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Facility Location and Strategic Supply Chain Management

Prof. Dr. Stefan Nickel

Location Theory

Chapter 5 – Discrete Location Problems

Contents

• Introduction

• Classification of Discrete Location Problems

• Warehouse Location Problem (WLP)

• Extensions of the WLP




Introduction

Topic of the chapterCost-oriented location of one or more than one new facilities out of a set of given locations to meet customer demands.

By far most realistic and most flexible class of location problems.

Almost all real-world constraints, demands and guidelines can be considered in these models.

However, these approaches involve an increasing complexity in modeling the problem and especially in solving the resulting model.



Introduction

Basic assumptionsCustomers and demands

Consider a set of customers with given demands.

CustomersE.g. retail stores, market areas, private individuals

DemandsE.g. products, servicesThey are independent from the location decision.

Potential locationsGiven a set of potential locations where

- new facilities can be opened or- already existing facilities can be closed.

FacilitiesE.g. central or regional warehouses, transshipment points, factories



IntroductionCosts

Occur as a result of - the satisfaction of customer demands and - location decisions

Satisfaction of customer demands- transportation, inventory and handling costs- material, raw material or production costs

Are quantity-dependent: variable costs.

Location decisions- estate- , land development- and building costs - closure costs- operating costs (personnel costs, unit costs)

Are both variable and fix (absolute or step-wise).



IntroductionExample

A company distributes their manufactured goods from its factories through the warehouses to the customers. Due to the increased demand the company has to restructure its distribution network.

Old warehouses which have to be closed and new ones which have to be openedin previously determined locations.

reorganization

Factory Ware-house

Customer Factory Ware-house

store

store

Customer

X



Location Theory


Contents

• Introduction

• Classification of discrete location problems






Classification of Discrete Location Problems

Time horizonNumber of time periods (e.g. months, years) within the planning horizon.Distinguish

• Static (one-period) modelsLocation decisions must be made at the beginning of the planning horizon.

• Dynamic (multi-period) modelsLocation decisions involve where and when, i.e. in which period, the facilities are located. One has to determine a sequence of changes to be made.

Facility typesDistinguish various specifications of facilities.E.g. regional as well as central warehouses can be located.



Chapter 5 – Discrete Location ProblemsProduct aggregation

Is there just one product to plan or several products with different characteristics?

Number of echelonsNumber of echelons in the distribution network.Differentiate

• Single-level modelsThere is just one transportation echelon (e.g. inventory retailer) to distribute the goods Location decisions only at one level (e.g. inventory).

• Multi-level (hierarchical) modelsThere are several transportation echelons (e.g. factory inventory retailer) to distribute the goods Location decisions only for

- one level possible (e.g. inventory)- several levels possible (e.g. inventory and factory)



Classification of Discrete Location ProblemsInteraction among facilities

Material can flow among facilities of the same level, e.g. transports between warehouses are possible. Optimal facility locations depend not only on the spatial distribution of product demand but also on the mutual position of the facilities.

Uncertainty• Deterministic models

all data and parameters are completely known and certain• Probabilistic (stochastic) models

some data or parameters are uncertain

Direction of flowDifferentiate

• Distribution LogisticsGoods flow from factories/suppliers to warehouses/customers

• Reverse LogisticsGoods flow from customers to recycling centers, waste disposals



Classification of Discrete Location ProblemsCapacity restrictions

Limits the quantity, which can be produced, transported, handled, etc.

Causes• technical-economical restrictions (size of the machine)• personal restrictions

Distinguish

• Uncapacitated modelsno or insignificant capacities

• Capacitated models- with single-sourcing (demand is uniquely assigned to the facilities)- without single-sourcing



Classification of Discrete Location ProblemsDelivery mode

• Route deliveryDeliveries from a facility on a higher level to lowerlevel facilities (or vice versa) will be bundled to tours. E.g. several retailers are visited per delivery trip, starting at the depot.

• Direct deliveryDeliveries from a facility on a higher levelto a lower one occur directly and without detours.E.g. a truck serves exactly one customer/retailerat a single tour.

The transportation mode has an influence on the locational decisions.



Location Theory


Contents

• Introduction

• Classification of discrete location problems

• Warehouse Location Problem (WLP)- Introduction- Heuristics- Dualoc Algorithm





Warehouse Location Problem (WLP)

Also called Uncapacitated Facility Location Problem (UFLP) or Simple Plant Location Problem (SPLP).

Assumptions• static, single-level, uncapacitated, deterministic• one product, direct delivery, one type of facilities, no interaction

Given• Set of customers I = {1, …, n} with demand bi , i = 1, …,n• Set of potential locations J = {1, …, m} for new facilities• Transportation costs tij per unit from the potential location j ∈ J to a customer i

∈ I. Include costs for material, production, storage, …

Total costs for serving customer i from location j: cij = bi · tij.



Warehouse Location ProblemCost matrix C = (cij )i=1,…,n, j=1,…,m

• Fixed costs fj for a new facility at location j.Include estate, building and fixed operating costs, but no variable costs.

Schematic illustration

1

2

n

1

2

m

b1

b2

bn

f1

f2

fm

c11

c1n

cnm

c2n

c12

Custo-mers

Pot. Locations



Warehouse Location Problem

Decisions• Number of new facilities• Locations for new facilities• Assignment of customers to the new facilities

Solution of the WLP consists of

p locations X = {x1, …, xp }X: Subset of potential sites J: X ⊆ J.Elements xk ∈ X are indices of potential sites.

Allocation of customers to the locations of new facilities

In uncapacitated problems a customer i ∈ I is always assigned to the most cost-effective new facility j ∈ X

cij = min {cik | k ∈ X}

X induces an allocation.



Warehouse Location ProblemNumber of new facilities

Trade-off between transportation costs and costs for the new facilities

Large number: low transportation costs but high costs for new facilitiesSmall number: high transportation costs but low costs for new facilities

# Facilities

Costs

Transpor-tation Costs

Costs for newfacilities

Total

Costs



Warehouse Location ProblemNumber of new facilities is sensitive to changes in costs.

Example: Daskin (1995)

„Average Distance“: average distance from the customers to the facilities.




Other real world examples for the WLPExample 1: supplier selection

Given• bi the demand for product i ∈ I per period• j ∈ J the feasible suppliers• the procurement costs cij of bi quantity units at supplier j• fixed costs fj of business connection

Possible taskChoose the suppliers with minimal costs

Alternative objectiveGiven: delivery time tij for bi quantity units at supplier jChoose at most p suppliers in order to minimize the total delivery time

p – median assignment problem



Warehouse Location ProblemExample 2: Physical Database Design, Index Selection

A database is a set of tables.Tables consists of data fields (columns) and data sets (rows).Data fields j ∈ J can be indexed in order to accelerates requests.

Given• Typical Workload (Set of SQL-Statements like Select, Update, lnsert, Delete)

determined over a Log-File of a period• Answer time gain cij from request i, if field j is indexed

• Maintenance time fj of index j (slowdown of updates at indexing field j)

Single field indexingChoose the fields which should be indexed in order to maximize ATG minus MT



Warehouse Location ProblemExample 3: Location of Bank Accounts

Problem• A company has to pay money to business partners in different cities every

month

• Time between cashing a check and debiting the account, depends on locationj of the debiting account and cashing location i.

Given• Locations i ∈ I of creditors and value bi of monthly payments in i.• Possible bank accounts j ∈ J with monthly account fees fj.• gij equivalent amount of the maturity of a check cut in j and cashed in i.

ObjectiveChoose bank accounts j ∈ J, so that the maturity of checks is maximizedregarding to account fees.




Formulation as a mixed integer linear program

Location decisionBinary variable yj, j ∈ J

Allocation decisionVariable xij, i ∈ I, j ∈ J



WLP as mixed integer linear programMathematical Model

Remark: xij ≥ 0 is sufficient.In uncapacitated problems the customer demand is always completely satisfied by the „cheapest“ new facility.

Each customer i is allocated to a new facility at location j

Minimize transportation costs and costs involved by the facilities

Customer i can only be allocated to location j, if a facility is locatedat j



WLP as mixed integer linear programInterpretation of the solution (x, y) of the linear program

• costs

• set of potential sites where facilities are located

X := { j | yj = 1, j = 1, …, m}

• customer i is allocated to the new facility at location j, if xij = 1 holds.

Vice versaEach solution X ⊆ J of the WLP with its corresponding allocations induces a solution (x, y) of the linear program

and



WLP as mixed integer linear programEnumeration of all subsets X from J yields an optimal solution.

HoweverThere are feasible subsets!

ResultThe Warehouse Location Problem is NP – hard.

Possible solution algorithms• Exact algorithms, which determine an optimal solution.

Special algorithms for small problemsDualoc – Algorithm

• HeuristicsGreedy and Interchange Heuristic



Location Theory


Contents

• Introduction







Heuristics

Algorithms to compute a heuristic (approximate) solution XH.

Characteristics• Highly flexible in order to consider alternative objectives or the incorporation

of complex planning restrictions.• Do not necessarily find an optimal solution.

• In general the quality of a solution is unknown.

Quality of an heuristic solution XH

Relative deviation (in percent), RPD(XH, X*), between the objective function value of XH and the objective function value of an optimal solution X*.



HeuristicsFormal

However: X* usually unknown⇒ lower bounds for the optimal objective function value

Lower bound LB for ƒ(X*), i.e. LB ≤ ƒ(X*), via• An (optimal) solution of the dual problem• Lagrange – Relaxation• LP – Relaxation (i.e. yj ≥ 0, instead of yj ∈ {0, 1})

Then




Location Allocation Heuristics

Distinguish between

Construction heuristicsCompute a first solution of the problem.Classical Algorithms:

- Greedy (ADD – Heuristic)- Stingy (DROP – Heuristic)

Are the basis for

Improvement heuristicsTry to improve an existing solution.Classical Algorithms:

- Interchange (ADD&DROP – Heuristic)- Metaheuristics: Variable Neighborhood/Tabu Search, Genetic Algorithms




Greedy Heuristic (ADD Heuristic)

Start with the empty solution (X = ∅) and construct a solutionX = { x1, …, xp } step by step.

For this purpose, add in each iteration the location to the solution, which decreases the objective function value as much as possible.

Stop, as soon as there is no such location any more.

Let• Xℓ = { x1, …, xℓ } denotes the current solution with ℓ facilities.

Note that x1, …, xℓ correspond to indices of potential locations.

• J‘ = J \ Xℓ denotes the set of the locations, which are not a part of the solution, yet.

• uiℓ = cij* = min { cij | j ∈ Xℓ } denotes the least costs to satisfy the demand of

customer i from a facility in Xℓ.A customer i is always assigned to the cheapest new facility j*.



Greedy HeuristicCosts of the solution Xℓ:

In iteration ℓ, we want to add one location to the solution Xℓ-1.Adding location j ∈ J‘ to the solution changes the costs by



Greedy HeuristicAdd the potential location j ∈ J‘ to the solution for which holds

Omega – Rule

Cost saving ωjℓ is monotonously decreasing

⇒ If for k ∈ J‘ : ωkℓ ≤ 0 holds, then delete k from J‘.

Because ωjℓ will not be positive anymore also in the following iterations



Greedy HeuristicGreedy Algorithm

1. Initialization• Set X0 = ∅, J‘ := J and ℓ = 1.• ui

0 = maxj = 1,…,m cij maximal costs to satisfy the demand of customer i ∈ Ifrom a potential location j ∈ J.

2. While J‘ ≠ ∅

If ωjℓ ≤ 0 for all j ∈ J‘, then STOP.

Determine a potential location j ∈ J‘ with

Set Xℓ := Xℓ-1 ∪ { j } and J‘ := J‘ \ { j }.

If ωjℓ ≤ 0 for j ∈ J‘, then delete the potential location j from J‘: J‘ := J‘ \ { j }

Set ℓ = ℓ + 1.

3. Xℓ is an heuristic solution of the problem.

Complexity (in this version): O(m2 n)



Example

Let the transportation costs and the costs for new facilities be

Iteration 1 ℓ = 1, J‘ = {1,2,3,4}

Maximal cost saving at location 2: ω21 = 13

⇒ X1 = X0 ∪ {2 }, J‘ := {1,3,4} and F(X1) = F(X0) – ω21 = 95 – 13 = 82.

Because ω41 < 0, delete location 4 from J‘ ⇒ J‘ = {1,3}

Greedy Heuristic

and



Iteration 2

ℓ = 2, J‘ = {1,3} and X1 = {2}.

Just location 3 allows new cost savings: ω31 = 1.

⇒ X2 = X1 ∪ {3 }, J‘ := {1} and F(X2) = 82 – 1 = 81.

Because ω11 < 0, delete location 1 out of J‘ ⇒ J‘ = ∅ stop

Heuristic solution: XH = X2 = {2, 3} and F(XH ) = 81.Value of the LP – Relaxation: 76 ⇒ XH max. 6.5% away from optimal solution.

Greedy Heuristic: Examplex1 = 2



Warehouse Location ProblemInterchange Heuristic

Given a solution X = { x1, …, xp }.Interchange locations step-by-step in order to improve the solution.

If xk ∈ X is a site where a facility is located and j ∈ J‘ = J \ X a site where no facility has been located yet, then obtain a new solution X‘ by

X‘ = X \ { xk } ∪ { j }

If the resulting solution X‘ is better than the last solution X: F(X‘) < F(X), then „accept“ this exchange, i.e. set X = X‘.

Otherwise try to interchange other solution-locations and non-solution-locations in order to improve the solution.

X

J‘



Interchange HeuristicStop the algorithm, if the solution cannot be improved by interchanges of locations.

Sequential choice of exchange pairs.Two possible strategies:

First ImprovementStart with the first location of the solution and interchange with all potential locations, which are not yet in the solution.Repeat this step for the second, third,… location of the solution.Stop, when an exchange provides a better solution.

Best ImprovementDetermine the cost saving for all possible interchanges and realize the interchange with maximal cost saving.



Interchange HeuristicCosts of the new solutionLet

• ui = cij* = min { cij | j ∈ X } the lowest costs to satisfy the demand of customer i from a facility located at a site in X.

• uik = min { cij | j ∈ X, j ≠ xk } the lowest costs to satisfy the demand of

customer i from a facility located at a site inX \ { xk }.

Change in costs by interchanging the locations k ∈ X and j ∈ J‘



Interchange HeuristicFirst Improvement

Do save the interchange of locations xk ∈ X and j ∈ J‘ as soon as holds.

Best ImprovementDo save the interchange for those xk‘ ∈ X and j‘ ∈ J‘, for which

Objective function value from solution X:

Write all information one needs for theinterchange of locations xk ∈ X and j ∈ J‘ in one table



Interchange HeuristicInterchange Algorithm

1. Input X = {x1, …, xp } and J‘ := J \ X.

2. Compute ui = min { cij | j ∈ X } for all i = 1,…,n.3. For k = 1,…,p

Compute uik = min { cij | j ∈ X, j ≠ xk } for all i = 1,…,n.

For all j ∈ J‘If ωj

k > 0, then interchange X = X \ {xk } ∪ { j }, update J‘ = J \ X and go to step 2.

4. X is a heuristic solution of the problem

ImplementationCollect the required data for step 2 and 3 in a single table.



Example: First Improvement StrategyLet the transportationcosts and the costs for new facilities be and

X = {2, 3 } solution of the Greedy-Algorithm, F(X) = 81.

Step 2: Compute ui for all i = 1,…,n.Step 3: Start with k = 1.

One gets the following table

The first interchange-pair provides immediately the first improvement: ωj1 > 0.

Interchange Heuristic

x1 = 2 x2 = 3



⇒ Interchange X = X \ { x1 = 2 } ∪ { 1 }, i.e. replace location 2 by 1 in the solution.

Improved solution X = {1, 3 } with F(X) = 76.

Begin with step 2 again. One gets the following table

No ωjk > 0 ⇒ the algorithm stops.

Heuristic solution XH = {1,3 } withF(XH) = 76.

Value of the LP – Relaxation: 76

⇒ The obtained heuristic solution XH is optimal.

Interchange Heuristic: Example

x2 = 3x1 = 1



Location Theory


Contents

• Introduction







Dualoc Algorithm

The Dualoc algorithm is due to Erlenkotter (1978).

It is based on the Branch & Bound algorithm and determines an optimal solution of the mixed integer linear program for the WLP.

It uses extremely efficient algorithms to compute the lower and upper bounds for the subproblems in the root and the nodes of the B&B – tree.

These bounds can be computed as follows:

• heuristics solution of the dual problem of the LP – relaxation lower bound

• construction of an integer solution of the WLPupper bound



Dualoc Algorithm

Efficiency of the Dualoc algorithm is based on the “nice” property of the WLP, that the LP – Relaxation of the mixed integer linear program for theWLP often provides already an integer solution.



Dualoc Algorithm

Overview of the Dualoc AlgorithmFormulate the WLP as a mixed integer LP

Formulate the dual problem to the LP – relaxation

Compute a heuristic solution of the dual LPlower bound for the WLP

Derive a primal solution from the dual oneupper bound for the WLP

lower bound = upper

bound ?Adjust dual solution

Stopprimal

solution optimal

yes no Modified dual solution ?

no

Branch in the B & B – tree

yes



Dualoc Algorithm

The WLP as a mixed integer linear program

LP – RelaxationRelaxation of all integer constraints



Dualoc Algorithm

The dual LP

LP – relaxation (LP) ofthe mixed integer linearprogram for the WLP …

… and the correspondingdual linear program (DP)with the dual variablesvi and wij.



The Dual LPElimination of variables wij

For a feasible choice of the vi , feasibleand as small as possible values wij canbe found easily.

The wij are bounded from below by constraints(1) and (2):

⇒ Replace wij by max {0, vi – cij } in the dual LP.

Simplified (Reduced) dual LP



The Dual LPIt holds

vi = 0, i = 1, …, n (i.e. v = 0) is a feasible solution of the (RDP).(Assuming that fj, cij ≥ 0, ∀ i, j)

Let (x‘, y‘) and (x, y) be optimal solutions of (LP) and (WLP), respectively. Then

Weak duality theorem of linear optimizationLet (x‘, y‘) be a feasible solution of (LP).Then for each feasible solution v of (RDP) holds

⇒ RD(v) is a lower bound for the optimal objective function value of (LP) and (WLP).



Dualoc Algorithm




Derive a primal solution from the Dual oneupper bound for the WLP

lower bound = upper


Stopprimal

solution optimal


no

Branch at B & B – tree

yes



Dualoc Algorithm

Dual Ascent Algorithm

(RDP) can be solved optimally using the simplex algorithm.Alternative:

Dual Ascent – AlgorithmSimple, fast and efficient heuristic, to determine a feasible, but not necessarilyoptimal solution of (RDP).

ObservationIn many cases, the solution of the Dual Ascent Algorithm is already optimalfor the dual problem.

ApproachStarting with an initial solution v, sequentially increase each vi by a small amount until no further increase is possible anymore.



Dual Ascent AlgorithmDefine

as the slack of the j-th constraint, i.e. the j-th facility of (RDP).

For each solution v of (RDP) the constraints sj ≥ 0 must hold.

If v = 0 holds, it follows sj = fj for the slack sj of each constraint.

Now sort the cost elements cij of each customer i∈I in non-decreasing order and denote them ci

k , k = 1,…,m.

Define for each i∈ I the index k(i) := min {k | cik ≥ vi } as the index of the

first cost element that is larger than vi.



Dual Ascent AlgorithmNow try to increase sequentially each vi to the next larger value ci

k(i).Only allowed, if no slack sj becomes negative.

Let ∆ := cik(i) - vi be the value of the intended increase of vi.

Effect of the increase on the slack sj , j∈ J:None, if vi < cij, because

and hence

∆, if vi ≥ cij, because

i.e. the slack sj decreases by ∆:



Dual Ascent AlgorithmDefine:

Set of all constraints j whose slack sj would decrease by an increase of vi.

For an intended increase of vi by ∆ it must hold:

and therewith:

⇒ increase vi by

Let I‘ ⊆ I be the set of customer indices i for which vi still can be increased.



Dual Ascent AlgorithmDual Ascent Algorithm

1. InitializationLet ci

1 ≤ ci2 ≤ … ≤ ci

m ≤ cim+1 := ∞ be an ordering of the cost elements, i∈ I

Set I‘ := I and vi := ci1 for all i∈ I.

Compute Ji, i∈ I and sj, j∈ J.Determine k(i) := min {k | ci

k ≥ vi }, i∈ I . If vi = cik(i) then k(i) := k(i) + 1.

2. IterationFor all i∈ I‘

Determine ∆i . If ∆i = 0 then set I‘ = I‘ \ { i } and choose the next i ∈ I‘.Set ∆ = min {∆i , ci

k(i) - vi }.If ∆ = ∆i then set I‘ = I‘ \ { i }, else set k(i) := k(i) + 1.Set vi := vi + ∆ and for all j ∈ Ji set sj := sj – ∆.Update Ji.

3. StopIf I‘ =∅, i.e. no vi can be increased anymore, Stop. Else Step 2



ExampleLet the transportationcosts and the costs for new facilities be and

I.e. 6 customers and 5 potential locations.

1. InitializationSort the cost elements: e.g. for i = 1

c11 = 0 ≤ c1

2 = 3 ≤ c13 = 4 ≤ c1

4 = 6 ≤ c15 = 12 ≤ ci

6 = ∞, and so on

I‘ := {1,…, 6} and v1 = v2 = v4 = v5 = v6 = 0, v3 = 4.

J1 = {1}, J2 = {5}, J3 = {1}, J4 = {2}, J5 = {3}, J6 = {4}.

s1 = 12, s2 = 6, s3 = 9, s4 = 9, s5 = 12.

k(1) = k(2) = k(3) = k(4) = k(5) = k(6) = 2.

Collect all these data in a table.

Dual Ascent Algorithm



For the sake of simplicity the sortingof the cost elements is not explicitlycarried out in the table.

The cost elements cik(i) are

marked by underlines.

2. Iteration: Start with i = 1∆1 = min {sj | j ∈ J1 } = s1 = 12.∆ = min {∆1 , c1

2 - v1 = 3 - 0} = 3.

Since ∆ = c12 – v1

set k(1) := k(1) + 1 = 2 + 1 = 3.

Set v1 := v1 + ∆ = 0 + 3 = 3 and for all j ∈ J1 = {1} set sj := sj – ∆, i.e. s1 – ∆ = 12 – 3 = 9.

Update J1: J1 = {1, 5}.

Dual Ascent Algorithm: Example



After i = 2 After i = 3,…,6


∆2 = min {sj | j ∈ J2 } = s5 = 12.∆ = min {∆i , c2

2 – v2 = 2 - 0} = 2.Since ∆ = c2

2 – v2 set k(2) := k(2) + 1 = 3.Set v2 := v2 + ∆ = 2 and for all j ∈ J2 = {5}

set sj := sj – ∆, i.e. s5 – ∆ = 12 – 2 = 10. Update J2: J2 = {1, 5}.



i = 2:∆2 = min {sj | j ∈ J2 } = s5 = 12.∆ = min {∆i , c2

2 – v2 = 2 - 0} = 2.Since ∆ = c2

2 – v2 set k(2) := k(2) + 1 = 3.Set v2 := v2 + ∆ = 2 and for all j ∈ J2 = {5} set

sj := sj – ∆, i.e. s5 – ∆ = 12 – 2 = 10. Update J2: J2 = {1, 5}.

After i = 3,…,6




I‘ ≠ ∅ ⇒ start a new iteration. Table after i = 1,…,6

Remarks• v5 could not be increased to the third largest cost value 8 in row 5, because

otherwise s4 would become negative.

• For v6 even ∆6 = 0.⇒ Change: I‘ = {1,…,5}




I‘ ≠ ∅ ⇒ start a new iteration. Table after i = 1,…,5

Merely v2 could be increased to the fourth largest cost value 4.Change: I‘ = {2 }

I‘ ≠ ∅ ⇒ start a new iteration. However, v2 can not be increased anymore (s2 = 0) ⇒ stop

Solution: v = (4,4,9,6,5,4) with RD(v) = 32.








Derive a primal solution from the Dual oneupper bound for the WLP

lower bound = upper


Stopprimal

solution optimal


no


yes



Dualoc Algorithm

Construction Algorithm

Based on a feasible solution v of (RDP), construct an integer solution (x, y)of (WLP).For this purpose consider the complementary slackness conditions.

TheoremLet (x, y) and (v, w) be feasible solutions of the LP – relaxation (P) and the dual LP (DP), respectively. (x, y) and (v, w) are optimal if:



Construction AlgorithmRemark: The fourth constraint is always satisfied for each solution.

Set and

into the complementary slackness conditions in order to get the following conditions for optimality.

Optimality conditionsLet v be a feasible solution of (RDP). If an integer feasible solution (x,y) of (WLP) can be found, so that the following conditions for optimalityhold,

then both solutions are optimal.



Construction AlgorithmProof

(x, y) is also a feasible solution for the LP – Relaxation (P)⇒ (x, y) optimal for (P) (complementary slackness conditions hold)⇒ F(x, y) = F‘(x, y) (objective function value identical for (P) and

(WLP))⇒ (x, y) optimal for (WLP)

Remark: the optimality conditions are just sufficient conditions.

Determination of the best integer solution for which (3) – (5) holds, is a combinatorial problem.Use a simple heuristic.

IdeaSelect a feasible solution (x, y) of (WLP),for which the conditions (3) and (4) hold.If (5) also holds, then (x, y) is optimal for (WLP).



Construction AlgorithmFor a feasible solution of (WLP) one has to determine

1. locations y of new facilities and2. allocations x of customers to these facility locations.

From (3) follows: yj > 0 just for sj = 0.

TherewithDefine J+ = { j ∈ J | sj = 0 } and set yj := 0 for all j ∉ J+.

J+ is the set of candidates of potential locations for the inclusion into the solution.

Actual selection of locations out of the set J+ is done by the analysis of the permitted allocations.



Construction AlgorithmCustomer i can only be assigned to alocation j ∈ J+, if cij ≤ vi holds.

If for customer i exists• only one location j ∈ J+ with cij ≤ vi, then xij = 1 must hold and therewith yj = 1

(because every customer must be allocated to one location).

• several locations j ∈ J+ with cij ≤ vi, then only one of these has to be a part of the solution. If none of them is a part of the solution yet, then choose the one with lowest costs j*: xij* = 1 and yj* = 1.

If for the solution (x, y) constraint (5) holds, then it is optimal.Otherwise one can not decide whether this solution is optimal (i.e. the solution may or may not be optimal!).

Denote X = { j ∈ J | yj = 1 } as the set of all potential locations, selected by the heuristic.




Construction AlgorithmGiven: feasible solution v of (RDP).1. Initialization

Set X = ∅. Compute J+ = { j ∈ J | sj = 0 }.

2. Selection of locationsFor all i∈ I

If there is exactly one location j ∈ J+ for customer i with cij ≤ vi and j ∉ X,then set X = X ∪ { j } and yj = 1.

For all i∈ IIf there are several locations j ∈ J+ for customer i with cij ≤ vi, but none of these j ∈ X, then set X = X ∪ { j* } and yj* = 1 for the j* with cij* = min { cij | cij ≤ vi}.

3. AllocationAllocate every customer i to the location j ∈ X with lowest costsxij* = 1 for the j* with cij* = min { cij | cij ≤ vi, j ∈ X }.



ExampleFeasible dual solution v = (4, 4, 9, 6, 5, 4)of the Dual Ascent – Algorithm, withs = (1, 0, 3, 0, 6) and RD(v) = 32.

1. InitializationX = ∅ and J+ = { j ∈ J | sj = 0 } = {2, 4}.

2. Selection of locationsExactly one location j ∈ J+ with cij ≤ vi for customer i ?

i = 1: solely c12 ≤ v1 ⇒ X := {2},i = 2: solely c24 ≤ v2 ⇒ X := {2, 4} = J+.

One can already stop, since all permitted locations are already selected.⇒ y2 = y4 = 1 and y1 = y3 = y5 = 0.

3. Assignment x12 = x24 = x32 = x42 = x54 = x64 = 1. All other xij = 0.




Value of the solution: F(x, y) = 35 > 32 = RD(v).

Optimality constraint (5) is not satisfied, sincev4 > c42 and v4 > c44

and no more than one x4∗ is allowed to be positive. Here: x42 = 1 and x44 = 0.

⇒ no information about optimality of the solution (x, y).

Construction Algorithm : Example







Construct primal solution out of the Dualupper bound for the WLP

lowerbound = upper


Stopprimal

solution optimal

yes noModified dual

solution ?

no


yes



Dual Adjustment Algorithm

The dual ascent algorithm does not always finds the optimal dual solution.⇒ It is possible to improve the solution.

If the dual ascent algorithm and the construction heuristic cannot find a verifiable optimal solution for (WLP), the following heuristic can be used to adjust the dual solution in order to eliminate all violations of the complementary slackness conditions.

If 2 or more of these constraints vi > cij are violated for one customer i, then decrease the value of vi, to reduce the number of violations.

Through this decrease it should be possible to increase at least one other dual variable vi by the same amount.

Search for an alternative solution, which is, if possible, better than the previous one and involves fewer violations of condition (5).

Dualoc Algorithm



Let

be the set of locations where customer i has to be allocated, to satisfy condition (5), i.e. xij = 1 ! (since yj = 1 for all j ∈ X.)

Condition (5) is violated, if | Jix | > 1.

Reduce | Jix | by decreasing vi to the next smaller cij.

Thereby at least one violation is eliminated.But the objective function value RD(v) of the dual problem is also reduced.




Example (cont.)

It holds: v4 > c42 and v4 > c44

⇒ J4x = { 2, 4 }.

Decrease v4 = 6 to the next smaller value,that is c44 = 3: v4 = 3

⇒ J4x = { 2 }.

The slack of s2 and s4 increases each by 3.Now it is possible to increase e.g. v5 by 3.

In generalDecreasing vi increases the slack of all constraints j ∈ Ji

x.

Use this slack to increase other dual variables vk. But only, if they are not blocked by other constraints.




Because | Jix | > 1, there exist j*, j** ∈ Ji

x , with

cij* = min { cij | j ∈ Jix } and cij** = min { cij | j ∈ Ji

x, j ≠ j* }.

Decreasing of vi increases the slack of the constraints j* and j**.

LetIjx = { i ∈ I | j ∈ X is the only location with vi ≥ cij }, for all j ∈ X

be the set of customers, which solely can be allocated to location j.

If Ijx ≠ ∅ holds, the dual variables vi, i ∈ Ijx, cannot be blocked by other constraints.

Dual variables vk, k ∈ Ij*x, Ij**x, are „good“ candidates, but no guaranty for a better dual solution.




Now determine an index i with | Jix | > 1, decrease vi to the next smaller

value cij and adjust the slacks of the constraints j ∈ Jix.

Thereafter use the dual ascent algorithm 3 times to increase the value of other dual variables.

But with the following changes:• use in the first step as set I‘ ⊆ I of customer indices i for which vi can

still be increasedI‘ := Ij*x ∪ Ij**x

• at the second time extend I‘ by i: I‘ := I‘ ∪ { i }• and at the last time to total I: I‘ := I.

If one gets another (better) dual solution v, use the construction heuristic to construct a new integer solution (x, y) for (WLP).





Dual Adjustment AlgorithmGiven: feasible solution v of (RDP) and (x, y) of (WLP).

IterationForall i∈ I

If | Jix | ≤ 1 holds or Ij*x ∪ Ij**x = ∅, then continue with the next i.

Determine the next smaller cij: ∆ := max { cij | cij < vi, j ∈ J }.Set sj := sj + (vi – ∆) for all j ∈ Ji

x .Set vi := ∆.

Perform the dual ascent – algorithm successively with- I‘ := Ij*x ∪ Ij**x,- I‘ := I‘ ∪ { i } and- I‘ := I



Example (cont.)Feasible dual solution v = (4, 4, 9, 6, 5, 4)with s = (1, 0, 3, 0, 6) and RD(v) = 32.

Feasible solution (x, y) of (WLP):• y2 = y4 = 1 and y1 = y3 = y5 = 0• x12 = x24 = x32 = x42 = x54 = x64 = 1.

All other xij = 0.

X = { 2, 4 }, F(x, y) = 35.

IterationFor i = 1,…, 3 it holds | Ji

x | = 1.

i = 4: | J4x | = 2 (J4

x = { 2, 4 }) and I2x ∪ I4x = { 1, 3 } ∪ { 2, 5, 6 } ≠ ∅.Next smaller c4j:

∆ := max { c4j | c4j < v4, j ∈ J } = c44 = 3.s2 := s2 + (v4 – ∆) = 0 + 6 – 3 = 3, s4 := 3, v4 := 3.


Ijx = { i ∈ I | j ∈ X is the onlylocation with vi ≥ cij }



Perform the dual ascent algorithmwith I‘ = { 1, 2, 3, 5, 6 } based on the following table:

One gets:

With a new dual solution v = (5, 4, 9, 3, 8, 4) and improved objective function value 33.Further steps yield no improvement.

Dual Adjustment Algorithm: Example



Perform the construction algorithm.

1. InitializationX = ∅ and J+ = { j ∈ J | sj = 0 } = {1, 3, 4}.

2. Selection of locationsExactly one location j ∈ J+ with cij ≤ vi for customer i ?

i = 1: solely c11 ≤ v1 ⇒ X := {1},i = 3: solely c31 ≤ v3, 1 already included in X. i = 4: solely c44 ≤ v4 ⇒ X := {1, 4}.

If there are several locations j ∈ J+ with cij ≤ vi for customer i, but none of thesej ∈ X, then set X = X ∪ { j* } and yj* = 1 for the j* for which holdscij* = min { cij | cij ≤ vi }.

The solution already consists of a location for each customers i.⇒ y1 = y4 = 1 and y2 = y3 = y5 = 0.

3. Assignmentx11 = x21 = x31 = x44 = x54 = x64 = 1. All other xij = 0.




Condition (5) holds⇒ (x, y) is optimal.

It also holds, thatthe objective function value F(x, y) of the solution (x, y) for the (WLP) and of the dual LP are equal.

F(x, y) = 33 = RD(v).

⇒ (x, y) is optimal.








Construct primal solution out of the Dualupper bound for the WLP

lower bound = upper


Stopprimal

solution optimal


no


yes



Branch & Bound Algorithm

If it is not possible to find a verifiable optimal solution of the (WLP), then the branch & bound algorithm can be used to• try to prove the optimality of the best found integer solution or• determine an other optimal solution.

Branching is done by fixing a binary variable yj to 0 or 1 respectively. It can be implicitly achieved by changing the fixed costs of the facility that has to be located:

• yj = 1: Set fixed costs fj := 0the location will be included in the solution.

• yj = 0: Set fixed costs fj := M (M a very large number)the location will not be in the solution.

Solve the subproblems in the nodes again by using a combination of the dual ascent, construction algorithm as well as dual adjustment.

Dualoc Algorithm



Location Theory


Contents

• Introduction







Extensions of the WLPClassical extensions of the Warehouse Location Problem:

• Capacitated WLP

• Closing of Facilities

• Budget restrictions

• Different facility types

• Multi-Product and Multi-Activity Models

• Multi-Echelon Models

• Multi-Period (dynamic) Models



Extensions of the WLP

The Capacitated WLP

Also known as Capacitated Facility Location Problem (CFLP).

Capacity restrictions for facilities limit the quantities, which can be• produced• transported• handled, etc..

The „single assignment“ property no longer holds for capacitated WLP:

In a solution of the problem, the demand of a customer is not necessarily completely satisfied by a single facility.

That is a customer can simultaneously be assigned pro-rata to several new facilities.



Extensions of the WLPRedefine the assignment variable xij as:

xij denotes the is a fraction of demand of customer i, which is satisfied by a new

facility at location j (i.e. transported from j to i).

Therewithbi xij denotes the amount of the product, which is delivered from a new facility at

location j to customer i.



Extensions of the WLPMaximal Capacity

Maximal capacity Kjmax of a new facility at the potential location j ∈ J.

Extension

The amount of goods, which is delivered from a new facility at location j to all customers, has to be smaller than the capacity of the facility.

Implications- Amount of goods equal to zero, if no facility was built.

- If goods are delivered, then the facility has to be located.



The Capacitated WLPMinimal Capacity Utilization

Often there is a minimal capacity utilization Kjmin of facilities required, e.g.,

because of profitability.

Extension

Implication- If a facility is located, then the outgoing amount of goods has to exceed a

minimal capacity utilization



The Capacitated WLPSingle-Sourcing

Often a unique allocation (called Single-Sourcing) of customers to facilities is required.

Causes for Single-Sourcing- closer customer loyalty, easier logistics- better traceability in case of deficiency

Model extension

Demand of a customer i is either completely satisfied by a new facility at location j, or not at all.



The Capacitated WLPMathematical model

This problem is much more difficult to solve than the WLP.In particular with Single-Sourcing.




Closing of facilities

Consideration of facilities, which already exist at the beginning of the planning period and may be closed.

DefineJo = Set of potential locations where a new facility can be built.

Jc = Set of locations where a facility is already in use, but which may be closed.

foj and fc

j denote costs for opening and closing, respectively, of facilities at location j ∈ Jo or Jc, respectively.

Costs for closing fcj may be positive or negative

Positive: costs for demolition, disposal, severance payments for staffNegative: sales profit



Closing of facilitiesLocation decisions

Opening

Closing

Extensions of the modelCosts for opening and closing

Supply is just possible from open facilities



Closing of facilities

Mathematical model




Budget restrictions

Only variable costs are to be minimized.

Fixed costs fj of facilities (e.g. for construction, infrastructure) are financed by limited budget B.

However, consider variable costs gj for operating the facilities.

Model extensionOperating costs for the facilities and the transportation of goods

Budget constraint



Budget restrictionsMathematical model




Different facility types

For the construction of a new facility at a location one can choose from several different building types.

Example• regional or central warehouses• production plants for temporary fabrication or finishing

In case of capacity facilities also differentiate between various sizes of facilities of the same type.Set K of possible facility types.

Location decision: binary variable yjk, j ∈ J, k ∈ K




Fixed costsFixed costs fjk for the opening of a facility from type k at location j.Model extension

Maximum capacity Max. capacity Kjk of a facility of type k at location j.Model extension

Amount of goods bounded by the capacity of type k of the new facility at locationj.



Different facility typesBuilding restriction

Model extension

At most one type of facility can be placed at location j.

Forbid the opening of a facility of type k ∈ K at location j ∈ J by fixing the corresponding decision variable in the model to zero: yjk = 0.

E.g. because infrastructure insufficient, needed personnel not on the spot,…




Mathematical model



Extensions of the WLPMulti-Product and Multi-Activity Models

Customers have a demand for different products, services or activities with various characteristics.Necessary, if product aggregation not reasonable.

E.g. due to • The case that transportation or inventory requirements are too different

(pallets refrigerating store house)• Completely different service levels (first aid emergency doctor)

Availability• All products / activities are available at each location without restrictions • Not all locations offer the complete range

Costs• Production / storage of a product or offering a service possibly generate

additional product-dependent fixed costs.



Multi-Product and Multi-Activity ModelsSet P of products.

Definexijp = Rate of the demand for product p of customer i, which is satisfied by a

new facility at location j.

CostsTransportation costs cijp and customer demands bip are product-dependent, p ∈ P.

Model extensionCosts for satisfying the customer demands

Customer demands for all products satisfied



Multi-Product and Multi-Activity ModelsProduct availability

Binary variable zjp, j ∈ J, p ∈ P

Fixed costs gjp for offering product p in a new facility at location j.

Model extensionCosts for making services and goods available



Multi-Product and Multi-Activity ModelsCustomers can only be served by facilities which offer the product

Locations can only offer products, if a facility has been located

Forbid that a product p ∈ P is not available at location j ∈ J by fixing the corresponding decision variable in the model to zero:

zjp = 0.

Reasons- infrastructure insufficient- economically not reasonable- no qualified staff on the spot



Multi-Product and Multi-Activity ModelsSingle-Sourcing

Total demand of a customer is satisfied from the same location.E.g. for better in case of deficiencies, closer customer loyalty, …

Auxiliary variables for Single-Sourcing

Model extensionIf customer i is (pro-rata) assigned to facility j, then he obtains products from

there

Customers can obtain products only from exactly one location



Multi-Product and Multi-Activity Models

Mathematical model




Multi-Level Models

(Production) distribution network often includes several stages of facilities each of the same or similar type.

Example: distribution of goods over three transport stages

Levels classically are arranged hierarchically, i.e. goods always flow just from one stage to the next one.

But also transportation links within levels or beyond levels possible.

factories centralstore customersregional

store



Multi-Stage ModelsLocation decisions on

• One stage possible (e.g. warehouses)• Several stages possible (e.g. warehouses and factories)

Define for each stage a corresponding set of location variables.

Two-stage problem with location decisions on one stage

first stage: customers i ∈ Isecond stage: potential locations j ∈ J for warehousesthird stage: factories k ∈ K

Transport variables and transportation costs factories warehousezjk = Amount of the product, which will be transported from factory k to warehouse jtjk = Transportation costs per quantity unit for factory k warehouse j



Multi-Stage ModelsExtensions

Transportation costs from factory to warehouse

Quantity of goods from factory to warehouse = transportation volume from factory to customer

Flow conservation:“What goes in, must come out”

zjk bi xij

j



Multi-Stage ModelsMathematical model




Multi-Period (dynamic) Models

Planning decisions about the opening (or closing) of facilities and their consequences often range over several years.Therefore it is important to know “where" but in particular also “when”facilities are to be opened.

Divide the planning period into several periods (e.g. months, years).

year t year t + 1factories customerswarehouse factories customerswarehouse



Multi-Period (Dynamic) ModelsApplications

Long-term constant increase in demand (of distribution).

When is the best moment for the opening of the new factory or inventory?

Stepwise reorganization of the (production) distribution network.

Do not „move“ all warehouses at once, but during a certain time period.

Consideration of high demand variabilitiySeasonal rental of facilities, if demand peaks exceed the capacity.



Multi-Period (Dynamic) ModelsVarious problem issues

Alternative 1If a facility is opened it remains in use for the rest of the planning period.Typically for facilities with high capital expenditure.

Alternative 2A facility remains open only for the duration of the period.E.g. when storage areas or production capacities were only rented.



Multi-Period (Dynamic) ModelsSet T of time periods.Transportation costs cijt and customer demands bit are time-dependent,t ∈ T.

Definexijt = Part of the demand from customer i, which is satisfied in period t from a

new facility at location j.

Model extensionCosts for satisfying the customer demands for the whole planning period

Customer demand satisfied in each period



Multi-Period (Dynamic) ModelsAlternative 1

Location decision: Binary variable yjt, j ∈ J, t ∈ T

Fixed costs fjt and operating costs gjt for the opening and the operation, respectively, of a facility at location j in period t.

Set

Costs for construction and operation of a facility at location j in period t for the rest of the planning period.



Multi-Period (Dynamic) ModelsExtensions

Costs for building a facility

Satisfaction of customer demands only from an already opened facility

Only open a facility at most once




Location decision: Binary variable yjt, j ∈ J, t ∈ T

Operating costs gjt of a facility at location j during period t (e.g. lease, personnel costs) .No fixed costs.

ExtensionsOperating costs

Just „active“ facilities can satisfy customer demands



Multi-Period (Dynamic) Models

Mathematical models

Alternative 1

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