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Teaching Suggestion 9.1: Transportation and Assignment Models.
The linear programming approach is used for both the transportation problem and the assignment problem. Students should be told that the transportation and assignment models are very common, and due to the special structure of these LP models, there are some specialized algorithms that can solve these more efficiently than using LP software. These are discussed in Module 8. The QM for Windows software allows students to solve these problems even without understanding the algorithms in Module 8. Particularly for large problems, using the software for the transportation and assignment models provides easier modeling and data input than does the linear programming approach.
Teaching Suggestion 9.2: Facility Location Problems.
These are an important application of the transportation model and make it easy to compare how a new city will fit into an existing shipping network. It is an application that has intuitive appeal. Both QM for Windows and Excel QM software are easy to run on these problems.
Teaching Suggestion 9.3: Minimal-Spanning Tree Problems
Have students solve the same minimal-spanning tree problem using different starting nodes. This will show students that they will get optimal solutions regardless of the starting point. In most cases, there will be alternate optimal solutions. Ask how students would recognize alternate optimal solutions for the minimal-spanning tree problem.
Teaching Suggestion 9.4: Maximal-Flow and Shortest-Route Problems.
The linear programming approach is used for both the maximal-flow problem and the shortestroute problem. However, there are special algorithms for these network problems that can be used for these types of problems. Module 8 discusses these methods. The QM for Windows software has the Network Module which is used for both of these as well as the minimal spanning tree. Data input for the network approach is very easy for these problems.
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9-1. The transportation model is an example of decision making under certainty where a decision maker knows beforehand exactly what state of nature will occur (see Chapter 3). In transportation problems, this means that the costs of each shipping route, the demand at each destination, and the supply at each source are all known with certainty.
9-2. In a transportation problem, each source-destination pair represents a decision variable. Hence, to determine the total number of decision variables one would need only multiply the number of sources by the number of destinations. Similarly, each source represents a capacity constraint and each destination represents a demand constraint. Hence, to determine the total number of constraints one would need only add the number of sources to the number of destinations.
9-3. A balanced assignment problem is one in which the number of people or sources is equal to the number of jobs or destinations. In a balanced problem, there is one person for each job and one job for each person.
9-4. The transshipment constraints will ensure that the number of units shipped from a transshipment point equal the number of units shipped into that point. Without this type of constraint, an optimal solution might have some units shipped from a transshipment point that doesn’t have any units to ship.
9-5. Planning a trip by automobile is a common example of a shortest-route problem.
9-6. A maximal-flow problem can be modeled as a capacitated transshipment problem in which every point between the source and the sink is a transshipment point with a limited capacity. The objective is to maximize the total flow into the sink.
9-7. Let AD, AE, AF, BD, BE, BF, CD, CE, and CF represent the amounts shipped to Albuquerque, Boston, and Cleveland from Des Moines, Evansville, and Fort Lauderdale respectively. The associated linear program can be formulated as follows:
Minimize costs: 5AD + 8AE + 9AF + 4BD + 4BE + 7BF + 3CD + 3CE + 5CF
subject to:
AD + AE + AF = 200
BD + BE + BF = 200
CD + CE + CF = 300
AD + BD + CD = 300
AE + BE + CE = 150
AF + BF + CF = 250
{All variables} ≥ 0
Solving with the computer we get: Total cost = 3200; AD= 200; AE= 0; AF= 0; BD= 50; BE= 150; BF= 0; CD= 50; CE= 0; and CF = 250.
9-2
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9-8. Let A1, A2, A3, B1, B2, B3, C1, C2, and C3 represent the amounts delivered to projects A, B, and C from plants 1, 2, and 3 respectively. The associated linear program is formulated as follows:
Minimize costs: 10A1 + 12A2 + 9A3 + 4B1 + 5B2 + 7B3 + 11C1 + 8C2 + 6C3
subject to:
A1 + A2 + A3 = 40
B1 + B2 + B3 = 50
C1 + C2 + C3 = 60
A1 + B1 + C1 < 70
A2 + B2 + C2 < 50
A3 + B3 + C3 < 30
{All variables} ≥ 0
The solution is: total cost = $1040; A1 = 20; A2 = 0; A3 = 20; B1 = 50; B2 = 0; B3 = 0; C1 = 0; C2 = 50; and C3 = 10. Multiple optimal solutions exist.
9-9. If the capacity at plant 3 is raised to 60, the only change to the linear program in Problem 9-8 is the right-hand-side of the last constraint. This becomes A3 + B3 + C3 ≤ 60. The new solution would be: total cost = $980; A1 = 40; A2 = 0; A3 = 0; B1 = 30; B2 = 20; B3 = 0; C1 = 0; C2 = 0; and C3 = 60. Multiple optimal solutions exist.
9-10. Let P1, P2, and P3 = the number of units shipped from Pineville to Supply House 1, 2, and 3 respectively. Let R1, R2, and R3 = the number of units shipped from Oak Ridge to Supply House 1, 2, and 3 respectively. Let M1, M2, and M3 = the number of units shipped from Mapletown to Supply House 1, 2, and 3 respectively.
Minimize cost = 3P1 + 3P2 + 2P3 + 4R1 + 2R2 + 3R3 + 3M1 + 2M2 + 3M3
Subject to P1 + P2 + P3 < 25
R1 + R2 + R3 < 40
M1 + M2 + M3 < 30
P1 + R1 + M1 = 30
P2 + R2 + M2 = 30
P3 + R3 + M3 = 35
All variables > 0
The optimal solution is P1 = 0, P2 = 0, P3 = 25, R1 = 0, R2 = 30, R3 = 10, M1 = 30, M2 = 0, M3 = 0, and total cost = $230.
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9-11. Let MCV, MCT, MCJ, MCB, YCV, YCT, YCJ, YCB, PCV, PCT, PCJ, and PCB represent the amounts shipped from Morgantown, Youngstown, and Pittsburgh to Coal Valley, Coaltown, Coal Junction, and Coalsburg respectively. The associated linear program is formulated as follows:
Minimize miles: 50MCV + 30MCT + 60MCJ + 70MCB + 20YCV + 80YCT + 10YCJ +
90YCB + 100PCV + 40PCT + 80PCJ + 30PCB
subject to:
MCV + YCV + PCV = 30
MCT + YCT + PCT = 45
MCJ + YCJ + PCJ = 25
MCB + YCB + PCB = 20
MCV + MCT + MCJ + MCB = 35
YCV + YCT + YCJ + YCB = 60
PCV + PCT + PCJ + PCB = 25
{All variables} ≥ 0
The last three constraints could have been expressed as less-than-or-equal-to constraints rather than equality constraints. Either way, the optimal solution found using the computer is:
total miles = 3100; MCV = 0; MCT = 35; MCJ = 0; MCB = 0; YCV = 30; YCT = 5; YCJ = 25; YCB = 0; PCV = ; 0 PCT = 5; PCJ = 0; and PCB = 20.
9-12. Let HDA, HAT, HDE, PDA, PAT, PDE, MDA, MAT, AND MDE represent the amounts shipped from Houston, Phoenix, and Memphis to Dallas, Atlanta, and Denver. The associated linear program can be formulated as follows:
Minimize costs: 8HDA + 12HAT + 10HDE + 10PDA + 14PAT + 9PDE + 11MDA +
8MAT + 12MDE
subject to:
HDA + PDA + MDA = 800
HAT + PAT + MAT = 600
HDE + PDE + MDE = 200
HDA + HAT + HDE < 850
PDA + PAT + PDE < 650
MDA + MAT + MDE < 300
{All variables} ≥ 0
The optimal solution found using the computer is: total cost = $14,700; HDA = 800; HAT = 50; HDE = 0; PDA = 0; PAT = 250; PDE = 200; MDA = 0; MAT = 300; and MDE = 0.
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9-13. Let RPH, RCL, RCO, DPH, DCL, DCH, PPH, PCL, and PCH represent the tables delivered from Reno, Denver, and Pittsburgh to Phoenix, Cleveland, and Chicago respectively. The associated linear program can then be formulated as follows:
Minimize costs: 10RPH + 16RCL + 19RCH +12DPH + 14DCL + 13DCH + 18PPH +
12PCL + 12PCH
subject to:
RPH + DPH + PPH = 140
RCL + DCL + PCL = 160
RCH + DCH + PCH = 180
RPH + RCL + RCH < 120
DPH + DCL + DCH < 200
PPH + PCL + PCH < 160
All variables ≥ 0 (Non-negativity constraints)
The optimal solution found using computer software is: total cost = $5,700; RPH = 120; RCL = 0; RCH = 0; DPH = 20; DCL = 0; DCH = 180; PPH = 0; PCL = 160; and PCH = 0. Thus, we ship 120 from Reno to Phoenix, 20 from Denver to Phoenix, 160 from Pittsburgh to Cleveland, and 180 from Denver to Chicago.
9-14. Let AW, AX, AY, AZ, BW, BX, BY, BZ, CW, CX, CY, CZ be the number of units shipped from sources A, B, and C to destinations W, X, Y, and Z. The linear program is:
+ 4CY + 7CZ
subject to:
AW + BW + CW = 40
AX + BX + CX = 20
AY + BY + CY = 50
AZ + BZ + CZ = 20
AW + AX + AY + AZ < 55
BW + BX + BY + BZ < 45
CW + CX + CY + CZ < 30
All variables ≥ 0 (Non-negativity constraints)
The optimal solution is AY = 35, AZ = 20, BW = 10, BX = 20, BY = 15, and CW = 30. This has a total cost = 635 cents or $6.35.
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9-15. Let B1H1, B1H2, B1H3, B1H4, B2H1, B2H2, B2H3, B2H4, B3H1, B3H2, B3H3, and B3H4 represent the containers of blood shipped from blood banks 1, 2, and 3 to hospitals 1, 2, 3, and 4 respectively. The linear program is:
Minimize costs: 8B1H1 + 9B1H2 +11B1H3 + 16B1H4 +12B2H1 + 7B2H2 + 5B2H3 +
8B2H4 + 14B3H1 + 10B3H2 + 6B3H3 + 7B3H4
subject to:
B1H1 + B2H1 + B3H1 = 90
B1H2 + B2H2 + B3H2 = 70
B1H3 + B2H3 + B3H3 = 40
B1H4 + B2H4 + B3H4 = 50
B1H1 + B1H2 + B1H3 + B1H4 < 50
B2H1 + B2H2 + B2H3 + B2H4 < 80
B3H1 + B3H2 + B3H3 + B3H4 < 120
{All variables} ≥ 0
Solving this on the computer we get: B1H1 = 50; B1H2 = 0; B1H3 = 0; B1H4 = 0; B2H1 = 10;
B2H2 = 70; B2H3 = 0; B2H4 = 0; B3H1 = 30; B3H2 = 0; B3H3 = 40; and B3H4 = 50. Total cost = $2020.
9-16. Let FH, FB, FP, FD, CH, CB, CP, CD, WH, WB, WP, and WD represent the dollar amounts lent to B. Hall Real Estate from First Homestead Bank, Commonwealth Bank, and Washington Federal Bank towards the purchase of the properties located at Hill Street, Banks Street, Park Avenue, and Drury Lane respectively. The associated linear program can then be formulated as follows:
+ 11WB + 10WP + 9WD
subject to:
FH + CH + WH = 60,000
FB + CB + WB = 40,000
FP + CP + WP = 130,000
FD + CD + WD = 70,000
FH + FB + FP + FD < 80,000
CH + CB + CP + CD < 100,000
WH + WB + WP + WD < 120,000
All variables ≥ 0
The optimal solution is: FH = 30,000; FB = 40,000; FP = 10,000; FD = 0; CH = 30,000; CB = 0; CP = 0; CD = 70,000; WH = 0; WB = 0; WP = 120,000 ; and WD = 0; interest cost = $28,300.
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9-17. There are several ways to formulate this as a linear program. Let RM1, RM2, RM3, and RM4, OM1, OM2, OM3, and OM4 represent the amount produced on regular time and on overtime in months 1-4 respectively. Let E0 = the amount in inventory at the beginning of month 1. Let E1, E2, E3, and E4 be the amount left in inventory at the end of month 1, 2, 3, and 4 respectively. Let B1, B2, B3, and B4 represent the amount bought in month 1, 2, 3, and 4 respectively. The associated linear program can then be formulated as follows:
Minimize cost: 100RM1 + 100RM2 + 100RM3 + 100RM4 + 130OM1 + 130OM2 +
130OM3 + 130OM4 + 10E0 +10E1 + 10E2 + 10E3 + 10E4 + 150B1 + 150B2+ 150B3+
150B4
subject to:
E0 = 40 Beginning inventory
RM1 < 100; RM2 < 100; RM3 < 100; RM4 < 100 regular production
OM1 < 50; OM2 < 50; OM3 < 50; OM4 < 50 overtime production
RM1 + OM1 +B1 + E0 – E1 = 120 demand month 1
RM2 + OM2 + B2 + E1 – E2 = 160 demand month 2
RM3 + OM3 + B3 + E2 – E3 = 240 demand month 3
RM4 + OM4 + B4 + E3 –E4 = 100 demand month 4
B1 + B2 + B3 + B4 < 450
All variables ≥ 0 (Non-negativity constraints)
maximum bought
One optimal solution (there are others) is: RM1 = 100; RM2 = 100; RM3 = 100; RM4 = 100; OM1 = 50; OM2 = 50; OM3 = 50; E0 = 40; E1 = 70; E2 = 60; B3 = 30; all other variables = 0. Total cost = $65700.
9-18. For this analysis, we must consider two linear programs. We first determine what the lowest cost would be if New Orleans is selected. Then we determine what the lowest cost would be if Houston is selected. For both of them let A1, A2, T1, T2, N1, N2, H1, and H2 represent the carriers delivered from Atlanta, Tulsa, New Orleans, and Houston to Los Angeles (1) and New York (2) respectively. The first linear program for the New Orleans option is formulated as follows:
Minimize costs: 14A1 + 11A2 + 9T1 +12T2 + 9N1 + 10N2
subject to:
A1 + A2 < 600
T1 + T2 < 900
N1 + N2 < 500
A1 + T1+ N1 = 800
A2 + T2 + N2 = 1200
All variables ≥ 0
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The optimal solution is: A1 = 0; A2 = 600; T1 = 800; T2 = 100; N1 = 0; N2 = 500; cost = $20,000.The second alternative for Houston is formulated as follows:
Minimize costs: 14A1 + 11A2 + 9T2 +12T2 + 7H1 + 9H2
subject to:
A1 + A2 < 600
T2 + T2 < 900
H1 + H2 < 500
A1 + T2 + H1 = 800
A2 + T2 + H2 = 1200
All variables ≥ 0
The optimal solution is: A1 = 0; A2 = 600 ; T1 = 800; T2 = 100; H1 = 0; H2 = 500; cost = 19,500. The cost for selecting Houston is $500 less than the cost if New Orleans is selected.
9-19. Two linear programs are formulated for this problem. The first will find the least cost solution if the Fontainebleau location is selected. The second will find the least cost solution if the Dublin location is selected. For both of these, the variables are defined as Xij = the number of units produced and shipped from plant i to market j where i = 1, 2, 3, 4, and 5 for Waterloo, Pusan, Bogota, Fontainebleau, and Dublin respectively,
and j = 1, 2, 3, and 4 for Canada, South America, Pacific Rim, and Europe respectively. For the Fontainebleau location, the linear program is
There are multiple optimal solutions to this linear program. All of these have a total cost of $1,530,000. One of these solutions is: X13 = 8000 (Waterloo to Pacific Rim); X23 = 2000 (Pusan to Pacific Rim); X32 = 5000 (Bogota to South America); X41 = 4000 (Fontainebleau to Canada); X44 = 5000 (Fontainebleau to Europe).
All other variables = 0
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For the Dublin location, the linear program is
X11 + X21 + X31 + X51 = 4000
X11 + X22 + X32 + X52 = 5000
X13 + X23 + X33 + X53 = 10000
X14 + X24 + X34 + X54 = 5000
X11 + X12 + X13 + X14 < 8000
X21 + X22 + X23 + X24 < 2000
X31 + X32 + X33 + X34 < 5000
X51 + X52 + X53 + X54 < 9000
All variables > 0
There are multiple optimal solutions to this linear program. All of these have a total cost of $1,535,000. One of these solutions is:
X13 = 8000 (Waterloo to Pacific Rim); X23 = 2000 (Pusan to Pacific Rim); X32 = 5000 (Bogota to South America); X41 = 4000 (Dublin to Canada); X44 = 5000 (Dublin to Europe). All other variables = 0
The Fontainebleau location is $5,000 less expensive than the Dublin location. As a practical matter, changes in exchange rates, subjective factors, or evaluation of future intangibles may overwhelm such a small difference in cost.
9-20. Two linear programs are formulated for this problem. The first will find the least cost solution if the East St. Louis location is selected. The second will find the least cost solution if the St. Louis location is selected. For both of these, the variables are defined as Let Xij = the number of units produced and shipped from plant i to destination j where i = 1, 2, 3, 4, and 5 for Decatur, Minneapolis, Carbondale, East St. Louis, and St. Louis respectively, and j = 1, 2, and 3 for Blue Earth, Ciro, and Des Moines respectively. For the East St. Louis location, the linear program is Minimize
29X41 + 30X42 + 30X43
Subject to
X11 + X21 + X31 + X41 = 250
X12 + X22 + X32 + X42 = 200
X13 + X23 + X33 + X43 = 350
X11 + X12 + X13 < 300
X21 + X22 + X23 < 200
X31 + X32 + X33 < 150
X41 + X42 + X43 < 150
All variables > 0
The optimal solution is: cost = $17,400, X11 = 50, X13 = 250, X21 = 200, X32 = 150, X42 = 50, X43 = 100, and all other variables = 0
For the St. Louis location, the linear program is
27X51 + 28X52 + 31X53
Subject to
X11 + X21 + X31 + X51 = 250
X12 + X22 + X32 + X52 = 200
X13 + X23 + X33 + X53 = 350
X11 + X12 + X13 < 300
X21 + X22 + X23 < 200
X31 + X32 + X33 < 150
X51 + X52 + X53 < 150
All variables > 0
The optimal solution is: cost = $17,250, X
= 100, and all other variables = 0
The cost for St. Louis is $150 lower than the cost using East St. Louis.
9-21. The two linear programs in Problem 9-20 are used again, with the objective functions being changed to include the production cost. The objective function for the East St. Louis
The East St. Louis location provides a lower cost.
9-22. This assignment problem can be solved as a linear program, and each variable in the solution will have a value of either 0 or 1. Numbering jobs A12, A15, B2, and B9 as job 1, 2, 3, and 4 respectively, let
Wi = 1 if machine W is assigned for job i for i = 1, 2, 3, and 4 = 0 otherwise
Xi = 1 if machine X is assigned for job i for i = 1, 2, 3, and 4 = 0 otherwise
Yi = 1 if machine Y is assigned for job i for i = 1, 2, 3, and 4 = 0 otherwise
Zi = 1 if machine Z is assigned for job i for i = 1, 2, 3, and 4 = 0 otherwise
9-10
12Y3 + 18Y4 + 13Z1 + 12Z2 + 11Z3 + 16Z4
Subject to
W1 + W2 + W3 + W4 = 1
X1 + X2 + X3 + X4 = 1
Y1 + Y2 + Y3 + Y4 = 1
Z1 + Z2 + Z3 + Z4 = 1
W1 + X1 + Y1 + Z1 = 1
W2 + X2 + Y2 + Z2 = 1
W3 + X3 + Y3 + Z3 = 1
W4 + X4 + Y4 + Z4 = 1
All variables = 0 or 1
There are multiple optimal solutions. One of these is W1 =1, Z2 = 1, Y3 = 1, X4 = 1, all other variables = 0, and the total time is 50 hours. Thus, the following assignments are made: Job A12 to machine W, Job A15 to machine Z, Job B2 to machine Y, and Job B9 to machine X.
9-23. This table provides the times for each person on each job in this assignment problem.
Let Xij = 1 if person i is assigned to job j where i = 1 for Billy, 2 for Taylor, 3 for Mark, 4 for John, and j = 1, 2, 3, and 4 for the job number, and . Xij = 0 otherwise.
1
All variables = 0 or 1
Solving this using linear programming software, there are multiple optimal solutions. All of these have a total time of 750 minutes. One solution is: X11 = X22 = X33 = X44 = 1, and all other variables equal 0. Thus, Billy is assigned to job 1, Taylor to job2, Mark is assigned to job 3, and John is assigned to job 4.
9-24. This table provides the miles for each possible assignment.
Kansas City (1) Chicago (2) Detroit (3) Toronto (4)
Seattle (1) 1500 1730 1940 2070
Arlington (2) 460 810 1020 1270
Oakland (3) 1500 1850 2080 X
Baltimore (4) 960 610 400 330
Let Xij = 1 if crew at city i is assigned to go to city j where i = 1 for Seattle, 2 for Arlington, 3 for Oakland, 4 for Baltimore, and j = 1, 2, 3, and 4 for KC, Chicago, Detroi, and Toronto respectively, and Xij = 0 otherwise. Because a crew cannot be sent from Oakland to Toronto, this variable is omitted from the formulation. It is also possible to include this variable with a very high cost (miles) so that it would never be selected in the optimal solution.
Minimize miles = 1500X11 + 1730X12 +
Subject to
X11 + X21 + X31 + X41 = 1
X11 + X22 + X32 + X42 = 1
X13 + X23 + X33 + X43 = 1
X14 + X24 + X44 = 1
X11 + X12 + X13 + X14 = 1
X21 + X22 + X23 + X24 = 1
X31 + X32 + X33 = 1
X41 + X42 + X43 + X44 = 1
All variables = 0 or 1
X
+ 460X21 + 810X22 + 1020X23 + 1270X24 + 1500X31 + 1850X32 +
X43 + 330X44
There are multiple optimal solutions to this problem, and all of these have total miles = 4,580. One solution is: X12 = X23 = X31 = X44 = 1, and all other variables equal 0.With this solution, the Seattle crew will go to Chicago, the Arlington crew will go to Detroit the Oakland crew will go to Kansas City, and the Baltimore crew will go to Toronto.
9-25. If the total distance is maximized, the total miles is 6,040. There are multiple optimal solutions, all with total miles of 6,040. One solution is: X12 = X24 = X33 = X41 = 1, and all other variables equal 0. This maximum distance is 1,460 miles more than the minimum distance (4,580).
9-26. Number the professors 1, 2, 3, and 4 for Anderson, Sweeney, Williams, and McKinney respectively, and number the courses similarly with 1, 2, 3, and 4 for statistics, management, finance, and economics respectively. Let Xij = 1 if professor i is assigned to teach course j, and . Xij = 0 otherwise. The linear program is
9-12
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Subject to X11 + X21 + X31 + X41 = 1
X11 + X22 + X32 + X42 = 1
X13 + X23 + X33 + X43 = 1
X14 + X24 + X34 + X44 = 1
X11 + X12 + X13 + X14 = 1
X21 + X22 + X23 + X24 = 1
X31 + X32 + X33 + X34 = 1
X41 + X42 + X43 + X44 = 1
All variables = 0 or 1
There are multiple optimal solutions to this problem, and all of these have total rating = 335. One solution is: X13 = X24 = X31 = X42 = 1, and all other variables equal 0. With this solution, Anderson is assigned to finance, Sweeney is assigned to economics, Williams is assigned to statistics, and McKinney is assigned to management.
9-27 Number the nurses 1, 2, 3, and 4 for Hawkins, Condriac, Bardot, and Hoolihan respectively, and number the departments similarly with 1, 2, 3, and 4 for urology, cardiology, orthopedics, and obstetrics respectively. Let Xij = 1 if nurse i is assigned to department j, and . Xij = 0 otherwise. The linear program is
X13 + X23 + X33 + X43 = 1
X14 + X24 + X34 + X44 = 1
X11 + X12 + X13 + X14 = 1
X21 + X22 + X23 + X24 = 1
X31 + X32 + X33 + X34 = 1
X41 + X42 + X43 + X44 = 1
All variables = 0 or 1
There are multiple optimal solutions to this problem, and all of these have total cost scale = 86. One solution is: X12 = X21 = X33 = X44 = 1, and all other variables equal 0. With this solution, Hawkins is assigned to cardiology, Condriac is assigned to urology, Bardot is assigned to orthopedics, and Hoolihan is assigned to obstetrics.
9-28. Number the time periods 1,
and 4 for the time at the beginning of the hour, and number the networks 1, 2, 3, and 4 for networks A, B, C, and independent respectively. Let Xij = 1 if time period i is scheduled for network j, and . Xij = 0 otherwise. The linear program is
Subject to
X11 + X21 + X31 + X41 = 1
X11 + X22 + X32 + X42 = 1
X13 + X23 + X33 + X43 = 1
X14 + X24 + X34 + X44 = 1
X11 + X12 + X13 + X14 = 1
X21 + X22 + X23 + X24 = 1
X31 + X32 + X33 + X34 = 1
X41 + X42 + X43 + X44 = 1
All variables = 0 or 1
There are multiple optimal solutions to this problem, and all of these have total rating = 75.5. One solution is: X11 = X23 = X32 = X44 = 1, and all other variables equal 0. With this solution, time period 1-2 is assigned to Network A, time period 2-3 is assigned to Network C, time period 3-4 is assigned to Network B, and time period 4-5 is assigned to the independent network.
9-29. If this problem is modeled as a linear program, there will be 7 x 8 = 56 variables and 7 + 8 = 15 constraints. It can also be solved using the Assignment Module in QM for Windows. The costs shown in the table are entered into a similar table in QM for Windows. The optimal solution is:
9-30. This can be modeled as a modified transportation problem with linear programming. Letting the months January through August be denoted as months 1, 2, …, 8, respectively, define the variables as:
Ri = number of units produced on regular time in month i on regular time, for i = 1, 2, …, 8.
Oi = number of units produced in month i on overtime, for i = 1, 2, …, 8.
Si = number of units produced in month i by a subcontractor, for i = 1, 2, …, 8.
Ei = extra units left at the end of month i, for i = 1, 2, …, 8.
Minimize cost = 1000(R1 + R2 + R3 + R4 + R5 + R6 + R7 + R8) + 1300(O1 + O2 + O3 + O4 + O5 +
O6 + O7 + O8) + 1500(S1 + S2 + S3 + S4 + S5 + S6 + S7 + S8)
9-14
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Subject to
R1 + O1 + S1 - E1 = 255
R2 + O2 + S2 + E1 - E2 = 294
R3 + O3 + S3 + E2 - E3 = 321
R4 + O4 + S4 + E3 - E4 = 301
R5 + O5 + S5 + E4 –E5 = 330
R6 + O6 + S6 + E5 –E6 = 320
R7 + O7 + S7 + E6 –E7 = 345
R8 + O8 + S8 + E7 –E8 = 340
R1 < 235; R2 < 255; R3 < 290; R4 < 300; R5 < 300; R6 < 290; R7 < 300; R8 < 290;
O1 < 20; O2 < 24; O3 < 26; O4 < 24; O5 < 30; O6 < 28; O7 < 30; O8 < 30;
S1 < 12; S2 < 15; S3 < 15; S4 < 17; S5 < 17; S6 < 19; S7 < 19; S8 < 20; All variables > 0
a. There are multiple optimal solutions with a cost of $2,591,200. One solution is shown in the table below. In this solution, the regular time production and the overtime production each month are at the limit except for overtime in month 4 (April) in which only 1 unit is produced. The subcontractor is needed in a few of the months.
b. Yes, the solution now costs $2,640,500 with 275 per month in regular time.
c. If overtime rises by $100 per unit to $1,400 per unit, the cost increases, from part a, to $2,610,100. The production plan remains the same as the one for part a. If overtime cost is $1,200 per unit, the total cost is $2,572,100.
9-31. The schedules are found using software for the assignment problem. Here are the three schedules.
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d. This is an easy way to make the assignments. Changes are easily accommodated. One weakness is that astronaut preferences are not initially included in the method. Also, certain combinations of astronauts might have benefits which are not recognized with this system. Factors such as these could be included in a modified assignment system.
9-32. This can be modeled as a maximization assignment problem. Using computer software we obtain the following solution: Erica to Austin/San Antonio, Louis to El Paso/West Texas, Maria to Dallas/Fort Worth, Paul to Houston/Galveston, and Orlando to Corpus Christi/Rio Grande Valley. The total of the rankings is 22.
9-33. There are multiple optimal solutions for this problem, and the total distance is 45. One solution is the following: connect 1–3, 1–4, 3–6, 6–7, 1–2, 4–5, 7–9, 9–8, 9–10, 10–11, 11–13, 13–14, and 14–12. Alternate solutions can be found by substituting 3–4 for 1–4 and substituting 9–12 for 13–14.
9-34. This maximal-flow problem can be modeled and solved as a linear program, or the QM for Windows software can be used for solving this as a maximal-flow network model. The linear program is:
Let Xij = number of units flowing from node i to node j
Maximize flow = X81 subject to
X12 < 2 X13 < 5 X14 < 1 Capacities for arcs from node 1
arcs from node 7
X84 < 0 X86 < 0 X87 < 2 Capacities for arcs from node 8
(X41 + X81) – (X12 + X13 + X14) = 0
(X12 + X52) – (X21 + X25) = 0
Flows into = flows out of node 1
Flows into = flows out of node 2
(X13 + X53 + X63) – (X31 + 0 X35 + X36) = 0 Flows into = flows out of node 3
X14 - (X41 + X46 + X48) = 0
Flows into = flows out of node 4
(X25 + X35 + X75) - (X52 + X53 + X75) = 0 Flows into = flows out of node 5
(X36 + X46) – (X63 + X64 + X68) = 0
(X57 + X87) – (X75 + X78) = 0
Flows into = flows out of node 6
Flows into = flows out of node 7
(X48 + X68 + X78) – (X81 + X84 + X87) = 0
All variables > 0
Flows into = flows out of node 8
The optimal solution with the current system is:
If the capacity increases are both made, the maximum flow would be 7, as the flow along the path 1-4-8 would be 3 instead of 1.
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9-35. The minimum required amount is 177 units of length. Using the minimal-spanning tree method, the following nodes are connected: 1-2, 2-3, 3-4, 3-5, and 5-6.
9-36. This maximal-flow problem can be modeled and solved as a linear program, or the QM for Windows software can be used for solving this as a maximal-flow network model. From the QM for Windows software, the maximum number of cars that can flow from the hotel complex to Disney World is 13 (1,300 cars per hour). The flow along each arc is shown below.
9-37. Solving this as a maximal-flow problem using the QM for Windows software, we find the optimal flow through the network is 3,000 gallons per hour. The flows are shown in the following table.
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If nodes 6 and 7 are shut down and cannot be used, all flow in and out of these nodes is 0. As a result, the flow from the origin to the final network node has been reduced to 2,000 gallons per hour (2). The solution is shown in the following table.
9-38. This is modeled as a shortest route problem and solved using QM for Windows. The shortest route from node 1 to node 16 is the following: 1-3-7-11-14-16. The total distance is 74 kilometers.
The impact of closing two nodes (nodes 7 and 8) is determined by using QM for Windows and either eliminating any arcs going into or out from these nodes, or by assigning an extremely high distance along the arcs. The net effect is that the shortest route from node 1 to node 16 is now the following: 1-2-6-9-13-16. The distance has increased to 76 kilometers.
9-39 a. Grey can use the minimal-spanning tree model to determine the least-cost approach to connect all houses to cable TV. Grey should use branches 1, 2, 3, 4, 6, 8, 9, and 11, which results in a total cost of $3400. When using the minimal-spanning tree method, if the first node selected is node 1, then the branches are selected in the following order: branch 1, 4, 8, 11, 2, 6, 9, 3.
b. With the changes indicated in the problem, the solutions would be to select the same branches as in part a, which would results in a cost of $2000. When using the minimal-spanning tree method, if the first node selected is node 1, then the branches are selected in the following order: branch 2, 3, 4, 8, 11, 1, 6, 9.
9-40. a. This is modeled as a shortest-route problem and solved using QM for Windows. The best way to go from Quincy (node 1) to Old Bainbridge (node 8) is to go through the following nodes: 1-3-5-7-6-8. The total distance of 12 (1,200 miles).
b. The new shortest-route goes through the same nodes as in part a, but the distance is now 10 (1,000 miles).
9-41. This maximal-flow problem can be modeled and solved as a linear program, or the QM for Windows software can be used for solving this as a maximal-flow network model. From the QM for Windows software, path 1-2-4-6-8 has a flow of 10, and path 1-3-5-7-8 has a flow of 5. The two paths provide a total flow rate of 15 hundred gallons.
b. The changes have no impact on the maximal flow or the paths generating this flow.
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9-42. Given the problem data, the network module in QM for Windows gives the following minimal-spanning tree results. The branches 1–3, 3–2, 3–5, 5–4 and 5–6 are used to connect the nodes, and the total distance is 40.
9-43. a. The network module in QM for Windows gives the following shortest route solution: 1-3-7-12-16-20-23-25 with a total distance of 49 (4,900 feet).
b. Eliminating the paths 6–11, 7–12, and 17–20 has changed the shortest route to 55 (5,500 feet). When using the software, these paths can be eliminated, or they can be given a very high value so that they would not be selected. The shortest route is 1-4-8-13-16-20-23-25.
c. In addition to eliminating paths 6-11, 7-12, and 17-20 from the network, the paths used in the solution presented in part b are also eliminated. Thus we eliminate the path 1-4-8-13-16-20-2325. Again, this is done in the software by increasing the distances along these paths to a very high relative value (10,000) to force them out of the solution. The new shortest path is 6400 feet (64). The shortest route is 1-2-5-1-14-18-22-25.
9-44. This transportation problem can be modeled as a linear program with 9 variables and 6 constraints. Let i = 1, 2, and 3 for dress shops A, B, and C respectively.
Wi = number of units shipped from factory W to dress shop i for i = 1, 2, and 3
Xi = number of units shipped from factory X to dress shop i for i = 1, 2, and 3
Yi = number of units shipped from factory Y to dress shop i for i = 1, 2, and 3
All variables > 0
Solving this as with software, we get W2 = 35, X
20, Y
= 10, Y3 = 40, all other variables = 0, with total cost = $505.
9-45. This assignment problem is modeled as a linear program. Let Xij = 1 if person i is assigned to office j where i = 1 for Jones, 2 for Smith, 3 for Wilson, and j = 1 for Omaha, 2 for Miami, and 3 for Dallas. Xij = 0 otherwise.
+ 1000X32 + 2300X33
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Subject to
X11 + X21 + X31 = 1
X11 + X22 + X32 = 1
X13 + X23 + X33 = 1
X11 + X12 + X13 = 1
X21 + X22 + X23 = 1
X31 + X32 + X33 = 1
All variables = 0 or 1
There are multiple optimal solutions to this problem, all with a cost of $2,700. One solution is
X13 = 1 (Jones to Dallas), X21 = 1 (Smith to Omaha), X32 = 1 (Wilson to Miami), all other variables = 0.
9-46. Formulating this assignment problem as a linear program, let
Ai = 1 if customer A is assigned to stand i for i = 1, 2, 3, and 4 = 0 otherwise
Bi = 1 if customer B is assigned to stand i for i = 1, 2, 3, and 4 = 0 otherwise
Ci = 1 if customer C is assigned to stand i for i = 1, 2, 3, and 4 = 0 otherwise
Di = 1 if customer D is assigned to stand i for i =
= 0 otherwise
and
There are multiple optimal solutions for this problem, all with a total distance of 18. One of these solutions is:
3
D to stand 4
All other variables = 0
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9-47. Formulating this assignment problem as a linear program, let
Ai = 1 if case A is assigned to squad i for i = 1, 2, 3, 4, and 5 = 0 otherwise
Bi = 1 if case B is assigned to squad i for i = 1, 2, 3, 4, and 5 = 0 otherwise
Ci = 1 if case C is assigned to squad i for i = 1, 2, 3, 4, and 5 = 0 otherwise
Di = 1 if case D is assigned to squad i for i = 1, 2, 3, 4, and 5 = 0 otherwise
Ei = 1 if case E is assigned to squad i for i = 1, 2, 3, 4, and 5 = 0 otherwise
A1 + B1 + C1 + D1 + E1 = 1
A2 + B2 + C2 + D2 + E2 = 1
A3 + B3 + C3 + D3 + E3 = 1
A4 + B4 + C4 + D4 + E4 = 1
A5 + B5 + C5 + D5 + E5 = 1
All variables = 0 or 1
There are multiple optimal solutions for this problem, all with a total time of 28. One of these solutions is:
A4 = 1 Assign Case A to squad 4
B3 = 1 Assign Case B to squad 3
C1 = 1 Assign Case C to squad 1
D2 = 1 Assign Case D to squad 2
E5 = 1 Assign Case E to squad 5
All other variables = 0
9-48. This assignment problem can be modeled as a linear program, and each variable in the solution will have a value of either 0 or 1. Numbering components C53, C81, D5, and D44 as component 1, 2, 3, and 4 respectively, let
Xij = 1 if component i is produced at plant j, and 0 otherwise for i, j = 1, 2, 3, 4
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X11 + X21 + X31 + X41 = 1
X11 + X22 + X32 + X42 = 1
X13 + X23 + X33 + X43 = 1
X14 + X24 + X34 + X44 = 1
X11 + X12 + X13 + X14 = 1
X21 + X22 + X23 + X24 = 1
X31 + X32 + X33 + X34 = 1
X41 + X42 + X43 + X44 = 1
All variables = 0 or 1
There are multiple optimal solutions to this problem, and all of these have total cost = $0.58. One solution is: X11 = X23 = X34 = X42 = 1, and all other variables equal 0.
9-49. This is modeled as a shortest-route problem and solved using QM for Windows. The shortest route from farm 1 to farm 6 is 1-3-5-6 with a total distance of 17 miles.
9-50. The minimal-spanning tree algorithm is used to find the minimum amount of cable to connect the 6 farms. There are multiple optimal solutions that yield a total distance of 22 miles. One of these solutions starts with node 1 and makes the connections in the following order: 1-2, 2-3, 3-5, 4-5, and 5-6.
This case presents some of the basic concepts of aggregate planning, and it involves solving a rather complex set of transportation problems. These can be modeled as linear programs or by using the transportation method discussed in Module 8. The QM for Windows software can be used to solve these problems.
The minimum cost solution for four different configurations of operating plants have to be determined. The four configurations are (1) all plants operating, (2) only plants 1 and 2 operating, (3) only plants 1 and 3 operating, and (4) only plants 2 and 3 operating. The linear programs for each of these can be developed and solved, although using the QM for Windows transportation module makes the data input and problem formulation much easier. The resulting costs are:
The lowest weekly total cost, operating plants 1 and 3 with 2 closed, is $217,430. This is $3,300 per week ($171,600 per year) or 1.5% less than the next most economical solution, operating all three plants. Closing a plant without expanding the capacity of the remaining plants means unemployment. The optimum solution, using plants 1 and 3, indicates overtime
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production of 4,000 units at plant 1 and 0 overtime at plant 3. The all-plant optima have no use of overtime and include substantial idle regular time capacity: 11,000 units (55%) in plant 2 and either 5,000 units in plant 1 (19% of capacity) or 5,000 in plant 3 (20% of capacity). The idled capacity versus unemployment question is an interesting, nonquantitative aspect of the case and could lead to a discussion of the forecasts for the housing market and thus the plant’s product.
The optimum producing and shipping pattern is
Plant 1 (R.T.) W2 (13,000); W4 (14,000)
Plant 3 (R.T.) W1 (5,000); W3 (11,000); W4 (1,000); W5 (8,000)
Plant 3 (O.T.) W1 (4,000)
There are three alternative optimal producing and shipping patterns, where R.T. = regular time, O.T. = overtime, and W = warehouse.
To find the maximal-spanning tree, a method similar to the minimal spanning tree is employed. At every iteration, choose the arc with the largest (maximum) value instead of the smallest (minimum) as in the minimal-spanning tree method.
The solution to the maximal-spanning tree problem for Northeastern Airlines is found by selecting the routes (arcs) listed below. The maximum total is $134 per passenger. The State College PA location was arbitrarily chosen as the starting point. The profit per passenger is shown in parentheses.
1. State College, PA to Hartford, CT (16)
2. Hartford, CT to Nashua, NH (22)
3. Nashua, NH to Boston, MA (18)
4. Boston, MA to Burlington, VT (21)
5. Hartford, CT to Newark, NJ (17)
6. State College to Syracuse, NY (15)
7. Nashua, NH to Orono, ME (13)
8. Hartford, CT to Providence, RI (12)
Note that this maximal-spanning tree flight service plan meets the FAA requirements of serving all of the nine cities with air service. However, it might not be the air service that everyone would want. For instance, to fly from Syracuse, NY to Burlington, VT on the maximal-spanningtree flight service plan would take five separate flights (Syracuse, NY to State College, PA to Hartford, CT to Nashua, NH to Boston, MA to Burlington, VT), whereas its only about a five hour drive through the Adirondack Mountains from Syracuse, NY to Burlington, VT!
As for Northeastern Airlines’ fleet of 16 planes, any spanning tree (minimal, maximal, or otherwise) for an underlying n node network will have (n-1) arcs. Hence, Northeastern’s maximal- spanning tree flight service plan has eight routes. If the demand for each of these routes is fairly uniform, then it would make sense to assign two planes per route. However, if the demands for these eight routes vary widely, then a more intricate approach may be needed. One
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such approach would be to assign one plane per route (to ensure that all eight routes are served) and then divvy up the remaining eight planes by the relative demands per route (e.g., the route with highest demand would have the most planes). Another approach one might take is to note the “hubs” inherent to the maximal-spanning tree service plan at Hartford, CT (which has flights to four cities) and Nashua, NH (which has flights to three cities). One could assign eight of the planes to each of them. The key here is for students to take into account the route demands in the allocation of the planes.
This is a maximal-flow problem, and the linear programming formulation can be used. However, using the network module in QM for Windows simplifies the problem considerably. This software provided the following results:
1. The capacity without any expansion is 28 (thousand) cars per hour. This would indicate that a serious problem will exist if there are 33,000 cars per hour leaving the stadium. The problem is not leaving node 1 or going into node 8. At node 2, the outflow is only 12 whereas the capacity from 1 to 2 is 15. Also, the capacity from 1 to 3 is only 12, but the capacity leaving 3 is 14. A similar problem exists at node 4. The total capacity from nodes 2, 3, and 4 is only 31. Thus, the problem is a mismatch of the capacity going into the nodes with the capacity leaving these nodes.
2. To get the capacity to 33, we must add an additional 5 units. You could add 3 units of capacity from node 1 to node 4. This matches the inflow to the outflow at node 4. Also, expanding the capacity from node 5 to node 8 by 2 will result in the total capacity being increased by 2. These changes will increase the total capacity to 33. To increase the capacity to 35, the capacity from node 5 to node 8 should be increased by an additional 2 units (for a total of 20). Also, the capacity from node 2 to node 5 should be increased by 2.
To determine whether the shipping pattern can be improved and where the two new plants should be located, the total costs for the entire transportation system for each combination of plants, as well as the existing shipping pattern costs, will have to be determined. In the headings identifying the combination being discussed, Gary and Fort Wayne will be omitted since they appear in every possible combination.
The costs and capacities for the current (Gary and Fort Wayne) and potential plants (Detroit, Madison, and Rockford) are shown below. The demands are also given. These are used in modeling this situation to find optimal shipping patterns.
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*Since a plant at Detroit could purchase a gallon of fiberglass for $2 less than any other plant, and one Shower-Rific takes 2 gallons of fiberglass, a systems approach to transportation warrants that $2(2), $4, be deducted from each price quoted in the case for shipments from Detroit.
**Since a plant at Madison could hire labor for $1 less per hour than the other plants, and one Shower-Rific takes 3 labor hours to build, $1(3) or $3 should be deducted from each price quoted for shipments from Madison.
The existing shipping pattern is shown in the table below.
With this, the cost is: 200(10) + 50(30) + 100(40) + 100(15) = $9,000
The optimum (lowest cost) shipping schedule is shown in the table below. Note that Milwaukee and Minneapolis would receive nothing.
With this, the cost is: 300($10) + 150($15) = $5,250.
The following tables provide the optimal shipping assignments in the three possible scenarios: (1) Detroit and Madison plants built; (2) Madison and Rockford plants built; and (3) Detroit and Rockford plants built. The costs are provided in each scenario.
Optimum Shipping Pattern with Detroit and Madison Plants
Cost = 200(10) + 100(20) + 100(2) + 100(40) + 50(22) + 50(15) + 150(1) = $10,200
Optimum Shipping Pattern with Madison and Rockford Plants
Cost = 250(10) + 50(5) + 100(2) + 50(22) + 100(30) + 50(25) + 150(15) = $10,550
Optimum Shipping Pattern with Detroit and Rockford Plants
costs = 200(10) +
+ 100(20) + 50(15) + 150(1) + 150(30) = $11,400
A summary of the total costs is listed below:
Detroit–Madison = $10,200
Madison–Rockford = $10,550
Detroit–Rockford = $11,400
Since the total cost is lowest in the Gary–Fort Wayne–Detroit–Madison combination ($10,200), the new plants should be located in Detroit and Madison. This system is also an improvement over the existing pattern, which costs $9,000, on a cost-per-unit basis.
Status quo: $9,000/450 units = $20/unit
Proposed: $10,200/750 units = $13.60/unit
Thus the two new plants would definitely be advantageous, both in satisfying demand and in minimizing transportation costs.
1. The minimum distance that will connect all houses to the water and sewer lines is 10,000 feet (100). The solution along with the final network follows:
2. Moving footprint number 16 to accommodate the expansion of the pond area has increased the minimum total distance to 10,100 feet (101). A decision now has to be made about whether the increased distance and cost for the water and sewer system is worth the additional expected property prices. The solution along with the final network follows.
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1. This assignment problem can be modeled using linear programming or by using the special purpose assignment algorithm in Module 8. The objective is to minimize the time to complete a table with each person assigned one task.
2. If Randy is used, then one person will be unassigned. The optimum assignment would be
This is a savings of 10 minutes with Cathy becoming the backup.
3. If Cathy is given the preparation task, the solution of the assignment with the remaining three workers assigned to the remaining three tasks is
If Cathy is assigned to the finishing task, the optimum assignment is
4. One possibility would be to combine the packaging operation with finishing. Then, George could build an entire table by himself (in 230 minutes) and Tom could do preparation (100 minutes), Randy the assembly (80 minutes), and Leon the finishing and packaging (90 minutes). This crew could build 4.8 tables in a 480-minute workday, while George himself could build 2.09 tables—a total of almost 7 tables per day.
To utilize all five workers, George and Tom could each build entire tables, 2.09 and 1.75 per day, respectively. Letting Randy do preparation (110 minutes), Cathy the assembly (70 minutes), and Leon the finishing and packaging (90 minutes) allows an additional 4.36 tables per day for a total of 8.2 per day.
Nine tables per day could be achieved by having Tom prepare and assemble 3 tables, George prepare and finish 3 tables, Cathy assemble 6 tables, Leon finish 6 tables, and Randy prepare 3 tables and package all 9. George, Cathy, and Randy would each have 60 minutes per day unutilized and could build 0.6 table having George do preparation (80 minutes), Cathy assembly and packaging (95 minutes), and Randy the finishing (100 minutes).
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This is a shortest-route problem. With the data given in the problem, the shortest-route model can be used to determine the minimum time in minutes required to go from the plant to the warehouse in east Denver. The results are on the next page. As you can see, the best route is to take North Street to I-70. At Exit 137, South Street is taken to the warehouse. This route takes one hour (60 minutes).
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