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
S.S. Heragu
Industrial Engineering Department
University of Louisville
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
12.4LOCATION-ALLOCATION MODELS
12.4.1Set Covering Model
12.4.2Uncapacitated Location-Allocation Model
12.4.3Comprehensive Location-Allocation Model
12.5SUMMARY
12.2LOCATIONMODELS
12.2.1Multiple-Facility Problemswith Rectilinear Distances
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
Thus, the transformed linear model is:
(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
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:
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
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
Consider the following objective for the euclidean distance problem.
Multiple-Facility Problemswith Euclidean Distances
Taking the partial derivatives, we get
Multiple-Facility Problemswith Euclidean Distances
Multiple-Facility Problemswith Euclidean Distances
To make sure the denominator is never 0, we add to it. We then get:
Multiple-Facility Problemswith Euclidean Distances
And …
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.
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)
Model 2:
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.
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.
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.
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:
1
1
1
2
2
3
2
3
4
q
p
r
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:
Model 3:
In the following discussion, the supply sources are assumed to have unlimited capacities and total plant capacity is more than total demand (case (i))
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.
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.
Example 4:
Raw Material SourceSupplyOutletDemand
A1000Dallas900
B800Phoenix800
C800Portland600
D700Montreal500
Orlando500
Example 4:
Raw Material Transportation Cost
TODenverAtlantaPittsburgh
FROM
Vancouver4139
Boston885
Miami1229
San Francisco111112
Example 4:
Chemical Trucking Cost
TODallasPhoenixPortlandMontrealOrlando
FROM
Den.2826123030
Atla.102223298
Pitts.1821231821
Da
V
D
Ph
B
Po
A
M
Mo
P
SF
Orl
Setup transportation tableau for Example 4
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.
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
Determine the number of vehicles required to:
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
Model 4:
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
Model 5:
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.
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.
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.
1234567
10412615108
2801560723
3501308659
4911809103
55084100227
6305793027
7859725270
[dij] =
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.
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.
1234567
11101001
21100111
30011110
40011001
50110110
60110110
71101001
[aij] =
The corresponding set covering model is:
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
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:
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
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.
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
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.
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
Modify Model 6 by transforming xij variables and the cij parameter
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
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
to the sum of the coefficients of the xij variables which have taken a value of 1
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.
Subproblem p+1
Subproblem p+2
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.
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.
12345 Fi
1
2
3
2012141210 6
151020815 5
1216251110 3
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.
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.
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.
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.
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
Step 4: Since all three y variables have fractional values, go to Step 5.
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}.
Step 6: Solution of subproblem 2 using minimum coefficient rule: Determine xij coefficients as follows.
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.
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.
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.
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
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.
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.
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.
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
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.
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
Decision Variables
Xijkl number of units of product i transported
from plant j via warehouse k to customer l
We can easily add more linear constraints not involving xijkl variables to model 10 to:
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
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)
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)
Dual of TPi, designated as DTPi, follows.
Combine p dual problems into one master problem MP
MinimizeT
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
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.
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.
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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.