Facilities Design

Facilities DesignFacilities Design

S.S. HeraguS.S. HeraguIndustrial Engineering DepartmentIndustrial Engineering Department

University of LouisvilleUniversity of Louisville

Chapter 12:Chapter 12:Advanced Location Advanced Location

ModelsModels

12.112.1 INTRODUCTIONINTRODUCTION12.212.2 LOCATION MODELSLOCATION MODELS

12.2.112.2.1 Multiple-Facility Problems with Multiple-Facility Problems with Rectilinear DistancesRectilinear Distances

12.2.212.2.2 Multiple-Facility Problems with Multiple-Facility Problems with Euclidean DistancesEuclidean Distances

12.312.3 ALLOCATION MODELSALLOCATION MODELS12.3.112.3.1 Network Flow ModelNetwork Flow Model12.3.212.3.2 Two-Stage Transportation ModelTwo-Stage Transportation Model12.3.312.3.3 Vehicle Routing ProblemVehicle Routing Problem

Chapter 12: Advanced Location Chapter 12: Advanced Location ModelsModels

12.412.4 LOCATION-ALLOCATION MODELSLOCATION-ALLOCATION MODELS12.4.112.4.1 Set Covering ModelSet Covering Model12.4.212.4.2 Uncapacitated Location-Allocation Uncapacitated Location-Allocation

ModelModel12.4.312.4.3 Comprehensive Location-Comprehensive Location-

Allocation ModelAllocation Model12.512.5 SUMMARYSUMMARY

Chapter 12: Advanced Location Chapter 12: Advanced Location ModelsModels

IntroductionIntroduction

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

- Where should they be located?Where should they be located?- How large should each new facility be? In How large should each new facility be? In

other words, what is the capacity of the new other words, what is the capacity of the new facility?facility?

IntroductionIntroduction

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

- Can more than one facility serve a customer?Can more than one facility serve a customer?

12.212.2LOCATIONLOCATION

MODELSMODELS

12.2.112.2.1Multiple-Facility ProblemsMultiple-Facility Problemswith Rectilinear Distanceswith Rectilinear Distances

Model 1:Model 1:

Minimize c ij f ij[| x i x j | | y i y j |]j 1

n

i1

n

dijgij[ | x i a j | | y i b j |]j 1

m

i1

n

Model 1:Model 1:

ijijji

ijijji

ji

jiijij

jijiij

xxxx

xxxx

xx

xxxxx

xxxxx

)(

and

0, or 0)( whether that,observecan Weotherwise 0

0 if )(

otherwise 0

0 if )(Define

Model 1:Model 1:

A similar definition of A similar definition of yy++ijij, , yy--

ijij, , xaxa++ijij, , xaxa--

ijij, yb, yb++ijij, and , and

ybyb--ijij yields yields

||yyii - y - yjj| = | = yy++ijij + y + y--

ijij

yyii - y - yjj = = yy++ijij - y - y--

ijij

||xxii - a - ajj| = | = xaxa++ijij + xa + xa--

ijij

xxii - a - ajj = xa = xa++ijij - xa - xa--

ijij

|y|yii - b - bjj| = yb| = yb++ijij + yb + yb--

ijij

yyii - b - bjj = yb = yb++ijij - yb - yb--

ijij

Model 1:Model 1:

Thus, the transformed linear model is:Thus, the transformed linear model is:

n

i

m

jijijijijijij

n

i

n

jijijijijijij

ybybxaxagd

yyxxfc

1 1

1 1

Minimize

Model 1:Model 1:Subject to:Subject to:

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

ij

yi - yj = y+ij - y-

ij

xi - aj = xa+ij - xa-

ij

yi - bj = yb+ij - yb-

ij

xx++ijij, x, x--

ijij, y, y++ijij , y , y--

ijij >> 0, 0, ii, , jj = 1, 2, ..., = 1, 2, ...,nnxaxa++

ijij, xa, xa--ijij, yb, yb++

ijij , yb , yb--ijij >> 0, 0, ii = 1, 2, ..., = 1, 2, ...,nn, , jj =1,2,..., =1,2,...,mm

xxii, , yyii unrestricted in sign, unrestricted in sign, ii = 1,2,..., = 1,2,...,nn

Model 1:Model 1:

• xxijij++ or or xxijij

--, but not both, can be greater than 0. (If , but not both, can be greater than 0. (If both are, then the values of both are, then the values of xxijij

++ and and xxijij-- do not do not

satisfy their definition in (2) and (3)). Similarly, satisfy their definition in (2) and (3)). Similarly, only one of (i) only one of (i) yyijij

++, , yyijij--, (ii) , (ii) xxijij

++, , xaxaijij--, (iii) , (iii) ybybijij

++, , ybybijij--, ,

must be greater than 0. must be greater than 0. • Model 1 can be simplified by noting that:Model 1 can be simplified by noting that:• xxii can be substituted as can be substituted as aaii + + xxijij

++ - - xaxaijij--

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

Example 1:Example 1:

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

Example 1: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.

Example 1: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.

SC1 SC2 SC3 SC4 W1

W2 7 7 5 4 2W3 3 2 4 5 2

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


• 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

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

12.2.212.2.2Multiple-Facility ProblemsMultiple-Facility Problemswith Euclidean Distanceswith Euclidean Distances

n

i

m

jjijiijij

n

i

n

jjijiijij

byaxgd

yyxxfc

1 1

22

1 1

22

)()(

)()( Minimize

Consider the following objective for the euclidean distance Consider the following objective for the euclidean distance problem.problem.

Multiple-Facility ProblemsMultiple-Facility Problemswith Euclidean Distanceswith Euclidean DistancesTaking the partial derivatives, we get

ni

)b(y)a(x

)a(xgd

)y(y)x(x

)x(xfc

m

j jiji

jiijij

n

j jiji

jiijij

1,2,...,

,0 1

1

22

22

Multiple-Facility ProblemsMultiple-Facility Problemswith Euclidean Distanceswith Euclidean Distances

ni

)b(y)a(x

)b(ygd

)y(y)x(x

)y(yfc

m

j jiji

jiijij

n

j jiji

jiijij

1,2,...,

,0 1

1

22

22

Multiple-Facility ProblemsMultiple-Facility Problemswith Euclidean Distanceswith Euclidean DistancesTo make sure the denominator is never 0, we add to it. We then get:

2 2 2 21 1

'

2 2 2 21 1

1, 2,...,

n mij ij j ij ij j

j ji j i j i j i j

i n mij ij ij ij

j ji j i j i j i j

c f x d g a

x x y y x a y bx i n

c f d g

x x y y x a y b

Multiple-Facility ProblemsMultiple-Facility Problemswith Euclidean Distanceswith Euclidean DistancesAnd …

2 2 2 21 1

'

2 2 2 21 1

1, 2,...,

n mij ij j ij ij j

j ji j i j i j i j

i n mij ij ij ij

j ji j i j i j i j

c f y d g b

x x y y x a y by i nc f d g

x x y y x a y b

Example 2Example 2

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

12.312.3ALLOCATIONALLOCATIONMODELMODEL

12.3.1 Network Flow Model12.3.1 Network Flow Model

Network Flow Model:Network Flow Model:Model 2Model 2Consider this notation:Consider this notation:ccijij cost of sending one unit of flow on arc (cost of sending one unit of flow on arc (ii, , jj))UUijij upper bound on the flow that can be sent upper bound on the flow that can be sent

on arc (on arc (ii, , jj), i.e., capacity of arc (), i.e., capacity of arc (ii,,jj))LLijij lower bound on the flow that can be sent lower bound on the flow that can be sent

on arc (on arc (ii, , jj))DDii net flow generated at node net flow generated at node iixxijij number of units of flow on arc (number of units of flow on arc (ii, , jj))

Model 2:Model 2:

1 1

Minimize n n

ij iji j

c x

1 1

Subject to 1,2,...,n n

ij ji ij i

x x D i n

, 1, 2,...,ij ij ijL x U i j n

Network Simplex Algorithm:Network Simplex Algorithm:Step 1:Step 1: Construct a spanning tree for the Construct a spanning tree for the nn nodes. The nodes. The

variables variables xxijij corresponding to the arcs ( corresponding to the arcs (ii,,jj) in the ) in the spanning tree are basic variables and the remaining are spanning tree are basic variables and the remaining are nonbasic. Find a basic feasible solution to the problem nonbasic. Find a basic feasible solution to the problem so that:so that:(1) the basic variables satisfy (1) the basic variables satisfy LLijij < < xxijij < < UUijij , and , and(2) the nonbasic variables take on a value of (2) the nonbasic variables take on a value of LLijij or or UUijij to to satisfy constraint (22).satisfy constraint (22).

Step 2:Step 2: Set Set uu11 =0 and find =0 and find uujj, , jj=2,...,=2,...,nn using the formula using the formula uuii - - uujj = = ccijij for all basic variables. for all basic variables.

Step 3:Step 3: If If uuii - - uujj - - ccijij << 0 for all nonbasic variables 0 for all nonbasic variables xxijij with a value with a value of of LLijij , and , and uuii - - uujj - - ccijij >> 0 for all nonbasic variables 0 for all nonbasic variables xxijij with with a value of a value of UUijij , then the current basic feasible solution is , then the current basic feasible solution is optimal; stop. Otherwise, go to step 4.optimal; stop. Otherwise, go to step 4.

Network Simplex Algorithm:Network Simplex Algorithm:Step 4:Step 4: Select the variable Select the variable xxi*j*i*j* that violates the optimality that violates the optimality

condition (in step 3) the most, i.e., the largest of the condition (in step 3) the most, i.e., the largest of the uuii - - uujj - - ccijij for those nonbasic variables with for those nonbasic variables with xxijij = = LLijij, , and the smallest of the and the smallest of the uuii - - uujj - - ccijij for those nonbasic for those nonbasic variables with variables with xxijij = = UUijij,. Make the arc (,. Make the arc (ii*,*,jj*) a basic *) a basic variable and add arc (variable and add arc (ii*,*,jj*) to the spanning tree. Make *) to the spanning tree. Make one of the other basic variables in the loop of basic one of the other basic variables in the loop of basic variables [formed by including arc (variables [formed by including arc (ii*,*,jj*)], a nonbasic *)], a nonbasic variable such that:variable such that:(1) (1) xxi*j*i*j* takes on the largest possible value, takes on the largest possible value,(2) constraint (21) is satisfied for all the (2) constraint (21) is satisfied for all the nn nodes, and nodes, and(3) constraint (22) is satisfied for all the arcs in the (3) constraint (22) is satisfied for all the arcs in the loop.loop.Remove the arc corresponding to the nonbasic Remove the arc corresponding to the nonbasic variable just identified so that we have a spanning variable just identified so that we have a spanning tree once again. Go to step 2. tree once again. Go to step 2.

Example 3Example 3The Fast Shipping Company manages the The Fast Shipping Company manages the distribution of lawnmowers from a company distribution of lawnmowers from a company that has two factories (that has two factories (FF1 and 1 and FF2) in the 2) in the Northeast to two large customer bases (Northeast to two large customer bases (CC1 and 1 and CC2) in the Southwest. For cost and freight 2) in the Southwest. For cost and freight consolidation reasons, Fast Shipping would consolidation reasons, Fast Shipping would like to route the shipments via three like to route the shipments via three intermediate nodes (intermediate nodes (TT1 - 1 - TT3) located in the 3) located in the midwest. The relevant data is provide in Tables midwest. The relevant data is provide in Tables 12.3-12.5. Setup a model to determine how the 12.3-12.5. Setup a model to determine how the shipment is to take place from the two factories shipment is to take place from the two factories to the two destinations via the three to the two destinations via the three intermediate shipment points. intermediate shipment points.

Supply and DemandSupply and Demand

Facility Supply Demand

F1 900 - F2 600 - C1 - 750 C2 - 750

Inbound and Outbound Transportation Inbound and Outbound Transportation Costs and Arc CapacitiesCosts and Arc Capacities

T1 T2 T3

F1 8 (500) 11 (1500) 5 (350)

F2 12 (1200) 8 (750) 5 (450)

C1 6 (1000) 12 (750) 9 (1000)

C2 3 (150) 1 (200) 19 (1500)

Example 3: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

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

1

2

q

1

2

3

p

1

2

3

4

r

12.3.2 Two-Stage Transportation 12.3.2 Two-Stage Transportation ModelModel

Two-Stage Transportation Model:Two-Stage Transportation Model:

Consider the following notation:Consider the following notation:SSii capacity of supply source capacity of supply source ii, , ii = 1, 2, ..., = 1, 2, ..., ppPPjj capacity of plant capacity of plant jj, , jj = 1, 2, ..., = 1, 2, ..., qqDDkk demand at customer demand at customer kk, , kk = 1, 2, ..., = 1, 2, ..., rrccijij cost of transporting one unit from supply cost of transporting one unit from supply

source source ii to plant to plant jj

The LP model is:The LP model is:

Model 3:Model 3:

1

Subject to 1,2,...,q

ij ij

x S i p

1

1, 2,...,p

ij ji

x P j q

1

1,2,...,q

jk kj

y D k r

1 1

1, 2,...,p r

ij jki k

x y j q

, 1, 2,..., ; 1, 2,..., ; 1, 2,...,ij jkx y i p j q k r

q

j

r

kjkjk

p

i

q

jijij ydxc

1 11 1

Minimize

Four cases arise:Four cases arise:

(i)(i) Supply source capacity is unlimited and total plant capacity Supply source capacity is unlimited and total plant capacity is more than total demandis more than total demand

(ii)(ii) Supply source capacity is unlimited and total demand Supply source capacity is unlimited and total demand exceeds total plant capacityexceeds total plant capacity

(iii)(iii)Plant capacity is unlimited and total supply source capacity Plant capacity is unlimited and total supply source capacity exceeds total demandexceeds total demand

(iv)(iv)Plant capacity is unlimited and total demand exceeds total Plant capacity is unlimited and total demand exceeds total supply source capacitysupply source capacity

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

Case 1:Case 1:Customer Dummy Plant Supply

Source Plant

1 2 r 1 2

q

Excess Plant

Capacity

Capacity

1 c111 c112 c11r 0 M

M 0 P1

2 c121 c122 c12r M 0

M 0 P2

1

q c1q1 c1q2 c1qr M M

0 0 Pq

1 c211 c212 c21r 0 M

M 0 P1

2 c221 c222 c22r M 0

M 0 P2

2

q c2q1 c2q2 c2qr M M

0 0 Pq

1 cp11 cp1r 0 M

M 0 P1

p 2 cp21 cp22 cp2r M 0

M 0 P2

q cpq1 cpq2 cpqr M M 0 0 Pq

Demand D1 D2 Dr (p 1)P1 (p 1)P2

(p 1)Pq

r

kk

q

jj DP

11

1

q

jj

p P

Example 4:Example 4:2-Stage Distribution Problem: RIFIN Company has 2-Stage Distribution Problem: RIFIN Company has recently developed a new method of manufacturing recently developed a new method of manufacturing a type of chemical. It involves refining a certain raw a type of chemical. It involves refining a certain raw material which can be obtained from four overseas material which can be obtained from four overseas suppliers A, B, C, D who have access to the four suppliers A, B, C, D who have access to the four ports at Vancouver, Boston, Miami, and San ports at Vancouver, Boston, Miami, and San Francisco, respectively. RIFIN wants to determine Francisco, respectively. RIFIN wants to determine the location of plants at which the chemical will be the location of plants at which the chemical will be refined. The chemical, once refined, will be refined. The chemical, once refined, will be transported via trucks to five outlets located at transported via trucks to five outlets located at Dallas, Phoenix, Portland, Montreal and Orlando. Dallas, Phoenix, Portland, Montreal and Orlando.

Example 4:Example 4:After an initial study, the choice of location After an initial study, the choice of location for RIFIN’s refineries has narrowed down to for RIFIN’s refineries has narrowed down to Denver, Atlanta and Pittsburgh. Assume that Denver, Atlanta and Pittsburgh. Assume that one unit of raw material is required to make one unit of raw material is required to make one unit of chemical. The amount of raw one unit of chemical. The amount of raw material from each potential refinery as well material from each potential refinery as well as the cost of trucking the chemical to outlets as the cost of trucking the chemical to outlets are also provided below. Determine the are also provided below. Determine the location of RIFIN’s refining plants, capacities location of RIFIN’s refining plants, capacities at these plants and distribution pattern for the at these plants and distribution pattern for the raw material and processed chemical.raw material and processed chemical.


Raw Material Source Supply Outlet DemandA 1000 Dallas 900B 800 Phoenix 800C 800 Portland 600D 700 Montreal 500

Orlando 500


TO Denver Atlanta PittsburghFROMVancouver 4 13 9Boston 8 8 5Miami 12 2 9San Francisco 11 11 12

Raw Material Transportation Cost


TO Dallas Phoenix Portland Montreal OrlandoFROMDen. 28 26 12 30 30Atla. 10 22 23 29 8Pitts. 18 21 23 18 21

Chemical Trucking Cost

Figure 12.3: Pictorial representation of Figure 12.3: Pictorial representation of RIFIN ExampleRIFIN Example

D

A

P

V

B

M

SF

Da

Ph

Po

Mo

Orl

Setup transportation tableau for Setup transportation tableau for Example 4Example 4

Customer Dummy Source Supply Source

Refinery

Dal Ph Port Mont Orl Van Bos Miami SanFr

Capacity

Denver 32 30 16 34 34 0 M M M 1000

Atlanta 23 35 36 42 41 0 M M M 1000 Vancouver (A)

Pittsburgh 47 30 32 27 30 0 M M M 1000

Denver 36 34 20 38 38 M 0 M M 800

Atlanta 18 30 31 37 16 M 0 M M 800 Boston (B)

Pittsburgh 23 26 28 23 26 M 0 M M 800

Denver 40 38 24 42 42 M M 0 M 800

Atlanta 12 24 25 31 10 M M 0 M 800 Miami (C)

Pittsburgh 27 30 32 27 30 M M 0 M 800

Denver 39 37 23 41 41 M M M 0 700

Atlanta 21 33 34 40 19 M M M 0 700

San Francisco (D) Pittsburgh 30 33 35 30 33 M M M 0 700

Demand 900 800 600 500 500 2000 1600 1600 1400 9900

Solution:Solution:

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

Figure 12.4:Figure 12.4:

D

A

P

V

B

M

SF

Da

Ph

Po

Mo

Orl

1000

1000

800

600

100

900

400

600400

500

500

12.3.3 Vehicle Routing Problem12.3.3 Vehicle Routing Problem

Determine the number of vehicles required to:Determine the number of vehicles required to:

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

(2)(2)the vehicle capacity is not exceeded, andthe vehicle capacity is not exceeded, and

(3)(3)the total travel time is minimizedthe total travel time is minimized

Vehicle Routing Problem:Vehicle Routing Problem:TTijij time to travel from customer time to travel from customer ii to to

customer customer jj, , ii,,jj=1,2, …, =1,2, …, nnDDii demand at customer demand at customer ii, , ii=1,2, …, =1,2, …, nnCCkk capacity of vehicle capacity of vehicle kk, , kk=1,2, …, =1,2, …, pp

1 if truck visits customer after visiting customer 0 otherwise ijk

k j ix

Model 4:Model 4:

1 1 1

Minimize p n n

ij ijkk i j

T x

1 1

Subject to 1 1, 2,..., p n

ijkk i

x j n

1 1

1 1,2,..., p n

ijkk j

x i n

1 1

1,2,..., n n

i ijk ki j

D x C k p

1 1

0 1, 2,..., ; 1,2,..., n n

ilk ljki j

x x l n k p

0 or 1 , 1,2,..., ; 1,2,..., ijkx i j n k p

12.412.4LOCATION-ALLOCATIONLOCATION-ALLOCATIONMODELSMODELS

12.4.112.4.1Set Set Covering Covering ModelModel

Set Covering ModelSet Covering ModelDefine:Define:ccjj cost of locating facility at site cost of locating facility at site jj

aaijij ==

xxjj ==

The set covering problem is to:The set covering problem is to:

{1 if facility located at site

j can cover customer i0 Otherwise

{ 1 if facility is located at site j0 Otherwise

Model 5:Model 5:

1

Minimize n

j jj

c x

0 or 1 1, 2,...,jx j n

mixan

jjij ,...,2,1 1 Subject to

1

Greedy Heuristic for Set Covering Greedy Heuristic for Set Covering Problem:Problem:Step 1:Step 1: If If ccjj = 0, for any = 0, for any jj = 1, 2, ..., = 1, 2, ..., nn, set , set xxjj = 1 and remove = 1 and remove

all constraints in which all constraints in which xxjj appears with a appears with a coefficient of +1.coefficient of +1.

Step 2:Step 2: If If ccjj > 0, for any > 0, for any jj = 1, 2, ..., = 1, 2, ..., nn and and xxjj does not does not appear with +1 coefficient in any of the remaining appear with +1 coefficient in any of the remaining constraints, set constraints, set xxjj = 0. = 0.

Step 3:Step 3: For each of the remaining variables, determine For each of the remaining variables, determine ccjj//ddjj, where , where ddjj is the number of constraints in is the number of constraints in which which xxjj appears with +1 coefficient. Select the appears with +1 coefficient. Select the variable variable kk for which for which cckk//ddkk is minimum, set is minimum, set xxkk = 1 = 1 and remove all constraints in which and remove all constraints in which xxjj appears appears with +1 coefficient. Examine the resulting model.with +1 coefficient. Examine the resulting model.

Greedy Heuristic for Set Covering Greedy Heuristic for Set Covering Problem:Problem:

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

We illustrate the above greedy heuristic We illustrate the above greedy heuristic with an example.with an example.


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

Figure 12.5:Figure 12.5:11 22 33 44 55 66 77

11 00 44 1212 66 1515 1010 8822 88 00 1515 6060 77 22 3333 5050 1313 00 88 66 55 9944 99 1111 88 00 99 1010 3355 5050 88 44 1010 00 22 272766 3030 55 77 99 33 00 272777 88 55 99 77 2525 2727 00

[dij] =


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

Solution:Solution:

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

Defining

aij = {1 if zone i’s center can be reached from

center of zone j within 8 minutes0 otherwise

Solution:Solution:1 2 3 4 5 6 7

1 1 1 0 1 0 0 12 1 1 0 0 1 1 13 0 0 1 1 1 1 04 0 0 1 1 0 0 15 0 1 1 0 1 1 06 0 1 1 0 1 1 07 1 1 0 1 0 0 1

[aij] =

The corresponding set covering model is:

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

Subject tox1 + x2 + x4 + x7 > 1x1 + x2 + x5 + x6 + x7 > 1

x3 + x4 + x5 + x6 > 1

x3 + x4 + x7 > 1

x2 + x3 + x5 + x6 > 1

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

Greedy HeuristicGreedy HeuristicStep 1: Since each Step 1: Since each ccjj > 0, > 0, jj = 1, 2, ..., 7, go to = 1, 2, ..., 7, go to

step 2.step 2.Step 2: Since Step 2: Since xxjj appears in each constraint with appears in each constraint with

+1 coefficient, go to step 3.+1 coefficient, go to step 3.Step 3:Step 3:

3 51 2 4

1 2 3 4 5

100 80 120 110 9033.3; 16; 30; 27.5; 22.5;3 5 4 4 4

c cc c cd d d d d

6 7

6 7

90 11022.5; 27.54 4

c cd d

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

Greedy Heuristic:Greedy Heuristic:Step 4: Since we have two constraints go to Step 4: Since we have two constraints go to

step 1.step 1.Step 1: Since Step 1: Since cc11 > 0, j = 1, 3, 4, ..., 7, go to step > 0, j = 1, 3, 4, ..., 7, go to step

22Step 2: Since Step 2: Since cc11 > 0 and > 0 and xx11 does not appear in does not appear in

any of the constraints with +1 any of the constraints with +1 coefficient, set coefficient, set xx1 1 = 0.= 0.

Greedy HeuristicGreedy HeuristicSince 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

Greedy Heuristic:Greedy Heuristic:

Step 4:Step 4: Since there are no constraints in the Since there are no constraints in the above model, set above model, set xx33 = = xx55 = = xx66 = = xx77 = 0 = 0 and stop.and stop.

The solution is The solution is xx22 = = xx44 = 1; = 1; xx11 = = xx33 = = xx55 = = xx66 = = xx77 = 0. Cost of locating = 0. Cost of locating emergency response units to meet emergency response units to meet the eight minute response service the eight minute response service level is $800,000 + $1,100,000 = level is $800,000 + $1,100,000 = $1,900,000.$1,900,000.

12.4.212.4.2Uncapacitated Uncapacitated Location-Allocation Location-Allocation ModelModel

Uncapacitated Location-Allocation Uncapacitated Location-Allocation ModelModelParametersParametersmm number of potential facilities number of potential facilitiesnn number of customers number of customersccijij cost of transporting one unit of product cost of transporting one unit of product from facility from facility ii to customer to customer jjFFii fixed cost of opening and operating facility fixed cost of opening and operating facility jjDDjj number of units demanded at customer number of units demanded at customer jj

Decision VariablesDecision Variablesxxijij number of units shipped from facility number of units shipped from facility ii to customer to customer jj

1 if facility is opened0 otherwisei

iy

Model 6Model 6

1 1 1

Minimize m m n

i i ij iji i j

F y c x

1

Subject to 1, 2,..., m

ij ji

x D j n

1 1

1, 2,...,n n

ij i jj j

x y D i m

0 1,2,..., ; 1, 2,...,ijx i m j n

0 or 1 1, 2,...,iy i m

Model 7Model 7Modify Model 6 by transforming xij variables and the cij parameter

' ' 1, 2,..., ; 1, 2,...,ijij ij ij j

j

xx c c D i m j n

D

' '

1 1 1

Minimize m m n

i i ij iji i j

F y c x

'

1

Subject to 1 1, 2,..., m

iji

x j n

'

1

1, 2,...,n

ij ij

x ny i m

' 0 1, 2,..., ; 1,2,...,ijx i m j n

0 or 1 1, 2,...,iy i m

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 , each term must also be < yi

Because Dj/ΣDj is a positive fraction for each j: x’ij < yi, j=1,2,…,n Adding we getAdding we get

'

1 1

1, 2,...,n n

ij j i jj j

x D y D i m

'

1

1

1 1,2,...,n

ij j inj

jj

x D y i mD

'

1

1, 2,...,ij jin

jj

x Dy i m

D

'

1

1, 2,...,n

ij ij

x ny i m

On solving Model 7On solving Model 7Although 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 7On solving Model 7The central idea of the branch-and-bound algorithm is

based on the following resultSuppose, 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 8Rewrite Model 7 as Model 8

1 1 2 21 1

Minimize n m n

i ij ij i i ij iji S i S j i S i S j

F c x F y c x

1


iji

x j n

1

1,2,...,n

ij ij

x ny i m

0 1, 2,..., ; 1, 2,...,ijx i m j n

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

Because max {Because max {xxijij} is 1, max {} is 1, max {yyii} is also 1 } is also 1

1

1 1, 2,...,n

i ijj

y x i mn

Rewrite Model 8 as Model 9Rewrite Model 8 as Model 9

1


iji

x j n

0 1, 2,..., ; 1,2,...,ijx i m j n

1 1 2 21 1 1

Minimize n n m n

iji ij ij i ij ij

i S i S j i S j i S j

xF c x F c x

n

1 1 21 1

Minimize n n

ii ij ij i ij ij

i S i S j i S j

FF c x F c xn

minyx i

n

jij ,...,2,1 ,

1

On solution of Model 9On solution of Model 9• Model 9 which is equivalent to model 6 without the Model 9 which is equivalent to model 6 without the

integer restrictions on the integer restrictions on the yy variables, is a half variables, is a half assignment problem. It can be proved (again, by assignment problem. It can be proved (again, by contradiction) that for each contradiction) that for each jj = 1, 2, ..., = 1, 2, ..., nn, only one of , only one of xx11jj, , xx22jj, ..., , ..., xxmjmj will take on a value of 1, due to will take on a value of 1, due to Σxij=1

• In fact, for each j, the xxijij 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 xxijij in the objective function, i=1,2,…,m, set the corresponding xxijij equal to 1 and all other xxijij 's to 0 as follows:

1

2

if

if

ij

iij

c i S

Fc i Sn

On solution of Model 9On solution of Model 9• Select the smallest Select the smallest ccijij from the list, set the from the list, set the

corresponding corresponding xxijij = 1 and all other = 1 and all other xxijij’s to 0. ’s to 0. This is the minimum coefficient rule.This is the minimum coefficient rule.

• We do not include facility We do not include facility ii SS22 in the min in the min coeff rule because these are closedcoeff rule because these are closed

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

tto the sum of the coefficients of the o the sum of the coefficients of the xxijij variables which have taken a value of 1variables which have taken a value of 1

1

ii S

F

Branch-and-BoundBranch-and-BoundAlgorithm Algorithm forforBasicBasicLocation-AllocationLocation-AllocationModelModel

1

1 1,2,...,n

i ijj

y x i mn

Branch-and-BoundBranch-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


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

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


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 yk value of 1 in S1;

facilities with yk value of 0 in S0; all other facilities in S2. If xkj = 1 for 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


• 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.

1

1 1, 2,...,n

i ijj

y x i mn

Example 6Example 6

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

Example 6Example 6

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

Example 6Example 6

11 22 33 44 5 F5 Fii

20 12 14 12 10 615 10 20 8 15 512 16 25 11 10 3

123

Solution:Solution:Step 1: Set UB = infinity; node counter Step 1: Set UB = infinity; node counter pp = 1; = 1; SS00 = = SS11 = { = {}; }; SS22 = =

{1, 2, 3}.{1, 2, 3}.Step 2: Minimum coefficient rule: Determine the Step 2: Minimum coefficient rule: Determine the xxijij coefficients coefficients

as follows.as follows.

Solution:Solution:

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

=21c11=20+65

15

; c21=15+ =1655

; c31=12+35

= 1235

=13c12=12+65

15

; c22=10+ =1155

; c32=16+35


= 1635

Solution:Solution:

=15c13=14+65

15

; c23=20+ =2155

; c32=25+35


= 2535

=13c14=12+65

15

; c24= 8+ = 955

; c34=11+35


= 1135

Solution:Solution:

=11c15=10+65

15

; c25=15+ =1655

; c35=10+35


= 1035

Solution:Solution:

[x11+x12+x13+x14+x15]=y1=15

[0+0+1+0+0]=15

15

[x21+x22+x23+x24+x25]=y2=15

[0+1+0+1+0]=15

25

[x31+x32+x33+x34+x35]=y3=15

[1+0+0+0+1]=15

25

Solution:Solution:Step 4: Since all three y variables have Step 4: Since all three y variables have

fractional values, go to Step 5.fractional values, go to Step 5.

Solution:Solution:Step 5: Lower bound of node 1 =

0 + 1235

15

+ 11 + 1535

+ 9 + 1025

= 58

Arbitrarily select variable yArbitrarily select variable y11 to branch on. to branch on.Create subproblems 2 and 3 as followsCreate subproblems 2 and 3 as followsSubproblem 2: SSubproblem 2: S00 = {1}; S = {1}; S11 = { = {}; S}; S22 = {2, 3}. = {2, 3}.Subproblem 3: SSubproblem 3: S00 = {1}; S = {1}; S11 = { = {}; S}; S22 = {2, 3}. = {2, 3}.

Solution:Solution:

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

Solution:

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

=16=16c21=15+ 5555

; ; c31=12++ =1235

3355


=11=11cc2222=10+=10+555

; c; c3232=16+=16+ =16=163355

3355

Solution:Solution:


= 9c24=8 +55

; c34=11+ =1135

35


=21c23=20+55

; c33=25+ =2535

35

Solution:Solution:


=16c24=15+55

; c35=10+ =1035

35

Solution:Solution:[x21+x22+x23+x24+x25]=y2=

15

[0+1+1+1+0]=15

35

[x31+x32+x33+x34+x35]=y3=15

[1+0+0+0+1]=15

25

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

Since xSince x1313 = 1 in the solution to subproblem 1, = 1 in the solution to subproblem 1, remove store 1 from consideration in node 3 and remove store 1 from consideration in node 3 and other nodes emanting from node 3. Reduce n by other nodes emanting from node 3. Reduce n by 1, n=5-1=4.1, n=5-1=4.


c11=20; c21 =15+34

; c31=12+=1654

14

Solution:Solution:

=1234


c12=12; c22 =10+34

; c32=16+=1154

14

=1634


c14=12; c24 =8 +34

; c34=11+= 954

14

Solution:Solution:

=1134


c15=10; c25 =15+34

; c35=10+=1654

14

=1034

Solution:Solution:

[x21+x22+x24+x25]=y2=14

[0+1+1+0]=14

24

=12

[x31+x32+x34+x35]=y2=14

[1+0+0+0]=14

14

Set p=1+2=3

Solution: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 12.8Figure 12.8

1

2 3

4 5 6 7

8 9

LB = 58.4y1 = 0.2y2 = y3 = 0.4

LB = 64.2y2 = 0.6; y3 = 0.4

LB = 68y2 = 1y3 = 1Pruned

LB = 77y3 = 1Pruned

LB = 63.25y1 = 1; y2 = 0.5y3 = 0.25

LB = 66.5y2 = 1; y3 = 0.5

LB = 68y1 = y2 = y3 = 1

LB = 74y1 = 1Pruned

LB = 68y1 = 1y3 = 1

12.5.312.5.3Comprehensive Location-Comprehensive Location-Allocation ModelAllocation Model

Comprehensive Location-Allocation Comprehensive Location-Allocation ModelModel

ParametersParametersSSijij production capacity of product production capacity of product ii at plant at plant jjDDilil demand for product demand for product ii at customer zone at customer zone llFFkk fixed cost of operating warehouse fixed cost of operating warehouse kkVVikik unit variable cost of handling product unit variable cost of handling product ii at warehouse at warehouse kkCCijkl ijkl average unit cost of producing and transportingaverage unit cost of producing and transporting product product ii from plant from plant jj via warehouse via warehouse kk to customer to customer llUCUCkk Upper bound on the capacity of warehouse Upper bound on the capacity of warehouse kkLCLCkk Lower bound on the capacity of warehouse Lower bound on the capacity of warehouse kk


Decision VariablesDecision VariablesXXijklijkl number of units of product number of units of product ii transported transported from plant from plant jj via warehouse via warehouse kk to customer to customer ll

1 if warehouse serves customer area 0 otherwisekl

i ly

1 if warehouse is opened at location 0 otherwisek

kz

Model 10Model 10

1 1 1 1 1 1 1 1

Minimize p q pr s s r r

ijkl ijkl il ik kl k ki j k l i l k k

c x D V Y F z

1 1

Subject to 1, 2,..., ; 1,2,...,r s

ijkl ijk l

x S i p j q

1

1, 2,..., ; 1, 2,..., ; 1, 2...,q

ijkl il klj

x D y i p k r l s

1

1 1, 2...,r

klk

y l s

1 1

1, 2,...,p s

il kl k ki l

D y LC z k r

Model 10Model 10

1 1

1, 2,...,p s

il kl k ki l

D y UC z k r

0 1, 2,..., ; 1,2,..., ; 1, 2,..., ; 1, 2,...,ijklx i p j q k r l s

, =0 or 1 1, 2,..., ; 1, 2,...,kl ky z k r l s

Comprehensive Location-Allocation Comprehensive Location-Allocation ModelModelWe can easily add more linear constraints not involving We can easily add more linear constraints not involving

xxijklijkl variables to model 10 to: variables to model 10 to:• Impose upper and lower limit on the number of Impose upper and lower limit on the number of

warehouses that can be opened;warehouses that can be opened;• Enforce precedence relations among warehouses Enforce precedence relations among warehouses

(e.g., open warehouse at location 1 only if another is (e.g., open warehouse at location 1 only if another is opened at location 3)opened at location 3)

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

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

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

Comprehensive Location-Allocation Comprehensive Location-Allocation ModelModelSuppose we fix the values of binary variables Suppose we fix the values of binary variables yyklkl and and zzkk temporarily at 0 temporarily at 0

or 1 so that corresponding constraints are satisfiedor 1 so that corresponding constraints are satisfiedThen, model 10 reduces to the following linear program which we will Then, model 10 reduces to the following linear program which we will

refer to as TPrefer to as TP

( ) ( )1 1 1

Minimize p q s

ijk l l ijk l li j l

c x K

( )1

Subject to 1, 2,..., ; 1, 2,...,s

ijk l l ijl

x S i p j q

( ) ( )1

1, 2,..., ; 1, 2...,q

ijk l l il k l lj

x D y i p l s

( ) 0 1,2,..., ; 1, 2,..., ; 1,2,...,ijk l lx i p j q l s

( ) ( )1 1 1

p s r

il ik l k l l k ki l k

K D V y F z

Comprehensive Location-Allocation Comprehensive Location-Allocation ModelModelTP can be decomposed into TP can be decomposed into ii separate transportation problems, separate transportation problems, TPTPii, as , as

follows because the variables pertaining to a specific product follows because the variables pertaining to a specific product appear only in the rows (constraints) corresponding to that product appear only in the rows (constraints) corresponding to that product and not elsewhere. (Notice that we have temporarily eliminated and not elsewhere. (Notice that we have temporarily eliminated KK from from TPTPii))

( ) ( )1 1

Minimize q s

ijk l l ijk l lj l

c x

( )1

Subject to 1, 2,...,s

ijk l l ijl

x S j q

( ) ( )1

1, 2...,q

ijk l l il k l lj

x D y l s

( ) 0 1,2,..., ; 1,2,...,ijk l lx j q l s

Comprehensive Location-Allocation Comprehensive Location-Allocation ModelModelTP can be decomposed into TP can be decomposed into ii separate transportation problems, separate transportation problems, TPTPii, as , as

follows because the variables pertaining to a specific product follows because the variables pertaining to a specific product appear only in the rows (constraints) corresponding to that product appear only in the rows (constraints) corresponding to that product and not elsewhere. (Notice that we have temporarily eliminated and not elsewhere. (Notice that we have temporarily eliminated KK from from TPTPii))

( ) ( )1 1

Minimize q s

ijk l l ijk l lj l

c x

( )1

Subject to 1, 2,...,s

ijk l l ijl

x S j q

( ) ( )1

1, 2...,q

ijk l l il k l lj

x D y l s

( ) 0 1,2,..., ; 1,2,...,ijk l lx j q l s

Comprehensive Location-Allocation Comprehensive Location-Allocation ModelModelDual of Dual of TPTPii, designated as D, designated as DTPTPii, follows., follows.

( )1 1

Maximize -q s

ij ij il il k l lj l

u S v D y

( )Subject to - 1,2,..., ; 1, 2,...,ij il ijk l lu v c j q l s

0 1,2,..., ; 1, 2,...,iju i p j q

0 1,2,..., ; 1, 2,...,ilv i p l s


Combine Combine pp dual problems into one master problem MP dual problems into one master problem MPMinimizeMinimize TT

( ) ( )1

1, 2,..., ; 1,2...,q

ijk l l il k l lj

x D y i p l s

1 1 1 1 1 1 1 1 1

Subject to p q p pr s s r r

ij ij il il kl il ik kl k ki j i k l i l k k

T u S v D y D V y F z

1

1 1, 2...,r

klk

y l s

1 1

1, 2,...,p s

il kl k ki l

D y LC z k r

1 1

1, 2,...,p s

il kl k ki l

D y UC z k r

, =0 or 1 1, 2,..., ; 1, 2,...,kl ky z k r l s

Modified Benders’ Decomposition Modified Benders’ Decomposition Algorithm for Comprehensive Algorithm for Comprehensive Location-Allocation ModelLocation-Allocation ModelStep 0:Step 0: Set upper bound Set upper bound UBUB=infinity, and convergence =infinity, and convergence

tolerance parameter tolerance parameter εε to a desired small, positive value. Set to a desired small, positive value. Set yyklkl, z, zkk = 0 or 1, for = 0 or 1, for kk=1,2,…,=1,2,…,rr, , ll=1,2,…,=1,2,…,s s so that the resulting so that the resulting values satisfy constraints with values satisfy constraints with yyklkl, z, zkk

Step 1:Step 1: Set up TPSet up TPii, , ii=1,2,…,=1,2,…,pp and determine and determine KK for the for the current values of current values of yyklkl, z, zkk kk=1,2,…,=1,2,…,rr, , ll=1,2,…,=1,2,…,ss. Set up . Set up corresponding dual model DTPcorresponding dual model DTPii for each for each ii. Solve each DTP. Solve each DTPii and add and add KK to the sum of the optimal objective function value to the sum of the optimal objective function value of each DTPof each DTPii. If this sum is less than or equal to . If this sum is less than or equal to UBUB, set , set UBUB = = KK + sum of original OFVs of each DTP + sum of original OFVs of each DTPii

Step 2:Step 2: Set up model MP for the current values of Set up model MP for the current values of uuijij, , vvilil, , ii=1,2,…=1,2,…pp; ; jj=1,2,…,=1,2,…,qq, , ll=1,2,…,=1,2,…,ss. Find a feasible solution to MP . Find a feasible solution to MP such that such that TT<<UB-εUB-ε. If there is no such feasible solution to the . If there is no such feasible solution to the current MP, stop. We have an current MP, stop. We have an εε-optimal solution. Otherwise, -optimal solution. Otherwise, go to step 1 with the current values of the go to step 1 with the current values of the yyklkl, and z, and zkk variablesvariables

Example 7Example 7

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

Example 7Example 7

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

Example 7Example 7Plant 1 2 3 4 W/H 1 2Capacity 200 100 50 500 FC 2000 1500

VC 10 15

Cust 1 2 3 UB 300 400

Demand 100 200 100 LB 0 0

1 2 1 2 3

1 20 10 W/H 1 5 8 3

Plant 2 15 12 2 6 9 10

3 8 16

4 10 14

Example 7 - MIPExample 7 - MIP MODEL:MODEL: [_1] MIN= 1000 * Y_1_1 + 2000 * Y_1_2 + 1000 * Y_1_3 + 1500 * Y_2_1 +[_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_33000 * 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 *+ 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 + 13X_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 +* 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_126 * 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 ;+ 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 ;[_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 ;[_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 ;[_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 ;[_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 ;[_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 ;[_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 ;[_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 ;[_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 ;[_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 ;[_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 ;[_12] Y_1_1 + Y_2_1 = 1 ; [_13] Y_1_2 + Y_2_2 = 1 ;[_13] Y_1_2 + Y_2_2 = 1 ; [_14] Y_1_3 + Y_2_3 = 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 ;[_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 ;[_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 ;[_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 ;[_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_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);@BIN( Y_2_2); @BIN( Y_2_3); @BIN( Z_1); @BIN( Z_2); ENDEND

Example 7 – MIP SolutionExample 7 – MIP Solution Global optimal solution found.Global optimal solution found. Objective value: 12300.00Objective value: 12300.00 Objective bound: 12300.00Objective bound: 12300.00 Infeasibilities: 0.000000Infeasibilities: 0.000000 Extended solver steps: 0Extended solver steps: 0 Total solver iterations: 13Total solver iterations: 13

Variable Value Reduced CostVariable Value Reduced Cost Q 4.000000 0.000000Q 4.000000 0.000000 R 2.000000 0.000000R 2.000000 0.000000 S 3.000000 0.000000S 3.000000 0.000000 CAPACITY( 1) 200.0000 0.000000CAPACITY( 1) 200.0000 0.000000 CAPACITY( 2) 100.0000 0.000000CAPACITY( 2) 100.0000 0.000000 CAPACITY( 3) 50.00000 0.000000CAPACITY( 3) 50.00000 0.000000 CAPACITY( 4) 500.0000 0.000000CAPACITY( 4) 500.0000 0.000000 FIXEDCOST( 1) 2000.000 0.000000FIXEDCOST( 1) 2000.000 0.000000 FIXEDCOST( 2) 1500.000 0.000000FIXEDCOST( 2) 1500.000 0.000000 VARIABLECOST( 1) 10.00000 0.000000VARIABLECOST( 1) 10.00000 0.000000 VARIABLECOST( 2) 15.00000 0.000000VARIABLECOST( 2) 15.00000 0.000000 UPPERBOUND( 1) 3000.000 0.000000UPPERBOUND( 1) 3000.000 0.000000 UPPERBOUND( 2) 4000.000 0.000000UPPERBOUND( 2) 4000.000 0.000000 LOWERBOUND( 1) 0.000000 0.000000LOWERBOUND( 1) 0.000000 0.000000 LOWERBOUND( 2) 0.000000 0.000000LOWERBOUND( 2) 0.000000 0.000000 Z( 1) 1.000000 2000.000Z( 1) 1.000000 2000.000 Z( 2) 0.000000 1500.000Z( 2) 0.000000 1500.000

Example 7 – LPExample 7 – LP MODEL:MODEL: [_1] MIN= 25 * X_1_1_1 + 28 * X_1_1_2 + 23 * X_1_1_3 + 16 * X_1_2_1 + 19[_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 +* 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_218 * 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 *+ 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 + 24X_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 ;* 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 ;[_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 ;[_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 ;[_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 ;[_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 ;[_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 ;[_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 ;[_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 ;[_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 ;[_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 ;[_11] X_1_2_3 + X_2_2_3 + X_3_2_3 + X_4_2_3 >= 100 ; [_12] 0 = 0 ;[_12] 0 = 0 ; [_13] 0 = 0 ;[_13] 0 = 0 ; [_14] 0 = 0 ;[_14] 0 = 0 ; [_15] 0 <= 0 ;[_15] 0 <= 0 ; [_16] 0 <= 3600 ;[_16] 0 <= 3600 ; [_17] 0 >= 0 ;[_17] 0 >= 0 ; [_18] 0 >= - 400 ;[_18] 0 >= - 400 ; ENDEND

Example 7 – LP SolutionExample 7 – LP Solution Global optimal solution found.Global optimal solution found. Objective value: 15500.00 Objective value: 15500.00 UPPER BOUNDUPPER BOUND Infeasibilities: 0.000000Infeasibilities: 0.000000 Total solver iterations: 6Total solver iterations: 6


Example 7 – DualExample 7 – Dual MODEL:MODEL: MAX = 200 * U_1 + 100 * U_2 + 50 * U_3 + 500 * U_4 + 100 * V_4MAX = 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;+ 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_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_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_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_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_1_3] U_2 + V_3 <= 18; [ X_2_2_1] U_2 + V_4 <= 18;[ X_2_2_1] U_2 + V_4 <= 18; [ X_2_2_2] U_2 + V_5 <= 21;[ X_2_2_2] U_2 + V_5 <= 21; [ X_2_2_3] U_2 + V_6 <= 22;[ X_2_2_3] U_2 + V_6 <= 22; [ X_3_1_1] U_3 + V_1 <= 13;[ X_3_1_1] U_3 + V_1 <= 13; [ X_3_1_2] U_3 + V_2 <= 16;[ X_3_1_2] U_3 + V_2 <= 16; [ X_3_1_3] U_3 + V_3 <= 11;[ X_3_1_3] U_3 + V_3 <= 11; [ X_3_2_1] U_3 + V_4 <= 22;[ X_3_2_1] U_3 + V_4 <= 22; [ X_3_2_2] U_3 + V_5 <= 25;[ X_3_2_2] U_3 + V_5 <= 25; [ X_3_2_3] U_3 + V_6 <= 26;[ X_3_2_3] U_3 + V_6 <= 26; [ X_4_1_1] U_4 + V_1 <= 15;[ X_4_1_1] U_4 + V_1 <= 15; [ X_4_1_2] U_4 + V_2 <= 18;[ X_4_1_2] U_4 + V_2 <= 18; [ X_4_1_3] U_4 + V_3 <= 13;[ X_4_1_3] U_4 + V_3 <= 13; [ X_4_2_1] U_4 + V_4 <= 20;[ X_4_2_1] U_4 + V_4 <= 20; [ X_4_2_2] U_4 + V_5 <= 23;[ X_4_2_2] U_4 + V_5 <= 23; [ X_4_2_3] U_4 + V_6 <= 24;[ 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_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);@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);@FREE( W_2); @FREE( W_3); @BND( -0.1E+31, T_1, 0); @BND( -0.1E+31, T_2, 0);@BND( -0.1E+31, T_2, 0); ENDEND

Example 7 – Dual SolutionExample 7 – Dual Solution Global optimal solution found.Global optimal solution found. Objective value: 8000.000 Objective value: 8000.000 + K=7500 =15,500 UPPER BOUND+ K=7500 =15,500 UPPER BOUND Infeasibilities: 0.000000Infeasibilities: 0.000000 Total solver iterations: 7Total solver iterations: 7

Variable Value Reduced CostVariable Value Reduced Cost U_1 -4.000000 0.000000U_1 -4.000000 0.000000 U_2 -2.000000 0.000000U_2 -2.000000 0.000000 U_3 0.000000 -50.00000U_3 0.000000 -50.00000 U_4 0.000000 -400.0000U_4 0.000000 -400.0000 V_4 20.00000 0.000000V_4 20.00000 0.000000 V_5 23.00000 0.000000V_5 23.00000 0.000000 V_6 24.00000 0.000000V_6 24.00000 0.000000 T_2 0.000000 -3600.000T_2 0.000000 -3600.000 T_4 0.000000 400.0000T_4 0.000000 400.0000 V_1 0.000000 0.000000V_1 0.000000 0.000000 V_2 0.000000 0.000000V_2 0.000000 0.000000 V_3 0.000000 0.000000V_3 0.000000 0.000000 W_1 0.000000 0.000000W_1 0.000000 0.000000 W_2 0.000000 0.000000W_2 0.000000 0.000000 W_3 0.000000 0.000000W_3 0.000000 0.000000 T_1 0.000000 0.000000T_1 0.000000 0.000000

Example 7 – Master ProblemExample 7 – Master Problem MODEL:MODEL: [_1] MIN= Z;[_1] MIN= Z; [_2] Z >= -1000 + 2000 * Y_2_1 + 4600 * Y_2_2 + 2400 * Y_2_3 + 2000 * Y_2_1 + [_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 + 4600 * Y_2_2 + 2400 * Y_2_3 + 1000 * Y_1_1 + 2000 * Y_1_2 + 1000 * Y_1_3 + 1500 * Y_2_1 +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 ;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 ;[_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 ;[_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 ;[_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 ;[_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 ;[_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 ;[_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 ;[_10] Y_1_1 + Y_2_1 = 1 ; [_11] Y_1_2 + Y_2_2 = 1 ;[_11] Y_1_2 + Y_2_2 = 1 ; [_12] Y_1_3 + Y_2_3 = 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 ;[_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 ;[_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 ;[_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 ;[_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_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);@BIN( Y_2_2); @BIN( Y_2_3); @BIN( Z_1); @BIN( Z_2); ENDEND

Example 7 – Master Problem SolutionExample 7 – Master Problem Solution Global optimal solution found.Global optimal solution found. Objective value: 5000.000 Objective value: 5000.000 LOWER BOUNDLOWER BOUND Objective bound: 5000.000Objective bound: 5000.000 Infeasibilities: 0.000000Infeasibilities: 0.000000 Extended solver steps: 0Extended solver steps: 0 Total solver iterations: 5Total solver iterations: 5

Variable Value Reduced CostVariable Value Reduced Cost Z 5000.000 0.000000Z 5000.000 0.000000 Y_2_1 0.000000 5500.000Y_2_1 0.000000 5500.000 Y_2_2 0.000000 12200.00Y_2_2 0.000000 12200.00 Y_2_3 0.000000 6300.000Y_2_3 0.000000 6300.000 Y_1_1 1.000000 1000.000Y_1_1 1.000000 1000.000 Y_1_2 1.000000 2000.000Y_1_2 1.000000 2000.000 Y_1_3 1.000000 1000.000Y_1_3 1.000000 1000.000 Z_1 1.000000 2000.000Z_1 1.000000 2000.000 Z_2 0.000000 1500.000Z_2 0.000000 1500.000

Example 7 – LPExample 7 – LP MODEL:MODEL: [_1] MIN= 25 * X_1_1_1 + 28 * X_1_1_2 + 23 * X_1_1_3 + 16 * X_1_2_1 + 19[_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 +* 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_218 * 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 *+ 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 + 24X_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 ;* 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 ;[_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 ;[_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 ;[_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 ;[_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 ;[_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 ;[_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 ;[_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 ;[_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 ;[_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 ;[_11] X_1_2_3 + X_2_2_3 + X_3_2_3 + X_4_2_3 >= 0 ; [_12] 0 = 0 ;[_12] 0 = 0 ; [_13] 0 = 0 ;[_13] 0 = 0 ; [_14] 0 = 0 ;[_14] 0 = 0 ; [_15] 0 <= 2600 ;[_15] 0 <= 2600 ; [_16] 0 <= 0 ;[_16] 0 <= 0 ; [_17] 0 >= - 400 ;[_17] 0 >= - 400 ; [_18] 0 >= 0 ;[_18] 0 >= 0 ; ENDEND

Example 7 – LP SolutionExample 7 – LP Solution Global optimal solution found.Global optimal solution found. Objective value: 12300.00 Objective value: 12300.00 UPPER BOUNDUPPER BOUND Infeasibilities: 0.000000Infeasibilities: 0.000000 Total solver iterations: 6Total solver iterations: 6


Example 7 – DualExample 7 – Dual MODEL:MODEL: MAX = 200 * U_1 + 100 * U_2 + 50 * U_3 + 500 * U_4 + 100 * V_1MAX = 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;+ 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_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_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_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_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_1_3] U_2 + V_3 <= 18; [ X_2_2_1] U_2 + V_4 <= 18;[ X_2_2_1] U_2 + V_4 <= 18; [ X_2_2_2] U_2 + V_5 <= 21;[ X_2_2_2] U_2 + V_5 <= 21; [ X_2_2_3] U_2 + V_6 <= 22;[ X_2_2_3] U_2 + V_6 <= 22; [ X_3_1_1] U_3 + V_1 <= 13;[ X_3_1_1] U_3 + V_1 <= 13; [ X_3_1_2] U_3 + V_2 <= 16;[ X_3_1_2] U_3 + V_2 <= 16; [ X_3_1_3] U_3 + V_3 <= 11;[ X_3_1_3] U_3 + V_3 <= 11; [ X_3_2_1] U_3 + V_4 <= 22;[ X_3_2_1] U_3 + V_4 <= 22; [ X_3_2_2] U_3 + V_5 <= 25;[ X_3_2_2] U_3 + V_5 <= 25; [ X_3_2_3] U_3 + V_6 <= 26;[ X_3_2_3] U_3 + V_6 <= 26; [ X_4_1_1] U_4 + V_1 <= 15;[ X_4_1_1] U_4 + V_1 <= 15; [ X_4_1_2] U_4 + V_2 <= 18;[ X_4_1_2] U_4 + V_2 <= 18; [ X_4_1_3] U_4 + V_3 <= 13;[ X_4_1_3] U_4 + V_3 <= 13; [ X_4_2_1] U_4 + V_4 <= 20;[ X_4_2_1] U_4 + V_4 <= 20; [ X_4_2_2] U_4 + V_5 <= 23;[ X_4_2_2] U_4 + V_5 <= 23; [ X_4_2_3] U_4 + V_6 <= 24;[ 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_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);@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);@FREE( W_2); @FREE( W_3); @BND( -0.1E+31, T_1, 0); @BND( -0.1E+31, T_2, 0);@BND( -0.1E+31, T_2, 0); ENDEND

Example 7 – Dual SolutionExample 7 – Dual Solution Global optimal solution found.Global optimal solution found. Objective value: 6300.000 + K = 6000 = 12300 Objective value: 6300.000 + K = 6000 = 12300 UPPPER BOUNDUPPPER BOUND Infeasibilities: 0.000000Infeasibilities: 0.000000 Total solver iterations: 10Total solver iterations: 10

Variable Value Reduced CostVariable Value Reduced Cost U_1 0.000000 -200.0000U_1 0.000000 -200.0000 U_2 0.000000 -100.0000U_2 0.000000 -100.0000 U_3 -2.000000 0.000000U_3 -2.000000 0.000000 U_4 0.000000 -150.0000U_4 0.000000 -150.0000 V_1 15.00000 0.000000V_1 15.00000 0.000000 V_2 18.00000 0.000000V_2 18.00000 0.000000 V_3 13.00000 0.000000V_3 13.00000 0.000000 T_1 0.000000 -2600.000T_1 0.000000 -2600.000 T_3 0.000000 400.0000T_3 0.000000 400.0000 V_4 0.000000 0.000000V_4 0.000000 0.000000 V_5 0.000000 0.000000V_5 0.000000 0.000000 V_6 0.000000 0.000000V_6 0.000000 0.000000 W_1 0.000000 0.000000W_1 0.000000 0.000000 W_2 0.000000 0.000000W_2 0.000000 0.000000 W_3 0.000000 0.000000W_3 0.000000 0.000000 T_2 0.000000 0.000000T_2 0.000000 0.000000

Example 7 – Master ProblemExample 7 – Master Problem MODEL:MODEL: [_1] MIN= Z;[_1] MIN= Z; [_2] Z >= -1000 + 2000 * Y_2_1 + 4600 * Y_2_2 + 2400 * Y_2_3 + [_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 +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 ;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 + [_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 +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 ;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 ;[_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 ;[_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 ;[_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 ;[_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 ;[_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 ;[_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 ;[_10] Y_1_1 + Y_2_1 = 1 ; [_11] Y_1_2 + Y_2_2 = 1 ;[_11] Y_1_2 + Y_2_2 = 1 ; [_12] Y_1_3 + Y_2_3 = 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 ;[_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 ;[_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 ;[_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 ;[_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_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);@BIN( Y_2_2); @BIN( Y_2_3); @BIN( Z_1); @BIN( Z_2); ENDEND

Example 7 – Master Problem SolutionExample 7 – Master Problem Solution Global optimal solution found.Global optimal solution found. Objective value: 12000.00 Objective value: 12000.00 LOWER BOUNDLOWER BOUND Objective bound: 12000.00Objective bound: 12000.00 Infeasibilities: 0.000000Infeasibilities: 0.000000 Extended solver steps: 0Extended solver steps: 0 Total solver iterations: 43Total solver iterations: 43

Variable Value Reduced CostVariable Value Reduced Cost Z 12000.00 0.000000Z 12000.00 0.000000 Y_2_1 1.000000 1500.000Y_2_1 1.000000 1500.000 Y_2_2 0.000000 3000.000Y_2_2 0.000000 3000.000 Y_2_3 1.000000 1500.000Y_2_3 1.000000 1500.000 Y_1_1 0.000000 2500.000Y_1_1 0.000000 2500.000 Y_1_2 1.000000 5600.000Y_1_2 1.000000 5600.000 Y_1_3 0.000000 2300.000Y_1_3 0.000000 2300.000 Z_1 1.000000 2000.000Z_1 1.000000 2000.000 Z_2 1.000000 1500.000Z_2 1.000000 1500.000

Example 7Example 7

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

Date post:	06-Feb-2016
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