
+ 119892(119896lowast119895)1119891(119890119906(119896lowast119895)1
) ((119896lowast119895)1)
(47)
Expanding (46) and (47) and equalizing them (this is possiblebecause 119901(119896lowast 119894)10 = 119901(119896lowast119895)10) leads to1Π119896lowast 119894
11990121198940 + Σ119896lowast119894 = 1Π119896lowast 119895
11990121198950 + Σ119896lowast119895 (48)
with Σ119896lowast119894 resp Σ119896lowast119895 defined before (48) directly implies (28)and (29) Using (48) together with (42) for V1198940 first timefor V1198950 second time and for V1198940 and V1198950 a third time theinequalities (30)ndash(35) follow directly So this part of the proofis complete
ldquolArr997904rdquo For this part we consider a vector 119887+ isin R119899+1
with 1119879119899+1119887+ = 0 which fulfills the inequalities (25)ndash(35) We
define the following sets (119894 = 2 1198982 + 1)11987511 fl [(119901119898119894119899
10 )2 (11990111989811988611990910 )2]
11987521 fl [ max119896=11198991
[(1199011198981198941198991119896 )2 + 1198921119896 (1)]
min119896=11198991
[(1199011198981198861199091119896 )2 + 1198921119896 (1)]]
1198751119894 fl [ 1Π1119894
(1199011198981198941198991198940 )2 + Σ1119894 1Π1198941
(1199011198981198861199091198940 )2 + Σ1119894]
1198752119894 fl [ 1Π1119894
max119896=1119899119894
[(119901119898119894119899119894119896 )2 + 119892119894119896 (119894)] + Σ1119894 1Π1198941
sdot min119896=1119899119894
[(119901119898119886119909119894119896 )2 + 119892119894119896 (119894)] + Σ1119894]
(49)
We set
1198750 fl 1198982+1⋂119896=1
1198751119896 cap 1198982+1⋂119896=1
1198752119896 (50)
and we will define a value 119901210 isin 1198750 later So to guarantee
such a value 119901210 exists we have to show first that 1198750 is not
empty Because 1198750 is a finite intersection of convex intervalswe have to make sure that all intervals are not empty andevery intersection between two intervals is not empty In onedimension this is sufficient that 1198750 is not empty For 119894 =1 1198982 + 11198751119894 = 0 according to requirement1198752119894 = 0 because of (26) (51)
Furthermore its 1198751119894 cap 1198752119894 = 0 because of (25) and (26) Themissing intersections are nonempty too for 119894 119895 isin 1 1198982+1 with 119894 lt 119895 as follows
1198751119894 cap 1198751119895 = 0 because of (28) and (29) 1198751119894 cap 1198752119895 = 0 because of (30) and (31) 1198752119894 cap 1198751119895 = 0 because of (32) and (33) 1198752119894 cap 1198752119895 = 0 because of (34) and (35)
(52)
So the set 1198750 is not empty We define the following values
119901210 isin 119875011990121198940 fl Π1119894 (1199012
10 minus Σ1119894) 119894 = 2 1198982 + 11199012119894119896 fl 1199012
1198940 minus 119892119894119896 (119894) 119894 = 1 1198982 + 1 119896 = 1 119899119894119902119894 fl 119860minus1119894 119894 119894 = 1 1198982 + 1
(53)
Now we show that our choice of pressures and flows isfeasible It is11990110 isin [119901119898119894119899
10 11990111989811988611990910 ] because of1199012
10 isin 11987511 Furtherbecause of 1199012
10 isin 1198751119894 for all 119894 = 2 1198982 + 1 it holds1Π1119894
11990121198940 + Σ1119894 le 1Π1119894
(1199011198981198861199091198940 )2 + Σ1119894 (54)
and 1Π1119894
11990121198940 + Σ1119894 ge 1Π1119894
(1199011198981198941198991198940 )2 + Σ1119894 (55)
Mathematical Problems in Engineering 9
Now we replace 11990121198940 in its definition with (1199012
119894119896 + 119892119894119896(119894)) andbecause of 1199012
1198940 isin 1198752119894 for all 119894 = 1 1198982 + 1 it follows1Π1119894
(1199012119894119896 + 119892119894119896 (119894)) + Σ1119894
le 1Π1119894
min119896=1119899119894
[(119901119898119886119909119894119896 )2 + 119892119894119896 (119894)] + Σ1119894
(56)
and 1Π1119894
(1199012119894119896 + 119892119894119896 (119894)) + Σ1119894
ge 1Π1119894
max119896=1119899119894
[(119901119898119894119899119894119896 )2 + 119892119894119896 (119894)] + Σ1119894 (57)
So for all 119894 = 1 1198982 + 1 and for all 119896 = 1 119899119894 it holds119901119894119896 isin [119901119898119894119899119894119896 119901119898119886119909
119894119896 ] which is equivalent to
119901+ isin [119901+119898119894119899 119901+119898119886119909] (58)
The equation for mass conservation follows directly from thedefinition of 119902119894 (119894 = 1 1198982 + 1) For the pressure loss weuse
1199012119894 = 1119899119894
11990121198940 minus 119892119894 (119894) (59)
And multiply this equation from left with 119860119879119894 and use (36)
With this it follows the equation for momentum conserva-tion
(119860+119894 )119879 (119901+
119894 )2 = minusΦ 10038161003816100381610038161199021198941003816100381610038161003816 119902119894 (60)
The last equation missing is the compressor property For acontrol 119895 isin 1 1198982 we fix a vertex V1198940 (119894 isin 1 1198982 +1)with ℎ(119890119906119895) = V1198940 We use 1199012
1198940 = Π1119894(119901210 minus Σ1119894) and solve the
formula for 119901210 in the following way
119901210 = 1119906(1119894)1
[sdot sdot [ 1119906(1119894)119898lowast1199012(1119894)119899lowast0
+ 119892(1119894)119899lowastminus1119891(119890119906(1119894)119899lowastminus1
) ((1119894)119899lowastminus1)] sdot sdot]+ 119892(1119894)1119891(119890119906(1119894)1 )
((1119894)1) (61)
Consider wlog that the subgraphs along the path from V10to V1198940 are numbered chronological st if ℎ(119890119906119895) isin 119866119894 then119891(119890119906119895) isin 119866119894minus1 We use 1199012
119894minus10 = Π1119894minus1(119901210 minus Σ1119894minus1) to get the
same representation for 119901210 depending on 1199012
(1119894minus1)119899lowast0 whichis equivalent to 119901(1119894)119899lowastminus10 We equalize these representationsdepending on 1199012
1198940 and 1199012119894minus10 and we get
1119906(1119894)119898lowast1199012(1119894)119899lowast0 + 119892(1119894)119899lowastminus1119891(119890119906
(1119894)119899lowastminus1) ((1119894)119899lowastminus1)
= 1199012(1119894)119899lowastminus10 (62)
We chose 119894 isin 1 1198982 + 1 st 119906119895 is the last control on thepath from V10 to V1198940 so it is 119906(1119894)119898lowast = 119906119895 Further we have
1199012(1119894)119899lowast 0 = 1199012
11989401199012(1119894)119899lowastminus10 = 1199012
119894minus10119892(1119894)119899lowastminus1119891(119890119906(1119894)119899lowastminus1
) (119887(1119894)119899lowastminus1) = 119892119894minus1119891(119890119906119895 )(119887119894minus1)
(63)
and we know that 11990121198940 resp 1199012
119894minus10 minus 119892119894minus1119891(119890119906119895 )(119887119894minus1) are the
pressures at the head resp the foot of the compressor edge119890119906119895 So it follows 1119906119895
1199012ℎ(119890119906119895 )
= 1199012119891(119890119906119895 )
(64)
which is equivalent to the compressor property This com-pletes the proof
With this theorem we have another characterization ofthe feasible set The last step in this intersection is theadaption of the results to the spheric-radial decompositionAs mentioned before the main step is to get a representationof the sets119872119894 in step (2) of Algorithm 3
Because we assumed that the graph has only one entry wedefine the following set
119872 fl 119887 isin R119899ge0 | (minus1119879
119899119887 119887) isin (65)
Consider a sampled point V119894 isin S119899minus1 of step (1) of Algorithm 3(119894 = 1 119873) We identify the load vector 119887 with the affinelinear function
119887119894 (119903) fl 119903119871V119894 + 120583 = 119903120596119894 + 120583 with 120596119894 = 119871V119894 (66)
where 120583 is the mean value of the Gaussian distribution and 119871is such that 119871119871119879 = Σ for positive definite covariance matrixΣ Note that the index 119894 here is for the sampled vector V119894 notfor the 119894-th subgraph Because 119887119894 is defined only for the exitnodes it must hold 119887 ge 0 which leads to the definition of theregular range
119877119903119890119892119894 fl 119903 ge 0 | 119887119894 (119903) ge 0 supe 119872119894 (67)
So the feasible set is
119872119894 = 119903 isin 119877119903119890119892119894 | 119887119894 (119903) fulfills (3) (5) and (7) (68)
and with Theorem 5 it holds
119872119894 = 119903 isin 119877119903119890119892119894 | 119894 (119903) fulfills (25) minus (34) (69)
where 119894(119903) is from (18) Because 119894 depends on 119903 now thefunction 119892 depends on 119903 too More exactly 119892 is quadratic in119903 so 119892119896(119887119894(119903)) can be represented as follows
1199032[[119899sum
119895=1
Ψ119896119895( 119899sumℓ=1
119860minus1119895ℓ120596119894ℓ)2]]⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
119886119894
10 Mathematical Problems in Engineering
+ 119903[[2119899sum
119895=1
Ψ119896119895( 119899sumℓ=1
119860minus1119895ℓ120596119894ℓ)( 119899sum
ℓ=1
119860minus1119895ℓ120583ℓ)]]⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
119894
+ [[119899sum
119895=1
Ψ119896119895( 119899sumℓ=1
119860minus1119895ℓ120583ℓ)2]]⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
119888119894
(70)
HereΨ is the product of119860 andΦ So the inequations whichcharacterize119872119894 are also quadratic in 119903 because the minimaland maximal pressures in the inequalities only count to 119888119894Now we can write the feasible sets as follows
119872119894 = 119903 isin 119877119903119890119892119894 | 0 le 1198861198941199051199032 + 119887119894119905119903 + 119888119894119905 (71)
where 119905 is an index for the total number of inequalities in119872119894Now the regular range can be cut with the positive intervals ofthe inequalities and the result is a union of disjoint intervals
119872119894 = 119904119894⋃119896=1
[119886119894119896 119886119894119896] 119904119894 isin N (72)
Here 119904119894 is the number of disjoint intervals in which allinequalities are fulfilled and 119886119894119896 and 119886119894119896 are the intervalbounds they depend on the pressure bounds and the con-trols So the third step of Algorithm 3 can be executed With
P (119887 isin 119872) asymp 1119873 119873sum119894=1
119904119894sum119896=1
F120594 (119886119894119896) minusF120594 (119886119894119896) (73)
the probability for a random load vector to be feasible can becomputed
4 Existence of Optimal Solutions
In this section we want to have a look at some optimizationproblems We distinguish between problems with constantloads (without uncertainty on the demand) and problemswith random loads The latter leads us to optimizationproblems with so called probabilistic constraints or chanceconstraints (see [21]) There are two aims in optimizationMinimize the maximal pressure bounds and minimize thecontrols For minimizing the maximal pressure bounds wedefine the following set
119875 fl119899⋃
119894=0
[119901+119898119894119899119894 infin) (74)
in which we look for them Because the maximal pressuresare in proportion to the cost of pipes an objective functionfor a optimization problem could be 119888119879119901+119898119886119909 for a cost vector119888 isin R119899+1
gt0 By minimizing the controls we use the Euclideannorm of the controls as an objective function
Optimization with Constant Loads Consider the two opti-mization problems
min119901+119898119886119909isin119875
119888119879119901+119898119886119909
subject to 119887+ (119901+119898119886119909) isin 119872 (75)
andmin119906isinR1198982ge1
1199062subject to 119887+ (119906) isin (76)
Lemma 7 There exists a unique solution for (75)
Proof We have a look at the side constraint 119887+(119901+119898119886119909) isin Using Theorem 5 we can write (75) as
min119901+119898119886119909isin119875
119888119879119901+119898119886119909
subject to (25) minus (35) are fulfilled (77)
For 119894 isin 1 1198982 + 1 we write inequality (25) (and allremaining analogous) in the following way
0 le 119892119894119896 (119894) minus (1199011198981198941198991198940 )2 + (119901119898119886119909
119894119896 )2 forall119896 = 1 119899119894 (78)
So every inequality depends only on one upper pressurebound (but more inequalities may depend on the samebound) Because the inequalities are not strict the values119901119896 fl min 119901119898119886119909
119896 isin [0infin) | (25) minus (35) are fulfilled (79)
exist for all 119896 = 1 119899 + 1 and we can define a value119901+119898119886119909119900119901119905119896 fl max 119901119896 119901119898119894119899
119896 (80)
which is obviously a solution of (75) Moreover this solutionis unique because reducing one component of the upperbounds would either hurt an inequality or it would not fulfill119901+119898119886119909 ge 119901+119898119894119899 anymore
For the optimization problem (76) we can formulate anecessary and a sufficient condition for optimal solutions
Lemma8 Consider pressure bounds119901+119898119894119899 119901+119898119886119909 isin R119899+1 st119901+119898119894119899 lt 119901+119898119886119909 and a suitable vector 119887+ isin R119899+1 (ie + fulfills(25)ndash(27)) For all 119896 = 1 1198982 we set 119894(119896) and 119895(119896) such that119891(119890119896) isin V+
119894(119896) and ℎ(119890119896) isin V+119895(119896)
(a) If 119906 isin R1198982 is a solution of (76) it holds for all 119896 =1 1198982 (119901119898119894119899119894(119896)0)2 minus 119892119894(119896)119891(119890119896)
(119894(119896)) le (119901119898119886119909119895(119896)0)2 (81)
(b) If the inequalities
(119901119898119886119909119894(119896)0)2 minus 119892119894(119896)119891(119890119896)
(119894(119896)) ge (119901119898119894119899119895(119896)0)2
and (119901119898119894119899119894(119896)0)2 minus 119892119894(119896)119891(119890119896)
(119894(119896)) le (119901119898119886119909119895(119896)0)2 (82)
are fulfilled for all 119896 = 1 1198982 an optimal solution 119906 isin R1198982ge1
of (76) exists
Mathematical Problems in Engineering 11
Note that a vector 119887+ isin R119899+1 is suitable here if the vector+ is feasible for the subgraphs ie the inequalities (25)-(27)are fulfilled for +Proof (a) For this part we use a proof by contradictionAssume that
(119901119898119894119899119894(119896)0)2 minus 119892119894(119896)119891(119890119896)
(119894(119896)) gt (119901119898119886119909119895(119896)0)2 (83)
Because 119906119896 ge 1 for all 119896 = 1 1198982 it follows
(119901119898119894119899119894(119896)0)2 minus 119892119894(119896)119891(119890119896)
(119894(119896)) gt 1119906119896
(119901119898119886119909119895(119896)0)2 (84)
It it Σ119896lowast119894(119896) = 0 and Π119896lowast 119894(119896) = 1 because the subgraph withindex 119896lowast is equal to the subgraphwith index 119894(119896) And becausethe subgraphs 119894(119896) and 119895(119896) are neighbours (ie directlyconnected) it follows Π119896lowast 119895(119896) = 119906119896 and
Σ119896lowast119895(119896) = 119892(119896lowast119895(119896))1119891(119890119906(119896lowast119895(119896))1
) (119887(119896lowast119895(119896))1)= 119892119894(119896)119891(119890119896)
(119894(119896)) (85)
Thus we have a contradiction to (28) and 119906119896 cannot beoptimal
(b) As mentioned before the vector 119887+ is such that isfeasible for the subgraphs So because of the conditions for119896 isin 1 1198982 there exists a 119901119894(119896)0 isin [119901119898119894119899
119894(119896)0 119901119898119886119909119894(119896)0] and a119901119895(119896) isin [119901119898119894119899
119895(119896) 119901119898119886119909119895(119896) ] so that1199012119894(119896)0 minus 119892119894(119896)119891(119890119896)
(119894(119896)) ge 1199012119895(119896)0 (86)
With 1199012119894(119896)0 minus 119892119894(119896)119891(119890119896)
(119894(119896)) = 1199012119891(119890119896)
and 1199012119895(119896)0 = 1199012
ℎ(119890119896)it
follows
(119901ℎ(119890119896)119901119891(119890119896)
) ge 1 (87)
Therefore we can find a 119906 isin R1198982ge1 that the inequalities (28)-
(35) are fulfilled Now we can use theWeierstraszlig theorem andget the existence of an optimal solution
Note that the sufficient condition in Lemma 8 is com-paratively strong there may exist optimal solutions with lessstrong conditions eg the following
Corollary 9 If the graph has only one compressor edge ie|E119862| = 1 then it holds If 119906 isin Rge1 is an optimal solution of(76) 119906 is unique
This conclusion follows directly from the strict mono-tonicity of the objective function
Optimization with Random Loads In this section we con-sider the load vector to be random as we mentioned inSection 22 In the optimization problems we want to makesure that the feasibility of a random load vector is guaranteedfor a probability 120572 isin (0 1) This leads us to optimizationproblems with so called chance or probabilistic constraints
In general problems with probabilistic constraints have theform
min 119891 (119909 120585)subject to P (ℎ (119909 120585) le 0) ge 120572 (88)
with an objective function 119891 and a vector of uncertainty 120585There exist many works about probabilistic constraints respstochastic programming eg [7] gives an excellent overviewabout chance constraints in theory and application In ourcase we consider the following problem
min119901+119898119886119909isin119875
119888119879119901+119898119886119909
subject to P (119887 (119901+119898119886119909) isin 119872) ge 120572 (89)
To handle this problem resp the probabilistic constraint thespheric-radial decomposition gives us an explicit representa-tion of the constraint for a Gaussian vector 119887 sim N(120583 Σ) withmean value 120583 and positive definite covariance Σ
P (119887 (119901+119898119886119909) isin 119872)= int
S119899minus1120583120594 119903 ge 0 | 119903119871V + 120583 isin 119872119889120583120578 (V) (90)
Further Algorithm 3 gives a way to approximate this integralin an efficient way In Section 3 we showed a way to character-ize the feasible set119872 and how to adapt this characterization tothe spheric-radial decomposition This changes our problemwith probabilistic constraints (89) to
min119901+119898119886119909isin119875
119888119879119901+119898119886119909
subject to 1119873 119873sum119894=1
119904119905sum119905=1
F120594 (119886119894119905) minusF120594 (119886119894119905) ge 120572 (91)
Of course the interval bounds depend on the pressure boundsand the control For this problem we formulate the nextlemma
Lemma 10 Let a sampling V1 V119873 (from step (1) ofAlgorithm 3) and 120572 isin (0 1) be given For a suitable 120583 isin R119899
problem (91) has at least one solution 119901+119898119886119909 isin 119875Proof For the proof we fix a V isin V1 V119873 from the givensampling From (25) and (26) it follows for all 119896 = 1 119899119894with 119894 isin 1 1198982 + 10 le 119892119894119896 (119894 (119903)) + (119901119898119886119909
119894119896 )2 minus (1199011198981198941198991198940 )2
0 le minus119892119894119896 (119894 (119903)) + (1199011198981198861199091198940 )2 minus (119901119898119894119899
119894119896 )2 (92)
With a random vector 119887 sim N(120583 Σ)119892119894119896(119894(119903)) are quadraticfunctions in 119903 and the upper pressure bounds only influencethe constant part with a positive sign
0 le 1198861198941198961199032 + 119887119894119896119903 + 119888119894119896 + (119901119898119886119909119894119896 )2 minus (119901119898119894119899
1198940 )2 0 le minus1198861198941198961199032 minus 119887119894119896119903 minus 119888119894119896 + (119901119898119886119909
1198940 )2 minus (119901119898119894119899119894119896 )2 (93)
12 Mathematical Problems in Engineering
It follows that a rise in 119901119898119886119909119894119896 for all 119896 = 0 119899119894 enlarges
the intervals in which the inequalities hold The same holdsfor all inequalities inTheorem 5The cumulative distributionfunction of theChi-Square-Distribution is strictlymonotonicincreasing and convergent to 1 Thus upper pressure bounds119901119898119886119909119894119896 (for 119894 = 1 119899) can be found so that
1119873 119873sum119895=1
119904119895sum119905=1
F120594 (119886119895119905) minusF120594 (119886119895119905) (94)
is higher that 120572 isin (0 1) The index119873 is for all sampled pointson the sphere and the index 119904119895 is for the union of intervals inwhich all inequalities hold The interval boundaries 119886119895119905 and119886119895119905 depend continuously on the pressure bounds and becausethe inequalities are not strict 119901119898119886119909 can be found st the sideconstraint is fulfilled with equality So there exists at least onesolution of (91)
5 Numerical Results
In this section we show a few results of implementation Atfirst we show the idea of the spheric-radial decomposition byusing an easy example Next we show a easy example with onecompressor edge and at last we use real data of the Greek gasnetworkThe focus of the implementation is on the theoremsproofed in Section 4
Example 1 Our first example is an easy graph without innercontrol (see Figure 6)
Assume that node 1 and 2 are gas consumers with meandemand 120583 = [05 05]119879 The pressure bounds are given by119901+119898119894119899 = [2 1 1]119879 and 119901+119898119886119909 = [3 2 2]119879 and Φ = E2 Thenthe incidence matrix is
119860+ = [[[minus1 01 minus10 1
]]] resp 119860 = [1 minus10 1 ] (95)
Consider the inequalities of Theorem 5 to characterize thefeasible set Because the network has no compressor edgesthe system of inequalities only contains the information (25)-(27) the other inequalities do not occur The function 119892 isgiven by
119892 (119887) = [1 01 1] [1 00 1] sdot1003816100381610038161003816100381610038161003816100381610038161003816[1 10 1](11988711198872)
1003816100381610038161003816100381610038161003816100381610038161003816∘ ([1 10 1](11988711198872)) = ( 11988721 + 211988711198872 + 1198872211988721 + 211988711198872 + 211988722) (96)
By inserting the pressure bounds one can see that inequality(25) is always fulfilled and inequality (27) implies1198872 le radic3 (97)
With this inequality (26) follows for 0 le 1198872 le radic30 le 1198871 le minus1198872 + radic8 minus 11988722 (98)
So the feasible set119872 is the following (see Figure 7)
119872 = 119887 isin R2ge0 | 1198872 le radic3 and 1198871 le minus1198872 + radic8 minus 11988722 (99)
Now consider the optimization problem
min119901+119898119886119909isin119875
11198793119901+119898119886119909
subject to 119887 isin 119872 (119901+119898119886119909) (100)
with 119887 = 120583 = [05 05]119879 Lemma 7 guarantees the existenceof a unique solution of this problem We use the fmincon-function inMATLAB which minimizes a constrained non-linear multivariable function The fmincon default settinguses a interior-point method For more information we referto the official MathWorks homepage (httpswwwmath-workscomhelpoptimugfminconhtml)The solver returnsthe following result
119901+119898119886119909119900119901119905 = [2 radic3radic275]119879 (101)
One can easily see that this is the correct solution The gasdemand at node 2 is 05 which is equal to the flow 1199022 sousing the equation of momentum implies a difference in thequadratic pressures between node 1 and node 2 of 025 Theflow through edge 1 is given by the two exit nodes and it is1199021 = 1 This implies a difference in the quadratic pressuresbetweennode 0 andnode 1 of 1 So the optimal solution119901+119898119886119909
119900119901119905
directly followsNext we use the spheric-radial decomposition especially
Algorithm 3 to calculate the probability P(119887 isin 119872) for arandom vector 119887 sim N(E2 [05 05]119879) Therefore we viewthe results of 8 tests with 1000 sampled points in each one(Table 1)
The probability in all eight tests is nearly the same Themean value is 03479 and the variance is 27723 sdot 10minus6 whichis very small This shows that the implementation is workingnearly exact The efficiency of the implementation of courseis not perfect but that is not part of this work
Last wewant to solve the following optimization problem
min119901+119898119886119909isin119875
11198793119901+119898119886119909
subject to 1119873 119873sum119894=1
119904sum119905=1
F120594 (119886119894119905) minusF120594 (119886119894119905) ge 120572 (102)
We use again the fmincon-function inMATLAB to solve theproblem eight times with 1000 sampled points in every caseBecause Lemma 8 needs a suitable 120583 isin R2
ge0 we assume 119887 simN(01sdot1198642 [15 15])We set120572 fl 09 and the results are shownin Table 2
Example 2 Now we add a compressor edge to the graph ofExample 1 (see Figure 8)
Assume that node 1 and 3 are consumers node 2 is ainner nodeThemeandemand120583 = [05 0 05]119879Thepressure
Mathematical Problems in Engineering 13
minus2 minus1 1 2 3
minus2
minus1
1
2
3
b2
b1
minus1 1 2
minus1
1
2
b2
b1
rLv1
rLv2
rLv3
Scheme of the spheric-radial decomposition with
sphere (yellow)and three points 1 2 3 (red) at the unit
the feasible set Mminus for Σ = E2 = [05 05]TFeasible set M for p+min = [2 1 1]T and
p+max = [3 2 2]T
Figure 7 Feasible set of Example 1
Table 1 Test results of easy example
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 8P(119887 isin 119872) 03486 03481 03450 03501 03484 03499 03459 03475
Table 2 Test results of Example 1
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 81199011198981198861199091199001199011199050 54514 54714 54758 54028 54767 53921 54888 551671199011198981198861199091199001199011199051 29802 30126 30225 29725 30300 29485 30336 300291199011198981198861199091199001199011199052 10084 10004 10776 10001 11221 11390 10285 10558
0 1 2 3q1 q2 q3
b0 b1 b2 b3
e1 isin ℰF e2 isin ℰC e3 isin ℰF
Figure 8 Graph of Example 2
0 1 2 3q1 q2 q2 q3
b0 b1 b2 b3
e1 isin ℰF e3 isin ℰF
Figure 9 Separated graph of Example 2
bounds are given by 119901+119898119894119899 = [2 1 1 1]119879 and 119901+119898119886119909 = [3 22 2]119879 and 1206011198901= 1 and 1206011198903
= 1 The incidence matrix is givenby
119860+ = [[[[[[
minus1 0 01 minus1 00 1 minus10 0 1]]]]]]
resp 119860 = [[[1 minus1 00 1 minus10 0 1
]]] (103)
Before we have a look at the inequalities of Theorem 5we have to remove the compressor edge from the graphThe flows can be computed by using the equation of massconservation and the new load vector is then
= (1198871 + 11990221198872 minus 11990221198873 ) with
119902 = (1198871 + 1198872 + 11988731198872 + 11988731198873)(104)
We assume the control to be switched off so it holds119906 = 1With this the control edge 1198902 is treated as a flow edge withoutfriction The resulting graph is shown in Figure 9 from theinequalities (25)ndash(35) it follows
3 le radic3 and 1 le radic8 minus 23 (105)
14 Mathematical Problems in Engineering
minus1 1 2 3
1
2
3
minus1 1 2 3
1
2
3
b3
b1
Hyperspace of the feasible set MHyperspace of the feasible set M
b1
b3
Figure 10 Feasible set of the Example 2
Table 3 Test results of control example
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 8P(119887 isin 119872) 01531 01500 01540 01536 01530 01550 01567 01528so the feasible set is (see Figure 10)
= + isin R119899+1 | 1119879
4 + = 0 and 0 le 3 le radic3 and 2 = 0and 0 le 1 le radic8 minus 22 (106)
This implies
119872 = 119887 isin R3ge0 | 0 le 1198873 le radic3 and 1198872 = 0 and 0 le 1198871 le minus1198873
+ radic8 minus 11988723 (107)
Solving the problem
min119901+119898119886119909isin119875
11198794119901+119898119886119909
subject to isin (108)
for 119887 = 120583 = [05 0 05]119879 leads us to the optimal solution
119901+119898119886119909119900119901119905 = [2 radic3radic3radic275]119879 (109)
The argumentation for the optimality of this vector is exactlylike before in the easy example Becausewe have a compressoredge in this graph we can minimize the control Thereforewe fix the pressure bounds as we did at the beginning of thisexample and we set 119887 = [1 0 15]119879 So inequality (31) is notfulfilled anymore for 119906 = 1 We use again the fmincon-solverinMATLAB to solve the following problem
min119906ge1
1199062subject to 119887 isin 119872(119906) (110)
The solver returns the optimal solution 119906119900119901119905 = 11818
Here one can see that the sufficient condition in Lemma 8is really strong Itrsquos not fulfilled with the given values but thereexists a solution anyway Of course the necessary conditionis fulfilled
Now we use again Algorithm 3 to compute the proba-bility P(119887 isin 119872) for a random load vector 119887 sim N(E3 [0505 05]119879) Table 3 shows us the results
Again we can see that the probability is nearly the samein all eight cases The mean probability is 01535 and thevariance is 32369 sdot 10minus6 For the optimization we reducethe covariance matrix and assume 119887 sim N(001 sdot E3 [05 0505]119879) For a probability 120572 = 09 Table 4 states the resultsExample 3 In this example we use a part of a real networkThe focus here is on realistic values for our model Weget these values from the GasLib (see [22]) a collection oftechnical gas network data
ldquoThegoal ofGasLib is to promote research on gasnetworks by providing a set of large and realisticbenchmark instancesrdquo (httpgaslibzibde)
We use a part of the Greek gas network (GasLib-134) (seeFigure 11)
For using real values we first need to know the constant120601119890 from (5)
120601119890 = (119877119878119879)2 120582119863 1198711198862 (111)
The value 119877119878 is the specific gas constant which can becalculated by 119877119878 = 119877119872 with the ideal gas constant 119877 =831451119896119892 sdot 11989821199042 sdot 119898119900119897 sdot 119870 (see [23]) and the molar mass119872 = 1662119896119892119896119898119900119897 (from real data) of the used natural gasAlso the temperature 119879 = 28915119870 is given from real dataFor the (constant) friction factor 120582 we use the law of Chen(see [24])
Mathematical Problems in Engineering 15
Table 4 Test results of control example
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 81199011198981198861199091199001199011199050 39013 39223 39039 38901 38879 39039 38901 390981199011198981198861199091199001199011199051 16568 16565 16365 16497 16496 16365 16497 164921199011198981198861199091199001199011199052 16568 16565 16365 16497 16496 16365 16497 164981199011198981198861199091199001199011199053 10000 10000 10000 10000 10002 10000 10000 10000
Table 5 Values based on real data
variable 119877119878 119879 120582 119863 119871 119886 120601value 50027 28915 00093 9144 178693 43862 198 sdot 107unit 11989821199042 sdot 119870 119870 minus 119898119898 119896119898 119898119904 11989821199042
Figure 11 Greek gas network from GasLib [22] red edges are usedin our example
1radic120582 = minus2 log10 ( 11989611986337065minus 50425
Relog10 [(119896119863)1109828257 + 58506
Re08981]) (112)
Diameter 119863 = 9144119898119898 and roughness value 119896 = 8 sdot 10minus6119898for the first pipe (see Figure 11) are known from real data TheReynolds-Number Re can be calculated by Re = (V119863120588)120578with the dynamic viscosity 120578 = 1 sdot 10minus5119875119886 sdot 119904 (see [1])the mean velocity V = 9548119898119904 (calculated with the meanvolume flow rate which is given by the real data) of the gasand the density 120588 = 074331198961198921198983 (from real data) So we getRe = 648959074 asymp 649 sdot 106 and with this the pipe frictioncoefficient after Chen is 120582 = 00093
The length of the first pipe is given by 119871 = 178693119896119898from the real data Last value missing is the sound speed innatural gas which we assume to be ideal The sound speed
119886 depends on the modulus of compressibility (see [9]) Forideal gases the modulus of compressibility only depends onthe adiabatic exponent and the pressure Using the ideal gasequation simplifies the equation for the sound speed to
119886 = radic120581119877119879119872 (113)
The adiabatic exponent for methan which is the main partof natural gas (gt 90) is about 133 (see [25]) Together allvalues are shown in Table 5 (example for the pipe next to thesource in Figure 11)
Nowwewant to optimize the inner control in the networkfor a fix load vector Therefor we choose two nominations2011-11-01 and 2016-02-17 In both cases the fmincon-functioninMATLAB returns
119906119900119901119905 = 1 (114)
as a local minimum which satisfies the constraint 119887 isin 119872This means for these nominations a compressor stationis not needed so it is switched off With this as a lastcalculation in this paper we calculate the probability for thegiven nominations to be feasible We use the values from thenominations as mean value 120583 and set 119887 sim N(120583 sdot 10minus3E 120583)The results are shown in Tables 6 and 7
Again one can see that the results are quite similarThe mean probability in Table 6 is 09860 and the meanprobability in Table 7 is 09760 Both values are nearly 1 sothere must happen something really unforeseeable that aGreek does not get his demanded gas
6 Conclusion
In this paper we have used an existing stationary modelfor gas transport based on the isothermal Euler equationsWe have extended this model by compressor stations andpresented a characterization of the set of feasible loadsEspecially Theorem 5 contains the central idea to get anaccess to the optimization in Section 4 and the spheric-radialdecomposition is our main tool to handle the uncertainty inthe loadsWewere able to show some results for optimizationwith probabilistic constraints in gas transport In a next stepthese results should be generalized
16 Mathematical Problems in Engineering
Table 6 Test results for nomination 2011-11-01
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 8P(119887 isin 119872) 09849 09849 09878 09885 09863 09845 09854 09853
Table 7 Test results for nomination 2016-02-17
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 8P(119887 isin 119872) 09744 09756 09788 09788 09754 09775 09734 09739
With the characterization of the feasible set we haveshown that the feasibility of a load depends significantly onthe pressure bounds These pressure bounds play a mainrole in our optimization problems The other main partsare the inner controls modeled by compressor edges whichshould lead in optimal control problems with probabilisticconstraints Here we have built a base for further works withthis topic
The idea of separating the graph and removing thecompressor edges can be used to extend themodel with othercomponents like eg control valves and resistorsThiswouldchange the inequalities inTheorem 5 but the approach worksalso for thesemodifications Further [8] showed a way to dealwith a networkwhich is a cycle Combining bothmethods weget an access to model gas networks in high detail
Finally the expansion to a dynamic model might bealso possible The idea is to use a space-mapping approachbecause one can use a lot of the analysis we did hereTherefore upcoming papers should deal with these problemsand present results of the turnpike-theory which describesa relation between stationary and instationary solutions Infact turnpike-theory implies that the optimal dynamic stateis close to the optimal static state in the interior of the timeintervals for sufficiently long time horizons AJ Zaslavskiexplains the turnpike-phenomenon in [26] This book alsogives an overview about the theory
Data Availability
The data used to support the findings of this study areavailable from the corresponding author upon request
Conflicts of Interest
The authors declare that there are no conflicts of interestregarding the publication of this paper
Acknowledgments
This work was supported by DFG in the framework of theCollaborative Research Centre CRCTransregio 154 Math-ematical Modeling Simulation and Optimization Using theExample of Gas Networks project C03
References
[1] P Domschke B Hiller J Lang and C Tischendorf ldquoModel-lierung von gasnetzwerken Eine Ubersichtrdquo Tech Rep 2717Technische Universitat Darmstadt 2017
[2] T Koch B Hiller M E Pfetsch and L Schewe Evaluating GasNetwork Capacities MOS-SIAM 2015
[3] M Gugat R Schultz and D Wintergerst ldquoNetworks of pipe-lines for gas with nonconstant compressibility factor stationarystatesrdquoComputational amp AppliedMathematics vol 37 no 2 pp1066ndash1097 2018
[4] M Gugat and M Herty ldquoExistence of classical solutions andfeedback stabilization for the flow in gas networksrdquo ESAIMControl Optimisation and Calculus of Variations vol 17 no 1pp 28ndash51 2011
[5] M K Banda M Herty and A Klar ldquoGas flow in pipeline net-worksrdquo Networks and Heterogeneous Media vol 1 no 1 pp 41ndash56 2006
[6] RM Colombo G GuerraM Herty and V Schleper ldquoOptimalcontrol in networks of pipes and canalsrdquo SIAM Journal onControl and Optimization vol 48 no 3 pp 2032ndash2050 2009
[7] A Shapiro D Dentcheva and A Ruszczynski ectures onStochastic Programming Modeling and Theory MPS-SIAM2009
[8] C Gotzes H Heitsch R Henrion and R Schultz ldquoOn thequantification of nomination feasibility in stationary gas net-works with random loadrdquoMathematical Methods of OperationsResearch vol 84 no 2 pp 427ndash457 2016
[9] D Halliday R Resnick and J Walker Physik Wiley-VCHBachelor edition edition 2007
[10] M Gugat F M Hante M Hirsch-Dick and G LeugeringldquoStationary states in gas networksrdquoNetworks andHeterogeneousMedia vol 10 no 2 pp 295ndash320 2015
[11] D RoseM SchmidtM C Steinbach and BMWillert ldquoCom-putational optimization of gas compressor stations MINLPmodels versus continuous reformulationsrdquoMathematical Meth-ods of Operations Research vol 83 no 3 pp 409ndash444 2016
[12] H Heitsch R Henrion H Leovey et al ldquoEmpirical observa-tions and statistical analysis of gas demandrdquo in Evaluating GasNetwork Capacities T data B KochM E Hiller et al Eds pp273ndash290 MOS-SIAM 2015
[13] A Genz and F Bretz Computation of Multivariate Normal andt Probabilities vol 195 of Lecture Notes in Statistics Springer2009
[14] M H Farshbaf-Shaker R Henrion and D Homberg ldquoProp-erties of Chance Constraints in Infinite Dimensions with anApplication to PDE Constrained Optimizationrdquo Set-Valued andVariational Analysis pp 1ndash21 2017
[15] T Gonzalez Grandon H Heitsch and R Henrion ldquoA jointmodel of probabilisticrobust constraints for gas transportman-agement in stationary networksrdquo Computational ManagementScience vol 14 no 3 pp 443ndash460 2017
[16] W van Ackooij I Aleksovska and M Munoz-Zuniga ldquo(Sub-)Differentiability of Probability Functionswith Elliptical Distri-butionsrdquo Set-Valued and Variational Analysis
Mathematical Problems in Engineering 17
[17] W van Ackooij and R Henrion ldquoGradient formulae for nonlin-ear probabilistic constraints with GAUssian and GAUssian-likedistributionsrdquo SIAM Journal on Optimization vol 24 no 4 pp1864ndash1889 2014
[18] W van Ackooij and R Henrion ldquo(Sub-)gradient formulaefor probability functions of random inequality systems underGaussian distributionrdquo SIAMASA Journal on UncertaintyQuantification vol 5 no 1 pp 63ndash87 2017
[19] W van Ackooij and J Malick ldquoSecond-order differentiabilityof probability functionsrdquo Optimization Letters vol 11 no 1 pp179ndash194 2017
[20] H Eckey R Kosfeld and M Turck Wahrscheinlichkeitsrech-nung und Induktive Statistik - Grundlagen -Methoden - BeispieleGabler Wiesbaden Germany 2005
[21] A Prekopa Stochastic Programming Springer 1st edition 1995[22] I Joormann M Pfetsch R Schwarz et al ldquoGasLibmdashA Library
of Gas Network Instancesrdquo 2017[23] G Wedler Lehrbuch der Physikalischen Chemie Wiley-VCH
5th edition 2004[24] N H Chen ldquoAn explicit equation for friction factor in piperdquo
Industrial amp Engineering Chemistry Fundamentals vol 18 no3 pp 296-297 1979
[25] W Schenk and F Kremer Physikalisches Praktikum Springer14th edition 2014
[26] A J Zaslavski ldquoTurnpike Properties in the Calculus of Varia-tions and Optimal Controlrdquo inNonconvex Optimization and ItsApplications vol 80 Springer 2006
Hindawiwwwhindawicom Volume 2018
MathematicsJournal of
Hindawiwwwhindawicom Volume 2018
Mathematical Problems in Engineering
Applied MathematicsJournal of
Hindawiwwwhindawicom Volume 2018
Probability and StatisticsHindawiwwwhindawicom Volume 2018
Journal of
Hindawiwwwhindawicom Volume 2018
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawiwwwhindawicom Volume 2018
OptimizationJournal of
Hindawiwwwhindawicom Volume 2018
Hindawiwwwhindawicom Volume 2018
Engineering Mathematics
International Journal of
Hindawiwwwhindawicom Volume 2018
Operations ResearchAdvances in
Journal of
Hindawiwwwhindawicom Volume 2018
Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018
International Journal of Mathematics and Mathematical Sciences
Hindawiwwwhindawicom Volume 2018
Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom
The Scientific World Journal
Volume 2018
Hindawiwwwhindawicom Volume 2018Volume 2018
Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in
Nature and SocietyHindawiwwwhindawicom Volume 2018
Hindawiwwwhindawicom
Dierential EquationsInternational Journal of
Volume 2018
Hindawiwwwhindawicom Volume 2018
Decision SciencesAdvances in
Hindawiwwwhindawicom Volume 2018
AnalysisInternational Journal of
Hindawiwwwhindawicom Volume 2018
Stochastic AnalysisInternational Journal of
Submit your manuscripts atwwwhindawicom
Mathematical Problems in Engineering 3
times104
times106
15
2
25
3
35
4
45
5
55
6
pres
sure
in P
a
1 2 3 4 5 6 7 80x in m
Figure 1 Pressure drop along a pipe for 119863 = 0762119898 119879 = 28915119870119877119878 = 50027119869119896119892119870 120582 = 0005 119886 = 42917119898119904 119902 = 7951198961198921198982119904 and119901(0) = 60 sdot 105119875119886frictionWe use the followingmodel equation (see [1] section747 [11] section 23)
119888120574 ((119901119900119906119905119901119894119899
)120574 minus 1) = (6)
Here 119901119900119906119905 and 119901119894119899 are the pressures at the end resp at thebeginning of one pipe 119888 and 120574 are constants and whichis our control is the specific change in adiabatic enthalpyMore information about the parameters can be found in[11] In addition [2] contains more information about thefunctionality of compressor stations and a deduction of theequation A short transformation of the equation leads toour representation of the pressure rise along the compressoredges
(119901ℎ(119890)119901119891(119890)
)2 = 119906119890 with 119906119890 = (120574119888 + 1)2120574 forall119890 isin E119862 (7)
It holds 119906119890 ge 1 where 119906119890 = 1 means that the compressorstation is switched off (see Figure 2)
Now we define the set of feasible loads We say that avector 119887+ isin R119899+1 is feasible if (3) (5) and (7) are fulfilledSo the feasible set (here called ) is defined as follows
= 119887+ isin R119899+1 | 1119879
119899+1119887+ = 0 and exist (119901+ 119902) isin R119899+1
times R119898 119901+
isin [119901+119898119894119899 119901+119898119886119909] and (3) (5) (7) are fulfilled (8)
22 Stochastic Uncertainty in the Loads In reality the futuredemand for gas can never be known exactly a priori Itdepends on various physical effects like eg the temperatureof the environment Also personal reasons can be responsiblefor a higher or lower use of gas Thus the load vector is
times105
times10755
05
1
15
2
25
3
35
4
45
5
pres
sure
in P
a
15 2 25 3 35 4 45 51u in Jkg
Figure 2 Compressor property for 119888 = 1446526119869119896119892 120574 = 0225and 119901119894119899 = 40 sdot 105119875119886considerd to be a random vector Many papers study howthe distribution functions of the load vector can be identifiedeg [12] This article also describes why a multivariate Gaus-sian distribution is a good choice for random gas demandSo we use some random vector 120585(120596) with 120585 sim N(120583 Σ) fora mean vector 120583 and a positive definite covariance matrix Σon a suitable probability space (ΩAP) and our aim is tocalculate the following probability
P 120596 isin Ω | 120585 (120596) isin 119872 (9)
for a suitable set119872In our case we assume that the tree-structured graph
has only one influx node which gets the number zeroall remaining nodes are efflux nodes Then the graph isnumerated with breadth-first search or depth-first search andevery edge 119890 isin E gets the number maxℎ(119890) 119891(119890) Our set119872is defined as follows
119872 fl 119887 isin R119899ge0 | (minus1119879
119899119887 119887) isin (10)
There are many studies for special sets 119872 eg for poly-hedral sets (see [13]) but our set is very general So we need avery general method for calculating the probability (9) Ourmethod of choice is the so called spheric-radial decompositionIt became popular for dealing with probabilistic constraints(eg see [8 14ndash19])
Theorem 2 (spheric-radial decomposition) Let 120585 sim N(0 119877)be the 119899-dimensional standard Gaussian distribution with zeromean and positive definite correlation matrix 119877 Then for anyBorel measurable subset119872 sube R119899 it holds that
P (120585 isin 119872) = intS119899minus1
120583120594 119903 ge 0 | 119903119871V isin 119872119889120583120578 (V) (11)
where S119899minus1 is the (119899 minus 1)-dimensional sphere in R119899 120583120578 is theuniform distribution on S119899minus1 120583120594 denotes the 120594-distribution
4 Mathematical Problems in Engineering
with 119899 degrees of freedom and 119871 is such that 119877 = 119871119871119879 (egCholesky decomposition)
Thedifference with our problem is that our load vector isnormal distributed with mean value 120583 and positive definitecovariance Σ We want to use the result of Theorem 2 togeneral Gaussian distributions (see [8]) Therefore we set120585lowast fl 119863minus1 (120585 minus 120583) sim N (0 119877) with
119863 fl diag (radicΣ119894119894)119894=1119899
and 119877 fl 119863minus1Σ119863minus1 (12)
and observe that P(120585 isin 119872) = P(120585lowast isin 119863minus1(119872 minus 120583)) So ouraim is to compute the following probability
P (120585lowast isin 119863minus1 (119872 minus 120583)) (13)
So the calculation of the probability (9) for a very generalset 119872 is equivalent to compute the integral in Theorem 2which is not easy to compute analytically too But the integralcan be computed numerically in a easy and nearly exactwayTherefore we formulate the following algorithm (see [8]Algorithm 4)
Algorithm 3 Let 120585 sim N(120583 Σ) and 119871 such that 119871119871119879 = Σ(1) Sample 119873 points V1 V119873 uniformly distributed
on the sphere S119899minus1(2) Compute the one-dimensional sets 119872119894 fl 119903 ge 0 |119903119871V119894 + 120583 isin 119872 for 119894 = 1 119873(3) Set P(120585 isin 119872) asymp (1119873)sum119873
119894=1 120583120594(119872119894)The first step is quite easy There are many possibilities
to sample a set of uniformly distributed points on the sphereS119899minus1 eg Monte-Carlo sampling or the randn-generator inMATLAB Normalizing these points leads us to the samplingwe are looking for Alternatively quasi Monte-Carlo methodscan be used In [8] the authors use a quasi Monte-Carlomethod for their computations and they show its advantagesin Figure 6
Themain challenge is to handle the one-dimensional setsof the second step If the set119872 is convex these sets are closedintervals but 119872 need not to be convex The aim is to get arepresentation of 119872119894 (119894 = 1 119873) as a union of disjointintervals With this the third step can be done efficiently Thenext section deals with this representation in detail
The third step is the approximation of the integral Thevalues 120583120594(119872119894) can be calculated easily with the distributionfunction F120594 of the distribution (see [20] for details of thatdistribution) With the representation 119872119894 = ⋃119905
119895=1[119886119895 119886119895] the120594-distribution can be calculated the following
120583120594 (119872119894) = 119905sum119895=1
F120594 (119886119895) minusF120594 (119886119895) (14)
This leads to the wanted probability The 120594-distributionfunction is given by the following (see [20] p 132)
F120594 (119909) = 121198732Γ (1198732) int119909
0V1198732minus1119890minusV2119889V (15)
2 4 6 8 10 12 14 16 18 200x
0010203040506070809
1
F(x)
Figure 3 Distribution function of the Chi-square distribution for 3(blue) 5 (red) 8 (green) and 12 (purple) degrees of freedom
with Γ(12) = 120587 Γ(1) = 1 and Γ(119903 + 1) = 119903Γ(119903) for all 119903 isin R+Note that there are very precise numerical approximationsfor the distribution function eg the chi2cdf -function inMATLAB (see Figure 3)
3 Explicit Characterization of the Gas Flow
In this section we want to characterize the feasible set to useAlgorithm 3 The aim is to get a representation of the one-dimensional sets of step (2) as a union of disjoint intervalsTherefore we need another representation of the feasible setas a system of inequalitiesThis is an idea from [8]Theorem 1In the proof the authors use a representation of the pressureloss for the full network graph
(119860+)119879 (119901+)2 = minusΦ 10038161003816100381610038161199021003816100381610038161003816 119902 (16)
which is equivalent to (5) if and only if E119862 = 0 In this paperwe also consider the case E119862 = 0 The matrix Φ is diagonalwith the values 120601119890 as defined in (5) at its diagonal Thisrepresentation is not possible yet because we extended ourmodel by compressor edges so this equation does not holdfor all edges Sowe have to find away to use a similar equationlike (16) to state a suitable theorem The idea that allowsto include compressor stations is to remove compressoredges from the graph and replace them by suitable algebraicconditions First we consider the equation of masssum
119890isin1198640(V)119902119890 = 119887+V lArrrArr
sum119890isin1198640(V)capE119865
119902119890 = 119887+V minus sum119890isin1198640(V)capE119862
119902119890 (17)
We want to treat the flows through the compressor edges asloads sowehave to adapt the load vectorThenew load vector+ is defined as follows+V
fl
119887+V + 119902119890 forall119890 isin 1198640 (V) with 119890 isin E119862 and 119891 (119890) = V119887+V minus 119902119890 forall119890 isin 1198640 (V) with 119890 isin E119862 and ℎ (119890) = V119887+V else
(18)
The resulting graph 119866 = (V+E119865) is no longer connected(scheme is shown in Figure 4)
Mathematical Problems in Engineering 5
p0
p1
p2
p3
p4
p5
p6
p7
p8
p9
q1
q2
q4
q6
q7
q8
q9
b0
b1
b2
b3
b4
b5
b6
b7
b8
b9
p0
p1
p2
p3
p4
p5
p6
p7
p8
p9
q1
q2
q3 q3
q4
q5 q5
q6
q7
q8
q9
b0
b1
b2
b3
b4
b5
b6
b7
b8
b9q5 isin ℰC
q3 isin ℰC
Figure 4 Network graph with two compressor edges (left) and new not connected network graph with three connected subgraphs afterremoving the compressor edges (right)
We set |E119865| fl 1198981 and |E119862| fl 1198982 with 1198981 + 1198982 = 119898so 119866 consists of 1198982 + 1 subgraphs which are all connectedNote that 119902 is a vector in1198771198981 nowWe determine that119866119894 is the
119894-th self-connected subgraph of119866 (depends on numbering ofthe graph) Analogously 119901+
119894119895 is the 119895-th component of the 119894-thsubgraph The terms 119902119894119895 and 119860 119894119895119896 are defined similarly Thefeasible set for 119866 is then
=+ isin R119899+1 |
1119879119899+1+ = 0 andexist (119901+ 119902) isin R119899+1
ge0 times R119898 with 119901+ isin [119901+119898119894119899 119901+119898119886119909] 119860 119894119902119894 = 119894 forall119894 = 1 1198982 + 1(119860+119894 )119879 (119901+
119894 )2 = minusΦ11989410038161003816100381610038161199021198941003816100381610038161003816 ∘ 119902119894 forall119894 = 1 1198982 + 1
(119901ℎ(119890)119901119891(119890)
)2 = 119906119890 forall119890 isin E119862
(19)
The last equation in is the same as in (8) that modelsthe pressure rise caused by the compressor With this we canpresent the equation for pressure loss in the form of (16) forevery subgraph Before we state a theorem for characterizingthe feasible set we show the equivalence of and
Lemma4 Let + be as defined in (18)The feasible sets definedin (8) and (19) are equivalent in the following sense
(i) For all 119887+ isin there is + isin (ii) For all + isin there is 119887+ isin (20)
Proof (i) Consider a vector 119887+ isin It holds 1119879119899+1119887+ = 0 and
because of (18) it follows directly 1119879119899+1+ = 0 After removing
the compressor edges from119860+119902 = 119887+ it follows with (17) that119860+119894 119902119894 = +119894 which is equivalent to 119860 119894119902119894 = 119894 The pressure loss
is fulfilled for all flux edges in and because the flow vector119902 doesnrsquot change in the relevant coefficients while removingthe compressor edges the pressure loss is fulfilled in tooThe compressor property is independent of 119887+ and 119902 so it alsoholds From this it follows + isin 119872
(ii) The other way around is quite similar We consider avector + isin From 1119879
119899+1+ = 0 it follows with (18) that1119879119899+1119887+ = 0 The equation for mass conservation is fulfilled
in every subgraph So with (17) it follows 119860+119902 = 119887+ whichimplies the mass conservation in119872 The pressure loss holds
in all self-connected parts of the graph too and because therelevant coefficients of 119902 doesnrsquot change while adding thecompressor edges the pressure loss is fulfilled in119872 too Lastas said before the compressor property is independent of +and 119902 it is also fulfilled in119872 This completes the proof
Because our network graph is a tree the mass conserva-tion is fulfilledwith a full rank incidencematrix119860of size 119899times119899so 119902 can be computed explicit by the following
119902 = 119860minus1119887 (21)
which makes us able to use Lemma 4With this result we canwrite the equation formomentum
conservation depending on the incidence matrices for theparts of the tree This makes us able to formulate the theoremfor characterizing the feasible set Therefore we will use thefollowing function
119892 R119899 997888rarr R
119909 997891997888rarr (119860119879)minus1Φ 10038161003816100381610038161003816119860minus111990910038161003816100381610038161003816 (119860minus1119909) (22)
which describes the pressure loss between the nodes We alsointroduce a new notation
(i) 119906(119894119895)119896 is the 119896-th control on the path from V1198940 to V1198950(ii) 119892(119894119895)119896 is the function 119892 for the 119896-th subgraph on the
path from V1198940 to V1198950 whereas 119892(119894119895)1 is the function 119892
6 Mathematical Problems in Engineering
of the graph 119866119894 According to this 119892(119894119895)119896ℓ is the ℓ-thcomponent of the function 119892(119894119895)119896
(iii) (119894119895)119896 is the load vector of the 119896-th subgraph in thepath from V1198940 to V1198950 whereas (119894119895)1 is the load vectorof the graph 119866119894 and (119894119895)119896ℓ is the ℓ-th component of(119894119896)119896
Analogously we define 119860 (119894119895)119896 119901(119894119895)119896 and 119902(119894119895)119896 Furthermorewe define the following values
(i) 119896lowast119894119895 fl max119896 isin 1 1198982 + 1 | V1198960 isin Π(V1198940) and V1198960 isinΠ(V1198950) as the largest index of all subgraph the pathsto V1198940 and V1198950 pass through
(ii) 119899lowast119894119895 |119896 isin 1 1198982 + 1 | V1198960 isin ΠV1198940(V1198950)| as thenumber of subgraph which are on the path from V1198940from V1198950 containing 119866119894 and 119866119895
(iii) 119898lowast119894119895 fl |119890 isin E119862 | ℎ(119890) isin ΠV1198940(V1198950)and119891(119890) isin ΠV1198940(V1198950)|
as the number of controls which are between V1198940 andV1198950
With this notation and these values we define a sum for 119894 119895 isin11198982 + 1 If V1198940 isin Π(V1198950) we set sum119894119894 fl 0 and else we set
Σ119896lowast119894 fl119899lowastminus2sum119896=1
1prod119898lowastminus119896ℓ=1 119906(119896lowast119894)ℓsdot 119892(119896lowast119894)119899lowastminus119896119891(119890119906
(119896lowast119894)119899lowastminus119896) ((119896lowast119894)119899lowastminus119896)
+ 119892(119896lowast119894)1119891(119890119906(119896lowast119894)1
) ((119896lowast119894)1) (23)
where we write only 119899lowast and 119898lowast instead of 119899lowast119896lowast119894 and 119898lowast119896lowast119894 for
better reading Last we define a productΠ119894119894 fl 1 andΠ119896lowast119894 fl
119898lowastprod119896=1
119906(119896lowast119894)119896 (24)
From a first point of view the notation seems to be confusingbut using this notation makes sense since we want toguarantee the feasibility of a load vector by comparing thepressure bounds at every node with each other We explainthis in a short example Figure 5 shows a graph after removingthe compressor edges
If we want to compare the pressure bounds of node 11 andnode 12 we need to know how the pressures change on thepath from 2 to 11 resp from 2 to 12 Node 11 is part of thesubgraph 1198665 and node 12 is part of the subgraph 1198666 so itfollows 119896lowast56 = 2 because node 2 is part of subgraph 2 Thenotation can be understood as a numbering along the pathIt is 119906(211)1 = 1199062 the first control on the path from node 2to node 11 and 119906(211)2 = 1199064 the second control on this pathFurther it follows 119899lowast211 = 3 (1198662 1198663 1198665) and119898lowast
211 = 2 (1199062 1199064)Theproduct defined before is the product of all controls alongthis path and the sum is the pressure loss between node 2 andnode 7 Using this notation makes us able to formulate thetheorem for characterizing the feasible set in a readable way
Theorem 5 For given pressure bounds 119901+119898119894119899 119901+119898119886119909 isin R119899+1
and controls 119906119894 isin R (119894 = 1 1198982) the following equivalenceholds
A vector + with 1119879119899+1+ = 0 is feasible ie + isin if and
only if the following inequalities hold For all 119894 = 1 1198982 + 1holds (feasibility inside the subgraphs)
(1199011198981198941198991198940 )2 le min
119896=1119899119894[(119901119898119886119909
119894119896 )2 + 119892119894119896 (119894)] (25)
(1199011198981198861199091198940 )2 ge max
119896=1119899119894[(119901119898119894119899
119894119896 )2 + 119892119894119896 (119894)] (26)
max119896=1119899119894
[(119901119898119894119899119894119896 )2 + 119892119894119896 (119894)]
le min119896=1119899119894
[(119901119898119886119909119894119896 )2 + 119892119894119896 (119894)] (27)
For all 119894 119895 = 1 1198982 + 1 with 119894 lt 119895 holds (feasibility betweenthe subgraphs)1Π119896lowast 119894
(1199011198981198941198991198940 )2 + Σ119896lowast119894 le 1Π119896lowast119895
(1199011198981198861199091198950 )2 + Σ119896lowast119895 (28)
1Π119896lowast 119894
(1199011198981198861199091198940 )2 + Σ119896lowast119894 ge 1Π119896lowast 119895
(1199011198981198941198991198950 )2 + Σ119896lowast119895 (29)
1Π119896lowast 119894
(1199011198981198941198991198940 )2 + Σ119896lowast119894
le 1Π119896lowast119895
min119896=1119899119895
[(119901119898119886119909119895119896 )2 + 119892119895119896 (119895)] + Σ119896lowast119895 (30)
1Π119896lowast 119894
(1199011198981198861199091198940 )2 + Σ119896lowast119895
ge 1Π119896lowast119895
max119896=1119899119895
[(119901119898119894119899119895119896 )2 + 119892119895119896 (119895)] + Σ119896lowast119895 (31)
1Π119896lowast 119894
max119896=1119899119894
[(119901119898119894119899119894119896 )2 + 119892119894119896 (119894)] + Σ119896lowast119894
le 1Π119896lowast119895
(1199011198981198861199091198950 )2 + Σ119896lowast119895 (32)
1Π119896lowast 119894
min119896=1119899119894
[(119901119898119886119909119894119896 )2 + 119892119894119896 (119894)] + Σ119896lowast119894
ge 1Π119896lowast119895
(1199011198981198941198991198950 )2 + Σ119896lowast119895 (33)
1Π119896lowast 119894
max119896=1119899119894
[(119901119898119894119899119894119896 )2 + 119892119894119896 (119894)] + Σ119896lowast119894
le 1Π119896lowast119895
min119896=1119899119895
[(119901119898119886119909119895119896 )2 + 119892119895119896 (119895)] + Σ119896lowast119895 (34)
1Π119896lowast 119894
min119896=1119899119894
[(119901119898119886119909119894119896 )2 + 119892119894119896 (119894)] + Σ119896lowast119894
ge 1Π119896lowast119895
max119896=1119899119895
[(119901119898119894119899119895119896 )2 + 119892119895119896 (119895)] + Σ119896lowast119895 (35)
Mathematical Problems in Engineering 7
0 1 2
3
4
5
6
7
8
9
10
11
12
u1
u2
u3
u4
u5
G1 G2 G3
G4
G5
G6
Figure 5 Example graph for illustration of the used notation
0 1 2q1 q2
b0 b1 b2
Figure 6 Graph of Example 1
Let us have a look at the inequalities Inequalities(25)ndash(27) are well-known from [8] and they describe thefeasibility inside a connected tree The other inequalities(28)ndash(35) guarantee the feasibility between the subgraphsTherefore we compare the pressures of the entries of two parts((28) (29)) of the entry of one part and the exit nodes of theother parts ((30)ndash(33)) and we compare the pressures of allexits ((34) (35)) Because we cannot compare the pressuresof119866119894 and 119866119895 directly (there may be no direct connection) wehave to compare them using the subgraph 119866119896lowast which is thelast part the paths to V1198940 and V1198950 pass Therefore we definedthe sumsΣ119896lowast 119894 andΣ119896lowast119895 which trace back the pressures to119866119896lowast
Remark 6 For an implementation it is wise to define a set119880(119894119895) containing all controls on the path from V1198940 to V1198950 anduse an index running over this set to define the sums and theproducts
Now we want to proof the theorem
Proof of Theorem 5 Because the columns of 119860+ sum up tozero it holdsminus1119879
119899119894119860 119894 = 119860+
1198940 (119894 = 1 1198982 + 1) (36)
ldquo997904rArrrdquo We consider a feasible vector 119887+ isin From theequation of mass conservation of every subgraph it follows
119902119894 = 119860minus1119894 119894 (37)
We take the equation of momentum conservation for the 119894-thsubgraph (119894 isin 1 1198982 +1) and separate 119860+
119894 to119860+1198940 and 119860 119894
(119860+1198940)119879 1199012
1198940 + 119860119879119894 1199012
119894 = minusΦ 10038161003816100381610038161199021198941003816100381610038161003816 119902119894 (38)
Next we insert (36) into (38) and get
minus119860119879119894 1119899119894
11990121198940 + 119860119879
119894 1199012119894 = minusΦ 10038161003816100381610038161199021198941003816100381610038161003816 119902119894 (39)
We multiply this equation from left with (119860119879119894 )minus1 and insert
(37) so it follows
111989911989411990121198940 = 1199012
119894 + (119860119879119894 )minus1Φ 10038161003816100381610038161003816119860minus1
119894 11989410038161003816100381610038161003816 119860minus1119894 119894 (40)
and using the definition of the function 119892 we have
111989911989411990121198940 = 1199012
119894 + 119892119894 (119894) (41)
Note here that 119901119894 is the vector of pressures of the subgraph119866119894
and 119892119894(119894) is the vector-valued function which describes thepressure loss in 119866119894 Equation (41) holds for all subgraphs 119866119894
with 119894 isin 1 1198982 + 1 From this the inequalities (25)ndash(27)follow directly with 119901+ isin [119901+119898119894119899 119901+119898119886119909] Next we use (41)component-by-component For fix 119894 119895 isin 1 1198982 + 1 with119894 lt 119895 it follows
1199012119896lowast0 = 1199012
119896lowast119896 + 119892119896lowast119896 (119896lowast) (119896 = 1 119899119896lowast) (42)
Remember that 119896lowast119894119895 is the largest index of all subgraphs thepaths to V1198940 and V1198950 pass through For better reading we onlywrite 119896lowast instead of 119896lowast119894119895 We use the equation of the compressorproperty for the first compressor on the path from V119896lowast0 to V1198940
1199012ℎ(119890119906(119896lowast119894)1
)1199012119891(119890119906(119896lowast119894)1
)
= 119906(119896lowast119894)1 (43)
and insert it into (42) for 119896 = 119891(119890119906(119896lowast119894)1 )1199012(119896lowast 119894)10 = 1119906(119896lowast119894)1
1199012(119896lowast119894)20
+ 119892(119896lowast 119894)1119891(119890119906(119896lowast119894)1
) ((119896lowast119894)1) (44)
Note that 1199012119896lowast 0 = 1199012
(119896lowast119894)10 119892119896lowast119896(119896lowast) = 119892(119896lowast119894)1119896((119896lowast119894)1) and1199012ℎ(119890119906(119896lowast119894)1
) = 1199012(119896lowast 119894)20 We use the compressor equation for
the second compressor on the path from V119896lowast0 to V1198940 and putit together with (42) into (44)
1199012(119896lowast119894)10 = 1119906(119896lowast119894)1
[ 1119906(119896lowast119894)2
1199012(119896lowast 119894)30
+ 119892(119896lowast119894)2119891(119890119906(119896lowast119894)2
) ((119896lowast119894)2)]+ 119892(119896lowast119894)1119891(119890119906
(119896lowast119894)1) ((119896lowast119894)1)
(45)
8 Mathematical Problems in Engineering
We repeat this procedure for all controls on the path fromV119896lowast0 to V1198940 Although we write 119899lowast and119898lowast instead of 119899lowast119896lowast119894 and119898lowast
119896lowast119894 Thus we get
1199012(119896lowast119894)10 = 1119906(119896lowast119894)1
[ 1119906(119896lowast119894)2
[sdot sdot sdot [ 1119906(119896lowast119894)119898lowast1199012(119896lowast119894)119899lowast0
+ 119892(119896lowast119894)119899lowastminus1119891(119890119906(119896lowast119894)119899lowastminus1
) ((119896lowast119894)119899lowastminus1)] sdot sdot sdot]+ 119892(119896lowast119894)2119891(119890119906
(119896lowast119894)2) ((119896lowast119894)2)]
+ 119892(119896lowast119894)1119891(119890119906(119896lowast119894)1
) ((119896lowast119894)1)
(46)
We take again (42) for the node V119896lowast0 and do the sameprocedure just like before for the path from V119896lowast0 to V1198950 With119899lowast = 119899lowast119896lowast119895 and119898lowast = 119898lowast
119896lowast119895 we get
1199012(119896lowast 119895)10
= 1119906(119896lowast119895)1
[ 1119906(119896lowast119895)2
[sdot sdot sdot [ 1119906(119896lowast119895)119898lowast1199012(119896lowast 119895)119899lowast0
+ 119892(119896lowast119895)119899lowastminus1119891(119890119906(119896lowast119895)119899lowastminus1
) ((119896lowast119895)119899lowastminus1)] sdot sdot sdot]+ 119892(119896lowast119895)2119891(119890119906
(119896lowast119895)2) ((119896lowast119895)2)]
+ 119892(119896lowast119895)1119891(119890119906(119896lowast119895)1
) ((119896lowast119895)1)
(47)
Expanding (46) and (47) and equalizing them (this is possiblebecause 119901(119896lowast 119894)10 = 119901(119896lowast119895)10) leads to1Π119896lowast 119894
11990121198940 + Σ119896lowast119894 = 1Π119896lowast 119895
11990121198950 + Σ119896lowast119895 (48)
with Σ119896lowast119894 resp Σ119896lowast119895 defined before (48) directly implies (28)and (29) Using (48) together with (42) for V1198940 first timefor V1198950 second time and for V1198940 and V1198950 a third time theinequalities (30)ndash(35) follow directly So this part of the proofis complete
ldquolArr997904rdquo For this part we consider a vector 119887+ isin R119899+1
with 1119879119899+1119887+ = 0 which fulfills the inequalities (25)ndash(35) We
define the following sets (119894 = 2 1198982 + 1)11987511 fl [(119901119898119894119899
10 )2 (11990111989811988611990910 )2]
11987521 fl [ max119896=11198991
[(1199011198981198941198991119896 )2 + 1198921119896 (1)]
min119896=11198991
[(1199011198981198861199091119896 )2 + 1198921119896 (1)]]
1198751119894 fl [ 1Π1119894
(1199011198981198941198991198940 )2 + Σ1119894 1Π1198941
(1199011198981198861199091198940 )2 + Σ1119894]
1198752119894 fl [ 1Π1119894
max119896=1119899119894
[(119901119898119894119899119894119896 )2 + 119892119894119896 (119894)] + Σ1119894 1Π1198941
sdot min119896=1119899119894
[(119901119898119886119909119894119896 )2 + 119892119894119896 (119894)] + Σ1119894]
(49)
We set
1198750 fl 1198982+1⋂119896=1
1198751119896 cap 1198982+1⋂119896=1
1198752119896 (50)
and we will define a value 119901210 isin 1198750 later So to guarantee
such a value 119901210 exists we have to show first that 1198750 is not
empty Because 1198750 is a finite intersection of convex intervalswe have to make sure that all intervals are not empty andevery intersection between two intervals is not empty In onedimension this is sufficient that 1198750 is not empty For 119894 =1 1198982 + 11198751119894 = 0 according to requirement1198752119894 = 0 because of (26) (51)
Furthermore its 1198751119894 cap 1198752119894 = 0 because of (25) and (26) Themissing intersections are nonempty too for 119894 119895 isin 1 1198982+1 with 119894 lt 119895 as follows
1198751119894 cap 1198751119895 = 0 because of (28) and (29) 1198751119894 cap 1198752119895 = 0 because of (30) and (31) 1198752119894 cap 1198751119895 = 0 because of (32) and (33) 1198752119894 cap 1198752119895 = 0 because of (34) and (35)
(52)
So the set 1198750 is not empty We define the following values
119901210 isin 119875011990121198940 fl Π1119894 (1199012
10 minus Σ1119894) 119894 = 2 1198982 + 11199012119894119896 fl 1199012
1198940 minus 119892119894119896 (119894) 119894 = 1 1198982 + 1 119896 = 1 119899119894119902119894 fl 119860minus1119894 119894 119894 = 1 1198982 + 1
(53)
Now we show that our choice of pressures and flows isfeasible It is11990110 isin [119901119898119894119899
10 11990111989811988611990910 ] because of1199012
10 isin 11987511 Furtherbecause of 1199012
10 isin 1198751119894 for all 119894 = 2 1198982 + 1 it holds1Π1119894
11990121198940 + Σ1119894 le 1Π1119894
(1199011198981198861199091198940 )2 + Σ1119894 (54)
and 1Π1119894
11990121198940 + Σ1119894 ge 1Π1119894
(1199011198981198941198991198940 )2 + Σ1119894 (55)
Mathematical Problems in Engineering 9
Now we replace 11990121198940 in its definition with (1199012
119894119896 + 119892119894119896(119894)) andbecause of 1199012
1198940 isin 1198752119894 for all 119894 = 1 1198982 + 1 it follows1Π1119894
(1199012119894119896 + 119892119894119896 (119894)) + Σ1119894
le 1Π1119894
min119896=1119899119894
[(119901119898119886119909119894119896 )2 + 119892119894119896 (119894)] + Σ1119894
(56)
and 1Π1119894
(1199012119894119896 + 119892119894119896 (119894)) + Σ1119894
ge 1Π1119894
max119896=1119899119894
[(119901119898119894119899119894119896 )2 + 119892119894119896 (119894)] + Σ1119894 (57)
So for all 119894 = 1 1198982 + 1 and for all 119896 = 1 119899119894 it holds119901119894119896 isin [119901119898119894119899119894119896 119901119898119886119909
119894119896 ] which is equivalent to
119901+ isin [119901+119898119894119899 119901+119898119886119909] (58)
The equation for mass conservation follows directly from thedefinition of 119902119894 (119894 = 1 1198982 + 1) For the pressure loss weuse
1199012119894 = 1119899119894
11990121198940 minus 119892119894 (119894) (59)
And multiply this equation from left with 119860119879119894 and use (36)
With this it follows the equation for momentum conserva-tion
(119860+119894 )119879 (119901+
119894 )2 = minusΦ 10038161003816100381610038161199021198941003816100381610038161003816 119902119894 (60)
The last equation missing is the compressor property For acontrol 119895 isin 1 1198982 we fix a vertex V1198940 (119894 isin 1 1198982 +1)with ℎ(119890119906119895) = V1198940 We use 1199012
1198940 = Π1119894(119901210 minus Σ1119894) and solve the
formula for 119901210 in the following way
119901210 = 1119906(1119894)1
[sdot sdot [ 1119906(1119894)119898lowast1199012(1119894)119899lowast0
+ 119892(1119894)119899lowastminus1119891(119890119906(1119894)119899lowastminus1
) ((1119894)119899lowastminus1)] sdot sdot]+ 119892(1119894)1119891(119890119906(1119894)1 )
((1119894)1) (61)
Consider wlog that the subgraphs along the path from V10to V1198940 are numbered chronological st if ℎ(119890119906119895) isin 119866119894 then119891(119890119906119895) isin 119866119894minus1 We use 1199012
119894minus10 = Π1119894minus1(119901210 minus Σ1119894minus1) to get the
same representation for 119901210 depending on 1199012
(1119894minus1)119899lowast0 whichis equivalent to 119901(1119894)119899lowastminus10 We equalize these representationsdepending on 1199012
1198940 and 1199012119894minus10 and we get
1119906(1119894)119898lowast1199012(1119894)119899lowast0 + 119892(1119894)119899lowastminus1119891(119890119906
(1119894)119899lowastminus1) ((1119894)119899lowastminus1)
= 1199012(1119894)119899lowastminus10 (62)
We chose 119894 isin 1 1198982 + 1 st 119906119895 is the last control on thepath from V10 to V1198940 so it is 119906(1119894)119898lowast = 119906119895 Further we have
1199012(1119894)119899lowast 0 = 1199012
11989401199012(1119894)119899lowastminus10 = 1199012
119894minus10119892(1119894)119899lowastminus1119891(119890119906(1119894)119899lowastminus1
) (119887(1119894)119899lowastminus1) = 119892119894minus1119891(119890119906119895 )(119887119894minus1)
(63)
and we know that 11990121198940 resp 1199012
119894minus10 minus 119892119894minus1119891(119890119906119895 )(119887119894minus1) are the
pressures at the head resp the foot of the compressor edge119890119906119895 So it follows 1119906119895
1199012ℎ(119890119906119895 )
= 1199012119891(119890119906119895 )
(64)
which is equivalent to the compressor property This com-pletes the proof
With this theorem we have another characterization ofthe feasible set The last step in this intersection is theadaption of the results to the spheric-radial decompositionAs mentioned before the main step is to get a representationof the sets119872119894 in step (2) of Algorithm 3
Because we assumed that the graph has only one entry wedefine the following set
119872 fl 119887 isin R119899ge0 | (minus1119879
119899119887 119887) isin (65)
Consider a sampled point V119894 isin S119899minus1 of step (1) of Algorithm 3(119894 = 1 119873) We identify the load vector 119887 with the affinelinear function
119887119894 (119903) fl 119903119871V119894 + 120583 = 119903120596119894 + 120583 with 120596119894 = 119871V119894 (66)
where 120583 is the mean value of the Gaussian distribution and 119871is such that 119871119871119879 = Σ for positive definite covariance matrixΣ Note that the index 119894 here is for the sampled vector V119894 notfor the 119894-th subgraph Because 119887119894 is defined only for the exitnodes it must hold 119887 ge 0 which leads to the definition of theregular range
119877119903119890119892119894 fl 119903 ge 0 | 119887119894 (119903) ge 0 supe 119872119894 (67)
So the feasible set is
119872119894 = 119903 isin 119877119903119890119892119894 | 119887119894 (119903) fulfills (3) (5) and (7) (68)
and with Theorem 5 it holds
119872119894 = 119903 isin 119877119903119890119892119894 | 119894 (119903) fulfills (25) minus (34) (69)
where 119894(119903) is from (18) Because 119894 depends on 119903 now thefunction 119892 depends on 119903 too More exactly 119892 is quadratic in119903 so 119892119896(119887119894(119903)) can be represented as follows
1199032[[119899sum
119895=1
Ψ119896119895( 119899sumℓ=1
119860minus1119895ℓ120596119894ℓ)2]]⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
119886119894
10 Mathematical Problems in Engineering
+ 119903[[2119899sum
119895=1
Ψ119896119895( 119899sumℓ=1
119860minus1119895ℓ120596119894ℓ)( 119899sum
ℓ=1
119860minus1119895ℓ120583ℓ)]]⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
119894
+ [[119899sum
119895=1
Ψ119896119895( 119899sumℓ=1
119860minus1119895ℓ120583ℓ)2]]⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
119888119894
(70)
HereΨ is the product of119860 andΦ So the inequations whichcharacterize119872119894 are also quadratic in 119903 because the minimaland maximal pressures in the inequalities only count to 119888119894Now we can write the feasible sets as follows
119872119894 = 119903 isin 119877119903119890119892119894 | 0 le 1198861198941199051199032 + 119887119894119905119903 + 119888119894119905 (71)
where 119905 is an index for the total number of inequalities in119872119894Now the regular range can be cut with the positive intervals ofthe inequalities and the result is a union of disjoint intervals
119872119894 = 119904119894⋃119896=1
[119886119894119896 119886119894119896] 119904119894 isin N (72)
Here 119904119894 is the number of disjoint intervals in which allinequalities are fulfilled and 119886119894119896 and 119886119894119896 are the intervalbounds they depend on the pressure bounds and the con-trols So the third step of Algorithm 3 can be executed With
P (119887 isin 119872) asymp 1119873 119873sum119894=1
119904119894sum119896=1
F120594 (119886119894119896) minusF120594 (119886119894119896) (73)
the probability for a random load vector to be feasible can becomputed
4 Existence of Optimal Solutions
In this section we want to have a look at some optimizationproblems We distinguish between problems with constantloads (without uncertainty on the demand) and problemswith random loads The latter leads us to optimizationproblems with so called probabilistic constraints or chanceconstraints (see [21]) There are two aims in optimizationMinimize the maximal pressure bounds and minimize thecontrols For minimizing the maximal pressure bounds wedefine the following set
119875 fl119899⋃
119894=0
[119901+119898119894119899119894 infin) (74)
in which we look for them Because the maximal pressuresare in proportion to the cost of pipes an objective functionfor a optimization problem could be 119888119879119901+119898119886119909 for a cost vector119888 isin R119899+1
gt0 By minimizing the controls we use the Euclideannorm of the controls as an objective function
Optimization with Constant Loads Consider the two opti-mization problems
min119901+119898119886119909isin119875
119888119879119901+119898119886119909
subject to 119887+ (119901+119898119886119909) isin 119872 (75)
andmin119906isinR1198982ge1
1199062subject to 119887+ (119906) isin (76)
Lemma 7 There exists a unique solution for (75)
Proof We have a look at the side constraint 119887+(119901+119898119886119909) isin Using Theorem 5 we can write (75) as
min119901+119898119886119909isin119875
119888119879119901+119898119886119909
subject to (25) minus (35) are fulfilled (77)
For 119894 isin 1 1198982 + 1 we write inequality (25) (and allremaining analogous) in the following way
0 le 119892119894119896 (119894) minus (1199011198981198941198991198940 )2 + (119901119898119886119909
119894119896 )2 forall119896 = 1 119899119894 (78)
So every inequality depends only on one upper pressurebound (but more inequalities may depend on the samebound) Because the inequalities are not strict the values119901119896 fl min 119901119898119886119909
119896 isin [0infin) | (25) minus (35) are fulfilled (79)
exist for all 119896 = 1 119899 + 1 and we can define a value119901+119898119886119909119900119901119905119896 fl max 119901119896 119901119898119894119899
119896 (80)
which is obviously a solution of (75) Moreover this solutionis unique because reducing one component of the upperbounds would either hurt an inequality or it would not fulfill119901+119898119886119909 ge 119901+119898119894119899 anymore
For the optimization problem (76) we can formulate anecessary and a sufficient condition for optimal solutions
Lemma8 Consider pressure bounds119901+119898119894119899 119901+119898119886119909 isin R119899+1 st119901+119898119894119899 lt 119901+119898119886119909 and a suitable vector 119887+ isin R119899+1 (ie + fulfills(25)ndash(27)) For all 119896 = 1 1198982 we set 119894(119896) and 119895(119896) such that119891(119890119896) isin V+
119894(119896) and ℎ(119890119896) isin V+119895(119896)
(a) If 119906 isin R1198982 is a solution of (76) it holds for all 119896 =1 1198982 (119901119898119894119899119894(119896)0)2 minus 119892119894(119896)119891(119890119896)
(119894(119896)) le (119901119898119886119909119895(119896)0)2 (81)
(b) If the inequalities
(119901119898119886119909119894(119896)0)2 minus 119892119894(119896)119891(119890119896)
(119894(119896)) ge (119901119898119894119899119895(119896)0)2
and (119901119898119894119899119894(119896)0)2 minus 119892119894(119896)119891(119890119896)
(119894(119896)) le (119901119898119886119909119895(119896)0)2 (82)
are fulfilled for all 119896 = 1 1198982 an optimal solution 119906 isin R1198982ge1
of (76) exists
Mathematical Problems in Engineering 11
Note that a vector 119887+ isin R119899+1 is suitable here if the vector+ is feasible for the subgraphs ie the inequalities (25)-(27)are fulfilled for +Proof (a) For this part we use a proof by contradictionAssume that
(119901119898119894119899119894(119896)0)2 minus 119892119894(119896)119891(119890119896)
(119894(119896)) gt (119901119898119886119909119895(119896)0)2 (83)
Because 119906119896 ge 1 for all 119896 = 1 1198982 it follows
(119901119898119894119899119894(119896)0)2 minus 119892119894(119896)119891(119890119896)
(119894(119896)) gt 1119906119896
(119901119898119886119909119895(119896)0)2 (84)
It it Σ119896lowast119894(119896) = 0 and Π119896lowast 119894(119896) = 1 because the subgraph withindex 119896lowast is equal to the subgraphwith index 119894(119896) And becausethe subgraphs 119894(119896) and 119895(119896) are neighbours (ie directlyconnected) it follows Π119896lowast 119895(119896) = 119906119896 and
Σ119896lowast119895(119896) = 119892(119896lowast119895(119896))1119891(119890119906(119896lowast119895(119896))1
) (119887(119896lowast119895(119896))1)= 119892119894(119896)119891(119890119896)
(119894(119896)) (85)
Thus we have a contradiction to (28) and 119906119896 cannot beoptimal
(b) As mentioned before the vector 119887+ is such that isfeasible for the subgraphs So because of the conditions for119896 isin 1 1198982 there exists a 119901119894(119896)0 isin [119901119898119894119899
119894(119896)0 119901119898119886119909119894(119896)0] and a119901119895(119896) isin [119901119898119894119899
119895(119896) 119901119898119886119909119895(119896) ] so that1199012119894(119896)0 minus 119892119894(119896)119891(119890119896)
(119894(119896)) ge 1199012119895(119896)0 (86)
With 1199012119894(119896)0 minus 119892119894(119896)119891(119890119896)
(119894(119896)) = 1199012119891(119890119896)
and 1199012119895(119896)0 = 1199012
ℎ(119890119896)it
follows
(119901ℎ(119890119896)119901119891(119890119896)
) ge 1 (87)
Therefore we can find a 119906 isin R1198982ge1 that the inequalities (28)-
(35) are fulfilled Now we can use theWeierstraszlig theorem andget the existence of an optimal solution
Note that the sufficient condition in Lemma 8 is com-paratively strong there may exist optimal solutions with lessstrong conditions eg the following
Corollary 9 If the graph has only one compressor edge ie|E119862| = 1 then it holds If 119906 isin Rge1 is an optimal solution of(76) 119906 is unique
This conclusion follows directly from the strict mono-tonicity of the objective function
Optimization with Random Loads In this section we con-sider the load vector to be random as we mentioned inSection 22 In the optimization problems we want to makesure that the feasibility of a random load vector is guaranteedfor a probability 120572 isin (0 1) This leads us to optimizationproblems with so called chance or probabilistic constraints
In general problems with probabilistic constraints have theform
min 119891 (119909 120585)subject to P (ℎ (119909 120585) le 0) ge 120572 (88)
with an objective function 119891 and a vector of uncertainty 120585There exist many works about probabilistic constraints respstochastic programming eg [7] gives an excellent overviewabout chance constraints in theory and application In ourcase we consider the following problem
min119901+119898119886119909isin119875
119888119879119901+119898119886119909
subject to P (119887 (119901+119898119886119909) isin 119872) ge 120572 (89)
To handle this problem resp the probabilistic constraint thespheric-radial decomposition gives us an explicit representa-tion of the constraint for a Gaussian vector 119887 sim N(120583 Σ) withmean value 120583 and positive definite covariance Σ
P (119887 (119901+119898119886119909) isin 119872)= int
S119899minus1120583120594 119903 ge 0 | 119903119871V + 120583 isin 119872119889120583120578 (V) (90)
Further Algorithm 3 gives a way to approximate this integralin an efficient way In Section 3 we showed a way to character-ize the feasible set119872 and how to adapt this characterization tothe spheric-radial decomposition This changes our problemwith probabilistic constraints (89) to
min119901+119898119886119909isin119875
119888119879119901+119898119886119909
subject to 1119873 119873sum119894=1
119904119905sum119905=1
F120594 (119886119894119905) minusF120594 (119886119894119905) ge 120572 (91)
Of course the interval bounds depend on the pressure boundsand the control For this problem we formulate the nextlemma
Lemma 10 Let a sampling V1 V119873 (from step (1) ofAlgorithm 3) and 120572 isin (0 1) be given For a suitable 120583 isin R119899
problem (91) has at least one solution 119901+119898119886119909 isin 119875Proof For the proof we fix a V isin V1 V119873 from the givensampling From (25) and (26) it follows for all 119896 = 1 119899119894with 119894 isin 1 1198982 + 10 le 119892119894119896 (119894 (119903)) + (119901119898119886119909
119894119896 )2 minus (1199011198981198941198991198940 )2
0 le minus119892119894119896 (119894 (119903)) + (1199011198981198861199091198940 )2 minus (119901119898119894119899
119894119896 )2 (92)
With a random vector 119887 sim N(120583 Σ)119892119894119896(119894(119903)) are quadraticfunctions in 119903 and the upper pressure bounds only influencethe constant part with a positive sign
0 le 1198861198941198961199032 + 119887119894119896119903 + 119888119894119896 + (119901119898119886119909119894119896 )2 minus (119901119898119894119899
1198940 )2 0 le minus1198861198941198961199032 minus 119887119894119896119903 minus 119888119894119896 + (119901119898119886119909
1198940 )2 minus (119901119898119894119899119894119896 )2 (93)
12 Mathematical Problems in Engineering
It follows that a rise in 119901119898119886119909119894119896 for all 119896 = 0 119899119894 enlarges
the intervals in which the inequalities hold The same holdsfor all inequalities inTheorem 5The cumulative distributionfunction of theChi-Square-Distribution is strictlymonotonicincreasing and convergent to 1 Thus upper pressure bounds119901119898119886119909119894119896 (for 119894 = 1 119899) can be found so that
1119873 119873sum119895=1
119904119895sum119905=1
F120594 (119886119895119905) minusF120594 (119886119895119905) (94)
is higher that 120572 isin (0 1) The index119873 is for all sampled pointson the sphere and the index 119904119895 is for the union of intervals inwhich all inequalities hold The interval boundaries 119886119895119905 and119886119895119905 depend continuously on the pressure bounds and becausethe inequalities are not strict 119901119898119886119909 can be found st the sideconstraint is fulfilled with equality So there exists at least onesolution of (91)
5 Numerical Results
In this section we show a few results of implementation Atfirst we show the idea of the spheric-radial decomposition byusing an easy example Next we show a easy example with onecompressor edge and at last we use real data of the Greek gasnetworkThe focus of the implementation is on the theoremsproofed in Section 4
Example 1 Our first example is an easy graph without innercontrol (see Figure 6)
Assume that node 1 and 2 are gas consumers with meandemand 120583 = [05 05]119879 The pressure bounds are given by119901+119898119894119899 = [2 1 1]119879 and 119901+119898119886119909 = [3 2 2]119879 and Φ = E2 Thenthe incidence matrix is
119860+ = [[[minus1 01 minus10 1
]]] resp 119860 = [1 minus10 1 ] (95)
Consider the inequalities of Theorem 5 to characterize thefeasible set Because the network has no compressor edgesthe system of inequalities only contains the information (25)-(27) the other inequalities do not occur The function 119892 isgiven by
119892 (119887) = [1 01 1] [1 00 1] sdot1003816100381610038161003816100381610038161003816100381610038161003816[1 10 1](11988711198872)
1003816100381610038161003816100381610038161003816100381610038161003816∘ ([1 10 1](11988711198872)) = ( 11988721 + 211988711198872 + 1198872211988721 + 211988711198872 + 211988722) (96)
By inserting the pressure bounds one can see that inequality(25) is always fulfilled and inequality (27) implies1198872 le radic3 (97)
With this inequality (26) follows for 0 le 1198872 le radic30 le 1198871 le minus1198872 + radic8 minus 11988722 (98)
So the feasible set119872 is the following (see Figure 7)
119872 = 119887 isin R2ge0 | 1198872 le radic3 and 1198871 le minus1198872 + radic8 minus 11988722 (99)
Now consider the optimization problem
min119901+119898119886119909isin119875
11198793119901+119898119886119909
subject to 119887 isin 119872 (119901+119898119886119909) (100)
with 119887 = 120583 = [05 05]119879 Lemma 7 guarantees the existenceof a unique solution of this problem We use the fmincon-function inMATLAB which minimizes a constrained non-linear multivariable function The fmincon default settinguses a interior-point method For more information we referto the official MathWorks homepage (httpswwwmath-workscomhelpoptimugfminconhtml)The solver returnsthe following result
119901+119898119886119909119900119901119905 = [2 radic3radic275]119879 (101)
One can easily see that this is the correct solution The gasdemand at node 2 is 05 which is equal to the flow 1199022 sousing the equation of momentum implies a difference in thequadratic pressures between node 1 and node 2 of 025 Theflow through edge 1 is given by the two exit nodes and it is1199021 = 1 This implies a difference in the quadratic pressuresbetweennode 0 andnode 1 of 1 So the optimal solution119901+119898119886119909
119900119901119905
directly followsNext we use the spheric-radial decomposition especially
Algorithm 3 to calculate the probability P(119887 isin 119872) for arandom vector 119887 sim N(E2 [05 05]119879) Therefore we viewthe results of 8 tests with 1000 sampled points in each one(Table 1)
The probability in all eight tests is nearly the same Themean value is 03479 and the variance is 27723 sdot 10minus6 whichis very small This shows that the implementation is workingnearly exact The efficiency of the implementation of courseis not perfect but that is not part of this work
Last wewant to solve the following optimization problem
min119901+119898119886119909isin119875
11198793119901+119898119886119909
subject to 1119873 119873sum119894=1
119904sum119905=1
F120594 (119886119894119905) minusF120594 (119886119894119905) ge 120572 (102)
We use again the fmincon-function inMATLAB to solve theproblem eight times with 1000 sampled points in every caseBecause Lemma 8 needs a suitable 120583 isin R2
ge0 we assume 119887 simN(01sdot1198642 [15 15])We set120572 fl 09 and the results are shownin Table 2
Example 2 Now we add a compressor edge to the graph ofExample 1 (see Figure 8)
Assume that node 1 and 3 are consumers node 2 is ainner nodeThemeandemand120583 = [05 0 05]119879Thepressure
Mathematical Problems in Engineering 13
minus2 minus1 1 2 3
minus2
minus1
1
2
3
b2
b1
minus1 1 2
minus1
1
2
b2
b1
rLv1
rLv2
rLv3
Scheme of the spheric-radial decomposition with
sphere (yellow)and three points 1 2 3 (red) at the unit
the feasible set Mminus for Σ = E2 = [05 05]TFeasible set M for p+min = [2 1 1]T and
p+max = [3 2 2]T
Figure 7 Feasible set of Example 1
Table 1 Test results of easy example
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 8P(119887 isin 119872) 03486 03481 03450 03501 03484 03499 03459 03475
Table 2 Test results of Example 1
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 81199011198981198861199091199001199011199050 54514 54714 54758 54028 54767 53921 54888 551671199011198981198861199091199001199011199051 29802 30126 30225 29725 30300 29485 30336 300291199011198981198861199091199001199011199052 10084 10004 10776 10001 11221 11390 10285 10558
0 1 2 3q1 q2 q3
b0 b1 b2 b3
e1 isin ℰF e2 isin ℰC e3 isin ℰF
Figure 8 Graph of Example 2
0 1 2 3q1 q2 q2 q3
b0 b1 b2 b3
e1 isin ℰF e3 isin ℰF
Figure 9 Separated graph of Example 2
bounds are given by 119901+119898119894119899 = [2 1 1 1]119879 and 119901+119898119886119909 = [3 22 2]119879 and 1206011198901= 1 and 1206011198903
= 1 The incidence matrix is givenby
119860+ = [[[[[[
minus1 0 01 minus1 00 1 minus10 0 1]]]]]]
resp 119860 = [[[1 minus1 00 1 minus10 0 1
]]] (103)
Before we have a look at the inequalities of Theorem 5we have to remove the compressor edge from the graphThe flows can be computed by using the equation of massconservation and the new load vector is then
= (1198871 + 11990221198872 minus 11990221198873 ) with
119902 = (1198871 + 1198872 + 11988731198872 + 11988731198873)(104)
We assume the control to be switched off so it holds119906 = 1With this the control edge 1198902 is treated as a flow edge withoutfriction The resulting graph is shown in Figure 9 from theinequalities (25)ndash(35) it follows
3 le radic3 and 1 le radic8 minus 23 (105)
14 Mathematical Problems in Engineering
minus1 1 2 3
1
2
3
minus1 1 2 3
1
2
3
b3
b1
Hyperspace of the feasible set MHyperspace of the feasible set M
b1
b3
Figure 10 Feasible set of the Example 2
Table 3 Test results of control example
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 8P(119887 isin 119872) 01531 01500 01540 01536 01530 01550 01567 01528so the feasible set is (see Figure 10)
= + isin R119899+1 | 1119879
4 + = 0 and 0 le 3 le radic3 and 2 = 0and 0 le 1 le radic8 minus 22 (106)
This implies
119872 = 119887 isin R3ge0 | 0 le 1198873 le radic3 and 1198872 = 0 and 0 le 1198871 le minus1198873
+ radic8 minus 11988723 (107)
Solving the problem
min119901+119898119886119909isin119875
11198794119901+119898119886119909
subject to isin (108)
for 119887 = 120583 = [05 0 05]119879 leads us to the optimal solution
119901+119898119886119909119900119901119905 = [2 radic3radic3radic275]119879 (109)
The argumentation for the optimality of this vector is exactlylike before in the easy example Becausewe have a compressoredge in this graph we can minimize the control Thereforewe fix the pressure bounds as we did at the beginning of thisexample and we set 119887 = [1 0 15]119879 So inequality (31) is notfulfilled anymore for 119906 = 1 We use again the fmincon-solverinMATLAB to solve the following problem
min119906ge1
1199062subject to 119887 isin 119872(119906) (110)
The solver returns the optimal solution 119906119900119901119905 = 11818
Here one can see that the sufficient condition in Lemma 8is really strong Itrsquos not fulfilled with the given values but thereexists a solution anyway Of course the necessary conditionis fulfilled
Now we use again Algorithm 3 to compute the proba-bility P(119887 isin 119872) for a random load vector 119887 sim N(E3 [0505 05]119879) Table 3 shows us the results
Again we can see that the probability is nearly the samein all eight cases The mean probability is 01535 and thevariance is 32369 sdot 10minus6 For the optimization we reducethe covariance matrix and assume 119887 sim N(001 sdot E3 [05 0505]119879) For a probability 120572 = 09 Table 4 states the resultsExample 3 In this example we use a part of a real networkThe focus here is on realistic values for our model Weget these values from the GasLib (see [22]) a collection oftechnical gas network data
ldquoThegoal ofGasLib is to promote research on gasnetworks by providing a set of large and realisticbenchmark instancesrdquo (httpgaslibzibde)
We use a part of the Greek gas network (GasLib-134) (seeFigure 11)
For using real values we first need to know the constant120601119890 from (5)
120601119890 = (119877119878119879)2 120582119863 1198711198862 (111)
The value 119877119878 is the specific gas constant which can becalculated by 119877119878 = 119877119872 with the ideal gas constant 119877 =831451119896119892 sdot 11989821199042 sdot 119898119900119897 sdot 119870 (see [23]) and the molar mass119872 = 1662119896119892119896119898119900119897 (from real data) of the used natural gasAlso the temperature 119879 = 28915119870 is given from real dataFor the (constant) friction factor 120582 we use the law of Chen(see [24])
Mathematical Problems in Engineering 15
Table 4 Test results of control example
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 81199011198981198861199091199001199011199050 39013 39223 39039 38901 38879 39039 38901 390981199011198981198861199091199001199011199051 16568 16565 16365 16497 16496 16365 16497 164921199011198981198861199091199001199011199052 16568 16565 16365 16497 16496 16365 16497 164981199011198981198861199091199001199011199053 10000 10000 10000 10000 10002 10000 10000 10000
Table 5 Values based on real data
variable 119877119878 119879 120582 119863 119871 119886 120601value 50027 28915 00093 9144 178693 43862 198 sdot 107unit 11989821199042 sdot 119870 119870 minus 119898119898 119896119898 119898119904 11989821199042
Figure 11 Greek gas network from GasLib [22] red edges are usedin our example
1radic120582 = minus2 log10 ( 11989611986337065minus 50425
Relog10 [(119896119863)1109828257 + 58506
Re08981]) (112)
Diameter 119863 = 9144119898119898 and roughness value 119896 = 8 sdot 10minus6119898for the first pipe (see Figure 11) are known from real data TheReynolds-Number Re can be calculated by Re = (V119863120588)120578with the dynamic viscosity 120578 = 1 sdot 10minus5119875119886 sdot 119904 (see [1])the mean velocity V = 9548119898119904 (calculated with the meanvolume flow rate which is given by the real data) of the gasand the density 120588 = 074331198961198921198983 (from real data) So we getRe = 648959074 asymp 649 sdot 106 and with this the pipe frictioncoefficient after Chen is 120582 = 00093
The length of the first pipe is given by 119871 = 178693119896119898from the real data Last value missing is the sound speed innatural gas which we assume to be ideal The sound speed
119886 depends on the modulus of compressibility (see [9]) Forideal gases the modulus of compressibility only depends onthe adiabatic exponent and the pressure Using the ideal gasequation simplifies the equation for the sound speed to
119886 = radic120581119877119879119872 (113)
The adiabatic exponent for methan which is the main partof natural gas (gt 90) is about 133 (see [25]) Together allvalues are shown in Table 5 (example for the pipe next to thesource in Figure 11)
Nowwewant to optimize the inner control in the networkfor a fix load vector Therefor we choose two nominations2011-11-01 and 2016-02-17 In both cases the fmincon-functioninMATLAB returns
119906119900119901119905 = 1 (114)
as a local minimum which satisfies the constraint 119887 isin 119872This means for these nominations a compressor stationis not needed so it is switched off With this as a lastcalculation in this paper we calculate the probability for thegiven nominations to be feasible We use the values from thenominations as mean value 120583 and set 119887 sim N(120583 sdot 10minus3E 120583)The results are shown in Tables 6 and 7
Again one can see that the results are quite similarThe mean probability in Table 6 is 09860 and the meanprobability in Table 7 is 09760 Both values are nearly 1 sothere must happen something really unforeseeable that aGreek does not get his demanded gas
6 Conclusion
In this paper we have used an existing stationary modelfor gas transport based on the isothermal Euler equationsWe have extended this model by compressor stations andpresented a characterization of the set of feasible loadsEspecially Theorem 5 contains the central idea to get anaccess to the optimization in Section 4 and the spheric-radialdecomposition is our main tool to handle the uncertainty inthe loadsWewere able to show some results for optimizationwith probabilistic constraints in gas transport In a next stepthese results should be generalized
16 Mathematical Problems in Engineering
Table 6 Test results for nomination 2011-11-01
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 8P(119887 isin 119872) 09849 09849 09878 09885 09863 09845 09854 09853
Table 7 Test results for nomination 2016-02-17
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 8P(119887 isin 119872) 09744 09756 09788 09788 09754 09775 09734 09739
With the characterization of the feasible set we haveshown that the feasibility of a load depends significantly onthe pressure bounds These pressure bounds play a mainrole in our optimization problems The other main partsare the inner controls modeled by compressor edges whichshould lead in optimal control problems with probabilisticconstraints Here we have built a base for further works withthis topic
The idea of separating the graph and removing thecompressor edges can be used to extend themodel with othercomponents like eg control valves and resistorsThiswouldchange the inequalities inTheorem 5 but the approach worksalso for thesemodifications Further [8] showed a way to dealwith a networkwhich is a cycle Combining bothmethods weget an access to model gas networks in high detail
Finally the expansion to a dynamic model might bealso possible The idea is to use a space-mapping approachbecause one can use a lot of the analysis we did hereTherefore upcoming papers should deal with these problemsand present results of the turnpike-theory which describesa relation between stationary and instationary solutions Infact turnpike-theory implies that the optimal dynamic stateis close to the optimal static state in the interior of the timeintervals for sufficiently long time horizons AJ Zaslavskiexplains the turnpike-phenomenon in [26] This book alsogives an overview about the theory
Data Availability
The data used to support the findings of this study areavailable from the corresponding author upon request
Conflicts of Interest
The authors declare that there are no conflicts of interestregarding the publication of this paper
Acknowledgments
This work was supported by DFG in the framework of theCollaborative Research Centre CRCTransregio 154 Math-ematical Modeling Simulation and Optimization Using theExample of Gas Networks project C03
References
[1] P Domschke B Hiller J Lang and C Tischendorf ldquoModel-lierung von gasnetzwerken Eine Ubersichtrdquo Tech Rep 2717Technische Universitat Darmstadt 2017
[2] T Koch B Hiller M E Pfetsch and L Schewe Evaluating GasNetwork Capacities MOS-SIAM 2015
[3] M Gugat R Schultz and D Wintergerst ldquoNetworks of pipe-lines for gas with nonconstant compressibility factor stationarystatesrdquoComputational amp AppliedMathematics vol 37 no 2 pp1066ndash1097 2018
[4] M Gugat and M Herty ldquoExistence of classical solutions andfeedback stabilization for the flow in gas networksrdquo ESAIMControl Optimisation and Calculus of Variations vol 17 no 1pp 28ndash51 2011
[5] M K Banda M Herty and A Klar ldquoGas flow in pipeline net-worksrdquo Networks and Heterogeneous Media vol 1 no 1 pp 41ndash56 2006
[6] RM Colombo G GuerraM Herty and V Schleper ldquoOptimalcontrol in networks of pipes and canalsrdquo SIAM Journal onControl and Optimization vol 48 no 3 pp 2032ndash2050 2009
[7] A Shapiro D Dentcheva and A Ruszczynski ectures onStochastic Programming Modeling and Theory MPS-SIAM2009
[8] C Gotzes H Heitsch R Henrion and R Schultz ldquoOn thequantification of nomination feasibility in stationary gas net-works with random loadrdquoMathematical Methods of OperationsResearch vol 84 no 2 pp 427ndash457 2016
[9] D Halliday R Resnick and J Walker Physik Wiley-VCHBachelor edition edition 2007
[10] M Gugat F M Hante M Hirsch-Dick and G LeugeringldquoStationary states in gas networksrdquoNetworks andHeterogeneousMedia vol 10 no 2 pp 295ndash320 2015
[11] D RoseM SchmidtM C Steinbach and BMWillert ldquoCom-putational optimization of gas compressor stations MINLPmodels versus continuous reformulationsrdquoMathematical Meth-ods of Operations Research vol 83 no 3 pp 409ndash444 2016
[12] H Heitsch R Henrion H Leovey et al ldquoEmpirical observa-tions and statistical analysis of gas demandrdquo in Evaluating GasNetwork Capacities T data B KochM E Hiller et al Eds pp273ndash290 MOS-SIAM 2015
[13] A Genz and F Bretz Computation of Multivariate Normal andt Probabilities vol 195 of Lecture Notes in Statistics Springer2009
[14] M H Farshbaf-Shaker R Henrion and D Homberg ldquoProp-erties of Chance Constraints in Infinite Dimensions with anApplication to PDE Constrained Optimizationrdquo Set-Valued andVariational Analysis pp 1ndash21 2017
[15] T Gonzalez Grandon H Heitsch and R Henrion ldquoA jointmodel of probabilisticrobust constraints for gas transportman-agement in stationary networksrdquo Computational ManagementScience vol 14 no 3 pp 443ndash460 2017
[16] W van Ackooij I Aleksovska and M Munoz-Zuniga ldquo(Sub-)Differentiability of Probability Functionswith Elliptical Distri-butionsrdquo Set-Valued and Variational Analysis
Mathematical Problems in Engineering 17
[17] W van Ackooij and R Henrion ldquoGradient formulae for nonlin-ear probabilistic constraints with GAUssian and GAUssian-likedistributionsrdquo SIAM Journal on Optimization vol 24 no 4 pp1864ndash1889 2014
[18] W van Ackooij and R Henrion ldquo(Sub-)gradient formulaefor probability functions of random inequality systems underGaussian distributionrdquo SIAMASA Journal on UncertaintyQuantification vol 5 no 1 pp 63ndash87 2017
[19] W van Ackooij and J Malick ldquoSecond-order differentiabilityof probability functionsrdquo Optimization Letters vol 11 no 1 pp179ndash194 2017
[20] H Eckey R Kosfeld and M Turck Wahrscheinlichkeitsrech-nung und Induktive Statistik - Grundlagen -Methoden - BeispieleGabler Wiesbaden Germany 2005
[21] A Prekopa Stochastic Programming Springer 1st edition 1995[22] I Joormann M Pfetsch R Schwarz et al ldquoGasLibmdashA Library
of Gas Network Instancesrdquo 2017[23] G Wedler Lehrbuch der Physikalischen Chemie Wiley-VCH
5th edition 2004[24] N H Chen ldquoAn explicit equation for friction factor in piperdquo
Industrial amp Engineering Chemistry Fundamentals vol 18 no3 pp 296-297 1979
[25] W Schenk and F Kremer Physikalisches Praktikum Springer14th edition 2014
[26] A J Zaslavski ldquoTurnpike Properties in the Calculus of Varia-tions and Optimal Controlrdquo inNonconvex Optimization and ItsApplications vol 80 Springer 2006
Hindawiwwwhindawicom Volume 2018
MathematicsJournal of
Hindawiwwwhindawicom Volume 2018
Mathematical Problems in Engineering
Applied MathematicsJournal of
Hindawiwwwhindawicom Volume 2018
Probability and StatisticsHindawiwwwhindawicom Volume 2018
Journal of
Hindawiwwwhindawicom Volume 2018
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawiwwwhindawicom Volume 2018
OptimizationJournal of
Hindawiwwwhindawicom Volume 2018
Hindawiwwwhindawicom Volume 2018
Engineering Mathematics
International Journal of
Hindawiwwwhindawicom Volume 2018
Operations ResearchAdvances in
Journal of
Hindawiwwwhindawicom Volume 2018
Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018
International Journal of Mathematics and Mathematical Sciences
Hindawiwwwhindawicom Volume 2018
Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom
The Scientific World Journal
Volume 2018
Hindawiwwwhindawicom Volume 2018Volume 2018
Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in
Nature and SocietyHindawiwwwhindawicom Volume 2018
Hindawiwwwhindawicom
Dierential EquationsInternational Journal of
Volume 2018
Hindawiwwwhindawicom Volume 2018
Decision SciencesAdvances in
Hindawiwwwhindawicom Volume 2018
AnalysisInternational Journal of
Hindawiwwwhindawicom Volume 2018
Stochastic AnalysisInternational Journal of
Submit your manuscripts atwwwhindawicom
4 Mathematical Problems in Engineering
with 119899 degrees of freedom and 119871 is such that 119877 = 119871119871119879 (egCholesky decomposition)
Thedifference with our problem is that our load vector isnormal distributed with mean value 120583 and positive definitecovariance Σ We want to use the result of Theorem 2 togeneral Gaussian distributions (see [8]) Therefore we set120585lowast fl 119863minus1 (120585 minus 120583) sim N (0 119877) with
119863 fl diag (radicΣ119894119894)119894=1119899
and 119877 fl 119863minus1Σ119863minus1 (12)
and observe that P(120585 isin 119872) = P(120585lowast isin 119863minus1(119872 minus 120583)) So ouraim is to compute the following probability
P (120585lowast isin 119863minus1 (119872 minus 120583)) (13)
So the calculation of the probability (9) for a very generalset 119872 is equivalent to compute the integral in Theorem 2which is not easy to compute analytically too But the integralcan be computed numerically in a easy and nearly exactwayTherefore we formulate the following algorithm (see [8]Algorithm 4)
Algorithm 3 Let 120585 sim N(120583 Σ) and 119871 such that 119871119871119879 = Σ(1) Sample 119873 points V1 V119873 uniformly distributed
on the sphere S119899minus1(2) Compute the one-dimensional sets 119872119894 fl 119903 ge 0 |119903119871V119894 + 120583 isin 119872 for 119894 = 1 119873(3) Set P(120585 isin 119872) asymp (1119873)sum119873
119894=1 120583120594(119872119894)The first step is quite easy There are many possibilities
to sample a set of uniformly distributed points on the sphereS119899minus1 eg Monte-Carlo sampling or the randn-generator inMATLAB Normalizing these points leads us to the samplingwe are looking for Alternatively quasi Monte-Carlo methodscan be used In [8] the authors use a quasi Monte-Carlomethod for their computations and they show its advantagesin Figure 6
Themain challenge is to handle the one-dimensional setsof the second step If the set119872 is convex these sets are closedintervals but 119872 need not to be convex The aim is to get arepresentation of 119872119894 (119894 = 1 119873) as a union of disjointintervals With this the third step can be done efficiently Thenext section deals with this representation in detail
The third step is the approximation of the integral Thevalues 120583120594(119872119894) can be calculated easily with the distributionfunction F120594 of the distribution (see [20] for details of thatdistribution) With the representation 119872119894 = ⋃119905
119895=1[119886119895 119886119895] the120594-distribution can be calculated the following
120583120594 (119872119894) = 119905sum119895=1
F120594 (119886119895) minusF120594 (119886119895) (14)
This leads to the wanted probability The 120594-distributionfunction is given by the following (see [20] p 132)
F120594 (119909) = 121198732Γ (1198732) int119909
0V1198732minus1119890minusV2119889V (15)
2 4 6 8 10 12 14 16 18 200x
0010203040506070809
1
F(x)
Figure 3 Distribution function of the Chi-square distribution for 3(blue) 5 (red) 8 (green) and 12 (purple) degrees of freedom
with Γ(12) = 120587 Γ(1) = 1 and Γ(119903 + 1) = 119903Γ(119903) for all 119903 isin R+Note that there are very precise numerical approximationsfor the distribution function eg the chi2cdf -function inMATLAB (see Figure 3)
3 Explicit Characterization of the Gas Flow
In this section we want to characterize the feasible set to useAlgorithm 3 The aim is to get a representation of the one-dimensional sets of step (2) as a union of disjoint intervalsTherefore we need another representation of the feasible setas a system of inequalitiesThis is an idea from [8]Theorem 1In the proof the authors use a representation of the pressureloss for the full network graph
(119860+)119879 (119901+)2 = minusΦ 10038161003816100381610038161199021003816100381610038161003816 119902 (16)
which is equivalent to (5) if and only if E119862 = 0 In this paperwe also consider the case E119862 = 0 The matrix Φ is diagonalwith the values 120601119890 as defined in (5) at its diagonal Thisrepresentation is not possible yet because we extended ourmodel by compressor edges so this equation does not holdfor all edges Sowe have to find away to use a similar equationlike (16) to state a suitable theorem The idea that allowsto include compressor stations is to remove compressoredges from the graph and replace them by suitable algebraicconditions First we consider the equation of masssum
119890isin1198640(V)119902119890 = 119887+V lArrrArr
sum119890isin1198640(V)capE119865
119902119890 = 119887+V minus sum119890isin1198640(V)capE119862
119902119890 (17)
We want to treat the flows through the compressor edges asloads sowehave to adapt the load vectorThenew load vector+ is defined as follows+V
fl
119887+V + 119902119890 forall119890 isin 1198640 (V) with 119890 isin E119862 and 119891 (119890) = V119887+V minus 119902119890 forall119890 isin 1198640 (V) with 119890 isin E119862 and ℎ (119890) = V119887+V else
(18)
The resulting graph 119866 = (V+E119865) is no longer connected(scheme is shown in Figure 4)
Mathematical Problems in Engineering 5
p0
p1
p2
p3
p4
p5
p6
p7
p8
p9
q1
q2
q4
q6
q7
q8
q9
b0
b1
b2
b3
b4
b5
b6
b7
b8
b9
p0
p1
p2
p3
p4
p5
p6
p7
p8
p9
q1
q2
q3 q3
q4
q5 q5
q6
q7
q8
q9
b0
b1
b2
b3
b4
b5
b6
b7
b8
b9q5 isin ℰC
q3 isin ℰC
Figure 4 Network graph with two compressor edges (left) and new not connected network graph with three connected subgraphs afterremoving the compressor edges (right)
We set |E119865| fl 1198981 and |E119862| fl 1198982 with 1198981 + 1198982 = 119898so 119866 consists of 1198982 + 1 subgraphs which are all connectedNote that 119902 is a vector in1198771198981 nowWe determine that119866119894 is the
119894-th self-connected subgraph of119866 (depends on numbering ofthe graph) Analogously 119901+
119894119895 is the 119895-th component of the 119894-thsubgraph The terms 119902119894119895 and 119860 119894119895119896 are defined similarly Thefeasible set for 119866 is then
=+ isin R119899+1 |
1119879119899+1+ = 0 andexist (119901+ 119902) isin R119899+1
ge0 times R119898 with 119901+ isin [119901+119898119894119899 119901+119898119886119909] 119860 119894119902119894 = 119894 forall119894 = 1 1198982 + 1(119860+119894 )119879 (119901+
119894 )2 = minusΦ11989410038161003816100381610038161199021198941003816100381610038161003816 ∘ 119902119894 forall119894 = 1 1198982 + 1
(119901ℎ(119890)119901119891(119890)
)2 = 119906119890 forall119890 isin E119862
(19)
The last equation in is the same as in (8) that modelsthe pressure rise caused by the compressor With this we canpresent the equation for pressure loss in the form of (16) forevery subgraph Before we state a theorem for characterizingthe feasible set we show the equivalence of and
Lemma4 Let + be as defined in (18)The feasible sets definedin (8) and (19) are equivalent in the following sense
(i) For all 119887+ isin there is + isin (ii) For all + isin there is 119887+ isin (20)
Proof (i) Consider a vector 119887+ isin It holds 1119879119899+1119887+ = 0 and
because of (18) it follows directly 1119879119899+1+ = 0 After removing
the compressor edges from119860+119902 = 119887+ it follows with (17) that119860+119894 119902119894 = +119894 which is equivalent to 119860 119894119902119894 = 119894 The pressure loss
is fulfilled for all flux edges in and because the flow vector119902 doesnrsquot change in the relevant coefficients while removingthe compressor edges the pressure loss is fulfilled in tooThe compressor property is independent of 119887+ and 119902 so it alsoholds From this it follows + isin 119872
(ii) The other way around is quite similar We consider avector + isin From 1119879
119899+1+ = 0 it follows with (18) that1119879119899+1119887+ = 0 The equation for mass conservation is fulfilled
in every subgraph So with (17) it follows 119860+119902 = 119887+ whichimplies the mass conservation in119872 The pressure loss holds
in all self-connected parts of the graph too and because therelevant coefficients of 119902 doesnrsquot change while adding thecompressor edges the pressure loss is fulfilled in119872 too Lastas said before the compressor property is independent of +and 119902 it is also fulfilled in119872 This completes the proof
Because our network graph is a tree the mass conserva-tion is fulfilledwith a full rank incidencematrix119860of size 119899times119899so 119902 can be computed explicit by the following
119902 = 119860minus1119887 (21)
which makes us able to use Lemma 4With this result we canwrite the equation formomentum
conservation depending on the incidence matrices for theparts of the tree This makes us able to formulate the theoremfor characterizing the feasible set Therefore we will use thefollowing function
119892 R119899 997888rarr R
119909 997891997888rarr (119860119879)minus1Φ 10038161003816100381610038161003816119860minus111990910038161003816100381610038161003816 (119860minus1119909) (22)
which describes the pressure loss between the nodes We alsointroduce a new notation
(i) 119906(119894119895)119896 is the 119896-th control on the path from V1198940 to V1198950(ii) 119892(119894119895)119896 is the function 119892 for the 119896-th subgraph on the
path from V1198940 to V1198950 whereas 119892(119894119895)1 is the function 119892
6 Mathematical Problems in Engineering
of the graph 119866119894 According to this 119892(119894119895)119896ℓ is the ℓ-thcomponent of the function 119892(119894119895)119896
(iii) (119894119895)119896 is the load vector of the 119896-th subgraph in thepath from V1198940 to V1198950 whereas (119894119895)1 is the load vectorof the graph 119866119894 and (119894119895)119896ℓ is the ℓ-th component of(119894119896)119896
Analogously we define 119860 (119894119895)119896 119901(119894119895)119896 and 119902(119894119895)119896 Furthermorewe define the following values
(i) 119896lowast119894119895 fl max119896 isin 1 1198982 + 1 | V1198960 isin Π(V1198940) and V1198960 isinΠ(V1198950) as the largest index of all subgraph the pathsto V1198940 and V1198950 pass through
(ii) 119899lowast119894119895 |119896 isin 1 1198982 + 1 | V1198960 isin ΠV1198940(V1198950)| as thenumber of subgraph which are on the path from V1198940from V1198950 containing 119866119894 and 119866119895
(iii) 119898lowast119894119895 fl |119890 isin E119862 | ℎ(119890) isin ΠV1198940(V1198950)and119891(119890) isin ΠV1198940(V1198950)|
as the number of controls which are between V1198940 andV1198950
With this notation and these values we define a sum for 119894 119895 isin11198982 + 1 If V1198940 isin Π(V1198950) we set sum119894119894 fl 0 and else we set
Σ119896lowast119894 fl119899lowastminus2sum119896=1
1prod119898lowastminus119896ℓ=1 119906(119896lowast119894)ℓsdot 119892(119896lowast119894)119899lowastminus119896119891(119890119906
(119896lowast119894)119899lowastminus119896) ((119896lowast119894)119899lowastminus119896)
+ 119892(119896lowast119894)1119891(119890119906(119896lowast119894)1
) ((119896lowast119894)1) (23)
where we write only 119899lowast and 119898lowast instead of 119899lowast119896lowast119894 and 119898lowast119896lowast119894 for
better reading Last we define a productΠ119894119894 fl 1 andΠ119896lowast119894 fl
119898lowastprod119896=1
119906(119896lowast119894)119896 (24)
From a first point of view the notation seems to be confusingbut using this notation makes sense since we want toguarantee the feasibility of a load vector by comparing thepressure bounds at every node with each other We explainthis in a short example Figure 5 shows a graph after removingthe compressor edges
If we want to compare the pressure bounds of node 11 andnode 12 we need to know how the pressures change on thepath from 2 to 11 resp from 2 to 12 Node 11 is part of thesubgraph 1198665 and node 12 is part of the subgraph 1198666 so itfollows 119896lowast56 = 2 because node 2 is part of subgraph 2 Thenotation can be understood as a numbering along the pathIt is 119906(211)1 = 1199062 the first control on the path from node 2to node 11 and 119906(211)2 = 1199064 the second control on this pathFurther it follows 119899lowast211 = 3 (1198662 1198663 1198665) and119898lowast
211 = 2 (1199062 1199064)Theproduct defined before is the product of all controls alongthis path and the sum is the pressure loss between node 2 andnode 7 Using this notation makes us able to formulate thetheorem for characterizing the feasible set in a readable way
Theorem 5 For given pressure bounds 119901+119898119894119899 119901+119898119886119909 isin R119899+1
and controls 119906119894 isin R (119894 = 1 1198982) the following equivalenceholds
A vector + with 1119879119899+1+ = 0 is feasible ie + isin if and
only if the following inequalities hold For all 119894 = 1 1198982 + 1holds (feasibility inside the subgraphs)
(1199011198981198941198991198940 )2 le min
119896=1119899119894[(119901119898119886119909
119894119896 )2 + 119892119894119896 (119894)] (25)
(1199011198981198861199091198940 )2 ge max
119896=1119899119894[(119901119898119894119899
119894119896 )2 + 119892119894119896 (119894)] (26)
max119896=1119899119894
[(119901119898119894119899119894119896 )2 + 119892119894119896 (119894)]
le min119896=1119899119894
[(119901119898119886119909119894119896 )2 + 119892119894119896 (119894)] (27)
For all 119894 119895 = 1 1198982 + 1 with 119894 lt 119895 holds (feasibility betweenthe subgraphs)1Π119896lowast 119894
(1199011198981198941198991198940 )2 + Σ119896lowast119894 le 1Π119896lowast119895
(1199011198981198861199091198950 )2 + Σ119896lowast119895 (28)
1Π119896lowast 119894
(1199011198981198861199091198940 )2 + Σ119896lowast119894 ge 1Π119896lowast 119895
(1199011198981198941198991198950 )2 + Σ119896lowast119895 (29)
1Π119896lowast 119894
(1199011198981198941198991198940 )2 + Σ119896lowast119894
le 1Π119896lowast119895
min119896=1119899119895
[(119901119898119886119909119895119896 )2 + 119892119895119896 (119895)] + Σ119896lowast119895 (30)
1Π119896lowast 119894
(1199011198981198861199091198940 )2 + Σ119896lowast119895
ge 1Π119896lowast119895
max119896=1119899119895
[(119901119898119894119899119895119896 )2 + 119892119895119896 (119895)] + Σ119896lowast119895 (31)
1Π119896lowast 119894
max119896=1119899119894
[(119901119898119894119899119894119896 )2 + 119892119894119896 (119894)] + Σ119896lowast119894
le 1Π119896lowast119895
(1199011198981198861199091198950 )2 + Σ119896lowast119895 (32)
1Π119896lowast 119894
min119896=1119899119894
[(119901119898119886119909119894119896 )2 + 119892119894119896 (119894)] + Σ119896lowast119894
ge 1Π119896lowast119895
(1199011198981198941198991198950 )2 + Σ119896lowast119895 (33)
1Π119896lowast 119894
max119896=1119899119894
[(119901119898119894119899119894119896 )2 + 119892119894119896 (119894)] + Σ119896lowast119894
le 1Π119896lowast119895
min119896=1119899119895
[(119901119898119886119909119895119896 )2 + 119892119895119896 (119895)] + Σ119896lowast119895 (34)
1Π119896lowast 119894
min119896=1119899119894
[(119901119898119886119909119894119896 )2 + 119892119894119896 (119894)] + Σ119896lowast119894
ge 1Π119896lowast119895
max119896=1119899119895
[(119901119898119894119899119895119896 )2 + 119892119895119896 (119895)] + Σ119896lowast119895 (35)
Mathematical Problems in Engineering 7
0 1 2
3
4
5
6
7
8
9
10
11
12
u1
u2
u3
u4
u5
G1 G2 G3
G4
G5
G6
Figure 5 Example graph for illustration of the used notation
0 1 2q1 q2
b0 b1 b2
Figure 6 Graph of Example 1
Let us have a look at the inequalities Inequalities(25)ndash(27) are well-known from [8] and they describe thefeasibility inside a connected tree The other inequalities(28)ndash(35) guarantee the feasibility between the subgraphsTherefore we compare the pressures of the entries of two parts((28) (29)) of the entry of one part and the exit nodes of theother parts ((30)ndash(33)) and we compare the pressures of allexits ((34) (35)) Because we cannot compare the pressuresof119866119894 and 119866119895 directly (there may be no direct connection) wehave to compare them using the subgraph 119866119896lowast which is thelast part the paths to V1198940 and V1198950 pass Therefore we definedthe sumsΣ119896lowast 119894 andΣ119896lowast119895 which trace back the pressures to119866119896lowast
Remark 6 For an implementation it is wise to define a set119880(119894119895) containing all controls on the path from V1198940 to V1198950 anduse an index running over this set to define the sums and theproducts
Now we want to proof the theorem
Proof of Theorem 5 Because the columns of 119860+ sum up tozero it holdsminus1119879
119899119894119860 119894 = 119860+
1198940 (119894 = 1 1198982 + 1) (36)
ldquo997904rArrrdquo We consider a feasible vector 119887+ isin From theequation of mass conservation of every subgraph it follows
119902119894 = 119860minus1119894 119894 (37)
We take the equation of momentum conservation for the 119894-thsubgraph (119894 isin 1 1198982 +1) and separate 119860+
119894 to119860+1198940 and 119860 119894
(119860+1198940)119879 1199012
1198940 + 119860119879119894 1199012
119894 = minusΦ 10038161003816100381610038161199021198941003816100381610038161003816 119902119894 (38)
Next we insert (36) into (38) and get
minus119860119879119894 1119899119894
11990121198940 + 119860119879
119894 1199012119894 = minusΦ 10038161003816100381610038161199021198941003816100381610038161003816 119902119894 (39)
We multiply this equation from left with (119860119879119894 )minus1 and insert
(37) so it follows
111989911989411990121198940 = 1199012
119894 + (119860119879119894 )minus1Φ 10038161003816100381610038161003816119860minus1
119894 11989410038161003816100381610038161003816 119860minus1119894 119894 (40)
and using the definition of the function 119892 we have
111989911989411990121198940 = 1199012
119894 + 119892119894 (119894) (41)
Note here that 119901119894 is the vector of pressures of the subgraph119866119894
and 119892119894(119894) is the vector-valued function which describes thepressure loss in 119866119894 Equation (41) holds for all subgraphs 119866119894
with 119894 isin 1 1198982 + 1 From this the inequalities (25)ndash(27)follow directly with 119901+ isin [119901+119898119894119899 119901+119898119886119909] Next we use (41)component-by-component For fix 119894 119895 isin 1 1198982 + 1 with119894 lt 119895 it follows
1199012119896lowast0 = 1199012
119896lowast119896 + 119892119896lowast119896 (119896lowast) (119896 = 1 119899119896lowast) (42)
Remember that 119896lowast119894119895 is the largest index of all subgraphs thepaths to V1198940 and V1198950 pass through For better reading we onlywrite 119896lowast instead of 119896lowast119894119895 We use the equation of the compressorproperty for the first compressor on the path from V119896lowast0 to V1198940
1199012ℎ(119890119906(119896lowast119894)1
)1199012119891(119890119906(119896lowast119894)1
)
= 119906(119896lowast119894)1 (43)
and insert it into (42) for 119896 = 119891(119890119906(119896lowast119894)1 )1199012(119896lowast 119894)10 = 1119906(119896lowast119894)1
1199012(119896lowast119894)20
+ 119892(119896lowast 119894)1119891(119890119906(119896lowast119894)1
) ((119896lowast119894)1) (44)
Note that 1199012119896lowast 0 = 1199012
(119896lowast119894)10 119892119896lowast119896(119896lowast) = 119892(119896lowast119894)1119896((119896lowast119894)1) and1199012ℎ(119890119906(119896lowast119894)1
) = 1199012(119896lowast 119894)20 We use the compressor equation for
the second compressor on the path from V119896lowast0 to V1198940 and putit together with (42) into (44)
1199012(119896lowast119894)10 = 1119906(119896lowast119894)1
[ 1119906(119896lowast119894)2
1199012(119896lowast 119894)30
+ 119892(119896lowast119894)2119891(119890119906(119896lowast119894)2
) ((119896lowast119894)2)]+ 119892(119896lowast119894)1119891(119890119906
(119896lowast119894)1) ((119896lowast119894)1)
(45)
8 Mathematical Problems in Engineering
We repeat this procedure for all controls on the path fromV119896lowast0 to V1198940 Although we write 119899lowast and119898lowast instead of 119899lowast119896lowast119894 and119898lowast
119896lowast119894 Thus we get
1199012(119896lowast119894)10 = 1119906(119896lowast119894)1
[ 1119906(119896lowast119894)2
[sdot sdot sdot [ 1119906(119896lowast119894)119898lowast1199012(119896lowast119894)119899lowast0
+ 119892(119896lowast119894)119899lowastminus1119891(119890119906(119896lowast119894)119899lowastminus1
) ((119896lowast119894)119899lowastminus1)] sdot sdot sdot]+ 119892(119896lowast119894)2119891(119890119906
(119896lowast119894)2) ((119896lowast119894)2)]
+ 119892(119896lowast119894)1119891(119890119906(119896lowast119894)1
) ((119896lowast119894)1)
(46)
We take again (42) for the node V119896lowast0 and do the sameprocedure just like before for the path from V119896lowast0 to V1198950 With119899lowast = 119899lowast119896lowast119895 and119898lowast = 119898lowast
119896lowast119895 we get
1199012(119896lowast 119895)10
= 1119906(119896lowast119895)1
[ 1119906(119896lowast119895)2
[sdot sdot sdot [ 1119906(119896lowast119895)119898lowast1199012(119896lowast 119895)119899lowast0
+ 119892(119896lowast119895)119899lowastminus1119891(119890119906(119896lowast119895)119899lowastminus1
) ((119896lowast119895)119899lowastminus1)] sdot sdot sdot]+ 119892(119896lowast119895)2119891(119890119906
(119896lowast119895)2) ((119896lowast119895)2)]
+ 119892(119896lowast119895)1119891(119890119906(119896lowast119895)1
) ((119896lowast119895)1)
(47)
Expanding (46) and (47) and equalizing them (this is possiblebecause 119901(119896lowast 119894)10 = 119901(119896lowast119895)10) leads to1Π119896lowast 119894
11990121198940 + Σ119896lowast119894 = 1Π119896lowast 119895
11990121198950 + Σ119896lowast119895 (48)
with Σ119896lowast119894 resp Σ119896lowast119895 defined before (48) directly implies (28)and (29) Using (48) together with (42) for V1198940 first timefor V1198950 second time and for V1198940 and V1198950 a third time theinequalities (30)ndash(35) follow directly So this part of the proofis complete
ldquolArr997904rdquo For this part we consider a vector 119887+ isin R119899+1
with 1119879119899+1119887+ = 0 which fulfills the inequalities (25)ndash(35) We
define the following sets (119894 = 2 1198982 + 1)11987511 fl [(119901119898119894119899
10 )2 (11990111989811988611990910 )2]
11987521 fl [ max119896=11198991
[(1199011198981198941198991119896 )2 + 1198921119896 (1)]
min119896=11198991
[(1199011198981198861199091119896 )2 + 1198921119896 (1)]]
1198751119894 fl [ 1Π1119894
(1199011198981198941198991198940 )2 + Σ1119894 1Π1198941
(1199011198981198861199091198940 )2 + Σ1119894]
1198752119894 fl [ 1Π1119894
max119896=1119899119894
[(119901119898119894119899119894119896 )2 + 119892119894119896 (119894)] + Σ1119894 1Π1198941
sdot min119896=1119899119894
[(119901119898119886119909119894119896 )2 + 119892119894119896 (119894)] + Σ1119894]
(49)
We set
1198750 fl 1198982+1⋂119896=1
1198751119896 cap 1198982+1⋂119896=1
1198752119896 (50)
and we will define a value 119901210 isin 1198750 later So to guarantee
such a value 119901210 exists we have to show first that 1198750 is not
empty Because 1198750 is a finite intersection of convex intervalswe have to make sure that all intervals are not empty andevery intersection between two intervals is not empty In onedimension this is sufficient that 1198750 is not empty For 119894 =1 1198982 + 11198751119894 = 0 according to requirement1198752119894 = 0 because of (26) (51)
Furthermore its 1198751119894 cap 1198752119894 = 0 because of (25) and (26) Themissing intersections are nonempty too for 119894 119895 isin 1 1198982+1 with 119894 lt 119895 as follows
1198751119894 cap 1198751119895 = 0 because of (28) and (29) 1198751119894 cap 1198752119895 = 0 because of (30) and (31) 1198752119894 cap 1198751119895 = 0 because of (32) and (33) 1198752119894 cap 1198752119895 = 0 because of (34) and (35)
(52)
So the set 1198750 is not empty We define the following values
119901210 isin 119875011990121198940 fl Π1119894 (1199012
10 minus Σ1119894) 119894 = 2 1198982 + 11199012119894119896 fl 1199012
1198940 minus 119892119894119896 (119894) 119894 = 1 1198982 + 1 119896 = 1 119899119894119902119894 fl 119860minus1119894 119894 119894 = 1 1198982 + 1
(53)
Now we show that our choice of pressures and flows isfeasible It is11990110 isin [119901119898119894119899
10 11990111989811988611990910 ] because of1199012
10 isin 11987511 Furtherbecause of 1199012
10 isin 1198751119894 for all 119894 = 2 1198982 + 1 it holds1Π1119894
11990121198940 + Σ1119894 le 1Π1119894
(1199011198981198861199091198940 )2 + Σ1119894 (54)
and 1Π1119894
11990121198940 + Σ1119894 ge 1Π1119894
(1199011198981198941198991198940 )2 + Σ1119894 (55)
Mathematical Problems in Engineering 9
Now we replace 11990121198940 in its definition with (1199012
119894119896 + 119892119894119896(119894)) andbecause of 1199012
1198940 isin 1198752119894 for all 119894 = 1 1198982 + 1 it follows1Π1119894
(1199012119894119896 + 119892119894119896 (119894)) + Σ1119894
le 1Π1119894
min119896=1119899119894
[(119901119898119886119909119894119896 )2 + 119892119894119896 (119894)] + Σ1119894
(56)
and 1Π1119894
(1199012119894119896 + 119892119894119896 (119894)) + Σ1119894
ge 1Π1119894
max119896=1119899119894
[(119901119898119894119899119894119896 )2 + 119892119894119896 (119894)] + Σ1119894 (57)
So for all 119894 = 1 1198982 + 1 and for all 119896 = 1 119899119894 it holds119901119894119896 isin [119901119898119894119899119894119896 119901119898119886119909
119894119896 ] which is equivalent to
119901+ isin [119901+119898119894119899 119901+119898119886119909] (58)
The equation for mass conservation follows directly from thedefinition of 119902119894 (119894 = 1 1198982 + 1) For the pressure loss weuse
1199012119894 = 1119899119894
11990121198940 minus 119892119894 (119894) (59)
And multiply this equation from left with 119860119879119894 and use (36)
With this it follows the equation for momentum conserva-tion
(119860+119894 )119879 (119901+
119894 )2 = minusΦ 10038161003816100381610038161199021198941003816100381610038161003816 119902119894 (60)
The last equation missing is the compressor property For acontrol 119895 isin 1 1198982 we fix a vertex V1198940 (119894 isin 1 1198982 +1)with ℎ(119890119906119895) = V1198940 We use 1199012
1198940 = Π1119894(119901210 minus Σ1119894) and solve the
formula for 119901210 in the following way
119901210 = 1119906(1119894)1
[sdot sdot [ 1119906(1119894)119898lowast1199012(1119894)119899lowast0
+ 119892(1119894)119899lowastminus1119891(119890119906(1119894)119899lowastminus1
) ((1119894)119899lowastminus1)] sdot sdot]+ 119892(1119894)1119891(119890119906(1119894)1 )
((1119894)1) (61)
Consider wlog that the subgraphs along the path from V10to V1198940 are numbered chronological st if ℎ(119890119906119895) isin 119866119894 then119891(119890119906119895) isin 119866119894minus1 We use 1199012
119894minus10 = Π1119894minus1(119901210 minus Σ1119894minus1) to get the
same representation for 119901210 depending on 1199012
(1119894minus1)119899lowast0 whichis equivalent to 119901(1119894)119899lowastminus10 We equalize these representationsdepending on 1199012
1198940 and 1199012119894minus10 and we get
1119906(1119894)119898lowast1199012(1119894)119899lowast0 + 119892(1119894)119899lowastminus1119891(119890119906
(1119894)119899lowastminus1) ((1119894)119899lowastminus1)
= 1199012(1119894)119899lowastminus10 (62)
We chose 119894 isin 1 1198982 + 1 st 119906119895 is the last control on thepath from V10 to V1198940 so it is 119906(1119894)119898lowast = 119906119895 Further we have
1199012(1119894)119899lowast 0 = 1199012
11989401199012(1119894)119899lowastminus10 = 1199012
119894minus10119892(1119894)119899lowastminus1119891(119890119906(1119894)119899lowastminus1
) (119887(1119894)119899lowastminus1) = 119892119894minus1119891(119890119906119895 )(119887119894minus1)
(63)
and we know that 11990121198940 resp 1199012
119894minus10 minus 119892119894minus1119891(119890119906119895 )(119887119894minus1) are the
pressures at the head resp the foot of the compressor edge119890119906119895 So it follows 1119906119895
1199012ℎ(119890119906119895 )
= 1199012119891(119890119906119895 )
(64)
which is equivalent to the compressor property This com-pletes the proof
With this theorem we have another characterization ofthe feasible set The last step in this intersection is theadaption of the results to the spheric-radial decompositionAs mentioned before the main step is to get a representationof the sets119872119894 in step (2) of Algorithm 3
Because we assumed that the graph has only one entry wedefine the following set
119872 fl 119887 isin R119899ge0 | (minus1119879
119899119887 119887) isin (65)
Consider a sampled point V119894 isin S119899minus1 of step (1) of Algorithm 3(119894 = 1 119873) We identify the load vector 119887 with the affinelinear function
119887119894 (119903) fl 119903119871V119894 + 120583 = 119903120596119894 + 120583 with 120596119894 = 119871V119894 (66)
where 120583 is the mean value of the Gaussian distribution and 119871is such that 119871119871119879 = Σ for positive definite covariance matrixΣ Note that the index 119894 here is for the sampled vector V119894 notfor the 119894-th subgraph Because 119887119894 is defined only for the exitnodes it must hold 119887 ge 0 which leads to the definition of theregular range
119877119903119890119892119894 fl 119903 ge 0 | 119887119894 (119903) ge 0 supe 119872119894 (67)
So the feasible set is
119872119894 = 119903 isin 119877119903119890119892119894 | 119887119894 (119903) fulfills (3) (5) and (7) (68)
and with Theorem 5 it holds
119872119894 = 119903 isin 119877119903119890119892119894 | 119894 (119903) fulfills (25) minus (34) (69)
where 119894(119903) is from (18) Because 119894 depends on 119903 now thefunction 119892 depends on 119903 too More exactly 119892 is quadratic in119903 so 119892119896(119887119894(119903)) can be represented as follows
1199032[[119899sum
119895=1
Ψ119896119895( 119899sumℓ=1
119860minus1119895ℓ120596119894ℓ)2]]⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
119886119894
10 Mathematical Problems in Engineering
+ 119903[[2119899sum
119895=1
Ψ119896119895( 119899sumℓ=1
119860minus1119895ℓ120596119894ℓ)( 119899sum
ℓ=1
119860minus1119895ℓ120583ℓ)]]⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
119894
+ [[119899sum
119895=1
Ψ119896119895( 119899sumℓ=1
119860minus1119895ℓ120583ℓ)2]]⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
119888119894
(70)
HereΨ is the product of119860 andΦ So the inequations whichcharacterize119872119894 are also quadratic in 119903 because the minimaland maximal pressures in the inequalities only count to 119888119894Now we can write the feasible sets as follows
119872119894 = 119903 isin 119877119903119890119892119894 | 0 le 1198861198941199051199032 + 119887119894119905119903 + 119888119894119905 (71)
where 119905 is an index for the total number of inequalities in119872119894Now the regular range can be cut with the positive intervals ofthe inequalities and the result is a union of disjoint intervals
119872119894 = 119904119894⋃119896=1
[119886119894119896 119886119894119896] 119904119894 isin N (72)
Here 119904119894 is the number of disjoint intervals in which allinequalities are fulfilled and 119886119894119896 and 119886119894119896 are the intervalbounds they depend on the pressure bounds and the con-trols So the third step of Algorithm 3 can be executed With
P (119887 isin 119872) asymp 1119873 119873sum119894=1
119904119894sum119896=1
F120594 (119886119894119896) minusF120594 (119886119894119896) (73)
the probability for a random load vector to be feasible can becomputed
4 Existence of Optimal Solutions
In this section we want to have a look at some optimizationproblems We distinguish between problems with constantloads (without uncertainty on the demand) and problemswith random loads The latter leads us to optimizationproblems with so called probabilistic constraints or chanceconstraints (see [21]) There are two aims in optimizationMinimize the maximal pressure bounds and minimize thecontrols For minimizing the maximal pressure bounds wedefine the following set
119875 fl119899⋃
119894=0
[119901+119898119894119899119894 infin) (74)
in which we look for them Because the maximal pressuresare in proportion to the cost of pipes an objective functionfor a optimization problem could be 119888119879119901+119898119886119909 for a cost vector119888 isin R119899+1
gt0 By minimizing the controls we use the Euclideannorm of the controls as an objective function
Optimization with Constant Loads Consider the two opti-mization problems
min119901+119898119886119909isin119875
119888119879119901+119898119886119909
subject to 119887+ (119901+119898119886119909) isin 119872 (75)
andmin119906isinR1198982ge1
1199062subject to 119887+ (119906) isin (76)
Lemma 7 There exists a unique solution for (75)
Proof We have a look at the side constraint 119887+(119901+119898119886119909) isin Using Theorem 5 we can write (75) as
min119901+119898119886119909isin119875
119888119879119901+119898119886119909
subject to (25) minus (35) are fulfilled (77)
For 119894 isin 1 1198982 + 1 we write inequality (25) (and allremaining analogous) in the following way
0 le 119892119894119896 (119894) minus (1199011198981198941198991198940 )2 + (119901119898119886119909
119894119896 )2 forall119896 = 1 119899119894 (78)
So every inequality depends only on one upper pressurebound (but more inequalities may depend on the samebound) Because the inequalities are not strict the values119901119896 fl min 119901119898119886119909
119896 isin [0infin) | (25) minus (35) are fulfilled (79)
exist for all 119896 = 1 119899 + 1 and we can define a value119901+119898119886119909119900119901119905119896 fl max 119901119896 119901119898119894119899
119896 (80)
which is obviously a solution of (75) Moreover this solutionis unique because reducing one component of the upperbounds would either hurt an inequality or it would not fulfill119901+119898119886119909 ge 119901+119898119894119899 anymore
For the optimization problem (76) we can formulate anecessary and a sufficient condition for optimal solutions
Lemma8 Consider pressure bounds119901+119898119894119899 119901+119898119886119909 isin R119899+1 st119901+119898119894119899 lt 119901+119898119886119909 and a suitable vector 119887+ isin R119899+1 (ie + fulfills(25)ndash(27)) For all 119896 = 1 1198982 we set 119894(119896) and 119895(119896) such that119891(119890119896) isin V+
119894(119896) and ℎ(119890119896) isin V+119895(119896)
(a) If 119906 isin R1198982 is a solution of (76) it holds for all 119896 =1 1198982 (119901119898119894119899119894(119896)0)2 minus 119892119894(119896)119891(119890119896)
(119894(119896)) le (119901119898119886119909119895(119896)0)2 (81)
(b) If the inequalities
(119901119898119886119909119894(119896)0)2 minus 119892119894(119896)119891(119890119896)
(119894(119896)) ge (119901119898119894119899119895(119896)0)2
and (119901119898119894119899119894(119896)0)2 minus 119892119894(119896)119891(119890119896)
(119894(119896)) le (119901119898119886119909119895(119896)0)2 (82)
are fulfilled for all 119896 = 1 1198982 an optimal solution 119906 isin R1198982ge1
of (76) exists
Mathematical Problems in Engineering 11
Note that a vector 119887+ isin R119899+1 is suitable here if the vector+ is feasible for the subgraphs ie the inequalities (25)-(27)are fulfilled for +Proof (a) For this part we use a proof by contradictionAssume that
(119901119898119894119899119894(119896)0)2 minus 119892119894(119896)119891(119890119896)
(119894(119896)) gt (119901119898119886119909119895(119896)0)2 (83)
Because 119906119896 ge 1 for all 119896 = 1 1198982 it follows
(119901119898119894119899119894(119896)0)2 minus 119892119894(119896)119891(119890119896)
(119894(119896)) gt 1119906119896
(119901119898119886119909119895(119896)0)2 (84)
It it Σ119896lowast119894(119896) = 0 and Π119896lowast 119894(119896) = 1 because the subgraph withindex 119896lowast is equal to the subgraphwith index 119894(119896) And becausethe subgraphs 119894(119896) and 119895(119896) are neighbours (ie directlyconnected) it follows Π119896lowast 119895(119896) = 119906119896 and
Σ119896lowast119895(119896) = 119892(119896lowast119895(119896))1119891(119890119906(119896lowast119895(119896))1
) (119887(119896lowast119895(119896))1)= 119892119894(119896)119891(119890119896)
(119894(119896)) (85)
Thus we have a contradiction to (28) and 119906119896 cannot beoptimal
(b) As mentioned before the vector 119887+ is such that isfeasible for the subgraphs So because of the conditions for119896 isin 1 1198982 there exists a 119901119894(119896)0 isin [119901119898119894119899
119894(119896)0 119901119898119886119909119894(119896)0] and a119901119895(119896) isin [119901119898119894119899
119895(119896) 119901119898119886119909119895(119896) ] so that1199012119894(119896)0 minus 119892119894(119896)119891(119890119896)
(119894(119896)) ge 1199012119895(119896)0 (86)
With 1199012119894(119896)0 minus 119892119894(119896)119891(119890119896)
(119894(119896)) = 1199012119891(119890119896)
and 1199012119895(119896)0 = 1199012
ℎ(119890119896)it
follows
(119901ℎ(119890119896)119901119891(119890119896)
) ge 1 (87)
Therefore we can find a 119906 isin R1198982ge1 that the inequalities (28)-
(35) are fulfilled Now we can use theWeierstraszlig theorem andget the existence of an optimal solution
Note that the sufficient condition in Lemma 8 is com-paratively strong there may exist optimal solutions with lessstrong conditions eg the following
Corollary 9 If the graph has only one compressor edge ie|E119862| = 1 then it holds If 119906 isin Rge1 is an optimal solution of(76) 119906 is unique
This conclusion follows directly from the strict mono-tonicity of the objective function
Optimization with Random Loads In this section we con-sider the load vector to be random as we mentioned inSection 22 In the optimization problems we want to makesure that the feasibility of a random load vector is guaranteedfor a probability 120572 isin (0 1) This leads us to optimizationproblems with so called chance or probabilistic constraints
In general problems with probabilistic constraints have theform
min 119891 (119909 120585)subject to P (ℎ (119909 120585) le 0) ge 120572 (88)
with an objective function 119891 and a vector of uncertainty 120585There exist many works about probabilistic constraints respstochastic programming eg [7] gives an excellent overviewabout chance constraints in theory and application In ourcase we consider the following problem
min119901+119898119886119909isin119875
119888119879119901+119898119886119909
subject to P (119887 (119901+119898119886119909) isin 119872) ge 120572 (89)
To handle this problem resp the probabilistic constraint thespheric-radial decomposition gives us an explicit representa-tion of the constraint for a Gaussian vector 119887 sim N(120583 Σ) withmean value 120583 and positive definite covariance Σ
P (119887 (119901+119898119886119909) isin 119872)= int
S119899minus1120583120594 119903 ge 0 | 119903119871V + 120583 isin 119872119889120583120578 (V) (90)
Further Algorithm 3 gives a way to approximate this integralin an efficient way In Section 3 we showed a way to character-ize the feasible set119872 and how to adapt this characterization tothe spheric-radial decomposition This changes our problemwith probabilistic constraints (89) to
min119901+119898119886119909isin119875
119888119879119901+119898119886119909
subject to 1119873 119873sum119894=1
119904119905sum119905=1
F120594 (119886119894119905) minusF120594 (119886119894119905) ge 120572 (91)
Of course the interval bounds depend on the pressure boundsand the control For this problem we formulate the nextlemma
Lemma 10 Let a sampling V1 V119873 (from step (1) ofAlgorithm 3) and 120572 isin (0 1) be given For a suitable 120583 isin R119899
problem (91) has at least one solution 119901+119898119886119909 isin 119875Proof For the proof we fix a V isin V1 V119873 from the givensampling From (25) and (26) it follows for all 119896 = 1 119899119894with 119894 isin 1 1198982 + 10 le 119892119894119896 (119894 (119903)) + (119901119898119886119909
119894119896 )2 minus (1199011198981198941198991198940 )2
0 le minus119892119894119896 (119894 (119903)) + (1199011198981198861199091198940 )2 minus (119901119898119894119899
119894119896 )2 (92)
With a random vector 119887 sim N(120583 Σ)119892119894119896(119894(119903)) are quadraticfunctions in 119903 and the upper pressure bounds only influencethe constant part with a positive sign
0 le 1198861198941198961199032 + 119887119894119896119903 + 119888119894119896 + (119901119898119886119909119894119896 )2 minus (119901119898119894119899
1198940 )2 0 le minus1198861198941198961199032 minus 119887119894119896119903 minus 119888119894119896 + (119901119898119886119909
1198940 )2 minus (119901119898119894119899119894119896 )2 (93)
12 Mathematical Problems in Engineering
It follows that a rise in 119901119898119886119909119894119896 for all 119896 = 0 119899119894 enlarges
the intervals in which the inequalities hold The same holdsfor all inequalities inTheorem 5The cumulative distributionfunction of theChi-Square-Distribution is strictlymonotonicincreasing and convergent to 1 Thus upper pressure bounds119901119898119886119909119894119896 (for 119894 = 1 119899) can be found so that
1119873 119873sum119895=1
119904119895sum119905=1
F120594 (119886119895119905) minusF120594 (119886119895119905) (94)
is higher that 120572 isin (0 1) The index119873 is for all sampled pointson the sphere and the index 119904119895 is for the union of intervals inwhich all inequalities hold The interval boundaries 119886119895119905 and119886119895119905 depend continuously on the pressure bounds and becausethe inequalities are not strict 119901119898119886119909 can be found st the sideconstraint is fulfilled with equality So there exists at least onesolution of (91)
5 Numerical Results
In this section we show a few results of implementation Atfirst we show the idea of the spheric-radial decomposition byusing an easy example Next we show a easy example with onecompressor edge and at last we use real data of the Greek gasnetworkThe focus of the implementation is on the theoremsproofed in Section 4
Example 1 Our first example is an easy graph without innercontrol (see Figure 6)
Assume that node 1 and 2 are gas consumers with meandemand 120583 = [05 05]119879 The pressure bounds are given by119901+119898119894119899 = [2 1 1]119879 and 119901+119898119886119909 = [3 2 2]119879 and Φ = E2 Thenthe incidence matrix is
119860+ = [[[minus1 01 minus10 1
]]] resp 119860 = [1 minus10 1 ] (95)
Consider the inequalities of Theorem 5 to characterize thefeasible set Because the network has no compressor edgesthe system of inequalities only contains the information (25)-(27) the other inequalities do not occur The function 119892 isgiven by
119892 (119887) = [1 01 1] [1 00 1] sdot1003816100381610038161003816100381610038161003816100381610038161003816[1 10 1](11988711198872)
1003816100381610038161003816100381610038161003816100381610038161003816∘ ([1 10 1](11988711198872)) = ( 11988721 + 211988711198872 + 1198872211988721 + 211988711198872 + 211988722) (96)
By inserting the pressure bounds one can see that inequality(25) is always fulfilled and inequality (27) implies1198872 le radic3 (97)
With this inequality (26) follows for 0 le 1198872 le radic30 le 1198871 le minus1198872 + radic8 minus 11988722 (98)
So the feasible set119872 is the following (see Figure 7)
119872 = 119887 isin R2ge0 | 1198872 le radic3 and 1198871 le minus1198872 + radic8 minus 11988722 (99)
Now consider the optimization problem
min119901+119898119886119909isin119875
11198793119901+119898119886119909
subject to 119887 isin 119872 (119901+119898119886119909) (100)
with 119887 = 120583 = [05 05]119879 Lemma 7 guarantees the existenceof a unique solution of this problem We use the fmincon-function inMATLAB which minimizes a constrained non-linear multivariable function The fmincon default settinguses a interior-point method For more information we referto the official MathWorks homepage (httpswwwmath-workscomhelpoptimugfminconhtml)The solver returnsthe following result
119901+119898119886119909119900119901119905 = [2 radic3radic275]119879 (101)
One can easily see that this is the correct solution The gasdemand at node 2 is 05 which is equal to the flow 1199022 sousing the equation of momentum implies a difference in thequadratic pressures between node 1 and node 2 of 025 Theflow through edge 1 is given by the two exit nodes and it is1199021 = 1 This implies a difference in the quadratic pressuresbetweennode 0 andnode 1 of 1 So the optimal solution119901+119898119886119909
119900119901119905
directly followsNext we use the spheric-radial decomposition especially
Algorithm 3 to calculate the probability P(119887 isin 119872) for arandom vector 119887 sim N(E2 [05 05]119879) Therefore we viewthe results of 8 tests with 1000 sampled points in each one(Table 1)
The probability in all eight tests is nearly the same Themean value is 03479 and the variance is 27723 sdot 10minus6 whichis very small This shows that the implementation is workingnearly exact The efficiency of the implementation of courseis not perfect but that is not part of this work
Last wewant to solve the following optimization problem
min119901+119898119886119909isin119875
11198793119901+119898119886119909
subject to 1119873 119873sum119894=1
119904sum119905=1
F120594 (119886119894119905) minusF120594 (119886119894119905) ge 120572 (102)
We use again the fmincon-function inMATLAB to solve theproblem eight times with 1000 sampled points in every caseBecause Lemma 8 needs a suitable 120583 isin R2
ge0 we assume 119887 simN(01sdot1198642 [15 15])We set120572 fl 09 and the results are shownin Table 2
Example 2 Now we add a compressor edge to the graph ofExample 1 (see Figure 8)
Assume that node 1 and 3 are consumers node 2 is ainner nodeThemeandemand120583 = [05 0 05]119879Thepressure
Mathematical Problems in Engineering 13
minus2 minus1 1 2 3
minus2
minus1
1
2
3
b2
b1
minus1 1 2
minus1
1
2
b2
b1
rLv1
rLv2
rLv3
Scheme of the spheric-radial decomposition with
sphere (yellow)and three points 1 2 3 (red) at the unit
the feasible set Mminus for Σ = E2 = [05 05]TFeasible set M for p+min = [2 1 1]T and
p+max = [3 2 2]T
Figure 7 Feasible set of Example 1
Table 1 Test results of easy example
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 8P(119887 isin 119872) 03486 03481 03450 03501 03484 03499 03459 03475
Table 2 Test results of Example 1
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 81199011198981198861199091199001199011199050 54514 54714 54758 54028 54767 53921 54888 551671199011198981198861199091199001199011199051 29802 30126 30225 29725 30300 29485 30336 300291199011198981198861199091199001199011199052 10084 10004 10776 10001 11221 11390 10285 10558
0 1 2 3q1 q2 q3
b0 b1 b2 b3
e1 isin ℰF e2 isin ℰC e3 isin ℰF
Figure 8 Graph of Example 2
0 1 2 3q1 q2 q2 q3
b0 b1 b2 b3
e1 isin ℰF e3 isin ℰF
Figure 9 Separated graph of Example 2
bounds are given by 119901+119898119894119899 = [2 1 1 1]119879 and 119901+119898119886119909 = [3 22 2]119879 and 1206011198901= 1 and 1206011198903
= 1 The incidence matrix is givenby
119860+ = [[[[[[
minus1 0 01 minus1 00 1 minus10 0 1]]]]]]
resp 119860 = [[[1 minus1 00 1 minus10 0 1
]]] (103)
Before we have a look at the inequalities of Theorem 5we have to remove the compressor edge from the graphThe flows can be computed by using the equation of massconservation and the new load vector is then
= (1198871 + 11990221198872 minus 11990221198873 ) with
119902 = (1198871 + 1198872 + 11988731198872 + 11988731198873)(104)
We assume the control to be switched off so it holds119906 = 1With this the control edge 1198902 is treated as a flow edge withoutfriction The resulting graph is shown in Figure 9 from theinequalities (25)ndash(35) it follows
3 le radic3 and 1 le radic8 minus 23 (105)
14 Mathematical Problems in Engineering
minus1 1 2 3
1
2
3
minus1 1 2 3
1
2
3
b3
b1
Hyperspace of the feasible set MHyperspace of the feasible set M
b1
b3
Figure 10 Feasible set of the Example 2
Table 3 Test results of control example
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 8P(119887 isin 119872) 01531 01500 01540 01536 01530 01550 01567 01528so the feasible set is (see Figure 10)
= + isin R119899+1 | 1119879
4 + = 0 and 0 le 3 le radic3 and 2 = 0and 0 le 1 le radic8 minus 22 (106)
This implies
119872 = 119887 isin R3ge0 | 0 le 1198873 le radic3 and 1198872 = 0 and 0 le 1198871 le minus1198873
+ radic8 minus 11988723 (107)
Solving the problem
min119901+119898119886119909isin119875
11198794119901+119898119886119909
subject to isin (108)
for 119887 = 120583 = [05 0 05]119879 leads us to the optimal solution
119901+119898119886119909119900119901119905 = [2 radic3radic3radic275]119879 (109)
The argumentation for the optimality of this vector is exactlylike before in the easy example Becausewe have a compressoredge in this graph we can minimize the control Thereforewe fix the pressure bounds as we did at the beginning of thisexample and we set 119887 = [1 0 15]119879 So inequality (31) is notfulfilled anymore for 119906 = 1 We use again the fmincon-solverinMATLAB to solve the following problem
min119906ge1
1199062subject to 119887 isin 119872(119906) (110)
The solver returns the optimal solution 119906119900119901119905 = 11818
Here one can see that the sufficient condition in Lemma 8is really strong Itrsquos not fulfilled with the given values but thereexists a solution anyway Of course the necessary conditionis fulfilled
Now we use again Algorithm 3 to compute the proba-bility P(119887 isin 119872) for a random load vector 119887 sim N(E3 [0505 05]119879) Table 3 shows us the results
Again we can see that the probability is nearly the samein all eight cases The mean probability is 01535 and thevariance is 32369 sdot 10minus6 For the optimization we reducethe covariance matrix and assume 119887 sim N(001 sdot E3 [05 0505]119879) For a probability 120572 = 09 Table 4 states the resultsExample 3 In this example we use a part of a real networkThe focus here is on realistic values for our model Weget these values from the GasLib (see [22]) a collection oftechnical gas network data
ldquoThegoal ofGasLib is to promote research on gasnetworks by providing a set of large and realisticbenchmark instancesrdquo (httpgaslibzibde)
We use a part of the Greek gas network (GasLib-134) (seeFigure 11)
For using real values we first need to know the constant120601119890 from (5)
120601119890 = (119877119878119879)2 120582119863 1198711198862 (111)
The value 119877119878 is the specific gas constant which can becalculated by 119877119878 = 119877119872 with the ideal gas constant 119877 =831451119896119892 sdot 11989821199042 sdot 119898119900119897 sdot 119870 (see [23]) and the molar mass119872 = 1662119896119892119896119898119900119897 (from real data) of the used natural gasAlso the temperature 119879 = 28915119870 is given from real dataFor the (constant) friction factor 120582 we use the law of Chen(see [24])
Mathematical Problems in Engineering 15
Table 4 Test results of control example
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 81199011198981198861199091199001199011199050 39013 39223 39039 38901 38879 39039 38901 390981199011198981198861199091199001199011199051 16568 16565 16365 16497 16496 16365 16497 164921199011198981198861199091199001199011199052 16568 16565 16365 16497 16496 16365 16497 164981199011198981198861199091199001199011199053 10000 10000 10000 10000 10002 10000 10000 10000
Table 5 Values based on real data
variable 119877119878 119879 120582 119863 119871 119886 120601value 50027 28915 00093 9144 178693 43862 198 sdot 107unit 11989821199042 sdot 119870 119870 minus 119898119898 119896119898 119898119904 11989821199042
Figure 11 Greek gas network from GasLib [22] red edges are usedin our example
1radic120582 = minus2 log10 ( 11989611986337065minus 50425
Relog10 [(119896119863)1109828257 + 58506
Re08981]) (112)
Diameter 119863 = 9144119898119898 and roughness value 119896 = 8 sdot 10minus6119898for the first pipe (see Figure 11) are known from real data TheReynolds-Number Re can be calculated by Re = (V119863120588)120578with the dynamic viscosity 120578 = 1 sdot 10minus5119875119886 sdot 119904 (see [1])the mean velocity V = 9548119898119904 (calculated with the meanvolume flow rate which is given by the real data) of the gasand the density 120588 = 074331198961198921198983 (from real data) So we getRe = 648959074 asymp 649 sdot 106 and with this the pipe frictioncoefficient after Chen is 120582 = 00093
The length of the first pipe is given by 119871 = 178693119896119898from the real data Last value missing is the sound speed innatural gas which we assume to be ideal The sound speed
119886 depends on the modulus of compressibility (see [9]) Forideal gases the modulus of compressibility only depends onthe adiabatic exponent and the pressure Using the ideal gasequation simplifies the equation for the sound speed to
119886 = radic120581119877119879119872 (113)
The adiabatic exponent for methan which is the main partof natural gas (gt 90) is about 133 (see [25]) Together allvalues are shown in Table 5 (example for the pipe next to thesource in Figure 11)
Nowwewant to optimize the inner control in the networkfor a fix load vector Therefor we choose two nominations2011-11-01 and 2016-02-17 In both cases the fmincon-functioninMATLAB returns
119906119900119901119905 = 1 (114)
as a local minimum which satisfies the constraint 119887 isin 119872This means for these nominations a compressor stationis not needed so it is switched off With this as a lastcalculation in this paper we calculate the probability for thegiven nominations to be feasible We use the values from thenominations as mean value 120583 and set 119887 sim N(120583 sdot 10minus3E 120583)The results are shown in Tables 6 and 7
Again one can see that the results are quite similarThe mean probability in Table 6 is 09860 and the meanprobability in Table 7 is 09760 Both values are nearly 1 sothere must happen something really unforeseeable that aGreek does not get his demanded gas
6 Conclusion
In this paper we have used an existing stationary modelfor gas transport based on the isothermal Euler equationsWe have extended this model by compressor stations andpresented a characterization of the set of feasible loadsEspecially Theorem 5 contains the central idea to get anaccess to the optimization in Section 4 and the spheric-radialdecomposition is our main tool to handle the uncertainty inthe loadsWewere able to show some results for optimizationwith probabilistic constraints in gas transport In a next stepthese results should be generalized
16 Mathematical Problems in Engineering
Table 6 Test results for nomination 2011-11-01
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 8P(119887 isin 119872) 09849 09849 09878 09885 09863 09845 09854 09853
Table 7 Test results for nomination 2016-02-17
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 8P(119887 isin 119872) 09744 09756 09788 09788 09754 09775 09734 09739
With the characterization of the feasible set we haveshown that the feasibility of a load depends significantly onthe pressure bounds These pressure bounds play a mainrole in our optimization problems The other main partsare the inner controls modeled by compressor edges whichshould lead in optimal control problems with probabilisticconstraints Here we have built a base for further works withthis topic
The idea of separating the graph and removing thecompressor edges can be used to extend themodel with othercomponents like eg control valves and resistorsThiswouldchange the inequalities inTheorem 5 but the approach worksalso for thesemodifications Further [8] showed a way to dealwith a networkwhich is a cycle Combining bothmethods weget an access to model gas networks in high detail
Finally the expansion to a dynamic model might bealso possible The idea is to use a space-mapping approachbecause one can use a lot of the analysis we did hereTherefore upcoming papers should deal with these problemsand present results of the turnpike-theory which describesa relation between stationary and instationary solutions Infact turnpike-theory implies that the optimal dynamic stateis close to the optimal static state in the interior of the timeintervals for sufficiently long time horizons AJ Zaslavskiexplains the turnpike-phenomenon in [26] This book alsogives an overview about the theory
Data Availability
The data used to support the findings of this study areavailable from the corresponding author upon request
Conflicts of Interest
The authors declare that there are no conflicts of interestregarding the publication of this paper
Acknowledgments
This work was supported by DFG in the framework of theCollaborative Research Centre CRCTransregio 154 Math-ematical Modeling Simulation and Optimization Using theExample of Gas Networks project C03
References
[1] P Domschke B Hiller J Lang and C Tischendorf ldquoModel-lierung von gasnetzwerken Eine Ubersichtrdquo Tech Rep 2717Technische Universitat Darmstadt 2017
[2] T Koch B Hiller M E Pfetsch and L Schewe Evaluating GasNetwork Capacities MOS-SIAM 2015
[3] M Gugat R Schultz and D Wintergerst ldquoNetworks of pipe-lines for gas with nonconstant compressibility factor stationarystatesrdquoComputational amp AppliedMathematics vol 37 no 2 pp1066ndash1097 2018
[4] M Gugat and M Herty ldquoExistence of classical solutions andfeedback stabilization for the flow in gas networksrdquo ESAIMControl Optimisation and Calculus of Variations vol 17 no 1pp 28ndash51 2011
[5] M K Banda M Herty and A Klar ldquoGas flow in pipeline net-worksrdquo Networks and Heterogeneous Media vol 1 no 1 pp 41ndash56 2006
[6] RM Colombo G GuerraM Herty and V Schleper ldquoOptimalcontrol in networks of pipes and canalsrdquo SIAM Journal onControl and Optimization vol 48 no 3 pp 2032ndash2050 2009
[7] A Shapiro D Dentcheva and A Ruszczynski ectures onStochastic Programming Modeling and Theory MPS-SIAM2009
[8] C Gotzes H Heitsch R Henrion and R Schultz ldquoOn thequantification of nomination feasibility in stationary gas net-works with random loadrdquoMathematical Methods of OperationsResearch vol 84 no 2 pp 427ndash457 2016
[9] D Halliday R Resnick and J Walker Physik Wiley-VCHBachelor edition edition 2007
[10] M Gugat F M Hante M Hirsch-Dick and G LeugeringldquoStationary states in gas networksrdquoNetworks andHeterogeneousMedia vol 10 no 2 pp 295ndash320 2015
[11] D RoseM SchmidtM C Steinbach and BMWillert ldquoCom-putational optimization of gas compressor stations MINLPmodels versus continuous reformulationsrdquoMathematical Meth-ods of Operations Research vol 83 no 3 pp 409ndash444 2016
[12] H Heitsch R Henrion H Leovey et al ldquoEmpirical observa-tions and statistical analysis of gas demandrdquo in Evaluating GasNetwork Capacities T data B KochM E Hiller et al Eds pp273ndash290 MOS-SIAM 2015
[13] A Genz and F Bretz Computation of Multivariate Normal andt Probabilities vol 195 of Lecture Notes in Statistics Springer2009
[14] M H Farshbaf-Shaker R Henrion and D Homberg ldquoProp-erties of Chance Constraints in Infinite Dimensions with anApplication to PDE Constrained Optimizationrdquo Set-Valued andVariational Analysis pp 1ndash21 2017
[15] T Gonzalez Grandon H Heitsch and R Henrion ldquoA jointmodel of probabilisticrobust constraints for gas transportman-agement in stationary networksrdquo Computational ManagementScience vol 14 no 3 pp 443ndash460 2017
[16] W van Ackooij I Aleksovska and M Munoz-Zuniga ldquo(Sub-)Differentiability of Probability Functionswith Elliptical Distri-butionsrdquo Set-Valued and Variational Analysis
Mathematical Problems in Engineering 17
[17] W van Ackooij and R Henrion ldquoGradient formulae for nonlin-ear probabilistic constraints with GAUssian and GAUssian-likedistributionsrdquo SIAM Journal on Optimization vol 24 no 4 pp1864ndash1889 2014
[18] W van Ackooij and R Henrion ldquo(Sub-)gradient formulaefor probability functions of random inequality systems underGaussian distributionrdquo SIAMASA Journal on UncertaintyQuantification vol 5 no 1 pp 63ndash87 2017
[19] W van Ackooij and J Malick ldquoSecond-order differentiabilityof probability functionsrdquo Optimization Letters vol 11 no 1 pp179ndash194 2017
[20] H Eckey R Kosfeld and M Turck Wahrscheinlichkeitsrech-nung und Induktive Statistik - Grundlagen -Methoden - BeispieleGabler Wiesbaden Germany 2005
[21] A Prekopa Stochastic Programming Springer 1st edition 1995[22] I Joormann M Pfetsch R Schwarz et al ldquoGasLibmdashA Library
of Gas Network Instancesrdquo 2017[23] G Wedler Lehrbuch der Physikalischen Chemie Wiley-VCH
5th edition 2004[24] N H Chen ldquoAn explicit equation for friction factor in piperdquo
Industrial amp Engineering Chemistry Fundamentals vol 18 no3 pp 296-297 1979
[25] W Schenk and F Kremer Physikalisches Praktikum Springer14th edition 2014
[26] A J Zaslavski ldquoTurnpike Properties in the Calculus of Varia-tions and Optimal Controlrdquo inNonconvex Optimization and ItsApplications vol 80 Springer 2006
Hindawiwwwhindawicom Volume 2018
MathematicsJournal of
Hindawiwwwhindawicom Volume 2018
Mathematical Problems in Engineering
Applied MathematicsJournal of
Hindawiwwwhindawicom Volume 2018
Probability and StatisticsHindawiwwwhindawicom Volume 2018
Journal of
Hindawiwwwhindawicom Volume 2018
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawiwwwhindawicom Volume 2018
OptimizationJournal of
Hindawiwwwhindawicom Volume 2018
Hindawiwwwhindawicom Volume 2018
Engineering Mathematics
International Journal of
Hindawiwwwhindawicom Volume 2018
Operations ResearchAdvances in
Journal of
Hindawiwwwhindawicom Volume 2018
Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018
International Journal of Mathematics and Mathematical Sciences
Hindawiwwwhindawicom Volume 2018
Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom
The Scientific World Journal
Volume 2018
Hindawiwwwhindawicom Volume 2018Volume 2018
Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in
Nature and SocietyHindawiwwwhindawicom Volume 2018
Hindawiwwwhindawicom
Dierential EquationsInternational Journal of
Volume 2018
Hindawiwwwhindawicom Volume 2018
Decision SciencesAdvances in
Hindawiwwwhindawicom Volume 2018
AnalysisInternational Journal of
Hindawiwwwhindawicom Volume 2018
Stochastic AnalysisInternational Journal of
Submit your manuscripts atwwwhindawicom
Mathematical Problems in Engineering 5
p0
p1
p2
p3
p4
p5
p6
p7
p8
p9
q1
q2
q4
q6
q7
q8
q9
b0
b1
b2
b3
b4
b5
b6
b7
b8
b9
p0
p1
p2
p3
p4
p5
p6
p7
p8
p9
q1
q2
q3 q3
q4
q5 q5
q6
q7
q8
q9
b0
b1
b2
b3
b4
b5
b6
b7
b8
b9q5 isin ℰC
q3 isin ℰC
Figure 4 Network graph with two compressor edges (left) and new not connected network graph with three connected subgraphs afterremoving the compressor edges (right)
We set |E119865| fl 1198981 and |E119862| fl 1198982 with 1198981 + 1198982 = 119898so 119866 consists of 1198982 + 1 subgraphs which are all connectedNote that 119902 is a vector in1198771198981 nowWe determine that119866119894 is the
119894-th self-connected subgraph of119866 (depends on numbering ofthe graph) Analogously 119901+
119894119895 is the 119895-th component of the 119894-thsubgraph The terms 119902119894119895 and 119860 119894119895119896 are defined similarly Thefeasible set for 119866 is then
=+ isin R119899+1 |
1119879119899+1+ = 0 andexist (119901+ 119902) isin R119899+1
ge0 times R119898 with 119901+ isin [119901+119898119894119899 119901+119898119886119909] 119860 119894119902119894 = 119894 forall119894 = 1 1198982 + 1(119860+119894 )119879 (119901+
119894 )2 = minusΦ11989410038161003816100381610038161199021198941003816100381610038161003816 ∘ 119902119894 forall119894 = 1 1198982 + 1
(119901ℎ(119890)119901119891(119890)
)2 = 119906119890 forall119890 isin E119862
(19)
The last equation in is the same as in (8) that modelsthe pressure rise caused by the compressor With this we canpresent the equation for pressure loss in the form of (16) forevery subgraph Before we state a theorem for characterizingthe feasible set we show the equivalence of and
Lemma4 Let + be as defined in (18)The feasible sets definedin (8) and (19) are equivalent in the following sense
(i) For all 119887+ isin there is + isin (ii) For all + isin there is 119887+ isin (20)
Proof (i) Consider a vector 119887+ isin It holds 1119879119899+1119887+ = 0 and
because of (18) it follows directly 1119879119899+1+ = 0 After removing
the compressor edges from119860+119902 = 119887+ it follows with (17) that119860+119894 119902119894 = +119894 which is equivalent to 119860 119894119902119894 = 119894 The pressure loss
is fulfilled for all flux edges in and because the flow vector119902 doesnrsquot change in the relevant coefficients while removingthe compressor edges the pressure loss is fulfilled in tooThe compressor property is independent of 119887+ and 119902 so it alsoholds From this it follows + isin 119872
(ii) The other way around is quite similar We consider avector + isin From 1119879
119899+1+ = 0 it follows with (18) that1119879119899+1119887+ = 0 The equation for mass conservation is fulfilled
in every subgraph So with (17) it follows 119860+119902 = 119887+ whichimplies the mass conservation in119872 The pressure loss holds
in all self-connected parts of the graph too and because therelevant coefficients of 119902 doesnrsquot change while adding thecompressor edges the pressure loss is fulfilled in119872 too Lastas said before the compressor property is independent of +and 119902 it is also fulfilled in119872 This completes the proof
Because our network graph is a tree the mass conserva-tion is fulfilledwith a full rank incidencematrix119860of size 119899times119899so 119902 can be computed explicit by the following
119902 = 119860minus1119887 (21)
which makes us able to use Lemma 4With this result we canwrite the equation formomentum
conservation depending on the incidence matrices for theparts of the tree This makes us able to formulate the theoremfor characterizing the feasible set Therefore we will use thefollowing function
119892 R119899 997888rarr R
119909 997891997888rarr (119860119879)minus1Φ 10038161003816100381610038161003816119860minus111990910038161003816100381610038161003816 (119860minus1119909) (22)
which describes the pressure loss between the nodes We alsointroduce a new notation
(i) 119906(119894119895)119896 is the 119896-th control on the path from V1198940 to V1198950(ii) 119892(119894119895)119896 is the function 119892 for the 119896-th subgraph on the
path from V1198940 to V1198950 whereas 119892(119894119895)1 is the function 119892
6 Mathematical Problems in Engineering
of the graph 119866119894 According to this 119892(119894119895)119896ℓ is the ℓ-thcomponent of the function 119892(119894119895)119896
(iii) (119894119895)119896 is the load vector of the 119896-th subgraph in thepath from V1198940 to V1198950 whereas (119894119895)1 is the load vectorof the graph 119866119894 and (119894119895)119896ℓ is the ℓ-th component of(119894119896)119896
Analogously we define 119860 (119894119895)119896 119901(119894119895)119896 and 119902(119894119895)119896 Furthermorewe define the following values
(i) 119896lowast119894119895 fl max119896 isin 1 1198982 + 1 | V1198960 isin Π(V1198940) and V1198960 isinΠ(V1198950) as the largest index of all subgraph the pathsto V1198940 and V1198950 pass through
(ii) 119899lowast119894119895 |119896 isin 1 1198982 + 1 | V1198960 isin ΠV1198940(V1198950)| as thenumber of subgraph which are on the path from V1198940from V1198950 containing 119866119894 and 119866119895
(iii) 119898lowast119894119895 fl |119890 isin E119862 | ℎ(119890) isin ΠV1198940(V1198950)and119891(119890) isin ΠV1198940(V1198950)|
as the number of controls which are between V1198940 andV1198950
With this notation and these values we define a sum for 119894 119895 isin11198982 + 1 If V1198940 isin Π(V1198950) we set sum119894119894 fl 0 and else we set
Σ119896lowast119894 fl119899lowastminus2sum119896=1
1prod119898lowastminus119896ℓ=1 119906(119896lowast119894)ℓsdot 119892(119896lowast119894)119899lowastminus119896119891(119890119906
(119896lowast119894)119899lowastminus119896) ((119896lowast119894)119899lowastminus119896)
+ 119892(119896lowast119894)1119891(119890119906(119896lowast119894)1
) ((119896lowast119894)1) (23)
where we write only 119899lowast and 119898lowast instead of 119899lowast119896lowast119894 and 119898lowast119896lowast119894 for
better reading Last we define a productΠ119894119894 fl 1 andΠ119896lowast119894 fl
119898lowastprod119896=1
119906(119896lowast119894)119896 (24)
From a first point of view the notation seems to be confusingbut using this notation makes sense since we want toguarantee the feasibility of a load vector by comparing thepressure bounds at every node with each other We explainthis in a short example Figure 5 shows a graph after removingthe compressor edges
If we want to compare the pressure bounds of node 11 andnode 12 we need to know how the pressures change on thepath from 2 to 11 resp from 2 to 12 Node 11 is part of thesubgraph 1198665 and node 12 is part of the subgraph 1198666 so itfollows 119896lowast56 = 2 because node 2 is part of subgraph 2 Thenotation can be understood as a numbering along the pathIt is 119906(211)1 = 1199062 the first control on the path from node 2to node 11 and 119906(211)2 = 1199064 the second control on this pathFurther it follows 119899lowast211 = 3 (1198662 1198663 1198665) and119898lowast
211 = 2 (1199062 1199064)Theproduct defined before is the product of all controls alongthis path and the sum is the pressure loss between node 2 andnode 7 Using this notation makes us able to formulate thetheorem for characterizing the feasible set in a readable way
Theorem 5 For given pressure bounds 119901+119898119894119899 119901+119898119886119909 isin R119899+1
and controls 119906119894 isin R (119894 = 1 1198982) the following equivalenceholds
A vector + with 1119879119899+1+ = 0 is feasible ie + isin if and
only if the following inequalities hold For all 119894 = 1 1198982 + 1holds (feasibility inside the subgraphs)
(1199011198981198941198991198940 )2 le min
119896=1119899119894[(119901119898119886119909
119894119896 )2 + 119892119894119896 (119894)] (25)
(1199011198981198861199091198940 )2 ge max
119896=1119899119894[(119901119898119894119899
119894119896 )2 + 119892119894119896 (119894)] (26)
max119896=1119899119894
[(119901119898119894119899119894119896 )2 + 119892119894119896 (119894)]
le min119896=1119899119894
[(119901119898119886119909119894119896 )2 + 119892119894119896 (119894)] (27)
For all 119894 119895 = 1 1198982 + 1 with 119894 lt 119895 holds (feasibility betweenthe subgraphs)1Π119896lowast 119894
(1199011198981198941198991198940 )2 + Σ119896lowast119894 le 1Π119896lowast119895
(1199011198981198861199091198950 )2 + Σ119896lowast119895 (28)
1Π119896lowast 119894
(1199011198981198861199091198940 )2 + Σ119896lowast119894 ge 1Π119896lowast 119895
(1199011198981198941198991198950 )2 + Σ119896lowast119895 (29)
1Π119896lowast 119894
(1199011198981198941198991198940 )2 + Σ119896lowast119894
le 1Π119896lowast119895
min119896=1119899119895
[(119901119898119886119909119895119896 )2 + 119892119895119896 (119895)] + Σ119896lowast119895 (30)
1Π119896lowast 119894
(1199011198981198861199091198940 )2 + Σ119896lowast119895
ge 1Π119896lowast119895
max119896=1119899119895
[(119901119898119894119899119895119896 )2 + 119892119895119896 (119895)] + Σ119896lowast119895 (31)
1Π119896lowast 119894
max119896=1119899119894
[(119901119898119894119899119894119896 )2 + 119892119894119896 (119894)] + Σ119896lowast119894
le 1Π119896lowast119895
(1199011198981198861199091198950 )2 + Σ119896lowast119895 (32)
1Π119896lowast 119894
min119896=1119899119894
[(119901119898119886119909119894119896 )2 + 119892119894119896 (119894)] + Σ119896lowast119894
ge 1Π119896lowast119895
(1199011198981198941198991198950 )2 + Σ119896lowast119895 (33)
1Π119896lowast 119894
max119896=1119899119894
[(119901119898119894119899119894119896 )2 + 119892119894119896 (119894)] + Σ119896lowast119894
le 1Π119896lowast119895
min119896=1119899119895
[(119901119898119886119909119895119896 )2 + 119892119895119896 (119895)] + Σ119896lowast119895 (34)
1Π119896lowast 119894
min119896=1119899119894
[(119901119898119886119909119894119896 )2 + 119892119894119896 (119894)] + Σ119896lowast119894
ge 1Π119896lowast119895
max119896=1119899119895
[(119901119898119894119899119895119896 )2 + 119892119895119896 (119895)] + Σ119896lowast119895 (35)
Mathematical Problems in Engineering 7
0 1 2
3
4
5
6
7
8
9
10
11
12
u1
u2
u3
u4
u5
G1 G2 G3
G4
G5
G6
Figure 5 Example graph for illustration of the used notation
0 1 2q1 q2
b0 b1 b2
Figure 6 Graph of Example 1
Let us have a look at the inequalities Inequalities(25)ndash(27) are well-known from [8] and they describe thefeasibility inside a connected tree The other inequalities(28)ndash(35) guarantee the feasibility between the subgraphsTherefore we compare the pressures of the entries of two parts((28) (29)) of the entry of one part and the exit nodes of theother parts ((30)ndash(33)) and we compare the pressures of allexits ((34) (35)) Because we cannot compare the pressuresof119866119894 and 119866119895 directly (there may be no direct connection) wehave to compare them using the subgraph 119866119896lowast which is thelast part the paths to V1198940 and V1198950 pass Therefore we definedthe sumsΣ119896lowast 119894 andΣ119896lowast119895 which trace back the pressures to119866119896lowast
Remark 6 For an implementation it is wise to define a set119880(119894119895) containing all controls on the path from V1198940 to V1198950 anduse an index running over this set to define the sums and theproducts
Now we want to proof the theorem
Proof of Theorem 5 Because the columns of 119860+ sum up tozero it holdsminus1119879
119899119894119860 119894 = 119860+
1198940 (119894 = 1 1198982 + 1) (36)
ldquo997904rArrrdquo We consider a feasible vector 119887+ isin From theequation of mass conservation of every subgraph it follows
119902119894 = 119860minus1119894 119894 (37)
We take the equation of momentum conservation for the 119894-thsubgraph (119894 isin 1 1198982 +1) and separate 119860+
119894 to119860+1198940 and 119860 119894
(119860+1198940)119879 1199012
1198940 + 119860119879119894 1199012
119894 = minusΦ 10038161003816100381610038161199021198941003816100381610038161003816 119902119894 (38)
Next we insert (36) into (38) and get
minus119860119879119894 1119899119894
11990121198940 + 119860119879
119894 1199012119894 = minusΦ 10038161003816100381610038161199021198941003816100381610038161003816 119902119894 (39)
We multiply this equation from left with (119860119879119894 )minus1 and insert
(37) so it follows
111989911989411990121198940 = 1199012
119894 + (119860119879119894 )minus1Φ 10038161003816100381610038161003816119860minus1
119894 11989410038161003816100381610038161003816 119860minus1119894 119894 (40)
and using the definition of the function 119892 we have
111989911989411990121198940 = 1199012
119894 + 119892119894 (119894) (41)
Note here that 119901119894 is the vector of pressures of the subgraph119866119894
and 119892119894(119894) is the vector-valued function which describes thepressure loss in 119866119894 Equation (41) holds for all subgraphs 119866119894
with 119894 isin 1 1198982 + 1 From this the inequalities (25)ndash(27)follow directly with 119901+ isin [119901+119898119894119899 119901+119898119886119909] Next we use (41)component-by-component For fix 119894 119895 isin 1 1198982 + 1 with119894 lt 119895 it follows
1199012119896lowast0 = 1199012
119896lowast119896 + 119892119896lowast119896 (119896lowast) (119896 = 1 119899119896lowast) (42)
Remember that 119896lowast119894119895 is the largest index of all subgraphs thepaths to V1198940 and V1198950 pass through For better reading we onlywrite 119896lowast instead of 119896lowast119894119895 We use the equation of the compressorproperty for the first compressor on the path from V119896lowast0 to V1198940
1199012ℎ(119890119906(119896lowast119894)1
)1199012119891(119890119906(119896lowast119894)1
)
= 119906(119896lowast119894)1 (43)
and insert it into (42) for 119896 = 119891(119890119906(119896lowast119894)1 )1199012(119896lowast 119894)10 = 1119906(119896lowast119894)1
1199012(119896lowast119894)20
+ 119892(119896lowast 119894)1119891(119890119906(119896lowast119894)1
) ((119896lowast119894)1) (44)
Note that 1199012119896lowast 0 = 1199012
(119896lowast119894)10 119892119896lowast119896(119896lowast) = 119892(119896lowast119894)1119896((119896lowast119894)1) and1199012ℎ(119890119906(119896lowast119894)1
) = 1199012(119896lowast 119894)20 We use the compressor equation for
the second compressor on the path from V119896lowast0 to V1198940 and putit together with (42) into (44)
1199012(119896lowast119894)10 = 1119906(119896lowast119894)1
[ 1119906(119896lowast119894)2
1199012(119896lowast 119894)30
+ 119892(119896lowast119894)2119891(119890119906(119896lowast119894)2
) ((119896lowast119894)2)]+ 119892(119896lowast119894)1119891(119890119906
(119896lowast119894)1) ((119896lowast119894)1)
(45)
8 Mathematical Problems in Engineering
We repeat this procedure for all controls on the path fromV119896lowast0 to V1198940 Although we write 119899lowast and119898lowast instead of 119899lowast119896lowast119894 and119898lowast
119896lowast119894 Thus we get
1199012(119896lowast119894)10 = 1119906(119896lowast119894)1
[ 1119906(119896lowast119894)2
[sdot sdot sdot [ 1119906(119896lowast119894)119898lowast1199012(119896lowast119894)119899lowast0
+ 119892(119896lowast119894)119899lowastminus1119891(119890119906(119896lowast119894)119899lowastminus1
) ((119896lowast119894)119899lowastminus1)] sdot sdot sdot]+ 119892(119896lowast119894)2119891(119890119906
(119896lowast119894)2) ((119896lowast119894)2)]
+ 119892(119896lowast119894)1119891(119890119906(119896lowast119894)1
) ((119896lowast119894)1)
(46)
We take again (42) for the node V119896lowast0 and do the sameprocedure just like before for the path from V119896lowast0 to V1198950 With119899lowast = 119899lowast119896lowast119895 and119898lowast = 119898lowast
119896lowast119895 we get
1199012(119896lowast 119895)10
= 1119906(119896lowast119895)1
[ 1119906(119896lowast119895)2
[sdot sdot sdot [ 1119906(119896lowast119895)119898lowast1199012(119896lowast 119895)119899lowast0
+ 119892(119896lowast119895)119899lowastminus1119891(119890119906(119896lowast119895)119899lowastminus1
) ((119896lowast119895)119899lowastminus1)] sdot sdot sdot]+ 119892(119896lowast119895)2119891(119890119906
(119896lowast119895)2) ((119896lowast119895)2)]
+ 119892(119896lowast119895)1119891(119890119906(119896lowast119895)1
) ((119896lowast119895)1)
(47)
Expanding (46) and (47) and equalizing them (this is possiblebecause 119901(119896lowast 119894)10 = 119901(119896lowast119895)10) leads to1Π119896lowast 119894
11990121198940 + Σ119896lowast119894 = 1Π119896lowast 119895
11990121198950 + Σ119896lowast119895 (48)
with Σ119896lowast119894 resp Σ119896lowast119895 defined before (48) directly implies (28)and (29) Using (48) together with (42) for V1198940 first timefor V1198950 second time and for V1198940 and V1198950 a third time theinequalities (30)ndash(35) follow directly So this part of the proofis complete
ldquolArr997904rdquo For this part we consider a vector 119887+ isin R119899+1
with 1119879119899+1119887+ = 0 which fulfills the inequalities (25)ndash(35) We
define the following sets (119894 = 2 1198982 + 1)11987511 fl [(119901119898119894119899
10 )2 (11990111989811988611990910 )2]
11987521 fl [ max119896=11198991
[(1199011198981198941198991119896 )2 + 1198921119896 (1)]
min119896=11198991
[(1199011198981198861199091119896 )2 + 1198921119896 (1)]]
1198751119894 fl [ 1Π1119894
(1199011198981198941198991198940 )2 + Σ1119894 1Π1198941
(1199011198981198861199091198940 )2 + Σ1119894]
1198752119894 fl [ 1Π1119894
max119896=1119899119894
[(119901119898119894119899119894119896 )2 + 119892119894119896 (119894)] + Σ1119894 1Π1198941
sdot min119896=1119899119894
[(119901119898119886119909119894119896 )2 + 119892119894119896 (119894)] + Σ1119894]
(49)
We set
1198750 fl 1198982+1⋂119896=1
1198751119896 cap 1198982+1⋂119896=1
1198752119896 (50)
and we will define a value 119901210 isin 1198750 later So to guarantee
such a value 119901210 exists we have to show first that 1198750 is not
empty Because 1198750 is a finite intersection of convex intervalswe have to make sure that all intervals are not empty andevery intersection between two intervals is not empty In onedimension this is sufficient that 1198750 is not empty For 119894 =1 1198982 + 11198751119894 = 0 according to requirement1198752119894 = 0 because of (26) (51)
Furthermore its 1198751119894 cap 1198752119894 = 0 because of (25) and (26) Themissing intersections are nonempty too for 119894 119895 isin 1 1198982+1 with 119894 lt 119895 as follows
1198751119894 cap 1198751119895 = 0 because of (28) and (29) 1198751119894 cap 1198752119895 = 0 because of (30) and (31) 1198752119894 cap 1198751119895 = 0 because of (32) and (33) 1198752119894 cap 1198752119895 = 0 because of (34) and (35)
(52)
So the set 1198750 is not empty We define the following values
119901210 isin 119875011990121198940 fl Π1119894 (1199012
10 minus Σ1119894) 119894 = 2 1198982 + 11199012119894119896 fl 1199012
1198940 minus 119892119894119896 (119894) 119894 = 1 1198982 + 1 119896 = 1 119899119894119902119894 fl 119860minus1119894 119894 119894 = 1 1198982 + 1
(53)
Now we show that our choice of pressures and flows isfeasible It is11990110 isin [119901119898119894119899
10 11990111989811988611990910 ] because of1199012
10 isin 11987511 Furtherbecause of 1199012
10 isin 1198751119894 for all 119894 = 2 1198982 + 1 it holds1Π1119894
11990121198940 + Σ1119894 le 1Π1119894
(1199011198981198861199091198940 )2 + Σ1119894 (54)
and 1Π1119894
11990121198940 + Σ1119894 ge 1Π1119894
(1199011198981198941198991198940 )2 + Σ1119894 (55)
Mathematical Problems in Engineering 9
Now we replace 11990121198940 in its definition with (1199012
119894119896 + 119892119894119896(119894)) andbecause of 1199012
1198940 isin 1198752119894 for all 119894 = 1 1198982 + 1 it follows1Π1119894
(1199012119894119896 + 119892119894119896 (119894)) + Σ1119894
le 1Π1119894
min119896=1119899119894
[(119901119898119886119909119894119896 )2 + 119892119894119896 (119894)] + Σ1119894
(56)
and 1Π1119894
(1199012119894119896 + 119892119894119896 (119894)) + Σ1119894
ge 1Π1119894
max119896=1119899119894
[(119901119898119894119899119894119896 )2 + 119892119894119896 (119894)] + Σ1119894 (57)
So for all 119894 = 1 1198982 + 1 and for all 119896 = 1 119899119894 it holds119901119894119896 isin [119901119898119894119899119894119896 119901119898119886119909
119894119896 ] which is equivalent to
119901+ isin [119901+119898119894119899 119901+119898119886119909] (58)
The equation for mass conservation follows directly from thedefinition of 119902119894 (119894 = 1 1198982 + 1) For the pressure loss weuse
1199012119894 = 1119899119894
11990121198940 minus 119892119894 (119894) (59)
And multiply this equation from left with 119860119879119894 and use (36)
With this it follows the equation for momentum conserva-tion
(119860+119894 )119879 (119901+
119894 )2 = minusΦ 10038161003816100381610038161199021198941003816100381610038161003816 119902119894 (60)
The last equation missing is the compressor property For acontrol 119895 isin 1 1198982 we fix a vertex V1198940 (119894 isin 1 1198982 +1)with ℎ(119890119906119895) = V1198940 We use 1199012
1198940 = Π1119894(119901210 minus Σ1119894) and solve the
formula for 119901210 in the following way
119901210 = 1119906(1119894)1
[sdot sdot [ 1119906(1119894)119898lowast1199012(1119894)119899lowast0
+ 119892(1119894)119899lowastminus1119891(119890119906(1119894)119899lowastminus1
) ((1119894)119899lowastminus1)] sdot sdot]+ 119892(1119894)1119891(119890119906(1119894)1 )
((1119894)1) (61)
Consider wlog that the subgraphs along the path from V10to V1198940 are numbered chronological st if ℎ(119890119906119895) isin 119866119894 then119891(119890119906119895) isin 119866119894minus1 We use 1199012
119894minus10 = Π1119894minus1(119901210 minus Σ1119894minus1) to get the
same representation for 119901210 depending on 1199012
(1119894minus1)119899lowast0 whichis equivalent to 119901(1119894)119899lowastminus10 We equalize these representationsdepending on 1199012
1198940 and 1199012119894minus10 and we get
1119906(1119894)119898lowast1199012(1119894)119899lowast0 + 119892(1119894)119899lowastminus1119891(119890119906
(1119894)119899lowastminus1) ((1119894)119899lowastminus1)
= 1199012(1119894)119899lowastminus10 (62)
We chose 119894 isin 1 1198982 + 1 st 119906119895 is the last control on thepath from V10 to V1198940 so it is 119906(1119894)119898lowast = 119906119895 Further we have
1199012(1119894)119899lowast 0 = 1199012
11989401199012(1119894)119899lowastminus10 = 1199012
119894minus10119892(1119894)119899lowastminus1119891(119890119906(1119894)119899lowastminus1
) (119887(1119894)119899lowastminus1) = 119892119894minus1119891(119890119906119895 )(119887119894minus1)
(63)
and we know that 11990121198940 resp 1199012
119894minus10 minus 119892119894minus1119891(119890119906119895 )(119887119894minus1) are the
pressures at the head resp the foot of the compressor edge119890119906119895 So it follows 1119906119895
1199012ℎ(119890119906119895 )
= 1199012119891(119890119906119895 )
(64)
which is equivalent to the compressor property This com-pletes the proof
With this theorem we have another characterization ofthe feasible set The last step in this intersection is theadaption of the results to the spheric-radial decompositionAs mentioned before the main step is to get a representationof the sets119872119894 in step (2) of Algorithm 3
Because we assumed that the graph has only one entry wedefine the following set
119872 fl 119887 isin R119899ge0 | (minus1119879
119899119887 119887) isin (65)
Consider a sampled point V119894 isin S119899minus1 of step (1) of Algorithm 3(119894 = 1 119873) We identify the load vector 119887 with the affinelinear function
119887119894 (119903) fl 119903119871V119894 + 120583 = 119903120596119894 + 120583 with 120596119894 = 119871V119894 (66)
where 120583 is the mean value of the Gaussian distribution and 119871is such that 119871119871119879 = Σ for positive definite covariance matrixΣ Note that the index 119894 here is for the sampled vector V119894 notfor the 119894-th subgraph Because 119887119894 is defined only for the exitnodes it must hold 119887 ge 0 which leads to the definition of theregular range
119877119903119890119892119894 fl 119903 ge 0 | 119887119894 (119903) ge 0 supe 119872119894 (67)
So the feasible set is
119872119894 = 119903 isin 119877119903119890119892119894 | 119887119894 (119903) fulfills (3) (5) and (7) (68)
and with Theorem 5 it holds
119872119894 = 119903 isin 119877119903119890119892119894 | 119894 (119903) fulfills (25) minus (34) (69)
where 119894(119903) is from (18) Because 119894 depends on 119903 now thefunction 119892 depends on 119903 too More exactly 119892 is quadratic in119903 so 119892119896(119887119894(119903)) can be represented as follows
1199032[[119899sum
119895=1
Ψ119896119895( 119899sumℓ=1
119860minus1119895ℓ120596119894ℓ)2]]⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
119886119894
10 Mathematical Problems in Engineering
+ 119903[[2119899sum
119895=1
Ψ119896119895( 119899sumℓ=1
119860minus1119895ℓ120596119894ℓ)( 119899sum
ℓ=1
119860minus1119895ℓ120583ℓ)]]⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
119894
+ [[119899sum
119895=1
Ψ119896119895( 119899sumℓ=1
119860minus1119895ℓ120583ℓ)2]]⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
119888119894
(70)
HereΨ is the product of119860 andΦ So the inequations whichcharacterize119872119894 are also quadratic in 119903 because the minimaland maximal pressures in the inequalities only count to 119888119894Now we can write the feasible sets as follows
119872119894 = 119903 isin 119877119903119890119892119894 | 0 le 1198861198941199051199032 + 119887119894119905119903 + 119888119894119905 (71)
where 119905 is an index for the total number of inequalities in119872119894Now the regular range can be cut with the positive intervals ofthe inequalities and the result is a union of disjoint intervals
119872119894 = 119904119894⋃119896=1
[119886119894119896 119886119894119896] 119904119894 isin N (72)
Here 119904119894 is the number of disjoint intervals in which allinequalities are fulfilled and 119886119894119896 and 119886119894119896 are the intervalbounds they depend on the pressure bounds and the con-trols So the third step of Algorithm 3 can be executed With
P (119887 isin 119872) asymp 1119873 119873sum119894=1
119904119894sum119896=1
F120594 (119886119894119896) minusF120594 (119886119894119896) (73)
the probability for a random load vector to be feasible can becomputed
4 Existence of Optimal Solutions
In this section we want to have a look at some optimizationproblems We distinguish between problems with constantloads (without uncertainty on the demand) and problemswith random loads The latter leads us to optimizationproblems with so called probabilistic constraints or chanceconstraints (see [21]) There are two aims in optimizationMinimize the maximal pressure bounds and minimize thecontrols For minimizing the maximal pressure bounds wedefine the following set
119875 fl119899⋃
119894=0
[119901+119898119894119899119894 infin) (74)
in which we look for them Because the maximal pressuresare in proportion to the cost of pipes an objective functionfor a optimization problem could be 119888119879119901+119898119886119909 for a cost vector119888 isin R119899+1
gt0 By minimizing the controls we use the Euclideannorm of the controls as an objective function
Optimization with Constant Loads Consider the two opti-mization problems
min119901+119898119886119909isin119875
119888119879119901+119898119886119909
subject to 119887+ (119901+119898119886119909) isin 119872 (75)
andmin119906isinR1198982ge1
1199062subject to 119887+ (119906) isin (76)
Lemma 7 There exists a unique solution for (75)
Proof We have a look at the side constraint 119887+(119901+119898119886119909) isin Using Theorem 5 we can write (75) as
min119901+119898119886119909isin119875
119888119879119901+119898119886119909
subject to (25) minus (35) are fulfilled (77)
For 119894 isin 1 1198982 + 1 we write inequality (25) (and allremaining analogous) in the following way
0 le 119892119894119896 (119894) minus (1199011198981198941198991198940 )2 + (119901119898119886119909
119894119896 )2 forall119896 = 1 119899119894 (78)
So every inequality depends only on one upper pressurebound (but more inequalities may depend on the samebound) Because the inequalities are not strict the values119901119896 fl min 119901119898119886119909
119896 isin [0infin) | (25) minus (35) are fulfilled (79)
exist for all 119896 = 1 119899 + 1 and we can define a value119901+119898119886119909119900119901119905119896 fl max 119901119896 119901119898119894119899
119896 (80)
which is obviously a solution of (75) Moreover this solutionis unique because reducing one component of the upperbounds would either hurt an inequality or it would not fulfill119901+119898119886119909 ge 119901+119898119894119899 anymore
For the optimization problem (76) we can formulate anecessary and a sufficient condition for optimal solutions
Lemma8 Consider pressure bounds119901+119898119894119899 119901+119898119886119909 isin R119899+1 st119901+119898119894119899 lt 119901+119898119886119909 and a suitable vector 119887+ isin R119899+1 (ie + fulfills(25)ndash(27)) For all 119896 = 1 1198982 we set 119894(119896) and 119895(119896) such that119891(119890119896) isin V+
119894(119896) and ℎ(119890119896) isin V+119895(119896)
(a) If 119906 isin R1198982 is a solution of (76) it holds for all 119896 =1 1198982 (119901119898119894119899119894(119896)0)2 minus 119892119894(119896)119891(119890119896)
(119894(119896)) le (119901119898119886119909119895(119896)0)2 (81)
(b) If the inequalities
(119901119898119886119909119894(119896)0)2 minus 119892119894(119896)119891(119890119896)
(119894(119896)) ge (119901119898119894119899119895(119896)0)2
and (119901119898119894119899119894(119896)0)2 minus 119892119894(119896)119891(119890119896)
(119894(119896)) le (119901119898119886119909119895(119896)0)2 (82)
are fulfilled for all 119896 = 1 1198982 an optimal solution 119906 isin R1198982ge1
of (76) exists
Mathematical Problems in Engineering 11
Note that a vector 119887+ isin R119899+1 is suitable here if the vector+ is feasible for the subgraphs ie the inequalities (25)-(27)are fulfilled for +Proof (a) For this part we use a proof by contradictionAssume that
(119901119898119894119899119894(119896)0)2 minus 119892119894(119896)119891(119890119896)
(119894(119896)) gt (119901119898119886119909119895(119896)0)2 (83)
Because 119906119896 ge 1 for all 119896 = 1 1198982 it follows
(119901119898119894119899119894(119896)0)2 minus 119892119894(119896)119891(119890119896)
(119894(119896)) gt 1119906119896
(119901119898119886119909119895(119896)0)2 (84)
It it Σ119896lowast119894(119896) = 0 and Π119896lowast 119894(119896) = 1 because the subgraph withindex 119896lowast is equal to the subgraphwith index 119894(119896) And becausethe subgraphs 119894(119896) and 119895(119896) are neighbours (ie directlyconnected) it follows Π119896lowast 119895(119896) = 119906119896 and
Σ119896lowast119895(119896) = 119892(119896lowast119895(119896))1119891(119890119906(119896lowast119895(119896))1
) (119887(119896lowast119895(119896))1)= 119892119894(119896)119891(119890119896)
(119894(119896)) (85)
Thus we have a contradiction to (28) and 119906119896 cannot beoptimal
(b) As mentioned before the vector 119887+ is such that isfeasible for the subgraphs So because of the conditions for119896 isin 1 1198982 there exists a 119901119894(119896)0 isin [119901119898119894119899
119894(119896)0 119901119898119886119909119894(119896)0] and a119901119895(119896) isin [119901119898119894119899
119895(119896) 119901119898119886119909119895(119896) ] so that1199012119894(119896)0 minus 119892119894(119896)119891(119890119896)
(119894(119896)) ge 1199012119895(119896)0 (86)
With 1199012119894(119896)0 minus 119892119894(119896)119891(119890119896)
(119894(119896)) = 1199012119891(119890119896)
and 1199012119895(119896)0 = 1199012
ℎ(119890119896)it
follows
(119901ℎ(119890119896)119901119891(119890119896)
) ge 1 (87)
Therefore we can find a 119906 isin R1198982ge1 that the inequalities (28)-
(35) are fulfilled Now we can use theWeierstraszlig theorem andget the existence of an optimal solution
Note that the sufficient condition in Lemma 8 is com-paratively strong there may exist optimal solutions with lessstrong conditions eg the following
Corollary 9 If the graph has only one compressor edge ie|E119862| = 1 then it holds If 119906 isin Rge1 is an optimal solution of(76) 119906 is unique
This conclusion follows directly from the strict mono-tonicity of the objective function
Optimization with Random Loads In this section we con-sider the load vector to be random as we mentioned inSection 22 In the optimization problems we want to makesure that the feasibility of a random load vector is guaranteedfor a probability 120572 isin (0 1) This leads us to optimizationproblems with so called chance or probabilistic constraints
In general problems with probabilistic constraints have theform
min 119891 (119909 120585)subject to P (ℎ (119909 120585) le 0) ge 120572 (88)
with an objective function 119891 and a vector of uncertainty 120585There exist many works about probabilistic constraints respstochastic programming eg [7] gives an excellent overviewabout chance constraints in theory and application In ourcase we consider the following problem
min119901+119898119886119909isin119875
119888119879119901+119898119886119909
subject to P (119887 (119901+119898119886119909) isin 119872) ge 120572 (89)
To handle this problem resp the probabilistic constraint thespheric-radial decomposition gives us an explicit representa-tion of the constraint for a Gaussian vector 119887 sim N(120583 Σ) withmean value 120583 and positive definite covariance Σ
P (119887 (119901+119898119886119909) isin 119872)= int
S119899minus1120583120594 119903 ge 0 | 119903119871V + 120583 isin 119872119889120583120578 (V) (90)
Further Algorithm 3 gives a way to approximate this integralin an efficient way In Section 3 we showed a way to character-ize the feasible set119872 and how to adapt this characterization tothe spheric-radial decomposition This changes our problemwith probabilistic constraints (89) to
min119901+119898119886119909isin119875
119888119879119901+119898119886119909
subject to 1119873 119873sum119894=1
119904119905sum119905=1
F120594 (119886119894119905) minusF120594 (119886119894119905) ge 120572 (91)
Of course the interval bounds depend on the pressure boundsand the control For this problem we formulate the nextlemma
Lemma 10 Let a sampling V1 V119873 (from step (1) ofAlgorithm 3) and 120572 isin (0 1) be given For a suitable 120583 isin R119899
problem (91) has at least one solution 119901+119898119886119909 isin 119875Proof For the proof we fix a V isin V1 V119873 from the givensampling From (25) and (26) it follows for all 119896 = 1 119899119894with 119894 isin 1 1198982 + 10 le 119892119894119896 (119894 (119903)) + (119901119898119886119909
119894119896 )2 minus (1199011198981198941198991198940 )2
0 le minus119892119894119896 (119894 (119903)) + (1199011198981198861199091198940 )2 minus (119901119898119894119899
119894119896 )2 (92)
With a random vector 119887 sim N(120583 Σ)119892119894119896(119894(119903)) are quadraticfunctions in 119903 and the upper pressure bounds only influencethe constant part with a positive sign
0 le 1198861198941198961199032 + 119887119894119896119903 + 119888119894119896 + (119901119898119886119909119894119896 )2 minus (119901119898119894119899
1198940 )2 0 le minus1198861198941198961199032 minus 119887119894119896119903 minus 119888119894119896 + (119901119898119886119909
1198940 )2 minus (119901119898119894119899119894119896 )2 (93)
12 Mathematical Problems in Engineering
It follows that a rise in 119901119898119886119909119894119896 for all 119896 = 0 119899119894 enlarges
the intervals in which the inequalities hold The same holdsfor all inequalities inTheorem 5The cumulative distributionfunction of theChi-Square-Distribution is strictlymonotonicincreasing and convergent to 1 Thus upper pressure bounds119901119898119886119909119894119896 (for 119894 = 1 119899) can be found so that
1119873 119873sum119895=1
119904119895sum119905=1
F120594 (119886119895119905) minusF120594 (119886119895119905) (94)
is higher that 120572 isin (0 1) The index119873 is for all sampled pointson the sphere and the index 119904119895 is for the union of intervals inwhich all inequalities hold The interval boundaries 119886119895119905 and119886119895119905 depend continuously on the pressure bounds and becausethe inequalities are not strict 119901119898119886119909 can be found st the sideconstraint is fulfilled with equality So there exists at least onesolution of (91)
5 Numerical Results
In this section we show a few results of implementation Atfirst we show the idea of the spheric-radial decomposition byusing an easy example Next we show a easy example with onecompressor edge and at last we use real data of the Greek gasnetworkThe focus of the implementation is on the theoremsproofed in Section 4
Example 1 Our first example is an easy graph without innercontrol (see Figure 6)
Assume that node 1 and 2 are gas consumers with meandemand 120583 = [05 05]119879 The pressure bounds are given by119901+119898119894119899 = [2 1 1]119879 and 119901+119898119886119909 = [3 2 2]119879 and Φ = E2 Thenthe incidence matrix is
119860+ = [[[minus1 01 minus10 1
]]] resp 119860 = [1 minus10 1 ] (95)
Consider the inequalities of Theorem 5 to characterize thefeasible set Because the network has no compressor edgesthe system of inequalities only contains the information (25)-(27) the other inequalities do not occur The function 119892 isgiven by
119892 (119887) = [1 01 1] [1 00 1] sdot1003816100381610038161003816100381610038161003816100381610038161003816[1 10 1](11988711198872)
1003816100381610038161003816100381610038161003816100381610038161003816∘ ([1 10 1](11988711198872)) = ( 11988721 + 211988711198872 + 1198872211988721 + 211988711198872 + 211988722) (96)
By inserting the pressure bounds one can see that inequality(25) is always fulfilled and inequality (27) implies1198872 le radic3 (97)
With this inequality (26) follows for 0 le 1198872 le radic30 le 1198871 le minus1198872 + radic8 minus 11988722 (98)
So the feasible set119872 is the following (see Figure 7)
119872 = 119887 isin R2ge0 | 1198872 le radic3 and 1198871 le minus1198872 + radic8 minus 11988722 (99)
Now consider the optimization problem
min119901+119898119886119909isin119875
11198793119901+119898119886119909
subject to 119887 isin 119872 (119901+119898119886119909) (100)
with 119887 = 120583 = [05 05]119879 Lemma 7 guarantees the existenceof a unique solution of this problem We use the fmincon-function inMATLAB which minimizes a constrained non-linear multivariable function The fmincon default settinguses a interior-point method For more information we referto the official MathWorks homepage (httpswwwmath-workscomhelpoptimugfminconhtml)The solver returnsthe following result
119901+119898119886119909119900119901119905 = [2 radic3radic275]119879 (101)
One can easily see that this is the correct solution The gasdemand at node 2 is 05 which is equal to the flow 1199022 sousing the equation of momentum implies a difference in thequadratic pressures between node 1 and node 2 of 025 Theflow through edge 1 is given by the two exit nodes and it is1199021 = 1 This implies a difference in the quadratic pressuresbetweennode 0 andnode 1 of 1 So the optimal solution119901+119898119886119909
119900119901119905
directly followsNext we use the spheric-radial decomposition especially
Algorithm 3 to calculate the probability P(119887 isin 119872) for arandom vector 119887 sim N(E2 [05 05]119879) Therefore we viewthe results of 8 tests with 1000 sampled points in each one(Table 1)
The probability in all eight tests is nearly the same Themean value is 03479 and the variance is 27723 sdot 10minus6 whichis very small This shows that the implementation is workingnearly exact The efficiency of the implementation of courseis not perfect but that is not part of this work
Last wewant to solve the following optimization problem
min119901+119898119886119909isin119875
11198793119901+119898119886119909
subject to 1119873 119873sum119894=1
119904sum119905=1
F120594 (119886119894119905) minusF120594 (119886119894119905) ge 120572 (102)
We use again the fmincon-function inMATLAB to solve theproblem eight times with 1000 sampled points in every caseBecause Lemma 8 needs a suitable 120583 isin R2
ge0 we assume 119887 simN(01sdot1198642 [15 15])We set120572 fl 09 and the results are shownin Table 2
Example 2 Now we add a compressor edge to the graph ofExample 1 (see Figure 8)
Assume that node 1 and 3 are consumers node 2 is ainner nodeThemeandemand120583 = [05 0 05]119879Thepressure
Mathematical Problems in Engineering 13
minus2 minus1 1 2 3
minus2
minus1
1
2
3
b2
b1
minus1 1 2
minus1
1
2
b2
b1
rLv1
rLv2
rLv3
Scheme of the spheric-radial decomposition with
sphere (yellow)and three points 1 2 3 (red) at the unit
the feasible set Mminus for Σ = E2 = [05 05]TFeasible set M for p+min = [2 1 1]T and
p+max = [3 2 2]T
Figure 7 Feasible set of Example 1
Table 1 Test results of easy example
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 8P(119887 isin 119872) 03486 03481 03450 03501 03484 03499 03459 03475
Table 2 Test results of Example 1
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 81199011198981198861199091199001199011199050 54514 54714 54758 54028 54767 53921 54888 551671199011198981198861199091199001199011199051 29802 30126 30225 29725 30300 29485 30336 300291199011198981198861199091199001199011199052 10084 10004 10776 10001 11221 11390 10285 10558
0 1 2 3q1 q2 q3
b0 b1 b2 b3
e1 isin ℰF e2 isin ℰC e3 isin ℰF
Figure 8 Graph of Example 2
0 1 2 3q1 q2 q2 q3
b0 b1 b2 b3
e1 isin ℰF e3 isin ℰF
Figure 9 Separated graph of Example 2
bounds are given by 119901+119898119894119899 = [2 1 1 1]119879 and 119901+119898119886119909 = [3 22 2]119879 and 1206011198901= 1 and 1206011198903
= 1 The incidence matrix is givenby
119860+ = [[[[[[
minus1 0 01 minus1 00 1 minus10 0 1]]]]]]
resp 119860 = [[[1 minus1 00 1 minus10 0 1
]]] (103)
Before we have a look at the inequalities of Theorem 5we have to remove the compressor edge from the graphThe flows can be computed by using the equation of massconservation and the new load vector is then
= (1198871 + 11990221198872 minus 11990221198873 ) with
119902 = (1198871 + 1198872 + 11988731198872 + 11988731198873)(104)
We assume the control to be switched off so it holds119906 = 1With this the control edge 1198902 is treated as a flow edge withoutfriction The resulting graph is shown in Figure 9 from theinequalities (25)ndash(35) it follows
3 le radic3 and 1 le radic8 minus 23 (105)
14 Mathematical Problems in Engineering
minus1 1 2 3
1
2
3
minus1 1 2 3
1
2
3
b3
b1
Hyperspace of the feasible set MHyperspace of the feasible set M
b1
b3
Figure 10 Feasible set of the Example 2
Table 3 Test results of control example
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 8P(119887 isin 119872) 01531 01500 01540 01536 01530 01550 01567 01528so the feasible set is (see Figure 10)
= + isin R119899+1 | 1119879
4 + = 0 and 0 le 3 le radic3 and 2 = 0and 0 le 1 le radic8 minus 22 (106)
This implies
119872 = 119887 isin R3ge0 | 0 le 1198873 le radic3 and 1198872 = 0 and 0 le 1198871 le minus1198873
+ radic8 minus 11988723 (107)
Solving the problem
min119901+119898119886119909isin119875
11198794119901+119898119886119909
subject to isin (108)
for 119887 = 120583 = [05 0 05]119879 leads us to the optimal solution
119901+119898119886119909119900119901119905 = [2 radic3radic3radic275]119879 (109)
The argumentation for the optimality of this vector is exactlylike before in the easy example Becausewe have a compressoredge in this graph we can minimize the control Thereforewe fix the pressure bounds as we did at the beginning of thisexample and we set 119887 = [1 0 15]119879 So inequality (31) is notfulfilled anymore for 119906 = 1 We use again the fmincon-solverinMATLAB to solve the following problem
min119906ge1
1199062subject to 119887 isin 119872(119906) (110)
The solver returns the optimal solution 119906119900119901119905 = 11818
Here one can see that the sufficient condition in Lemma 8is really strong Itrsquos not fulfilled with the given values but thereexists a solution anyway Of course the necessary conditionis fulfilled
Now we use again Algorithm 3 to compute the proba-bility P(119887 isin 119872) for a random load vector 119887 sim N(E3 [0505 05]119879) Table 3 shows us the results
Again we can see that the probability is nearly the samein all eight cases The mean probability is 01535 and thevariance is 32369 sdot 10minus6 For the optimization we reducethe covariance matrix and assume 119887 sim N(001 sdot E3 [05 0505]119879) For a probability 120572 = 09 Table 4 states the resultsExample 3 In this example we use a part of a real networkThe focus here is on realistic values for our model Weget these values from the GasLib (see [22]) a collection oftechnical gas network data
ldquoThegoal ofGasLib is to promote research on gasnetworks by providing a set of large and realisticbenchmark instancesrdquo (httpgaslibzibde)
We use a part of the Greek gas network (GasLib-134) (seeFigure 11)
For using real values we first need to know the constant120601119890 from (5)
120601119890 = (119877119878119879)2 120582119863 1198711198862 (111)
The value 119877119878 is the specific gas constant which can becalculated by 119877119878 = 119877119872 with the ideal gas constant 119877 =831451119896119892 sdot 11989821199042 sdot 119898119900119897 sdot 119870 (see [23]) and the molar mass119872 = 1662119896119892119896119898119900119897 (from real data) of the used natural gasAlso the temperature 119879 = 28915119870 is given from real dataFor the (constant) friction factor 120582 we use the law of Chen(see [24])
Mathematical Problems in Engineering 15
Table 4 Test results of control example
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 81199011198981198861199091199001199011199050 39013 39223 39039 38901 38879 39039 38901 390981199011198981198861199091199001199011199051 16568 16565 16365 16497 16496 16365 16497 164921199011198981198861199091199001199011199052 16568 16565 16365 16497 16496 16365 16497 164981199011198981198861199091199001199011199053 10000 10000 10000 10000 10002 10000 10000 10000
Table 5 Values based on real data
variable 119877119878 119879 120582 119863 119871 119886 120601value 50027 28915 00093 9144 178693 43862 198 sdot 107unit 11989821199042 sdot 119870 119870 minus 119898119898 119896119898 119898119904 11989821199042
Figure 11 Greek gas network from GasLib [22] red edges are usedin our example
1radic120582 = minus2 log10 ( 11989611986337065minus 50425
Relog10 [(119896119863)1109828257 + 58506
Re08981]) (112)
Diameter 119863 = 9144119898119898 and roughness value 119896 = 8 sdot 10minus6119898for the first pipe (see Figure 11) are known from real data TheReynolds-Number Re can be calculated by Re = (V119863120588)120578with the dynamic viscosity 120578 = 1 sdot 10minus5119875119886 sdot 119904 (see [1])the mean velocity V = 9548119898119904 (calculated with the meanvolume flow rate which is given by the real data) of the gasand the density 120588 = 074331198961198921198983 (from real data) So we getRe = 648959074 asymp 649 sdot 106 and with this the pipe frictioncoefficient after Chen is 120582 = 00093
The length of the first pipe is given by 119871 = 178693119896119898from the real data Last value missing is the sound speed innatural gas which we assume to be ideal The sound speed
119886 depends on the modulus of compressibility (see [9]) Forideal gases the modulus of compressibility only depends onthe adiabatic exponent and the pressure Using the ideal gasequation simplifies the equation for the sound speed to
119886 = radic120581119877119879119872 (113)
The adiabatic exponent for methan which is the main partof natural gas (gt 90) is about 133 (see [25]) Together allvalues are shown in Table 5 (example for the pipe next to thesource in Figure 11)
Nowwewant to optimize the inner control in the networkfor a fix load vector Therefor we choose two nominations2011-11-01 and 2016-02-17 In both cases the fmincon-functioninMATLAB returns
119906119900119901119905 = 1 (114)
as a local minimum which satisfies the constraint 119887 isin 119872This means for these nominations a compressor stationis not needed so it is switched off With this as a lastcalculation in this paper we calculate the probability for thegiven nominations to be feasible We use the values from thenominations as mean value 120583 and set 119887 sim N(120583 sdot 10minus3E 120583)The results are shown in Tables 6 and 7
Again one can see that the results are quite similarThe mean probability in Table 6 is 09860 and the meanprobability in Table 7 is 09760 Both values are nearly 1 sothere must happen something really unforeseeable that aGreek does not get his demanded gas
6 Conclusion
In this paper we have used an existing stationary modelfor gas transport based on the isothermal Euler equationsWe have extended this model by compressor stations andpresented a characterization of the set of feasible loadsEspecially Theorem 5 contains the central idea to get anaccess to the optimization in Section 4 and the spheric-radialdecomposition is our main tool to handle the uncertainty inthe loadsWewere able to show some results for optimizationwith probabilistic constraints in gas transport In a next stepthese results should be generalized
16 Mathematical Problems in Engineering
Table 6 Test results for nomination 2011-11-01
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 8P(119887 isin 119872) 09849 09849 09878 09885 09863 09845 09854 09853
Table 7 Test results for nomination 2016-02-17
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 8P(119887 isin 119872) 09744 09756 09788 09788 09754 09775 09734 09739
With the characterization of the feasible set we haveshown that the feasibility of a load depends significantly onthe pressure bounds These pressure bounds play a mainrole in our optimization problems The other main partsare the inner controls modeled by compressor edges whichshould lead in optimal control problems with probabilisticconstraints Here we have built a base for further works withthis topic
The idea of separating the graph and removing thecompressor edges can be used to extend themodel with othercomponents like eg control valves and resistorsThiswouldchange the inequalities inTheorem 5 but the approach worksalso for thesemodifications Further [8] showed a way to dealwith a networkwhich is a cycle Combining bothmethods weget an access to model gas networks in high detail
Finally the expansion to a dynamic model might bealso possible The idea is to use a space-mapping approachbecause one can use a lot of the analysis we did hereTherefore upcoming papers should deal with these problemsand present results of the turnpike-theory which describesa relation between stationary and instationary solutions Infact turnpike-theory implies that the optimal dynamic stateis close to the optimal static state in the interior of the timeintervals for sufficiently long time horizons AJ Zaslavskiexplains the turnpike-phenomenon in [26] This book alsogives an overview about the theory
Data Availability
The data used to support the findings of this study areavailable from the corresponding author upon request
Conflicts of Interest
The authors declare that there are no conflicts of interestregarding the publication of this paper
Acknowledgments
This work was supported by DFG in the framework of theCollaborative Research Centre CRCTransregio 154 Math-ematical Modeling Simulation and Optimization Using theExample of Gas Networks project C03
References
[1] P Domschke B Hiller J Lang and C Tischendorf ldquoModel-lierung von gasnetzwerken Eine Ubersichtrdquo Tech Rep 2717Technische Universitat Darmstadt 2017
[2] T Koch B Hiller M E Pfetsch and L Schewe Evaluating GasNetwork Capacities MOS-SIAM 2015
[3] M Gugat R Schultz and D Wintergerst ldquoNetworks of pipe-lines for gas with nonconstant compressibility factor stationarystatesrdquoComputational amp AppliedMathematics vol 37 no 2 pp1066ndash1097 2018
[4] M Gugat and M Herty ldquoExistence of classical solutions andfeedback stabilization for the flow in gas networksrdquo ESAIMControl Optimisation and Calculus of Variations vol 17 no 1pp 28ndash51 2011
[5] M K Banda M Herty and A Klar ldquoGas flow in pipeline net-worksrdquo Networks and Heterogeneous Media vol 1 no 1 pp 41ndash56 2006
[6] RM Colombo G GuerraM Herty and V Schleper ldquoOptimalcontrol in networks of pipes and canalsrdquo SIAM Journal onControl and Optimization vol 48 no 3 pp 2032ndash2050 2009
[7] A Shapiro D Dentcheva and A Ruszczynski ectures onStochastic Programming Modeling and Theory MPS-SIAM2009
[8] C Gotzes H Heitsch R Henrion and R Schultz ldquoOn thequantification of nomination feasibility in stationary gas net-works with random loadrdquoMathematical Methods of OperationsResearch vol 84 no 2 pp 427ndash457 2016
[9] D Halliday R Resnick and J Walker Physik Wiley-VCHBachelor edition edition 2007
[10] M Gugat F M Hante M Hirsch-Dick and G LeugeringldquoStationary states in gas networksrdquoNetworks andHeterogeneousMedia vol 10 no 2 pp 295ndash320 2015
[11] D RoseM SchmidtM C Steinbach and BMWillert ldquoCom-putational optimization of gas compressor stations MINLPmodels versus continuous reformulationsrdquoMathematical Meth-ods of Operations Research vol 83 no 3 pp 409ndash444 2016
[12] H Heitsch R Henrion H Leovey et al ldquoEmpirical observa-tions and statistical analysis of gas demandrdquo in Evaluating GasNetwork Capacities T data B KochM E Hiller et al Eds pp273ndash290 MOS-SIAM 2015
[13] A Genz and F Bretz Computation of Multivariate Normal andt Probabilities vol 195 of Lecture Notes in Statistics Springer2009
[14] M H Farshbaf-Shaker R Henrion and D Homberg ldquoProp-erties of Chance Constraints in Infinite Dimensions with anApplication to PDE Constrained Optimizationrdquo Set-Valued andVariational Analysis pp 1ndash21 2017
[15] T Gonzalez Grandon H Heitsch and R Henrion ldquoA jointmodel of probabilisticrobust constraints for gas transportman-agement in stationary networksrdquo Computational ManagementScience vol 14 no 3 pp 443ndash460 2017
[16] W van Ackooij I Aleksovska and M Munoz-Zuniga ldquo(Sub-)Differentiability of Probability Functionswith Elliptical Distri-butionsrdquo Set-Valued and Variational Analysis
Mathematical Problems in Engineering 17
[17] W van Ackooij and R Henrion ldquoGradient formulae for nonlin-ear probabilistic constraints with GAUssian and GAUssian-likedistributionsrdquo SIAM Journal on Optimization vol 24 no 4 pp1864ndash1889 2014
[18] W van Ackooij and R Henrion ldquo(Sub-)gradient formulaefor probability functions of random inequality systems underGaussian distributionrdquo SIAMASA Journal on UncertaintyQuantification vol 5 no 1 pp 63ndash87 2017
[19] W van Ackooij and J Malick ldquoSecond-order differentiabilityof probability functionsrdquo Optimization Letters vol 11 no 1 pp179ndash194 2017
[20] H Eckey R Kosfeld and M Turck Wahrscheinlichkeitsrech-nung und Induktive Statistik - Grundlagen -Methoden - BeispieleGabler Wiesbaden Germany 2005
[21] A Prekopa Stochastic Programming Springer 1st edition 1995[22] I Joormann M Pfetsch R Schwarz et al ldquoGasLibmdashA Library
of Gas Network Instancesrdquo 2017[23] G Wedler Lehrbuch der Physikalischen Chemie Wiley-VCH
5th edition 2004[24] N H Chen ldquoAn explicit equation for friction factor in piperdquo
Industrial amp Engineering Chemistry Fundamentals vol 18 no3 pp 296-297 1979
[25] W Schenk and F Kremer Physikalisches Praktikum Springer14th edition 2014
[26] A J Zaslavski ldquoTurnpike Properties in the Calculus of Varia-tions and Optimal Controlrdquo inNonconvex Optimization and ItsApplications vol 80 Springer 2006
Hindawiwwwhindawicom Volume 2018
MathematicsJournal of
Hindawiwwwhindawicom Volume 2018
Mathematical Problems in Engineering
Applied MathematicsJournal of
Hindawiwwwhindawicom Volume 2018
Probability and StatisticsHindawiwwwhindawicom Volume 2018
Journal of
Hindawiwwwhindawicom Volume 2018
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawiwwwhindawicom Volume 2018
OptimizationJournal of
Hindawiwwwhindawicom Volume 2018
Hindawiwwwhindawicom Volume 2018
Engineering Mathematics
International Journal of
Hindawiwwwhindawicom Volume 2018
Operations ResearchAdvances in
Journal of
Hindawiwwwhindawicom Volume 2018
Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018
International Journal of Mathematics and Mathematical Sciences
Hindawiwwwhindawicom Volume 2018
Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom
The Scientific World Journal
Volume 2018
Hindawiwwwhindawicom Volume 2018Volume 2018
Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in
Nature and SocietyHindawiwwwhindawicom Volume 2018
Hindawiwwwhindawicom
Dierential EquationsInternational Journal of
Volume 2018
Hindawiwwwhindawicom Volume 2018
Decision SciencesAdvances in
Hindawiwwwhindawicom Volume 2018
AnalysisInternational Journal of
Hindawiwwwhindawicom Volume 2018
Stochastic AnalysisInternational Journal of
Submit your manuscripts atwwwhindawicom
6 Mathematical Problems in Engineering
of the graph 119866119894 According to this 119892(119894119895)119896ℓ is the ℓ-thcomponent of the function 119892(119894119895)119896
(iii) (119894119895)119896 is the load vector of the 119896-th subgraph in thepath from V1198940 to V1198950 whereas (119894119895)1 is the load vectorof the graph 119866119894 and (119894119895)119896ℓ is the ℓ-th component of(119894119896)119896
Analogously we define 119860 (119894119895)119896 119901(119894119895)119896 and 119902(119894119895)119896 Furthermorewe define the following values
(i) 119896lowast119894119895 fl max119896 isin 1 1198982 + 1 | V1198960 isin Π(V1198940) and V1198960 isinΠ(V1198950) as the largest index of all subgraph the pathsto V1198940 and V1198950 pass through
(ii) 119899lowast119894119895 |119896 isin 1 1198982 + 1 | V1198960 isin ΠV1198940(V1198950)| as thenumber of subgraph which are on the path from V1198940from V1198950 containing 119866119894 and 119866119895
(iii) 119898lowast119894119895 fl |119890 isin E119862 | ℎ(119890) isin ΠV1198940(V1198950)and119891(119890) isin ΠV1198940(V1198950)|
as the number of controls which are between V1198940 andV1198950
With this notation and these values we define a sum for 119894 119895 isin11198982 + 1 If V1198940 isin Π(V1198950) we set sum119894119894 fl 0 and else we set
Σ119896lowast119894 fl119899lowastminus2sum119896=1
1prod119898lowastminus119896ℓ=1 119906(119896lowast119894)ℓsdot 119892(119896lowast119894)119899lowastminus119896119891(119890119906
(119896lowast119894)119899lowastminus119896) ((119896lowast119894)119899lowastminus119896)
+ 119892(119896lowast119894)1119891(119890119906(119896lowast119894)1
) ((119896lowast119894)1) (23)
where we write only 119899lowast and 119898lowast instead of 119899lowast119896lowast119894 and 119898lowast119896lowast119894 for
better reading Last we define a productΠ119894119894 fl 1 andΠ119896lowast119894 fl
119898lowastprod119896=1
119906(119896lowast119894)119896 (24)
From a first point of view the notation seems to be confusingbut using this notation makes sense since we want toguarantee the feasibility of a load vector by comparing thepressure bounds at every node with each other We explainthis in a short example Figure 5 shows a graph after removingthe compressor edges
If we want to compare the pressure bounds of node 11 andnode 12 we need to know how the pressures change on thepath from 2 to 11 resp from 2 to 12 Node 11 is part of thesubgraph 1198665 and node 12 is part of the subgraph 1198666 so itfollows 119896lowast56 = 2 because node 2 is part of subgraph 2 Thenotation can be understood as a numbering along the pathIt is 119906(211)1 = 1199062 the first control on the path from node 2to node 11 and 119906(211)2 = 1199064 the second control on this pathFurther it follows 119899lowast211 = 3 (1198662 1198663 1198665) and119898lowast
211 = 2 (1199062 1199064)Theproduct defined before is the product of all controls alongthis path and the sum is the pressure loss between node 2 andnode 7 Using this notation makes us able to formulate thetheorem for characterizing the feasible set in a readable way
Theorem 5 For given pressure bounds 119901+119898119894119899 119901+119898119886119909 isin R119899+1
and controls 119906119894 isin R (119894 = 1 1198982) the following equivalenceholds
A vector + with 1119879119899+1+ = 0 is feasible ie + isin if and
only if the following inequalities hold For all 119894 = 1 1198982 + 1holds (feasibility inside the subgraphs)
(1199011198981198941198991198940 )2 le min
119896=1119899119894[(119901119898119886119909
119894119896 )2 + 119892119894119896 (119894)] (25)
(1199011198981198861199091198940 )2 ge max
119896=1119899119894[(119901119898119894119899
119894119896 )2 + 119892119894119896 (119894)] (26)
max119896=1119899119894
[(119901119898119894119899119894119896 )2 + 119892119894119896 (119894)]
le min119896=1119899119894
[(119901119898119886119909119894119896 )2 + 119892119894119896 (119894)] (27)
For all 119894 119895 = 1 1198982 + 1 with 119894 lt 119895 holds (feasibility betweenthe subgraphs)1Π119896lowast 119894
(1199011198981198941198991198940 )2 + Σ119896lowast119894 le 1Π119896lowast119895
(1199011198981198861199091198950 )2 + Σ119896lowast119895 (28)
1Π119896lowast 119894
(1199011198981198861199091198940 )2 + Σ119896lowast119894 ge 1Π119896lowast 119895
(1199011198981198941198991198950 )2 + Σ119896lowast119895 (29)
1Π119896lowast 119894
(1199011198981198941198991198940 )2 + Σ119896lowast119894
le 1Π119896lowast119895
min119896=1119899119895
[(119901119898119886119909119895119896 )2 + 119892119895119896 (119895)] + Σ119896lowast119895 (30)
1Π119896lowast 119894
(1199011198981198861199091198940 )2 + Σ119896lowast119895
ge 1Π119896lowast119895
max119896=1119899119895
[(119901119898119894119899119895119896 )2 + 119892119895119896 (119895)] + Σ119896lowast119895 (31)
1Π119896lowast 119894
max119896=1119899119894
[(119901119898119894119899119894119896 )2 + 119892119894119896 (119894)] + Σ119896lowast119894
le 1Π119896lowast119895
(1199011198981198861199091198950 )2 + Σ119896lowast119895 (32)
1Π119896lowast 119894
min119896=1119899119894
[(119901119898119886119909119894119896 )2 + 119892119894119896 (119894)] + Σ119896lowast119894
ge 1Π119896lowast119895
(1199011198981198941198991198950 )2 + Σ119896lowast119895 (33)
1Π119896lowast 119894
max119896=1119899119894
[(119901119898119894119899119894119896 )2 + 119892119894119896 (119894)] + Σ119896lowast119894
le 1Π119896lowast119895
min119896=1119899119895
[(119901119898119886119909119895119896 )2 + 119892119895119896 (119895)] + Σ119896lowast119895 (34)
1Π119896lowast 119894
min119896=1119899119894
[(119901119898119886119909119894119896 )2 + 119892119894119896 (119894)] + Σ119896lowast119894
ge 1Π119896lowast119895
max119896=1119899119895
[(119901119898119894119899119895119896 )2 + 119892119895119896 (119895)] + Σ119896lowast119895 (35)
Mathematical Problems in Engineering 7
0 1 2
3
4
5
6
7
8
9
10
11
12
u1
u2
u3
u4
u5
G1 G2 G3
G4
G5
G6
Figure 5 Example graph for illustration of the used notation
0 1 2q1 q2
b0 b1 b2
Figure 6 Graph of Example 1
Let us have a look at the inequalities Inequalities(25)ndash(27) are well-known from [8] and they describe thefeasibility inside a connected tree The other inequalities(28)ndash(35) guarantee the feasibility between the subgraphsTherefore we compare the pressures of the entries of two parts((28) (29)) of the entry of one part and the exit nodes of theother parts ((30)ndash(33)) and we compare the pressures of allexits ((34) (35)) Because we cannot compare the pressuresof119866119894 and 119866119895 directly (there may be no direct connection) wehave to compare them using the subgraph 119866119896lowast which is thelast part the paths to V1198940 and V1198950 pass Therefore we definedthe sumsΣ119896lowast 119894 andΣ119896lowast119895 which trace back the pressures to119866119896lowast
Remark 6 For an implementation it is wise to define a set119880(119894119895) containing all controls on the path from V1198940 to V1198950 anduse an index running over this set to define the sums and theproducts
Now we want to proof the theorem
Proof of Theorem 5 Because the columns of 119860+ sum up tozero it holdsminus1119879
119899119894119860 119894 = 119860+
1198940 (119894 = 1 1198982 + 1) (36)
ldquo997904rArrrdquo We consider a feasible vector 119887+ isin From theequation of mass conservation of every subgraph it follows
119902119894 = 119860minus1119894 119894 (37)
We take the equation of momentum conservation for the 119894-thsubgraph (119894 isin 1 1198982 +1) and separate 119860+
119894 to119860+1198940 and 119860 119894
(119860+1198940)119879 1199012
1198940 + 119860119879119894 1199012
119894 = minusΦ 10038161003816100381610038161199021198941003816100381610038161003816 119902119894 (38)
Next we insert (36) into (38) and get
minus119860119879119894 1119899119894
11990121198940 + 119860119879
119894 1199012119894 = minusΦ 10038161003816100381610038161199021198941003816100381610038161003816 119902119894 (39)
We multiply this equation from left with (119860119879119894 )minus1 and insert
(37) so it follows
111989911989411990121198940 = 1199012
119894 + (119860119879119894 )minus1Φ 10038161003816100381610038161003816119860minus1
119894 11989410038161003816100381610038161003816 119860minus1119894 119894 (40)
and using the definition of the function 119892 we have
111989911989411990121198940 = 1199012
119894 + 119892119894 (119894) (41)
Note here that 119901119894 is the vector of pressures of the subgraph119866119894
and 119892119894(119894) is the vector-valued function which describes thepressure loss in 119866119894 Equation (41) holds for all subgraphs 119866119894
with 119894 isin 1 1198982 + 1 From this the inequalities (25)ndash(27)follow directly with 119901+ isin [119901+119898119894119899 119901+119898119886119909] Next we use (41)component-by-component For fix 119894 119895 isin 1 1198982 + 1 with119894 lt 119895 it follows
1199012119896lowast0 = 1199012
119896lowast119896 + 119892119896lowast119896 (119896lowast) (119896 = 1 119899119896lowast) (42)
Remember that 119896lowast119894119895 is the largest index of all subgraphs thepaths to V1198940 and V1198950 pass through For better reading we onlywrite 119896lowast instead of 119896lowast119894119895 We use the equation of the compressorproperty for the first compressor on the path from V119896lowast0 to V1198940
1199012ℎ(119890119906(119896lowast119894)1
)1199012119891(119890119906(119896lowast119894)1
)
= 119906(119896lowast119894)1 (43)
and insert it into (42) for 119896 = 119891(119890119906(119896lowast119894)1 )1199012(119896lowast 119894)10 = 1119906(119896lowast119894)1
1199012(119896lowast119894)20
+ 119892(119896lowast 119894)1119891(119890119906(119896lowast119894)1
) ((119896lowast119894)1) (44)
Note that 1199012119896lowast 0 = 1199012
(119896lowast119894)10 119892119896lowast119896(119896lowast) = 119892(119896lowast119894)1119896((119896lowast119894)1) and1199012ℎ(119890119906(119896lowast119894)1
) = 1199012(119896lowast 119894)20 We use the compressor equation for
the second compressor on the path from V119896lowast0 to V1198940 and putit together with (42) into (44)
1199012(119896lowast119894)10 = 1119906(119896lowast119894)1
[ 1119906(119896lowast119894)2
1199012(119896lowast 119894)30
+ 119892(119896lowast119894)2119891(119890119906(119896lowast119894)2
) ((119896lowast119894)2)]+ 119892(119896lowast119894)1119891(119890119906
(119896lowast119894)1) ((119896lowast119894)1)
(45)
8 Mathematical Problems in Engineering
We repeat this procedure for all controls on the path fromV119896lowast0 to V1198940 Although we write 119899lowast and119898lowast instead of 119899lowast119896lowast119894 and119898lowast
119896lowast119894 Thus we get
1199012(119896lowast119894)10 = 1119906(119896lowast119894)1
[ 1119906(119896lowast119894)2
[sdot sdot sdot [ 1119906(119896lowast119894)119898lowast1199012(119896lowast119894)119899lowast0
+ 119892(119896lowast119894)119899lowastminus1119891(119890119906(119896lowast119894)119899lowastminus1
) ((119896lowast119894)119899lowastminus1)] sdot sdot sdot]+ 119892(119896lowast119894)2119891(119890119906
(119896lowast119894)2) ((119896lowast119894)2)]
+ 119892(119896lowast119894)1119891(119890119906(119896lowast119894)1
) ((119896lowast119894)1)
(46)
We take again (42) for the node V119896lowast0 and do the sameprocedure just like before for the path from V119896lowast0 to V1198950 With119899lowast = 119899lowast119896lowast119895 and119898lowast = 119898lowast
119896lowast119895 we get
1199012(119896lowast 119895)10
= 1119906(119896lowast119895)1
[ 1119906(119896lowast119895)2
[sdot sdot sdot [ 1119906(119896lowast119895)119898lowast1199012(119896lowast 119895)119899lowast0
+ 119892(119896lowast119895)119899lowastminus1119891(119890119906(119896lowast119895)119899lowastminus1
) ((119896lowast119895)119899lowastminus1)] sdot sdot sdot]+ 119892(119896lowast119895)2119891(119890119906
(119896lowast119895)2) ((119896lowast119895)2)]
+ 119892(119896lowast119895)1119891(119890119906(119896lowast119895)1
) ((119896lowast119895)1)
(47)
Expanding (46) and (47) and equalizing them (this is possiblebecause 119901(119896lowast 119894)10 = 119901(119896lowast119895)10) leads to1Π119896lowast 119894
11990121198940 + Σ119896lowast119894 = 1Π119896lowast 119895
11990121198950 + Σ119896lowast119895 (48)
with Σ119896lowast119894 resp Σ119896lowast119895 defined before (48) directly implies (28)and (29) Using (48) together with (42) for V1198940 first timefor V1198950 second time and for V1198940 and V1198950 a third time theinequalities (30)ndash(35) follow directly So this part of the proofis complete
ldquolArr997904rdquo For this part we consider a vector 119887+ isin R119899+1
with 1119879119899+1119887+ = 0 which fulfills the inequalities (25)ndash(35) We
define the following sets (119894 = 2 1198982 + 1)11987511 fl [(119901119898119894119899
10 )2 (11990111989811988611990910 )2]
11987521 fl [ max119896=11198991
[(1199011198981198941198991119896 )2 + 1198921119896 (1)]
min119896=11198991
[(1199011198981198861199091119896 )2 + 1198921119896 (1)]]
1198751119894 fl [ 1Π1119894
(1199011198981198941198991198940 )2 + Σ1119894 1Π1198941
(1199011198981198861199091198940 )2 + Σ1119894]
1198752119894 fl [ 1Π1119894
max119896=1119899119894
[(119901119898119894119899119894119896 )2 + 119892119894119896 (119894)] + Σ1119894 1Π1198941
sdot min119896=1119899119894
[(119901119898119886119909119894119896 )2 + 119892119894119896 (119894)] + Σ1119894]
(49)
We set
1198750 fl 1198982+1⋂119896=1
1198751119896 cap 1198982+1⋂119896=1
1198752119896 (50)
and we will define a value 119901210 isin 1198750 later So to guarantee
such a value 119901210 exists we have to show first that 1198750 is not
empty Because 1198750 is a finite intersection of convex intervalswe have to make sure that all intervals are not empty andevery intersection between two intervals is not empty In onedimension this is sufficient that 1198750 is not empty For 119894 =1 1198982 + 11198751119894 = 0 according to requirement1198752119894 = 0 because of (26) (51)
Furthermore its 1198751119894 cap 1198752119894 = 0 because of (25) and (26) Themissing intersections are nonempty too for 119894 119895 isin 1 1198982+1 with 119894 lt 119895 as follows
1198751119894 cap 1198751119895 = 0 because of (28) and (29) 1198751119894 cap 1198752119895 = 0 because of (30) and (31) 1198752119894 cap 1198751119895 = 0 because of (32) and (33) 1198752119894 cap 1198752119895 = 0 because of (34) and (35)
(52)
So the set 1198750 is not empty We define the following values
119901210 isin 119875011990121198940 fl Π1119894 (1199012
10 minus Σ1119894) 119894 = 2 1198982 + 11199012119894119896 fl 1199012
1198940 minus 119892119894119896 (119894) 119894 = 1 1198982 + 1 119896 = 1 119899119894119902119894 fl 119860minus1119894 119894 119894 = 1 1198982 + 1
(53)
Now we show that our choice of pressures and flows isfeasible It is11990110 isin [119901119898119894119899
10 11990111989811988611990910 ] because of1199012
10 isin 11987511 Furtherbecause of 1199012
10 isin 1198751119894 for all 119894 = 2 1198982 + 1 it holds1Π1119894
11990121198940 + Σ1119894 le 1Π1119894
(1199011198981198861199091198940 )2 + Σ1119894 (54)
and 1Π1119894
11990121198940 + Σ1119894 ge 1Π1119894
(1199011198981198941198991198940 )2 + Σ1119894 (55)
Mathematical Problems in Engineering 9
Now we replace 11990121198940 in its definition with (1199012
119894119896 + 119892119894119896(119894)) andbecause of 1199012
1198940 isin 1198752119894 for all 119894 = 1 1198982 + 1 it follows1Π1119894
(1199012119894119896 + 119892119894119896 (119894)) + Σ1119894
le 1Π1119894
min119896=1119899119894
[(119901119898119886119909119894119896 )2 + 119892119894119896 (119894)] + Σ1119894
(56)
and 1Π1119894
(1199012119894119896 + 119892119894119896 (119894)) + Σ1119894
ge 1Π1119894
max119896=1119899119894
[(119901119898119894119899119894119896 )2 + 119892119894119896 (119894)] + Σ1119894 (57)
So for all 119894 = 1 1198982 + 1 and for all 119896 = 1 119899119894 it holds119901119894119896 isin [119901119898119894119899119894119896 119901119898119886119909
119894119896 ] which is equivalent to
119901+ isin [119901+119898119894119899 119901+119898119886119909] (58)
The equation for mass conservation follows directly from thedefinition of 119902119894 (119894 = 1 1198982 + 1) For the pressure loss weuse
1199012119894 = 1119899119894
11990121198940 minus 119892119894 (119894) (59)
And multiply this equation from left with 119860119879119894 and use (36)
With this it follows the equation for momentum conserva-tion
(119860+119894 )119879 (119901+
119894 )2 = minusΦ 10038161003816100381610038161199021198941003816100381610038161003816 119902119894 (60)
The last equation missing is the compressor property For acontrol 119895 isin 1 1198982 we fix a vertex V1198940 (119894 isin 1 1198982 +1)with ℎ(119890119906119895) = V1198940 We use 1199012
1198940 = Π1119894(119901210 minus Σ1119894) and solve the
formula for 119901210 in the following way
119901210 = 1119906(1119894)1
[sdot sdot [ 1119906(1119894)119898lowast1199012(1119894)119899lowast0
+ 119892(1119894)119899lowastminus1119891(119890119906(1119894)119899lowastminus1
) ((1119894)119899lowastminus1)] sdot sdot]+ 119892(1119894)1119891(119890119906(1119894)1 )
((1119894)1) (61)
Consider wlog that the subgraphs along the path from V10to V1198940 are numbered chronological st if ℎ(119890119906119895) isin 119866119894 then119891(119890119906119895) isin 119866119894minus1 We use 1199012
119894minus10 = Π1119894minus1(119901210 minus Σ1119894minus1) to get the
same representation for 119901210 depending on 1199012
(1119894minus1)119899lowast0 whichis equivalent to 119901(1119894)119899lowastminus10 We equalize these representationsdepending on 1199012
1198940 and 1199012119894minus10 and we get
1119906(1119894)119898lowast1199012(1119894)119899lowast0 + 119892(1119894)119899lowastminus1119891(119890119906
(1119894)119899lowastminus1) ((1119894)119899lowastminus1)
= 1199012(1119894)119899lowastminus10 (62)
We chose 119894 isin 1 1198982 + 1 st 119906119895 is the last control on thepath from V10 to V1198940 so it is 119906(1119894)119898lowast = 119906119895 Further we have
1199012(1119894)119899lowast 0 = 1199012
11989401199012(1119894)119899lowastminus10 = 1199012
119894minus10119892(1119894)119899lowastminus1119891(119890119906(1119894)119899lowastminus1
) (119887(1119894)119899lowastminus1) = 119892119894minus1119891(119890119906119895 )(119887119894minus1)
(63)
and we know that 11990121198940 resp 1199012
119894minus10 minus 119892119894minus1119891(119890119906119895 )(119887119894minus1) are the
pressures at the head resp the foot of the compressor edge119890119906119895 So it follows 1119906119895
1199012ℎ(119890119906119895 )
= 1199012119891(119890119906119895 )
(64)
which is equivalent to the compressor property This com-pletes the proof
With this theorem we have another characterization ofthe feasible set The last step in this intersection is theadaption of the results to the spheric-radial decompositionAs mentioned before the main step is to get a representationof the sets119872119894 in step (2) of Algorithm 3
Because we assumed that the graph has only one entry wedefine the following set
119872 fl 119887 isin R119899ge0 | (minus1119879
119899119887 119887) isin (65)
Consider a sampled point V119894 isin S119899minus1 of step (1) of Algorithm 3(119894 = 1 119873) We identify the load vector 119887 with the affinelinear function
119887119894 (119903) fl 119903119871V119894 + 120583 = 119903120596119894 + 120583 with 120596119894 = 119871V119894 (66)
where 120583 is the mean value of the Gaussian distribution and 119871is such that 119871119871119879 = Σ for positive definite covariance matrixΣ Note that the index 119894 here is for the sampled vector V119894 notfor the 119894-th subgraph Because 119887119894 is defined only for the exitnodes it must hold 119887 ge 0 which leads to the definition of theregular range
119877119903119890119892119894 fl 119903 ge 0 | 119887119894 (119903) ge 0 supe 119872119894 (67)
So the feasible set is
119872119894 = 119903 isin 119877119903119890119892119894 | 119887119894 (119903) fulfills (3) (5) and (7) (68)
and with Theorem 5 it holds
119872119894 = 119903 isin 119877119903119890119892119894 | 119894 (119903) fulfills (25) minus (34) (69)
where 119894(119903) is from (18) Because 119894 depends on 119903 now thefunction 119892 depends on 119903 too More exactly 119892 is quadratic in119903 so 119892119896(119887119894(119903)) can be represented as follows
1199032[[119899sum
119895=1
Ψ119896119895( 119899sumℓ=1
119860minus1119895ℓ120596119894ℓ)2]]⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
119886119894
10 Mathematical Problems in Engineering
+ 119903[[2119899sum
119895=1
Ψ119896119895( 119899sumℓ=1
119860minus1119895ℓ120596119894ℓ)( 119899sum
ℓ=1
119860minus1119895ℓ120583ℓ)]]⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
119894
+ [[119899sum
119895=1
Ψ119896119895( 119899sumℓ=1
119860minus1119895ℓ120583ℓ)2]]⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
119888119894
(70)
HereΨ is the product of119860 andΦ So the inequations whichcharacterize119872119894 are also quadratic in 119903 because the minimaland maximal pressures in the inequalities only count to 119888119894Now we can write the feasible sets as follows
119872119894 = 119903 isin 119877119903119890119892119894 | 0 le 1198861198941199051199032 + 119887119894119905119903 + 119888119894119905 (71)
where 119905 is an index for the total number of inequalities in119872119894Now the regular range can be cut with the positive intervals ofthe inequalities and the result is a union of disjoint intervals
119872119894 = 119904119894⋃119896=1
[119886119894119896 119886119894119896] 119904119894 isin N (72)
Here 119904119894 is the number of disjoint intervals in which allinequalities are fulfilled and 119886119894119896 and 119886119894119896 are the intervalbounds they depend on the pressure bounds and the con-trols So the third step of Algorithm 3 can be executed With
P (119887 isin 119872) asymp 1119873 119873sum119894=1
119904119894sum119896=1
F120594 (119886119894119896) minusF120594 (119886119894119896) (73)
the probability for a random load vector to be feasible can becomputed
4 Existence of Optimal Solutions
In this section we want to have a look at some optimizationproblems We distinguish between problems with constantloads (without uncertainty on the demand) and problemswith random loads The latter leads us to optimizationproblems with so called probabilistic constraints or chanceconstraints (see [21]) There are two aims in optimizationMinimize the maximal pressure bounds and minimize thecontrols For minimizing the maximal pressure bounds wedefine the following set
119875 fl119899⋃
119894=0
[119901+119898119894119899119894 infin) (74)
in which we look for them Because the maximal pressuresare in proportion to the cost of pipes an objective functionfor a optimization problem could be 119888119879119901+119898119886119909 for a cost vector119888 isin R119899+1
gt0 By minimizing the controls we use the Euclideannorm of the controls as an objective function
Optimization with Constant Loads Consider the two opti-mization problems
min119901+119898119886119909isin119875
119888119879119901+119898119886119909
subject to 119887+ (119901+119898119886119909) isin 119872 (75)
andmin119906isinR1198982ge1
1199062subject to 119887+ (119906) isin (76)
Lemma 7 There exists a unique solution for (75)
Proof We have a look at the side constraint 119887+(119901+119898119886119909) isin Using Theorem 5 we can write (75) as
min119901+119898119886119909isin119875
119888119879119901+119898119886119909
subject to (25) minus (35) are fulfilled (77)
For 119894 isin 1 1198982 + 1 we write inequality (25) (and allremaining analogous) in the following way
0 le 119892119894119896 (119894) minus (1199011198981198941198991198940 )2 + (119901119898119886119909
119894119896 )2 forall119896 = 1 119899119894 (78)
So every inequality depends only on one upper pressurebound (but more inequalities may depend on the samebound) Because the inequalities are not strict the values119901119896 fl min 119901119898119886119909
119896 isin [0infin) | (25) minus (35) are fulfilled (79)
exist for all 119896 = 1 119899 + 1 and we can define a value119901+119898119886119909119900119901119905119896 fl max 119901119896 119901119898119894119899
119896 (80)
which is obviously a solution of (75) Moreover this solutionis unique because reducing one component of the upperbounds would either hurt an inequality or it would not fulfill119901+119898119886119909 ge 119901+119898119894119899 anymore
For the optimization problem (76) we can formulate anecessary and a sufficient condition for optimal solutions
Lemma8 Consider pressure bounds119901+119898119894119899 119901+119898119886119909 isin R119899+1 st119901+119898119894119899 lt 119901+119898119886119909 and a suitable vector 119887+ isin R119899+1 (ie + fulfills(25)ndash(27)) For all 119896 = 1 1198982 we set 119894(119896) and 119895(119896) such that119891(119890119896) isin V+
119894(119896) and ℎ(119890119896) isin V+119895(119896)
(a) If 119906 isin R1198982 is a solution of (76) it holds for all 119896 =1 1198982 (119901119898119894119899119894(119896)0)2 minus 119892119894(119896)119891(119890119896)
(119894(119896)) le (119901119898119886119909119895(119896)0)2 (81)
(b) If the inequalities
(119901119898119886119909119894(119896)0)2 minus 119892119894(119896)119891(119890119896)
(119894(119896)) ge (119901119898119894119899119895(119896)0)2
and (119901119898119894119899119894(119896)0)2 minus 119892119894(119896)119891(119890119896)
(119894(119896)) le (119901119898119886119909119895(119896)0)2 (82)
are fulfilled for all 119896 = 1 1198982 an optimal solution 119906 isin R1198982ge1
of (76) exists
Mathematical Problems in Engineering 11
Note that a vector 119887+ isin R119899+1 is suitable here if the vector+ is feasible for the subgraphs ie the inequalities (25)-(27)are fulfilled for +Proof (a) For this part we use a proof by contradictionAssume that
(119901119898119894119899119894(119896)0)2 minus 119892119894(119896)119891(119890119896)
(119894(119896)) gt (119901119898119886119909119895(119896)0)2 (83)
Because 119906119896 ge 1 for all 119896 = 1 1198982 it follows
(119901119898119894119899119894(119896)0)2 minus 119892119894(119896)119891(119890119896)
(119894(119896)) gt 1119906119896
(119901119898119886119909119895(119896)0)2 (84)
It it Σ119896lowast119894(119896) = 0 and Π119896lowast 119894(119896) = 1 because the subgraph withindex 119896lowast is equal to the subgraphwith index 119894(119896) And becausethe subgraphs 119894(119896) and 119895(119896) are neighbours (ie directlyconnected) it follows Π119896lowast 119895(119896) = 119906119896 and
Σ119896lowast119895(119896) = 119892(119896lowast119895(119896))1119891(119890119906(119896lowast119895(119896))1
) (119887(119896lowast119895(119896))1)= 119892119894(119896)119891(119890119896)
(119894(119896)) (85)
Thus we have a contradiction to (28) and 119906119896 cannot beoptimal
(b) As mentioned before the vector 119887+ is such that isfeasible for the subgraphs So because of the conditions for119896 isin 1 1198982 there exists a 119901119894(119896)0 isin [119901119898119894119899
119894(119896)0 119901119898119886119909119894(119896)0] and a119901119895(119896) isin [119901119898119894119899
119895(119896) 119901119898119886119909119895(119896) ] so that1199012119894(119896)0 minus 119892119894(119896)119891(119890119896)
(119894(119896)) ge 1199012119895(119896)0 (86)
With 1199012119894(119896)0 minus 119892119894(119896)119891(119890119896)
(119894(119896)) = 1199012119891(119890119896)
and 1199012119895(119896)0 = 1199012
ℎ(119890119896)it
follows
(119901ℎ(119890119896)119901119891(119890119896)
) ge 1 (87)
Therefore we can find a 119906 isin R1198982ge1 that the inequalities (28)-
(35) are fulfilled Now we can use theWeierstraszlig theorem andget the existence of an optimal solution
Note that the sufficient condition in Lemma 8 is com-paratively strong there may exist optimal solutions with lessstrong conditions eg the following
Corollary 9 If the graph has only one compressor edge ie|E119862| = 1 then it holds If 119906 isin Rge1 is an optimal solution of(76) 119906 is unique
This conclusion follows directly from the strict mono-tonicity of the objective function
Optimization with Random Loads In this section we con-sider the load vector to be random as we mentioned inSection 22 In the optimization problems we want to makesure that the feasibility of a random load vector is guaranteedfor a probability 120572 isin (0 1) This leads us to optimizationproblems with so called chance or probabilistic constraints
In general problems with probabilistic constraints have theform
min 119891 (119909 120585)subject to P (ℎ (119909 120585) le 0) ge 120572 (88)
with an objective function 119891 and a vector of uncertainty 120585There exist many works about probabilistic constraints respstochastic programming eg [7] gives an excellent overviewabout chance constraints in theory and application In ourcase we consider the following problem
min119901+119898119886119909isin119875
119888119879119901+119898119886119909
subject to P (119887 (119901+119898119886119909) isin 119872) ge 120572 (89)
To handle this problem resp the probabilistic constraint thespheric-radial decomposition gives us an explicit representa-tion of the constraint for a Gaussian vector 119887 sim N(120583 Σ) withmean value 120583 and positive definite covariance Σ
P (119887 (119901+119898119886119909) isin 119872)= int
S119899minus1120583120594 119903 ge 0 | 119903119871V + 120583 isin 119872119889120583120578 (V) (90)
Further Algorithm 3 gives a way to approximate this integralin an efficient way In Section 3 we showed a way to character-ize the feasible set119872 and how to adapt this characterization tothe spheric-radial decomposition This changes our problemwith probabilistic constraints (89) to
min119901+119898119886119909isin119875
119888119879119901+119898119886119909
subject to 1119873 119873sum119894=1
119904119905sum119905=1
F120594 (119886119894119905) minusF120594 (119886119894119905) ge 120572 (91)
Of course the interval bounds depend on the pressure boundsand the control For this problem we formulate the nextlemma
Lemma 10 Let a sampling V1 V119873 (from step (1) ofAlgorithm 3) and 120572 isin (0 1) be given For a suitable 120583 isin R119899
problem (91) has at least one solution 119901+119898119886119909 isin 119875Proof For the proof we fix a V isin V1 V119873 from the givensampling From (25) and (26) it follows for all 119896 = 1 119899119894with 119894 isin 1 1198982 + 10 le 119892119894119896 (119894 (119903)) + (119901119898119886119909
119894119896 )2 minus (1199011198981198941198991198940 )2
0 le minus119892119894119896 (119894 (119903)) + (1199011198981198861199091198940 )2 minus (119901119898119894119899
119894119896 )2 (92)
With a random vector 119887 sim N(120583 Σ)119892119894119896(119894(119903)) are quadraticfunctions in 119903 and the upper pressure bounds only influencethe constant part with a positive sign
0 le 1198861198941198961199032 + 119887119894119896119903 + 119888119894119896 + (119901119898119886119909119894119896 )2 minus (119901119898119894119899
1198940 )2 0 le minus1198861198941198961199032 minus 119887119894119896119903 minus 119888119894119896 + (119901119898119886119909
1198940 )2 minus (119901119898119894119899119894119896 )2 (93)
12 Mathematical Problems in Engineering
It follows that a rise in 119901119898119886119909119894119896 for all 119896 = 0 119899119894 enlarges
the intervals in which the inequalities hold The same holdsfor all inequalities inTheorem 5The cumulative distributionfunction of theChi-Square-Distribution is strictlymonotonicincreasing and convergent to 1 Thus upper pressure bounds119901119898119886119909119894119896 (for 119894 = 1 119899) can be found so that
1119873 119873sum119895=1
119904119895sum119905=1
F120594 (119886119895119905) minusF120594 (119886119895119905) (94)
is higher that 120572 isin (0 1) The index119873 is for all sampled pointson the sphere and the index 119904119895 is for the union of intervals inwhich all inequalities hold The interval boundaries 119886119895119905 and119886119895119905 depend continuously on the pressure bounds and becausethe inequalities are not strict 119901119898119886119909 can be found st the sideconstraint is fulfilled with equality So there exists at least onesolution of (91)
5 Numerical Results
In this section we show a few results of implementation Atfirst we show the idea of the spheric-radial decomposition byusing an easy example Next we show a easy example with onecompressor edge and at last we use real data of the Greek gasnetworkThe focus of the implementation is on the theoremsproofed in Section 4
Example 1 Our first example is an easy graph without innercontrol (see Figure 6)
Assume that node 1 and 2 are gas consumers with meandemand 120583 = [05 05]119879 The pressure bounds are given by119901+119898119894119899 = [2 1 1]119879 and 119901+119898119886119909 = [3 2 2]119879 and Φ = E2 Thenthe incidence matrix is
119860+ = [[[minus1 01 minus10 1
]]] resp 119860 = [1 minus10 1 ] (95)
Consider the inequalities of Theorem 5 to characterize thefeasible set Because the network has no compressor edgesthe system of inequalities only contains the information (25)-(27) the other inequalities do not occur The function 119892 isgiven by
119892 (119887) = [1 01 1] [1 00 1] sdot1003816100381610038161003816100381610038161003816100381610038161003816[1 10 1](11988711198872)
1003816100381610038161003816100381610038161003816100381610038161003816∘ ([1 10 1](11988711198872)) = ( 11988721 + 211988711198872 + 1198872211988721 + 211988711198872 + 211988722) (96)
By inserting the pressure bounds one can see that inequality(25) is always fulfilled and inequality (27) implies1198872 le radic3 (97)
With this inequality (26) follows for 0 le 1198872 le radic30 le 1198871 le minus1198872 + radic8 minus 11988722 (98)
So the feasible set119872 is the following (see Figure 7)
119872 = 119887 isin R2ge0 | 1198872 le radic3 and 1198871 le minus1198872 + radic8 minus 11988722 (99)
Now consider the optimization problem
min119901+119898119886119909isin119875
11198793119901+119898119886119909
subject to 119887 isin 119872 (119901+119898119886119909) (100)
with 119887 = 120583 = [05 05]119879 Lemma 7 guarantees the existenceof a unique solution of this problem We use the fmincon-function inMATLAB which minimizes a constrained non-linear multivariable function The fmincon default settinguses a interior-point method For more information we referto the official MathWorks homepage (httpswwwmath-workscomhelpoptimugfminconhtml)The solver returnsthe following result
119901+119898119886119909119900119901119905 = [2 radic3radic275]119879 (101)
One can easily see that this is the correct solution The gasdemand at node 2 is 05 which is equal to the flow 1199022 sousing the equation of momentum implies a difference in thequadratic pressures between node 1 and node 2 of 025 Theflow through edge 1 is given by the two exit nodes and it is1199021 = 1 This implies a difference in the quadratic pressuresbetweennode 0 andnode 1 of 1 So the optimal solution119901+119898119886119909
119900119901119905
directly followsNext we use the spheric-radial decomposition especially
Algorithm 3 to calculate the probability P(119887 isin 119872) for arandom vector 119887 sim N(E2 [05 05]119879) Therefore we viewthe results of 8 tests with 1000 sampled points in each one(Table 1)
The probability in all eight tests is nearly the same Themean value is 03479 and the variance is 27723 sdot 10minus6 whichis very small This shows that the implementation is workingnearly exact The efficiency of the implementation of courseis not perfect but that is not part of this work
Last wewant to solve the following optimization problem
min119901+119898119886119909isin119875
11198793119901+119898119886119909
subject to 1119873 119873sum119894=1
119904sum119905=1
F120594 (119886119894119905) minusF120594 (119886119894119905) ge 120572 (102)
We use again the fmincon-function inMATLAB to solve theproblem eight times with 1000 sampled points in every caseBecause Lemma 8 needs a suitable 120583 isin R2
ge0 we assume 119887 simN(01sdot1198642 [15 15])We set120572 fl 09 and the results are shownin Table 2
Example 2 Now we add a compressor edge to the graph ofExample 1 (see Figure 8)
Assume that node 1 and 3 are consumers node 2 is ainner nodeThemeandemand120583 = [05 0 05]119879Thepressure
Mathematical Problems in Engineering 13
minus2 minus1 1 2 3
minus2
minus1
1
2
3
b2
b1
minus1 1 2
minus1
1
2
b2
b1
rLv1
rLv2
rLv3
Scheme of the spheric-radial decomposition with
sphere (yellow)and three points 1 2 3 (red) at the unit
the feasible set Mminus for Σ = E2 = [05 05]TFeasible set M for p+min = [2 1 1]T and
p+max = [3 2 2]T
Figure 7 Feasible set of Example 1
Table 1 Test results of easy example
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 8P(119887 isin 119872) 03486 03481 03450 03501 03484 03499 03459 03475
Table 2 Test results of Example 1
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 81199011198981198861199091199001199011199050 54514 54714 54758 54028 54767 53921 54888 551671199011198981198861199091199001199011199051 29802 30126 30225 29725 30300 29485 30336 300291199011198981198861199091199001199011199052 10084 10004 10776 10001 11221 11390 10285 10558
0 1 2 3q1 q2 q3
b0 b1 b2 b3
e1 isin ℰF e2 isin ℰC e3 isin ℰF
Figure 8 Graph of Example 2
0 1 2 3q1 q2 q2 q3
b0 b1 b2 b3
e1 isin ℰF e3 isin ℰF
Figure 9 Separated graph of Example 2
bounds are given by 119901+119898119894119899 = [2 1 1 1]119879 and 119901+119898119886119909 = [3 22 2]119879 and 1206011198901= 1 and 1206011198903
= 1 The incidence matrix is givenby
119860+ = [[[[[[
minus1 0 01 minus1 00 1 minus10 0 1]]]]]]
resp 119860 = [[[1 minus1 00 1 minus10 0 1
]]] (103)
Before we have a look at the inequalities of Theorem 5we have to remove the compressor edge from the graphThe flows can be computed by using the equation of massconservation and the new load vector is then
= (1198871 + 11990221198872 minus 11990221198873 ) with
119902 = (1198871 + 1198872 + 11988731198872 + 11988731198873)(104)
We assume the control to be switched off so it holds119906 = 1With this the control edge 1198902 is treated as a flow edge withoutfriction The resulting graph is shown in Figure 9 from theinequalities (25)ndash(35) it follows
3 le radic3 and 1 le radic8 minus 23 (105)
14 Mathematical Problems in Engineering
minus1 1 2 3
1
2
3
minus1 1 2 3
1
2
3
b3
b1
Hyperspace of the feasible set MHyperspace of the feasible set M
b1
b3
Figure 10 Feasible set of the Example 2
Table 3 Test results of control example
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 8P(119887 isin 119872) 01531 01500 01540 01536 01530 01550 01567 01528so the feasible set is (see Figure 10)
= + isin R119899+1 | 1119879
4 + = 0 and 0 le 3 le radic3 and 2 = 0and 0 le 1 le radic8 minus 22 (106)
This implies
119872 = 119887 isin R3ge0 | 0 le 1198873 le radic3 and 1198872 = 0 and 0 le 1198871 le minus1198873
+ radic8 minus 11988723 (107)
Solving the problem
min119901+119898119886119909isin119875
11198794119901+119898119886119909
subject to isin (108)
for 119887 = 120583 = [05 0 05]119879 leads us to the optimal solution
119901+119898119886119909119900119901119905 = [2 radic3radic3radic275]119879 (109)
The argumentation for the optimality of this vector is exactlylike before in the easy example Becausewe have a compressoredge in this graph we can minimize the control Thereforewe fix the pressure bounds as we did at the beginning of thisexample and we set 119887 = [1 0 15]119879 So inequality (31) is notfulfilled anymore for 119906 = 1 We use again the fmincon-solverinMATLAB to solve the following problem
min119906ge1
1199062subject to 119887 isin 119872(119906) (110)
The solver returns the optimal solution 119906119900119901119905 = 11818
Here one can see that the sufficient condition in Lemma 8is really strong Itrsquos not fulfilled with the given values but thereexists a solution anyway Of course the necessary conditionis fulfilled
Now we use again Algorithm 3 to compute the proba-bility P(119887 isin 119872) for a random load vector 119887 sim N(E3 [0505 05]119879) Table 3 shows us the results
Again we can see that the probability is nearly the samein all eight cases The mean probability is 01535 and thevariance is 32369 sdot 10minus6 For the optimization we reducethe covariance matrix and assume 119887 sim N(001 sdot E3 [05 0505]119879) For a probability 120572 = 09 Table 4 states the resultsExample 3 In this example we use a part of a real networkThe focus here is on realistic values for our model Weget these values from the GasLib (see [22]) a collection oftechnical gas network data
ldquoThegoal ofGasLib is to promote research on gasnetworks by providing a set of large and realisticbenchmark instancesrdquo (httpgaslibzibde)
We use a part of the Greek gas network (GasLib-134) (seeFigure 11)
For using real values we first need to know the constant120601119890 from (5)
120601119890 = (119877119878119879)2 120582119863 1198711198862 (111)
The value 119877119878 is the specific gas constant which can becalculated by 119877119878 = 119877119872 with the ideal gas constant 119877 =831451119896119892 sdot 11989821199042 sdot 119898119900119897 sdot 119870 (see [23]) and the molar mass119872 = 1662119896119892119896119898119900119897 (from real data) of the used natural gasAlso the temperature 119879 = 28915119870 is given from real dataFor the (constant) friction factor 120582 we use the law of Chen(see [24])
Mathematical Problems in Engineering 15
Table 4 Test results of control example
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 81199011198981198861199091199001199011199050 39013 39223 39039 38901 38879 39039 38901 390981199011198981198861199091199001199011199051 16568 16565 16365 16497 16496 16365 16497 164921199011198981198861199091199001199011199052 16568 16565 16365 16497 16496 16365 16497 164981199011198981198861199091199001199011199053 10000 10000 10000 10000 10002 10000 10000 10000
Table 5 Values based on real data
variable 119877119878 119879 120582 119863 119871 119886 120601value 50027 28915 00093 9144 178693 43862 198 sdot 107unit 11989821199042 sdot 119870 119870 minus 119898119898 119896119898 119898119904 11989821199042
Figure 11 Greek gas network from GasLib [22] red edges are usedin our example
1radic120582 = minus2 log10 ( 11989611986337065minus 50425
Relog10 [(119896119863)1109828257 + 58506
Re08981]) (112)
Diameter 119863 = 9144119898119898 and roughness value 119896 = 8 sdot 10minus6119898for the first pipe (see Figure 11) are known from real data TheReynolds-Number Re can be calculated by Re = (V119863120588)120578with the dynamic viscosity 120578 = 1 sdot 10minus5119875119886 sdot 119904 (see [1])the mean velocity V = 9548119898119904 (calculated with the meanvolume flow rate which is given by the real data) of the gasand the density 120588 = 074331198961198921198983 (from real data) So we getRe = 648959074 asymp 649 sdot 106 and with this the pipe frictioncoefficient after Chen is 120582 = 00093
The length of the first pipe is given by 119871 = 178693119896119898from the real data Last value missing is the sound speed innatural gas which we assume to be ideal The sound speed
119886 depends on the modulus of compressibility (see [9]) Forideal gases the modulus of compressibility only depends onthe adiabatic exponent and the pressure Using the ideal gasequation simplifies the equation for the sound speed to
119886 = radic120581119877119879119872 (113)
The adiabatic exponent for methan which is the main partof natural gas (gt 90) is about 133 (see [25]) Together allvalues are shown in Table 5 (example for the pipe next to thesource in Figure 11)
Nowwewant to optimize the inner control in the networkfor a fix load vector Therefor we choose two nominations2011-11-01 and 2016-02-17 In both cases the fmincon-functioninMATLAB returns
119906119900119901119905 = 1 (114)
as a local minimum which satisfies the constraint 119887 isin 119872This means for these nominations a compressor stationis not needed so it is switched off With this as a lastcalculation in this paper we calculate the probability for thegiven nominations to be feasible We use the values from thenominations as mean value 120583 and set 119887 sim N(120583 sdot 10minus3E 120583)The results are shown in Tables 6 and 7
Again one can see that the results are quite similarThe mean probability in Table 6 is 09860 and the meanprobability in Table 7 is 09760 Both values are nearly 1 sothere must happen something really unforeseeable that aGreek does not get his demanded gas
6 Conclusion
In this paper we have used an existing stationary modelfor gas transport based on the isothermal Euler equationsWe have extended this model by compressor stations andpresented a characterization of the set of feasible loadsEspecially Theorem 5 contains the central idea to get anaccess to the optimization in Section 4 and the spheric-radialdecomposition is our main tool to handle the uncertainty inthe loadsWewere able to show some results for optimizationwith probabilistic constraints in gas transport In a next stepthese results should be generalized
16 Mathematical Problems in Engineering
Table 6 Test results for nomination 2011-11-01
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 8P(119887 isin 119872) 09849 09849 09878 09885 09863 09845 09854 09853
Table 7 Test results for nomination 2016-02-17
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 8P(119887 isin 119872) 09744 09756 09788 09788 09754 09775 09734 09739
With the characterization of the feasible set we haveshown that the feasibility of a load depends significantly onthe pressure bounds These pressure bounds play a mainrole in our optimization problems The other main partsare the inner controls modeled by compressor edges whichshould lead in optimal control problems with probabilisticconstraints Here we have built a base for further works withthis topic
The idea of separating the graph and removing thecompressor edges can be used to extend themodel with othercomponents like eg control valves and resistorsThiswouldchange the inequalities inTheorem 5 but the approach worksalso for thesemodifications Further [8] showed a way to dealwith a networkwhich is a cycle Combining bothmethods weget an access to model gas networks in high detail
Finally the expansion to a dynamic model might bealso possible The idea is to use a space-mapping approachbecause one can use a lot of the analysis we did hereTherefore upcoming papers should deal with these problemsand present results of the turnpike-theory which describesa relation between stationary and instationary solutions Infact turnpike-theory implies that the optimal dynamic stateis close to the optimal static state in the interior of the timeintervals for sufficiently long time horizons AJ Zaslavskiexplains the turnpike-phenomenon in [26] This book alsogives an overview about the theory
Data Availability
The data used to support the findings of this study areavailable from the corresponding author upon request
Conflicts of Interest
The authors declare that there are no conflicts of interestregarding the publication of this paper
Acknowledgments
This work was supported by DFG in the framework of theCollaborative Research Centre CRCTransregio 154 Math-ematical Modeling Simulation and Optimization Using theExample of Gas Networks project C03
References
[1] P Domschke B Hiller J Lang and C Tischendorf ldquoModel-lierung von gasnetzwerken Eine Ubersichtrdquo Tech Rep 2717Technische Universitat Darmstadt 2017
[2] T Koch B Hiller M E Pfetsch and L Schewe Evaluating GasNetwork Capacities MOS-SIAM 2015
[3] M Gugat R Schultz and D Wintergerst ldquoNetworks of pipe-lines for gas with nonconstant compressibility factor stationarystatesrdquoComputational amp AppliedMathematics vol 37 no 2 pp1066ndash1097 2018
[4] M Gugat and M Herty ldquoExistence of classical solutions andfeedback stabilization for the flow in gas networksrdquo ESAIMControl Optimisation and Calculus of Variations vol 17 no 1pp 28ndash51 2011
[5] M K Banda M Herty and A Klar ldquoGas flow in pipeline net-worksrdquo Networks and Heterogeneous Media vol 1 no 1 pp 41ndash56 2006
[6] RM Colombo G GuerraM Herty and V Schleper ldquoOptimalcontrol in networks of pipes and canalsrdquo SIAM Journal onControl and Optimization vol 48 no 3 pp 2032ndash2050 2009
[7] A Shapiro D Dentcheva and A Ruszczynski ectures onStochastic Programming Modeling and Theory MPS-SIAM2009
[8] C Gotzes H Heitsch R Henrion and R Schultz ldquoOn thequantification of nomination feasibility in stationary gas net-works with random loadrdquoMathematical Methods of OperationsResearch vol 84 no 2 pp 427ndash457 2016
[9] D Halliday R Resnick and J Walker Physik Wiley-VCHBachelor edition edition 2007
[10] M Gugat F M Hante M Hirsch-Dick and G LeugeringldquoStationary states in gas networksrdquoNetworks andHeterogeneousMedia vol 10 no 2 pp 295ndash320 2015
[11] D RoseM SchmidtM C Steinbach and BMWillert ldquoCom-putational optimization of gas compressor stations MINLPmodels versus continuous reformulationsrdquoMathematical Meth-ods of Operations Research vol 83 no 3 pp 409ndash444 2016
[12] H Heitsch R Henrion H Leovey et al ldquoEmpirical observa-tions and statistical analysis of gas demandrdquo in Evaluating GasNetwork Capacities T data B KochM E Hiller et al Eds pp273ndash290 MOS-SIAM 2015
[13] A Genz and F Bretz Computation of Multivariate Normal andt Probabilities vol 195 of Lecture Notes in Statistics Springer2009
[14] M H Farshbaf-Shaker R Henrion and D Homberg ldquoProp-erties of Chance Constraints in Infinite Dimensions with anApplication to PDE Constrained Optimizationrdquo Set-Valued andVariational Analysis pp 1ndash21 2017
[15] T Gonzalez Grandon H Heitsch and R Henrion ldquoA jointmodel of probabilisticrobust constraints for gas transportman-agement in stationary networksrdquo Computational ManagementScience vol 14 no 3 pp 443ndash460 2017
[16] W van Ackooij I Aleksovska and M Munoz-Zuniga ldquo(Sub-)Differentiability of Probability Functionswith Elliptical Distri-butionsrdquo Set-Valued and Variational Analysis
Mathematical Problems in Engineering 17
[17] W van Ackooij and R Henrion ldquoGradient formulae for nonlin-ear probabilistic constraints with GAUssian and GAUssian-likedistributionsrdquo SIAM Journal on Optimization vol 24 no 4 pp1864ndash1889 2014
[18] W van Ackooij and R Henrion ldquo(Sub-)gradient formulaefor probability functions of random inequality systems underGaussian distributionrdquo SIAMASA Journal on UncertaintyQuantification vol 5 no 1 pp 63ndash87 2017
[19] W van Ackooij and J Malick ldquoSecond-order differentiabilityof probability functionsrdquo Optimization Letters vol 11 no 1 pp179ndash194 2017
[20] H Eckey R Kosfeld and M Turck Wahrscheinlichkeitsrech-nung und Induktive Statistik - Grundlagen -Methoden - BeispieleGabler Wiesbaden Germany 2005
[21] A Prekopa Stochastic Programming Springer 1st edition 1995[22] I Joormann M Pfetsch R Schwarz et al ldquoGasLibmdashA Library
of Gas Network Instancesrdquo 2017[23] G Wedler Lehrbuch der Physikalischen Chemie Wiley-VCH
5th edition 2004[24] N H Chen ldquoAn explicit equation for friction factor in piperdquo
Industrial amp Engineering Chemistry Fundamentals vol 18 no3 pp 296-297 1979
[25] W Schenk and F Kremer Physikalisches Praktikum Springer14th edition 2014
[26] A J Zaslavski ldquoTurnpike Properties in the Calculus of Varia-tions and Optimal Controlrdquo inNonconvex Optimization and ItsApplications vol 80 Springer 2006
Hindawiwwwhindawicom Volume 2018
MathematicsJournal of
Hindawiwwwhindawicom Volume 2018
Mathematical Problems in Engineering
Applied MathematicsJournal of
Hindawiwwwhindawicom Volume 2018
Probability and StatisticsHindawiwwwhindawicom Volume 2018
Journal of
Hindawiwwwhindawicom Volume 2018
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawiwwwhindawicom Volume 2018
OptimizationJournal of
Hindawiwwwhindawicom Volume 2018
Hindawiwwwhindawicom Volume 2018
Engineering Mathematics
International Journal of
Hindawiwwwhindawicom Volume 2018
Operations ResearchAdvances in
Journal of
Hindawiwwwhindawicom Volume 2018
Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018
International Journal of Mathematics and Mathematical Sciences
Hindawiwwwhindawicom Volume 2018
Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom
The Scientific World Journal
Volume 2018
Hindawiwwwhindawicom Volume 2018Volume 2018
Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in
Nature and SocietyHindawiwwwhindawicom Volume 2018
Hindawiwwwhindawicom
Dierential EquationsInternational Journal of
Volume 2018
Hindawiwwwhindawicom Volume 2018
Decision SciencesAdvances in
Hindawiwwwhindawicom Volume 2018
AnalysisInternational Journal of
Hindawiwwwhindawicom Volume 2018
Stochastic AnalysisInternational Journal of
Submit your manuscripts atwwwhindawicom
Mathematical Problems in Engineering 7
0 1 2
3
4
5
6
7
8
9
10
11
12
u1
u2
u3
u4
u5
G1 G2 G3
G4
G5
G6
Figure 5 Example graph for illustration of the used notation
0 1 2q1 q2
b0 b1 b2
Figure 6 Graph of Example 1
Let us have a look at the inequalities Inequalities(25)ndash(27) are well-known from [8] and they describe thefeasibility inside a connected tree The other inequalities(28)ndash(35) guarantee the feasibility between the subgraphsTherefore we compare the pressures of the entries of two parts((28) (29)) of the entry of one part and the exit nodes of theother parts ((30)ndash(33)) and we compare the pressures of allexits ((34) (35)) Because we cannot compare the pressuresof119866119894 and 119866119895 directly (there may be no direct connection) wehave to compare them using the subgraph 119866119896lowast which is thelast part the paths to V1198940 and V1198950 pass Therefore we definedthe sumsΣ119896lowast 119894 andΣ119896lowast119895 which trace back the pressures to119866119896lowast
Remark 6 For an implementation it is wise to define a set119880(119894119895) containing all controls on the path from V1198940 to V1198950 anduse an index running over this set to define the sums and theproducts
Now we want to proof the theorem
Proof of Theorem 5 Because the columns of 119860+ sum up tozero it holdsminus1119879
119899119894119860 119894 = 119860+
1198940 (119894 = 1 1198982 + 1) (36)
ldquo997904rArrrdquo We consider a feasible vector 119887+ isin From theequation of mass conservation of every subgraph it follows
119902119894 = 119860minus1119894 119894 (37)
We take the equation of momentum conservation for the 119894-thsubgraph (119894 isin 1 1198982 +1) and separate 119860+
119894 to119860+1198940 and 119860 119894
(119860+1198940)119879 1199012
1198940 + 119860119879119894 1199012
119894 = minusΦ 10038161003816100381610038161199021198941003816100381610038161003816 119902119894 (38)
Next we insert (36) into (38) and get
minus119860119879119894 1119899119894
11990121198940 + 119860119879
119894 1199012119894 = minusΦ 10038161003816100381610038161199021198941003816100381610038161003816 119902119894 (39)
We multiply this equation from left with (119860119879119894 )minus1 and insert
(37) so it follows
111989911989411990121198940 = 1199012
119894 + (119860119879119894 )minus1Φ 10038161003816100381610038161003816119860minus1
119894 11989410038161003816100381610038161003816 119860minus1119894 119894 (40)
and using the definition of the function 119892 we have
111989911989411990121198940 = 1199012
119894 + 119892119894 (119894) (41)
Note here that 119901119894 is the vector of pressures of the subgraph119866119894
and 119892119894(119894) is the vector-valued function which describes thepressure loss in 119866119894 Equation (41) holds for all subgraphs 119866119894
with 119894 isin 1 1198982 + 1 From this the inequalities (25)ndash(27)follow directly with 119901+ isin [119901+119898119894119899 119901+119898119886119909] Next we use (41)component-by-component For fix 119894 119895 isin 1 1198982 + 1 with119894 lt 119895 it follows
1199012119896lowast0 = 1199012
119896lowast119896 + 119892119896lowast119896 (119896lowast) (119896 = 1 119899119896lowast) (42)
Remember that 119896lowast119894119895 is the largest index of all subgraphs thepaths to V1198940 and V1198950 pass through For better reading we onlywrite 119896lowast instead of 119896lowast119894119895 We use the equation of the compressorproperty for the first compressor on the path from V119896lowast0 to V1198940
1199012ℎ(119890119906(119896lowast119894)1
)1199012119891(119890119906(119896lowast119894)1
)
= 119906(119896lowast119894)1 (43)
and insert it into (42) for 119896 = 119891(119890119906(119896lowast119894)1 )1199012(119896lowast 119894)10 = 1119906(119896lowast119894)1
1199012(119896lowast119894)20
+ 119892(119896lowast 119894)1119891(119890119906(119896lowast119894)1
) ((119896lowast119894)1) (44)
Note that 1199012119896lowast 0 = 1199012
(119896lowast119894)10 119892119896lowast119896(119896lowast) = 119892(119896lowast119894)1119896((119896lowast119894)1) and1199012ℎ(119890119906(119896lowast119894)1
) = 1199012(119896lowast 119894)20 We use the compressor equation for
the second compressor on the path from V119896lowast0 to V1198940 and putit together with (42) into (44)
1199012(119896lowast119894)10 = 1119906(119896lowast119894)1
[ 1119906(119896lowast119894)2
1199012(119896lowast 119894)30
+ 119892(119896lowast119894)2119891(119890119906(119896lowast119894)2
) ((119896lowast119894)2)]+ 119892(119896lowast119894)1119891(119890119906
(119896lowast119894)1) ((119896lowast119894)1)
(45)
8 Mathematical Problems in Engineering
We repeat this procedure for all controls on the path fromV119896lowast0 to V1198940 Although we write 119899lowast and119898lowast instead of 119899lowast119896lowast119894 and119898lowast
119896lowast119894 Thus we get
1199012(119896lowast119894)10 = 1119906(119896lowast119894)1
[ 1119906(119896lowast119894)2
[sdot sdot sdot [ 1119906(119896lowast119894)119898lowast1199012(119896lowast119894)119899lowast0
+ 119892(119896lowast119894)119899lowastminus1119891(119890119906(119896lowast119894)119899lowastminus1
) ((119896lowast119894)119899lowastminus1)] sdot sdot sdot]+ 119892(119896lowast119894)2119891(119890119906
(119896lowast119894)2) ((119896lowast119894)2)]
+ 119892(119896lowast119894)1119891(119890119906(119896lowast119894)1
) ((119896lowast119894)1)
(46)
We take again (42) for the node V119896lowast0 and do the sameprocedure just like before for the path from V119896lowast0 to V1198950 With119899lowast = 119899lowast119896lowast119895 and119898lowast = 119898lowast
119896lowast119895 we get
1199012(119896lowast 119895)10
= 1119906(119896lowast119895)1
[ 1119906(119896lowast119895)2
[sdot sdot sdot [ 1119906(119896lowast119895)119898lowast1199012(119896lowast 119895)119899lowast0
+ 119892(119896lowast119895)119899lowastminus1119891(119890119906(119896lowast119895)119899lowastminus1
) ((119896lowast119895)119899lowastminus1)] sdot sdot sdot]+ 119892(119896lowast119895)2119891(119890119906
(119896lowast119895)2) ((119896lowast119895)2)]
+ 119892(119896lowast119895)1119891(119890119906(119896lowast119895)1
) ((119896lowast119895)1)
(47)
Expanding (46) and (47) and equalizing them (this is possiblebecause 119901(119896lowast 119894)10 = 119901(119896lowast119895)10) leads to1Π119896lowast 119894
11990121198940 + Σ119896lowast119894 = 1Π119896lowast 119895
11990121198950 + Σ119896lowast119895 (48)
with Σ119896lowast119894 resp Σ119896lowast119895 defined before (48) directly implies (28)and (29) Using (48) together with (42) for V1198940 first timefor V1198950 second time and for V1198940 and V1198950 a third time theinequalities (30)ndash(35) follow directly So this part of the proofis complete
ldquolArr997904rdquo For this part we consider a vector 119887+ isin R119899+1
with 1119879119899+1119887+ = 0 which fulfills the inequalities (25)ndash(35) We
define the following sets (119894 = 2 1198982 + 1)11987511 fl [(119901119898119894119899
10 )2 (11990111989811988611990910 )2]
11987521 fl [ max119896=11198991
[(1199011198981198941198991119896 )2 + 1198921119896 (1)]
min119896=11198991
[(1199011198981198861199091119896 )2 + 1198921119896 (1)]]
1198751119894 fl [ 1Π1119894
(1199011198981198941198991198940 )2 + Σ1119894 1Π1198941
(1199011198981198861199091198940 )2 + Σ1119894]
1198752119894 fl [ 1Π1119894
max119896=1119899119894
[(119901119898119894119899119894119896 )2 + 119892119894119896 (119894)] + Σ1119894 1Π1198941
sdot min119896=1119899119894
[(119901119898119886119909119894119896 )2 + 119892119894119896 (119894)] + Σ1119894]
(49)
We set
1198750 fl 1198982+1⋂119896=1
1198751119896 cap 1198982+1⋂119896=1
1198752119896 (50)
and we will define a value 119901210 isin 1198750 later So to guarantee
such a value 119901210 exists we have to show first that 1198750 is not
empty Because 1198750 is a finite intersection of convex intervalswe have to make sure that all intervals are not empty andevery intersection between two intervals is not empty In onedimension this is sufficient that 1198750 is not empty For 119894 =1 1198982 + 11198751119894 = 0 according to requirement1198752119894 = 0 because of (26) (51)
Furthermore its 1198751119894 cap 1198752119894 = 0 because of (25) and (26) Themissing intersections are nonempty too for 119894 119895 isin 1 1198982+1 with 119894 lt 119895 as follows
1198751119894 cap 1198751119895 = 0 because of (28) and (29) 1198751119894 cap 1198752119895 = 0 because of (30) and (31) 1198752119894 cap 1198751119895 = 0 because of (32) and (33) 1198752119894 cap 1198752119895 = 0 because of (34) and (35)
(52)
So the set 1198750 is not empty We define the following values
119901210 isin 119875011990121198940 fl Π1119894 (1199012
10 minus Σ1119894) 119894 = 2 1198982 + 11199012119894119896 fl 1199012
1198940 minus 119892119894119896 (119894) 119894 = 1 1198982 + 1 119896 = 1 119899119894119902119894 fl 119860minus1119894 119894 119894 = 1 1198982 + 1
(53)
Now we show that our choice of pressures and flows isfeasible It is11990110 isin [119901119898119894119899
10 11990111989811988611990910 ] because of1199012
10 isin 11987511 Furtherbecause of 1199012
10 isin 1198751119894 for all 119894 = 2 1198982 + 1 it holds1Π1119894
11990121198940 + Σ1119894 le 1Π1119894
(1199011198981198861199091198940 )2 + Σ1119894 (54)
and 1Π1119894
11990121198940 + Σ1119894 ge 1Π1119894
(1199011198981198941198991198940 )2 + Σ1119894 (55)
Mathematical Problems in Engineering 9
Now we replace 11990121198940 in its definition with (1199012
119894119896 + 119892119894119896(119894)) andbecause of 1199012
1198940 isin 1198752119894 for all 119894 = 1 1198982 + 1 it follows1Π1119894
(1199012119894119896 + 119892119894119896 (119894)) + Σ1119894
le 1Π1119894
min119896=1119899119894
[(119901119898119886119909119894119896 )2 + 119892119894119896 (119894)] + Σ1119894
(56)
and 1Π1119894
(1199012119894119896 + 119892119894119896 (119894)) + Σ1119894
ge 1Π1119894
max119896=1119899119894
[(119901119898119894119899119894119896 )2 + 119892119894119896 (119894)] + Σ1119894 (57)
So for all 119894 = 1 1198982 + 1 and for all 119896 = 1 119899119894 it holds119901119894119896 isin [119901119898119894119899119894119896 119901119898119886119909
119894119896 ] which is equivalent to
119901+ isin [119901+119898119894119899 119901+119898119886119909] (58)
The equation for mass conservation follows directly from thedefinition of 119902119894 (119894 = 1 1198982 + 1) For the pressure loss weuse
1199012119894 = 1119899119894
11990121198940 minus 119892119894 (119894) (59)
And multiply this equation from left with 119860119879119894 and use (36)
With this it follows the equation for momentum conserva-tion
(119860+119894 )119879 (119901+
119894 )2 = minusΦ 10038161003816100381610038161199021198941003816100381610038161003816 119902119894 (60)
The last equation missing is the compressor property For acontrol 119895 isin 1 1198982 we fix a vertex V1198940 (119894 isin 1 1198982 +1)with ℎ(119890119906119895) = V1198940 We use 1199012
1198940 = Π1119894(119901210 minus Σ1119894) and solve the
formula for 119901210 in the following way
119901210 = 1119906(1119894)1
[sdot sdot [ 1119906(1119894)119898lowast1199012(1119894)119899lowast0
+ 119892(1119894)119899lowastminus1119891(119890119906(1119894)119899lowastminus1
) ((1119894)119899lowastminus1)] sdot sdot]+ 119892(1119894)1119891(119890119906(1119894)1 )
((1119894)1) (61)
Consider wlog that the subgraphs along the path from V10to V1198940 are numbered chronological st if ℎ(119890119906119895) isin 119866119894 then119891(119890119906119895) isin 119866119894minus1 We use 1199012
119894minus10 = Π1119894minus1(119901210 minus Σ1119894minus1) to get the
same representation for 119901210 depending on 1199012
(1119894minus1)119899lowast0 whichis equivalent to 119901(1119894)119899lowastminus10 We equalize these representationsdepending on 1199012
1198940 and 1199012119894minus10 and we get
1119906(1119894)119898lowast1199012(1119894)119899lowast0 + 119892(1119894)119899lowastminus1119891(119890119906
(1119894)119899lowastminus1) ((1119894)119899lowastminus1)
= 1199012(1119894)119899lowastminus10 (62)
We chose 119894 isin 1 1198982 + 1 st 119906119895 is the last control on thepath from V10 to V1198940 so it is 119906(1119894)119898lowast = 119906119895 Further we have
1199012(1119894)119899lowast 0 = 1199012
11989401199012(1119894)119899lowastminus10 = 1199012
119894minus10119892(1119894)119899lowastminus1119891(119890119906(1119894)119899lowastminus1
) (119887(1119894)119899lowastminus1) = 119892119894minus1119891(119890119906119895 )(119887119894minus1)
(63)
and we know that 11990121198940 resp 1199012
119894minus10 minus 119892119894minus1119891(119890119906119895 )(119887119894minus1) are the
pressures at the head resp the foot of the compressor edge119890119906119895 So it follows 1119906119895
1199012ℎ(119890119906119895 )
= 1199012119891(119890119906119895 )
(64)
which is equivalent to the compressor property This com-pletes the proof
With this theorem we have another characterization ofthe feasible set The last step in this intersection is theadaption of the results to the spheric-radial decompositionAs mentioned before the main step is to get a representationof the sets119872119894 in step (2) of Algorithm 3
Because we assumed that the graph has only one entry wedefine the following set
119872 fl 119887 isin R119899ge0 | (minus1119879
119899119887 119887) isin (65)
Consider a sampled point V119894 isin S119899minus1 of step (1) of Algorithm 3(119894 = 1 119873) We identify the load vector 119887 with the affinelinear function
119887119894 (119903) fl 119903119871V119894 + 120583 = 119903120596119894 + 120583 with 120596119894 = 119871V119894 (66)
where 120583 is the mean value of the Gaussian distribution and 119871is such that 119871119871119879 = Σ for positive definite covariance matrixΣ Note that the index 119894 here is for the sampled vector V119894 notfor the 119894-th subgraph Because 119887119894 is defined only for the exitnodes it must hold 119887 ge 0 which leads to the definition of theregular range
119877119903119890119892119894 fl 119903 ge 0 | 119887119894 (119903) ge 0 supe 119872119894 (67)
So the feasible set is
119872119894 = 119903 isin 119877119903119890119892119894 | 119887119894 (119903) fulfills (3) (5) and (7) (68)
and with Theorem 5 it holds
119872119894 = 119903 isin 119877119903119890119892119894 | 119894 (119903) fulfills (25) minus (34) (69)
where 119894(119903) is from (18) Because 119894 depends on 119903 now thefunction 119892 depends on 119903 too More exactly 119892 is quadratic in119903 so 119892119896(119887119894(119903)) can be represented as follows
1199032[[119899sum
119895=1
Ψ119896119895( 119899sumℓ=1
119860minus1119895ℓ120596119894ℓ)2]]⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
119886119894
10 Mathematical Problems in Engineering
+ 119903[[2119899sum
119895=1
Ψ119896119895( 119899sumℓ=1
119860minus1119895ℓ120596119894ℓ)( 119899sum
ℓ=1
119860minus1119895ℓ120583ℓ)]]⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
119894
+ [[119899sum
119895=1
Ψ119896119895( 119899sumℓ=1
119860minus1119895ℓ120583ℓ)2]]⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
119888119894
(70)
HereΨ is the product of119860 andΦ So the inequations whichcharacterize119872119894 are also quadratic in 119903 because the minimaland maximal pressures in the inequalities only count to 119888119894Now we can write the feasible sets as follows
119872119894 = 119903 isin 119877119903119890119892119894 | 0 le 1198861198941199051199032 + 119887119894119905119903 + 119888119894119905 (71)
where 119905 is an index for the total number of inequalities in119872119894Now the regular range can be cut with the positive intervals ofthe inequalities and the result is a union of disjoint intervals
119872119894 = 119904119894⋃119896=1
[119886119894119896 119886119894119896] 119904119894 isin N (72)
Here 119904119894 is the number of disjoint intervals in which allinequalities are fulfilled and 119886119894119896 and 119886119894119896 are the intervalbounds they depend on the pressure bounds and the con-trols So the third step of Algorithm 3 can be executed With
P (119887 isin 119872) asymp 1119873 119873sum119894=1
119904119894sum119896=1
F120594 (119886119894119896) minusF120594 (119886119894119896) (73)
the probability for a random load vector to be feasible can becomputed
4 Existence of Optimal Solutions
In this section we want to have a look at some optimizationproblems We distinguish between problems with constantloads (without uncertainty on the demand) and problemswith random loads The latter leads us to optimizationproblems with so called probabilistic constraints or chanceconstraints (see [21]) There are two aims in optimizationMinimize the maximal pressure bounds and minimize thecontrols For minimizing the maximal pressure bounds wedefine the following set
119875 fl119899⋃
119894=0
[119901+119898119894119899119894 infin) (74)
in which we look for them Because the maximal pressuresare in proportion to the cost of pipes an objective functionfor a optimization problem could be 119888119879119901+119898119886119909 for a cost vector119888 isin R119899+1
gt0 By minimizing the controls we use the Euclideannorm of the controls as an objective function
Optimization with Constant Loads Consider the two opti-mization problems
min119901+119898119886119909isin119875
119888119879119901+119898119886119909
subject to 119887+ (119901+119898119886119909) isin 119872 (75)
andmin119906isinR1198982ge1
1199062subject to 119887+ (119906) isin (76)
Lemma 7 There exists a unique solution for (75)
Proof We have a look at the side constraint 119887+(119901+119898119886119909) isin Using Theorem 5 we can write (75) as
min119901+119898119886119909isin119875
119888119879119901+119898119886119909
subject to (25) minus (35) are fulfilled (77)
For 119894 isin 1 1198982 + 1 we write inequality (25) (and allremaining analogous) in the following way
0 le 119892119894119896 (119894) minus (1199011198981198941198991198940 )2 + (119901119898119886119909
119894119896 )2 forall119896 = 1 119899119894 (78)
So every inequality depends only on one upper pressurebound (but more inequalities may depend on the samebound) Because the inequalities are not strict the values119901119896 fl min 119901119898119886119909
119896 isin [0infin) | (25) minus (35) are fulfilled (79)
exist for all 119896 = 1 119899 + 1 and we can define a value119901+119898119886119909119900119901119905119896 fl max 119901119896 119901119898119894119899
119896 (80)
which is obviously a solution of (75) Moreover this solutionis unique because reducing one component of the upperbounds would either hurt an inequality or it would not fulfill119901+119898119886119909 ge 119901+119898119894119899 anymore
For the optimization problem (76) we can formulate anecessary and a sufficient condition for optimal solutions
Lemma8 Consider pressure bounds119901+119898119894119899 119901+119898119886119909 isin R119899+1 st119901+119898119894119899 lt 119901+119898119886119909 and a suitable vector 119887+ isin R119899+1 (ie + fulfills(25)ndash(27)) For all 119896 = 1 1198982 we set 119894(119896) and 119895(119896) such that119891(119890119896) isin V+
119894(119896) and ℎ(119890119896) isin V+119895(119896)
(a) If 119906 isin R1198982 is a solution of (76) it holds for all 119896 =1 1198982 (119901119898119894119899119894(119896)0)2 minus 119892119894(119896)119891(119890119896)
(119894(119896)) le (119901119898119886119909119895(119896)0)2 (81)
(b) If the inequalities
(119901119898119886119909119894(119896)0)2 minus 119892119894(119896)119891(119890119896)
(119894(119896)) ge (119901119898119894119899119895(119896)0)2
and (119901119898119894119899119894(119896)0)2 minus 119892119894(119896)119891(119890119896)
(119894(119896)) le (119901119898119886119909119895(119896)0)2 (82)
are fulfilled for all 119896 = 1 1198982 an optimal solution 119906 isin R1198982ge1
of (76) exists
Mathematical Problems in Engineering 11
Note that a vector 119887+ isin R119899+1 is suitable here if the vector+ is feasible for the subgraphs ie the inequalities (25)-(27)are fulfilled for +Proof (a) For this part we use a proof by contradictionAssume that
(119901119898119894119899119894(119896)0)2 minus 119892119894(119896)119891(119890119896)
(119894(119896)) gt (119901119898119886119909119895(119896)0)2 (83)
Because 119906119896 ge 1 for all 119896 = 1 1198982 it follows
(119901119898119894119899119894(119896)0)2 minus 119892119894(119896)119891(119890119896)
(119894(119896)) gt 1119906119896
(119901119898119886119909119895(119896)0)2 (84)
It it Σ119896lowast119894(119896) = 0 and Π119896lowast 119894(119896) = 1 because the subgraph withindex 119896lowast is equal to the subgraphwith index 119894(119896) And becausethe subgraphs 119894(119896) and 119895(119896) are neighbours (ie directlyconnected) it follows Π119896lowast 119895(119896) = 119906119896 and
Σ119896lowast119895(119896) = 119892(119896lowast119895(119896))1119891(119890119906(119896lowast119895(119896))1
) (119887(119896lowast119895(119896))1)= 119892119894(119896)119891(119890119896)
(119894(119896)) (85)
Thus we have a contradiction to (28) and 119906119896 cannot beoptimal
(b) As mentioned before the vector 119887+ is such that isfeasible for the subgraphs So because of the conditions for119896 isin 1 1198982 there exists a 119901119894(119896)0 isin [119901119898119894119899
119894(119896)0 119901119898119886119909119894(119896)0] and a119901119895(119896) isin [119901119898119894119899
119895(119896) 119901119898119886119909119895(119896) ] so that1199012119894(119896)0 minus 119892119894(119896)119891(119890119896)
(119894(119896)) ge 1199012119895(119896)0 (86)
With 1199012119894(119896)0 minus 119892119894(119896)119891(119890119896)
(119894(119896)) = 1199012119891(119890119896)
and 1199012119895(119896)0 = 1199012
ℎ(119890119896)it
follows
(119901ℎ(119890119896)119901119891(119890119896)
) ge 1 (87)
Therefore we can find a 119906 isin R1198982ge1 that the inequalities (28)-
(35) are fulfilled Now we can use theWeierstraszlig theorem andget the existence of an optimal solution
Note that the sufficient condition in Lemma 8 is com-paratively strong there may exist optimal solutions with lessstrong conditions eg the following
Corollary 9 If the graph has only one compressor edge ie|E119862| = 1 then it holds If 119906 isin Rge1 is an optimal solution of(76) 119906 is unique
This conclusion follows directly from the strict mono-tonicity of the objective function
Optimization with Random Loads In this section we con-sider the load vector to be random as we mentioned inSection 22 In the optimization problems we want to makesure that the feasibility of a random load vector is guaranteedfor a probability 120572 isin (0 1) This leads us to optimizationproblems with so called chance or probabilistic constraints
In general problems with probabilistic constraints have theform
min 119891 (119909 120585)subject to P (ℎ (119909 120585) le 0) ge 120572 (88)
with an objective function 119891 and a vector of uncertainty 120585There exist many works about probabilistic constraints respstochastic programming eg [7] gives an excellent overviewabout chance constraints in theory and application In ourcase we consider the following problem
min119901+119898119886119909isin119875
119888119879119901+119898119886119909
subject to P (119887 (119901+119898119886119909) isin 119872) ge 120572 (89)
To handle this problem resp the probabilistic constraint thespheric-radial decomposition gives us an explicit representa-tion of the constraint for a Gaussian vector 119887 sim N(120583 Σ) withmean value 120583 and positive definite covariance Σ
P (119887 (119901+119898119886119909) isin 119872)= int
S119899minus1120583120594 119903 ge 0 | 119903119871V + 120583 isin 119872119889120583120578 (V) (90)
Further Algorithm 3 gives a way to approximate this integralin an efficient way In Section 3 we showed a way to character-ize the feasible set119872 and how to adapt this characterization tothe spheric-radial decomposition This changes our problemwith probabilistic constraints (89) to
min119901+119898119886119909isin119875
119888119879119901+119898119886119909
subject to 1119873 119873sum119894=1
119904119905sum119905=1
F120594 (119886119894119905) minusF120594 (119886119894119905) ge 120572 (91)
Of course the interval bounds depend on the pressure boundsand the control For this problem we formulate the nextlemma
Lemma 10 Let a sampling V1 V119873 (from step (1) ofAlgorithm 3) and 120572 isin (0 1) be given For a suitable 120583 isin R119899
problem (91) has at least one solution 119901+119898119886119909 isin 119875Proof For the proof we fix a V isin V1 V119873 from the givensampling From (25) and (26) it follows for all 119896 = 1 119899119894with 119894 isin 1 1198982 + 10 le 119892119894119896 (119894 (119903)) + (119901119898119886119909
119894119896 )2 minus (1199011198981198941198991198940 )2
0 le minus119892119894119896 (119894 (119903)) + (1199011198981198861199091198940 )2 minus (119901119898119894119899
119894119896 )2 (92)
With a random vector 119887 sim N(120583 Σ)119892119894119896(119894(119903)) are quadraticfunctions in 119903 and the upper pressure bounds only influencethe constant part with a positive sign
0 le 1198861198941198961199032 + 119887119894119896119903 + 119888119894119896 + (119901119898119886119909119894119896 )2 minus (119901119898119894119899
1198940 )2 0 le minus1198861198941198961199032 minus 119887119894119896119903 minus 119888119894119896 + (119901119898119886119909
1198940 )2 minus (119901119898119894119899119894119896 )2 (93)
12 Mathematical Problems in Engineering
It follows that a rise in 119901119898119886119909119894119896 for all 119896 = 0 119899119894 enlarges
the intervals in which the inequalities hold The same holdsfor all inequalities inTheorem 5The cumulative distributionfunction of theChi-Square-Distribution is strictlymonotonicincreasing and convergent to 1 Thus upper pressure bounds119901119898119886119909119894119896 (for 119894 = 1 119899) can be found so that
1119873 119873sum119895=1
119904119895sum119905=1
F120594 (119886119895119905) minusF120594 (119886119895119905) (94)
is higher that 120572 isin (0 1) The index119873 is for all sampled pointson the sphere and the index 119904119895 is for the union of intervals inwhich all inequalities hold The interval boundaries 119886119895119905 and119886119895119905 depend continuously on the pressure bounds and becausethe inequalities are not strict 119901119898119886119909 can be found st the sideconstraint is fulfilled with equality So there exists at least onesolution of (91)
5 Numerical Results
In this section we show a few results of implementation Atfirst we show the idea of the spheric-radial decomposition byusing an easy example Next we show a easy example with onecompressor edge and at last we use real data of the Greek gasnetworkThe focus of the implementation is on the theoremsproofed in Section 4
Example 1 Our first example is an easy graph without innercontrol (see Figure 6)
Assume that node 1 and 2 are gas consumers with meandemand 120583 = [05 05]119879 The pressure bounds are given by119901+119898119894119899 = [2 1 1]119879 and 119901+119898119886119909 = [3 2 2]119879 and Φ = E2 Thenthe incidence matrix is
119860+ = [[[minus1 01 minus10 1
]]] resp 119860 = [1 minus10 1 ] (95)
Consider the inequalities of Theorem 5 to characterize thefeasible set Because the network has no compressor edgesthe system of inequalities only contains the information (25)-(27) the other inequalities do not occur The function 119892 isgiven by
119892 (119887) = [1 01 1] [1 00 1] sdot1003816100381610038161003816100381610038161003816100381610038161003816[1 10 1](11988711198872)
1003816100381610038161003816100381610038161003816100381610038161003816∘ ([1 10 1](11988711198872)) = ( 11988721 + 211988711198872 + 1198872211988721 + 211988711198872 + 211988722) (96)
By inserting the pressure bounds one can see that inequality(25) is always fulfilled and inequality (27) implies1198872 le radic3 (97)
With this inequality (26) follows for 0 le 1198872 le radic30 le 1198871 le minus1198872 + radic8 minus 11988722 (98)
So the feasible set119872 is the following (see Figure 7)
119872 = 119887 isin R2ge0 | 1198872 le radic3 and 1198871 le minus1198872 + radic8 minus 11988722 (99)
Now consider the optimization problem
min119901+119898119886119909isin119875
11198793119901+119898119886119909
subject to 119887 isin 119872 (119901+119898119886119909) (100)
with 119887 = 120583 = [05 05]119879 Lemma 7 guarantees the existenceof a unique solution of this problem We use the fmincon-function inMATLAB which minimizes a constrained non-linear multivariable function The fmincon default settinguses a interior-point method For more information we referto the official MathWorks homepage (httpswwwmath-workscomhelpoptimugfminconhtml)The solver returnsthe following result
119901+119898119886119909119900119901119905 = [2 radic3radic275]119879 (101)
One can easily see that this is the correct solution The gasdemand at node 2 is 05 which is equal to the flow 1199022 sousing the equation of momentum implies a difference in thequadratic pressures between node 1 and node 2 of 025 Theflow through edge 1 is given by the two exit nodes and it is1199021 = 1 This implies a difference in the quadratic pressuresbetweennode 0 andnode 1 of 1 So the optimal solution119901+119898119886119909
119900119901119905
directly followsNext we use the spheric-radial decomposition especially
Algorithm 3 to calculate the probability P(119887 isin 119872) for arandom vector 119887 sim N(E2 [05 05]119879) Therefore we viewthe results of 8 tests with 1000 sampled points in each one(Table 1)
The probability in all eight tests is nearly the same Themean value is 03479 and the variance is 27723 sdot 10minus6 whichis very small This shows that the implementation is workingnearly exact The efficiency of the implementation of courseis not perfect but that is not part of this work
Last wewant to solve the following optimization problem
min119901+119898119886119909isin119875
11198793119901+119898119886119909
subject to 1119873 119873sum119894=1
119904sum119905=1
F120594 (119886119894119905) minusF120594 (119886119894119905) ge 120572 (102)
We use again the fmincon-function inMATLAB to solve theproblem eight times with 1000 sampled points in every caseBecause Lemma 8 needs a suitable 120583 isin R2
ge0 we assume 119887 simN(01sdot1198642 [15 15])We set120572 fl 09 and the results are shownin Table 2
Example 2 Now we add a compressor edge to the graph ofExample 1 (see Figure 8)
Assume that node 1 and 3 are consumers node 2 is ainner nodeThemeandemand120583 = [05 0 05]119879Thepressure
Mathematical Problems in Engineering 13
minus2 minus1 1 2 3
minus2
minus1
1
2
3
b2
b1
minus1 1 2
minus1
1
2
b2
b1
rLv1
rLv2
rLv3
Scheme of the spheric-radial decomposition with
sphere (yellow)and three points 1 2 3 (red) at the unit
the feasible set Mminus for Σ = E2 = [05 05]TFeasible set M for p+min = [2 1 1]T and
p+max = [3 2 2]T
Figure 7 Feasible set of Example 1
Table 1 Test results of easy example
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 8P(119887 isin 119872) 03486 03481 03450 03501 03484 03499 03459 03475
Table 2 Test results of Example 1
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 81199011198981198861199091199001199011199050 54514 54714 54758 54028 54767 53921 54888 551671199011198981198861199091199001199011199051 29802 30126 30225 29725 30300 29485 30336 300291199011198981198861199091199001199011199052 10084 10004 10776 10001 11221 11390 10285 10558
0 1 2 3q1 q2 q3
b0 b1 b2 b3
e1 isin ℰF e2 isin ℰC e3 isin ℰF
Figure 8 Graph of Example 2
0 1 2 3q1 q2 q2 q3
b0 b1 b2 b3
e1 isin ℰF e3 isin ℰF
Figure 9 Separated graph of Example 2
bounds are given by 119901+119898119894119899 = [2 1 1 1]119879 and 119901+119898119886119909 = [3 22 2]119879 and 1206011198901= 1 and 1206011198903
= 1 The incidence matrix is givenby
119860+ = [[[[[[
minus1 0 01 minus1 00 1 minus10 0 1]]]]]]
resp 119860 = [[[1 minus1 00 1 minus10 0 1
]]] (103)
Before we have a look at the inequalities of Theorem 5we have to remove the compressor edge from the graphThe flows can be computed by using the equation of massconservation and the new load vector is then
= (1198871 + 11990221198872 minus 11990221198873 ) with
119902 = (1198871 + 1198872 + 11988731198872 + 11988731198873)(104)
We assume the control to be switched off so it holds119906 = 1With this the control edge 1198902 is treated as a flow edge withoutfriction The resulting graph is shown in Figure 9 from theinequalities (25)ndash(35) it follows
3 le radic3 and 1 le radic8 minus 23 (105)
14 Mathematical Problems in Engineering
minus1 1 2 3
1
2
3
minus1 1 2 3
1
2
3
b3
b1
Hyperspace of the feasible set MHyperspace of the feasible set M
b1
b3
Figure 10 Feasible set of the Example 2
Table 3 Test results of control example
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 8P(119887 isin 119872) 01531 01500 01540 01536 01530 01550 01567 01528so the feasible set is (see Figure 10)
= + isin R119899+1 | 1119879
4 + = 0 and 0 le 3 le radic3 and 2 = 0and 0 le 1 le radic8 minus 22 (106)
This implies
119872 = 119887 isin R3ge0 | 0 le 1198873 le radic3 and 1198872 = 0 and 0 le 1198871 le minus1198873
+ radic8 minus 11988723 (107)
Solving the problem
min119901+119898119886119909isin119875
11198794119901+119898119886119909
subject to isin (108)
for 119887 = 120583 = [05 0 05]119879 leads us to the optimal solution
119901+119898119886119909119900119901119905 = [2 radic3radic3radic275]119879 (109)
The argumentation for the optimality of this vector is exactlylike before in the easy example Becausewe have a compressoredge in this graph we can minimize the control Thereforewe fix the pressure bounds as we did at the beginning of thisexample and we set 119887 = [1 0 15]119879 So inequality (31) is notfulfilled anymore for 119906 = 1 We use again the fmincon-solverinMATLAB to solve the following problem
min119906ge1
1199062subject to 119887 isin 119872(119906) (110)
The solver returns the optimal solution 119906119900119901119905 = 11818
Here one can see that the sufficient condition in Lemma 8is really strong Itrsquos not fulfilled with the given values but thereexists a solution anyway Of course the necessary conditionis fulfilled
Now we use again Algorithm 3 to compute the proba-bility P(119887 isin 119872) for a random load vector 119887 sim N(E3 [0505 05]119879) Table 3 shows us the results
Again we can see that the probability is nearly the samein all eight cases The mean probability is 01535 and thevariance is 32369 sdot 10minus6 For the optimization we reducethe covariance matrix and assume 119887 sim N(001 sdot E3 [05 0505]119879) For a probability 120572 = 09 Table 4 states the resultsExample 3 In this example we use a part of a real networkThe focus here is on realistic values for our model Weget these values from the GasLib (see [22]) a collection oftechnical gas network data
ldquoThegoal ofGasLib is to promote research on gasnetworks by providing a set of large and realisticbenchmark instancesrdquo (httpgaslibzibde)
We use a part of the Greek gas network (GasLib-134) (seeFigure 11)
For using real values we first need to know the constant120601119890 from (5)
120601119890 = (119877119878119879)2 120582119863 1198711198862 (111)
The value 119877119878 is the specific gas constant which can becalculated by 119877119878 = 119877119872 with the ideal gas constant 119877 =831451119896119892 sdot 11989821199042 sdot 119898119900119897 sdot 119870 (see [23]) and the molar mass119872 = 1662119896119892119896119898119900119897 (from real data) of the used natural gasAlso the temperature 119879 = 28915119870 is given from real dataFor the (constant) friction factor 120582 we use the law of Chen(see [24])
Mathematical Problems in Engineering 15
Table 4 Test results of control example
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 81199011198981198861199091199001199011199050 39013 39223 39039 38901 38879 39039 38901 390981199011198981198861199091199001199011199051 16568 16565 16365 16497 16496 16365 16497 164921199011198981198861199091199001199011199052 16568 16565 16365 16497 16496 16365 16497 164981199011198981198861199091199001199011199053 10000 10000 10000 10000 10002 10000 10000 10000
Table 5 Values based on real data
variable 119877119878 119879 120582 119863 119871 119886 120601value 50027 28915 00093 9144 178693 43862 198 sdot 107unit 11989821199042 sdot 119870 119870 minus 119898119898 119896119898 119898119904 11989821199042
Figure 11 Greek gas network from GasLib [22] red edges are usedin our example
1radic120582 = minus2 log10 ( 11989611986337065minus 50425
Relog10 [(119896119863)1109828257 + 58506
Re08981]) (112)
Diameter 119863 = 9144119898119898 and roughness value 119896 = 8 sdot 10minus6119898for the first pipe (see Figure 11) are known from real data TheReynolds-Number Re can be calculated by Re = (V119863120588)120578with the dynamic viscosity 120578 = 1 sdot 10minus5119875119886 sdot 119904 (see [1])the mean velocity V = 9548119898119904 (calculated with the meanvolume flow rate which is given by the real data) of the gasand the density 120588 = 074331198961198921198983 (from real data) So we getRe = 648959074 asymp 649 sdot 106 and with this the pipe frictioncoefficient after Chen is 120582 = 00093
The length of the first pipe is given by 119871 = 178693119896119898from the real data Last value missing is the sound speed innatural gas which we assume to be ideal The sound speed
119886 depends on the modulus of compressibility (see [9]) Forideal gases the modulus of compressibility only depends onthe adiabatic exponent and the pressure Using the ideal gasequation simplifies the equation for the sound speed to
119886 = radic120581119877119879119872 (113)
The adiabatic exponent for methan which is the main partof natural gas (gt 90) is about 133 (see [25]) Together allvalues are shown in Table 5 (example for the pipe next to thesource in Figure 11)
Nowwewant to optimize the inner control in the networkfor a fix load vector Therefor we choose two nominations2011-11-01 and 2016-02-17 In both cases the fmincon-functioninMATLAB returns
119906119900119901119905 = 1 (114)
as a local minimum which satisfies the constraint 119887 isin 119872This means for these nominations a compressor stationis not needed so it is switched off With this as a lastcalculation in this paper we calculate the probability for thegiven nominations to be feasible We use the values from thenominations as mean value 120583 and set 119887 sim N(120583 sdot 10minus3E 120583)The results are shown in Tables 6 and 7
Again one can see that the results are quite similarThe mean probability in Table 6 is 09860 and the meanprobability in Table 7 is 09760 Both values are nearly 1 sothere must happen something really unforeseeable that aGreek does not get his demanded gas
6 Conclusion
In this paper we have used an existing stationary modelfor gas transport based on the isothermal Euler equationsWe have extended this model by compressor stations andpresented a characterization of the set of feasible loadsEspecially Theorem 5 contains the central idea to get anaccess to the optimization in Section 4 and the spheric-radialdecomposition is our main tool to handle the uncertainty inthe loadsWewere able to show some results for optimizationwith probabilistic constraints in gas transport In a next stepthese results should be generalized
16 Mathematical Problems in Engineering
Table 6 Test results for nomination 2011-11-01
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 8P(119887 isin 119872) 09849 09849 09878 09885 09863 09845 09854 09853
Table 7 Test results for nomination 2016-02-17
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 8P(119887 isin 119872) 09744 09756 09788 09788 09754 09775 09734 09739
With the characterization of the feasible set we haveshown that the feasibility of a load depends significantly onthe pressure bounds These pressure bounds play a mainrole in our optimization problems The other main partsare the inner controls modeled by compressor edges whichshould lead in optimal control problems with probabilisticconstraints Here we have built a base for further works withthis topic
The idea of separating the graph and removing thecompressor edges can be used to extend themodel with othercomponents like eg control valves and resistorsThiswouldchange the inequalities inTheorem 5 but the approach worksalso for thesemodifications Further [8] showed a way to dealwith a networkwhich is a cycle Combining bothmethods weget an access to model gas networks in high detail
Finally the expansion to a dynamic model might bealso possible The idea is to use a space-mapping approachbecause one can use a lot of the analysis we did hereTherefore upcoming papers should deal with these problemsand present results of the turnpike-theory which describesa relation between stationary and instationary solutions Infact turnpike-theory implies that the optimal dynamic stateis close to the optimal static state in the interior of the timeintervals for sufficiently long time horizons AJ Zaslavskiexplains the turnpike-phenomenon in [26] This book alsogives an overview about the theory
Data Availability
The data used to support the findings of this study areavailable from the corresponding author upon request
Conflicts of Interest
The authors declare that there are no conflicts of interestregarding the publication of this paper
Acknowledgments
This work was supported by DFG in the framework of theCollaborative Research Centre CRCTransregio 154 Math-ematical Modeling Simulation and Optimization Using theExample of Gas Networks project C03
References
[1] P Domschke B Hiller J Lang and C Tischendorf ldquoModel-lierung von gasnetzwerken Eine Ubersichtrdquo Tech Rep 2717Technische Universitat Darmstadt 2017
[2] T Koch B Hiller M E Pfetsch and L Schewe Evaluating GasNetwork Capacities MOS-SIAM 2015
[3] M Gugat R Schultz and D Wintergerst ldquoNetworks of pipe-lines for gas with nonconstant compressibility factor stationarystatesrdquoComputational amp AppliedMathematics vol 37 no 2 pp1066ndash1097 2018
[4] M Gugat and M Herty ldquoExistence of classical solutions andfeedback stabilization for the flow in gas networksrdquo ESAIMControl Optimisation and Calculus of Variations vol 17 no 1pp 28ndash51 2011
[5] M K Banda M Herty and A Klar ldquoGas flow in pipeline net-worksrdquo Networks and Heterogeneous Media vol 1 no 1 pp 41ndash56 2006
[6] RM Colombo G GuerraM Herty and V Schleper ldquoOptimalcontrol in networks of pipes and canalsrdquo SIAM Journal onControl and Optimization vol 48 no 3 pp 2032ndash2050 2009
[7] A Shapiro D Dentcheva and A Ruszczynski ectures onStochastic Programming Modeling and Theory MPS-SIAM2009
[8] C Gotzes H Heitsch R Henrion and R Schultz ldquoOn thequantification of nomination feasibility in stationary gas net-works with random loadrdquoMathematical Methods of OperationsResearch vol 84 no 2 pp 427ndash457 2016
[9] D Halliday R Resnick and J Walker Physik Wiley-VCHBachelor edition edition 2007
[10] M Gugat F M Hante M Hirsch-Dick and G LeugeringldquoStationary states in gas networksrdquoNetworks andHeterogeneousMedia vol 10 no 2 pp 295ndash320 2015
[11] D RoseM SchmidtM C Steinbach and BMWillert ldquoCom-putational optimization of gas compressor stations MINLPmodels versus continuous reformulationsrdquoMathematical Meth-ods of Operations Research vol 83 no 3 pp 409ndash444 2016
[12] H Heitsch R Henrion H Leovey et al ldquoEmpirical observa-tions and statistical analysis of gas demandrdquo in Evaluating GasNetwork Capacities T data B KochM E Hiller et al Eds pp273ndash290 MOS-SIAM 2015
[13] A Genz and F Bretz Computation of Multivariate Normal andt Probabilities vol 195 of Lecture Notes in Statistics Springer2009
[14] M H Farshbaf-Shaker R Henrion and D Homberg ldquoProp-erties of Chance Constraints in Infinite Dimensions with anApplication to PDE Constrained Optimizationrdquo Set-Valued andVariational Analysis pp 1ndash21 2017
[15] T Gonzalez Grandon H Heitsch and R Henrion ldquoA jointmodel of probabilisticrobust constraints for gas transportman-agement in stationary networksrdquo Computational ManagementScience vol 14 no 3 pp 443ndash460 2017
[16] W van Ackooij I Aleksovska and M Munoz-Zuniga ldquo(Sub-)Differentiability of Probability Functionswith Elliptical Distri-butionsrdquo Set-Valued and Variational Analysis
Mathematical Problems in Engineering 17
[17] W van Ackooij and R Henrion ldquoGradient formulae for nonlin-ear probabilistic constraints with GAUssian and GAUssian-likedistributionsrdquo SIAM Journal on Optimization vol 24 no 4 pp1864ndash1889 2014
[18] W van Ackooij and R Henrion ldquo(Sub-)gradient formulaefor probability functions of random inequality systems underGaussian distributionrdquo SIAMASA Journal on UncertaintyQuantification vol 5 no 1 pp 63ndash87 2017
[19] W van Ackooij and J Malick ldquoSecond-order differentiabilityof probability functionsrdquo Optimization Letters vol 11 no 1 pp179ndash194 2017
[20] H Eckey R Kosfeld and M Turck Wahrscheinlichkeitsrech-nung und Induktive Statistik - Grundlagen -Methoden - BeispieleGabler Wiesbaden Germany 2005
[21] A Prekopa Stochastic Programming Springer 1st edition 1995[22] I Joormann M Pfetsch R Schwarz et al ldquoGasLibmdashA Library
of Gas Network Instancesrdquo 2017[23] G Wedler Lehrbuch der Physikalischen Chemie Wiley-VCH
5th edition 2004[24] N H Chen ldquoAn explicit equation for friction factor in piperdquo
Industrial amp Engineering Chemistry Fundamentals vol 18 no3 pp 296-297 1979
[25] W Schenk and F Kremer Physikalisches Praktikum Springer14th edition 2014
[26] A J Zaslavski ldquoTurnpike Properties in the Calculus of Varia-tions and Optimal Controlrdquo inNonconvex Optimization and ItsApplications vol 80 Springer 2006
Hindawiwwwhindawicom Volume 2018
MathematicsJournal of
Hindawiwwwhindawicom Volume 2018
Mathematical Problems in Engineering
Applied MathematicsJournal of
Hindawiwwwhindawicom Volume 2018
Probability and StatisticsHindawiwwwhindawicom Volume 2018
Journal of
Hindawiwwwhindawicom Volume 2018
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawiwwwhindawicom Volume 2018
OptimizationJournal of
Hindawiwwwhindawicom Volume 2018
Hindawiwwwhindawicom Volume 2018
Engineering Mathematics
International Journal of
Hindawiwwwhindawicom Volume 2018
Operations ResearchAdvances in
Journal of
Hindawiwwwhindawicom Volume 2018
Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018
International Journal of Mathematics and Mathematical Sciences
Hindawiwwwhindawicom Volume 2018
Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom
The Scientific World Journal
Volume 2018
Hindawiwwwhindawicom Volume 2018Volume 2018
Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in
Nature and SocietyHindawiwwwhindawicom Volume 2018
Hindawiwwwhindawicom
Dierential EquationsInternational Journal of
Volume 2018
Hindawiwwwhindawicom Volume 2018
Decision SciencesAdvances in
Hindawiwwwhindawicom Volume 2018
AnalysisInternational Journal of
Hindawiwwwhindawicom Volume 2018
Stochastic AnalysisInternational Journal of
Submit your manuscripts atwwwhindawicom
8 Mathematical Problems in Engineering
We repeat this procedure for all controls on the path fromV119896lowast0 to V1198940 Although we write 119899lowast and119898lowast instead of 119899lowast119896lowast119894 and119898lowast
119896lowast119894 Thus we get
1199012(119896lowast119894)10 = 1119906(119896lowast119894)1
[ 1119906(119896lowast119894)2
[sdot sdot sdot [ 1119906(119896lowast119894)119898lowast1199012(119896lowast119894)119899lowast0
+ 119892(119896lowast119894)119899lowastminus1119891(119890119906(119896lowast119894)119899lowastminus1
) ((119896lowast119894)119899lowastminus1)] sdot sdot sdot]+ 119892(119896lowast119894)2119891(119890119906
(119896lowast119894)2) ((119896lowast119894)2)]
+ 119892(119896lowast119894)1119891(119890119906(119896lowast119894)1
) ((119896lowast119894)1)
(46)
We take again (42) for the node V119896lowast0 and do the sameprocedure just like before for the path from V119896lowast0 to V1198950 With119899lowast = 119899lowast119896lowast119895 and119898lowast = 119898lowast
119896lowast119895 we get
1199012(119896lowast 119895)10
= 1119906(119896lowast119895)1
[ 1119906(119896lowast119895)2
[sdot sdot sdot [ 1119906(119896lowast119895)119898lowast1199012(119896lowast 119895)119899lowast0
+ 119892(119896lowast119895)119899lowastminus1119891(119890119906(119896lowast119895)119899lowastminus1
) ((119896lowast119895)119899lowastminus1)] sdot sdot sdot]+ 119892(119896lowast119895)2119891(119890119906
(119896lowast119895)2) ((119896lowast119895)2)]
+ 119892(119896lowast119895)1119891(119890119906(119896lowast119895)1
) ((119896lowast119895)1)
(47)
Expanding (46) and (47) and equalizing them (this is possiblebecause 119901(119896lowast 119894)10 = 119901(119896lowast119895)10) leads to1Π119896lowast 119894
11990121198940 + Σ119896lowast119894 = 1Π119896lowast 119895
11990121198950 + Σ119896lowast119895 (48)
with Σ119896lowast119894 resp Σ119896lowast119895 defined before (48) directly implies (28)and (29) Using (48) together with (42) for V1198940 first timefor V1198950 second time and for V1198940 and V1198950 a third time theinequalities (30)ndash(35) follow directly So this part of the proofis complete
ldquolArr997904rdquo For this part we consider a vector 119887+ isin R119899+1
with 1119879119899+1119887+ = 0 which fulfills the inequalities (25)ndash(35) We
define the following sets (119894 = 2 1198982 + 1)11987511 fl [(119901119898119894119899
10 )2 (11990111989811988611990910 )2]
11987521 fl [ max119896=11198991
[(1199011198981198941198991119896 )2 + 1198921119896 (1)]
min119896=11198991
[(1199011198981198861199091119896 )2 + 1198921119896 (1)]]
1198751119894 fl [ 1Π1119894
(1199011198981198941198991198940 )2 + Σ1119894 1Π1198941
(1199011198981198861199091198940 )2 + Σ1119894]
1198752119894 fl [ 1Π1119894
max119896=1119899119894
[(119901119898119894119899119894119896 )2 + 119892119894119896 (119894)] + Σ1119894 1Π1198941
sdot min119896=1119899119894
[(119901119898119886119909119894119896 )2 + 119892119894119896 (119894)] + Σ1119894]
(49)
We set
1198750 fl 1198982+1⋂119896=1
1198751119896 cap 1198982+1⋂119896=1
1198752119896 (50)
and we will define a value 119901210 isin 1198750 later So to guarantee
such a value 119901210 exists we have to show first that 1198750 is not
empty Because 1198750 is a finite intersection of convex intervalswe have to make sure that all intervals are not empty andevery intersection between two intervals is not empty In onedimension this is sufficient that 1198750 is not empty For 119894 =1 1198982 + 11198751119894 = 0 according to requirement1198752119894 = 0 because of (26) (51)
Furthermore its 1198751119894 cap 1198752119894 = 0 because of (25) and (26) Themissing intersections are nonempty too for 119894 119895 isin 1 1198982+1 with 119894 lt 119895 as follows
1198751119894 cap 1198751119895 = 0 because of (28) and (29) 1198751119894 cap 1198752119895 = 0 because of (30) and (31) 1198752119894 cap 1198751119895 = 0 because of (32) and (33) 1198752119894 cap 1198752119895 = 0 because of (34) and (35)
(52)
So the set 1198750 is not empty We define the following values
119901210 isin 119875011990121198940 fl Π1119894 (1199012
10 minus Σ1119894) 119894 = 2 1198982 + 11199012119894119896 fl 1199012
1198940 minus 119892119894119896 (119894) 119894 = 1 1198982 + 1 119896 = 1 119899119894119902119894 fl 119860minus1119894 119894 119894 = 1 1198982 + 1
(53)
Now we show that our choice of pressures and flows isfeasible It is11990110 isin [119901119898119894119899
10 11990111989811988611990910 ] because of1199012
10 isin 11987511 Furtherbecause of 1199012
10 isin 1198751119894 for all 119894 = 2 1198982 + 1 it holds1Π1119894
11990121198940 + Σ1119894 le 1Π1119894
(1199011198981198861199091198940 )2 + Σ1119894 (54)
and 1Π1119894
11990121198940 + Σ1119894 ge 1Π1119894
(1199011198981198941198991198940 )2 + Σ1119894 (55)
Mathematical Problems in Engineering 9
Now we replace 11990121198940 in its definition with (1199012
119894119896 + 119892119894119896(119894)) andbecause of 1199012
1198940 isin 1198752119894 for all 119894 = 1 1198982 + 1 it follows1Π1119894
(1199012119894119896 + 119892119894119896 (119894)) + Σ1119894
le 1Π1119894
min119896=1119899119894
[(119901119898119886119909119894119896 )2 + 119892119894119896 (119894)] + Σ1119894
(56)
and 1Π1119894
(1199012119894119896 + 119892119894119896 (119894)) + Σ1119894
ge 1Π1119894
max119896=1119899119894
[(119901119898119894119899119894119896 )2 + 119892119894119896 (119894)] + Σ1119894 (57)
So for all 119894 = 1 1198982 + 1 and for all 119896 = 1 119899119894 it holds119901119894119896 isin [119901119898119894119899119894119896 119901119898119886119909
119894119896 ] which is equivalent to
119901+ isin [119901+119898119894119899 119901+119898119886119909] (58)
The equation for mass conservation follows directly from thedefinition of 119902119894 (119894 = 1 1198982 + 1) For the pressure loss weuse
1199012119894 = 1119899119894
11990121198940 minus 119892119894 (119894) (59)
And multiply this equation from left with 119860119879119894 and use (36)
With this it follows the equation for momentum conserva-tion
(119860+119894 )119879 (119901+
119894 )2 = minusΦ 10038161003816100381610038161199021198941003816100381610038161003816 119902119894 (60)
The last equation missing is the compressor property For acontrol 119895 isin 1 1198982 we fix a vertex V1198940 (119894 isin 1 1198982 +1)with ℎ(119890119906119895) = V1198940 We use 1199012
1198940 = Π1119894(119901210 minus Σ1119894) and solve the
formula for 119901210 in the following way
119901210 = 1119906(1119894)1
[sdot sdot [ 1119906(1119894)119898lowast1199012(1119894)119899lowast0
+ 119892(1119894)119899lowastminus1119891(119890119906(1119894)119899lowastminus1
) ((1119894)119899lowastminus1)] sdot sdot]+ 119892(1119894)1119891(119890119906(1119894)1 )
((1119894)1) (61)
Consider wlog that the subgraphs along the path from V10to V1198940 are numbered chronological st if ℎ(119890119906119895) isin 119866119894 then119891(119890119906119895) isin 119866119894minus1 We use 1199012
119894minus10 = Π1119894minus1(119901210 minus Σ1119894minus1) to get the
same representation for 119901210 depending on 1199012
(1119894minus1)119899lowast0 whichis equivalent to 119901(1119894)119899lowastminus10 We equalize these representationsdepending on 1199012
1198940 and 1199012119894minus10 and we get
1119906(1119894)119898lowast1199012(1119894)119899lowast0 + 119892(1119894)119899lowastminus1119891(119890119906
(1119894)119899lowastminus1) ((1119894)119899lowastminus1)
= 1199012(1119894)119899lowastminus10 (62)
We chose 119894 isin 1 1198982 + 1 st 119906119895 is the last control on thepath from V10 to V1198940 so it is 119906(1119894)119898lowast = 119906119895 Further we have
1199012(1119894)119899lowast 0 = 1199012
11989401199012(1119894)119899lowastminus10 = 1199012
119894minus10119892(1119894)119899lowastminus1119891(119890119906(1119894)119899lowastminus1
) (119887(1119894)119899lowastminus1) = 119892119894minus1119891(119890119906119895 )(119887119894minus1)
(63)
and we know that 11990121198940 resp 1199012
119894minus10 minus 119892119894minus1119891(119890119906119895 )(119887119894minus1) are the
pressures at the head resp the foot of the compressor edge119890119906119895 So it follows 1119906119895
1199012ℎ(119890119906119895 )
= 1199012119891(119890119906119895 )
(64)
which is equivalent to the compressor property This com-pletes the proof
With this theorem we have another characterization ofthe feasible set The last step in this intersection is theadaption of the results to the spheric-radial decompositionAs mentioned before the main step is to get a representationof the sets119872119894 in step (2) of Algorithm 3
Because we assumed that the graph has only one entry wedefine the following set
119872 fl 119887 isin R119899ge0 | (minus1119879
119899119887 119887) isin (65)
Consider a sampled point V119894 isin S119899minus1 of step (1) of Algorithm 3(119894 = 1 119873) We identify the load vector 119887 with the affinelinear function
119887119894 (119903) fl 119903119871V119894 + 120583 = 119903120596119894 + 120583 with 120596119894 = 119871V119894 (66)
where 120583 is the mean value of the Gaussian distribution and 119871is such that 119871119871119879 = Σ for positive definite covariance matrixΣ Note that the index 119894 here is for the sampled vector V119894 notfor the 119894-th subgraph Because 119887119894 is defined only for the exitnodes it must hold 119887 ge 0 which leads to the definition of theregular range
119877119903119890119892119894 fl 119903 ge 0 | 119887119894 (119903) ge 0 supe 119872119894 (67)
So the feasible set is
119872119894 = 119903 isin 119877119903119890119892119894 | 119887119894 (119903) fulfills (3) (5) and (7) (68)
and with Theorem 5 it holds
119872119894 = 119903 isin 119877119903119890119892119894 | 119894 (119903) fulfills (25) minus (34) (69)
where 119894(119903) is from (18) Because 119894 depends on 119903 now thefunction 119892 depends on 119903 too More exactly 119892 is quadratic in119903 so 119892119896(119887119894(119903)) can be represented as follows
1199032[[119899sum
119895=1
Ψ119896119895( 119899sumℓ=1
119860minus1119895ℓ120596119894ℓ)2]]⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
119886119894
10 Mathematical Problems in Engineering
+ 119903[[2119899sum
119895=1
Ψ119896119895( 119899sumℓ=1
119860minus1119895ℓ120596119894ℓ)( 119899sum
ℓ=1
119860minus1119895ℓ120583ℓ)]]⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
119894
+ [[119899sum
119895=1
Ψ119896119895( 119899sumℓ=1
119860minus1119895ℓ120583ℓ)2]]⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
119888119894
(70)
HereΨ is the product of119860 andΦ So the inequations whichcharacterize119872119894 are also quadratic in 119903 because the minimaland maximal pressures in the inequalities only count to 119888119894Now we can write the feasible sets as follows
119872119894 = 119903 isin 119877119903119890119892119894 | 0 le 1198861198941199051199032 + 119887119894119905119903 + 119888119894119905 (71)
where 119905 is an index for the total number of inequalities in119872119894Now the regular range can be cut with the positive intervals ofthe inequalities and the result is a union of disjoint intervals
119872119894 = 119904119894⋃119896=1
[119886119894119896 119886119894119896] 119904119894 isin N (72)
Here 119904119894 is the number of disjoint intervals in which allinequalities are fulfilled and 119886119894119896 and 119886119894119896 are the intervalbounds they depend on the pressure bounds and the con-trols So the third step of Algorithm 3 can be executed With
P (119887 isin 119872) asymp 1119873 119873sum119894=1
119904119894sum119896=1
F120594 (119886119894119896) minusF120594 (119886119894119896) (73)
the probability for a random load vector to be feasible can becomputed
4 Existence of Optimal Solutions
In this section we want to have a look at some optimizationproblems We distinguish between problems with constantloads (without uncertainty on the demand) and problemswith random loads The latter leads us to optimizationproblems with so called probabilistic constraints or chanceconstraints (see [21]) There are two aims in optimizationMinimize the maximal pressure bounds and minimize thecontrols For minimizing the maximal pressure bounds wedefine the following set
119875 fl119899⋃
119894=0
[119901+119898119894119899119894 infin) (74)
in which we look for them Because the maximal pressuresare in proportion to the cost of pipes an objective functionfor a optimization problem could be 119888119879119901+119898119886119909 for a cost vector119888 isin R119899+1
gt0 By minimizing the controls we use the Euclideannorm of the controls as an objective function
Optimization with Constant Loads Consider the two opti-mization problems
min119901+119898119886119909isin119875
119888119879119901+119898119886119909
subject to 119887+ (119901+119898119886119909) isin 119872 (75)
andmin119906isinR1198982ge1
1199062subject to 119887+ (119906) isin (76)
Lemma 7 There exists a unique solution for (75)
Proof We have a look at the side constraint 119887+(119901+119898119886119909) isin Using Theorem 5 we can write (75) as
min119901+119898119886119909isin119875
119888119879119901+119898119886119909
subject to (25) minus (35) are fulfilled (77)
For 119894 isin 1 1198982 + 1 we write inequality (25) (and allremaining analogous) in the following way
0 le 119892119894119896 (119894) minus (1199011198981198941198991198940 )2 + (119901119898119886119909
119894119896 )2 forall119896 = 1 119899119894 (78)
So every inequality depends only on one upper pressurebound (but more inequalities may depend on the samebound) Because the inequalities are not strict the values119901119896 fl min 119901119898119886119909
119896 isin [0infin) | (25) minus (35) are fulfilled (79)
exist for all 119896 = 1 119899 + 1 and we can define a value119901+119898119886119909119900119901119905119896 fl max 119901119896 119901119898119894119899
119896 (80)
which is obviously a solution of (75) Moreover this solutionis unique because reducing one component of the upperbounds would either hurt an inequality or it would not fulfill119901+119898119886119909 ge 119901+119898119894119899 anymore
For the optimization problem (76) we can formulate anecessary and a sufficient condition for optimal solutions
Lemma8 Consider pressure bounds119901+119898119894119899 119901+119898119886119909 isin R119899+1 st119901+119898119894119899 lt 119901+119898119886119909 and a suitable vector 119887+ isin R119899+1 (ie + fulfills(25)ndash(27)) For all 119896 = 1 1198982 we set 119894(119896) and 119895(119896) such that119891(119890119896) isin V+
119894(119896) and ℎ(119890119896) isin V+119895(119896)
(a) If 119906 isin R1198982 is a solution of (76) it holds for all 119896 =1 1198982 (119901119898119894119899119894(119896)0)2 minus 119892119894(119896)119891(119890119896)
(119894(119896)) le (119901119898119886119909119895(119896)0)2 (81)
(b) If the inequalities
(119901119898119886119909119894(119896)0)2 minus 119892119894(119896)119891(119890119896)
(119894(119896)) ge (119901119898119894119899119895(119896)0)2
and (119901119898119894119899119894(119896)0)2 minus 119892119894(119896)119891(119890119896)
(119894(119896)) le (119901119898119886119909119895(119896)0)2 (82)
are fulfilled for all 119896 = 1 1198982 an optimal solution 119906 isin R1198982ge1
of (76) exists
Mathematical Problems in Engineering 11
Note that a vector 119887+ isin R119899+1 is suitable here if the vector+ is feasible for the subgraphs ie the inequalities (25)-(27)are fulfilled for +Proof (a) For this part we use a proof by contradictionAssume that
(119901119898119894119899119894(119896)0)2 minus 119892119894(119896)119891(119890119896)
(119894(119896)) gt (119901119898119886119909119895(119896)0)2 (83)
Because 119906119896 ge 1 for all 119896 = 1 1198982 it follows
(119901119898119894119899119894(119896)0)2 minus 119892119894(119896)119891(119890119896)
(119894(119896)) gt 1119906119896
(119901119898119886119909119895(119896)0)2 (84)
It it Σ119896lowast119894(119896) = 0 and Π119896lowast 119894(119896) = 1 because the subgraph withindex 119896lowast is equal to the subgraphwith index 119894(119896) And becausethe subgraphs 119894(119896) and 119895(119896) are neighbours (ie directlyconnected) it follows Π119896lowast 119895(119896) = 119906119896 and
Σ119896lowast119895(119896) = 119892(119896lowast119895(119896))1119891(119890119906(119896lowast119895(119896))1
) (119887(119896lowast119895(119896))1)= 119892119894(119896)119891(119890119896)
(119894(119896)) (85)
Thus we have a contradiction to (28) and 119906119896 cannot beoptimal
(b) As mentioned before the vector 119887+ is such that isfeasible for the subgraphs So because of the conditions for119896 isin 1 1198982 there exists a 119901119894(119896)0 isin [119901119898119894119899
119894(119896)0 119901119898119886119909119894(119896)0] and a119901119895(119896) isin [119901119898119894119899
119895(119896) 119901119898119886119909119895(119896) ] so that1199012119894(119896)0 minus 119892119894(119896)119891(119890119896)
(119894(119896)) ge 1199012119895(119896)0 (86)
With 1199012119894(119896)0 minus 119892119894(119896)119891(119890119896)
(119894(119896)) = 1199012119891(119890119896)
and 1199012119895(119896)0 = 1199012
ℎ(119890119896)it
follows
(119901ℎ(119890119896)119901119891(119890119896)
) ge 1 (87)
Therefore we can find a 119906 isin R1198982ge1 that the inequalities (28)-
(35) are fulfilled Now we can use theWeierstraszlig theorem andget the existence of an optimal solution
Note that the sufficient condition in Lemma 8 is com-paratively strong there may exist optimal solutions with lessstrong conditions eg the following
Corollary 9 If the graph has only one compressor edge ie|E119862| = 1 then it holds If 119906 isin Rge1 is an optimal solution of(76) 119906 is unique
This conclusion follows directly from the strict mono-tonicity of the objective function
Optimization with Random Loads In this section we con-sider the load vector to be random as we mentioned inSection 22 In the optimization problems we want to makesure that the feasibility of a random load vector is guaranteedfor a probability 120572 isin (0 1) This leads us to optimizationproblems with so called chance or probabilistic constraints
In general problems with probabilistic constraints have theform
min 119891 (119909 120585)subject to P (ℎ (119909 120585) le 0) ge 120572 (88)
with an objective function 119891 and a vector of uncertainty 120585There exist many works about probabilistic constraints respstochastic programming eg [7] gives an excellent overviewabout chance constraints in theory and application In ourcase we consider the following problem
min119901+119898119886119909isin119875
119888119879119901+119898119886119909
subject to P (119887 (119901+119898119886119909) isin 119872) ge 120572 (89)
To handle this problem resp the probabilistic constraint thespheric-radial decomposition gives us an explicit representa-tion of the constraint for a Gaussian vector 119887 sim N(120583 Σ) withmean value 120583 and positive definite covariance Σ
P (119887 (119901+119898119886119909) isin 119872)= int
S119899minus1120583120594 119903 ge 0 | 119903119871V + 120583 isin 119872119889120583120578 (V) (90)
Further Algorithm 3 gives a way to approximate this integralin an efficient way In Section 3 we showed a way to character-ize the feasible set119872 and how to adapt this characterization tothe spheric-radial decomposition This changes our problemwith probabilistic constraints (89) to
min119901+119898119886119909isin119875
119888119879119901+119898119886119909
subject to 1119873 119873sum119894=1
119904119905sum119905=1
F120594 (119886119894119905) minusF120594 (119886119894119905) ge 120572 (91)
Of course the interval bounds depend on the pressure boundsand the control For this problem we formulate the nextlemma
Lemma 10 Let a sampling V1 V119873 (from step (1) ofAlgorithm 3) and 120572 isin (0 1) be given For a suitable 120583 isin R119899
problem (91) has at least one solution 119901+119898119886119909 isin 119875Proof For the proof we fix a V isin V1 V119873 from the givensampling From (25) and (26) it follows for all 119896 = 1 119899119894with 119894 isin 1 1198982 + 10 le 119892119894119896 (119894 (119903)) + (119901119898119886119909
119894119896 )2 minus (1199011198981198941198991198940 )2
0 le minus119892119894119896 (119894 (119903)) + (1199011198981198861199091198940 )2 minus (119901119898119894119899
119894119896 )2 (92)
With a random vector 119887 sim N(120583 Σ)119892119894119896(119894(119903)) are quadraticfunctions in 119903 and the upper pressure bounds only influencethe constant part with a positive sign
0 le 1198861198941198961199032 + 119887119894119896119903 + 119888119894119896 + (119901119898119886119909119894119896 )2 minus (119901119898119894119899
1198940 )2 0 le minus1198861198941198961199032 minus 119887119894119896119903 minus 119888119894119896 + (119901119898119886119909
1198940 )2 minus (119901119898119894119899119894119896 )2 (93)
12 Mathematical Problems in Engineering
It follows that a rise in 119901119898119886119909119894119896 for all 119896 = 0 119899119894 enlarges
the intervals in which the inequalities hold The same holdsfor all inequalities inTheorem 5The cumulative distributionfunction of theChi-Square-Distribution is strictlymonotonicincreasing and convergent to 1 Thus upper pressure bounds119901119898119886119909119894119896 (for 119894 = 1 119899) can be found so that
1119873 119873sum119895=1
119904119895sum119905=1
F120594 (119886119895119905) minusF120594 (119886119895119905) (94)
is higher that 120572 isin (0 1) The index119873 is for all sampled pointson the sphere and the index 119904119895 is for the union of intervals inwhich all inequalities hold The interval boundaries 119886119895119905 and119886119895119905 depend continuously on the pressure bounds and becausethe inequalities are not strict 119901119898119886119909 can be found st the sideconstraint is fulfilled with equality So there exists at least onesolution of (91)
5 Numerical Results
In this section we show a few results of implementation Atfirst we show the idea of the spheric-radial decomposition byusing an easy example Next we show a easy example with onecompressor edge and at last we use real data of the Greek gasnetworkThe focus of the implementation is on the theoremsproofed in Section 4
Example 1 Our first example is an easy graph without innercontrol (see Figure 6)
Assume that node 1 and 2 are gas consumers with meandemand 120583 = [05 05]119879 The pressure bounds are given by119901+119898119894119899 = [2 1 1]119879 and 119901+119898119886119909 = [3 2 2]119879 and Φ = E2 Thenthe incidence matrix is
119860+ = [[[minus1 01 minus10 1
]]] resp 119860 = [1 minus10 1 ] (95)
Consider the inequalities of Theorem 5 to characterize thefeasible set Because the network has no compressor edgesthe system of inequalities only contains the information (25)-(27) the other inequalities do not occur The function 119892 isgiven by
119892 (119887) = [1 01 1] [1 00 1] sdot1003816100381610038161003816100381610038161003816100381610038161003816[1 10 1](11988711198872)
1003816100381610038161003816100381610038161003816100381610038161003816∘ ([1 10 1](11988711198872)) = ( 11988721 + 211988711198872 + 1198872211988721 + 211988711198872 + 211988722) (96)
By inserting the pressure bounds one can see that inequality(25) is always fulfilled and inequality (27) implies1198872 le radic3 (97)
With this inequality (26) follows for 0 le 1198872 le radic30 le 1198871 le minus1198872 + radic8 minus 11988722 (98)
So the feasible set119872 is the following (see Figure 7)
119872 = 119887 isin R2ge0 | 1198872 le radic3 and 1198871 le minus1198872 + radic8 minus 11988722 (99)
Now consider the optimization problem
min119901+119898119886119909isin119875
11198793119901+119898119886119909
subject to 119887 isin 119872 (119901+119898119886119909) (100)
with 119887 = 120583 = [05 05]119879 Lemma 7 guarantees the existenceof a unique solution of this problem We use the fmincon-function inMATLAB which minimizes a constrained non-linear multivariable function The fmincon default settinguses a interior-point method For more information we referto the official MathWorks homepage (httpswwwmath-workscomhelpoptimugfminconhtml)The solver returnsthe following result
119901+119898119886119909119900119901119905 = [2 radic3radic275]119879 (101)
One can easily see that this is the correct solution The gasdemand at node 2 is 05 which is equal to the flow 1199022 sousing the equation of momentum implies a difference in thequadratic pressures between node 1 and node 2 of 025 Theflow through edge 1 is given by the two exit nodes and it is1199021 = 1 This implies a difference in the quadratic pressuresbetweennode 0 andnode 1 of 1 So the optimal solution119901+119898119886119909
119900119901119905
directly followsNext we use the spheric-radial decomposition especially
Algorithm 3 to calculate the probability P(119887 isin 119872) for arandom vector 119887 sim N(E2 [05 05]119879) Therefore we viewthe results of 8 tests with 1000 sampled points in each one(Table 1)
The probability in all eight tests is nearly the same Themean value is 03479 and the variance is 27723 sdot 10minus6 whichis very small This shows that the implementation is workingnearly exact The efficiency of the implementation of courseis not perfect but that is not part of this work
Last wewant to solve the following optimization problem
min119901+119898119886119909isin119875
11198793119901+119898119886119909
subject to 1119873 119873sum119894=1
119904sum119905=1
F120594 (119886119894119905) minusF120594 (119886119894119905) ge 120572 (102)
We use again the fmincon-function inMATLAB to solve theproblem eight times with 1000 sampled points in every caseBecause Lemma 8 needs a suitable 120583 isin R2
ge0 we assume 119887 simN(01sdot1198642 [15 15])We set120572 fl 09 and the results are shownin Table 2
Example 2 Now we add a compressor edge to the graph ofExample 1 (see Figure 8)
Assume that node 1 and 3 are consumers node 2 is ainner nodeThemeandemand120583 = [05 0 05]119879Thepressure
Mathematical Problems in Engineering 13
minus2 minus1 1 2 3
minus2
minus1
1
2
3
b2
b1
minus1 1 2
minus1
1
2
b2
b1
rLv1
rLv2
rLv3
Scheme of the spheric-radial decomposition with
sphere (yellow)and three points 1 2 3 (red) at the unit
the feasible set Mminus for Σ = E2 = [05 05]TFeasible set M for p+min = [2 1 1]T and
p+max = [3 2 2]T
Figure 7 Feasible set of Example 1
Table 1 Test results of easy example
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 8P(119887 isin 119872) 03486 03481 03450 03501 03484 03499 03459 03475
Table 2 Test results of Example 1
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 81199011198981198861199091199001199011199050 54514 54714 54758 54028 54767 53921 54888 551671199011198981198861199091199001199011199051 29802 30126 30225 29725 30300 29485 30336 300291199011198981198861199091199001199011199052 10084 10004 10776 10001 11221 11390 10285 10558
0 1 2 3q1 q2 q3
b0 b1 b2 b3
e1 isin ℰF e2 isin ℰC e3 isin ℰF
Figure 8 Graph of Example 2
0 1 2 3q1 q2 q2 q3
b0 b1 b2 b3
e1 isin ℰF e3 isin ℰF
Figure 9 Separated graph of Example 2
bounds are given by 119901+119898119894119899 = [2 1 1 1]119879 and 119901+119898119886119909 = [3 22 2]119879 and 1206011198901= 1 and 1206011198903
= 1 The incidence matrix is givenby
119860+ = [[[[[[
minus1 0 01 minus1 00 1 minus10 0 1]]]]]]
resp 119860 = [[[1 minus1 00 1 minus10 0 1
]]] (103)
Before we have a look at the inequalities of Theorem 5we have to remove the compressor edge from the graphThe flows can be computed by using the equation of massconservation and the new load vector is then
= (1198871 + 11990221198872 minus 11990221198873 ) with
119902 = (1198871 + 1198872 + 11988731198872 + 11988731198873)(104)
We assume the control to be switched off so it holds119906 = 1With this the control edge 1198902 is treated as a flow edge withoutfriction The resulting graph is shown in Figure 9 from theinequalities (25)ndash(35) it follows
3 le radic3 and 1 le radic8 minus 23 (105)
14 Mathematical Problems in Engineering
minus1 1 2 3
1
2
3
minus1 1 2 3
1
2
3
b3
b1
Hyperspace of the feasible set MHyperspace of the feasible set M
b1
b3
Figure 10 Feasible set of the Example 2
Table 3 Test results of control example
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 8P(119887 isin 119872) 01531 01500 01540 01536 01530 01550 01567 01528so the feasible set is (see Figure 10)
= + isin R119899+1 | 1119879
4 + = 0 and 0 le 3 le radic3 and 2 = 0and 0 le 1 le radic8 minus 22 (106)
This implies
119872 = 119887 isin R3ge0 | 0 le 1198873 le radic3 and 1198872 = 0 and 0 le 1198871 le minus1198873
+ radic8 minus 11988723 (107)
Solving the problem
min119901+119898119886119909isin119875
11198794119901+119898119886119909
subject to isin (108)
for 119887 = 120583 = [05 0 05]119879 leads us to the optimal solution
119901+119898119886119909119900119901119905 = [2 radic3radic3radic275]119879 (109)
The argumentation for the optimality of this vector is exactlylike before in the easy example Becausewe have a compressoredge in this graph we can minimize the control Thereforewe fix the pressure bounds as we did at the beginning of thisexample and we set 119887 = [1 0 15]119879 So inequality (31) is notfulfilled anymore for 119906 = 1 We use again the fmincon-solverinMATLAB to solve the following problem
min119906ge1
1199062subject to 119887 isin 119872(119906) (110)
The solver returns the optimal solution 119906119900119901119905 = 11818
Here one can see that the sufficient condition in Lemma 8is really strong Itrsquos not fulfilled with the given values but thereexists a solution anyway Of course the necessary conditionis fulfilled
Now we use again Algorithm 3 to compute the proba-bility P(119887 isin 119872) for a random load vector 119887 sim N(E3 [0505 05]119879) Table 3 shows us the results
Again we can see that the probability is nearly the samein all eight cases The mean probability is 01535 and thevariance is 32369 sdot 10minus6 For the optimization we reducethe covariance matrix and assume 119887 sim N(001 sdot E3 [05 0505]119879) For a probability 120572 = 09 Table 4 states the resultsExample 3 In this example we use a part of a real networkThe focus here is on realistic values for our model Weget these values from the GasLib (see [22]) a collection oftechnical gas network data
ldquoThegoal ofGasLib is to promote research on gasnetworks by providing a set of large and realisticbenchmark instancesrdquo (httpgaslibzibde)
We use a part of the Greek gas network (GasLib-134) (seeFigure 11)
For using real values we first need to know the constant120601119890 from (5)
120601119890 = (119877119878119879)2 120582119863 1198711198862 (111)
The value 119877119878 is the specific gas constant which can becalculated by 119877119878 = 119877119872 with the ideal gas constant 119877 =831451119896119892 sdot 11989821199042 sdot 119898119900119897 sdot 119870 (see [23]) and the molar mass119872 = 1662119896119892119896119898119900119897 (from real data) of the used natural gasAlso the temperature 119879 = 28915119870 is given from real dataFor the (constant) friction factor 120582 we use the law of Chen(see [24])
Mathematical Problems in Engineering 15
Table 4 Test results of control example
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 81199011198981198861199091199001199011199050 39013 39223 39039 38901 38879 39039 38901 390981199011198981198861199091199001199011199051 16568 16565 16365 16497 16496 16365 16497 164921199011198981198861199091199001199011199052 16568 16565 16365 16497 16496 16365 16497 164981199011198981198861199091199001199011199053 10000 10000 10000 10000 10002 10000 10000 10000
Table 5 Values based on real data
variable 119877119878 119879 120582 119863 119871 119886 120601value 50027 28915 00093 9144 178693 43862 198 sdot 107unit 11989821199042 sdot 119870 119870 minus 119898119898 119896119898 119898119904 11989821199042
Figure 11 Greek gas network from GasLib [22] red edges are usedin our example
1radic120582 = minus2 log10 ( 11989611986337065minus 50425
Relog10 [(119896119863)1109828257 + 58506
Re08981]) (112)
Diameter 119863 = 9144119898119898 and roughness value 119896 = 8 sdot 10minus6119898for the first pipe (see Figure 11) are known from real data TheReynolds-Number Re can be calculated by Re = (V119863120588)120578with the dynamic viscosity 120578 = 1 sdot 10minus5119875119886 sdot 119904 (see [1])the mean velocity V = 9548119898119904 (calculated with the meanvolume flow rate which is given by the real data) of the gasand the density 120588 = 074331198961198921198983 (from real data) So we getRe = 648959074 asymp 649 sdot 106 and with this the pipe frictioncoefficient after Chen is 120582 = 00093
The length of the first pipe is given by 119871 = 178693119896119898from the real data Last value missing is the sound speed innatural gas which we assume to be ideal The sound speed
119886 depends on the modulus of compressibility (see [9]) Forideal gases the modulus of compressibility only depends onthe adiabatic exponent and the pressure Using the ideal gasequation simplifies the equation for the sound speed to
119886 = radic120581119877119879119872 (113)
The adiabatic exponent for methan which is the main partof natural gas (gt 90) is about 133 (see [25]) Together allvalues are shown in Table 5 (example for the pipe next to thesource in Figure 11)
Nowwewant to optimize the inner control in the networkfor a fix load vector Therefor we choose two nominations2011-11-01 and 2016-02-17 In both cases the fmincon-functioninMATLAB returns
119906119900119901119905 = 1 (114)
as a local minimum which satisfies the constraint 119887 isin 119872This means for these nominations a compressor stationis not needed so it is switched off With this as a lastcalculation in this paper we calculate the probability for thegiven nominations to be feasible We use the values from thenominations as mean value 120583 and set 119887 sim N(120583 sdot 10minus3E 120583)The results are shown in Tables 6 and 7
Again one can see that the results are quite similarThe mean probability in Table 6 is 09860 and the meanprobability in Table 7 is 09760 Both values are nearly 1 sothere must happen something really unforeseeable that aGreek does not get his demanded gas
6 Conclusion
In this paper we have used an existing stationary modelfor gas transport based on the isothermal Euler equationsWe have extended this model by compressor stations andpresented a characterization of the set of feasible loadsEspecially Theorem 5 contains the central idea to get anaccess to the optimization in Section 4 and the spheric-radialdecomposition is our main tool to handle the uncertainty inthe loadsWewere able to show some results for optimizationwith probabilistic constraints in gas transport In a next stepthese results should be generalized
16 Mathematical Problems in Engineering
Table 6 Test results for nomination 2011-11-01
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 8P(119887 isin 119872) 09849 09849 09878 09885 09863 09845 09854 09853
Table 7 Test results for nomination 2016-02-17
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 8P(119887 isin 119872) 09744 09756 09788 09788 09754 09775 09734 09739
With the characterization of the feasible set we haveshown that the feasibility of a load depends significantly onthe pressure bounds These pressure bounds play a mainrole in our optimization problems The other main partsare the inner controls modeled by compressor edges whichshould lead in optimal control problems with probabilisticconstraints Here we have built a base for further works withthis topic
The idea of separating the graph and removing thecompressor edges can be used to extend themodel with othercomponents like eg control valves and resistorsThiswouldchange the inequalities inTheorem 5 but the approach worksalso for thesemodifications Further [8] showed a way to dealwith a networkwhich is a cycle Combining bothmethods weget an access to model gas networks in high detail
Finally the expansion to a dynamic model might bealso possible The idea is to use a space-mapping approachbecause one can use a lot of the analysis we did hereTherefore upcoming papers should deal with these problemsand present results of the turnpike-theory which describesa relation between stationary and instationary solutions Infact turnpike-theory implies that the optimal dynamic stateis close to the optimal static state in the interior of the timeintervals for sufficiently long time horizons AJ Zaslavskiexplains the turnpike-phenomenon in [26] This book alsogives an overview about the theory
Data Availability
The data used to support the findings of this study areavailable from the corresponding author upon request
Conflicts of Interest
The authors declare that there are no conflicts of interestregarding the publication of this paper
Acknowledgments
This work was supported by DFG in the framework of theCollaborative Research Centre CRCTransregio 154 Math-ematical Modeling Simulation and Optimization Using theExample of Gas Networks project C03
References
[1] P Domschke B Hiller J Lang and C Tischendorf ldquoModel-lierung von gasnetzwerken Eine Ubersichtrdquo Tech Rep 2717Technische Universitat Darmstadt 2017
[2] T Koch B Hiller M E Pfetsch and L Schewe Evaluating GasNetwork Capacities MOS-SIAM 2015
[3] M Gugat R Schultz and D Wintergerst ldquoNetworks of pipe-lines for gas with nonconstant compressibility factor stationarystatesrdquoComputational amp AppliedMathematics vol 37 no 2 pp1066ndash1097 2018
[4] M Gugat and M Herty ldquoExistence of classical solutions andfeedback stabilization for the flow in gas networksrdquo ESAIMControl Optimisation and Calculus of Variations vol 17 no 1pp 28ndash51 2011
[5] M K Banda M Herty and A Klar ldquoGas flow in pipeline net-worksrdquo Networks and Heterogeneous Media vol 1 no 1 pp 41ndash56 2006
[6] RM Colombo G GuerraM Herty and V Schleper ldquoOptimalcontrol in networks of pipes and canalsrdquo SIAM Journal onControl and Optimization vol 48 no 3 pp 2032ndash2050 2009
[7] A Shapiro D Dentcheva and A Ruszczynski ectures onStochastic Programming Modeling and Theory MPS-SIAM2009
[8] C Gotzes H Heitsch R Henrion and R Schultz ldquoOn thequantification of nomination feasibility in stationary gas net-works with random loadrdquoMathematical Methods of OperationsResearch vol 84 no 2 pp 427ndash457 2016
[9] D Halliday R Resnick and J Walker Physik Wiley-VCHBachelor edition edition 2007
[10] M Gugat F M Hante M Hirsch-Dick and G LeugeringldquoStationary states in gas networksrdquoNetworks andHeterogeneousMedia vol 10 no 2 pp 295ndash320 2015
[11] D RoseM SchmidtM C Steinbach and BMWillert ldquoCom-putational optimization of gas compressor stations MINLPmodels versus continuous reformulationsrdquoMathematical Meth-ods of Operations Research vol 83 no 3 pp 409ndash444 2016
[12] H Heitsch R Henrion H Leovey et al ldquoEmpirical observa-tions and statistical analysis of gas demandrdquo in Evaluating GasNetwork Capacities T data B KochM E Hiller et al Eds pp273ndash290 MOS-SIAM 2015
[13] A Genz and F Bretz Computation of Multivariate Normal andt Probabilities vol 195 of Lecture Notes in Statistics Springer2009
[14] M H Farshbaf-Shaker R Henrion and D Homberg ldquoProp-erties of Chance Constraints in Infinite Dimensions with anApplication to PDE Constrained Optimizationrdquo Set-Valued andVariational Analysis pp 1ndash21 2017
[15] T Gonzalez Grandon H Heitsch and R Henrion ldquoA jointmodel of probabilisticrobust constraints for gas transportman-agement in stationary networksrdquo Computational ManagementScience vol 14 no 3 pp 443ndash460 2017
[16] W van Ackooij I Aleksovska and M Munoz-Zuniga ldquo(Sub-)Differentiability of Probability Functionswith Elliptical Distri-butionsrdquo Set-Valued and Variational Analysis
Mathematical Problems in Engineering 17
[17] W van Ackooij and R Henrion ldquoGradient formulae for nonlin-ear probabilistic constraints with GAUssian and GAUssian-likedistributionsrdquo SIAM Journal on Optimization vol 24 no 4 pp1864ndash1889 2014
[18] W van Ackooij and R Henrion ldquo(Sub-)gradient formulaefor probability functions of random inequality systems underGaussian distributionrdquo SIAMASA Journal on UncertaintyQuantification vol 5 no 1 pp 63ndash87 2017
[19] W van Ackooij and J Malick ldquoSecond-order differentiabilityof probability functionsrdquo Optimization Letters vol 11 no 1 pp179ndash194 2017
[20] H Eckey R Kosfeld and M Turck Wahrscheinlichkeitsrech-nung und Induktive Statistik - Grundlagen -Methoden - BeispieleGabler Wiesbaden Germany 2005
[21] A Prekopa Stochastic Programming Springer 1st edition 1995[22] I Joormann M Pfetsch R Schwarz et al ldquoGasLibmdashA Library
of Gas Network Instancesrdquo 2017[23] G Wedler Lehrbuch der Physikalischen Chemie Wiley-VCH
5th edition 2004[24] N H Chen ldquoAn explicit equation for friction factor in piperdquo
Industrial amp Engineering Chemistry Fundamentals vol 18 no3 pp 296-297 1979
[25] W Schenk and F Kremer Physikalisches Praktikum Springer14th edition 2014
[26] A J Zaslavski ldquoTurnpike Properties in the Calculus of Varia-tions and Optimal Controlrdquo inNonconvex Optimization and ItsApplications vol 80 Springer 2006
Hindawiwwwhindawicom Volume 2018
MathematicsJournal of
Hindawiwwwhindawicom Volume 2018
Mathematical Problems in Engineering
Applied MathematicsJournal of
Hindawiwwwhindawicom Volume 2018
Probability and StatisticsHindawiwwwhindawicom Volume 2018
Journal of
Hindawiwwwhindawicom Volume 2018
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawiwwwhindawicom Volume 2018
OptimizationJournal of
Hindawiwwwhindawicom Volume 2018
Hindawiwwwhindawicom Volume 2018
Engineering Mathematics
International Journal of
Hindawiwwwhindawicom Volume 2018
Operations ResearchAdvances in
Journal of
Hindawiwwwhindawicom Volume 2018
Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018
International Journal of Mathematics and Mathematical Sciences
Hindawiwwwhindawicom Volume 2018
Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom
The Scientific World Journal
Volume 2018
Hindawiwwwhindawicom Volume 2018Volume 2018
Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in
Nature and SocietyHindawiwwwhindawicom Volume 2018
Hindawiwwwhindawicom
Dierential EquationsInternational Journal of
Volume 2018
Hindawiwwwhindawicom Volume 2018
Decision SciencesAdvances in
Hindawiwwwhindawicom Volume 2018
AnalysisInternational Journal of
Hindawiwwwhindawicom Volume 2018
Stochastic AnalysisInternational Journal of
Submit your manuscripts atwwwhindawicom
Mathematical Problems in Engineering 9
Now we replace 11990121198940 in its definition with (1199012
119894119896 + 119892119894119896(119894)) andbecause of 1199012
1198940 isin 1198752119894 for all 119894 = 1 1198982 + 1 it follows1Π1119894
(1199012119894119896 + 119892119894119896 (119894)) + Σ1119894
le 1Π1119894
min119896=1119899119894
[(119901119898119886119909119894119896 )2 + 119892119894119896 (119894)] + Σ1119894
(56)
and 1Π1119894
(1199012119894119896 + 119892119894119896 (119894)) + Σ1119894
ge 1Π1119894
max119896=1119899119894
[(119901119898119894119899119894119896 )2 + 119892119894119896 (119894)] + Σ1119894 (57)
So for all 119894 = 1 1198982 + 1 and for all 119896 = 1 119899119894 it holds119901119894119896 isin [119901119898119894119899119894119896 119901119898119886119909
119894119896 ] which is equivalent to
119901+ isin [119901+119898119894119899 119901+119898119886119909] (58)
The equation for mass conservation follows directly from thedefinition of 119902119894 (119894 = 1 1198982 + 1) For the pressure loss weuse
1199012119894 = 1119899119894
11990121198940 minus 119892119894 (119894) (59)
And multiply this equation from left with 119860119879119894 and use (36)
With this it follows the equation for momentum conserva-tion
(119860+119894 )119879 (119901+
119894 )2 = minusΦ 10038161003816100381610038161199021198941003816100381610038161003816 119902119894 (60)
The last equation missing is the compressor property For acontrol 119895 isin 1 1198982 we fix a vertex V1198940 (119894 isin 1 1198982 +1)with ℎ(119890119906119895) = V1198940 We use 1199012
1198940 = Π1119894(119901210 minus Σ1119894) and solve the
formula for 119901210 in the following way
119901210 = 1119906(1119894)1
[sdot sdot [ 1119906(1119894)119898lowast1199012(1119894)119899lowast0
+ 119892(1119894)119899lowastminus1119891(119890119906(1119894)119899lowastminus1
) ((1119894)119899lowastminus1)] sdot sdot]+ 119892(1119894)1119891(119890119906(1119894)1 )
((1119894)1) (61)
Consider wlog that the subgraphs along the path from V10to V1198940 are numbered chronological st if ℎ(119890119906119895) isin 119866119894 then119891(119890119906119895) isin 119866119894minus1 We use 1199012
119894minus10 = Π1119894minus1(119901210 minus Σ1119894minus1) to get the
same representation for 119901210 depending on 1199012
(1119894minus1)119899lowast0 whichis equivalent to 119901(1119894)119899lowastminus10 We equalize these representationsdepending on 1199012
1198940 and 1199012119894minus10 and we get
1119906(1119894)119898lowast1199012(1119894)119899lowast0 + 119892(1119894)119899lowastminus1119891(119890119906
(1119894)119899lowastminus1) ((1119894)119899lowastminus1)
= 1199012(1119894)119899lowastminus10 (62)
We chose 119894 isin 1 1198982 + 1 st 119906119895 is the last control on thepath from V10 to V1198940 so it is 119906(1119894)119898lowast = 119906119895 Further we have
1199012(1119894)119899lowast 0 = 1199012
11989401199012(1119894)119899lowastminus10 = 1199012
119894minus10119892(1119894)119899lowastminus1119891(119890119906(1119894)119899lowastminus1
) (119887(1119894)119899lowastminus1) = 119892119894minus1119891(119890119906119895 )(119887119894minus1)
(63)
and we know that 11990121198940 resp 1199012
119894minus10 minus 119892119894minus1119891(119890119906119895 )(119887119894minus1) are the
pressures at the head resp the foot of the compressor edge119890119906119895 So it follows 1119906119895
1199012ℎ(119890119906119895 )
= 1199012119891(119890119906119895 )
(64)
which is equivalent to the compressor property This com-pletes the proof
With this theorem we have another characterization ofthe feasible set The last step in this intersection is theadaption of the results to the spheric-radial decompositionAs mentioned before the main step is to get a representationof the sets119872119894 in step (2) of Algorithm 3
Because we assumed that the graph has only one entry wedefine the following set
119872 fl 119887 isin R119899ge0 | (minus1119879
119899119887 119887) isin (65)
Consider a sampled point V119894 isin S119899minus1 of step (1) of Algorithm 3(119894 = 1 119873) We identify the load vector 119887 with the affinelinear function
119887119894 (119903) fl 119903119871V119894 + 120583 = 119903120596119894 + 120583 with 120596119894 = 119871V119894 (66)
where 120583 is the mean value of the Gaussian distribution and 119871is such that 119871119871119879 = Σ for positive definite covariance matrixΣ Note that the index 119894 here is for the sampled vector V119894 notfor the 119894-th subgraph Because 119887119894 is defined only for the exitnodes it must hold 119887 ge 0 which leads to the definition of theregular range
119877119903119890119892119894 fl 119903 ge 0 | 119887119894 (119903) ge 0 supe 119872119894 (67)
So the feasible set is
119872119894 = 119903 isin 119877119903119890119892119894 | 119887119894 (119903) fulfills (3) (5) and (7) (68)
and with Theorem 5 it holds
119872119894 = 119903 isin 119877119903119890119892119894 | 119894 (119903) fulfills (25) minus (34) (69)
where 119894(119903) is from (18) Because 119894 depends on 119903 now thefunction 119892 depends on 119903 too More exactly 119892 is quadratic in119903 so 119892119896(119887119894(119903)) can be represented as follows
1199032[[119899sum
119895=1
Ψ119896119895( 119899sumℓ=1
119860minus1119895ℓ120596119894ℓ)2]]⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
119886119894
10 Mathematical Problems in Engineering
+ 119903[[2119899sum
119895=1
Ψ119896119895( 119899sumℓ=1
119860minus1119895ℓ120596119894ℓ)( 119899sum
ℓ=1
119860minus1119895ℓ120583ℓ)]]⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
119894
+ [[119899sum
119895=1
Ψ119896119895( 119899sumℓ=1
119860minus1119895ℓ120583ℓ)2]]⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
119888119894
(70)
HereΨ is the product of119860 andΦ So the inequations whichcharacterize119872119894 are also quadratic in 119903 because the minimaland maximal pressures in the inequalities only count to 119888119894Now we can write the feasible sets as follows
119872119894 = 119903 isin 119877119903119890119892119894 | 0 le 1198861198941199051199032 + 119887119894119905119903 + 119888119894119905 (71)
where 119905 is an index for the total number of inequalities in119872119894Now the regular range can be cut with the positive intervals ofthe inequalities and the result is a union of disjoint intervals
119872119894 = 119904119894⋃119896=1
[119886119894119896 119886119894119896] 119904119894 isin N (72)
Here 119904119894 is the number of disjoint intervals in which allinequalities are fulfilled and 119886119894119896 and 119886119894119896 are the intervalbounds they depend on the pressure bounds and the con-trols So the third step of Algorithm 3 can be executed With
P (119887 isin 119872) asymp 1119873 119873sum119894=1
119904119894sum119896=1
F120594 (119886119894119896) minusF120594 (119886119894119896) (73)
the probability for a random load vector to be feasible can becomputed
4 Existence of Optimal Solutions
In this section we want to have a look at some optimizationproblems We distinguish between problems with constantloads (without uncertainty on the demand) and problemswith random loads The latter leads us to optimizationproblems with so called probabilistic constraints or chanceconstraints (see [21]) There are two aims in optimizationMinimize the maximal pressure bounds and minimize thecontrols For minimizing the maximal pressure bounds wedefine the following set
119875 fl119899⋃
119894=0
[119901+119898119894119899119894 infin) (74)
in which we look for them Because the maximal pressuresare in proportion to the cost of pipes an objective functionfor a optimization problem could be 119888119879119901+119898119886119909 for a cost vector119888 isin R119899+1
gt0 By minimizing the controls we use the Euclideannorm of the controls as an objective function
Optimization with Constant Loads Consider the two opti-mization problems
min119901+119898119886119909isin119875
119888119879119901+119898119886119909
subject to 119887+ (119901+119898119886119909) isin 119872 (75)
andmin119906isinR1198982ge1
1199062subject to 119887+ (119906) isin (76)
Lemma 7 There exists a unique solution for (75)
Proof We have a look at the side constraint 119887+(119901+119898119886119909) isin Using Theorem 5 we can write (75) as
min119901+119898119886119909isin119875
119888119879119901+119898119886119909
subject to (25) minus (35) are fulfilled (77)
For 119894 isin 1 1198982 + 1 we write inequality (25) (and allremaining analogous) in the following way
0 le 119892119894119896 (119894) minus (1199011198981198941198991198940 )2 + (119901119898119886119909
119894119896 )2 forall119896 = 1 119899119894 (78)
So every inequality depends only on one upper pressurebound (but more inequalities may depend on the samebound) Because the inequalities are not strict the values119901119896 fl min 119901119898119886119909
119896 isin [0infin) | (25) minus (35) are fulfilled (79)
exist for all 119896 = 1 119899 + 1 and we can define a value119901+119898119886119909119900119901119905119896 fl max 119901119896 119901119898119894119899
119896 (80)
which is obviously a solution of (75) Moreover this solutionis unique because reducing one component of the upperbounds would either hurt an inequality or it would not fulfill119901+119898119886119909 ge 119901+119898119894119899 anymore
For the optimization problem (76) we can formulate anecessary and a sufficient condition for optimal solutions
Lemma8 Consider pressure bounds119901+119898119894119899 119901+119898119886119909 isin R119899+1 st119901+119898119894119899 lt 119901+119898119886119909 and a suitable vector 119887+ isin R119899+1 (ie + fulfills(25)ndash(27)) For all 119896 = 1 1198982 we set 119894(119896) and 119895(119896) such that119891(119890119896) isin V+
119894(119896) and ℎ(119890119896) isin V+119895(119896)
(a) If 119906 isin R1198982 is a solution of (76) it holds for all 119896 =1 1198982 (119901119898119894119899119894(119896)0)2 minus 119892119894(119896)119891(119890119896)
(119894(119896)) le (119901119898119886119909119895(119896)0)2 (81)
(b) If the inequalities
(119901119898119886119909119894(119896)0)2 minus 119892119894(119896)119891(119890119896)
(119894(119896)) ge (119901119898119894119899119895(119896)0)2
and (119901119898119894119899119894(119896)0)2 minus 119892119894(119896)119891(119890119896)
(119894(119896)) le (119901119898119886119909119895(119896)0)2 (82)
are fulfilled for all 119896 = 1 1198982 an optimal solution 119906 isin R1198982ge1
of (76) exists
Mathematical Problems in Engineering 11
Note that a vector 119887+ isin R119899+1 is suitable here if the vector+ is feasible for the subgraphs ie the inequalities (25)-(27)are fulfilled for +Proof (a) For this part we use a proof by contradictionAssume that
(119901119898119894119899119894(119896)0)2 minus 119892119894(119896)119891(119890119896)
(119894(119896)) gt (119901119898119886119909119895(119896)0)2 (83)
Because 119906119896 ge 1 for all 119896 = 1 1198982 it follows
(119901119898119894119899119894(119896)0)2 minus 119892119894(119896)119891(119890119896)
(119894(119896)) gt 1119906119896
(119901119898119886119909119895(119896)0)2 (84)
It it Σ119896lowast119894(119896) = 0 and Π119896lowast 119894(119896) = 1 because the subgraph withindex 119896lowast is equal to the subgraphwith index 119894(119896) And becausethe subgraphs 119894(119896) and 119895(119896) are neighbours (ie directlyconnected) it follows Π119896lowast 119895(119896) = 119906119896 and
Σ119896lowast119895(119896) = 119892(119896lowast119895(119896))1119891(119890119906(119896lowast119895(119896))1
) (119887(119896lowast119895(119896))1)= 119892119894(119896)119891(119890119896)
(119894(119896)) (85)
Thus we have a contradiction to (28) and 119906119896 cannot beoptimal
(b) As mentioned before the vector 119887+ is such that isfeasible for the subgraphs So because of the conditions for119896 isin 1 1198982 there exists a 119901119894(119896)0 isin [119901119898119894119899
119894(119896)0 119901119898119886119909119894(119896)0] and a119901119895(119896) isin [119901119898119894119899
119895(119896) 119901119898119886119909119895(119896) ] so that1199012119894(119896)0 minus 119892119894(119896)119891(119890119896)
(119894(119896)) ge 1199012119895(119896)0 (86)
With 1199012119894(119896)0 minus 119892119894(119896)119891(119890119896)
(119894(119896)) = 1199012119891(119890119896)
and 1199012119895(119896)0 = 1199012
ℎ(119890119896)it
follows
(119901ℎ(119890119896)119901119891(119890119896)
) ge 1 (87)
Therefore we can find a 119906 isin R1198982ge1 that the inequalities (28)-
(35) are fulfilled Now we can use theWeierstraszlig theorem andget the existence of an optimal solution
Note that the sufficient condition in Lemma 8 is com-paratively strong there may exist optimal solutions with lessstrong conditions eg the following
Corollary 9 If the graph has only one compressor edge ie|E119862| = 1 then it holds If 119906 isin Rge1 is an optimal solution of(76) 119906 is unique
This conclusion follows directly from the strict mono-tonicity of the objective function
Optimization with Random Loads In this section we con-sider the load vector to be random as we mentioned inSection 22 In the optimization problems we want to makesure that the feasibility of a random load vector is guaranteedfor a probability 120572 isin (0 1) This leads us to optimizationproblems with so called chance or probabilistic constraints
In general problems with probabilistic constraints have theform
min 119891 (119909 120585)subject to P (ℎ (119909 120585) le 0) ge 120572 (88)
with an objective function 119891 and a vector of uncertainty 120585There exist many works about probabilistic constraints respstochastic programming eg [7] gives an excellent overviewabout chance constraints in theory and application In ourcase we consider the following problem
min119901+119898119886119909isin119875
119888119879119901+119898119886119909
subject to P (119887 (119901+119898119886119909) isin 119872) ge 120572 (89)
To handle this problem resp the probabilistic constraint thespheric-radial decomposition gives us an explicit representa-tion of the constraint for a Gaussian vector 119887 sim N(120583 Σ) withmean value 120583 and positive definite covariance Σ
P (119887 (119901+119898119886119909) isin 119872)= int
S119899minus1120583120594 119903 ge 0 | 119903119871V + 120583 isin 119872119889120583120578 (V) (90)
Further Algorithm 3 gives a way to approximate this integralin an efficient way In Section 3 we showed a way to character-ize the feasible set119872 and how to adapt this characterization tothe spheric-radial decomposition This changes our problemwith probabilistic constraints (89) to
min119901+119898119886119909isin119875
119888119879119901+119898119886119909
subject to 1119873 119873sum119894=1
119904119905sum119905=1
F120594 (119886119894119905) minusF120594 (119886119894119905) ge 120572 (91)
Of course the interval bounds depend on the pressure boundsand the control For this problem we formulate the nextlemma
Lemma 10 Let a sampling V1 V119873 (from step (1) ofAlgorithm 3) and 120572 isin (0 1) be given For a suitable 120583 isin R119899
problem (91) has at least one solution 119901+119898119886119909 isin 119875Proof For the proof we fix a V isin V1 V119873 from the givensampling From (25) and (26) it follows for all 119896 = 1 119899119894with 119894 isin 1 1198982 + 10 le 119892119894119896 (119894 (119903)) + (119901119898119886119909
119894119896 )2 minus (1199011198981198941198991198940 )2
0 le minus119892119894119896 (119894 (119903)) + (1199011198981198861199091198940 )2 minus (119901119898119894119899
119894119896 )2 (92)
With a random vector 119887 sim N(120583 Σ)119892119894119896(119894(119903)) are quadraticfunctions in 119903 and the upper pressure bounds only influencethe constant part with a positive sign
0 le 1198861198941198961199032 + 119887119894119896119903 + 119888119894119896 + (119901119898119886119909119894119896 )2 minus (119901119898119894119899
1198940 )2 0 le minus1198861198941198961199032 minus 119887119894119896119903 minus 119888119894119896 + (119901119898119886119909
1198940 )2 minus (119901119898119894119899119894119896 )2 (93)
12 Mathematical Problems in Engineering
It follows that a rise in 119901119898119886119909119894119896 for all 119896 = 0 119899119894 enlarges
the intervals in which the inequalities hold The same holdsfor all inequalities inTheorem 5The cumulative distributionfunction of theChi-Square-Distribution is strictlymonotonicincreasing and convergent to 1 Thus upper pressure bounds119901119898119886119909119894119896 (for 119894 = 1 119899) can be found so that
1119873 119873sum119895=1
119904119895sum119905=1
F120594 (119886119895119905) minusF120594 (119886119895119905) (94)
is higher that 120572 isin (0 1) The index119873 is for all sampled pointson the sphere and the index 119904119895 is for the union of intervals inwhich all inequalities hold The interval boundaries 119886119895119905 and119886119895119905 depend continuously on the pressure bounds and becausethe inequalities are not strict 119901119898119886119909 can be found st the sideconstraint is fulfilled with equality So there exists at least onesolution of (91)
5 Numerical Results
In this section we show a few results of implementation Atfirst we show the idea of the spheric-radial decomposition byusing an easy example Next we show a easy example with onecompressor edge and at last we use real data of the Greek gasnetworkThe focus of the implementation is on the theoremsproofed in Section 4
Example 1 Our first example is an easy graph without innercontrol (see Figure 6)
Assume that node 1 and 2 are gas consumers with meandemand 120583 = [05 05]119879 The pressure bounds are given by119901+119898119894119899 = [2 1 1]119879 and 119901+119898119886119909 = [3 2 2]119879 and Φ = E2 Thenthe incidence matrix is
119860+ = [[[minus1 01 minus10 1
]]] resp 119860 = [1 minus10 1 ] (95)
Consider the inequalities of Theorem 5 to characterize thefeasible set Because the network has no compressor edgesthe system of inequalities only contains the information (25)-(27) the other inequalities do not occur The function 119892 isgiven by
119892 (119887) = [1 01 1] [1 00 1] sdot1003816100381610038161003816100381610038161003816100381610038161003816[1 10 1](11988711198872)
1003816100381610038161003816100381610038161003816100381610038161003816∘ ([1 10 1](11988711198872)) = ( 11988721 + 211988711198872 + 1198872211988721 + 211988711198872 + 211988722) (96)
By inserting the pressure bounds one can see that inequality(25) is always fulfilled and inequality (27) implies1198872 le radic3 (97)
With this inequality (26) follows for 0 le 1198872 le radic30 le 1198871 le minus1198872 + radic8 minus 11988722 (98)
So the feasible set119872 is the following (see Figure 7)
119872 = 119887 isin R2ge0 | 1198872 le radic3 and 1198871 le minus1198872 + radic8 minus 11988722 (99)
Now consider the optimization problem
min119901+119898119886119909isin119875
11198793119901+119898119886119909
subject to 119887 isin 119872 (119901+119898119886119909) (100)
with 119887 = 120583 = [05 05]119879 Lemma 7 guarantees the existenceof a unique solution of this problem We use the fmincon-function inMATLAB which minimizes a constrained non-linear multivariable function The fmincon default settinguses a interior-point method For more information we referto the official MathWorks homepage (httpswwwmath-workscomhelpoptimugfminconhtml)The solver returnsthe following result
119901+119898119886119909119900119901119905 = [2 radic3radic275]119879 (101)
One can easily see that this is the correct solution The gasdemand at node 2 is 05 which is equal to the flow 1199022 sousing the equation of momentum implies a difference in thequadratic pressures between node 1 and node 2 of 025 Theflow through edge 1 is given by the two exit nodes and it is1199021 = 1 This implies a difference in the quadratic pressuresbetweennode 0 andnode 1 of 1 So the optimal solution119901+119898119886119909
119900119901119905
directly followsNext we use the spheric-radial decomposition especially
Algorithm 3 to calculate the probability P(119887 isin 119872) for arandom vector 119887 sim N(E2 [05 05]119879) Therefore we viewthe results of 8 tests with 1000 sampled points in each one(Table 1)
The probability in all eight tests is nearly the same Themean value is 03479 and the variance is 27723 sdot 10minus6 whichis very small This shows that the implementation is workingnearly exact The efficiency of the implementation of courseis not perfect but that is not part of this work
Last wewant to solve the following optimization problem
min119901+119898119886119909isin119875
11198793119901+119898119886119909
subject to 1119873 119873sum119894=1
119904sum119905=1
F120594 (119886119894119905) minusF120594 (119886119894119905) ge 120572 (102)
We use again the fmincon-function inMATLAB to solve theproblem eight times with 1000 sampled points in every caseBecause Lemma 8 needs a suitable 120583 isin R2
ge0 we assume 119887 simN(01sdot1198642 [15 15])We set120572 fl 09 and the results are shownin Table 2
Example 2 Now we add a compressor edge to the graph ofExample 1 (see Figure 8)
Assume that node 1 and 3 are consumers node 2 is ainner nodeThemeandemand120583 = [05 0 05]119879Thepressure
Mathematical Problems in Engineering 13
minus2 minus1 1 2 3
minus2
minus1
1
2
3
b2
b1
minus1 1 2
minus1
1
2
b2
b1
rLv1
rLv2
rLv3
Scheme of the spheric-radial decomposition with
sphere (yellow)and three points 1 2 3 (red) at the unit
the feasible set Mminus for Σ = E2 = [05 05]TFeasible set M for p+min = [2 1 1]T and
p+max = [3 2 2]T
Figure 7 Feasible set of Example 1
Table 1 Test results of easy example
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 8P(119887 isin 119872) 03486 03481 03450 03501 03484 03499 03459 03475
Table 2 Test results of Example 1
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 81199011198981198861199091199001199011199050 54514 54714 54758 54028 54767 53921 54888 551671199011198981198861199091199001199011199051 29802 30126 30225 29725 30300 29485 30336 300291199011198981198861199091199001199011199052 10084 10004 10776 10001 11221 11390 10285 10558
0 1 2 3q1 q2 q3
b0 b1 b2 b3
e1 isin ℰF e2 isin ℰC e3 isin ℰF
Figure 8 Graph of Example 2
0 1 2 3q1 q2 q2 q3
b0 b1 b2 b3
e1 isin ℰF e3 isin ℰF
Figure 9 Separated graph of Example 2
bounds are given by 119901+119898119894119899 = [2 1 1 1]119879 and 119901+119898119886119909 = [3 22 2]119879 and 1206011198901= 1 and 1206011198903
= 1 The incidence matrix is givenby
119860+ = [[[[[[
minus1 0 01 minus1 00 1 minus10 0 1]]]]]]
resp 119860 = [[[1 minus1 00 1 minus10 0 1
]]] (103)
Before we have a look at the inequalities of Theorem 5we have to remove the compressor edge from the graphThe flows can be computed by using the equation of massconservation and the new load vector is then
= (1198871 + 11990221198872 minus 11990221198873 ) with
119902 = (1198871 + 1198872 + 11988731198872 + 11988731198873)(104)
We assume the control to be switched off so it holds119906 = 1With this the control edge 1198902 is treated as a flow edge withoutfriction The resulting graph is shown in Figure 9 from theinequalities (25)ndash(35) it follows
3 le radic3 and 1 le radic8 minus 23 (105)
14 Mathematical Problems in Engineering
minus1 1 2 3
1
2
3
minus1 1 2 3
1
2
3
b3
b1
Hyperspace of the feasible set MHyperspace of the feasible set M
b1
b3
Figure 10 Feasible set of the Example 2
Table 3 Test results of control example
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 8P(119887 isin 119872) 01531 01500 01540 01536 01530 01550 01567 01528so the feasible set is (see Figure 10)
= + isin R119899+1 | 1119879
4 + = 0 and 0 le 3 le radic3 and 2 = 0and 0 le 1 le radic8 minus 22 (106)
This implies
119872 = 119887 isin R3ge0 | 0 le 1198873 le radic3 and 1198872 = 0 and 0 le 1198871 le minus1198873
+ radic8 minus 11988723 (107)
Solving the problem
min119901+119898119886119909isin119875
11198794119901+119898119886119909
subject to isin (108)
for 119887 = 120583 = [05 0 05]119879 leads us to the optimal solution
119901+119898119886119909119900119901119905 = [2 radic3radic3radic275]119879 (109)
The argumentation for the optimality of this vector is exactlylike before in the easy example Becausewe have a compressoredge in this graph we can minimize the control Thereforewe fix the pressure bounds as we did at the beginning of thisexample and we set 119887 = [1 0 15]119879 So inequality (31) is notfulfilled anymore for 119906 = 1 We use again the fmincon-solverinMATLAB to solve the following problem
min119906ge1
1199062subject to 119887 isin 119872(119906) (110)
The solver returns the optimal solution 119906119900119901119905 = 11818
Here one can see that the sufficient condition in Lemma 8is really strong Itrsquos not fulfilled with the given values but thereexists a solution anyway Of course the necessary conditionis fulfilled
Now we use again Algorithm 3 to compute the proba-bility P(119887 isin 119872) for a random load vector 119887 sim N(E3 [0505 05]119879) Table 3 shows us the results
Again we can see that the probability is nearly the samein all eight cases The mean probability is 01535 and thevariance is 32369 sdot 10minus6 For the optimization we reducethe covariance matrix and assume 119887 sim N(001 sdot E3 [05 0505]119879) For a probability 120572 = 09 Table 4 states the resultsExample 3 In this example we use a part of a real networkThe focus here is on realistic values for our model Weget these values from the GasLib (see [22]) a collection oftechnical gas network data
ldquoThegoal ofGasLib is to promote research on gasnetworks by providing a set of large and realisticbenchmark instancesrdquo (httpgaslibzibde)
We use a part of the Greek gas network (GasLib-134) (seeFigure 11)
For using real values we first need to know the constant120601119890 from (5)
120601119890 = (119877119878119879)2 120582119863 1198711198862 (111)
The value 119877119878 is the specific gas constant which can becalculated by 119877119878 = 119877119872 with the ideal gas constant 119877 =831451119896119892 sdot 11989821199042 sdot 119898119900119897 sdot 119870 (see [23]) and the molar mass119872 = 1662119896119892119896119898119900119897 (from real data) of the used natural gasAlso the temperature 119879 = 28915119870 is given from real dataFor the (constant) friction factor 120582 we use the law of Chen(see [24])
Mathematical Problems in Engineering 15
Table 4 Test results of control example
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 81199011198981198861199091199001199011199050 39013 39223 39039 38901 38879 39039 38901 390981199011198981198861199091199001199011199051 16568 16565 16365 16497 16496 16365 16497 164921199011198981198861199091199001199011199052 16568 16565 16365 16497 16496 16365 16497 164981199011198981198861199091199001199011199053 10000 10000 10000 10000 10002 10000 10000 10000
Table 5 Values based on real data
variable 119877119878 119879 120582 119863 119871 119886 120601value 50027 28915 00093 9144 178693 43862 198 sdot 107unit 11989821199042 sdot 119870 119870 minus 119898119898 119896119898 119898119904 11989821199042
Figure 11 Greek gas network from GasLib [22] red edges are usedin our example
1radic120582 = minus2 log10 ( 11989611986337065minus 50425
Relog10 [(119896119863)1109828257 + 58506
Re08981]) (112)
Diameter 119863 = 9144119898119898 and roughness value 119896 = 8 sdot 10minus6119898for the first pipe (see Figure 11) are known from real data TheReynolds-Number Re can be calculated by Re = (V119863120588)120578with the dynamic viscosity 120578 = 1 sdot 10minus5119875119886 sdot 119904 (see [1])the mean velocity V = 9548119898119904 (calculated with the meanvolume flow rate which is given by the real data) of the gasand the density 120588 = 074331198961198921198983 (from real data) So we getRe = 648959074 asymp 649 sdot 106 and with this the pipe frictioncoefficient after Chen is 120582 = 00093
The length of the first pipe is given by 119871 = 178693119896119898from the real data Last value missing is the sound speed innatural gas which we assume to be ideal The sound speed
119886 depends on the modulus of compressibility (see [9]) Forideal gases the modulus of compressibility only depends onthe adiabatic exponent and the pressure Using the ideal gasequation simplifies the equation for the sound speed to
119886 = radic120581119877119879119872 (113)
The adiabatic exponent for methan which is the main partof natural gas (gt 90) is about 133 (see [25]) Together allvalues are shown in Table 5 (example for the pipe next to thesource in Figure 11)
Nowwewant to optimize the inner control in the networkfor a fix load vector Therefor we choose two nominations2011-11-01 and 2016-02-17 In both cases the fmincon-functioninMATLAB returns
119906119900119901119905 = 1 (114)
as a local minimum which satisfies the constraint 119887 isin 119872This means for these nominations a compressor stationis not needed so it is switched off With this as a lastcalculation in this paper we calculate the probability for thegiven nominations to be feasible We use the values from thenominations as mean value 120583 and set 119887 sim N(120583 sdot 10minus3E 120583)The results are shown in Tables 6 and 7
Again one can see that the results are quite similarThe mean probability in Table 6 is 09860 and the meanprobability in Table 7 is 09760 Both values are nearly 1 sothere must happen something really unforeseeable that aGreek does not get his demanded gas
6 Conclusion
In this paper we have used an existing stationary modelfor gas transport based on the isothermal Euler equationsWe have extended this model by compressor stations andpresented a characterization of the set of feasible loadsEspecially Theorem 5 contains the central idea to get anaccess to the optimization in Section 4 and the spheric-radialdecomposition is our main tool to handle the uncertainty inthe loadsWewere able to show some results for optimizationwith probabilistic constraints in gas transport In a next stepthese results should be generalized
16 Mathematical Problems in Engineering
Table 6 Test results for nomination 2011-11-01
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 8P(119887 isin 119872) 09849 09849 09878 09885 09863 09845 09854 09853
Table 7 Test results for nomination 2016-02-17
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 8P(119887 isin 119872) 09744 09756 09788 09788 09754 09775 09734 09739
With the characterization of the feasible set we haveshown that the feasibility of a load depends significantly onthe pressure bounds These pressure bounds play a mainrole in our optimization problems The other main partsare the inner controls modeled by compressor edges whichshould lead in optimal control problems with probabilisticconstraints Here we have built a base for further works withthis topic
The idea of separating the graph and removing thecompressor edges can be used to extend themodel with othercomponents like eg control valves and resistorsThiswouldchange the inequalities inTheorem 5 but the approach worksalso for thesemodifications Further [8] showed a way to dealwith a networkwhich is a cycle Combining bothmethods weget an access to model gas networks in high detail
Finally the expansion to a dynamic model might bealso possible The idea is to use a space-mapping approachbecause one can use a lot of the analysis we did hereTherefore upcoming papers should deal with these problemsand present results of the turnpike-theory which describesa relation between stationary and instationary solutions Infact turnpike-theory implies that the optimal dynamic stateis close to the optimal static state in the interior of the timeintervals for sufficiently long time horizons AJ Zaslavskiexplains the turnpike-phenomenon in [26] This book alsogives an overview about the theory
Data Availability
The data used to support the findings of this study areavailable from the corresponding author upon request
Conflicts of Interest
The authors declare that there are no conflicts of interestregarding the publication of this paper
Acknowledgments
This work was supported by DFG in the framework of theCollaborative Research Centre CRCTransregio 154 Math-ematical Modeling Simulation and Optimization Using theExample of Gas Networks project C03
References
[1] P Domschke B Hiller J Lang and C Tischendorf ldquoModel-lierung von gasnetzwerken Eine Ubersichtrdquo Tech Rep 2717Technische Universitat Darmstadt 2017
[2] T Koch B Hiller M E Pfetsch and L Schewe Evaluating GasNetwork Capacities MOS-SIAM 2015
[3] M Gugat R Schultz and D Wintergerst ldquoNetworks of pipe-lines for gas with nonconstant compressibility factor stationarystatesrdquoComputational amp AppliedMathematics vol 37 no 2 pp1066ndash1097 2018
[4] M Gugat and M Herty ldquoExistence of classical solutions andfeedback stabilization for the flow in gas networksrdquo ESAIMControl Optimisation and Calculus of Variations vol 17 no 1pp 28ndash51 2011
[5] M K Banda M Herty and A Klar ldquoGas flow in pipeline net-worksrdquo Networks and Heterogeneous Media vol 1 no 1 pp 41ndash56 2006
[6] RM Colombo G GuerraM Herty and V Schleper ldquoOptimalcontrol in networks of pipes and canalsrdquo SIAM Journal onControl and Optimization vol 48 no 3 pp 2032ndash2050 2009
[7] A Shapiro D Dentcheva and A Ruszczynski ectures onStochastic Programming Modeling and Theory MPS-SIAM2009
[8] C Gotzes H Heitsch R Henrion and R Schultz ldquoOn thequantification of nomination feasibility in stationary gas net-works with random loadrdquoMathematical Methods of OperationsResearch vol 84 no 2 pp 427ndash457 2016
[9] D Halliday R Resnick and J Walker Physik Wiley-VCHBachelor edition edition 2007
[10] M Gugat F M Hante M Hirsch-Dick and G LeugeringldquoStationary states in gas networksrdquoNetworks andHeterogeneousMedia vol 10 no 2 pp 295ndash320 2015
[11] D RoseM SchmidtM C Steinbach and BMWillert ldquoCom-putational optimization of gas compressor stations MINLPmodels versus continuous reformulationsrdquoMathematical Meth-ods of Operations Research vol 83 no 3 pp 409ndash444 2016
[12] H Heitsch R Henrion H Leovey et al ldquoEmpirical observa-tions and statistical analysis of gas demandrdquo in Evaluating GasNetwork Capacities T data B KochM E Hiller et al Eds pp273ndash290 MOS-SIAM 2015
[13] A Genz and F Bretz Computation of Multivariate Normal andt Probabilities vol 195 of Lecture Notes in Statistics Springer2009
[14] M H Farshbaf-Shaker R Henrion and D Homberg ldquoProp-erties of Chance Constraints in Infinite Dimensions with anApplication to PDE Constrained Optimizationrdquo Set-Valued andVariational Analysis pp 1ndash21 2017
[15] T Gonzalez Grandon H Heitsch and R Henrion ldquoA jointmodel of probabilisticrobust constraints for gas transportman-agement in stationary networksrdquo Computational ManagementScience vol 14 no 3 pp 443ndash460 2017
[16] W van Ackooij I Aleksovska and M Munoz-Zuniga ldquo(Sub-)Differentiability of Probability Functionswith Elliptical Distri-butionsrdquo Set-Valued and Variational Analysis
Mathematical Problems in Engineering 17
[17] W van Ackooij and R Henrion ldquoGradient formulae for nonlin-ear probabilistic constraints with GAUssian and GAUssian-likedistributionsrdquo SIAM Journal on Optimization vol 24 no 4 pp1864ndash1889 2014
[18] W van Ackooij and R Henrion ldquo(Sub-)gradient formulaefor probability functions of random inequality systems underGaussian distributionrdquo SIAMASA Journal on UncertaintyQuantification vol 5 no 1 pp 63ndash87 2017
[19] W van Ackooij and J Malick ldquoSecond-order differentiabilityof probability functionsrdquo Optimization Letters vol 11 no 1 pp179ndash194 2017
[20] H Eckey R Kosfeld and M Turck Wahrscheinlichkeitsrech-nung und Induktive Statistik - Grundlagen -Methoden - BeispieleGabler Wiesbaden Germany 2005
[21] A Prekopa Stochastic Programming Springer 1st edition 1995[22] I Joormann M Pfetsch R Schwarz et al ldquoGasLibmdashA Library
of Gas Network Instancesrdquo 2017[23] G Wedler Lehrbuch der Physikalischen Chemie Wiley-VCH
5th edition 2004[24] N H Chen ldquoAn explicit equation for friction factor in piperdquo
Industrial amp Engineering Chemistry Fundamentals vol 18 no3 pp 296-297 1979
[25] W Schenk and F Kremer Physikalisches Praktikum Springer14th edition 2014
[26] A J Zaslavski ldquoTurnpike Properties in the Calculus of Varia-tions and Optimal Controlrdquo inNonconvex Optimization and ItsApplications vol 80 Springer 2006
Hindawiwwwhindawicom Volume 2018
MathematicsJournal of
Hindawiwwwhindawicom Volume 2018
Mathematical Problems in Engineering
Applied MathematicsJournal of
Hindawiwwwhindawicom Volume 2018
Probability and StatisticsHindawiwwwhindawicom Volume 2018
Journal of
Hindawiwwwhindawicom Volume 2018
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawiwwwhindawicom Volume 2018
OptimizationJournal of
Hindawiwwwhindawicom Volume 2018
Hindawiwwwhindawicom Volume 2018
Engineering Mathematics
International Journal of
Hindawiwwwhindawicom Volume 2018
Operations ResearchAdvances in
Journal of
Hindawiwwwhindawicom Volume 2018
Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018
International Journal of Mathematics and Mathematical Sciences
Hindawiwwwhindawicom Volume 2018
Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom
The Scientific World Journal
Volume 2018
Hindawiwwwhindawicom Volume 2018Volume 2018
Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in
Nature and SocietyHindawiwwwhindawicom Volume 2018
Hindawiwwwhindawicom
Dierential EquationsInternational Journal of
Volume 2018
Hindawiwwwhindawicom Volume 2018
Decision SciencesAdvances in
Hindawiwwwhindawicom Volume 2018
AnalysisInternational Journal of
Hindawiwwwhindawicom Volume 2018
Stochastic AnalysisInternational Journal of
Submit your manuscripts atwwwhindawicom
10 Mathematical Problems in Engineering
+ 119903[[2119899sum
119895=1
Ψ119896119895( 119899sumℓ=1
119860minus1119895ℓ120596119894ℓ)( 119899sum
ℓ=1
119860minus1119895ℓ120583ℓ)]]⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
119894
+ [[119899sum
119895=1
Ψ119896119895( 119899sumℓ=1
119860minus1119895ℓ120583ℓ)2]]⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
119888119894
(70)
HereΨ is the product of119860 andΦ So the inequations whichcharacterize119872119894 are also quadratic in 119903 because the minimaland maximal pressures in the inequalities only count to 119888119894Now we can write the feasible sets as follows
119872119894 = 119903 isin 119877119903119890119892119894 | 0 le 1198861198941199051199032 + 119887119894119905119903 + 119888119894119905 (71)
where 119905 is an index for the total number of inequalities in119872119894Now the regular range can be cut with the positive intervals ofthe inequalities and the result is a union of disjoint intervals
119872119894 = 119904119894⋃119896=1
[119886119894119896 119886119894119896] 119904119894 isin N (72)
Here 119904119894 is the number of disjoint intervals in which allinequalities are fulfilled and 119886119894119896 and 119886119894119896 are the intervalbounds they depend on the pressure bounds and the con-trols So the third step of Algorithm 3 can be executed With
P (119887 isin 119872) asymp 1119873 119873sum119894=1
119904119894sum119896=1
F120594 (119886119894119896) minusF120594 (119886119894119896) (73)
the probability for a random load vector to be feasible can becomputed
4 Existence of Optimal Solutions
In this section we want to have a look at some optimizationproblems We distinguish between problems with constantloads (without uncertainty on the demand) and problemswith random loads The latter leads us to optimizationproblems with so called probabilistic constraints or chanceconstraints (see [21]) There are two aims in optimizationMinimize the maximal pressure bounds and minimize thecontrols For minimizing the maximal pressure bounds wedefine the following set
119875 fl119899⋃
119894=0
[119901+119898119894119899119894 infin) (74)
in which we look for them Because the maximal pressuresare in proportion to the cost of pipes an objective functionfor a optimization problem could be 119888119879119901+119898119886119909 for a cost vector119888 isin R119899+1
gt0 By minimizing the controls we use the Euclideannorm of the controls as an objective function
Optimization with Constant Loads Consider the two opti-mization problems
min119901+119898119886119909isin119875
119888119879119901+119898119886119909
subject to 119887+ (119901+119898119886119909) isin 119872 (75)
andmin119906isinR1198982ge1
1199062subject to 119887+ (119906) isin (76)
Lemma 7 There exists a unique solution for (75)
Proof We have a look at the side constraint 119887+(119901+119898119886119909) isin Using Theorem 5 we can write (75) as
min119901+119898119886119909isin119875
119888119879119901+119898119886119909
subject to (25) minus (35) are fulfilled (77)
For 119894 isin 1 1198982 + 1 we write inequality (25) (and allremaining analogous) in the following way
0 le 119892119894119896 (119894) minus (1199011198981198941198991198940 )2 + (119901119898119886119909
119894119896 )2 forall119896 = 1 119899119894 (78)
So every inequality depends only on one upper pressurebound (but more inequalities may depend on the samebound) Because the inequalities are not strict the values119901119896 fl min 119901119898119886119909
119896 isin [0infin) | (25) minus (35) are fulfilled (79)
exist for all 119896 = 1 119899 + 1 and we can define a value119901+119898119886119909119900119901119905119896 fl max 119901119896 119901119898119894119899
119896 (80)
which is obviously a solution of (75) Moreover this solutionis unique because reducing one component of the upperbounds would either hurt an inequality or it would not fulfill119901+119898119886119909 ge 119901+119898119894119899 anymore
For the optimization problem (76) we can formulate anecessary and a sufficient condition for optimal solutions
Lemma8 Consider pressure bounds119901+119898119894119899 119901+119898119886119909 isin R119899+1 st119901+119898119894119899 lt 119901+119898119886119909 and a suitable vector 119887+ isin R119899+1 (ie + fulfills(25)ndash(27)) For all 119896 = 1 1198982 we set 119894(119896) and 119895(119896) such that119891(119890119896) isin V+
119894(119896) and ℎ(119890119896) isin V+119895(119896)
(a) If 119906 isin R1198982 is a solution of (76) it holds for all 119896 =1 1198982 (119901119898119894119899119894(119896)0)2 minus 119892119894(119896)119891(119890119896)
(119894(119896)) le (119901119898119886119909119895(119896)0)2 (81)
(b) If the inequalities
(119901119898119886119909119894(119896)0)2 minus 119892119894(119896)119891(119890119896)
(119894(119896)) ge (119901119898119894119899119895(119896)0)2
and (119901119898119894119899119894(119896)0)2 minus 119892119894(119896)119891(119890119896)
(119894(119896)) le (119901119898119886119909119895(119896)0)2 (82)
are fulfilled for all 119896 = 1 1198982 an optimal solution 119906 isin R1198982ge1
of (76) exists
Mathematical Problems in Engineering 11
Note that a vector 119887+ isin R119899+1 is suitable here if the vector+ is feasible for the subgraphs ie the inequalities (25)-(27)are fulfilled for +Proof (a) For this part we use a proof by contradictionAssume that
(119901119898119894119899119894(119896)0)2 minus 119892119894(119896)119891(119890119896)
(119894(119896)) gt (119901119898119886119909119895(119896)0)2 (83)
Because 119906119896 ge 1 for all 119896 = 1 1198982 it follows
(119901119898119894119899119894(119896)0)2 minus 119892119894(119896)119891(119890119896)
(119894(119896)) gt 1119906119896
(119901119898119886119909119895(119896)0)2 (84)
It it Σ119896lowast119894(119896) = 0 and Π119896lowast 119894(119896) = 1 because the subgraph withindex 119896lowast is equal to the subgraphwith index 119894(119896) And becausethe subgraphs 119894(119896) and 119895(119896) are neighbours (ie directlyconnected) it follows Π119896lowast 119895(119896) = 119906119896 and
Σ119896lowast119895(119896) = 119892(119896lowast119895(119896))1119891(119890119906(119896lowast119895(119896))1
) (119887(119896lowast119895(119896))1)= 119892119894(119896)119891(119890119896)
(119894(119896)) (85)
Thus we have a contradiction to (28) and 119906119896 cannot beoptimal
(b) As mentioned before the vector 119887+ is such that isfeasible for the subgraphs So because of the conditions for119896 isin 1 1198982 there exists a 119901119894(119896)0 isin [119901119898119894119899
119894(119896)0 119901119898119886119909119894(119896)0] and a119901119895(119896) isin [119901119898119894119899
119895(119896) 119901119898119886119909119895(119896) ] so that1199012119894(119896)0 minus 119892119894(119896)119891(119890119896)
(119894(119896)) ge 1199012119895(119896)0 (86)
With 1199012119894(119896)0 minus 119892119894(119896)119891(119890119896)
(119894(119896)) = 1199012119891(119890119896)
and 1199012119895(119896)0 = 1199012
ℎ(119890119896)it
follows
(119901ℎ(119890119896)119901119891(119890119896)
) ge 1 (87)
Therefore we can find a 119906 isin R1198982ge1 that the inequalities (28)-
(35) are fulfilled Now we can use theWeierstraszlig theorem andget the existence of an optimal solution
Note that the sufficient condition in Lemma 8 is com-paratively strong there may exist optimal solutions with lessstrong conditions eg the following
Corollary 9 If the graph has only one compressor edge ie|E119862| = 1 then it holds If 119906 isin Rge1 is an optimal solution of(76) 119906 is unique
This conclusion follows directly from the strict mono-tonicity of the objective function
Optimization with Random Loads In this section we con-sider the load vector to be random as we mentioned inSection 22 In the optimization problems we want to makesure that the feasibility of a random load vector is guaranteedfor a probability 120572 isin (0 1) This leads us to optimizationproblems with so called chance or probabilistic constraints
In general problems with probabilistic constraints have theform
min 119891 (119909 120585)subject to P (ℎ (119909 120585) le 0) ge 120572 (88)
with an objective function 119891 and a vector of uncertainty 120585There exist many works about probabilistic constraints respstochastic programming eg [7] gives an excellent overviewabout chance constraints in theory and application In ourcase we consider the following problem
min119901+119898119886119909isin119875
119888119879119901+119898119886119909
subject to P (119887 (119901+119898119886119909) isin 119872) ge 120572 (89)
To handle this problem resp the probabilistic constraint thespheric-radial decomposition gives us an explicit representa-tion of the constraint for a Gaussian vector 119887 sim N(120583 Σ) withmean value 120583 and positive definite covariance Σ
P (119887 (119901+119898119886119909) isin 119872)= int
S119899minus1120583120594 119903 ge 0 | 119903119871V + 120583 isin 119872119889120583120578 (V) (90)
Further Algorithm 3 gives a way to approximate this integralin an efficient way In Section 3 we showed a way to character-ize the feasible set119872 and how to adapt this characterization tothe spheric-radial decomposition This changes our problemwith probabilistic constraints (89) to
min119901+119898119886119909isin119875
119888119879119901+119898119886119909
subject to 1119873 119873sum119894=1
119904119905sum119905=1
F120594 (119886119894119905) minusF120594 (119886119894119905) ge 120572 (91)
Of course the interval bounds depend on the pressure boundsand the control For this problem we formulate the nextlemma
Lemma 10 Let a sampling V1 V119873 (from step (1) ofAlgorithm 3) and 120572 isin (0 1) be given For a suitable 120583 isin R119899
problem (91) has at least one solution 119901+119898119886119909 isin 119875Proof For the proof we fix a V isin V1 V119873 from the givensampling From (25) and (26) it follows for all 119896 = 1 119899119894with 119894 isin 1 1198982 + 10 le 119892119894119896 (119894 (119903)) + (119901119898119886119909
119894119896 )2 minus (1199011198981198941198991198940 )2
0 le minus119892119894119896 (119894 (119903)) + (1199011198981198861199091198940 )2 minus (119901119898119894119899
119894119896 )2 (92)
With a random vector 119887 sim N(120583 Σ)119892119894119896(119894(119903)) are quadraticfunctions in 119903 and the upper pressure bounds only influencethe constant part with a positive sign
0 le 1198861198941198961199032 + 119887119894119896119903 + 119888119894119896 + (119901119898119886119909119894119896 )2 minus (119901119898119894119899
1198940 )2 0 le minus1198861198941198961199032 minus 119887119894119896119903 minus 119888119894119896 + (119901119898119886119909
1198940 )2 minus (119901119898119894119899119894119896 )2 (93)
12 Mathematical Problems in Engineering
It follows that a rise in 119901119898119886119909119894119896 for all 119896 = 0 119899119894 enlarges
the intervals in which the inequalities hold The same holdsfor all inequalities inTheorem 5The cumulative distributionfunction of theChi-Square-Distribution is strictlymonotonicincreasing and convergent to 1 Thus upper pressure bounds119901119898119886119909119894119896 (for 119894 = 1 119899) can be found so that
1119873 119873sum119895=1
119904119895sum119905=1
F120594 (119886119895119905) minusF120594 (119886119895119905) (94)
is higher that 120572 isin (0 1) The index119873 is for all sampled pointson the sphere and the index 119904119895 is for the union of intervals inwhich all inequalities hold The interval boundaries 119886119895119905 and119886119895119905 depend continuously on the pressure bounds and becausethe inequalities are not strict 119901119898119886119909 can be found st the sideconstraint is fulfilled with equality So there exists at least onesolution of (91)
5 Numerical Results
In this section we show a few results of implementation Atfirst we show the idea of the spheric-radial decomposition byusing an easy example Next we show a easy example with onecompressor edge and at last we use real data of the Greek gasnetworkThe focus of the implementation is on the theoremsproofed in Section 4
Example 1 Our first example is an easy graph without innercontrol (see Figure 6)
Assume that node 1 and 2 are gas consumers with meandemand 120583 = [05 05]119879 The pressure bounds are given by119901+119898119894119899 = [2 1 1]119879 and 119901+119898119886119909 = [3 2 2]119879 and Φ = E2 Thenthe incidence matrix is
119860+ = [[[minus1 01 minus10 1
]]] resp 119860 = [1 minus10 1 ] (95)
Consider the inequalities of Theorem 5 to characterize thefeasible set Because the network has no compressor edgesthe system of inequalities only contains the information (25)-(27) the other inequalities do not occur The function 119892 isgiven by
119892 (119887) = [1 01 1] [1 00 1] sdot1003816100381610038161003816100381610038161003816100381610038161003816[1 10 1](11988711198872)
1003816100381610038161003816100381610038161003816100381610038161003816∘ ([1 10 1](11988711198872)) = ( 11988721 + 211988711198872 + 1198872211988721 + 211988711198872 + 211988722) (96)
By inserting the pressure bounds one can see that inequality(25) is always fulfilled and inequality (27) implies1198872 le radic3 (97)
With this inequality (26) follows for 0 le 1198872 le radic30 le 1198871 le minus1198872 + radic8 minus 11988722 (98)
So the feasible set119872 is the following (see Figure 7)
119872 = 119887 isin R2ge0 | 1198872 le radic3 and 1198871 le minus1198872 + radic8 minus 11988722 (99)
Now consider the optimization problem
min119901+119898119886119909isin119875
11198793119901+119898119886119909
subject to 119887 isin 119872 (119901+119898119886119909) (100)
with 119887 = 120583 = [05 05]119879 Lemma 7 guarantees the existenceof a unique solution of this problem We use the fmincon-function inMATLAB which minimizes a constrained non-linear multivariable function The fmincon default settinguses a interior-point method For more information we referto the official MathWorks homepage (httpswwwmath-workscomhelpoptimugfminconhtml)The solver returnsthe following result
119901+119898119886119909119900119901119905 = [2 radic3radic275]119879 (101)
One can easily see that this is the correct solution The gasdemand at node 2 is 05 which is equal to the flow 1199022 sousing the equation of momentum implies a difference in thequadratic pressures between node 1 and node 2 of 025 Theflow through edge 1 is given by the two exit nodes and it is1199021 = 1 This implies a difference in the quadratic pressuresbetweennode 0 andnode 1 of 1 So the optimal solution119901+119898119886119909
119900119901119905
directly followsNext we use the spheric-radial decomposition especially
Algorithm 3 to calculate the probability P(119887 isin 119872) for arandom vector 119887 sim N(E2 [05 05]119879) Therefore we viewthe results of 8 tests with 1000 sampled points in each one(Table 1)
The probability in all eight tests is nearly the same Themean value is 03479 and the variance is 27723 sdot 10minus6 whichis very small This shows that the implementation is workingnearly exact The efficiency of the implementation of courseis not perfect but that is not part of this work
Last wewant to solve the following optimization problem
min119901+119898119886119909isin119875
11198793119901+119898119886119909
subject to 1119873 119873sum119894=1
119904sum119905=1
F120594 (119886119894119905) minusF120594 (119886119894119905) ge 120572 (102)
We use again the fmincon-function inMATLAB to solve theproblem eight times with 1000 sampled points in every caseBecause Lemma 8 needs a suitable 120583 isin R2
ge0 we assume 119887 simN(01sdot1198642 [15 15])We set120572 fl 09 and the results are shownin Table 2
Example 2 Now we add a compressor edge to the graph ofExample 1 (see Figure 8)
Assume that node 1 and 3 are consumers node 2 is ainner nodeThemeandemand120583 = [05 0 05]119879Thepressure
Mathematical Problems in Engineering 13
minus2 minus1 1 2 3
minus2
minus1
1
2
3
b2
b1
minus1 1 2
minus1
1
2
b2
b1
rLv1
rLv2
rLv3
Scheme of the spheric-radial decomposition with
sphere (yellow)and three points 1 2 3 (red) at the unit
the feasible set Mminus for Σ = E2 = [05 05]TFeasible set M for p+min = [2 1 1]T and
p+max = [3 2 2]T
Figure 7 Feasible set of Example 1
Table 1 Test results of easy example
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 8P(119887 isin 119872) 03486 03481 03450 03501 03484 03499 03459 03475
Table 2 Test results of Example 1
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 81199011198981198861199091199001199011199050 54514 54714 54758 54028 54767 53921 54888 551671199011198981198861199091199001199011199051 29802 30126 30225 29725 30300 29485 30336 300291199011198981198861199091199001199011199052 10084 10004 10776 10001 11221 11390 10285 10558
0 1 2 3q1 q2 q3
b0 b1 b2 b3
e1 isin ℰF e2 isin ℰC e3 isin ℰF
Figure 8 Graph of Example 2
0 1 2 3q1 q2 q2 q3
b0 b1 b2 b3
e1 isin ℰF e3 isin ℰF
Figure 9 Separated graph of Example 2
bounds are given by 119901+119898119894119899 = [2 1 1 1]119879 and 119901+119898119886119909 = [3 22 2]119879 and 1206011198901= 1 and 1206011198903
= 1 The incidence matrix is givenby
119860+ = [[[[[[
minus1 0 01 minus1 00 1 minus10 0 1]]]]]]
resp 119860 = [[[1 minus1 00 1 minus10 0 1
]]] (103)
Before we have a look at the inequalities of Theorem 5we have to remove the compressor edge from the graphThe flows can be computed by using the equation of massconservation and the new load vector is then
= (1198871 + 11990221198872 minus 11990221198873 ) with
119902 = (1198871 + 1198872 + 11988731198872 + 11988731198873)(104)
We assume the control to be switched off so it holds119906 = 1With this the control edge 1198902 is treated as a flow edge withoutfriction The resulting graph is shown in Figure 9 from theinequalities (25)ndash(35) it follows
3 le radic3 and 1 le radic8 minus 23 (105)
14 Mathematical Problems in Engineering
minus1 1 2 3
1
2
3
minus1 1 2 3
1
2
3
b3
b1
Hyperspace of the feasible set MHyperspace of the feasible set M
b1
b3
Figure 10 Feasible set of the Example 2
Table 3 Test results of control example
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 8P(119887 isin 119872) 01531 01500 01540 01536 01530 01550 01567 01528so the feasible set is (see Figure 10)
= + isin R119899+1 | 1119879
4 + = 0 and 0 le 3 le radic3 and 2 = 0and 0 le 1 le radic8 minus 22 (106)
This implies
119872 = 119887 isin R3ge0 | 0 le 1198873 le radic3 and 1198872 = 0 and 0 le 1198871 le minus1198873
+ radic8 minus 11988723 (107)
Solving the problem
min119901+119898119886119909isin119875
11198794119901+119898119886119909
subject to isin (108)
for 119887 = 120583 = [05 0 05]119879 leads us to the optimal solution
119901+119898119886119909119900119901119905 = [2 radic3radic3radic275]119879 (109)
The argumentation for the optimality of this vector is exactlylike before in the easy example Becausewe have a compressoredge in this graph we can minimize the control Thereforewe fix the pressure bounds as we did at the beginning of thisexample and we set 119887 = [1 0 15]119879 So inequality (31) is notfulfilled anymore for 119906 = 1 We use again the fmincon-solverinMATLAB to solve the following problem
min119906ge1
1199062subject to 119887 isin 119872(119906) (110)
The solver returns the optimal solution 119906119900119901119905 = 11818
Here one can see that the sufficient condition in Lemma 8is really strong Itrsquos not fulfilled with the given values but thereexists a solution anyway Of course the necessary conditionis fulfilled
Now we use again Algorithm 3 to compute the proba-bility P(119887 isin 119872) for a random load vector 119887 sim N(E3 [0505 05]119879) Table 3 shows us the results
Again we can see that the probability is nearly the samein all eight cases The mean probability is 01535 and thevariance is 32369 sdot 10minus6 For the optimization we reducethe covariance matrix and assume 119887 sim N(001 sdot E3 [05 0505]119879) For a probability 120572 = 09 Table 4 states the resultsExample 3 In this example we use a part of a real networkThe focus here is on realistic values for our model Weget these values from the GasLib (see [22]) a collection oftechnical gas network data
ldquoThegoal ofGasLib is to promote research on gasnetworks by providing a set of large and realisticbenchmark instancesrdquo (httpgaslibzibde)
We use a part of the Greek gas network (GasLib-134) (seeFigure 11)
For using real values we first need to know the constant120601119890 from (5)
120601119890 = (119877119878119879)2 120582119863 1198711198862 (111)
The value 119877119878 is the specific gas constant which can becalculated by 119877119878 = 119877119872 with the ideal gas constant 119877 =831451119896119892 sdot 11989821199042 sdot 119898119900119897 sdot 119870 (see [23]) and the molar mass119872 = 1662119896119892119896119898119900119897 (from real data) of the used natural gasAlso the temperature 119879 = 28915119870 is given from real dataFor the (constant) friction factor 120582 we use the law of Chen(see [24])
Mathematical Problems in Engineering 15
Table 4 Test results of control example
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 81199011198981198861199091199001199011199050 39013 39223 39039 38901 38879 39039 38901 390981199011198981198861199091199001199011199051 16568 16565 16365 16497 16496 16365 16497 164921199011198981198861199091199001199011199052 16568 16565 16365 16497 16496 16365 16497 164981199011198981198861199091199001199011199053 10000 10000 10000 10000 10002 10000 10000 10000
Table 5 Values based on real data
variable 119877119878 119879 120582 119863 119871 119886 120601value 50027 28915 00093 9144 178693 43862 198 sdot 107unit 11989821199042 sdot 119870 119870 minus 119898119898 119896119898 119898119904 11989821199042
Figure 11 Greek gas network from GasLib [22] red edges are usedin our example
1radic120582 = minus2 log10 ( 11989611986337065minus 50425
Relog10 [(119896119863)1109828257 + 58506
Re08981]) (112)
Diameter 119863 = 9144119898119898 and roughness value 119896 = 8 sdot 10minus6119898for the first pipe (see Figure 11) are known from real data TheReynolds-Number Re can be calculated by Re = (V119863120588)120578with the dynamic viscosity 120578 = 1 sdot 10minus5119875119886 sdot 119904 (see [1])the mean velocity V = 9548119898119904 (calculated with the meanvolume flow rate which is given by the real data) of the gasand the density 120588 = 074331198961198921198983 (from real data) So we getRe = 648959074 asymp 649 sdot 106 and with this the pipe frictioncoefficient after Chen is 120582 = 00093
The length of the first pipe is given by 119871 = 178693119896119898from the real data Last value missing is the sound speed innatural gas which we assume to be ideal The sound speed
119886 depends on the modulus of compressibility (see [9]) Forideal gases the modulus of compressibility only depends onthe adiabatic exponent and the pressure Using the ideal gasequation simplifies the equation for the sound speed to
119886 = radic120581119877119879119872 (113)
The adiabatic exponent for methan which is the main partof natural gas (gt 90) is about 133 (see [25]) Together allvalues are shown in Table 5 (example for the pipe next to thesource in Figure 11)
Nowwewant to optimize the inner control in the networkfor a fix load vector Therefor we choose two nominations2011-11-01 and 2016-02-17 In both cases the fmincon-functioninMATLAB returns
119906119900119901119905 = 1 (114)
as a local minimum which satisfies the constraint 119887 isin 119872This means for these nominations a compressor stationis not needed so it is switched off With this as a lastcalculation in this paper we calculate the probability for thegiven nominations to be feasible We use the values from thenominations as mean value 120583 and set 119887 sim N(120583 sdot 10minus3E 120583)The results are shown in Tables 6 and 7
Again one can see that the results are quite similarThe mean probability in Table 6 is 09860 and the meanprobability in Table 7 is 09760 Both values are nearly 1 sothere must happen something really unforeseeable that aGreek does not get his demanded gas
6 Conclusion
In this paper we have used an existing stationary modelfor gas transport based on the isothermal Euler equationsWe have extended this model by compressor stations andpresented a characterization of the set of feasible loadsEspecially Theorem 5 contains the central idea to get anaccess to the optimization in Section 4 and the spheric-radialdecomposition is our main tool to handle the uncertainty inthe loadsWewere able to show some results for optimizationwith probabilistic constraints in gas transport In a next stepthese results should be generalized
16 Mathematical Problems in Engineering
Table 6 Test results for nomination 2011-11-01
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 8P(119887 isin 119872) 09849 09849 09878 09885 09863 09845 09854 09853
Table 7 Test results for nomination 2016-02-17
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 8P(119887 isin 119872) 09744 09756 09788 09788 09754 09775 09734 09739
With the characterization of the feasible set we haveshown that the feasibility of a load depends significantly onthe pressure bounds These pressure bounds play a mainrole in our optimization problems The other main partsare the inner controls modeled by compressor edges whichshould lead in optimal control problems with probabilisticconstraints Here we have built a base for further works withthis topic
The idea of separating the graph and removing thecompressor edges can be used to extend themodel with othercomponents like eg control valves and resistorsThiswouldchange the inequalities inTheorem 5 but the approach worksalso for thesemodifications Further [8] showed a way to dealwith a networkwhich is a cycle Combining bothmethods weget an access to model gas networks in high detail
Finally the expansion to a dynamic model might bealso possible The idea is to use a space-mapping approachbecause one can use a lot of the analysis we did hereTherefore upcoming papers should deal with these problemsand present results of the turnpike-theory which describesa relation between stationary and instationary solutions Infact turnpike-theory implies that the optimal dynamic stateis close to the optimal static state in the interior of the timeintervals for sufficiently long time horizons AJ Zaslavskiexplains the turnpike-phenomenon in [26] This book alsogives an overview about the theory
Data Availability
The data used to support the findings of this study areavailable from the corresponding author upon request
Conflicts of Interest
The authors declare that there are no conflicts of interestregarding the publication of this paper
Acknowledgments
This work was supported by DFG in the framework of theCollaborative Research Centre CRCTransregio 154 Math-ematical Modeling Simulation and Optimization Using theExample of Gas Networks project C03
References
[1] P Domschke B Hiller J Lang and C Tischendorf ldquoModel-lierung von gasnetzwerken Eine Ubersichtrdquo Tech Rep 2717Technische Universitat Darmstadt 2017
[2] T Koch B Hiller M E Pfetsch and L Schewe Evaluating GasNetwork Capacities MOS-SIAM 2015
[3] M Gugat R Schultz and D Wintergerst ldquoNetworks of pipe-lines for gas with nonconstant compressibility factor stationarystatesrdquoComputational amp AppliedMathematics vol 37 no 2 pp1066ndash1097 2018
[4] M Gugat and M Herty ldquoExistence of classical solutions andfeedback stabilization for the flow in gas networksrdquo ESAIMControl Optimisation and Calculus of Variations vol 17 no 1pp 28ndash51 2011
[5] M K Banda M Herty and A Klar ldquoGas flow in pipeline net-worksrdquo Networks and Heterogeneous Media vol 1 no 1 pp 41ndash56 2006
[6] RM Colombo G GuerraM Herty and V Schleper ldquoOptimalcontrol in networks of pipes and canalsrdquo SIAM Journal onControl and Optimization vol 48 no 3 pp 2032ndash2050 2009
[7] A Shapiro D Dentcheva and A Ruszczynski ectures onStochastic Programming Modeling and Theory MPS-SIAM2009
[8] C Gotzes H Heitsch R Henrion and R Schultz ldquoOn thequantification of nomination feasibility in stationary gas net-works with random loadrdquoMathematical Methods of OperationsResearch vol 84 no 2 pp 427ndash457 2016
[9] D Halliday R Resnick and J Walker Physik Wiley-VCHBachelor edition edition 2007
[10] M Gugat F M Hante M Hirsch-Dick and G LeugeringldquoStationary states in gas networksrdquoNetworks andHeterogeneousMedia vol 10 no 2 pp 295ndash320 2015
[11] D RoseM SchmidtM C Steinbach and BMWillert ldquoCom-putational optimization of gas compressor stations MINLPmodels versus continuous reformulationsrdquoMathematical Meth-ods of Operations Research vol 83 no 3 pp 409ndash444 2016
[12] H Heitsch R Henrion H Leovey et al ldquoEmpirical observa-tions and statistical analysis of gas demandrdquo in Evaluating GasNetwork Capacities T data B KochM E Hiller et al Eds pp273ndash290 MOS-SIAM 2015
[13] A Genz and F Bretz Computation of Multivariate Normal andt Probabilities vol 195 of Lecture Notes in Statistics Springer2009
[14] M H Farshbaf-Shaker R Henrion and D Homberg ldquoProp-erties of Chance Constraints in Infinite Dimensions with anApplication to PDE Constrained Optimizationrdquo Set-Valued andVariational Analysis pp 1ndash21 2017
[15] T Gonzalez Grandon H Heitsch and R Henrion ldquoA jointmodel of probabilisticrobust constraints for gas transportman-agement in stationary networksrdquo Computational ManagementScience vol 14 no 3 pp 443ndash460 2017
[16] W van Ackooij I Aleksovska and M Munoz-Zuniga ldquo(Sub-)Differentiability of Probability Functionswith Elliptical Distri-butionsrdquo Set-Valued and Variational Analysis
Mathematical Problems in Engineering 17
[17] W van Ackooij and R Henrion ldquoGradient formulae for nonlin-ear probabilistic constraints with GAUssian and GAUssian-likedistributionsrdquo SIAM Journal on Optimization vol 24 no 4 pp1864ndash1889 2014
[18] W van Ackooij and R Henrion ldquo(Sub-)gradient formulaefor probability functions of random inequality systems underGaussian distributionrdquo SIAMASA Journal on UncertaintyQuantification vol 5 no 1 pp 63ndash87 2017
[19] W van Ackooij and J Malick ldquoSecond-order differentiabilityof probability functionsrdquo Optimization Letters vol 11 no 1 pp179ndash194 2017
[20] H Eckey R Kosfeld and M Turck Wahrscheinlichkeitsrech-nung und Induktive Statistik - Grundlagen -Methoden - BeispieleGabler Wiesbaden Germany 2005
[21] A Prekopa Stochastic Programming Springer 1st edition 1995[22] I Joormann M Pfetsch R Schwarz et al ldquoGasLibmdashA Library
of Gas Network Instancesrdquo 2017[23] G Wedler Lehrbuch der Physikalischen Chemie Wiley-VCH
5th edition 2004[24] N H Chen ldquoAn explicit equation for friction factor in piperdquo
Industrial amp Engineering Chemistry Fundamentals vol 18 no3 pp 296-297 1979
[25] W Schenk and F Kremer Physikalisches Praktikum Springer14th edition 2014
[26] A J Zaslavski ldquoTurnpike Properties in the Calculus of Varia-tions and Optimal Controlrdquo inNonconvex Optimization and ItsApplications vol 80 Springer 2006
Hindawiwwwhindawicom Volume 2018
MathematicsJournal of
Hindawiwwwhindawicom Volume 2018
Mathematical Problems in Engineering
Applied MathematicsJournal of
Hindawiwwwhindawicom Volume 2018
Probability and StatisticsHindawiwwwhindawicom Volume 2018
Journal of
Hindawiwwwhindawicom Volume 2018
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawiwwwhindawicom Volume 2018
OptimizationJournal of
Hindawiwwwhindawicom Volume 2018
Hindawiwwwhindawicom Volume 2018
Engineering Mathematics
International Journal of
Hindawiwwwhindawicom Volume 2018
Operations ResearchAdvances in
Journal of
Hindawiwwwhindawicom Volume 2018
Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018
International Journal of Mathematics and Mathematical Sciences
Hindawiwwwhindawicom Volume 2018
Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom
The Scientific World Journal
Volume 2018
Hindawiwwwhindawicom Volume 2018Volume 2018
Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in
Nature and SocietyHindawiwwwhindawicom Volume 2018
Hindawiwwwhindawicom
Dierential EquationsInternational Journal of
Volume 2018
Hindawiwwwhindawicom Volume 2018
Decision SciencesAdvances in
Hindawiwwwhindawicom Volume 2018
AnalysisInternational Journal of
Hindawiwwwhindawicom Volume 2018
Stochastic AnalysisInternational Journal of
Submit your manuscripts atwwwhindawicom
Mathematical Problems in Engineering 11
Note that a vector 119887+ isin R119899+1 is suitable here if the vector+ is feasible for the subgraphs ie the inequalities (25)-(27)are fulfilled for +Proof (a) For this part we use a proof by contradictionAssume that
(119901119898119894119899119894(119896)0)2 minus 119892119894(119896)119891(119890119896)
(119894(119896)) gt (119901119898119886119909119895(119896)0)2 (83)
Because 119906119896 ge 1 for all 119896 = 1 1198982 it follows
(119901119898119894119899119894(119896)0)2 minus 119892119894(119896)119891(119890119896)
(119894(119896)) gt 1119906119896
(119901119898119886119909119895(119896)0)2 (84)
It it Σ119896lowast119894(119896) = 0 and Π119896lowast 119894(119896) = 1 because the subgraph withindex 119896lowast is equal to the subgraphwith index 119894(119896) And becausethe subgraphs 119894(119896) and 119895(119896) are neighbours (ie directlyconnected) it follows Π119896lowast 119895(119896) = 119906119896 and
Σ119896lowast119895(119896) = 119892(119896lowast119895(119896))1119891(119890119906(119896lowast119895(119896))1
) (119887(119896lowast119895(119896))1)= 119892119894(119896)119891(119890119896)
(119894(119896)) (85)
Thus we have a contradiction to (28) and 119906119896 cannot beoptimal
(b) As mentioned before the vector 119887+ is such that isfeasible for the subgraphs So because of the conditions for119896 isin 1 1198982 there exists a 119901119894(119896)0 isin [119901119898119894119899
119894(119896)0 119901119898119886119909119894(119896)0] and a119901119895(119896) isin [119901119898119894119899
119895(119896) 119901119898119886119909119895(119896) ] so that1199012119894(119896)0 minus 119892119894(119896)119891(119890119896)
(119894(119896)) ge 1199012119895(119896)0 (86)
With 1199012119894(119896)0 minus 119892119894(119896)119891(119890119896)
(119894(119896)) = 1199012119891(119890119896)
and 1199012119895(119896)0 = 1199012
ℎ(119890119896)it
follows
(119901ℎ(119890119896)119901119891(119890119896)
) ge 1 (87)
Therefore we can find a 119906 isin R1198982ge1 that the inequalities (28)-
(35) are fulfilled Now we can use theWeierstraszlig theorem andget the existence of an optimal solution
Note that the sufficient condition in Lemma 8 is com-paratively strong there may exist optimal solutions with lessstrong conditions eg the following
Corollary 9 If the graph has only one compressor edge ie|E119862| = 1 then it holds If 119906 isin Rge1 is an optimal solution of(76) 119906 is unique
This conclusion follows directly from the strict mono-tonicity of the objective function
Optimization with Random Loads In this section we con-sider the load vector to be random as we mentioned inSection 22 In the optimization problems we want to makesure that the feasibility of a random load vector is guaranteedfor a probability 120572 isin (0 1) This leads us to optimizationproblems with so called chance or probabilistic constraints
In general problems with probabilistic constraints have theform
min 119891 (119909 120585)subject to P (ℎ (119909 120585) le 0) ge 120572 (88)
with an objective function 119891 and a vector of uncertainty 120585There exist many works about probabilistic constraints respstochastic programming eg [7] gives an excellent overviewabout chance constraints in theory and application In ourcase we consider the following problem
min119901+119898119886119909isin119875
119888119879119901+119898119886119909
subject to P (119887 (119901+119898119886119909) isin 119872) ge 120572 (89)
To handle this problem resp the probabilistic constraint thespheric-radial decomposition gives us an explicit representa-tion of the constraint for a Gaussian vector 119887 sim N(120583 Σ) withmean value 120583 and positive definite covariance Σ
P (119887 (119901+119898119886119909) isin 119872)= int
S119899minus1120583120594 119903 ge 0 | 119903119871V + 120583 isin 119872119889120583120578 (V) (90)
Further Algorithm 3 gives a way to approximate this integralin an efficient way In Section 3 we showed a way to character-ize the feasible set119872 and how to adapt this characterization tothe spheric-radial decomposition This changes our problemwith probabilistic constraints (89) to
min119901+119898119886119909isin119875
119888119879119901+119898119886119909
subject to 1119873 119873sum119894=1
119904119905sum119905=1
F120594 (119886119894119905) minusF120594 (119886119894119905) ge 120572 (91)
Of course the interval bounds depend on the pressure boundsand the control For this problem we formulate the nextlemma
Lemma 10 Let a sampling V1 V119873 (from step (1) ofAlgorithm 3) and 120572 isin (0 1) be given For a suitable 120583 isin R119899
problem (91) has at least one solution 119901+119898119886119909 isin 119875Proof For the proof we fix a V isin V1 V119873 from the givensampling From (25) and (26) it follows for all 119896 = 1 119899119894with 119894 isin 1 1198982 + 10 le 119892119894119896 (119894 (119903)) + (119901119898119886119909
119894119896 )2 minus (1199011198981198941198991198940 )2
0 le minus119892119894119896 (119894 (119903)) + (1199011198981198861199091198940 )2 minus (119901119898119894119899
119894119896 )2 (92)
With a random vector 119887 sim N(120583 Σ)119892119894119896(119894(119903)) are quadraticfunctions in 119903 and the upper pressure bounds only influencethe constant part with a positive sign
0 le 1198861198941198961199032 + 119887119894119896119903 + 119888119894119896 + (119901119898119886119909119894119896 )2 minus (119901119898119894119899
1198940 )2 0 le minus1198861198941198961199032 minus 119887119894119896119903 minus 119888119894119896 + (119901119898119886119909
1198940 )2 minus (119901119898119894119899119894119896 )2 (93)
12 Mathematical Problems in Engineering
It follows that a rise in 119901119898119886119909119894119896 for all 119896 = 0 119899119894 enlarges
the intervals in which the inequalities hold The same holdsfor all inequalities inTheorem 5The cumulative distributionfunction of theChi-Square-Distribution is strictlymonotonicincreasing and convergent to 1 Thus upper pressure bounds119901119898119886119909119894119896 (for 119894 = 1 119899) can be found so that
1119873 119873sum119895=1
119904119895sum119905=1
F120594 (119886119895119905) minusF120594 (119886119895119905) (94)
is higher that 120572 isin (0 1) The index119873 is for all sampled pointson the sphere and the index 119904119895 is for the union of intervals inwhich all inequalities hold The interval boundaries 119886119895119905 and119886119895119905 depend continuously on the pressure bounds and becausethe inequalities are not strict 119901119898119886119909 can be found st the sideconstraint is fulfilled with equality So there exists at least onesolution of (91)
5 Numerical Results
In this section we show a few results of implementation Atfirst we show the idea of the spheric-radial decomposition byusing an easy example Next we show a easy example with onecompressor edge and at last we use real data of the Greek gasnetworkThe focus of the implementation is on the theoremsproofed in Section 4
Example 1 Our first example is an easy graph without innercontrol (see Figure 6)
Assume that node 1 and 2 are gas consumers with meandemand 120583 = [05 05]119879 The pressure bounds are given by119901+119898119894119899 = [2 1 1]119879 and 119901+119898119886119909 = [3 2 2]119879 and Φ = E2 Thenthe incidence matrix is
119860+ = [[[minus1 01 minus10 1
]]] resp 119860 = [1 minus10 1 ] (95)
Consider the inequalities of Theorem 5 to characterize thefeasible set Because the network has no compressor edgesthe system of inequalities only contains the information (25)-(27) the other inequalities do not occur The function 119892 isgiven by
119892 (119887) = [1 01 1] [1 00 1] sdot1003816100381610038161003816100381610038161003816100381610038161003816[1 10 1](11988711198872)
1003816100381610038161003816100381610038161003816100381610038161003816∘ ([1 10 1](11988711198872)) = ( 11988721 + 211988711198872 + 1198872211988721 + 211988711198872 + 211988722) (96)
By inserting the pressure bounds one can see that inequality(25) is always fulfilled and inequality (27) implies1198872 le radic3 (97)
With this inequality (26) follows for 0 le 1198872 le radic30 le 1198871 le minus1198872 + radic8 minus 11988722 (98)
So the feasible set119872 is the following (see Figure 7)
119872 = 119887 isin R2ge0 | 1198872 le radic3 and 1198871 le minus1198872 + radic8 minus 11988722 (99)
Now consider the optimization problem
min119901+119898119886119909isin119875
11198793119901+119898119886119909
subject to 119887 isin 119872 (119901+119898119886119909) (100)
with 119887 = 120583 = [05 05]119879 Lemma 7 guarantees the existenceof a unique solution of this problem We use the fmincon-function inMATLAB which minimizes a constrained non-linear multivariable function The fmincon default settinguses a interior-point method For more information we referto the official MathWorks homepage (httpswwwmath-workscomhelpoptimugfminconhtml)The solver returnsthe following result
119901+119898119886119909119900119901119905 = [2 radic3radic275]119879 (101)
One can easily see that this is the correct solution The gasdemand at node 2 is 05 which is equal to the flow 1199022 sousing the equation of momentum implies a difference in thequadratic pressures between node 1 and node 2 of 025 Theflow through edge 1 is given by the two exit nodes and it is1199021 = 1 This implies a difference in the quadratic pressuresbetweennode 0 andnode 1 of 1 So the optimal solution119901+119898119886119909
119900119901119905
directly followsNext we use the spheric-radial decomposition especially
Algorithm 3 to calculate the probability P(119887 isin 119872) for arandom vector 119887 sim N(E2 [05 05]119879) Therefore we viewthe results of 8 tests with 1000 sampled points in each one(Table 1)
The probability in all eight tests is nearly the same Themean value is 03479 and the variance is 27723 sdot 10minus6 whichis very small This shows that the implementation is workingnearly exact The efficiency of the implementation of courseis not perfect but that is not part of this work
Last wewant to solve the following optimization problem
min119901+119898119886119909isin119875
11198793119901+119898119886119909
subject to 1119873 119873sum119894=1
119904sum119905=1
F120594 (119886119894119905) minusF120594 (119886119894119905) ge 120572 (102)
We use again the fmincon-function inMATLAB to solve theproblem eight times with 1000 sampled points in every caseBecause Lemma 8 needs a suitable 120583 isin R2
ge0 we assume 119887 simN(01sdot1198642 [15 15])We set120572 fl 09 and the results are shownin Table 2
Example 2 Now we add a compressor edge to the graph ofExample 1 (see Figure 8)
Assume that node 1 and 3 are consumers node 2 is ainner nodeThemeandemand120583 = [05 0 05]119879Thepressure
Mathematical Problems in Engineering 13
minus2 minus1 1 2 3
minus2
minus1
1
2
3
b2
b1
minus1 1 2
minus1
1
2
b2
b1
rLv1
rLv2
rLv3
Scheme of the spheric-radial decomposition with
sphere (yellow)and three points 1 2 3 (red) at the unit
the feasible set Mminus for Σ = E2 = [05 05]TFeasible set M for p+min = [2 1 1]T and
p+max = [3 2 2]T
Figure 7 Feasible set of Example 1
Table 1 Test results of easy example
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 8P(119887 isin 119872) 03486 03481 03450 03501 03484 03499 03459 03475
Table 2 Test results of Example 1
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 81199011198981198861199091199001199011199050 54514 54714 54758 54028 54767 53921 54888 551671199011198981198861199091199001199011199051 29802 30126 30225 29725 30300 29485 30336 300291199011198981198861199091199001199011199052 10084 10004 10776 10001 11221 11390 10285 10558
0 1 2 3q1 q2 q3
b0 b1 b2 b3
e1 isin ℰF e2 isin ℰC e3 isin ℰF
Figure 8 Graph of Example 2
0 1 2 3q1 q2 q2 q3
b0 b1 b2 b3
e1 isin ℰF e3 isin ℰF
Figure 9 Separated graph of Example 2
bounds are given by 119901+119898119894119899 = [2 1 1 1]119879 and 119901+119898119886119909 = [3 22 2]119879 and 1206011198901= 1 and 1206011198903
= 1 The incidence matrix is givenby
119860+ = [[[[[[
minus1 0 01 minus1 00 1 minus10 0 1]]]]]]
resp 119860 = [[[1 minus1 00 1 minus10 0 1
]]] (103)
Before we have a look at the inequalities of Theorem 5we have to remove the compressor edge from the graphThe flows can be computed by using the equation of massconservation and the new load vector is then
= (1198871 + 11990221198872 minus 11990221198873 ) with
119902 = (1198871 + 1198872 + 11988731198872 + 11988731198873)(104)
We assume the control to be switched off so it holds119906 = 1With this the control edge 1198902 is treated as a flow edge withoutfriction The resulting graph is shown in Figure 9 from theinequalities (25)ndash(35) it follows
3 le radic3 and 1 le radic8 minus 23 (105)
14 Mathematical Problems in Engineering
minus1 1 2 3
1
2
3
minus1 1 2 3
1
2
3
b3
b1
Hyperspace of the feasible set MHyperspace of the feasible set M
b1
b3
Figure 10 Feasible set of the Example 2
Table 3 Test results of control example
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 8P(119887 isin 119872) 01531 01500 01540 01536 01530 01550 01567 01528so the feasible set is (see Figure 10)
= + isin R119899+1 | 1119879
4 + = 0 and 0 le 3 le radic3 and 2 = 0and 0 le 1 le radic8 minus 22 (106)
This implies
119872 = 119887 isin R3ge0 | 0 le 1198873 le radic3 and 1198872 = 0 and 0 le 1198871 le minus1198873
+ radic8 minus 11988723 (107)
Solving the problem
min119901+119898119886119909isin119875
11198794119901+119898119886119909
subject to isin (108)
for 119887 = 120583 = [05 0 05]119879 leads us to the optimal solution
119901+119898119886119909119900119901119905 = [2 radic3radic3radic275]119879 (109)
The argumentation for the optimality of this vector is exactlylike before in the easy example Becausewe have a compressoredge in this graph we can minimize the control Thereforewe fix the pressure bounds as we did at the beginning of thisexample and we set 119887 = [1 0 15]119879 So inequality (31) is notfulfilled anymore for 119906 = 1 We use again the fmincon-solverinMATLAB to solve the following problem
min119906ge1
1199062subject to 119887 isin 119872(119906) (110)
The solver returns the optimal solution 119906119900119901119905 = 11818
Here one can see that the sufficient condition in Lemma 8is really strong Itrsquos not fulfilled with the given values but thereexists a solution anyway Of course the necessary conditionis fulfilled
Now we use again Algorithm 3 to compute the proba-bility P(119887 isin 119872) for a random load vector 119887 sim N(E3 [0505 05]119879) Table 3 shows us the results
Again we can see that the probability is nearly the samein all eight cases The mean probability is 01535 and thevariance is 32369 sdot 10minus6 For the optimization we reducethe covariance matrix and assume 119887 sim N(001 sdot E3 [05 0505]119879) For a probability 120572 = 09 Table 4 states the resultsExample 3 In this example we use a part of a real networkThe focus here is on realistic values for our model Weget these values from the GasLib (see [22]) a collection oftechnical gas network data
ldquoThegoal ofGasLib is to promote research on gasnetworks by providing a set of large and realisticbenchmark instancesrdquo (httpgaslibzibde)
We use a part of the Greek gas network (GasLib-134) (seeFigure 11)
For using real values we first need to know the constant120601119890 from (5)
120601119890 = (119877119878119879)2 120582119863 1198711198862 (111)
The value 119877119878 is the specific gas constant which can becalculated by 119877119878 = 119877119872 with the ideal gas constant 119877 =831451119896119892 sdot 11989821199042 sdot 119898119900119897 sdot 119870 (see [23]) and the molar mass119872 = 1662119896119892119896119898119900119897 (from real data) of the used natural gasAlso the temperature 119879 = 28915119870 is given from real dataFor the (constant) friction factor 120582 we use the law of Chen(see [24])
Mathematical Problems in Engineering 15
Table 4 Test results of control example
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 81199011198981198861199091199001199011199050 39013 39223 39039 38901 38879 39039 38901 390981199011198981198861199091199001199011199051 16568 16565 16365 16497 16496 16365 16497 164921199011198981198861199091199001199011199052 16568 16565 16365 16497 16496 16365 16497 164981199011198981198861199091199001199011199053 10000 10000 10000 10000 10002 10000 10000 10000
Table 5 Values based on real data
variable 119877119878 119879 120582 119863 119871 119886 120601value 50027 28915 00093 9144 178693 43862 198 sdot 107unit 11989821199042 sdot 119870 119870 minus 119898119898 119896119898 119898119904 11989821199042
Figure 11 Greek gas network from GasLib [22] red edges are usedin our example
1radic120582 = minus2 log10 ( 11989611986337065minus 50425
Relog10 [(119896119863)1109828257 + 58506
Re08981]) (112)
Diameter 119863 = 9144119898119898 and roughness value 119896 = 8 sdot 10minus6119898for the first pipe (see Figure 11) are known from real data TheReynolds-Number Re can be calculated by Re = (V119863120588)120578with the dynamic viscosity 120578 = 1 sdot 10minus5119875119886 sdot 119904 (see [1])the mean velocity V = 9548119898119904 (calculated with the meanvolume flow rate which is given by the real data) of the gasand the density 120588 = 074331198961198921198983 (from real data) So we getRe = 648959074 asymp 649 sdot 106 and with this the pipe frictioncoefficient after Chen is 120582 = 00093
The length of the first pipe is given by 119871 = 178693119896119898from the real data Last value missing is the sound speed innatural gas which we assume to be ideal The sound speed
119886 depends on the modulus of compressibility (see [9]) Forideal gases the modulus of compressibility only depends onthe adiabatic exponent and the pressure Using the ideal gasequation simplifies the equation for the sound speed to
119886 = radic120581119877119879119872 (113)
The adiabatic exponent for methan which is the main partof natural gas (gt 90) is about 133 (see [25]) Together allvalues are shown in Table 5 (example for the pipe next to thesource in Figure 11)
Nowwewant to optimize the inner control in the networkfor a fix load vector Therefor we choose two nominations2011-11-01 and 2016-02-17 In both cases the fmincon-functioninMATLAB returns
119906119900119901119905 = 1 (114)
as a local minimum which satisfies the constraint 119887 isin 119872This means for these nominations a compressor stationis not needed so it is switched off With this as a lastcalculation in this paper we calculate the probability for thegiven nominations to be feasible We use the values from thenominations as mean value 120583 and set 119887 sim N(120583 sdot 10minus3E 120583)The results are shown in Tables 6 and 7
Again one can see that the results are quite similarThe mean probability in Table 6 is 09860 and the meanprobability in Table 7 is 09760 Both values are nearly 1 sothere must happen something really unforeseeable that aGreek does not get his demanded gas
6 Conclusion
In this paper we have used an existing stationary modelfor gas transport based on the isothermal Euler equationsWe have extended this model by compressor stations andpresented a characterization of the set of feasible loadsEspecially Theorem 5 contains the central idea to get anaccess to the optimization in Section 4 and the spheric-radialdecomposition is our main tool to handle the uncertainty inthe loadsWewere able to show some results for optimizationwith probabilistic constraints in gas transport In a next stepthese results should be generalized
16 Mathematical Problems in Engineering
Table 6 Test results for nomination 2011-11-01
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 8P(119887 isin 119872) 09849 09849 09878 09885 09863 09845 09854 09853
Table 7 Test results for nomination 2016-02-17
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 8P(119887 isin 119872) 09744 09756 09788 09788 09754 09775 09734 09739
With the characterization of the feasible set we haveshown that the feasibility of a load depends significantly onthe pressure bounds These pressure bounds play a mainrole in our optimization problems The other main partsare the inner controls modeled by compressor edges whichshould lead in optimal control problems with probabilisticconstraints Here we have built a base for further works withthis topic
The idea of separating the graph and removing thecompressor edges can be used to extend themodel with othercomponents like eg control valves and resistorsThiswouldchange the inequalities inTheorem 5 but the approach worksalso for thesemodifications Further [8] showed a way to dealwith a networkwhich is a cycle Combining bothmethods weget an access to model gas networks in high detail
Finally the expansion to a dynamic model might bealso possible The idea is to use a space-mapping approachbecause one can use a lot of the analysis we did hereTherefore upcoming papers should deal with these problemsand present results of the turnpike-theory which describesa relation between stationary and instationary solutions Infact turnpike-theory implies that the optimal dynamic stateis close to the optimal static state in the interior of the timeintervals for sufficiently long time horizons AJ Zaslavskiexplains the turnpike-phenomenon in [26] This book alsogives an overview about the theory
Data Availability
The data used to support the findings of this study areavailable from the corresponding author upon request
Conflicts of Interest
The authors declare that there are no conflicts of interestregarding the publication of this paper
Acknowledgments
This work was supported by DFG in the framework of theCollaborative Research Centre CRCTransregio 154 Math-ematical Modeling Simulation and Optimization Using theExample of Gas Networks project C03
References
[1] P Domschke B Hiller J Lang and C Tischendorf ldquoModel-lierung von gasnetzwerken Eine Ubersichtrdquo Tech Rep 2717Technische Universitat Darmstadt 2017
[2] T Koch B Hiller M E Pfetsch and L Schewe Evaluating GasNetwork Capacities MOS-SIAM 2015
[3] M Gugat R Schultz and D Wintergerst ldquoNetworks of pipe-lines for gas with nonconstant compressibility factor stationarystatesrdquoComputational amp AppliedMathematics vol 37 no 2 pp1066ndash1097 2018
[4] M Gugat and M Herty ldquoExistence of classical solutions andfeedback stabilization for the flow in gas networksrdquo ESAIMControl Optimisation and Calculus of Variations vol 17 no 1pp 28ndash51 2011
[5] M K Banda M Herty and A Klar ldquoGas flow in pipeline net-worksrdquo Networks and Heterogeneous Media vol 1 no 1 pp 41ndash56 2006
[6] RM Colombo G GuerraM Herty and V Schleper ldquoOptimalcontrol in networks of pipes and canalsrdquo SIAM Journal onControl and Optimization vol 48 no 3 pp 2032ndash2050 2009
[7] A Shapiro D Dentcheva and A Ruszczynski ectures onStochastic Programming Modeling and Theory MPS-SIAM2009
[8] C Gotzes H Heitsch R Henrion and R Schultz ldquoOn thequantification of nomination feasibility in stationary gas net-works with random loadrdquoMathematical Methods of OperationsResearch vol 84 no 2 pp 427ndash457 2016
[9] D Halliday R Resnick and J Walker Physik Wiley-VCHBachelor edition edition 2007
[10] M Gugat F M Hante M Hirsch-Dick and G LeugeringldquoStationary states in gas networksrdquoNetworks andHeterogeneousMedia vol 10 no 2 pp 295ndash320 2015
[11] D RoseM SchmidtM C Steinbach and BMWillert ldquoCom-putational optimization of gas compressor stations MINLPmodels versus continuous reformulationsrdquoMathematical Meth-ods of Operations Research vol 83 no 3 pp 409ndash444 2016
[12] H Heitsch R Henrion H Leovey et al ldquoEmpirical observa-tions and statistical analysis of gas demandrdquo in Evaluating GasNetwork Capacities T data B KochM E Hiller et al Eds pp273ndash290 MOS-SIAM 2015
[13] A Genz and F Bretz Computation of Multivariate Normal andt Probabilities vol 195 of Lecture Notes in Statistics Springer2009
[14] M H Farshbaf-Shaker R Henrion and D Homberg ldquoProp-erties of Chance Constraints in Infinite Dimensions with anApplication to PDE Constrained Optimizationrdquo Set-Valued andVariational Analysis pp 1ndash21 2017
[15] T Gonzalez Grandon H Heitsch and R Henrion ldquoA jointmodel of probabilisticrobust constraints for gas transportman-agement in stationary networksrdquo Computational ManagementScience vol 14 no 3 pp 443ndash460 2017
[16] W van Ackooij I Aleksovska and M Munoz-Zuniga ldquo(Sub-)Differentiability of Probability Functionswith Elliptical Distri-butionsrdquo Set-Valued and Variational Analysis
Mathematical Problems in Engineering 17
[17] W van Ackooij and R Henrion ldquoGradient formulae for nonlin-ear probabilistic constraints with GAUssian and GAUssian-likedistributionsrdquo SIAM Journal on Optimization vol 24 no 4 pp1864ndash1889 2014
[18] W van Ackooij and R Henrion ldquo(Sub-)gradient formulaefor probability functions of random inequality systems underGaussian distributionrdquo SIAMASA Journal on UncertaintyQuantification vol 5 no 1 pp 63ndash87 2017
[19] W van Ackooij and J Malick ldquoSecond-order differentiabilityof probability functionsrdquo Optimization Letters vol 11 no 1 pp179ndash194 2017
[20] H Eckey R Kosfeld and M Turck Wahrscheinlichkeitsrech-nung und Induktive Statistik - Grundlagen -Methoden - BeispieleGabler Wiesbaden Germany 2005
[21] A Prekopa Stochastic Programming Springer 1st edition 1995[22] I Joormann M Pfetsch R Schwarz et al ldquoGasLibmdashA Library
of Gas Network Instancesrdquo 2017[23] G Wedler Lehrbuch der Physikalischen Chemie Wiley-VCH
5th edition 2004[24] N H Chen ldquoAn explicit equation for friction factor in piperdquo
Industrial amp Engineering Chemistry Fundamentals vol 18 no3 pp 296-297 1979
[25] W Schenk and F Kremer Physikalisches Praktikum Springer14th edition 2014
[26] A J Zaslavski ldquoTurnpike Properties in the Calculus of Varia-tions and Optimal Controlrdquo inNonconvex Optimization and ItsApplications vol 80 Springer 2006
Hindawiwwwhindawicom Volume 2018
MathematicsJournal of
Hindawiwwwhindawicom Volume 2018
Mathematical Problems in Engineering
Applied MathematicsJournal of
Hindawiwwwhindawicom Volume 2018
Probability and StatisticsHindawiwwwhindawicom Volume 2018
Journal of
Hindawiwwwhindawicom Volume 2018
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawiwwwhindawicom Volume 2018
OptimizationJournal of
Hindawiwwwhindawicom Volume 2018
Hindawiwwwhindawicom Volume 2018
Engineering Mathematics
International Journal of
Hindawiwwwhindawicom Volume 2018
Operations ResearchAdvances in
Journal of
Hindawiwwwhindawicom Volume 2018
Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018
International Journal of Mathematics and Mathematical Sciences
Hindawiwwwhindawicom Volume 2018
Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom
The Scientific World Journal
Volume 2018
Hindawiwwwhindawicom Volume 2018Volume 2018
Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in
Nature and SocietyHindawiwwwhindawicom Volume 2018
Hindawiwwwhindawicom
Dierential EquationsInternational Journal of
Volume 2018
Hindawiwwwhindawicom Volume 2018
Decision SciencesAdvances in
Hindawiwwwhindawicom Volume 2018
AnalysisInternational Journal of
Hindawiwwwhindawicom Volume 2018
Stochastic AnalysisInternational Journal of
Submit your manuscripts atwwwhindawicom
12 Mathematical Problems in Engineering
It follows that a rise in 119901119898119886119909119894119896 for all 119896 = 0 119899119894 enlarges
the intervals in which the inequalities hold The same holdsfor all inequalities inTheorem 5The cumulative distributionfunction of theChi-Square-Distribution is strictlymonotonicincreasing and convergent to 1 Thus upper pressure bounds119901119898119886119909119894119896 (for 119894 = 1 119899) can be found so that
1119873 119873sum119895=1
119904119895sum119905=1
F120594 (119886119895119905) minusF120594 (119886119895119905) (94)
is higher that 120572 isin (0 1) The index119873 is for all sampled pointson the sphere and the index 119904119895 is for the union of intervals inwhich all inequalities hold The interval boundaries 119886119895119905 and119886119895119905 depend continuously on the pressure bounds and becausethe inequalities are not strict 119901119898119886119909 can be found st the sideconstraint is fulfilled with equality So there exists at least onesolution of (91)
5 Numerical Results
In this section we show a few results of implementation Atfirst we show the idea of the spheric-radial decomposition byusing an easy example Next we show a easy example with onecompressor edge and at last we use real data of the Greek gasnetworkThe focus of the implementation is on the theoremsproofed in Section 4
Example 1 Our first example is an easy graph without innercontrol (see Figure 6)
Assume that node 1 and 2 are gas consumers with meandemand 120583 = [05 05]119879 The pressure bounds are given by119901+119898119894119899 = [2 1 1]119879 and 119901+119898119886119909 = [3 2 2]119879 and Φ = E2 Thenthe incidence matrix is
119860+ = [[[minus1 01 minus10 1
]]] resp 119860 = [1 minus10 1 ] (95)
Consider the inequalities of Theorem 5 to characterize thefeasible set Because the network has no compressor edgesthe system of inequalities only contains the information (25)-(27) the other inequalities do not occur The function 119892 isgiven by
119892 (119887) = [1 01 1] [1 00 1] sdot1003816100381610038161003816100381610038161003816100381610038161003816[1 10 1](11988711198872)
1003816100381610038161003816100381610038161003816100381610038161003816∘ ([1 10 1](11988711198872)) = ( 11988721 + 211988711198872 + 1198872211988721 + 211988711198872 + 211988722) (96)
By inserting the pressure bounds one can see that inequality(25) is always fulfilled and inequality (27) implies1198872 le radic3 (97)
With this inequality (26) follows for 0 le 1198872 le radic30 le 1198871 le minus1198872 + radic8 minus 11988722 (98)
So the feasible set119872 is the following (see Figure 7)
119872 = 119887 isin R2ge0 | 1198872 le radic3 and 1198871 le minus1198872 + radic8 minus 11988722 (99)
Now consider the optimization problem
min119901+119898119886119909isin119875
11198793119901+119898119886119909
subject to 119887 isin 119872 (119901+119898119886119909) (100)
with 119887 = 120583 = [05 05]119879 Lemma 7 guarantees the existenceof a unique solution of this problem We use the fmincon-function inMATLAB which minimizes a constrained non-linear multivariable function The fmincon default settinguses a interior-point method For more information we referto the official MathWorks homepage (httpswwwmath-workscomhelpoptimugfminconhtml)The solver returnsthe following result
119901+119898119886119909119900119901119905 = [2 radic3radic275]119879 (101)
One can easily see that this is the correct solution The gasdemand at node 2 is 05 which is equal to the flow 1199022 sousing the equation of momentum implies a difference in thequadratic pressures between node 1 and node 2 of 025 Theflow through edge 1 is given by the two exit nodes and it is1199021 = 1 This implies a difference in the quadratic pressuresbetweennode 0 andnode 1 of 1 So the optimal solution119901+119898119886119909
119900119901119905
directly followsNext we use the spheric-radial decomposition especially
Algorithm 3 to calculate the probability P(119887 isin 119872) for arandom vector 119887 sim N(E2 [05 05]119879) Therefore we viewthe results of 8 tests with 1000 sampled points in each one(Table 1)
The probability in all eight tests is nearly the same Themean value is 03479 and the variance is 27723 sdot 10minus6 whichis very small This shows that the implementation is workingnearly exact The efficiency of the implementation of courseis not perfect but that is not part of this work
Last wewant to solve the following optimization problem
min119901+119898119886119909isin119875
11198793119901+119898119886119909
subject to 1119873 119873sum119894=1
119904sum119905=1
F120594 (119886119894119905) minusF120594 (119886119894119905) ge 120572 (102)
We use again the fmincon-function inMATLAB to solve theproblem eight times with 1000 sampled points in every caseBecause Lemma 8 needs a suitable 120583 isin R2
ge0 we assume 119887 simN(01sdot1198642 [15 15])We set120572 fl 09 and the results are shownin Table 2
Example 2 Now we add a compressor edge to the graph ofExample 1 (see Figure 8)
Assume that node 1 and 3 are consumers node 2 is ainner nodeThemeandemand120583 = [05 0 05]119879Thepressure
Mathematical Problems in Engineering 13
minus2 minus1 1 2 3
minus2
minus1
1
2
3
b2
b1
minus1 1 2
minus1
1
2
b2
b1
rLv1
rLv2
rLv3
Scheme of the spheric-radial decomposition with
sphere (yellow)and three points 1 2 3 (red) at the unit
the feasible set Mminus for Σ = E2 = [05 05]TFeasible set M for p+min = [2 1 1]T and
p+max = [3 2 2]T
Figure 7 Feasible set of Example 1
Table 1 Test results of easy example
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 8P(119887 isin 119872) 03486 03481 03450 03501 03484 03499 03459 03475
Table 2 Test results of Example 1
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 81199011198981198861199091199001199011199050 54514 54714 54758 54028 54767 53921 54888 551671199011198981198861199091199001199011199051 29802 30126 30225 29725 30300 29485 30336 300291199011198981198861199091199001199011199052 10084 10004 10776 10001 11221 11390 10285 10558
0 1 2 3q1 q2 q3
b0 b1 b2 b3
e1 isin ℰF e2 isin ℰC e3 isin ℰF
Figure 8 Graph of Example 2
0 1 2 3q1 q2 q2 q3
b0 b1 b2 b3
e1 isin ℰF e3 isin ℰF
Figure 9 Separated graph of Example 2
bounds are given by 119901+119898119894119899 = [2 1 1 1]119879 and 119901+119898119886119909 = [3 22 2]119879 and 1206011198901= 1 and 1206011198903
= 1 The incidence matrix is givenby
119860+ = [[[[[[
minus1 0 01 minus1 00 1 minus10 0 1]]]]]]
resp 119860 = [[[1 minus1 00 1 minus10 0 1
]]] (103)
Before we have a look at the inequalities of Theorem 5we have to remove the compressor edge from the graphThe flows can be computed by using the equation of massconservation and the new load vector is then
= (1198871 + 11990221198872 minus 11990221198873 ) with
119902 = (1198871 + 1198872 + 11988731198872 + 11988731198873)(104)
We assume the control to be switched off so it holds119906 = 1With this the control edge 1198902 is treated as a flow edge withoutfriction The resulting graph is shown in Figure 9 from theinequalities (25)ndash(35) it follows
3 le radic3 and 1 le radic8 minus 23 (105)
14 Mathematical Problems in Engineering
minus1 1 2 3
1
2
3
minus1 1 2 3
1
2
3
b3
b1
Hyperspace of the feasible set MHyperspace of the feasible set M
b1
b3
Figure 10 Feasible set of the Example 2
Table 3 Test results of control example
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 8P(119887 isin 119872) 01531 01500 01540 01536 01530 01550 01567 01528so the feasible set is (see Figure 10)
= + isin R119899+1 | 1119879
4 + = 0 and 0 le 3 le radic3 and 2 = 0and 0 le 1 le radic8 minus 22 (106)
This implies
119872 = 119887 isin R3ge0 | 0 le 1198873 le radic3 and 1198872 = 0 and 0 le 1198871 le minus1198873
+ radic8 minus 11988723 (107)
Solving the problem
min119901+119898119886119909isin119875
11198794119901+119898119886119909
subject to isin (108)
for 119887 = 120583 = [05 0 05]119879 leads us to the optimal solution
119901+119898119886119909119900119901119905 = [2 radic3radic3radic275]119879 (109)
The argumentation for the optimality of this vector is exactlylike before in the easy example Becausewe have a compressoredge in this graph we can minimize the control Thereforewe fix the pressure bounds as we did at the beginning of thisexample and we set 119887 = [1 0 15]119879 So inequality (31) is notfulfilled anymore for 119906 = 1 We use again the fmincon-solverinMATLAB to solve the following problem
min119906ge1
1199062subject to 119887 isin 119872(119906) (110)
The solver returns the optimal solution 119906119900119901119905 = 11818
Here one can see that the sufficient condition in Lemma 8is really strong Itrsquos not fulfilled with the given values but thereexists a solution anyway Of course the necessary conditionis fulfilled
Now we use again Algorithm 3 to compute the proba-bility P(119887 isin 119872) for a random load vector 119887 sim N(E3 [0505 05]119879) Table 3 shows us the results
Again we can see that the probability is nearly the samein all eight cases The mean probability is 01535 and thevariance is 32369 sdot 10minus6 For the optimization we reducethe covariance matrix and assume 119887 sim N(001 sdot E3 [05 0505]119879) For a probability 120572 = 09 Table 4 states the resultsExample 3 In this example we use a part of a real networkThe focus here is on realistic values for our model Weget these values from the GasLib (see [22]) a collection oftechnical gas network data
ldquoThegoal ofGasLib is to promote research on gasnetworks by providing a set of large and realisticbenchmark instancesrdquo (httpgaslibzibde)
We use a part of the Greek gas network (GasLib-134) (seeFigure 11)
For using real values we first need to know the constant120601119890 from (5)
120601119890 = (119877119878119879)2 120582119863 1198711198862 (111)
The value 119877119878 is the specific gas constant which can becalculated by 119877119878 = 119877119872 with the ideal gas constant 119877 =831451119896119892 sdot 11989821199042 sdot 119898119900119897 sdot 119870 (see [23]) and the molar mass119872 = 1662119896119892119896119898119900119897 (from real data) of the used natural gasAlso the temperature 119879 = 28915119870 is given from real dataFor the (constant) friction factor 120582 we use the law of Chen(see [24])
Mathematical Problems in Engineering 15
Table 4 Test results of control example
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 81199011198981198861199091199001199011199050 39013 39223 39039 38901 38879 39039 38901 390981199011198981198861199091199001199011199051 16568 16565 16365 16497 16496 16365 16497 164921199011198981198861199091199001199011199052 16568 16565 16365 16497 16496 16365 16497 164981199011198981198861199091199001199011199053 10000 10000 10000 10000 10002 10000 10000 10000
Table 5 Values based on real data
variable 119877119878 119879 120582 119863 119871 119886 120601value 50027 28915 00093 9144 178693 43862 198 sdot 107unit 11989821199042 sdot 119870 119870 minus 119898119898 119896119898 119898119904 11989821199042
Figure 11 Greek gas network from GasLib [22] red edges are usedin our example
1radic120582 = minus2 log10 ( 11989611986337065minus 50425
Relog10 [(119896119863)1109828257 + 58506
Re08981]) (112)
Diameter 119863 = 9144119898119898 and roughness value 119896 = 8 sdot 10minus6119898for the first pipe (see Figure 11) are known from real data TheReynolds-Number Re can be calculated by Re = (V119863120588)120578with the dynamic viscosity 120578 = 1 sdot 10minus5119875119886 sdot 119904 (see [1])the mean velocity V = 9548119898119904 (calculated with the meanvolume flow rate which is given by the real data) of the gasand the density 120588 = 074331198961198921198983 (from real data) So we getRe = 648959074 asymp 649 sdot 106 and with this the pipe frictioncoefficient after Chen is 120582 = 00093
The length of the first pipe is given by 119871 = 178693119896119898from the real data Last value missing is the sound speed innatural gas which we assume to be ideal The sound speed
119886 depends on the modulus of compressibility (see [9]) Forideal gases the modulus of compressibility only depends onthe adiabatic exponent and the pressure Using the ideal gasequation simplifies the equation for the sound speed to
119886 = radic120581119877119879119872 (113)
The adiabatic exponent for methan which is the main partof natural gas (gt 90) is about 133 (see [25]) Together allvalues are shown in Table 5 (example for the pipe next to thesource in Figure 11)
Nowwewant to optimize the inner control in the networkfor a fix load vector Therefor we choose two nominations2011-11-01 and 2016-02-17 In both cases the fmincon-functioninMATLAB returns
119906119900119901119905 = 1 (114)
as a local minimum which satisfies the constraint 119887 isin 119872This means for these nominations a compressor stationis not needed so it is switched off With this as a lastcalculation in this paper we calculate the probability for thegiven nominations to be feasible We use the values from thenominations as mean value 120583 and set 119887 sim N(120583 sdot 10minus3E 120583)The results are shown in Tables 6 and 7
Again one can see that the results are quite similarThe mean probability in Table 6 is 09860 and the meanprobability in Table 7 is 09760 Both values are nearly 1 sothere must happen something really unforeseeable that aGreek does not get his demanded gas
6 Conclusion
In this paper we have used an existing stationary modelfor gas transport based on the isothermal Euler equationsWe have extended this model by compressor stations andpresented a characterization of the set of feasible loadsEspecially Theorem 5 contains the central idea to get anaccess to the optimization in Section 4 and the spheric-radialdecomposition is our main tool to handle the uncertainty inthe loadsWewere able to show some results for optimizationwith probabilistic constraints in gas transport In a next stepthese results should be generalized
16 Mathematical Problems in Engineering
Table 6 Test results for nomination 2011-11-01
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 8P(119887 isin 119872) 09849 09849 09878 09885 09863 09845 09854 09853
Table 7 Test results for nomination 2016-02-17
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 8P(119887 isin 119872) 09744 09756 09788 09788 09754 09775 09734 09739
With the characterization of the feasible set we haveshown that the feasibility of a load depends significantly onthe pressure bounds These pressure bounds play a mainrole in our optimization problems The other main partsare the inner controls modeled by compressor edges whichshould lead in optimal control problems with probabilisticconstraints Here we have built a base for further works withthis topic
The idea of separating the graph and removing thecompressor edges can be used to extend themodel with othercomponents like eg control valves and resistorsThiswouldchange the inequalities inTheorem 5 but the approach worksalso for thesemodifications Further [8] showed a way to dealwith a networkwhich is a cycle Combining bothmethods weget an access to model gas networks in high detail
Finally the expansion to a dynamic model might bealso possible The idea is to use a space-mapping approachbecause one can use a lot of the analysis we did hereTherefore upcoming papers should deal with these problemsand present results of the turnpike-theory which describesa relation between stationary and instationary solutions Infact turnpike-theory implies that the optimal dynamic stateis close to the optimal static state in the interior of the timeintervals for sufficiently long time horizons AJ Zaslavskiexplains the turnpike-phenomenon in [26] This book alsogives an overview about the theory
Data Availability
The data used to support the findings of this study areavailable from the corresponding author upon request
Conflicts of Interest
The authors declare that there are no conflicts of interestregarding the publication of this paper
Acknowledgments
This work was supported by DFG in the framework of theCollaborative Research Centre CRCTransregio 154 Math-ematical Modeling Simulation and Optimization Using theExample of Gas Networks project C03
References
[1] P Domschke B Hiller J Lang and C Tischendorf ldquoModel-lierung von gasnetzwerken Eine Ubersichtrdquo Tech Rep 2717Technische Universitat Darmstadt 2017
[2] T Koch B Hiller M E Pfetsch and L Schewe Evaluating GasNetwork Capacities MOS-SIAM 2015
[3] M Gugat R Schultz and D Wintergerst ldquoNetworks of pipe-lines for gas with nonconstant compressibility factor stationarystatesrdquoComputational amp AppliedMathematics vol 37 no 2 pp1066ndash1097 2018
[4] M Gugat and M Herty ldquoExistence of classical solutions andfeedback stabilization for the flow in gas networksrdquo ESAIMControl Optimisation and Calculus of Variations vol 17 no 1pp 28ndash51 2011
[5] M K Banda M Herty and A Klar ldquoGas flow in pipeline net-worksrdquo Networks and Heterogeneous Media vol 1 no 1 pp 41ndash56 2006
[6] RM Colombo G GuerraM Herty and V Schleper ldquoOptimalcontrol in networks of pipes and canalsrdquo SIAM Journal onControl and Optimization vol 48 no 3 pp 2032ndash2050 2009
[7] A Shapiro D Dentcheva and A Ruszczynski ectures onStochastic Programming Modeling and Theory MPS-SIAM2009
[8] C Gotzes H Heitsch R Henrion and R Schultz ldquoOn thequantification of nomination feasibility in stationary gas net-works with random loadrdquoMathematical Methods of OperationsResearch vol 84 no 2 pp 427ndash457 2016
[9] D Halliday R Resnick and J Walker Physik Wiley-VCHBachelor edition edition 2007
[10] M Gugat F M Hante M Hirsch-Dick and G LeugeringldquoStationary states in gas networksrdquoNetworks andHeterogeneousMedia vol 10 no 2 pp 295ndash320 2015
[11] D RoseM SchmidtM C Steinbach and BMWillert ldquoCom-putational optimization of gas compressor stations MINLPmodels versus continuous reformulationsrdquoMathematical Meth-ods of Operations Research vol 83 no 3 pp 409ndash444 2016
[12] H Heitsch R Henrion H Leovey et al ldquoEmpirical observa-tions and statistical analysis of gas demandrdquo in Evaluating GasNetwork Capacities T data B KochM E Hiller et al Eds pp273ndash290 MOS-SIAM 2015
[13] A Genz and F Bretz Computation of Multivariate Normal andt Probabilities vol 195 of Lecture Notes in Statistics Springer2009
[14] M H Farshbaf-Shaker R Henrion and D Homberg ldquoProp-erties of Chance Constraints in Infinite Dimensions with anApplication to PDE Constrained Optimizationrdquo Set-Valued andVariational Analysis pp 1ndash21 2017
[15] T Gonzalez Grandon H Heitsch and R Henrion ldquoA jointmodel of probabilisticrobust constraints for gas transportman-agement in stationary networksrdquo Computational ManagementScience vol 14 no 3 pp 443ndash460 2017
[16] W van Ackooij I Aleksovska and M Munoz-Zuniga ldquo(Sub-)Differentiability of Probability Functionswith Elliptical Distri-butionsrdquo Set-Valued and Variational Analysis
Mathematical Problems in Engineering 17
[17] W van Ackooij and R Henrion ldquoGradient formulae for nonlin-ear probabilistic constraints with GAUssian and GAUssian-likedistributionsrdquo SIAM Journal on Optimization vol 24 no 4 pp1864ndash1889 2014
[18] W van Ackooij and R Henrion ldquo(Sub-)gradient formulaefor probability functions of random inequality systems underGaussian distributionrdquo SIAMASA Journal on UncertaintyQuantification vol 5 no 1 pp 63ndash87 2017
[19] W van Ackooij and J Malick ldquoSecond-order differentiabilityof probability functionsrdquo Optimization Letters vol 11 no 1 pp179ndash194 2017
[20] H Eckey R Kosfeld and M Turck Wahrscheinlichkeitsrech-nung und Induktive Statistik - Grundlagen -Methoden - BeispieleGabler Wiesbaden Germany 2005
[21] A Prekopa Stochastic Programming Springer 1st edition 1995[22] I Joormann M Pfetsch R Schwarz et al ldquoGasLibmdashA Library
of Gas Network Instancesrdquo 2017[23] G Wedler Lehrbuch der Physikalischen Chemie Wiley-VCH
5th edition 2004[24] N H Chen ldquoAn explicit equation for friction factor in piperdquo
Industrial amp Engineering Chemistry Fundamentals vol 18 no3 pp 296-297 1979
[25] W Schenk and F Kremer Physikalisches Praktikum Springer14th edition 2014
[26] A J Zaslavski ldquoTurnpike Properties in the Calculus of Varia-tions and Optimal Controlrdquo inNonconvex Optimization and ItsApplications vol 80 Springer 2006
Hindawiwwwhindawicom Volume 2018
MathematicsJournal of
Hindawiwwwhindawicom Volume 2018
Mathematical Problems in Engineering
Applied MathematicsJournal of
Hindawiwwwhindawicom Volume 2018
Probability and StatisticsHindawiwwwhindawicom Volume 2018
Journal of
Hindawiwwwhindawicom Volume 2018
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawiwwwhindawicom Volume 2018
OptimizationJournal of
Hindawiwwwhindawicom Volume 2018
Hindawiwwwhindawicom Volume 2018
Engineering Mathematics
International Journal of
Hindawiwwwhindawicom Volume 2018
Operations ResearchAdvances in
Journal of
Hindawiwwwhindawicom Volume 2018
Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018
International Journal of Mathematics and Mathematical Sciences
Hindawiwwwhindawicom Volume 2018
Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom
The Scientific World Journal
Volume 2018
Hindawiwwwhindawicom Volume 2018Volume 2018
Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in
Nature and SocietyHindawiwwwhindawicom Volume 2018
Hindawiwwwhindawicom
Dierential EquationsInternational Journal of
Volume 2018
Hindawiwwwhindawicom Volume 2018
Decision SciencesAdvances in
Hindawiwwwhindawicom Volume 2018
AnalysisInternational Journal of
Hindawiwwwhindawicom Volume 2018
Stochastic AnalysisInternational Journal of
Submit your manuscripts atwwwhindawicom
Mathematical Problems in Engineering 13
minus2 minus1 1 2 3
minus2
minus1
1
2
3
b2
b1
minus1 1 2
minus1
1
2
b2
b1
rLv1
rLv2
rLv3
Scheme of the spheric-radial decomposition with
sphere (yellow)and three points 1 2 3 (red) at the unit
the feasible set Mminus for Σ = E2 = [05 05]TFeasible set M for p+min = [2 1 1]T and
p+max = [3 2 2]T
Figure 7 Feasible set of Example 1
Table 1 Test results of easy example
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 8P(119887 isin 119872) 03486 03481 03450 03501 03484 03499 03459 03475
Table 2 Test results of Example 1
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 81199011198981198861199091199001199011199050 54514 54714 54758 54028 54767 53921 54888 551671199011198981198861199091199001199011199051 29802 30126 30225 29725 30300 29485 30336 300291199011198981198861199091199001199011199052 10084 10004 10776 10001 11221 11390 10285 10558
0 1 2 3q1 q2 q3
b0 b1 b2 b3
e1 isin ℰF e2 isin ℰC e3 isin ℰF
Figure 8 Graph of Example 2
0 1 2 3q1 q2 q2 q3
b0 b1 b2 b3
e1 isin ℰF e3 isin ℰF
Figure 9 Separated graph of Example 2
bounds are given by 119901+119898119894119899 = [2 1 1 1]119879 and 119901+119898119886119909 = [3 22 2]119879 and 1206011198901= 1 and 1206011198903
= 1 The incidence matrix is givenby
119860+ = [[[[[[
minus1 0 01 minus1 00 1 minus10 0 1]]]]]]
resp 119860 = [[[1 minus1 00 1 minus10 0 1
]]] (103)
Before we have a look at the inequalities of Theorem 5we have to remove the compressor edge from the graphThe flows can be computed by using the equation of massconservation and the new load vector is then
= (1198871 + 11990221198872 minus 11990221198873 ) with
119902 = (1198871 + 1198872 + 11988731198872 + 11988731198873)(104)
We assume the control to be switched off so it holds119906 = 1With this the control edge 1198902 is treated as a flow edge withoutfriction The resulting graph is shown in Figure 9 from theinequalities (25)ndash(35) it follows
3 le radic3 and 1 le radic8 minus 23 (105)
14 Mathematical Problems in Engineering
minus1 1 2 3
1
2
3
minus1 1 2 3
1
2
3
b3
b1
Hyperspace of the feasible set MHyperspace of the feasible set M
b1
b3
Figure 10 Feasible set of the Example 2
Table 3 Test results of control example
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 8P(119887 isin 119872) 01531 01500 01540 01536 01530 01550 01567 01528so the feasible set is (see Figure 10)
= + isin R119899+1 | 1119879
4 + = 0 and 0 le 3 le radic3 and 2 = 0and 0 le 1 le radic8 minus 22 (106)
This implies
119872 = 119887 isin R3ge0 | 0 le 1198873 le radic3 and 1198872 = 0 and 0 le 1198871 le minus1198873
+ radic8 minus 11988723 (107)
Solving the problem
min119901+119898119886119909isin119875
11198794119901+119898119886119909
subject to isin (108)
for 119887 = 120583 = [05 0 05]119879 leads us to the optimal solution
119901+119898119886119909119900119901119905 = [2 radic3radic3radic275]119879 (109)
The argumentation for the optimality of this vector is exactlylike before in the easy example Becausewe have a compressoredge in this graph we can minimize the control Thereforewe fix the pressure bounds as we did at the beginning of thisexample and we set 119887 = [1 0 15]119879 So inequality (31) is notfulfilled anymore for 119906 = 1 We use again the fmincon-solverinMATLAB to solve the following problem
min119906ge1
1199062subject to 119887 isin 119872(119906) (110)
The solver returns the optimal solution 119906119900119901119905 = 11818
Here one can see that the sufficient condition in Lemma 8is really strong Itrsquos not fulfilled with the given values but thereexists a solution anyway Of course the necessary conditionis fulfilled
Now we use again Algorithm 3 to compute the proba-bility P(119887 isin 119872) for a random load vector 119887 sim N(E3 [0505 05]119879) Table 3 shows us the results
Again we can see that the probability is nearly the samein all eight cases The mean probability is 01535 and thevariance is 32369 sdot 10minus6 For the optimization we reducethe covariance matrix and assume 119887 sim N(001 sdot E3 [05 0505]119879) For a probability 120572 = 09 Table 4 states the resultsExample 3 In this example we use a part of a real networkThe focus here is on realistic values for our model Weget these values from the GasLib (see [22]) a collection oftechnical gas network data
ldquoThegoal ofGasLib is to promote research on gasnetworks by providing a set of large and realisticbenchmark instancesrdquo (httpgaslibzibde)
We use a part of the Greek gas network (GasLib-134) (seeFigure 11)
For using real values we first need to know the constant120601119890 from (5)
120601119890 = (119877119878119879)2 120582119863 1198711198862 (111)
The value 119877119878 is the specific gas constant which can becalculated by 119877119878 = 119877119872 with the ideal gas constant 119877 =831451119896119892 sdot 11989821199042 sdot 119898119900119897 sdot 119870 (see [23]) and the molar mass119872 = 1662119896119892119896119898119900119897 (from real data) of the used natural gasAlso the temperature 119879 = 28915119870 is given from real dataFor the (constant) friction factor 120582 we use the law of Chen(see [24])
Mathematical Problems in Engineering 15
Table 4 Test results of control example
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 81199011198981198861199091199001199011199050 39013 39223 39039 38901 38879 39039 38901 390981199011198981198861199091199001199011199051 16568 16565 16365 16497 16496 16365 16497 164921199011198981198861199091199001199011199052 16568 16565 16365 16497 16496 16365 16497 164981199011198981198861199091199001199011199053 10000 10000 10000 10000 10002 10000 10000 10000
Table 5 Values based on real data
variable 119877119878 119879 120582 119863 119871 119886 120601value 50027 28915 00093 9144 178693 43862 198 sdot 107unit 11989821199042 sdot 119870 119870 minus 119898119898 119896119898 119898119904 11989821199042
Figure 11 Greek gas network from GasLib [22] red edges are usedin our example
1radic120582 = minus2 log10 ( 11989611986337065minus 50425
Relog10 [(119896119863)1109828257 + 58506
Re08981]) (112)
Diameter 119863 = 9144119898119898 and roughness value 119896 = 8 sdot 10minus6119898for the first pipe (see Figure 11) are known from real data TheReynolds-Number Re can be calculated by Re = (V119863120588)120578with the dynamic viscosity 120578 = 1 sdot 10minus5119875119886 sdot 119904 (see [1])the mean velocity V = 9548119898119904 (calculated with the meanvolume flow rate which is given by the real data) of the gasand the density 120588 = 074331198961198921198983 (from real data) So we getRe = 648959074 asymp 649 sdot 106 and with this the pipe frictioncoefficient after Chen is 120582 = 00093
The length of the first pipe is given by 119871 = 178693119896119898from the real data Last value missing is the sound speed innatural gas which we assume to be ideal The sound speed
119886 depends on the modulus of compressibility (see [9]) Forideal gases the modulus of compressibility only depends onthe adiabatic exponent and the pressure Using the ideal gasequation simplifies the equation for the sound speed to
119886 = radic120581119877119879119872 (113)
The adiabatic exponent for methan which is the main partof natural gas (gt 90) is about 133 (see [25]) Together allvalues are shown in Table 5 (example for the pipe next to thesource in Figure 11)
Nowwewant to optimize the inner control in the networkfor a fix load vector Therefor we choose two nominations2011-11-01 and 2016-02-17 In both cases the fmincon-functioninMATLAB returns
119906119900119901119905 = 1 (114)
as a local minimum which satisfies the constraint 119887 isin 119872This means for these nominations a compressor stationis not needed so it is switched off With this as a lastcalculation in this paper we calculate the probability for thegiven nominations to be feasible We use the values from thenominations as mean value 120583 and set 119887 sim N(120583 sdot 10minus3E 120583)The results are shown in Tables 6 and 7
Again one can see that the results are quite similarThe mean probability in Table 6 is 09860 and the meanprobability in Table 7 is 09760 Both values are nearly 1 sothere must happen something really unforeseeable that aGreek does not get his demanded gas
6 Conclusion
In this paper we have used an existing stationary modelfor gas transport based on the isothermal Euler equationsWe have extended this model by compressor stations andpresented a characterization of the set of feasible loadsEspecially Theorem 5 contains the central idea to get anaccess to the optimization in Section 4 and the spheric-radialdecomposition is our main tool to handle the uncertainty inthe loadsWewere able to show some results for optimizationwith probabilistic constraints in gas transport In a next stepthese results should be generalized
16 Mathematical Problems in Engineering
Table 6 Test results for nomination 2011-11-01
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 8P(119887 isin 119872) 09849 09849 09878 09885 09863 09845 09854 09853
Table 7 Test results for nomination 2016-02-17
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 8P(119887 isin 119872) 09744 09756 09788 09788 09754 09775 09734 09739
With the characterization of the feasible set we haveshown that the feasibility of a load depends significantly onthe pressure bounds These pressure bounds play a mainrole in our optimization problems The other main partsare the inner controls modeled by compressor edges whichshould lead in optimal control problems with probabilisticconstraints Here we have built a base for further works withthis topic
The idea of separating the graph and removing thecompressor edges can be used to extend themodel with othercomponents like eg control valves and resistorsThiswouldchange the inequalities inTheorem 5 but the approach worksalso for thesemodifications Further [8] showed a way to dealwith a networkwhich is a cycle Combining bothmethods weget an access to model gas networks in high detail
Finally the expansion to a dynamic model might bealso possible The idea is to use a space-mapping approachbecause one can use a lot of the analysis we did hereTherefore upcoming papers should deal with these problemsand present results of the turnpike-theory which describesa relation between stationary and instationary solutions Infact turnpike-theory implies that the optimal dynamic stateis close to the optimal static state in the interior of the timeintervals for sufficiently long time horizons AJ Zaslavskiexplains the turnpike-phenomenon in [26] This book alsogives an overview about the theory
Data Availability
The data used to support the findings of this study areavailable from the corresponding author upon request
Conflicts of Interest
The authors declare that there are no conflicts of interestregarding the publication of this paper
Acknowledgments
This work was supported by DFG in the framework of theCollaborative Research Centre CRCTransregio 154 Math-ematical Modeling Simulation and Optimization Using theExample of Gas Networks project C03
References
[1] P Domschke B Hiller J Lang and C Tischendorf ldquoModel-lierung von gasnetzwerken Eine Ubersichtrdquo Tech Rep 2717Technische Universitat Darmstadt 2017
[2] T Koch B Hiller M E Pfetsch and L Schewe Evaluating GasNetwork Capacities MOS-SIAM 2015
[3] M Gugat R Schultz and D Wintergerst ldquoNetworks of pipe-lines for gas with nonconstant compressibility factor stationarystatesrdquoComputational amp AppliedMathematics vol 37 no 2 pp1066ndash1097 2018
[4] M Gugat and M Herty ldquoExistence of classical solutions andfeedback stabilization for the flow in gas networksrdquo ESAIMControl Optimisation and Calculus of Variations vol 17 no 1pp 28ndash51 2011
[5] M K Banda M Herty and A Klar ldquoGas flow in pipeline net-worksrdquo Networks and Heterogeneous Media vol 1 no 1 pp 41ndash56 2006
[6] RM Colombo G GuerraM Herty and V Schleper ldquoOptimalcontrol in networks of pipes and canalsrdquo SIAM Journal onControl and Optimization vol 48 no 3 pp 2032ndash2050 2009
[7] A Shapiro D Dentcheva and A Ruszczynski ectures onStochastic Programming Modeling and Theory MPS-SIAM2009
[8] C Gotzes H Heitsch R Henrion and R Schultz ldquoOn thequantification of nomination feasibility in stationary gas net-works with random loadrdquoMathematical Methods of OperationsResearch vol 84 no 2 pp 427ndash457 2016
[9] D Halliday R Resnick and J Walker Physik Wiley-VCHBachelor edition edition 2007
[10] M Gugat F M Hante M Hirsch-Dick and G LeugeringldquoStationary states in gas networksrdquoNetworks andHeterogeneousMedia vol 10 no 2 pp 295ndash320 2015
[11] D RoseM SchmidtM C Steinbach and BMWillert ldquoCom-putational optimization of gas compressor stations MINLPmodels versus continuous reformulationsrdquoMathematical Meth-ods of Operations Research vol 83 no 3 pp 409ndash444 2016
[12] H Heitsch R Henrion H Leovey et al ldquoEmpirical observa-tions and statistical analysis of gas demandrdquo in Evaluating GasNetwork Capacities T data B KochM E Hiller et al Eds pp273ndash290 MOS-SIAM 2015
[13] A Genz and F Bretz Computation of Multivariate Normal andt Probabilities vol 195 of Lecture Notes in Statistics Springer2009
[14] M H Farshbaf-Shaker R Henrion and D Homberg ldquoProp-erties of Chance Constraints in Infinite Dimensions with anApplication to PDE Constrained Optimizationrdquo Set-Valued andVariational Analysis pp 1ndash21 2017
[15] T Gonzalez Grandon H Heitsch and R Henrion ldquoA jointmodel of probabilisticrobust constraints for gas transportman-agement in stationary networksrdquo Computational ManagementScience vol 14 no 3 pp 443ndash460 2017
[16] W van Ackooij I Aleksovska and M Munoz-Zuniga ldquo(Sub-)Differentiability of Probability Functionswith Elliptical Distri-butionsrdquo Set-Valued and Variational Analysis
Mathematical Problems in Engineering 17
[17] W van Ackooij and R Henrion ldquoGradient formulae for nonlin-ear probabilistic constraints with GAUssian and GAUssian-likedistributionsrdquo SIAM Journal on Optimization vol 24 no 4 pp1864ndash1889 2014
[18] W van Ackooij and R Henrion ldquo(Sub-)gradient formulaefor probability functions of random inequality systems underGaussian distributionrdquo SIAMASA Journal on UncertaintyQuantification vol 5 no 1 pp 63ndash87 2017
[19] W van Ackooij and J Malick ldquoSecond-order differentiabilityof probability functionsrdquo Optimization Letters vol 11 no 1 pp179ndash194 2017
[20] H Eckey R Kosfeld and M Turck Wahrscheinlichkeitsrech-nung und Induktive Statistik - Grundlagen -Methoden - BeispieleGabler Wiesbaden Germany 2005
[21] A Prekopa Stochastic Programming Springer 1st edition 1995[22] I Joormann M Pfetsch R Schwarz et al ldquoGasLibmdashA Library
of Gas Network Instancesrdquo 2017[23] G Wedler Lehrbuch der Physikalischen Chemie Wiley-VCH
5th edition 2004[24] N H Chen ldquoAn explicit equation for friction factor in piperdquo
Industrial amp Engineering Chemistry Fundamentals vol 18 no3 pp 296-297 1979
[25] W Schenk and F Kremer Physikalisches Praktikum Springer14th edition 2014
[26] A J Zaslavski ldquoTurnpike Properties in the Calculus of Varia-tions and Optimal Controlrdquo inNonconvex Optimization and ItsApplications vol 80 Springer 2006
Hindawiwwwhindawicom Volume 2018
MathematicsJournal of
Hindawiwwwhindawicom Volume 2018
Mathematical Problems in Engineering
Applied MathematicsJournal of
Hindawiwwwhindawicom Volume 2018
Probability and StatisticsHindawiwwwhindawicom Volume 2018
Journal of
Hindawiwwwhindawicom Volume 2018
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawiwwwhindawicom Volume 2018
OptimizationJournal of
Hindawiwwwhindawicom Volume 2018
Hindawiwwwhindawicom Volume 2018
Engineering Mathematics
International Journal of
Hindawiwwwhindawicom Volume 2018
Operations ResearchAdvances in
Journal of
Hindawiwwwhindawicom Volume 2018
Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018
International Journal of Mathematics and Mathematical Sciences
Hindawiwwwhindawicom Volume 2018
Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom
The Scientific World Journal
Volume 2018
Hindawiwwwhindawicom Volume 2018Volume 2018
Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in
Nature and SocietyHindawiwwwhindawicom Volume 2018
Hindawiwwwhindawicom
Dierential EquationsInternational Journal of
Volume 2018
Hindawiwwwhindawicom Volume 2018
Decision SciencesAdvances in
Hindawiwwwhindawicom Volume 2018
AnalysisInternational Journal of
Hindawiwwwhindawicom Volume 2018
Stochastic AnalysisInternational Journal of
Submit your manuscripts atwwwhindawicom
14 Mathematical Problems in Engineering
minus1 1 2 3
1
2
3
minus1 1 2 3
1
2
3
b3
b1
Hyperspace of the feasible set MHyperspace of the feasible set M
b1
b3
Figure 10 Feasible set of the Example 2
Table 3 Test results of control example
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 8P(119887 isin 119872) 01531 01500 01540 01536 01530 01550 01567 01528so the feasible set is (see Figure 10)
= + isin R119899+1 | 1119879
4 + = 0 and 0 le 3 le radic3 and 2 = 0and 0 le 1 le radic8 minus 22 (106)
This implies
119872 = 119887 isin R3ge0 | 0 le 1198873 le radic3 and 1198872 = 0 and 0 le 1198871 le minus1198873
+ radic8 minus 11988723 (107)
Solving the problem
min119901+119898119886119909isin119875
11198794119901+119898119886119909
subject to isin (108)
for 119887 = 120583 = [05 0 05]119879 leads us to the optimal solution
119901+119898119886119909119900119901119905 = [2 radic3radic3radic275]119879 (109)
The argumentation for the optimality of this vector is exactlylike before in the easy example Becausewe have a compressoredge in this graph we can minimize the control Thereforewe fix the pressure bounds as we did at the beginning of thisexample and we set 119887 = [1 0 15]119879 So inequality (31) is notfulfilled anymore for 119906 = 1 We use again the fmincon-solverinMATLAB to solve the following problem
min119906ge1
1199062subject to 119887 isin 119872(119906) (110)
The solver returns the optimal solution 119906119900119901119905 = 11818
Here one can see that the sufficient condition in Lemma 8is really strong Itrsquos not fulfilled with the given values but thereexists a solution anyway Of course the necessary conditionis fulfilled
Now we use again Algorithm 3 to compute the proba-bility P(119887 isin 119872) for a random load vector 119887 sim N(E3 [0505 05]119879) Table 3 shows us the results
Again we can see that the probability is nearly the samein all eight cases The mean probability is 01535 and thevariance is 32369 sdot 10minus6 For the optimization we reducethe covariance matrix and assume 119887 sim N(001 sdot E3 [05 0505]119879) For a probability 120572 = 09 Table 4 states the resultsExample 3 In this example we use a part of a real networkThe focus here is on realistic values for our model Weget these values from the GasLib (see [22]) a collection oftechnical gas network data
ldquoThegoal ofGasLib is to promote research on gasnetworks by providing a set of large and realisticbenchmark instancesrdquo (httpgaslibzibde)
We use a part of the Greek gas network (GasLib-134) (seeFigure 11)
For using real values we first need to know the constant120601119890 from (5)
120601119890 = (119877119878119879)2 120582119863 1198711198862 (111)
The value 119877119878 is the specific gas constant which can becalculated by 119877119878 = 119877119872 with the ideal gas constant 119877 =831451119896119892 sdot 11989821199042 sdot 119898119900119897 sdot 119870 (see [23]) and the molar mass119872 = 1662119896119892119896119898119900119897 (from real data) of the used natural gasAlso the temperature 119879 = 28915119870 is given from real dataFor the (constant) friction factor 120582 we use the law of Chen(see [24])
Mathematical Problems in Engineering 15
Table 4 Test results of control example
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 81199011198981198861199091199001199011199050 39013 39223 39039 38901 38879 39039 38901 390981199011198981198861199091199001199011199051 16568 16565 16365 16497 16496 16365 16497 164921199011198981198861199091199001199011199052 16568 16565 16365 16497 16496 16365 16497 164981199011198981198861199091199001199011199053 10000 10000 10000 10000 10002 10000 10000 10000
Table 5 Values based on real data
variable 119877119878 119879 120582 119863 119871 119886 120601value 50027 28915 00093 9144 178693 43862 198 sdot 107unit 11989821199042 sdot 119870 119870 minus 119898119898 119896119898 119898119904 11989821199042
Figure 11 Greek gas network from GasLib [22] red edges are usedin our example
1radic120582 = minus2 log10 ( 11989611986337065minus 50425
Relog10 [(119896119863)1109828257 + 58506
Re08981]) (112)
Diameter 119863 = 9144119898119898 and roughness value 119896 = 8 sdot 10minus6119898for the first pipe (see Figure 11) are known from real data TheReynolds-Number Re can be calculated by Re = (V119863120588)120578with the dynamic viscosity 120578 = 1 sdot 10minus5119875119886 sdot 119904 (see [1])the mean velocity V = 9548119898119904 (calculated with the meanvolume flow rate which is given by the real data) of the gasand the density 120588 = 074331198961198921198983 (from real data) So we getRe = 648959074 asymp 649 sdot 106 and with this the pipe frictioncoefficient after Chen is 120582 = 00093
The length of the first pipe is given by 119871 = 178693119896119898from the real data Last value missing is the sound speed innatural gas which we assume to be ideal The sound speed
119886 depends on the modulus of compressibility (see [9]) Forideal gases the modulus of compressibility only depends onthe adiabatic exponent and the pressure Using the ideal gasequation simplifies the equation for the sound speed to
119886 = radic120581119877119879119872 (113)
The adiabatic exponent for methan which is the main partof natural gas (gt 90) is about 133 (see [25]) Together allvalues are shown in Table 5 (example for the pipe next to thesource in Figure 11)
Nowwewant to optimize the inner control in the networkfor a fix load vector Therefor we choose two nominations2011-11-01 and 2016-02-17 In both cases the fmincon-functioninMATLAB returns
119906119900119901119905 = 1 (114)
as a local minimum which satisfies the constraint 119887 isin 119872This means for these nominations a compressor stationis not needed so it is switched off With this as a lastcalculation in this paper we calculate the probability for thegiven nominations to be feasible We use the values from thenominations as mean value 120583 and set 119887 sim N(120583 sdot 10minus3E 120583)The results are shown in Tables 6 and 7
Again one can see that the results are quite similarThe mean probability in Table 6 is 09860 and the meanprobability in Table 7 is 09760 Both values are nearly 1 sothere must happen something really unforeseeable that aGreek does not get his demanded gas
6 Conclusion
In this paper we have used an existing stationary modelfor gas transport based on the isothermal Euler equationsWe have extended this model by compressor stations andpresented a characterization of the set of feasible loadsEspecially Theorem 5 contains the central idea to get anaccess to the optimization in Section 4 and the spheric-radialdecomposition is our main tool to handle the uncertainty inthe loadsWewere able to show some results for optimizationwith probabilistic constraints in gas transport In a next stepthese results should be generalized
16 Mathematical Problems in Engineering
Table 6 Test results for nomination 2011-11-01
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 8P(119887 isin 119872) 09849 09849 09878 09885 09863 09845 09854 09853
Table 7 Test results for nomination 2016-02-17
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 8P(119887 isin 119872) 09744 09756 09788 09788 09754 09775 09734 09739
With the characterization of the feasible set we haveshown that the feasibility of a load depends significantly onthe pressure bounds These pressure bounds play a mainrole in our optimization problems The other main partsare the inner controls modeled by compressor edges whichshould lead in optimal control problems with probabilisticconstraints Here we have built a base for further works withthis topic
The idea of separating the graph and removing thecompressor edges can be used to extend themodel with othercomponents like eg control valves and resistorsThiswouldchange the inequalities inTheorem 5 but the approach worksalso for thesemodifications Further [8] showed a way to dealwith a networkwhich is a cycle Combining bothmethods weget an access to model gas networks in high detail
Finally the expansion to a dynamic model might bealso possible The idea is to use a space-mapping approachbecause one can use a lot of the analysis we did hereTherefore upcoming papers should deal with these problemsand present results of the turnpike-theory which describesa relation between stationary and instationary solutions Infact turnpike-theory implies that the optimal dynamic stateis close to the optimal static state in the interior of the timeintervals for sufficiently long time horizons AJ Zaslavskiexplains the turnpike-phenomenon in [26] This book alsogives an overview about the theory
Data Availability
The data used to support the findings of this study areavailable from the corresponding author upon request
Conflicts of Interest
The authors declare that there are no conflicts of interestregarding the publication of this paper
Acknowledgments
This work was supported by DFG in the framework of theCollaborative Research Centre CRCTransregio 154 Math-ematical Modeling Simulation and Optimization Using theExample of Gas Networks project C03
References
[1] P Domschke B Hiller J Lang and C Tischendorf ldquoModel-lierung von gasnetzwerken Eine Ubersichtrdquo Tech Rep 2717Technische Universitat Darmstadt 2017
[2] T Koch B Hiller M E Pfetsch and L Schewe Evaluating GasNetwork Capacities MOS-SIAM 2015
[3] M Gugat R Schultz and D Wintergerst ldquoNetworks of pipe-lines for gas with nonconstant compressibility factor stationarystatesrdquoComputational amp AppliedMathematics vol 37 no 2 pp1066ndash1097 2018
[4] M Gugat and M Herty ldquoExistence of classical solutions andfeedback stabilization for the flow in gas networksrdquo ESAIMControl Optimisation and Calculus of Variations vol 17 no 1pp 28ndash51 2011
[5] M K Banda M Herty and A Klar ldquoGas flow in pipeline net-worksrdquo Networks and Heterogeneous Media vol 1 no 1 pp 41ndash56 2006
[6] RM Colombo G GuerraM Herty and V Schleper ldquoOptimalcontrol in networks of pipes and canalsrdquo SIAM Journal onControl and Optimization vol 48 no 3 pp 2032ndash2050 2009
[7] A Shapiro D Dentcheva and A Ruszczynski ectures onStochastic Programming Modeling and Theory MPS-SIAM2009
[8] C Gotzes H Heitsch R Henrion and R Schultz ldquoOn thequantification of nomination feasibility in stationary gas net-works with random loadrdquoMathematical Methods of OperationsResearch vol 84 no 2 pp 427ndash457 2016
[9] D Halliday R Resnick and J Walker Physik Wiley-VCHBachelor edition edition 2007
[10] M Gugat F M Hante M Hirsch-Dick and G LeugeringldquoStationary states in gas networksrdquoNetworks andHeterogeneousMedia vol 10 no 2 pp 295ndash320 2015
[11] D RoseM SchmidtM C Steinbach and BMWillert ldquoCom-putational optimization of gas compressor stations MINLPmodels versus continuous reformulationsrdquoMathematical Meth-ods of Operations Research vol 83 no 3 pp 409ndash444 2016
[12] H Heitsch R Henrion H Leovey et al ldquoEmpirical observa-tions and statistical analysis of gas demandrdquo in Evaluating GasNetwork Capacities T data B KochM E Hiller et al Eds pp273ndash290 MOS-SIAM 2015
[13] A Genz and F Bretz Computation of Multivariate Normal andt Probabilities vol 195 of Lecture Notes in Statistics Springer2009
[14] M H Farshbaf-Shaker R Henrion and D Homberg ldquoProp-erties of Chance Constraints in Infinite Dimensions with anApplication to PDE Constrained Optimizationrdquo Set-Valued andVariational Analysis pp 1ndash21 2017
[15] T Gonzalez Grandon H Heitsch and R Henrion ldquoA jointmodel of probabilisticrobust constraints for gas transportman-agement in stationary networksrdquo Computational ManagementScience vol 14 no 3 pp 443ndash460 2017
[16] W van Ackooij I Aleksovska and M Munoz-Zuniga ldquo(Sub-)Differentiability of Probability Functionswith Elliptical Distri-butionsrdquo Set-Valued and Variational Analysis
Mathematical Problems in Engineering 17
[17] W van Ackooij and R Henrion ldquoGradient formulae for nonlin-ear probabilistic constraints with GAUssian and GAUssian-likedistributionsrdquo SIAM Journal on Optimization vol 24 no 4 pp1864ndash1889 2014
[18] W van Ackooij and R Henrion ldquo(Sub-)gradient formulaefor probability functions of random inequality systems underGaussian distributionrdquo SIAMASA Journal on UncertaintyQuantification vol 5 no 1 pp 63ndash87 2017
[19] W van Ackooij and J Malick ldquoSecond-order differentiabilityof probability functionsrdquo Optimization Letters vol 11 no 1 pp179ndash194 2017
[20] H Eckey R Kosfeld and M Turck Wahrscheinlichkeitsrech-nung und Induktive Statistik - Grundlagen -Methoden - BeispieleGabler Wiesbaden Germany 2005
[21] A Prekopa Stochastic Programming Springer 1st edition 1995[22] I Joormann M Pfetsch R Schwarz et al ldquoGasLibmdashA Library
of Gas Network Instancesrdquo 2017[23] G Wedler Lehrbuch der Physikalischen Chemie Wiley-VCH
5th edition 2004[24] N H Chen ldquoAn explicit equation for friction factor in piperdquo
Industrial amp Engineering Chemistry Fundamentals vol 18 no3 pp 296-297 1979
[25] W Schenk and F Kremer Physikalisches Praktikum Springer14th edition 2014
[26] A J Zaslavski ldquoTurnpike Properties in the Calculus of Varia-tions and Optimal Controlrdquo inNonconvex Optimization and ItsApplications vol 80 Springer 2006
Hindawiwwwhindawicom Volume 2018
MathematicsJournal of
Hindawiwwwhindawicom Volume 2018
Mathematical Problems in Engineering
Applied MathematicsJournal of
Hindawiwwwhindawicom Volume 2018
Probability and StatisticsHindawiwwwhindawicom Volume 2018
Journal of
Hindawiwwwhindawicom Volume 2018
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawiwwwhindawicom Volume 2018
OptimizationJournal of
Hindawiwwwhindawicom Volume 2018
Hindawiwwwhindawicom Volume 2018
Engineering Mathematics
International Journal of
Hindawiwwwhindawicom Volume 2018
Operations ResearchAdvances in
Journal of
Hindawiwwwhindawicom Volume 2018
Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018
International Journal of Mathematics and Mathematical Sciences
Hindawiwwwhindawicom Volume 2018
Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom
The Scientific World Journal
Volume 2018
Hindawiwwwhindawicom Volume 2018Volume 2018
Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in
Nature and SocietyHindawiwwwhindawicom Volume 2018
Hindawiwwwhindawicom
Dierential EquationsInternational Journal of
Volume 2018
Hindawiwwwhindawicom Volume 2018
Decision SciencesAdvances in
Hindawiwwwhindawicom Volume 2018
AnalysisInternational Journal of
Hindawiwwwhindawicom Volume 2018
Stochastic AnalysisInternational Journal of
Submit your manuscripts atwwwhindawicom
Mathematical Problems in Engineering 15
Table 4 Test results of control example
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 81199011198981198861199091199001199011199050 39013 39223 39039 38901 38879 39039 38901 390981199011198981198861199091199001199011199051 16568 16565 16365 16497 16496 16365 16497 164921199011198981198861199091199001199011199052 16568 16565 16365 16497 16496 16365 16497 164981199011198981198861199091199001199011199053 10000 10000 10000 10000 10002 10000 10000 10000
Table 5 Values based on real data
variable 119877119878 119879 120582 119863 119871 119886 120601value 50027 28915 00093 9144 178693 43862 198 sdot 107unit 11989821199042 sdot 119870 119870 minus 119898119898 119896119898 119898119904 11989821199042
Figure 11 Greek gas network from GasLib [22] red edges are usedin our example
1radic120582 = minus2 log10 ( 11989611986337065minus 50425
Relog10 [(119896119863)1109828257 + 58506
Re08981]) (112)
Diameter 119863 = 9144119898119898 and roughness value 119896 = 8 sdot 10minus6119898for the first pipe (see Figure 11) are known from real data TheReynolds-Number Re can be calculated by Re = (V119863120588)120578with the dynamic viscosity 120578 = 1 sdot 10minus5119875119886 sdot 119904 (see [1])the mean velocity V = 9548119898119904 (calculated with the meanvolume flow rate which is given by the real data) of the gasand the density 120588 = 074331198961198921198983 (from real data) So we getRe = 648959074 asymp 649 sdot 106 and with this the pipe frictioncoefficient after Chen is 120582 = 00093
The length of the first pipe is given by 119871 = 178693119896119898from the real data Last value missing is the sound speed innatural gas which we assume to be ideal The sound speed
119886 depends on the modulus of compressibility (see [9]) Forideal gases the modulus of compressibility only depends onthe adiabatic exponent and the pressure Using the ideal gasequation simplifies the equation for the sound speed to
119886 = radic120581119877119879119872 (113)
The adiabatic exponent for methan which is the main partof natural gas (gt 90) is about 133 (see [25]) Together allvalues are shown in Table 5 (example for the pipe next to thesource in Figure 11)
Nowwewant to optimize the inner control in the networkfor a fix load vector Therefor we choose two nominations2011-11-01 and 2016-02-17 In both cases the fmincon-functioninMATLAB returns
119906119900119901119905 = 1 (114)
as a local minimum which satisfies the constraint 119887 isin 119872This means for these nominations a compressor stationis not needed so it is switched off With this as a lastcalculation in this paper we calculate the probability for thegiven nominations to be feasible We use the values from thenominations as mean value 120583 and set 119887 sim N(120583 sdot 10minus3E 120583)The results are shown in Tables 6 and 7
Again one can see that the results are quite similarThe mean probability in Table 6 is 09860 and the meanprobability in Table 7 is 09760 Both values are nearly 1 sothere must happen something really unforeseeable that aGreek does not get his demanded gas
6 Conclusion
In this paper we have used an existing stationary modelfor gas transport based on the isothermal Euler equationsWe have extended this model by compressor stations andpresented a characterization of the set of feasible loadsEspecially Theorem 5 contains the central idea to get anaccess to the optimization in Section 4 and the spheric-radialdecomposition is our main tool to handle the uncertainty inthe loadsWewere able to show some results for optimizationwith probabilistic constraints in gas transport In a next stepthese results should be generalized
16 Mathematical Problems in Engineering
Table 6 Test results for nomination 2011-11-01
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 8P(119887 isin 119872) 09849 09849 09878 09885 09863 09845 09854 09853
Table 7 Test results for nomination 2016-02-17
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 8P(119887 isin 119872) 09744 09756 09788 09788 09754 09775 09734 09739
With the characterization of the feasible set we haveshown that the feasibility of a load depends significantly onthe pressure bounds These pressure bounds play a mainrole in our optimization problems The other main partsare the inner controls modeled by compressor edges whichshould lead in optimal control problems with probabilisticconstraints Here we have built a base for further works withthis topic
The idea of separating the graph and removing thecompressor edges can be used to extend themodel with othercomponents like eg control valves and resistorsThiswouldchange the inequalities inTheorem 5 but the approach worksalso for thesemodifications Further [8] showed a way to dealwith a networkwhich is a cycle Combining bothmethods weget an access to model gas networks in high detail
Finally the expansion to a dynamic model might bealso possible The idea is to use a space-mapping approachbecause one can use a lot of the analysis we did hereTherefore upcoming papers should deal with these problemsand present results of the turnpike-theory which describesa relation between stationary and instationary solutions Infact turnpike-theory implies that the optimal dynamic stateis close to the optimal static state in the interior of the timeintervals for sufficiently long time horizons AJ Zaslavskiexplains the turnpike-phenomenon in [26] This book alsogives an overview about the theory
Data Availability
The data used to support the findings of this study areavailable from the corresponding author upon request
Conflicts of Interest
The authors declare that there are no conflicts of interestregarding the publication of this paper
Acknowledgments
This work was supported by DFG in the framework of theCollaborative Research Centre CRCTransregio 154 Math-ematical Modeling Simulation and Optimization Using theExample of Gas Networks project C03
References
[1] P Domschke B Hiller J Lang and C Tischendorf ldquoModel-lierung von gasnetzwerken Eine Ubersichtrdquo Tech Rep 2717Technische Universitat Darmstadt 2017
[2] T Koch B Hiller M E Pfetsch and L Schewe Evaluating GasNetwork Capacities MOS-SIAM 2015
[3] M Gugat R Schultz and D Wintergerst ldquoNetworks of pipe-lines for gas with nonconstant compressibility factor stationarystatesrdquoComputational amp AppliedMathematics vol 37 no 2 pp1066ndash1097 2018
[4] M Gugat and M Herty ldquoExistence of classical solutions andfeedback stabilization for the flow in gas networksrdquo ESAIMControl Optimisation and Calculus of Variations vol 17 no 1pp 28ndash51 2011
[5] M K Banda M Herty and A Klar ldquoGas flow in pipeline net-worksrdquo Networks and Heterogeneous Media vol 1 no 1 pp 41ndash56 2006
[6] RM Colombo G GuerraM Herty and V Schleper ldquoOptimalcontrol in networks of pipes and canalsrdquo SIAM Journal onControl and Optimization vol 48 no 3 pp 2032ndash2050 2009
[7] A Shapiro D Dentcheva and A Ruszczynski ectures onStochastic Programming Modeling and Theory MPS-SIAM2009
[8] C Gotzes H Heitsch R Henrion and R Schultz ldquoOn thequantification of nomination feasibility in stationary gas net-works with random loadrdquoMathematical Methods of OperationsResearch vol 84 no 2 pp 427ndash457 2016
[9] D Halliday R Resnick and J Walker Physik Wiley-VCHBachelor edition edition 2007
[10] M Gugat F M Hante M Hirsch-Dick and G LeugeringldquoStationary states in gas networksrdquoNetworks andHeterogeneousMedia vol 10 no 2 pp 295ndash320 2015
[11] D RoseM SchmidtM C Steinbach and BMWillert ldquoCom-putational optimization of gas compressor stations MINLPmodels versus continuous reformulationsrdquoMathematical Meth-ods of Operations Research vol 83 no 3 pp 409ndash444 2016
[12] H Heitsch R Henrion H Leovey et al ldquoEmpirical observa-tions and statistical analysis of gas demandrdquo in Evaluating GasNetwork Capacities T data B KochM E Hiller et al Eds pp273ndash290 MOS-SIAM 2015
[13] A Genz and F Bretz Computation of Multivariate Normal andt Probabilities vol 195 of Lecture Notes in Statistics Springer2009
[14] M H Farshbaf-Shaker R Henrion and D Homberg ldquoProp-erties of Chance Constraints in Infinite Dimensions with anApplication to PDE Constrained Optimizationrdquo Set-Valued andVariational Analysis pp 1ndash21 2017
[15] T Gonzalez Grandon H Heitsch and R Henrion ldquoA jointmodel of probabilisticrobust constraints for gas transportman-agement in stationary networksrdquo Computational ManagementScience vol 14 no 3 pp 443ndash460 2017
[16] W van Ackooij I Aleksovska and M Munoz-Zuniga ldquo(Sub-)Differentiability of Probability Functionswith Elliptical Distri-butionsrdquo Set-Valued and Variational Analysis
Mathematical Problems in Engineering 17
[17] W van Ackooij and R Henrion ldquoGradient formulae for nonlin-ear probabilistic constraints with GAUssian and GAUssian-likedistributionsrdquo SIAM Journal on Optimization vol 24 no 4 pp1864ndash1889 2014
[18] W van Ackooij and R Henrion ldquo(Sub-)gradient formulaefor probability functions of random inequality systems underGaussian distributionrdquo SIAMASA Journal on UncertaintyQuantification vol 5 no 1 pp 63ndash87 2017
[19] W van Ackooij and J Malick ldquoSecond-order differentiabilityof probability functionsrdquo Optimization Letters vol 11 no 1 pp179ndash194 2017
[20] H Eckey R Kosfeld and M Turck Wahrscheinlichkeitsrech-nung und Induktive Statistik - Grundlagen -Methoden - BeispieleGabler Wiesbaden Germany 2005
[21] A Prekopa Stochastic Programming Springer 1st edition 1995[22] I Joormann M Pfetsch R Schwarz et al ldquoGasLibmdashA Library
of Gas Network Instancesrdquo 2017[23] G Wedler Lehrbuch der Physikalischen Chemie Wiley-VCH
5th edition 2004[24] N H Chen ldquoAn explicit equation for friction factor in piperdquo
Industrial amp Engineering Chemistry Fundamentals vol 18 no3 pp 296-297 1979
[25] W Schenk and F Kremer Physikalisches Praktikum Springer14th edition 2014
[26] A J Zaslavski ldquoTurnpike Properties in the Calculus of Varia-tions and Optimal Controlrdquo inNonconvex Optimization and ItsApplications vol 80 Springer 2006
Hindawiwwwhindawicom Volume 2018
MathematicsJournal of
Hindawiwwwhindawicom Volume 2018
Mathematical Problems in Engineering
Applied MathematicsJournal of
Hindawiwwwhindawicom Volume 2018
Probability and StatisticsHindawiwwwhindawicom Volume 2018
Journal of
Hindawiwwwhindawicom Volume 2018
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawiwwwhindawicom Volume 2018
OptimizationJournal of
Hindawiwwwhindawicom Volume 2018
Hindawiwwwhindawicom Volume 2018
Engineering Mathematics
International Journal of
Hindawiwwwhindawicom Volume 2018
Operations ResearchAdvances in
Journal of
Hindawiwwwhindawicom Volume 2018
Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018
International Journal of Mathematics and Mathematical Sciences
Hindawiwwwhindawicom Volume 2018
Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom
The Scientific World Journal
Volume 2018
Hindawiwwwhindawicom Volume 2018Volume 2018
Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in
Nature and SocietyHindawiwwwhindawicom Volume 2018
Hindawiwwwhindawicom
Dierential EquationsInternational Journal of
Volume 2018
Hindawiwwwhindawicom Volume 2018
Decision SciencesAdvances in
Hindawiwwwhindawicom Volume 2018
AnalysisInternational Journal of
Hindawiwwwhindawicom Volume 2018
Stochastic AnalysisInternational Journal of
Submit your manuscripts atwwwhindawicom
16 Mathematical Problems in Engineering
Table 6 Test results for nomination 2011-11-01
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 8P(119887 isin 119872) 09849 09849 09878 09885 09863 09845 09854 09853
Table 7 Test results for nomination 2016-02-17
Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 8P(119887 isin 119872) 09744 09756 09788 09788 09754 09775 09734 09739
With the characterization of the feasible set we haveshown that the feasibility of a load depends significantly onthe pressure bounds These pressure bounds play a mainrole in our optimization problems The other main partsare the inner controls modeled by compressor edges whichshould lead in optimal control problems with probabilisticconstraints Here we have built a base for further works withthis topic
The idea of separating the graph and removing thecompressor edges can be used to extend themodel with othercomponents like eg control valves and resistorsThiswouldchange the inequalities inTheorem 5 but the approach worksalso for thesemodifications Further [8] showed a way to dealwith a networkwhich is a cycle Combining bothmethods weget an access to model gas networks in high detail
Finally the expansion to a dynamic model might bealso possible The idea is to use a space-mapping approachbecause one can use a lot of the analysis we did hereTherefore upcoming papers should deal with these problemsand present results of the turnpike-theory which describesa relation between stationary and instationary solutions Infact turnpike-theory implies that the optimal dynamic stateis close to the optimal static state in the interior of the timeintervals for sufficiently long time horizons AJ Zaslavskiexplains the turnpike-phenomenon in [26] This book alsogives an overview about the theory
Data Availability
The data used to support the findings of this study areavailable from the corresponding author upon request
Conflicts of Interest
The authors declare that there are no conflicts of interestregarding the publication of this paper
Acknowledgments
This work was supported by DFG in the framework of theCollaborative Research Centre CRCTransregio 154 Math-ematical Modeling Simulation and Optimization Using theExample of Gas Networks project C03
References
[1] P Domschke B Hiller J Lang and C Tischendorf ldquoModel-lierung von gasnetzwerken Eine Ubersichtrdquo Tech Rep 2717Technische Universitat Darmstadt 2017
[2] T Koch B Hiller M E Pfetsch and L Schewe Evaluating GasNetwork Capacities MOS-SIAM 2015
[3] M Gugat R Schultz and D Wintergerst ldquoNetworks of pipe-lines for gas with nonconstant compressibility factor stationarystatesrdquoComputational amp AppliedMathematics vol 37 no 2 pp1066ndash1097 2018
[4] M Gugat and M Herty ldquoExistence of classical solutions andfeedback stabilization for the flow in gas networksrdquo ESAIMControl Optimisation and Calculus of Variations vol 17 no 1pp 28ndash51 2011
[5] M K Banda M Herty and A Klar ldquoGas flow in pipeline net-worksrdquo Networks and Heterogeneous Media vol 1 no 1 pp 41ndash56 2006
[6] RM Colombo G GuerraM Herty and V Schleper ldquoOptimalcontrol in networks of pipes and canalsrdquo SIAM Journal onControl and Optimization vol 48 no 3 pp 2032ndash2050 2009
[7] A Shapiro D Dentcheva and A Ruszczynski ectures onStochastic Programming Modeling and Theory MPS-SIAM2009
[8] C Gotzes H Heitsch R Henrion and R Schultz ldquoOn thequantification of nomination feasibility in stationary gas net-works with random loadrdquoMathematical Methods of OperationsResearch vol 84 no 2 pp 427ndash457 2016
[9] D Halliday R Resnick and J Walker Physik Wiley-VCHBachelor edition edition 2007
[10] M Gugat F M Hante M Hirsch-Dick and G LeugeringldquoStationary states in gas networksrdquoNetworks andHeterogeneousMedia vol 10 no 2 pp 295ndash320 2015
[11] D RoseM SchmidtM C Steinbach and BMWillert ldquoCom-putational optimization of gas compressor stations MINLPmodels versus continuous reformulationsrdquoMathematical Meth-ods of Operations Research vol 83 no 3 pp 409ndash444 2016
[12] H Heitsch R Henrion H Leovey et al ldquoEmpirical observa-tions and statistical analysis of gas demandrdquo in Evaluating GasNetwork Capacities T data B KochM E Hiller et al Eds pp273ndash290 MOS-SIAM 2015
[13] A Genz and F Bretz Computation of Multivariate Normal andt Probabilities vol 195 of Lecture Notes in Statistics Springer2009
[14] M H Farshbaf-Shaker R Henrion and D Homberg ldquoProp-erties of Chance Constraints in Infinite Dimensions with anApplication to PDE Constrained Optimizationrdquo Set-Valued andVariational Analysis pp 1ndash21 2017
[15] T Gonzalez Grandon H Heitsch and R Henrion ldquoA jointmodel of probabilisticrobust constraints for gas transportman-agement in stationary networksrdquo Computational ManagementScience vol 14 no 3 pp 443ndash460 2017
[16] W van Ackooij I Aleksovska and M Munoz-Zuniga ldquo(Sub-)Differentiability of Probability Functionswith Elliptical Distri-butionsrdquo Set-Valued and Variational Analysis
Mathematical Problems in Engineering 17
[17] W van Ackooij and R Henrion ldquoGradient formulae for nonlin-ear probabilistic constraints with GAUssian and GAUssian-likedistributionsrdquo SIAM Journal on Optimization vol 24 no 4 pp1864ndash1889 2014
[18] W van Ackooij and R Henrion ldquo(Sub-)gradient formulaefor probability functions of random inequality systems underGaussian distributionrdquo SIAMASA Journal on UncertaintyQuantification vol 5 no 1 pp 63ndash87 2017
[19] W van Ackooij and J Malick ldquoSecond-order differentiabilityof probability functionsrdquo Optimization Letters vol 11 no 1 pp179ndash194 2017
[20] H Eckey R Kosfeld and M Turck Wahrscheinlichkeitsrech-nung und Induktive Statistik - Grundlagen -Methoden - BeispieleGabler Wiesbaden Germany 2005
[21] A Prekopa Stochastic Programming Springer 1st edition 1995[22] I Joormann M Pfetsch R Schwarz et al ldquoGasLibmdashA Library
of Gas Network Instancesrdquo 2017[23] G Wedler Lehrbuch der Physikalischen Chemie Wiley-VCH
5th edition 2004[24] N H Chen ldquoAn explicit equation for friction factor in piperdquo
Industrial amp Engineering Chemistry Fundamentals vol 18 no3 pp 296-297 1979
[25] W Schenk and F Kremer Physikalisches Praktikum Springer14th edition 2014
[26] A J Zaslavski ldquoTurnpike Properties in the Calculus of Varia-tions and Optimal Controlrdquo inNonconvex Optimization and ItsApplications vol 80 Springer 2006
Hindawiwwwhindawicom Volume 2018
MathematicsJournal of
Hindawiwwwhindawicom Volume 2018
Mathematical Problems in Engineering
Applied MathematicsJournal of
Hindawiwwwhindawicom Volume 2018
Probability and StatisticsHindawiwwwhindawicom Volume 2018
Journal of
Hindawiwwwhindawicom Volume 2018
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawiwwwhindawicom Volume 2018
OptimizationJournal of
Hindawiwwwhindawicom Volume 2018
Hindawiwwwhindawicom Volume 2018
Engineering Mathematics
International Journal of
Hindawiwwwhindawicom Volume 2018
Operations ResearchAdvances in
Journal of
Hindawiwwwhindawicom Volume 2018
Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018
International Journal of Mathematics and Mathematical Sciences
Hindawiwwwhindawicom Volume 2018
Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom
The Scientific World Journal
Volume 2018
Hindawiwwwhindawicom Volume 2018Volume 2018
Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in
Nature and SocietyHindawiwwwhindawicom Volume 2018
Hindawiwwwhindawicom
Dierential EquationsInternational Journal of
Volume 2018
Hindawiwwwhindawicom Volume 2018
Decision SciencesAdvances in
Hindawiwwwhindawicom Volume 2018
AnalysisInternational Journal of
Hindawiwwwhindawicom Volume 2018
Stochastic AnalysisInternational Journal of
Submit your manuscripts atwwwhindawicom
Mathematical Problems in Engineering 17
[17] W van Ackooij and R Henrion ldquoGradient formulae for nonlin-ear probabilistic constraints with GAUssian and GAUssian-likedistributionsrdquo SIAM Journal on Optimization vol 24 no 4 pp1864ndash1889 2014
[18] W van Ackooij and R Henrion ldquo(Sub-)gradient formulaefor probability functions of random inequality systems underGaussian distributionrdquo SIAMASA Journal on UncertaintyQuantification vol 5 no 1 pp 63ndash87 2017
[19] W van Ackooij and J Malick ldquoSecond-order differentiabilityof probability functionsrdquo Optimization Letters vol 11 no 1 pp179ndash194 2017
[20] H Eckey R Kosfeld and M Turck Wahrscheinlichkeitsrech-nung und Induktive Statistik - Grundlagen -Methoden - BeispieleGabler Wiesbaden Germany 2005
[21] A Prekopa Stochastic Programming Springer 1st edition 1995[22] I Joormann M Pfetsch R Schwarz et al ldquoGasLibmdashA Library
of Gas Network Instancesrdquo 2017[23] G Wedler Lehrbuch der Physikalischen Chemie Wiley-VCH
5th edition 2004[24] N H Chen ldquoAn explicit equation for friction factor in piperdquo
Industrial amp Engineering Chemistry Fundamentals vol 18 no3 pp 296-297 1979
[25] W Schenk and F Kremer Physikalisches Praktikum Springer14th edition 2014
[26] A J Zaslavski ldquoTurnpike Properties in the Calculus of Varia-tions and Optimal Controlrdquo inNonconvex Optimization and ItsApplications vol 80 Springer 2006
Hindawiwwwhindawicom Volume 2018
MathematicsJournal of
Hindawiwwwhindawicom Volume 2018
Mathematical Problems in Engineering
Applied MathematicsJournal of
Hindawiwwwhindawicom Volume 2018
Probability and StatisticsHindawiwwwhindawicom Volume 2018
Journal of
Hindawiwwwhindawicom Volume 2018
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawiwwwhindawicom Volume 2018
OptimizationJournal of
Hindawiwwwhindawicom Volume 2018
Hindawiwwwhindawicom Volume 2018
Engineering Mathematics
International Journal of
Hindawiwwwhindawicom Volume 2018
Operations ResearchAdvances in
Journal of
Hindawiwwwhindawicom Volume 2018
Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018
International Journal of Mathematics and Mathematical Sciences
Hindawiwwwhindawicom Volume 2018
Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom
The Scientific World Journal
Volume 2018
Hindawiwwwhindawicom Volume 2018Volume 2018
Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in
Nature and SocietyHindawiwwwhindawicom Volume 2018
Hindawiwwwhindawicom
Dierential EquationsInternational Journal of
Volume 2018
Hindawiwwwhindawicom Volume 2018
Decision SciencesAdvances in
Hindawiwwwhindawicom Volume 2018
AnalysisInternational Journal of
Hindawiwwwhindawicom Volume 2018
Stochastic AnalysisInternational Journal of
Submit your manuscripts atwwwhindawicom
Hindawiwwwhindawicom Volume 2018
MathematicsJournal of
Hindawiwwwhindawicom Volume 2018
Mathematical Problems in Engineering
Applied MathematicsJournal of
Hindawiwwwhindawicom Volume 2018
Probability and StatisticsHindawiwwwhindawicom Volume 2018
Journal of
Hindawiwwwhindawicom Volume 2018
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawiwwwhindawicom Volume 2018
OptimizationJournal of
Hindawiwwwhindawicom Volume 2018
Hindawiwwwhindawicom Volume 2018
Engineering Mathematics
International Journal of
Hindawiwwwhindawicom Volume 2018
Operations ResearchAdvances in
Journal of
Hindawiwwwhindawicom Volume 2018
Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018
International Journal of Mathematics and Mathematical Sciences
Hindawiwwwhindawicom Volume 2018
Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom
The Scientific World Journal
Volume 2018
Hindawiwwwhindawicom Volume 2018Volume 2018
Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in
Nature and SocietyHindawiwwwhindawicom Volume 2018
Hindawiwwwhindawicom
Dierential EquationsInternational Journal of
Volume 2018
Hindawiwwwhindawicom Volume 2018
Decision SciencesAdvances in
Hindawiwwwhindawicom Volume 2018
AnalysisInternational Journal of
Hindawiwwwhindawicom Volume 2018
Stochastic AnalysisInternational Journal of
Submit your manuscripts atwwwhindawicom