1 network optimization chapter 2 transshipment problem

1

Network Optimization

Chapter 2Transshipment Problem

2

G = (V, E) is a digraph. |V|=n, |E|=m. A type of commodity is shipped on the network. For each arc , is the cost of sending

one unit of commodity from vertex i to vertex j

. Vector c = has m components, and is

called the cost vector. There are three kinds of vertices:

2.1 The Network Simplex Method

ijc

0ijc

ji,

ijc

3


(1) Source If the supply at source i is ( >0 ),

then let bi =- . (2) Sink If the demand at sink j is ( >0 ),

then let bj = . (3) Intermediate vertex Neither a source, nor a sink. If k is an

intermediate vertex, then let bk = 0.

ii

jj

4


S denotes supply and D denotes demand

5


Balanced transshipment problem

Total supply = total demand: , i.e.

ji

01

n

iib

6


inflow at vertex i ~ the total commodity sent to i;

outflow at vertex i ~ the total commodity sent out from i;

netflow at vertex i ~ inflow at i - outflow at i; flow vector x ~ , where is the

amount of commodity from i to j ; feasible flow x ~ a flow x that makes the

netflow at each vertex i equal bi .

ijx mij Rxx

0ijx

7


A Feasible Flow

8


Single commodity uncapacitated transshipment problem

Find a feasible flow such that its cost is as low as possible.

Let A be the incidence matrix of the network.

The supply-demand can be expressed by

Ax=b

0x

9netflow at i


For example, if at vertex i:

then

n

i

b

b

b

x

x

x

i

1

3

2

1

0...0111

e1 e2 e3

A x b

ibxxx 321

10


The transshipment problem is now formulated as

min cx s.t. Ax=b

c ~ row vector, x & b ~ column vector c, , uncapacitated ~ all xij have no upper

bound

0x

mRx nb R

11


Example 2.1 Consider the network shown below with 7

vertices and 10 arcs in which any flow vector x is of the form

xT = [x13 x21 x32 x51 x52 x61 x64 x67 x74 x75]. The cost vector is given by c = [1 1 1 1 5 11 20 5 5 6] and the supply-demand vector by bT = [0 2 3 5 -2 -3 -5].

12


13


111

111

111

11

11

111

1111

A

7

6

5

4

3

2

1

1 2 3 4 5 6 7 8 9 10e e e e e e e e e e

14


Note that the rows of A are linearly dependent because the sum of all rows of A is a zero vector.

So, rank (A) < n. On the other hand, we will see later that A

contains an nonsingular submatrix. Therefore, rank (A) = n-1. We may delete one equation from Ax = b

without changing the solution set.

111

111

111

11

11

111

1111

A

11 nn

15

Do you still remember?

What is a basic? What is a basic solution? What is a basic feasible solution? What is the main idea of Simplex

method? From a vertex (feasible basic sol) to another vertex, obj. value keep decreasing.

16


Feasible tree solution (FTS) A feasible flow x is called an FTS if

there is a spanning tree T such that if . As each spanning tree has n-1

edges, an FTS x has at most n-1 nonzero components xij.

0ijx Tji ,

17


If an FTS x has n-1 nonzero components, then the associated spanning tree is unique. It must be the tree consisting of the n-1 edges eij that correspond to the n-1 nonzero xij.

We call such an FTS a nondegenerate FTS. An FTS x is called a degenerate FTS if the

number of nonzero components is less than n-1.

A degenerate FTS may associate with more than one spanning tree.

18


Example

b = (-3, 3, 0, 0)T

11

11

11

11

A

It is easy to see that x = ( 3, 0, 0, 0 )T is a feasible solution of the equation Ax = b.

19


In fact x is a degenerate FTS, and the associated spanning trees are not unique. They can be one of the following trees:

Conversely, every spanning tree T is associated with at most one feasible tree solution.

20

The above few pages told you

FS can be classified into FTS/not FTS. FTS can be further classified into DE/non-DE. Given non-DE FTS, we can defined a unique

ST. Given DE FTS, there will be many ST.

The following few pages tell: Given a ST, we can solve for a unique solution. The solution may be a FS or not FS.

21


By relabeling vertices and arcs, we may let the part of A corresponding to the arcs in T contain a nonsingular matrix B.

For example, if we consider the following spanning tree T (formed by solid lines) in Example 2.1:

11 nn

22


It is not clear if in A, the columns corresponding to the arcs in T contain an nonsingular B.

111

111

111

11

11

111

1111

A

11 nn

23


Relabel vertices and arcs as follows:

24


How to relabel?First pick up an arbitrary vertex and label it as th

e first one. In general suppose i vertices and i-1 arcs in the tr

ee are already relabeled ( ). Then from all unlabeled vertices, choose one (say w) which is adjacent to a relabeled vertex (say u), and label w as the i+1th vertex, and also label the arc between w and u in the tree (either (u,w), or (w,u)) as the ith arc.

Repeat the procedure until we have labeled all n vertices and n-1 arcs in the tree.

1 1 ni

25


a7, a8, a9, a10 are the four arcs not in T, now the incidence matrix

7

6

5

4

3

2

1

....1

....11

....1

....11

....11

....111

1

v

v

v

v

v

v

v

A

1 2 3 4 5 6 7 8 9 10a a a a a a a a a a

26


Let x = (x1, … , x10)T , where xi corresponds to arc ai. Rows 2-7 and columns 1-6 form a matrix B, which is upper triangular, and the diagonal elements are 1 or -1. So, B is a nonsingular matrix.

11 nn

7

6

5

4

3

2

1

....1

....11

....1

....11

....11

....111

1

v

v

v

v

v

v

v

A

27


The feasible tree solution corresponding to T must have

x7 = x8 = x9 = x10 = 0. For xB= ( x1, x2, x3, x4, x5, x6 )T, it can be

decided by

bi ~ the supply or demand, or zero corresponding to vi.

7

3

2

b

b

b

BxB def

b

28


Therefore, xB is uniquely determined:

(1) If , then is an FTS, i.e., T is associated with a unique FTS.

(2) If some components of xB are negative, then is not a feasible solution, i.e., T is not associated with any FTS.

bBxB1

0Bx 0,0,0,0,TB

T xx

0,0,0,0,TB

T xx

29


Summary of the relationship between FTS and ST

30


From the above analysis, we know that a Tree Solution is actually a Basic Solution of the system Ax = b, and the base B can be determined by an associated spanning tree. An FTS is a feasible basic solution (FBS).

In simplex method we move from one FBS to another FBS. Now in network simplex method, we move from one FTS to another FTS, and geometrically, from one spanning tree to another spanning tree.

31


The dual solution defined by a spanning tree

We know that the dual problem of min cx s.t. Ax=b

is max yb s.t. where y is a row vector: . Also, the complementarity slackness condition

holds.

0x

cyA

nn yyyy ,,...,, 121

32


Now for a nondegenerate FTS x, the associated spanning tree T is unique, and for any arc in T, . Let the column in A which corresponds to arc be , then by the complementarity slackness condition,

ijijij cyAx 0

i.e.

ijn cj

iyy

1

1,,1

(*)

ji,0ijx

ji, ijA

Tjicyy ijji ),( ,

33


The above set (*) has n-1 equations for n variables. So, the solutions are not unique. If we fix yn = 0, then y1, … , yn-1 can be uniquely determined. (If we fix yn to another value t, then in the solution, each shall increase its value by t.)

We call the row vector , determined by the equations (*), the dual solution defined by the spanning tree T.

ijji cyy Tji , (*)

0,,, 11 nyyy

iy

34


Profitable arcs Suppose x is an FTS that defines a spanning

tree T(x) and a dual solution y. An arc (i, j) which is not in T(x) is said to be profitable with respect to the FTS x if .

Note that if we change the value of yn in deciding y, as all shall change by the same value, the value of every shall remain unchanged. So, the determination of profitable arcs is independent of the value of yn chosen.

ijij cyy

iyij yy

35


Theorem 2.2 An FTS u with no profitable arcs is an optimal solution.

Proof Suppose u is associated with tree T(u) and dual solution y. Define

. We know that . In fact (1) If arc , then by the definition of y, (2) If arc , then . So, for any arc , .

yAcc 0uc

uTji ,

0 jiijjiijij yycyycc

uTji , 0iju

ji, 00 ucuc ijij

36


Now for any feasible solution v: Av= b, ,

. (1) In particular, when v = u, . (2) (1) - (2): .

0v

ybvcvyAccv

ybybuccu vccucv

37


Since there is no profitable arc, for any arc ,

. But , and hence

u is an optimal solution.

ji,

0 jiijijijij yycccyy

0vcucvvc 0

38

2.1 The network simplex method

The above theorem provides an optimality criterion.

Now we consider what happens if there is a profitable arc . Of course e is not in T, and . If we can increase the flow on

from to , meanwhile keep for other arcs not in T, then as , we may have

(will explain in detail later), i.e. we obtain a better solution v.

qpe ,0pqc qp,

0pqu 0tv pq 0 ijij uv ji, 0 pqpqvc

0 tcvccucv pq

39


When arc e is added to the spanning tree T, a unique cycle C(e) will be obtained.

40


In the cycle, we call each arc with the same direction as e a forward arc (arcs e1 & e2 in the above graph) and each arc with the opposite direction a backward arc (arcs e3 & e4 in the above graph).

If the flow along arc e is increased from 0 to t, then the flow along each forward arc ei is from ui to ui+t, and the flow along each backward arc ej is from uj to uj-t.

41


There must be some backward arcs in the cycle. Why? Because if not, we may increase t infinitely (Can this problem be unbounded?).

But when , the new feasible flow v shall let , which is impossible as and .

t tccucv pq cv

0c 0v

42


Now since cycle C(e) has backward arcs, when t is increased to certain amount, the flow along a backward arc becomes zero, and we cannot further increase t (for otherwise some flow shall become negative, violating the feasibility).

43


In the above example, the largest value of t to maintain a non-negative flow is .

Suppose t = u3. In this case, call e3 the leaving arc. Generally, if an arc f in C(e) is the leaving arc, we

obtain a new tree by deleting and adjoining .

The updated flow vector v is associated with , i.e., v is a new FTS, and the associated spanning tree is .

43 ,min uut

T uTf uTe

T

TvT

44


Theorem 2.3 If u is an FTS with a profitable arc e and if v is the updated flow using this arc e, then v is an FTS and .

Proof We first show that v is still an FTS, because

(1) When we change from flow u to v, the net-flow at each vertex remains unaffected, and hence all supply / demand requirements are still satisfied, i.e. Av = b.

(2) Obviously, .

cucv

0v

45


We next show that . Let , where y is the dual solution of u. By Theorem 2.2, . Consider two types of arcs: (a) If arc is in the tree T(u)

(b) If arc is not in the T(u) and other than the profitable arc .

cucv yAcc

vccucv

ji,

0ijc 0 ijij vc

ji, qp,

0ijv 0 ijij vc

46


Therefore, . In particular, if u is a nondegenerate FTS , arc t = min { uij | is a backward arc of

T(u) } >0 . If all calculated FTS are non-degenerate,

the computation must be terminated after a finite number of iterations (i.e. no cycling).

0pqpqijij vcvcvc cucv

0 iju uTji ,

ji,0 tv pq cucv uv

47


We now summarize the steps. Initial Step: find an initial FTS and an

associated spanning tree . Let k = 1. Iterative Steps: for k = 1, 2, … Step 1 Calculate dual solution on the

tree . Step 2 If , i.e., the

network has no profitable arc, terminate, is the optimal FTS. Otherwise obtain a profitable arc with .

)1(x)1(T

)(ky)(kT

ijk

ik

j cyy )()( )(, kTji )(kx

)(, kTqpe pqk

pk

q cyy )()(

48


Step 3 Form the cycle and determine t = min { | is a backward arc in C(e) } Suppose the minimum is attained at arc f. Let If (i, j) is a forward arc of C(e) If (i, j) is a backward arc of C(e) If (i, j) = (p, q) For other arcs and . and return to Step 1.

)()( kTeeC )(k

ijx ji,

,

,

,

,

1

kij

kij

kij

kij

x

t

tx

tx

x

feTT kk )()1(

1 kk

49


Let us now use Example 2.1 to explain the algorithm.

S = 3

S = 5 S = 2

D = 5

D = 2

D = 3

50

Tradition about arc labeling

In the above example, there are totally 10 arcs: (1,3), (2,1), (3,2) (5,1), (5,2), (6,1), (6,4) (6,7) (7,4) (7,5).

We normally, label them according to the first number and then the second.

e1 = (1,3), e2 = (2,1), …

51


Iteration 1 We have given an initial feasible tree solution

with which we start our first iteration. Step 1. The initial FTS is xT = [ 3 0 0 0 2 3 0 0 5 0 ] and its cost is z = cx = 71. The arcs of a

spanning tree T (not unique since x is degenerate) with which x can be associated are (1, 3), (2, 1), (5, 2), (6, 1), (6, 7) and (7, 4).

52


For each arc in T ji,

ijij cyy or ijji cyy Start with

07 y

5507474 cyy

5506776 cyy

61156161 cyy

7161313 cyy

5162112 cyy

0555225 cyy

.

So, y = [ 6, 5, 7, 5, 0, -5, 0 ] .

53


Step 2. We test whether there is a profitable arc e with respect to x. In the present case we see that e = (5, 1) is the only profitable arc which comes into T(x) as an entering arc.

In the degenerate case, cycling might occur, but we can use the methods learned in Linear Programming to avoid cycling.

ji, jiijij yycc

05601155151 yycc

For each arc not in T, calculate

.

~ entering arc.

54


Step 3. The cycle C(e) is

in which (5, 2) and (2, 1) are both backward arcs with existing flow 2 and 0, respectively. We revise the flow in this cycle as t units along (5, 1), 2 - t units along (5, 2) and 0 - t units along (2, 1). Therefore t = 0. Since (5, 1) has just entered it does not make sense to take it as a leaving arc. Thus (2, 1) becomes the leaving arc. Since t = 0, the cost remains unchanged.

55


Iteration 2 Step 1. The current FTS is

xT = [ 3 0 0 0 2 3 0 0 5 0 ] with cost z = 71. The arcs of the

spanning tree are (1, 3), (5, 1), (5, 2), (6, 1), (6, 7) and (7, 4). From the defining equations,

y = [ 6 10 7 5 5 -5 0 ] Step 2. We see that e = (3, 2) is the

only profitable arc which enters T(x).

56


Step 3. The cycle C(e) is

We get t = 2 and (5, 2) leaves the tree.

57


Iteration 3 Step 1. Our FTS is now

xT = [ 5 0 2 2 0 3 0 0 5 0 ] and z = 67. The spanning tree has arcs (1, 3),

(3, 2), (5, 1), (6, 1), (6, 7) and (7, 4). Also y = [6 8 7 5 5 -5 0 ] Step 2. There are no profitable arcs for the

current FTS. At this stage the algorithm halts after reporting that the current FTS is an optimal solution.

58

Again to avoid having many different answer

If there are two profitable arc, add the arc having prior (smallest) label.

For example, if there are 3 profitable arc (2,3), (3,4) and (4,1), Then please add (2,3). Otherwise, I may mark your answer wrong.

59

2.2 Initialization As we have seen, the network simplex

method need to have a feasible tree solution (FTS) to start with.

How to obtain the initial FTS and the associated spanning tree?

Use artificial arcs.

60

2.2 Initialization

The way to obtain an initial spanning tree T

1. choose an arbitrary vertex v in the graph.(Always pick vertex 1 in exam)

2. for each source i, if there is an arc from i to v, then this arc (i, v) is an arc of T;

if there is no arc from i to v, construct an artificial arc (i, v) and take it as an arc of T;

61

2.2 Initialization 3. for each sink j, if there is an arc from v to j,

then this arc (v, j) is an arc of T; if there is no arc from v to j, construct an

artificial arc (v, j) and take it as an arc of T; 4. for each intermediate vertex k, if there is an

arc from v to k or from k to v, we take that arc as an arc of T;

if there is no such an arc, construct an artificial arc in either direction and take it as an arc of T.

This T must be a spanning tree.

62

2.2 Initialization Suppose there are r artificial arcs, we

denote the enlarged network by which has m + r arcs, but still n vertices.

Define a new cost vector by taking

G

1

0ijc

If (i, j) is an arc of the original network G

If (i, j) is an artificial arc

m rc R

63

2.2 Initialization

If the incidence matrix of is , we solve the problem:

min s.t.

Note that the vector .

G A

P xcbxA

0xrmRx

64

2.2 Initialization

Obviously, , for all sources i; , for all sinks j; , for other arcs in form a feasible tree solution (FTS)

corresponding to the tree T. Starting with this (initial) FTS, we use the

network simplex method to solve problem .

iiv bx

jvj bx 0ijx G

P

65

2.2 Initialization Suppose x* is an optimal FTS with an

associated spanning tree T*. There are three possibilities: Case 1 T* contains no artificial arc. In this case T* is a spanning tree of the

original network G and the components of x* that correspond to the artificial arcs must be zero. Thus the original transshipment problem is feasible. We delete all artificial arcs, replace cost vector by c and use T* as the initial tree to solve the original network problem.

c

66

2.2 Initialization

Case 2 T* has at least one artificial arc along which the flow is positive.

In this case, the optimal value of problem is positive which means that the original problem has no feasible solution.

P

67

2.2 Initialization Case 3 T* has at least one artificial arc, but

the flow in every artificial arc is zero. The original problem has a feasible solution

x obtained by the restriction of x* to the set of original arcs. After deleting the artificial arcs in T*, the remaining part of T* has less than n-1 arcs, which cannot form a spanning tree. So, it must consist of several disconnected components, which implies that the original problem can be decomposed into several smaller sub-problems which may be solved separately.

68

2.2 Initialization

Example 2.3

T675746453525231412 ,,,,,,,, xxxxxxxxxx

T7,11,0,0,8,4,6 b

69

2.2 Initialization Iteration 1. Introduce arcs a10 =(1,5) and a11 =(5,6). Initial spanning tree T(1):

S = 4

S = 6

S = 8

D = 7

D = 11

70

2.2 Initialization Let = [ 0 0 0 0 0 0 0 0 0 | 1 1 ] . Initial FTS x(1) = [ 0 0 0 4 8 0 0 7 0 | 6 11 ]T . , , , , , , . .

c

07 y

05775 cyy 15656 cyy

04554 cyy 03553 cyy

02552 cyy 11551 cyy

0,1,0,0,0,0,1)1( y

71

2.2 Initialization

For the arc (1, 2) not in T(1), . arc is profitable.

01010211212 yycc

2,11 a

1)1( aT )1(C contains cycle

.

72

2.2 Initialization

The only backward arc is which is the leaving arc. t = 6.

5,110 a

x(2) = [6 0 0 10 8 0 0 7 0 0 11 ]T

73

2.2 Initialization Iteration 2. y2 y7 don’t change, . So, y(2) = [ 0 0 0 0 0 1 0 ]. For arc , . is a profitable arc.

0001221 cyy

6,47 a01100644646 yycc

6,47 a

)2(C

5,46 a

Backward arcs in : (4, 5) and (5, 6)

is the leaving arc.

Since min {x45, x56} = min {0, 11} = 0,

74

2.2 Initialization

, but T(3) is different from T(2).)2()3( xx

75

2.2 Initialization

Iteration 3. don’t change, . So, y(3) = [ 0 0 0 1 0 1 0 ]. For arc , is a profitable arc.

123567 ,,,,, yyyyyy

1014664 cyy

4,12 a 01100411414 yycc

4,12 a

76

2.2 Initialization

Since min {x12, x25, x56} = min {6, 10, 11} = 6,

is the leaving arc. x(4) = [ 0 6 0 4 8 0 6 7 0 0 5 ]T.

2,11 a

77

2.2 Initialization

Iteration 4.y2 y7 don’t change, 1011441 cyy

.

So, y(4) = [ 1 0 0 1 0 1 0 ].

10 2112 yyc

00 3223 yyc

20 5115 yyc

10 7667 yyc

10 5445 yyc

For the 5 arcs not in T(4),So, there are no profitable arcs, and x(4) is the optimal solution of problem . PBut the flow on the artificial arc (5, 6) is positive, which means the original problem is infeasible.

78

2.2 Initialization

Integer Solutions If all supplies / demands are integer

values, i.e. b is an integer vector, then the optimal solution x* of the transshipment problem, obtained by the network simplex method, must be an integer vector.

Why? because the incidence matrix A is unimodular.

79

2.2 Initialization Let be the matrix obtained by deleting the first

row of A; and be the vector obtained by deleting the first component of b.

Then the equation is equivalent to (since rank(A) = n-1). Any feasible tree solution x corresponds to a non-singular submatrix B of such that , where and .

So, each component of has the form , where p is an integer. Since A is unimodular, as a

nonsingular square submatrix of A, B must satisfy . So, must be an integer.

A mn 1

11 n

bAx bxA

b

)1(1 nn A TN

TB

T xxx ,0Nx bBxB

ix BxB

pxi det

1det B

ix

80

2.2 Initialization Unbalanced Transshipment Problem So far we consider balanced transshipment

problem, i.e. total supply = total demand. Now if total supply > total demand we let r = total supply - total demand and introduce one dummy sink v with the

demand r and an arc from each source vertex to v with zero cost.

Hence the unbalanced problem becomes a balanced problem in the enlarged network.

is

81

2.2 Initialization

Note that if in the optimal solution, the supply of the source i is and the flow from i to the dummy vertex v is , then the source i sends out only units of the commodity and remains units at its place.

ib

0ivx

ivi xb

ivx

82

2.4 Transportation Problems

Motivating Example: A number of warehouses must ship a

single commodity to a number of manufacturing plants.

Shipping costs are proportional to distance and the amount shipped.

What shipping pattern will minimize total shipment costs?

83


84

2.4 Transportation Problems A transportation problem is a special

transshipment problem in which (1) there is no intermediate vertices (i.e.

every vertex is either a source, or a sink).

(2) every arc is from a source to a sink. The underlying graph of a transportation

problem is a bipartite graph.

85

2.4 Transportation Problems G = (S, D, E). S ~ the set of sources, denoted by (i

= 1, … , m); D ~ the set of sinks, denoted by (j =

1, …, n); E ~ the set of arcs from a source

to a sink . Flow matrix: matrix ~ is the

flow on the arc ;

iS

jD

ji, iSjD

nm ijx ijx ji,

86


Cost matrix: matrix ~ is the cost to send one unit of the commodity from

to . Note that if actually there is no arc from

to , in order to express the problem easily, we assume that there is an arc , but the cost is very expensive, i.e. let , where M is a very large number.

nm ijc ijc

iS jD

iSjD

ji,

Mcij

87

2.4 Transportation Problems Now the transportation problem can be

expressed mathematically as min

s.t. , for i = 1, … , m (1)

, for j = 1, … , n (2)

for i = 1, … , m; j = 1, … , n (3)

m

i

n

jijij xc

1 1

n

jiij bx

1

m

ijij bx

1

0ijx

88


The constraint equations (1) and (2) can be jointly expressed as:

A x = b,where A is an (m+n) x mn matrix, which is in fact the

incidence matrix of the associated bipartite graph (hence is a unimodular one according to Corollary 1.3b);

b is an m+n vector by combining the right-hand-side of the equations (1) and (2); and

x is an m∙n vector consists of all . ijx

89

2.4 Transportation Problems Transportation problem can be solved

on a particularly designed tableau, called the transportation tableau.

90

2.4 Transportation Problems Finding an initial FTS ~ the

northwest corner method.

91


Note that the m+n-1 = 6 filled entries, together with other , form an initial FTS.

The arcs corresponding to the filled entries form a spanning tree T.

ijx

0ijx

ijx

92

2.4 Transportation Problems Calculating dual solution y

associated with the tree T We use the equation for any arc

to determine y. In the above example, let .

We fill in y1 ~ y3 to the right-most column and y4 ~ y7 to the bottom row.

jiji ycy Tji ,

07 y

88037373 yycy

19866363 yycy

67126262 yycy

...5 y

...1 y

...4 y

93

2.4 Transportation Problems Checking if there is a profitable arc For each arc , calculate and

check if all such . If yes, an optimal solution is obtained. But is profitable.

Tji , jiijij yycd 0ijd

6,106183611616 yycd

94

2.4 Transportation Problems Deciding the leaving arc Starting from the cell (1, 6) in the

tableau, move horizontally or vertically until return to the cell (1, 6) to form a cycle. 6,15,15,26,26,1

It is seen that in the cycle

15261

the arcs are forward and backward alternatively.

95


Put a + in the tableau for every forward arc;

Put a - in the tableau for every backward arc;

Decide t = min { | is a backward arc}.

ijx ji,

96

2.4 Transportation Problems Moving from a non-optimal

tableau to the next tableau

Let If (i, j) is a forward arc; If (i, j) is a backward arc;

Other arcs

ij

ij

ij

ij

x

tx

tx

xnew

97

2.4 Transportation Problems Introduce the entering (profitable) arc into the

new tree; remove the leaving arc from the tree. Note that in the example the two backward arcs (1, 5) and (2, 6) both have .

In this case, we can choose either one as a leaving arc (in the example (2,6) is chosen to leave the tree), but not both, because to maintain a spanning tree need to have m+n-1 = 6 arcs.

So far we completed one iteration and switch to a new tableau.

102615 txx

98

2.4 Transportation Problems Example 2.7 If an FTS is given by the

circled numbers in the tableau below, and suppose the entering cell is (3, 7), how to form a cycle in the tableau?

99

2.4 Transportation Problems Note that when we are at (1, 8) and move

horizontally, we have to skip the cell (1, 6) and reach (1, 4). Then we move vertically. If we change the moving direction at (1, 6), no cycle can be formed.

100

2.5 Some Applications to

Combinatorics Definition A matching in a graph is a

set M of edges of the graph such that no two edges in M have a vertex in common.

graph G a matching of G

101

2.5 Some Applications to Combinatorics

In a bipartite graph G = (X, Y, E) where X and Y both have n vertices (it is possible that between some pair of vertices and , there exists no edges, i.e., G may not be a complete bipartite graph), a matching M is called a perfect matching if there are n edges in M.

Xx Yy

102


This bipartite graph has no perfect matching

This graph has a perfect matching

132231 ,,,,, vuvuvuM

103


Theorem 2.5 (Konig’s marriage theorem) A regular bipartite graph G(U, V, E) of

degree k (where k is a positive integer) with has a perfect matching.

A graph is called a regular graph of degree k if the degree of each vertex is k.

2-regular graph

nVU

104


Proof We formulate a transportation problem as follows:

Let the vertices in U be sources; the vertices in V be sinks; For each edge , give an arc from to . Let the supply at each source be 1; the demand at each sink be 1; the unit cost on each arc (u, v) be 1.

Uu Vv

uvc

Evu },{

105


This problem is feasible because if we let the flow along each arc be , it is a feasible flow.

Also, the supply-demand vector is an integer vector as the incident matrix of a bipartite graph is unimodular. Therefore, the transportation problem must have an integer optimal solution x*.

In fact as all supplies / demands are 1, x* must be a 0-1 vector.

Let M = { (u, v) | , u, v = 1, … , n }. It is easy to see that |M|=n, i.e., M must be a

perfect matching.

k1

1* uvx

106


Example Does the following bipartite graph have a perfect matching? If yes, give a perfect matching. As this is a regular

graph of degree 2, it has a perfect matching.

Find a perfect matching by yourself.

107


Definition A covering in a graph is a set W of vertices such that every edge of the graph is incident to at least one of the vertices in W.

v6, v1, v3, v4 (or v5) is a covering of the left graph

108


For any matching M and any covering C of a graph, suppose

the graph has k edges , ,… , , such that all and are

different to cover these k edges, we need to

use at least k vertices from a covering C must have at least k vertices

kM

111 ,vue 222 ,vue

kkk vue , iu iv

,...,,,..., 11 kk vvuu

kC

109

Therefore, for any matching M and any covering C

This result can be strengthened to become an equality if bipartite graphs are concerned. (Just like Primal and Dual)


CM

} allfor : |min{|

} allfor : |max{|

CC

MM

110


Theorem 2.8 (Konig’s Theorem) In any bipartite graph, the size of a

maximum matching equals to the size of a minimum covering.

111


Konig’s Theorem can be translated equivalently to the context of matrices. We use the term a line to denote either a row, or a column.

Suppose P is a property that an element of a matrix may or may not have. We call an element of a matrix a P-element if it has the property.

A set of P-elements is said to be P-independent if no two of them lie on the same line.

The size of the largest P-independent set is called the P-rank of the matrix.

112


Example 2.10 Let matrix

An element of A is said to have property P if it is positive, i.e., if its value is 1.

Obviously, the P-rank of A is 3 as a largest P-independent set contains 3 elements, for example, the (1, 1)th, (2, 3)th and (5, 4)th element of A.

1100

0100

0100

0100

0111

A

113


Theorem 2.9 (Konig-Egervary Theorem)

The P-rank of a matrix A is equal to the minimum number of lines that contain all P-elements of the matrix.

114


Proof We construct a bipartite graph. Let each row i of A correspond to a left vertex i; each column j of A correspond to a right vertex j. Introduce edge if and only if the element of A is a P-element.

ji, th, ji

1100

0100

0100

0100

0111

A

115


By Theorem 2.8, Size of a max. matching = Size of a min.

covering. Now, a matching is a set of edges, all with

different (left and right) endpoints. (correspond to) a set of P-elements in different

rows/columns, i.e. a P-independent set. So, Size of a max. matching = Size of the largest P-

independent set = P-rank of A.

116


A covering is a set of vertices such that every edge of the graph is incident to at least one of these vertices.

(correspond to) a set of rows and columns such that each P-element is contained in one of these lines.

So, Size of a min. covering = minimum number

lines of A that contain all P-elements of the matrix.

Thus, the theorem is proved.

117

2.6 Assignment Problem The assignment problem is a

special kind of transportation problem in which

(1) the supply at each source and the demand at each sink are equal to 1.

(2) the number of sources = the number of sinks.

118

2.6 Assignment Problem Let be the cost to send one unit of

product (or sevice) from source i to sink j.

The mathematical formulation of the assignment problem is

min s.t. , , ,

ijc

n

jiijij xc

1,

n

jijx

1

1 ni ,...,1

n

iijx

1

1 nj ,...,1

0ijx nji ,...,1,

119

2.6 Assignment Problem

Note that actually we are interested in integer solutions, i.e. should be 0 or 1, but this request does not need to be imposed explicitly, because the unimodular property for bipartite graph guarantees an integer optimal solution.

ijx

120

2.6 Assignment Problem Example A mill has

4 machines and 4 jobs to be completed. Each machine must be assigned to complete one job. The times required to complete each job on different machines are shown in the table:

Time (hours) Job 1 Job 2 Job 3 Job 4

Machine 1 14 5 8 7

Machine 2 2 12 6 5

Machine 3 7 8 3 9

Machine 4 2 4 6 10

Decide an assignment scheme to minimize the total time needed to complete the 4 jobs.

121


Method 1 use the method for transportation problems.

Method 2 the Hungarian method.

Theorem 2.17 Suppose C is the cost matrix and is the

matrix obtained by adding a number t to each element in a row or to each element in a column. Then a solution is optimal with respect to if and only if it is optimal with respect to C.

C'

C'

122

2.6 Assignment Problem Proof Suppose the row is added by

t, i.e. , and , for and

Then for any feasible solution , = =

= = Therefore, under C and C’ , the

optimal solutions are the same.

thktcc kjkj nj ,...,1 ijij cc ki

nj ,...,1

[ ]ijx

n

i

n

jijij xc

1 1

ki

n

j

n

jkjkjijij xcxc

1 1

ki

n

j

n

jkjkjijij xtcxc

1 1

n

i

n

j

n

jkjijij xtxc

1 1 1

n

i

n

jijij txc

1 1

123


Based on this theorem, we can change C to C’ such that each row and each column of C’ has at least one zero entry by the following operations:

Step 1 Subtract each row by its smallest entry; Subtract each column by its smallest

entry.

124

2.6 Assignment Problem2 examples

00234

03214

00320

32004

31203

)3(

)1(

)1(

)2(

)1(

33567

14325

11431

54226

42314

AA

32022

20005

24003

46035

07040

43043

31026

35024

57056

18061

)7(

)3(

)3(

)3(

)3(

111071110

64359

68357

810389

411394

BB

(-1) (-2) (-1) (-1)

125

If in such modified matrix C’ , there are n zero entries and not two of them in the same row or same column, for example

then obviously , other is an optimal solution, because for

any feasible solution x,

0

0

0

0

y y y

y y yc

y y y

y y y


(all y are non-negative numbers)

1*44

*32

*21

*13 xxxx 0* ijx

4,3,2,1, ji

0* xc

126


Question: How to check if C’ has zero entries at all different rows and columns?

This is in the method of step 2. Step 2 Check whether C’ has n zero entries

at different rows and columns. If the answer is yes, then an optimal assignment is obtained by taking the assignment corresponding to the n zero entries and we stop.

Otherwise we go to Step 3.

0, ' 0 , ' ijijijij xcxc but 0. ' * ijij xc

127

2.6 Assignment Problem This step can be fulfilled by the following

sub-steps. Step 2a Locate a row or a column in C’

with exactly one 0, and the 0 is selected. If every line (row or column) has more

than one zero, a line with the minimum number of 0s is located, and we select anyone of the 0s in this line. This line should not be located again (we can put a check mark beside it)

128


Put a circle around the selected 0, and draw a line through (covering) the 0. This line should be

a vertical line if the line located is a row, and

a horizontal line if the line located is a column.

129


Step 2b Repeat the previous step (but do not consider any 0 which has been covered by a line) till every 0 has at least one line through it.

Step 2c If the number of lines drawn is n, the n selected (circled) 0s will point out an optimal solution. If the number of lines is less than n, we go to Step 3.

130

2.6 Assignment Problem Examples

A

00234

03213

00320

32004

31203

00234

03213

00320

32004

31203

00234

03213

00320

32004

31203

00234

03213

00320

32004

31203

00234

03213

00320

32004

31203

00234

03213

00320

32004

31203

1*31

*54

*45

*23

*12 xxxxx 0* ijx 5,...1, ji

n = 5 (lines), we obtain an optimal solution which is

and other

131


B

32022

20005

24003

46035

07040

32022

20005

24003

46035

07040

32022

20005

24003

46035

07040

32022

20005

24003

46035

07040

32022

20005

24003

46035

07040

n = 4 < 5, so, we are unable to obtain an optimal assignment by using the modified matrix .

B

132


Generally, if the number of lines needed to cover the zeros is k and k<n, we call the 0 element in the matrix as P-element and by Th. 2.9, we know that the P-rank of the matrix is k, i.e., the maximal number of 0 elements in different rows and columns is k.

In this case, we need to modify the matrix further to create more 0s in the right place.

Let m be the smallest value of the uncovered elements.

We subtract m from all uncovered rows ~ so that more 0s appear in the previously uncovered part.

c

133

2.6 Assignment Problem Example m = 2

10200

20005

02221

24213

07040

)2(

)2(

)2(

32022

20005

24003

46035

07040

B

However, now in the covered column, some 0 become –m which are negative.We now add m to the covered columns.

B

10020

20225

02001

24033

07260 (+2) (+2)

134


The above operations can be further simplified. We divide all elements in the matrix obtained at Step 2 into 3 groups.

(1) Uncovered elements. (2) Singly covered elements ~ covered

by only one line (a row or a column) (3) Doubly covered elements ~ covered

by two lines: a row and a column.

135


Step 3a Find the value of the smallest uncovered element. Suppose it is m. We

(a) subtract m from all uncovered elements;

(b) add m to all doubly covered elements. (and keep the singly covered elements

unchanged)

136


For example, if we apply the operations (a) and (b) to :B

10020

20225

02001

24033

07260

32022

20005

24003

46035

07040

B

which is just B

.

137


Note that the 0s to be selected to draw the lines (i.e. the 0s with ) are singly covered, hence they are still 0.

But some elements in the uncovered part become 0 ~ new 0, and thus need to use more lines to cover all 0.

138


Step 3b Delete all covering lines and go back to Step 2.

10020

20225

02001

24033

07260

10020

20225

02001

24033

07260

B

139


Now the number of lines to cover all 0s = 5 = n.

So, we obtain an optimal assignment:

and other , i.e. source 1 sink 5, source 2 sink 3, source 3 sink 2, source 4 sink 4, source 5 sink 1.

1*51

*44

*32

*23

*15 xxxxx 0* ijx

1 network optimization chapter 2 transshipment problem

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