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Facilities Design. S.S. Heragu Industrial Engineering Department University of Louisville. Chapter 12: Advanced Location Models. Chapter 12: Advanced Location Models. 12.1INTRODUCTION 12.2LOCATION MODELS 12.2.1Multiple-Facility Problems with Rectilinear Distances

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Facilities design

Facilities Design

S.S. Heragu

Industrial Engineering Department

University of Louisville


Chapter 12 advanced location models

Chapter 12:Advanced Location Models


Chapter 12 advanced location models1

Chapter 12: Advanced Location Models

12.1INTRODUCTION

12.2LOCATION MODELS

12.2.1Multiple-Facility Problems with

Rectilinear Distances

12.2.2Multiple-Facility Problems with

Euclidean Distances

12.3ALLOCATION MODELS

12.3.1Network Flow Model

12.3.2Two-Stage Transportation Model

12.3.3Vehicle Routing Problem


Chapter 12 advanced location models2

Chapter 12: Advanced Location Models

12.4LOCATION-ALLOCATION MODELS

12.4.1Set Covering Model

12.4.2Uncapacitated Location-Allocation Model

12.4.3Comprehensive Location-Allocation Model

12.5SUMMARY


Introduction

Introduction

  • How many new facilities are to be located in the distribution network consisting of previously established facilities and customers?

  • Where should they be located?

  • How large should each new facility be? In other words, what is the capacity of the new facility?


Introduction1

Introduction

  • How should customers be assigned to the new and existing facilities? More specifically, which facilities should be serving each customer?

  • Can more than one facility serve a customer?


12 2 location models

12.2LOCATIONMODELS


12 2 1 multiple facility problems with rectilinear distances

12.2.1Multiple-Facility Problemswith Rectilinear Distances


Model 1

Model 1:


Model 11

Model 1:


Model 12

Model 1:

A similar definition of y+ij, y-ij, xa+ij, xa-ij, yb+ij, and yb-ij yields

|yi - yj| = y+ij + y-ij

yi - yj = y+ij - y-ij

|xi - aj| = xa+ij + xa-ij

xi - aj = xa+ij - xa-ij

|yi - bj| = yb+ij + yb-ij

yi - bj = yb+ij - yb-ij


Model 13

Model 1:

Thus, the transformed linear model is:


Model 14

Model 1:

(xi - xj) = x+ij - x-ij

Subject to:

yi - yj = y+ij - y-ij

xi - aj = xa+ij - xa-ij

yi - bj = yb+ij - yb-ij

x+ij, x-ij, y+ij , y-ij> 0, i, j = 1, 2, ...,n

xa+ij, xa-ij, yb+ij , yb-ij> 0, i = 1, 2, ...,n, j =1,2,...,m

xi, yi unrestricted in sign, i = 1,2,...,n


Model 15

Model 1:

  • xij+ or xij-, but not both, can be greater than 0. (If both are, then the values of xij+ and xij- do not satisfy their definition in (2) and (3)). Similarly, only one of (i) yij+, yij-, (ii) xij+, xaij-, (iii) ybij+, ybij-, must be greater than 0.

  • Model 1 can be simplified by noting that:

  • xi can be substituted as ai + xij+ - xaij-

  • yi may also be substituted similarly, resulting in a model with 2n fewer constraints and variables than model 1.


Example 1

Example 1:

  • Tires and Brakes, Inc., is an automobile service company specializing in tire and brake replacement. It has four service centers in a metropolitan area and a warehouse that supplies tires, brakes and other components to the service centers. The company manager has determined that he needs to add two more warehouses so as to improve component delivery service.


Facilities design

Example 1:

At the time, he wants to ensure that the location of the two new warehouses is such that the cost of delivery components from the new warehouse to the existing facilities (four service centers and existing warehouse) as well as between the new warehouses is minimal. The four service centers and warehouse are located at the following coordinate locations - (8, 20), (8, 10), (10, 20), (16, 30), and (35, 20). It is anticipated that there will be one trip per day between the new warehouses.


Facilities design

Example 1:

In addition, the number of trips between the new warehouses and four service centers as well as the existing warehouse is provided below.

SC1SC2SC3SC4W1

W277542

W332452

Develop a mode similar to the transformed Model 1 to minimize distribution cost and solve it using LINGO, LINDO or the LP Solver in Excel


Facilities design

SUBJECT TO

2) - XP12 + XN12 + X1 - X2 = 0

3) - XP21 + XN21 - X1 + X2 = 0

4) - YP12 + YN12 + Y1 - Y2 = 0

5) - YP21 + YN21 - Y1 + Y2 = 0

6) - XAP11 + XAN11 + X1 = 8

7) - XAP12 + XAN12 + X1 = 8

8) - XAP13 + XAN13 + X1 = 10

9) - XAP14 + XAN14 + X1 = 16

10) - XAP15 + XAN15 + X1 = 35

11) - XAP21 + XAN21 + X2 = 8

12) - XAP22 + XAN22 + X2 = 8

13) - XAP23 + XAN23 + X2 = 10

14) - XAP24 + XAN24 + X2 = 16

15) - XAP25 + XAN25 + X2 = 35

16) - YBP11 + YBN11 + Y1 = 20

17) - YBP12 + YBN12 + Y1 = 10

18) - YBP13 + YBN13 + Y1 = 20

19) - YBP14 + YBN14 + Y1 = 30

20) - YBP15 + YBN15 + Y1 = 20

21) - YBP21 + YBN21 + Y2 = 20

22) - YBP22 + YBN22 + Y2 = 10

23) - YBP23 + YBN23 + Y2 = 20

24) - YBP24 + YBN24 + Y2 = 30

25) - YBP25 + YBN25 + Y2 = 20

END

FREE X1

FREE X2

FREE Y1

FREE Y2

Example 1:

MIN XP12 + XN12 + YP12 + YN12 + XP21 + XN21 + YP21 + YN21 + 7 XAP11 + 7 XAN11 + 7 YBP11 + 7 YBN11 + 7 XAP12 + 7 XAN12 + 7 YBP12 + 7 YBN12 + 5 XAP13 + 5 XAN13 + 5 YBP13 + 5 YBN13 + 4 XAP14 + 4 XAN14 + 4 YBP14 + 4 YBN14 + 2 XAP15 + 2 XAN15 + 2 YBP15 + 2 YBN15 + 3 XAP21 + 3 XAN21 + 3 YBP21 + 3 YBN21 + 2 XAP22 + 2 XAN22 + 2 YBP22 + 2 YBN22 + 4 XAP23 + 4 XAN23 + 4 YBP23 + 4 YBN23 + 5 XAP24 + 5 XAN24 + 5 YBP24 + 5 YBN24 + 2 XAP25 + 2 XAN25 + 2 YBP25 + 2 YBN25


12 2 2 multiple facility problems with euclidean distances

12.2.2Multiple-Facility Problemswith Euclidean Distances

Consider the following objective for the euclidean distance problem.


Facilities design

Multiple-Facility Problemswith Euclidean Distances

Taking the partial derivatives, we get


Facilities design

Multiple-Facility Problemswith Euclidean Distances


Facilities design

Multiple-Facility Problemswith Euclidean Distances

To make sure the denominator is never 0, we add  to it. We then get:


Facilities design

Multiple-Facility Problemswith Euclidean Distances

And …


Example 2

Example 2

Consider Example 1. Assuming the Euclidean metric is more appropriate and that Tire and Brakes, Inc. does not currently have a warehouse, determine where the two new warehouses are to be located.


12 3 allocation model

12.3ALLOCATIONMODEL


12 3 1 network flow model

12.3.1 Network Flow Model


Network flow model

Network Flow Model:

Model 2

Consider this notation:

cijcost of sending one unit of flow on arc (i, j)

Uijupper bound on the flow that can be sent on arc (i, j), i.e., capacity of arc (i,j)

Lijlower bound on the flow that can be sent on arc (i, j)

Dinet flow generated at node i

xijnumber of units of flow on arc (i, j)


Facilities design

Model 2:


Network simplex algorithm

Network Simplex Algorithm:

Step 1:Construct a spanning tree for the n nodes. The variables xij corresponding to the arcs (i,j) in the spanning tree are basic variables and the remaining are nonbasic. Find a basic feasible solution to the problem so that:

(1) the basic variables satisfy Lij < xij < Uij , and

(2) the nonbasic variables take on a value of Lij or Uij to satisfy constraint (22).

Step 2:Set u1 =0 and find uj, j=2,...,n using the formula ui - uj = cij for all basic variables.

Step 3:If ui - uj - cij< 0 for all nonbasic variables xij with a value of Lij , and ui - uj - cij> 0 for all nonbasic variables xij with a value of Uij , then the current basic feasible solution is optimal; stop. Otherwise, go to step 4.


Network simplex algorithm1

Network Simplex Algorithm:

Step 4:Select the variable xi*j* that violates the optimality condition (in step 3) the most, i.e., the largest of the ui - uj - cij for those nonbasic variables with xij = Lij, and the smallest of the ui - uj - cij for those nonbasic variables with xij = Uij,. Make the arc (i*,j*) a basic variable and add arc (i*,j*) to the spanning tree. Make one of the other basic variables in the loop of basic variables [formed by including arc (i*,j*)], a nonbasic variable such that:

(1) xi*j* takes on the largest possible value,

(2) constraint (21) is satisfied for all the n nodes, and

(3) constraint (22) is satisfied for all the arcs in the loop.

Remove the arc corresponding to the nonbasic variable just identified so that we have a spanning tree once again. Go to step 2.


Example 3

Example 3

The Fast Shipping Company manages the distribution of lawnmowers from a company that has two factories (F1 and F2) in the Northeast to two large customer bases (C1 and C2) in the Southwest. For cost and freight consolidation reasons, Fast Shipping would like to route the shipments via three intermediate nodes (T1 - T3) located in the midwest. The relevant data is provide in Tables 12.3-12.5. Setup a model to determine how the shipment is to take place from the two factories to the two destinations via the three intermediate shipment points.


Supply and demand

Supply and Demand


Inbound and outbound transportation costs and arc capacities

Inbound and Outbound Transportation Costs and Arc Capacities


Min 8 x11 11 x12 5 x13 12 x21 8 x22 5 x23 6 y11 12 y21 9 y31 3 y12 y22 19 y32

SUBJECT TO

2) X11 + X12 + X13 = 900

3) X21 + X22 + X23 = 600

4) - Y11 - Y21 - Y31 = - 750

5) - Y12 - Y22 - Y32 = - 750

6) X11 + X21 - Y11 - Y12 = 0

7) X12 + X22 - Y21 - Y22 = 0

8) X13 + X23 - Y31 - Y32 = 0

9) X11 <= 500

10) X12 <= 1500

11) X13 <= 350

12) X21 <= 1200

13) X22 <= 750

14) X23 <= 450

15) Y11 <= 1000

16) Y21 <= 750

17) Y31 <= 1000

18) Y12 <= 150

19) Y22 <= 200

20) Y32 <= 1500

END

Example 3:

MIN 8 X11 + 11 X12 + 5 X13 + 12 X21 + 8 X22 + 5 X23 + 6 Y11 + 12 Y21 + 9 Y31 + 3 Y12 + Y22 + 19 Y32


12 3 2 two stage transportation model

12.3.2 Two-Stage Transportation Model

1

1

1

2

2

3

2

3

4

q

p

r


Two stage transportation model

Two-Stage Transportation Model:

Consider the following notation:

Sicapacity of supply source i, i = 1, 2, ..., p

Pjcapacity of plant j, j = 1, 2, ..., q

Dkdemand at customer k, k = 1, 2, ..., r

cijcost of transporting one unit from supply source i to plant j

The LP model is:


Facilities design

Model 3:


Four cases arise

Four cases arise:

  • Supply source capacity is unlimited and total plant capacity is more than total demand

  • Supply source capacity is unlimited and total demand exceeds total plant capacity

  • Plant capacity is unlimited and total supply source capacity exceeds total demand

  • Plant capacity is unlimited and total demand exceeds total supply source capacity

    In the following discussion, the supply sources are assumed to have unlimited capacities and total plant capacity is more than total demand (case (i))


Case 1

Case 1:


Example 4

Example 4:

2-Stage Distribution Problem: RIFIN Company has recently developed a new method of manufacturing a type of chemical. It involves refining a certain raw material which can be obtained from four overseas suppliers A, B, C, D who have access to the four ports at Vancouver, Boston, Miami, and San Francisco, respectively. RIFIN wants to determine the location of plants at which the chemical will be refined. The chemical, once refined, will be transported via trucks to five outlets located at Dallas, Phoenix, Portland, Montreal and Orlando.


Example 41

Example 4:

After an initial study, the choice of location for RIFIN’s refineries has narrowed down to Denver, Atlanta and Pittsburgh. Assume that one unit of raw material is required to make one unit of chemical. The amount of raw material from each potential refinery as well as the cost of trucking the chemical to outlets are also provided below. Determine the location of RIFIN’s refining plants, capacities at these plants and distribution pattern for the raw material and processed chemical.


Facilities design

Example 4:

Raw Material SourceSupplyOutletDemand

A1000Dallas900

B800Phoenix800

C800Portland600

D700Montreal500

Orlando500


Facilities design

Example 4:

Raw Material Transportation Cost

TODenverAtlantaPittsburgh

FROM

Vancouver4139

Boston885

Miami1229

San Francisco111112


Facilities design

Example 4:

Chemical Trucking Cost

TODallasPhoenixPortlandMontrealOrlando

FROM

Den.2826123030

Atla.102223298

Pitts.1821231821


Figure 12 3 pictorial representation of rifin example

Figure 12.3: Pictorial representation of RIFIN Example

Da

V

D

Ph

B

Po

A

M

Mo

P

SF

Orl


Facilities design

Setup transportation tableau for Example 4


Solution

Solution:

The transportation problem may be solved to yield the solution indicated in the following figure. Notice that the solution indicates that refineries be built at all locations.


Facilities design

Figure 12.4:

Da

V

900

1000

400

D

Ph

600

B

400

1000

Po

A

800

M

500

Mo

600

P

500

SF

100

Orl


12 3 3 vehicle routing problem

12.3.3 Vehicle Routing Problem

Determine the number of vehicles required to:

  • serve its customers (pick-up or deliver parcels) so that each customer is visited once and only once per day

  • the vehicle capacity is not exceeded, and

  • the total travel time is minimized


Vehicle routing problem

Vehicle Routing Problem:

Tijtime to travel from customer i to customer j, i,j=1,2, …, n

Didemand at customer i, i=1,2, …, n

Ckcapacity of vehicle k, k=1,2, …, p


Facilities design

Model 4:


12 4 location allocation models

12.4LOCATION-ALLOCATIONMODELS


12 4 1 set covering model

12.4.1Set Covering Model


Set covering model

Set Covering Model

Define:

cjcost of locating facility at site j

aij=

xj =

{

1 if facility located at site j can cover customer i

0 Otherwise

{

1 if facility is located at site j

The set covering problem is to:

0 Otherwise


Facilities design

Model 5:


Greedy heuristic for set covering problem

Greedy Heuristic for Set Covering Problem:

Step 1:If cj = 0, for any j = 1, 2, ..., n, set xj = 1 and remove all constraints in which xj appears with a coefficient of +1.

Step 2:If cj > 0, for any j = 1, 2, ..., n and xj does not appear with +1 coefficient in any of the remaining constraints, set xj = 0.

Step 3:For each of the remaining variables, determine cj/dj, where dj is the number of constraints in which xj appears with +1 coefficient. Select the variable k for which ck/dk is minimum, set xk = 1 and remove all constraints in which xj appears with +1 coefficient. Examine the resulting model.


Greedy heuristic for set covering problem1

Greedy Heuristic for Set Covering Problem:

Step 4If there are no more constraints, set all the remaining variables to 0 and stop. Otherwise go to step 1.

We illustrate the above greedy heuristic with an example.


Example 5

Example 5:

A rural country administration wants to locate several medical emergency response units so that it can respond to calls within the county within eight minutes of the call. The county is divided into seven population zones. The distance between the centers of each pair of zones is known and is given in the matrix below.


Figure 12 5

Figure 12.5:

1234567

10412615108

2801560723

3501308659

4911809103

55084100227

6305793027

7859725270

[dij] =


Example 51

Example 5:

The response units can be located in the center of population zones 1 through 7 at a cost (in hundreds of thousands of dollars) of 100, 80, 120 110, 90, 90, and 110 respectively. Assuming the average travel speed during an emergency to be 60 miles per hour, formulate an appropriate set covering model to determine where the units are to be located and how the population zones are to be covered and solve the model using the greedy heuristic.


Solution1

Solution:

Defining

{

1if zone i’s center can be reached from center of zone j within 8 minutes

aij =

0otherwise

and noting that dij > 8, dij< 8 would yield aij values of 0, 1, respectively the following [aij] matrix can be set up.


Solution2

Solution:

1234567

11101001

21100111

30011110

40011001

50110110

60110110

71101001

[aij] =

The corresponding set covering model is:


Facilities design

Solution:

Minimize 100x1+80x2+120x3+110x4+90x5+90x6+110x7

Subject to

x1+x2+ x4 + x7> 1

x1+x2+ x5 +x6+x7> 1

x3+x4 +x5 +x6> 1

x3+x4 +x7> 1

x2+ x3+x5 +x6 > 1

x2+ x3+x5 +x6 > 1

x1 +x2+ x4 +x7 > 1

x1,x2,x3,x4, x5,x6,x7> 0 or 1


Greedy heuristic

Greedy Heuristic

Step 1: Since each cj > 0, j = 1, 2, ..., 7, go to step 2.

Step 2: Since xj appears in each constraint with +1 coefficient, go to step 3.

Step 3:


Facilities design

Solution:

Since the minimum ck/dk occurs for k = 2, set x2 = 1 and remove the first two and the last three constraints. The resulting model is shown below.

Minimize 100x1+80x2+120x3+110x4+90x5+90x6+110x7

Subject to

x3+x4 +x5 +x6> 1

x3+x4 +x7> 1

x1,x2,x3,x4, x5,x6,x7 = 0 or 1


Greedy heuristic1

Greedy Heuristic:

Step 4: Since we have two constraints go to step 1.

Step 1: Since c1 > 0, j = 1, 3, 4, ..., 7, go to step 2

Step 2: Since c1 > 0 and x1 does not appear in any of the constraints with +1 coefficient, set x1 = 0.


Greedy heuristic2

Greedy Heuristic

Since the minimum ck/dk occurs for k = 4, set x4 = 1 and remove both constraints in the above model since x4 has a +1 coefficient in each. The resulting model is shown below.

Minimize:

120x3+90x5+90x6+110x7

Subject to

x3 , x5 , x6 , x7 > 0


Facilities design

Greedy Heuristic:

Step 4:Since there are no constraints in the above model, set x3 = x5 = x6 = x7 = 0 and stop.

The solution is x2 = x4 = 1; x1 = x3 = x5 = x6 = x7 = 0. Cost of locating emergency response units to meet the eight minute response service level is $800,000 + $1,100,000 = $1,900,000.


12 4 2 uncapacitated location allocation model

12.4.2Uncapacitated Location-Allocation Model


Uncapacitated location allocation model

Uncapacitated Location-Allocation Model

Parameters

m number of potential facilities

n number of customers

cij cost of transporting one unit of product

from facility i to customer j

Fi fixed cost of opening and operating facility j

Dj number of units demanded at customer j

Decision Variables

xijnumber of units shipped from facility i to customer j


Model 6

Model 6


Model 7

Model 7

Modify Model 6 by transforming xij variables and the cij parameter


Is model 7 equivalent to model 6

Is Model 7 equivalent to Model 6?

Substitute x’ij = xij/Dj, we get

Divide LHS and RHS by ΣDj, we get

Because the sum of LHS terms is <yi, each term must also be <yi

Because Dj/ΣDjis a positive fraction for each j:

x’ij< yi, j=1,2,…,n

Adding we get


On solving model 7

On solving Model 7

  • Although a general purpose branch-and-bound technique can be applied to solve model 5, it is not very efficient since we have to solve several subproblems, one at each node, using the Simplex algorithm. In what follows, we discuss a very efficient way of solving the subproblems that does not use the Simplex algorithm

  • To facilitate its discussion, it is convenient to refer to x’ij, the fraction of customer j’s demand met by facility i in model 7, as simply xij. Thus, xij in the remainder of this section does not refer to the number of units, rather a fraction. Similarly cij now refers to cij


On solving model 71

On solving Model 7

  • The central idea of the branch-and-bound algorithm is based on the following result

    • Suppose, at some stage of the branch-and-bound solution process, we are at a node where some facilities are closed (corresponding yi = 0), and some are open (yi = 1) and the remaining are free, i.e., a decision whether to open or close has not yet been taken (0< yi <1). Let us define:

      • S0 as the set of facilities whose yi value is equal to 0; {i: yi = 0}

      • S1 as the set of facilities whose yi value is equal to 1; {i: yi = 1)

      • S2 as the set of facilities whose yi value is greater than 0 but less than 1; {i: 0 < yi < 1}


Rewrite model 7 as model 8

Rewrite Model 7 as Model 8

Note: The inequality in the second constraint above can be converted to an equality because in the optimal solution LHS will be equal to RHS. Thus,

Because max {xij} is 1, max {yi} is also 1


Rewrite model 8 as model 9

Rewrite Model 8 as Model 9


On solution of model 9

On solution of Model 9

  • Model 9 which is equivalent to model 6 without the integer restrictions on the y variables, is a half assignment problem. It can be proved (again, by contradiction) that for each j = 1, 2, ..., n, only one of x1j, x2j, ..., xmj will take on a value of 1, due to Σxij=1

  • In fact, for each j, the xij taking on a value of 1 will be the one that has the smallest coefficient in the objective function

  • Thus, to solve model 9, we only need to find for a specific j, the smallest coefficient of xij in the objective function, i=1,2,…,m, set the corresponding xij equal to 1 and all other xij 's to 0 as follows:


On solution of model 91

On solution of Model 9

  • Select the smallest cij from the list, set the corresponding xij = 1 and all other xij’s to 0. This is the minimum coefficient rule.

  • We do not include facility iS2 in the min coeff rule because these are closed

  • Moreover, a lower bound on the partial solution of the node under consideration can be obtained by adding

    to the sum of the coefficients of the xij variables which have taken a value of 1


Branch and bound algorithm for basic location allocation model

Branch-and-BoundAlgorithm forBasicLocation-AllocationModel


Branch and bound

Branch-and-Bound

  • Step 1: Set best known upper bound UB = infinity; node counter, p = 1; S0 = S1 = {}; S2 = {1, 2, ..., m}

  • Step 2: Construct a subproblem (node) p with the current values of the y variables.

  • Step 3: Solve subproblem corresponding to the node under consideration using the minimum coefficient rule and


Branch and bound1

Branch-and-Bound

Step 4: If the solution is such that all y variables take on integer (0 or 1) values, go to step 7. Otherwise go to step 5.

Step 5: Determine the lower bound of node p using model 7. Arbitrarily select one of the facilities, say k, which has taken on a fractional value for yk, i.e., 0 < yk < 1 and create two subproblems (nodes) p+1 and p+2 as follows.


Branch and bound2

Branch-and-Bound

Subproblem p+1

  • Include facility k and others with a yk value of 0 in So; facilities with yk value of 1 in S1; all other facilities in S2

    Subproblem p+2

  • Include facility k and others with a ykvalue of 1 in S1; facilities with yk value of 0 in S0; all other facilities in S2. If xkj = 1for j = 1, 2, ..., n, in the solution to subproblem p, remove each such customer j from consideration in subproblem p+2, and reduce n by the number of j’s for which xkj = 1


Branch and bound3

Branch-and-Bound

  • Step 6: Solve subproblem p+1 using the minimum coefficient rule and

  • Set p = p+2. Go to step 4.

  • Step 7: Determine the lower bound of node p using model 9. If it is greater than UB, set UB = lower bound of node p. Prune node p as well as any other node whose bound is greater than or equal to UB. If there are no more nodes to be pruned, stop. Otherwise consider any unpruned node and go to step 3.


Example 6

Example 6

The nation’s leading retailer Sam-Mart wants to establish its presence in the Northeast by opening five department stores. In order to serve the stores (whose locations have already been determined), the retailer wants to have a maximum of three distribution warehouses. The potential locations for these warehouses have already been selected and there are no practical limits on the size of the warehouses. The fixed cost (in hundreds of thousands of dollars) of building and operating the warehouse at each location is 6, 5, and 3, respectively.


Example 61

Example 6

The variable cost of serving each warehouse from each of the potential warehouse locations is given below (again in hundreds of thousands of dollars). Determine how many warehouses are to be built and in what locations. Also determine how the customers (departmental stores) are to be served.


Example 62

Example 6

12345 Fi

1

2

3

2012141210 6

151020815 5

1216251110 3


Solution3

Solution:

Step 1: Set UB = infinity; node counter p = 1; S0 = S1 = {}; S2 = {1, 2, 3}.

Step 2: Minimum coefficient rule: Determine the xij coefficients as follows.


Facilities design

Solution:

6

1

5

3

3

c11=20+

=21

; c21=15+

=16

; c31=12+

= 12

5

5

5

5

5

Since the minimum occurs for cij = c31, set x31 = 1 and x11 = x21 = 0.

6

1

5

3

3

c12=12+

=13

; c22=10+

=11

; c32=16+

= 16

5

5

5

5

5

Since the minimum occurs for cij = c22, set x22 = 1 and x12 = x32 = 0.


Facilities design

Solution:

6

1

5

3

3

c13=14+

=15

; c23=20+

=21

; c32=25+

= 25

5

5

5

5

5

Since the minimum occurs for cij = c13, set x13 = 1 and x23 = x33 = 0.

6

1

5

3

3

c14=12+

=13

; c24= 8+

= 9

; c34=11+

= 11

5

5

5

5

5

Since the minimum occurs for cij = c24, set x24 = 1 and x14 = x34 = 0.


Facilities design

Solution:

6

1

5

3

3

c15=10+

=11

; c25=15+

=16

; c35=10+

= 10

5

5

5

5

5

Since the minimum occurs for cij = c35, set x35 = 1 and x15 = x25 = 0.


Facilities design

Solution:

1

1

1

y1=

[x11+x12+x13+x14+x15]=

[0+0+1+0+0]=

5

5

5

1

1

2

y2=

[x21+x22+x23+x24+x25]=

[0+1+0+1+0]=

5

5

5

1

1

2

y3=

[x31+x32+x33+x34+x35]=

[1+0+0+0+1]=

5

5

5


Solution4

Solution:

Step 4: Since all three y variables have fractional values, go to Step 5.


Facilities design

Solution:

Step 5: Lower bound of node 1 =

3

1

3

2

0 + 12

+ 11 + 15

+ 9 + 10

= 58

5

5

5

5

Arbitrarily select variable y1 to branch on.

Create subproblems 2 and 3 as follows

Subproblem 2: S0 = {1}; S1 = {}; S2 = {2, 3}.

Subproblem 3: S0 = {1}; S1 = {}; S2 = {2, 3}.


Solution5

Solution:

Step 6: Solution of subproblem 2 using minimum coefficient rule: Determine xij coefficients as follows.


Facilities design

Solution:

5

3

3

c21=15+

=16

; c31=12+

=12

5

5

5

Since the minimum occurs for cij = c31, set x31 = 1, x21 = 0.

5

3

3

c22=10+

=11

; c32=16+

=16

5

5

5

Since the minimum occurs for cij = c22, set x22 = 1, x32 = 0.


Facilities design

Solution:

5

3

3

c23=20+

=21

; c33=25+

=25

5

5

5

Since the minimum occurs for cij = c23, set x23 = 1, x33 = 0.

5

3

3

c24=8 +

= 9

; c34=11+

=11

5

5

5

Since the minimum occurs for cij = c24, set x24 = 1, x34 = 0.


Facilities design

Solution:

5

3

3

c24=15+

=16

; c35=10+

=10

5

5

5

Since the minimum occurs for cij = c35, set x35 = 1, x25 = 0.


Solution6

Solution:

1

1

3

y2=

[x21+x22+x23+x24+x25]=

[0+1+1+1+0]=

5

5

5

1

1

2

y3=

[x31+x32+x33+x34+x35]=

[1+0+0+0+1]=

5

5

5


Solution7

Solution:

Solution of subproblem 3 using minimum coefficient rule: Determine xij coefficients as follows:

Since x13 = 1 in the solution to subproblem 1, remove store 1 from consideration in node 3 and other nodes emanting from node 3. Reduce n by 1, n=5-1=4.


Facilities design

Solution:

5

1

3

3

c11=20; c21 =15+

=16

; c31=12+

=12

4

4

4

4

Since the minimum occurs for cij = c31, set x31 = 1 and x11 = x21 = 0.

5

1

3

3

c12=12; c22 =10+

=11

; c32=16+

=16

4

4

4

4

Since the minimum occurs for cij = c22, set x22 = 1 and x12 = x32 = 0.


Facilities design

Solution:

5

1

3

3

c14=12; c24 =8 +

= 9

; c34=11+

=11

4

4

4

4

Since the minimum occurs for cij = c24, set x24 = 1 and x24 = x34 = 0.

5

1

3

3

c15=10; c25 =15+

=16

; c35=10+

=10

4

4

4

4

Since the minimum occurs for cij = c15, set x15 = 1 and x25 = x35 = 0.


Facilities design

Solution:

1

1

2

1

y2=

[x21+x22+x24+x25]=

[0+1+1+0]=

=

4

4

4

2

1

1

1

y2=

[x31+x32+x34+x35]=

[1+0+0+0]=

4

4

4

Set p=1+2=3


Solution8

Solution:

Step 4: Since the solution for subproblem 2 is not all integer, go to step 5.

We repeat steps 3, 4, 5, 6, and 7 until all the nodes are pruned. We then have an optimal solution. These steps are summarized in Table 2.


Figure 12 8

Figure 12.8


12 5 3 comprehensive location allocation model

12.5.3Comprehensive Location-Allocation Model


Comprehensive location allocation model

Comprehensive Location-Allocation Model

Parameters

Sijproduction capacity of product i at plant j

Dildemand for product i at customer zone l

Fkfixed cost of operating warehouse k

Vikunit variable cost of handling product i at warehouse k

Cijklaverage unit cost of producing and transporting

product i from plant j via warehouse k to customer l

UCkUpper bound on the capacity of warehouse k

LCkLower bound on the capacity of warehouse k


Comprehensive location allocation model1

Comprehensive Location-Allocation Model

Decision Variables

Xijkl number of units of product i transported

from plant j via warehouse k to customer l


Model 10

Model 10


Model 101

Model 10


Comprehensive location allocation model2

Comprehensive Location-Allocation Model

We can easily add more linear constraints not involving xijkl variables to model 10 to:

  • Impose upper and lower limit on the number of warehouses that can be opened;

  • Enforce precedence relations among warehouses (e.g., open warehouse at location 1 only if another is opened at location 3)

  • Enforce service constraints (e.g., if it is decided to open a certain warehouse, then a specific customer area must be served by it)

  • Other constraints that can be added are discussed further in Geoffrion and Graves (1974).

  • Many of these constraints reduce the solution space, so they allow quicker solution of the model while giving the modeler much flexibility


Comprehensive location allocation model3

Comprehensive Location-Allocation Model

Suppose we fix the values of binary variables ykl and zk temporarily at 0 or 1 so that corresponding constraints are satisfied

Then, model 10 reduces to the following linear program which we will refer to as TP


Comprehensive location allocation model4

Comprehensive Location-Allocation Model

TP can be decomposed into i separate transportation problems, TPi, as follows because the variables pertaining to a specific product appear only in the rows (constraints) corresponding to that product and not elsewhere. (Notice that we have temporarily eliminated K from TPi)


Comprehensive location allocation model5

Comprehensive Location-Allocation Model

TP can be decomposed into i separate transportation problems, TPi, as follows because the variables pertaining to a specific product appear only in the rows (constraints) corresponding to that product and not elsewhere. (Notice that we have temporarily eliminated K from TPi)


Comprehensive location allocation model6

Comprehensive Location-Allocation Model

Dual of TPi, designated as DTPi, follows.


Comprehensive location allocation model7

Comprehensive Location-Allocation Model

Combine p dual problems into one master problem MP

MinimizeT


Modified benders decomposition algorithm for comprehensive location allocation model

Modified Benders’ Decomposition Algorithm for Comprehensive Location-Allocation Model

Step 0:Set upper bound UB=infinity, and convergence tolerance parameter ε to a desired small, positive value. Set ykl, zk = 0 or 1, for k=1,2,…,r, l=1,2,…,s so that the resulting values satisfy constraints with ykl, zk

Step 1:Set up TPi, i=1,2,…,p and determine K for the current values of ykl, zkk=1,2,…,r, l=1,2,…,s. Set up corresponding dual model DTPi for each i. Solve each DTPi and add K to the sum of the optimal objective function value of each DTPi. If this sum is less than or equal to UB, set UB = K + sum of original OFVs of each DTPi

Step 2:Set up model MP for the current values of uij, vil, i=1,2,…p; j=1,2,…,q, l=1,2,…,s. Find a feasible solution to MP such that T<UB-ε. If there is no such feasible solution to the current MP, stop. We have an ε-optimal solution. Otherwise, go to step 1 with the current values of the ykl, and zk variables


Example 7

Example 7

The nation’s leading grocer, Myers, wants to determine how to source the highest margin product, and also determine the warehouses through which to serve three of its largest stores in Louisville. In order to serve the stores (whose locations have already been determined), the grocer wants to utilize one or two distribution warehouses which will receive the product from one or more of four plants which produce the product. The potential locations for these warehouses have already been selected and there are no practical limits on the size of the warehouses. The fixed cost (in hundreds of thousands of dollars) of building and operating the warehouse at each location is 6, 5, and 3, respectively.


Example 71

Example 7

The variable cost of serving each warehouse from each of the potential warehouse locations is given below (again in hundreds of thousands of dollars). Determine how many warehouses are to be built and in what locations. Also determine how the customers (departmental stores) are to be served.


Example 72

Example 7


Example 7 mip

Example 7 - MIP

MODEL:

[_1] MIN= 1000 * Y_1_1 + 2000 * Y_1_2 + 1000 * Y_1_3 + 1500 * Y_2_1 +

3000 * Y_2_2 + 1500 * Y_2_3 + 25 * X_1_1_1 + 28 * X_1_1_2 + 23 * X_1_1_3

+ 16 * X_1_2_1 + 19 * X_1_2_2 + 20 * X_1_2_3 + 20 * X_2_1_1 + 23 *

X_2_1_2 + 18 * X_2_1_3 + 18 * X_2_2_1 + 21 * X_2_2_2 + 22 * X_2_2_3 + 13

* X_3_1_1 + 16 * X_3_1_2 + 11 * X_3_1_3 + 22 * X_3_2_1 + 25 * X_3_2_2 +

26 * X_3_2_3 + 15 * X_4_1_1 + 18 * X_4_1_2 + 13 * X_4_1_3 + 20 * X_4_2_1

+ 23 * X_4_2_2 + 24 * X_4_2_3 + 2000 * Z_1 + 1500 * Z_2 ;

[_2] X_1_1_1 + X_1_1_2 + X_1_1_3 + X_1_2_1 + X_1_2_2 + X_1_2_3 <= 200 ;

[_3] X_2_1_1 + X_2_1_2 + X_2_1_3 + X_2_2_1 + X_2_2_2 + X_2_2_3 <= 100 ;

[_4] X_3_1_1 + X_3_1_2 + X_3_1_3 + X_3_2_1 + X_3_2_2 + X_3_2_3 <= 50 ;

[_5] X_4_1_1 + X_4_1_2 + X_4_1_3 + X_4_2_1 + X_4_2_2 + X_4_2_3 <= 500 ;

[_6] - 100 * Y_1_1 + X_1_1_1 + X_2_1_1 + X_3_1_1 + X_4_1_1 >= 0 ;

[_7] - 200 * Y_1_2 + X_1_1_2 + X_2_1_2 + X_3_1_2 + X_4_1_2 >= 0 ;

[_8] - 100 * Y_1_3 + X_1_1_3 + X_2_1_3 + X_3_1_3 + X_4_1_3 >= 0 ;

[_9] - 100 * Y_2_1 + X_1_2_1 + X_2_2_1 + X_3_2_1 + X_4_2_1 >= 0 ;

[_10] - 200 * Y_2_2 + X_1_2_2 + X_2_2_2 + X_3_2_2 + X_4_2_2 >= 0 ;

[_11] - 100 * Y_2_3 + X_1_2_3 + X_2_2_3 + X_3_2_3 + X_4_2_3 >= 0 ;

[_12] Y_1_1 + Y_2_1 = 1 ;

[_13] Y_1_2 + Y_2_2 = 1 ;

[_14] Y_1_3 + Y_2_3 = 1 ;

[_15] 100 * Y_1_1 + 200 * Y_1_2 + 100 * Y_1_3 - 3000 * Z_1 <= 0 ;

[_16] 100 * Y_2_1 + 200 * Y_2_2 + 100 * Y_2_3 - 4000 * Z_2 <= 0 ;

[_17] 100 * Y_1_1 + 200 * Y_1_2 + 100 * Y_1_3 >= 0 ;

[_18] 100 * Y_2_1 + 200 * Y_2_2 + 100 * Y_2_3 >= 0 ;

@BIN( Y_1_1); @BIN( Y_1_2); @BIN( Y_1_3); @BIN( Y_2_1);

@BIN( Y_2_2); @BIN( Y_2_3); @BIN( Z_1); @BIN( Z_2);

END


Example 7 mip solution

Example 7 – MIP Solution

Global optimal solution found.

Objective value: 12300.00

Objective bound: 12300.00

Infeasibilities: 0.000000

Extended solver steps: 0

Total solver iterations: 13

Variable Value Reduced Cost

Q 4.000000 0.000000

R 2.000000 0.000000

S 3.000000 0.000000

CAPACITY( 1) 200.0000 0.000000

CAPACITY( 2) 100.0000 0.000000

CAPACITY( 3) 50.00000 0.000000

CAPACITY( 4) 500.0000 0.000000

FIXEDCOST( 1) 2000.000 0.000000

FIXEDCOST( 2) 1500.000 0.000000

VARIABLECOST( 1) 10.00000 0.000000

VARIABLECOST( 2) 15.00000 0.000000

UPPERBOUND( 1) 3000.000 0.000000

UPPERBOUND( 2) 4000.000 0.000000

LOWERBOUND( 1) 0.000000 0.000000

LOWERBOUND( 2) 0.000000 0.000000

Z( 1) 1.000000 2000.000

Z( 2) 0.000000 1500.000


Example 7 lp

Example 7 – LP

MODEL:

[_1] MIN= 25 * X_1_1_1 + 28 * X_1_1_2 + 23 * X_1_1_3 + 16 * X_1_2_1 + 19

* X_1_2_2 + 20 * X_1_2_3 + 20 * X_2_1_1 + 23 * X_2_1_2 + 18 * X_2_1_3 +

18 * X_2_2_1 + 21 * X_2_2_2 + 22 * X_2_2_3 + 13 * X_3_1_1 + 16 * X_3_1_2

+ 11 * X_3_1_3 + 22 * X_3_2_1 + 25 * X_3_2_2 + 26 * X_3_2_3 + 15 *

X_4_1_1 + 18 * X_4_1_2 + 13 * X_4_1_3 + 20 * X_4_2_1 + 23 * X_4_2_2 + 24

* X_4_2_3 + 7500 ;

[_2] X_1_1_1 + X_1_1_2 + X_1_1_3 + X_1_2_1 + X_1_2_2 + X_1_2_3 <= 200 ;

[_3] X_2_1_1 + X_2_1_2 + X_2_1_3 + X_2_2_1 + X_2_2_2 + X_2_2_3 <= 100 ;

[_4] X_3_1_1 + X_3_1_2 + X_3_1_3 + X_3_2_1 + X_3_2_2 + X_3_2_3 <= 50 ;

[_5] X_4_1_1 + X_4_1_2 + X_4_1_3 + X_4_2_1 + X_4_2_2 + X_4_2_3 <= 500 ;

[_6] X_1_1_1 + X_2_1_1 + X_3_1_1 + X_4_1_1 >= 0 ;

[_7] X_1_1_2 + X_2_1_2 + X_3_1_2 + X_4_1_2 >= 0 ;

[_8] X_1_1_3 + X_2_1_3 + X_3_1_3 + X_4_1_3 >= 0 ;

[_9] X_1_2_1 + X_2_2_1 + X_3_2_1 + X_4_2_1 >= 100 ;

[_10] X_1_2_2 + X_2_2_2 + X_3_2_2 + X_4_2_2 >= 200 ;

[_11] X_1_2_3 + X_2_2_3 + X_3_2_3 + X_4_2_3 >= 100 ;

[_12] 0 = 0 ;

[_13] 0 = 0 ;

[_14] 0 = 0 ;

[_15] 0 <= 0 ;

[_16] 0 <= 3600 ;

[_17] 0 >= 0 ;

[_18] 0 >= - 400 ;

END


Example 7 lp solution

Example 7 – LP Solution

Global optimal solution found.

Objective value: 15500.00 UPPER BOUND

Infeasibilities: 0.000000

Total solver iterations: 6

Variable Value Reduced Cost

Q 4.000000 0.000000

R 2.000000 0.000000

S 3.000000 0.000000

CAPACITY( 1) 200.0000 0.000000

CAPACITY( 2) 100.0000 0.000000

CAPACITY( 3) 50.00000 0.000000

CAPACITY( 4) 500.0000 0.000000

FIXEDCOST( 1) 2000.000 0.000000

FIXEDCOST( 2) 1500.000 0.000000

VARIABLECOST( 1) 10.00000 0.000000

VARIABLECOST( 2) 15.00000 0.000000

UPPERBOUND( 1) 3000.000 0.000000

UPPERBOUND( 2) 4000.000 0.000000

LOWERBOUND( 1) 0.000000 0.000000

LOWERBOUND( 2) 0.000000 0.000000

Z( 1) 0.000000 0.000000

Z( 2) 1.000000 0.000000


Example 7 dual

Example 7 – Dual

MODEL:

MAX = 200 * U_1 + 100 * U_2 + 50 * U_3 + 500 * U_4 + 100 * V_4

+ 200 * V_5 + 100 * V_6 + 3600 * T_2 - 400 * T_4;

[ X_1_1_1] U_1 + V_1 <= 25; [ X_1_1_2] U_1 + V_2 <= 28;

[ X_1_1_3] U_1 + V_3 <= 23; [ X_1_2_1] U_1 + V_4 <= 16;

[ X_1_2_2] U_1 + V_5 <= 19; [ X_1_2_3] U_1 + V_6 <= 20;

[ X_2_1_1] U_2 + V_1 <= 20; [ X_2_1_2] U_2 + V_2 <= 23;

[ X_2_1_3] U_2 + V_3 <= 18;

[ X_2_2_1] U_2 + V_4 <= 18;

[ X_2_2_2] U_2 + V_5 <= 21;

[ X_2_2_3] U_2 + V_6 <= 22;

[ X_3_1_1] U_3 + V_1 <= 13;

[ X_3_1_2] U_3 + V_2 <= 16;

[ X_3_1_3] U_3 + V_3 <= 11;

[ X_3_2_1] U_3 + V_4 <= 22;

[ X_3_2_2] U_3 + V_5 <= 25;

[ X_3_2_3] U_3 + V_6 <= 26;

[ X_4_1_1] U_4 + V_1 <= 15;

[ X_4_1_2] U_4 + V_2 <= 18;

[ X_4_1_3] U_4 + V_3 <= 13;

[ X_4_2_1] U_4 + V_4 <= 20;

[ X_4_2_2] U_4 + V_5 <= 23;

[ X_4_2_3] U_4 + V_6 <= 24;

@BND( -0.1E+31, U_1, 0); @BND( -0.1E+31, U_2, 0);

@BND( -0.1E+31, U_3, 0); @BND( -0.1E+31, U_4, 0); @FREE( W_1);

@FREE( W_2); @FREE( W_3); @BND( -0.1E+31, T_1, 0);

@BND( -0.1E+31, T_2, 0);

END


Example 7 dual solution

Example 7 – Dual Solution

Global optimal solution found.

Objective value: 8000.000 + K=7500 =15,500 UPPER BOUND

Infeasibilities: 0.000000

Total solver iterations: 7

Variable Value Reduced Cost

U_1 -4.000000 0.000000

U_2 -2.000000 0.000000

U_3 0.000000 -50.00000

U_4 0.000000 -400.0000

V_4 20.00000 0.000000

V_5 23.00000 0.000000

V_6 24.00000 0.000000

T_2 0.000000 -3600.000

T_4 0.000000 400.0000

V_1 0.000000 0.000000

V_2 0.000000 0.000000

V_3 0.000000 0.000000

W_1 0.000000 0.000000

W_2 0.000000 0.000000

W_3 0.000000 0.000000

T_1 0.000000 0.000000


Example 7 master problem

Example 7 – Master Problem

MODEL:

[_1] MIN= Z;

[_2] Z >= -1000 + 2000 * Y_2_1 + 4600 * Y_2_2 + 2400 * Y_2_3 + 2000 * Y_2_1 + 4600 * Y_2_2 + 2400 * Y_2_3 +

1000 * Y_1_1 + 2000 * Y_1_2 + 1000 * Y_1_3 + 1500 * Y_2_1 +

3000 * Y_2_2 + 1500 * Y_2_3 + 2000 * Z_1 + 1500 * Z_2 ;

[_4] - 100 * Y_1_1 + X_1_1_1 + X_2_1_1 + X_3_1_1 + X_4_1_1 >= 0 ;

[_5] - 200 * Y_1_2 + X_1_1_2 + X_2_1_2 + X_3_1_2 + X_4_1_2 >= 0 ;

[_6] - 100 * Y_1_3 + X_1_1_3 + X_2_1_3 + X_3_1_3 + X_4_1_3 >= 0 ;

[_7] - 100 * Y_2_1 + X_1_2_1 + X_2_2_1 + X_3_2_1 + X_4_2_1 >= 0 ;

[_8] - 200 * Y_2_2 + X_1_2_2 + X_2_2_2 + X_3_2_2 + X_4_2_2 >= 0 ;

[_9] - 100 * Y_2_3 + X_1_2_3 + X_2_2_3 + X_3_2_3 + X_4_2_3 >= 0 ;

[_10] Y_1_1 + Y_2_1 = 1 ;

[_11] Y_1_2 + Y_2_2 = 1 ;

[_12] Y_1_3 + Y_2_3 = 1 ;

[_13] 100 * Y_1_1 + 200 * Y_1_2 + 100 * Y_1_3 - 3000 * Z_1 <= 0 ;

[_14] 100 * Y_2_1 + 200 * Y_2_2 + 100 * Y_2_3 - 4000 * Z_2 <= 0 ;

[_15] 100 * Y_1_1 + 200 * Y_1_2 + 100 * Y_1_3 >= 0 ;

[_16] 100 * Y_2_1 + 200 * Y_2_2 + 100 * Y_2_3 >= 0 ;

@BIN( Y_1_1); @BIN( Y_1_2); @BIN( Y_1_3); @BIN( Y_2_1);

@BIN( Y_2_2); @BIN( Y_2_3); @BIN( Z_1); @BIN( Z_2);

END


Example 7 master problem solution

Example 7 – Master Problem Solution

Global optimal solution found.

Objective value: 5000.000 LOWER BOUND

Objective bound: 5000.000

Infeasibilities: 0.000000

Extended solver steps: 0

Total solver iterations: 5

Variable Value Reduced Cost

Z 5000.000 0.000000

Y_2_1 0.000000 5500.000

Y_2_2 0.000000 12200.00

Y_2_3 0.000000 6300.000

Y_1_1 1.000000 1000.000

Y_1_2 1.000000 2000.000

Y_1_3 1.000000 1000.000

Z_1 1.000000 2000.000

Z_2 0.000000 1500.000


Example 7 lp1

Example 7 – LP

MODEL:

[_1] MIN= 25 * X_1_1_1 + 28 * X_1_1_2 + 23 * X_1_1_3 + 16 * X_1_2_1 + 19

* X_1_2_2 + 20 * X_1_2_3 + 20 * X_2_1_1 + 23 * X_2_1_2 + 18 * X_2_1_3 +

18 * X_2_2_1 + 21 * X_2_2_2 + 22 * X_2_2_3 + 13 * X_3_1_1 + 16 * X_3_1_2

+ 11 * X_3_1_3 + 22 * X_3_2_1 + 25 * X_3_2_2 + 26 * X_3_2_3 + 15 *

X_4_1_1 + 18 * X_4_1_2 + 13 * X_4_1_3 + 20 * X_4_2_1 + 23 * X_4_2_2 + 24

* X_4_2_3 + 6000 ;

[_2] X_1_1_1 + X_1_1_2 + X_1_1_3 + X_1_2_1 + X_1_2_2 + X_1_2_3 <= 200 ;

[_3] X_2_1_1 + X_2_1_2 + X_2_1_3 + X_2_2_1 + X_2_2_2 + X_2_2_3 <= 100 ;

[_4] X_3_1_1 + X_3_1_2 + X_3_1_3 + X_3_2_1 + X_3_2_2 + X_3_2_3 <= 50 ;

[_5] X_4_1_1 + X_4_1_2 + X_4_1_3 + X_4_2_1 + X_4_2_2 + X_4_2_3 <= 500 ;

[_6] X_1_1_1 + X_2_1_1 + X_3_1_1 + X_4_1_1 >= 100 ;

[_7] X_1_1_2 + X_2_1_2 + X_3_1_2 + X_4_1_2 >= 200 ;

[_8] X_1_1_3 + X_2_1_3 + X_3_1_3 + X_4_1_3 >= 100 ;

[_9] X_1_2_1 + X_2_2_1 + X_3_2_1 + X_4_2_1 >= 0 ;

[_10] X_1_2_2 + X_2_2_2 + X_3_2_2 + X_4_2_2 >= 0 ;

[_11] X_1_2_3 + X_2_2_3 + X_3_2_3 + X_4_2_3 >= 0 ;

[_12] 0 = 0 ;

[_13] 0 = 0 ;

[_14] 0 = 0 ;

[_15] 0 <= 2600 ;

[_16] 0 <= 0 ;

[_17] 0 >= - 400 ;

[_18] 0 >= 0 ;

END


Example 7 lp solution1

Example 7 – LP Solution

Global optimal solution found.

Objective value: 12300.00 UPPER BOUND

Infeasibilities: 0.000000

Total solver iterations: 6

Variable Value Reduced Cost

Q 4.000000 0.000000

R 2.000000 0.000000

S 3.000000 0.000000

CAPACITY( 1) 200.0000 0.000000

CAPACITY( 2) 100.0000 0.000000

CAPACITY( 3) 50.00000 0.000000

CAPACITY( 4) 500.0000 0.000000

FIXEDCOST( 1) 2000.000 0.000000

FIXEDCOST( 2) 1500.000 0.000000

VARIABLECOST( 1) 10.00000 0.000000

VARIABLECOST( 2) 15.00000 0.000000

UPPERBOUND( 1) 3000.000 0.000000

UPPERBOUND( 2) 4000.000 0.000000

LOWERBOUND( 1) 0.000000 0.000000

LOWERBOUND( 2) 0.000000 0.000000

Z( 1) 1.000000 0.000000

Z( 2) 0.000000 0.000000


Example 7 dual1

Example 7 – Dual

MODEL:

MAX = 200 * U_1 + 100 * U_2 + 50 * U_3 + 500 * U_4 + 100 * V_1

+ 200 * V_2 + 100 * V_3 + 2600 * T_1 - 400 * T_3;

[ X_1_1_1] U_1 + V_1 <= 25; [ X_1_1_2] U_1 + V_2 <= 28;

[ X_1_1_3] U_1 + V_3 <= 23; [ X_1_2_1] U_1 + V_4 <= 16;

[ X_1_2_2] U_1 + V_5 <= 19; [ X_1_2_3] U_1 + V_6 <= 20;

[ X_2_1_1] U_2 + V_1 <= 20; [ X_2_1_2] U_2 + V_2 <= 23;

[ X_2_1_3] U_2 + V_3 <= 18;

[ X_2_2_1] U_2 + V_4 <= 18;

[ X_2_2_2] U_2 + V_5 <= 21;

[ X_2_2_3] U_2 + V_6 <= 22;

[ X_3_1_1] U_3 + V_1 <= 13;

[ X_3_1_2] U_3 + V_2 <= 16;

[ X_3_1_3] U_3 + V_3 <= 11;

[ X_3_2_1] U_3 + V_4 <= 22;

[ X_3_2_2] U_3 + V_5 <= 25;

[ X_3_2_3] U_3 + V_6 <= 26;

[ X_4_1_1] U_4 + V_1 <= 15;

[ X_4_1_2] U_4 + V_2 <= 18;

[ X_4_1_3] U_4 + V_3 <= 13;

[ X_4_2_1] U_4 + V_4 <= 20;

[ X_4_2_2] U_4 + V_5 <= 23;

[ X_4_2_3] U_4 + V_6 <= 24;

@BND( -0.1E+31,U_1, 0); @BND( -0.1E+31,U_2, 0);

@BND( -0.1E+31,U_3, 0); @BND( -0.1E+31,U_4, 0); @FREE( W_1);

@FREE( W_2); @FREE( W_3); @BND( -0.1E+31, T_1, 0);

@BND( -0.1E+31, T_2, 0);

END


Example 7 dual solution1

Example 7 – Dual Solution

Global optimal solution found.

Objective value: 6300.000 + K = 6000 = 12300 UPPPER BOUND

Infeasibilities: 0.000000

Total solver iterations: 10

Variable Value Reduced Cost

U_1 0.000000 -200.0000

U_2 0.000000 -100.0000

U_3 -2.000000 0.000000

U_4 0.000000 -150.0000

V_1 15.00000 0.000000

V_2 18.00000 0.000000

V_3 13.00000 0.000000

T_1 0.000000 -2600.000

T_3 0.000000 400.0000

V_4 0.000000 0.000000

V_5 0.000000 0.000000

V_6 0.000000 0.000000

W_1 0.000000 0.000000

W_2 0.000000 0.000000

W_3 0.000000 0.000000

T_2 0.000000 0.000000


Example 7 master problem1

Example 7 – Master Problem

MODEL:

[_1] MIN= Z;

[_2] Z >= -1000 + 2000 * Y_2_1 + 4600 * Y_2_2 + 2400 * Y_2_3 +

1000 * Y_1_1 + 2000 * Y_1_2 + 1000 * Y_1_3 + 1500 * Y_2_1 +

3000 * Y_2_2 + 1500 * Y_2_3 + 2000 * Z_1 + 1500 * Z_2 ;

[_3] Z >= -100 + 1500 * Y_1_1 + 3600 * Y_1_2 + 1300 * Y_1_3 +

1000 * Y_1_1 + 2000 * Y_1_2 + 1000 * Y_1_3 + 1500 * Y_2_1 +

3000 * Y_2_2 + 1500 * Y_2_3 + 2000 * Z_1 + 1500 * Z_2 ;

[_4] - 100 * Y_1_1 + X_1_1_1 + X_2_1_1 + X_3_1_1 + X_4_1_1 >= 0 ;

[_5] - 200 * Y_1_2 + X_1_1_2 + X_2_1_2 + X_3_1_2 + X_4_1_2 >= 0 ;

[_6] - 100 * Y_1_3 + X_1_1_3 + X_2_1_3 + X_3_1_3 + X_4_1_3 >= 0 ;

[_7] - 100 * Y_2_1 + X_1_2_1 + X_2_2_1 + X_3_2_1 + X_4_2_1 >= 0 ;

[_8] - 200 * Y_2_2 + X_1_2_2 + X_2_2_2 + X_3_2_2 + X_4_2_2 >= 0 ;

[_9] - 100 * Y_2_3 + X_1_2_3 + X_2_2_3 + X_3_2_3 + X_4_2_3 >= 0 ;

[_10] Y_1_1 + Y_2_1 = 1 ;

[_11] Y_1_2 + Y_2_2 = 1 ;

[_12] Y_1_3 + Y_2_3 = 1 ;

[_13] 100 * Y_1_1 + 200 * Y_1_2 + 100 * Y_1_3 - 3000 * Z_1 <= 0 ;

[_14] 100 * Y_2_1 + 200 * Y_2_2 + 100 * Y_2_3 - 4000 * Z_2 <= 0 ;

[_15] 100 * Y_1_1 + 200 * Y_1_2 + 100 * Y_1_3 >= 0 ;

[_16] 100 * Y_2_1 + 200 * Y_2_2 + 100 * Y_2_3 >= 0 ;

@BIN( Y_1_1); @BIN( Y_1_2); @BIN( Y_1_3); @BIN( Y_2_1);

@BIN( Y_2_2); @BIN( Y_2_3); @BIN( Z_1); @BIN( Z_2);

END


Example 7 master problem solution1

Example 7 – Master Problem Solution

Global optimal solution found.

Objective value: 12000.00 LOWER BOUND

Objective bound: 12000.00

Infeasibilities: 0.000000

Extended solver steps: 0

Total solver iterations: 43

Variable Value Reduced Cost

Z 12000.00 0.000000

Y_2_1 1.000000 1500.000

Y_2_2 0.000000 3000.000

Y_2_3 1.000000 1500.000

Y_1_1 0.000000 2500.000

Y_1_2 1.000000 5600.000

Y_1_3 0.000000 2300.000

Z_1 1.000000 2000.000

Z_2 1.000000 1500.000


Example 73

Example 7

Lower Bound of 12,000 is close to Upper Bound. So, optimal solution must be between the two. The student is encouraged to carry Benders’ decomposition algorithm one more time to ensure LB=UB in the third iteration.


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