ch05_ism

26

SOLUTIONS TO DISCUSSION QUESTIONS

AND PROBLEMS

5-1. The transportation model is an example of decision makingunder certainty since the costs of each shipping route, the demandat each destination, and the supply at each source are all knownwith certainty.

5-2. A balanced transportation problem is one in which total de-mand (from all destinations) is exactly equal to total supply (fromall sources). If a problem is unbalanced, either the demand or thesupply constraints must be inequalities.

5-3. The enumeration method is not a practical means of solving5 � 5 or 7 � 7 problems because of the number of possible as-signments to be considered. In the 5 � 5 case, 5! (� 5 � 4 � 3 �2 � 1) � 120 alternatives need to be evaluated. In the 7 � 7 case,there are 7! � 5,040 alternatives.

5-4. The minimal-spanning model is one that will find the bestway to connect all the nodes in a network together while minimiz-ing the total distance between nodes or the total cost of connectingthe nodes together. A number of decision modeling problems canbe solved using this model: an example was given connectingwater and power to a real estate development project. This modelcan also be used to determine the best way to deliver cable TV tohouseholds, connect computers on a computer network, install anoil pipeline, develop a natural gas network, and more.

5-5. The maximal-flow model can be used to determine the max-imum number of cars that can flow through a road system, thenumber of gallons of chemicals that can flow through a chemicalprocessing plant, the barrels of oil that can go through a pipelinenetwork, the number of people that can use public transportationto get to work, the number of pieces of mail that can go through amail service, and more. Any time that material or items flowthrough a network, the maximal-flow model can be used.

5-6. The shortest-route model can be used to find the best way toinstall a phone cable between two major cities. Any time itemsmust be moved from one place to another or something, like acable, must be used to connect two points, the shortest-routemodel can be used.

5-7. A flow balance constraint calculates the net flow at a node(that is, the difference between the total flow on all arcs entering thenode and the total flow on all arcs leaving the node). At each sourcenode, the net flow is expressed as a negative quantity, and representsthe amount of flow created at the node. At each destination node,

the net flow is expressed as a positive quantity, and represents theamount of flow consumed at the node. At each pure transshipmentnode, the net flow is zero.

5-8. To set up a maximal-flow problem as an LP problem, wecreate a unidirectional dummy arc going from the destination nodeto the source node, and set the capacity of this arc at infinity.

5-9. For many network models, the number of arcs (each ofwhich corresponds to a decision variables) could be quite large. It may therefore be convenient to model these problems in Excelin such a way that decision variables are in a tabular form. For ex-ample, rows in the table could denote starting nodes for the arcs,and columns could denote ending nodes.

5-10. To specify the entire table as the Changing Cells in Solverfor a maximal-flow network model, we assign a capacity of zerofor all arcs that do not actually exist. This will prevent any flow onthose arcs.

5-11. To specify the entire table as the Changing Cells in Solverfor a shortest-path network model, we assign a unit flow cost of in-finity (or some arbitrarily high number) for all arcs that do not ac-tually exist. This will prevent any flow on those arcs.

5-12. Let Xij � number of students bused from sector i to school j

Objective: minimize total travel miles �

5XAB � 8XAC � 6XAE

� 0XBB � 4XBC � 12XBE

� 4XCB � 0XCC � 7XCE

� 7XDB � 2XDC � 5XDE

� 12XEB � 7XEC � 0XEE

subject to

XAB � XAC � XAE � 700 (number students in sector A)

XBB � XBC � XBE � 500 (number students in sector B)

XCB � XCC � XCE � 100 (number students in sector C)

XDB � XDC � XDE � 800 (number students in sector D)

XEB � XEC � XEE � 400 (number students in sector E)

XAB � XBB � XCB � XDB � XEB � 900 (school B capacity)

XAC � XBC � XCC � XDC � XEC � 900 (school C capacity)

XAE � XBE � XCE � XDE � XEE � 900 (school E capacity)

All Xij � 0

Note that this is an unbalanced problem since total students �total school capacity

5C H A P T E R

Transportation, Assignment, and Network Models

6634 CH05 UG 8/23/02 2:10 PM Page 26

CHAPTER 5 TRANSPORTAT ION ASSIGNMENT AND NETWORK MODELS 27

Solution: XAB � 400 (See file P5-12.XLS)

XAE � 300

XBB � 500

XCC � 100

XDC � 800

XEE � 400

Distance � 5,400 “student miles”

5-13(a). Minimize 9X11 � 8X12 � 7X13 � 7X21 � 11X22 �6X23 � 4X31 � 3X32 � 12X33

subject to

X11 � X12 � X13 � 1,500

X21 � X22 � X23 � 1,750

X31 � X32 � X33 � 2,750

X11 � X21 � X31 � 2,000

X12 � X22 � X32 � 3,000

X13 � X23 � X33 � 1,000

All Xij � 0

The solution (see worksheet “a” in file P5-13.XLS) is:

Dubuque to Job 1 – 250 tons

Dubuque to Job 2 – 250 tons

Dubuque to Job 3 – 1,000 tons

Davenport to Job 1 – 1,750 tons

Des Moines to Job 2 – 2,750 tons

Total cost $31,750

(b) The formulation should be expanded to include two addi-tional decision variables: Davenport to Des Moines, andDubuque to Des Moines. The revised solution (see work-sheet “b” in file P5-13.XLS) is:

Dubuque to Des Moines – 1,500 tons

Davenport to Job 3 – 1,000 tons

Davenport to Des Moines – 750 tons



Total cost $27,500. Marc Smith saves $4,250 byconsolidating shipping at Des Moines.

5-14. Krampf’s problem is a balanced transportation problemsince total supply of cars equals the total demand. The Excel layoutand solution is shown in file P5-14.xls. The optimal solution is:

Morgantown to Coaltown – 35 cars

Youngstown to Coal Valley – 30 cars

Youngstown to Coaltown – 5 cars

Youngstown to Coal Junction – 25 cars

Pittsburgh to Coaltown – 5 cars

Pittsburgh to Coalsburg – 20 cars

Total distance � 3,100 miles

5-15. The optimal solution to the Hall Real Estate decision isshown in the table below. (See file P5-15.XLS)

The total interest cost would be $28,300, or an average rate of9.43%. An alternative optimal solution exists. It is

First Homestead–Hill Street 30,000First Homestead–Banks Street 40,000First Homestead–Park Avenue 10,000Commonwealth–Hill Street 30,000Commonwealth–Drury Lane 70,000Washington Federal–Park Avenue 120,000

5-16. Mehta’s production smoothing problem is a good exercisein the formulation of transportation problems and applying themto real-world issues. The problem may be set up as a transporta-tion model as shown in the table. All squares with X’s representnonfeasible (backorder) solutions. In applying an LP model tosolve such a problem, a very large cost (say about $5,000) wouldbe assigned to each of these squares. This would assure that theywould not appear in the final solution.

The optimal solution has a cost of $65,700. (See file P5-16.XLS)

TO Drury Max.FROM Hill St. Banks St. Park Ave. Lane Avail.

8% 8% 10% 11%First Homestead $40,000 $40,000 $ 80,000

9% 10% 12% 10%Commonwealth $60,000 $40,000 $100,000

9% 11% 10% 9%Washington Federal $90,000 $30,000 $120,000

Loan Needed $60,000 $40,000 $130,000 $70,000 $300,000

6634 CH05 UG 8/23/02 2:10 PM Page 27

28 CHAPTER 5 TRANSPORTAT ION ASSIGNMENT AND NETWORK MODELS

5-17. To determine which new plant will yield the lowest costfor Ashley in combination with the existing plants, we need tosolve two transportation problems. We begin by setting up a trans-portation table that represents the opening of the third plant inNew Orleans (see the table). You should note that the cost of eachindividual “plant to distribution center” route is found by addingthe distribution costs to the respective unit production costs. Thus the total production plus shipping cost of one auto top carrierfrom Atlanta to Los Angeles is $14 ($8 for shipping plus $6 forproduction).

Table for Problem 5-17

If Ashley selects to open the New Orleans plant, the firm’s totaldistribution system cost will be $20,000. (See file P5-17.XLS)

If the Houston plant site is chosen, the table is as follows:

If Ashley selects to open the Houston plant, the total cost willbe $19,500 (See file P5-17.XLS).

Upon comparing total costs for the Houston option ($19,500)to those for the New Orleans option ($20,000), we would recom-mend to Ashley that all factors being equal, the Houston siteshould be selected.

TO ProductionFROM Los Angeles New York Capacity

$14 $11Atlanta 600

$9 $12Tulsa 900

$9 $10New Orleans 500

Demand 800 1,200 2,000

Table for Problem 5-16

Destination (Month)

Sources 1 2 3 4 Capacity

10 20 30 40Beginning inventory 40 40

100 110 120 130Regular prod. (month 1) 80 20 100

130 140 150 160Overtime (month 1) 50 50

100 110 120Regular prod. (month 2) 90 10 100

130 140 150Overtime (month 2) 50 50

100 110Regular prod. (month 3) 100 100

130 140Overtime (month 3) 50 50

100Regular prod. (month 4) 100 100

130Overtime (month 4) 50

150 150 150 150Outside purchases 30 450

Demand 120 160 240 100 1,090

TO ProductionFROM Los Angeles New York Capacity

$14 $11Atlanta 600

$9 $12Tulsa 900

$7 $9Houston 500

Demand 800 1,200 2,000

unit cost

unit cost

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5-18. Considering Fontainebleau, we have (See file P5-18.XLS)

Optimal cost � $1,530,000.

Considering Dublin, we have the following final solution: (See file P5-18.XLS)

Optimal cost � $1,535,000. (See file P5-18.XLS)

There is no difference in the routing of shipments, but theFontainebleau location is $5,000 less expensive than the Dublinlocation. As a practical matter, changes in exchange rates, subjec-tive factors, or evaluation of future intangibles may overwhelmsuch a small difference in cost.

5-19. See file P5-19.XLS.

If we open the new plant in East St. Louis, the optimal solution is:

Decatur to Blue Earth – 50 units

Decatur to Des Moines – 250 units

Minneapolis to Blue Earth – 200 units

Carbondale to Ciro – 150 units

East St. Louis to Ciro – 50 units

East St. Louis to Des Moines – 100 units

Total cost = $17,400

If we open the new plant in St. Louis, the optimal solution is:




Carbondale to Des Moines – 50 units

St. Louis to Blue Earth – 50 units

St. Louis to Ciro – 100 units


Therefore, St. Louis is $150 per week cheaper than East St. Louis.

5-20. The Excel set up and solution for the revised problem is shown in file P5-20.XLS. The revised solution for the East St. Louis plant is:

Decatur to Blue Earth – 50 units




Carbondale to Des Moines – 100 units

East St. Louis to Carbondale – 150 units


Therefore, with the new shipping option, East St. Louis is $400per week cheaper than St. Louis.

South PacificCanada America Rim Europe Capacity

60 70 75 75Waterloo 4,000 4,000 8,000

55 55 40 70Pusan 2,000 2,000

60 50 65 70Bogota 5,000 5,000

75 80 90 60Fontainebleau 4,000 5,000 9,000

Market Demand 4,000 5,000 10,000 5,000 24,000

South PacificCanada America Rim Europe Capacity

60 70 75 75Waterloo 4,000 4,000 8,000

55 55 40 70Pusan 2,000 2,000

60 50 65 70Bogota 5,000 5,000

70 75 85 65Dublin 4,000 5,000 9,000

Market Demand 4,000 5,000 10,000 5,000 24,000

shipping quantity unit cost

shipping quantity unit cost

6634 CH05 UG 8/23/02 2:10 PM Page 29


5-21. Assignment can be made as follows (See file P5-21.XLS)

Job A12 to machine WJob A15 to machine ZJob B2 to machine YJob B9 to machine XTime � 10 � 12 � 12 � 16 � 50 hours

5-22. Let Xij � 1 if pitcher i is scheduled to go against opponent j,0 otherwise

where i � 1, 2, 3, 4 stands for Jones, Baker, Parker, andWilson, respectively, and

j � 1, 2, 3, 4 stands for Des Moines, Davenport,Omaha, and Peoria, respectively.

Objective: maximize overall probability of winning � sum ofprobability of winning each game �

0.6X11 � 0.8X12 � 0.5X13 � 0.4X14

� 0.7X21 � 0.4X22 � 0.8X23 � 0.3X24

� 0.9X31 � 0.8X32 � 0.7X33 � 0.8X34

� 0.5X41 � 0.3X42 � 0.4X43 � 0.2X44

subject to

X11 � X12 � X13 � X14 � 1 (“Dead-Arm” Jones)

X21 � X22 � X23 � X24 � 1 (“Spitball” Baker)

X31 � X32 � X33 � X34 � 1 (“Ace” Parker)

X41 � X42 � X43 � X44 � 1 (“Gutter” Wilson)

X11 � X21 � X31 � X41 � 1 (Des Moines)

X12 � X22 � X32 � X42 � 1 (Davenport)

X13 � X23 � X33 � X43 � 1 (Omaha)

X14 � X24 � X34 � X44 � 1 (Peoria)

All Xij � 0

Solution: X12 � 1, X23 � 1, X34 � 1, X41 � 1, Total P � 2.9 (See file P5-22.XLS) 5-27 Minimal-spanning tree model. The optimal solution is

shown by the bold arcs. Total length � 4500 feet.

5-23 Optimal Solution: (See file P5-23.XLS)

5-24. Optimal assignment: (See file P5-24.XLS)

taxi at post 1 to customer Ctaxi at post 2 to customer Btaxi at post 3 to customer Ataxi at post 4 to customer D

Total distance traveled � 4 � 4 � 6 � 4 � 18 miles.

5-25. Optimal assignment: (See file P5-25.XLS)

squad 1 to case Csquad 2 to case Dsquad 3 to case Bsquad 4 to case Asquad 5 to case E

Total person-days projected using this assignment � 3 � 6 � 3 �8 � 8 � 28 days.

5-26 Optimal Solution: (See file P5-26.XLS)

5-28 Optimal solution is 500 cars per hour. See file P5-28.XLS.(Assuming all figures are in hundreds of cars)

Assignment Rating

Anderson—finance 95Sweeney—economics 75Williams—statistics 85McKinney—management 380

Total rating 335

Assignment Rating

Hawkins to cardiology 18Condriac to urology 32Bardot to orthopedics 24Hoolihan to obstetrics 12

Total “cost scale” 86

31

2

4

5

4

2

3

3

5

6

7

4

3 3

6 5

4

77

3 5

6 8 12

1

2

47 9

14

13

11

105

6

FlowPath (Cars/Hour)

1–2–5–7–8 2001–3–6–8 2001–4–8 100Total 500

6634 CH05 UG 8/23/02 2:10 PM Page 30


5-29 The shortest route is 1–3–5–7–10–13. The distance is 430miles. See file P5-29.XLS.

5-30. Minimal-spanning tree model. This is the only optimumsolution to this problem (177 units of length).

5-31. There are several possible solutions for this maximal flowproblem: One solution is presented in file P5-31.XLS. The solu-tion may be interpreted as:

1–4–6 401–2–5–6 551–3–5–6 451–4–5–6 27

167 widgets per day

Alternative solutions: Substitute 1–2–4–6 for 32 in lieu of 1–4–6or 1–4–5–6 (or for some portion of the 32).

5-32. See file P5-32.XLS. No, the changes do not have an im-pact on the final solution. With the changes, the optimal solutionstill has a shortest distance of 430 miles. The final network isgiven below. Note that we have increased the value for the paths 6–9 and 8–9 to a very high relative number (5,000 in ourExcel model) to ensure that these paths are forced out of the finalsolution.

5-33. The maximum number of cars that can flow from the hotelcomplex to Disney World is 13 (1,300 cars per hour). See file P5-33.XLS for the Excel solution. The solution is:

5-34. The impact of the construction project to increase the roadcapacity around the outside roads from International Drive to Dis-ney World would increase the number of cars per hour to 1,700per hour (17). The increase is 400 cars per hour as would be ex-pected. The solution shown in file P5-34.XLS, is as follows:

5-35. Solving this maximal flow problem results in a situationwhere 3,000 gallons per hour (3) will be flowing from the origin tothe final network node. The solution is shown in file P5-35.XLSand is as follows:

1 3 5

4

62

50

37

26

41

23

Flow

1–2 31–3 81–4 22–6 33–7 84–8 26–9 37–10 88–10 29–11 3

10–11 10

Total maximum flow: 13.

Flow

1–2 11–3 11–4 12–5 13–6 14–8 15–9 16–10 18–11 19–10 1

10–12 211–13 113–14 3

Total maximum flow: 3.Alternate solutions are possible.

Flow

1–2 51–3 81–4 21–5 22–6 53–7 84–8 25–8 26–9 57–10 88–10 49–11 5

10–11 12


6634 CH05 UG 8/23/02 2:10 PM Page 31


5-36. The impact of the emergency repair is that nodes 6 and 7cannot 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 re-duced to 2,000 gallons per hour (2). The solution is shown in thefollowing table. Note that flows leading to and from nodes 6 and 7have been changed to 0. See file P5-36.XLS.

5-37. The shortest route from node 1 to node 16 is 74 kilometers.The solution along with the final network is shown in the follow-ing table and in file P5-37.XLS.

5-38. The impact of closing two nodes (nodes 7 and 8) is to in-crease the shortest route from 74 to 76 kilometers. Note that allpaths into and from nodes 7 and 8 have their values changed to avery high relative number (100) to force these paths out of thefinal solution. The solution along with the final network is given inthe following table and in file P5-38.XLS.

5-39. Grey can use the minimal-spanning tree model to deter-mine the least-cost approach to connect all houses the cable TV.The bold arcs in the figure indicate the selected arcs. The total costis $3,400.

5-40. The solution to the minimal-spanning tree problem resultsin a minimum distance of 21 (2,100 yards). The final network fol-lows. Arcs in bold should be selected.

SOLUTION TO OLD OREGON WOOD STORE CASE

1. The assignment algorithm can be utilized to yield the fastesttime to complete a table with each person assigned one task. Seesheet #1 in file P5-Oregon.XLS.

2. If Randy is used, the optimum assignment would be (see sheet#2 in file P5-Oregon.XLS)

This is a savings of 10 minutes with Cathy becoming the backup.

Flow

1–2 11–4 12–5 14–8 15–9 18–11 19–10 1

10–13 111–13 113–14 2


Distance

1–3 153–7 117–11 18

11–14 1614–16 14

Shortest path:1–3–7–11–14–16

Total shortest distance: 74.

Distance

1–2 202–6 106–9 129–13 16

13–16 18

Shortest path:1–2–6–9–13–16

Total shortest distance: 76.

1

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7

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7

5

6

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7 9

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8 93

3

3

34

44

4

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8

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2 2

TimePerson Job (Minutes)

Tom Preparation 100Cathy Assembly 70George Finishing 60Leon Packaging 210

Total time 240


George Preparation 80Tom Assembly 60Leon Finishing 80Randy Packaging 210

Total time 230

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3. If Cathy is given the preparation task, the solution of the as-signment with the remaining three workers assigned to the remain-ing three tasks is (see sheet #3A in file P5-Oregon.XLS)

If Cathy is assigned to the finishing task, the optimum assign-ment is (See sheet #3B in file P5-Oregon.XLS)

4. One possibility would be to combine the packaging operationwith 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 andpackaging (90 minutes). This crew could build 4.8 tables in a 480-minute workday, while George himself could build 2.09 tables—atotal of almost 7 tables per day.

To utilize all five workers, George and Tom could each buildentire tables, 2.09 and 1.75 per day, respectively. Letting Randydo preparation (110 minutes), Cathy the assembly (70 minutes),and Leon the finishing and packaging (90 minutes) allows an addi-tional 4.36 tables per day for a total of 8.2 per day.

Nine tables per day could be achieved by having Tom pre-pare 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 eachhave 60 minutes per day unutilized and could build 0.6 table hav-ing George do preparation (80 minutes), Cathy assembly andpackaging (95 minutes), and Randy the finishing (100 minutes).

SOLUTION TO CUSTOM VANS, INC. CASE

To determine whether the shipping pattern can be improved andwhere the two new plants should be located, the total costs for theentire transportation system for each combination of plants, aswell as the existing shipping pattern costs, will have to be deter-mined. In the headings identifying the combination being dis-cussed, Gary and Fort Wayne will be omitted since they appear inevery possible combination.

Total costs and optimal solutions for each combination aregiven on succeeding pages. A summary of the total costs and therespective systems is listed below:

Detroit–Madison � $10,200Madison–Rockford � $10,550Detroit–Rockford � $11,400

Since the total cost is lowest in the Gary–Fort Wayne–Detroit–Madison combination ($10,200), the new plants should belocated in Detroit and Madison. This system is also an improve-ment over the existing pattern, which costs $9,000, on a cost-per-unit basis.

Status quo: $9,000/450 units � $20/unitProposed: $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.

Existing Shipping Pattern (See sheet #1 in file P5-Custom.XLS)

Total costs � 200(10) � 50(30) � 100(40) � 100(15)

� $9,000

The costs for the additional plants are shown on the next page.


Cathy Preparation 120Tom Assembly 60George Finishing 60Leon or Randy Packaging 210

Total time 250


George Preparation 80Tom Assembly 60Cathy Finishing 100Leon or Randy Packaging 210

Total time 250

SHOP

PLANT Chicago Milwaukee Minneapolis Detroit Capacity

10 20 40 25Gary 200 100 300

20 30 50 15Fort Wayne 50 100 150

Demand 300 100 150 200 750

6634 CH05 UG 8/23/02 2:10 PM Page 33


Cost Table for Custom Vans, Inc.

Detroit–Madison (see sheet #2 in file P5-Custom.XLS)

Total cost � $10,200

Madison–Rockford (See sheet #3 in file P5-Custom.XLS)

SHOP


Gary 10 20 40 25 300Existing

Fort Wayne 20 30 50 15 150

Detroit* 26 36 56 1 150

Proposed Madison** 7 2 22 37 150

Rockford 5 10 30 35 150

Forecast Demand 300 100 150 200

*Since a plant at Detroit could purchase a gallon of fiberglass for $2 less than any other plant, and one Shower-Rific takes 2 gallonsof fiberglass, a systems approach to transportation warrants that $2(2), $4, be deducted from each price quoted in the case for ship-ments 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 tobuild, $1(3) or $3 should be deducted from each price quoted for shipments from Madison.

SHOP


10 20 40 25Gary 200 100 300

20 30 50 15Fort Wayne 100 50 150

26 36 56 1Detroit 150 150

7 2 22 37Madison 150 150

Demand 300 100 150 200 750

SHOP


10 20 40 25Gary 250 50 300

20 30 50 15Fort Wayne 150 150

7 2 22 37Madison 50 100 150

5 10 30 35Rockford 150 150

Demand 300 100 150 200 750

Total Cost � $10,550

6634 CH05 UG 8/23/02 2:10 PM Page 34


SOLUTION TO BINDER’S BEVERAGE CASE

This is a shortest-route problem. With the data given in the problem,the shortest-route model can be used to determine the minimumtime in minutes required to go from the plant to the warehouse ineast Denver. The results are shown in the table, and in file P5-Binders.XLS. As you can see, the best route is to take North Streetto I-70. At Exit 137, South Street is taken to the warehouse. Thisroute takes one hour (60 minutes).

INTERNET CASE STUDIESNorthwest General Hospital

ANDREW–CARTER, INC.This case presents some of the basic concepts of aggregate plan-ning by the transportation method. The case involves solving arather complex set of transportation problems. Four different con-figurations of operating plants have to be tested. The costs are:(See file PS-Andrew.XLS)

The lowest weekly total cost, operating plants 1 and 3 with 2closed, is $217,430. This is $3,300 per week ($171,600 per year)or 1.5% less than the next most economical solution, operating allthree plants. Closing a plant without expanding the capacity of theremaining plants means unemployment. The optimum solution,using plants 1 and 3, indicates overtime production of 4,000 unitsat plant 1 and 0 overtime at plant 3. The all-plant optima have nouse of overtime and include substantial idle regular time capacity:11,000 units (55%) in plant 2 and either 5,000 units in plant 1

Detroit–Rockford (see sheet #4 in file P5-Custom.XLS)

10 30 40 5

SHOP


10 20 40 25Gary 200 100 300

20 30 50 15Fort Wayne 100 50 150

26 36 56 1Detroit 150 150

5 10 30 35Rockford 150 150

Demand 300 100 150 200 750

Total cost � $11,400

Data

Start Node End Node Distance

North Street 1 2 20I 70—A 2 4 5I 70—B 4 8 10High Street—A 1 3 20High Street—B 3 4 20Columbine Street 1 5 30West Street—A 3 5 15West Street—B 5 7 20West Street—C 7 9 156 Ave—A 4 5 156 Ave—B 5 6 256 Ave—C 6 10 40Rose Street—A 6 7 20Rose Street—B 7 8 20South Ave—A 8 9 10South Ave—B 9 10 15

Shortest PathTotal distance � 60

Start End CumulativeNode Node Distance Distance

North Street 1 2 20 20I 70—A 2 4 5 25I 70—B 4 8 10 35South Ave—A 8 9 10 45South Ave—B 9 10 15 60

Optimal Solution (See file P5-Northwest.XLS)

Source Destination Number of Trays

From: Station 5A To: Wing 2 555A 3 655A 5 803G 1 803G 3 853G 5 601S 2 651S 4 210

Optimal Cost: 4,825 minutes. Alternal solutions are possible.

Total Total Variable Fixed Total

Configuration Cost Cost Cost

All plants operating $179,730 $41,000 $220,7301 and 2 operating, 3 closed 188,930 33,500 222,4301 and 3 operating, 2 closed 183,430 34,000 217,4302 and 3 operating, 1 closed 188,360 33,000 221,360

6634 CH05 UG 8/23/02 2:10 PM Page 35


(19% of capacity) or 5,000 in plant 3 (20% of capacity). The idledcapacity versus unemployment question is an interesting, non-quantitative aspect of the case and could lead to a discussion of theforecasts for the housing market and thus the plant’s product.

The optimum producing and shipping pattern is

There are three alternative optimal producing and shipping pat-terns, where R.T. � regular time, O.T. � overtime, and W �warehouse.

SOLUTION TO RANCH DEVELOPMENT PROJECT CASE

1. The minimum distance that will connect all houses to thewater and sewer lines is 10,000 feet (100). The solution along withthe final network follows:

The figure for (1) is shown on the next page.

2. Moving footprint number 16 to accommodate the expansion of the pond area has increased the minimum total distance to 10,100feet (101). A decision now has to be made about whether the in-creased distance and cost for the water and sewer system is worththe additional expected property prices. The solution follows.

From To (Amount)

Plant 1 (R.T.) W2 (13,000); W4 (14,000)Plant 3 (R.T.) W1 (9,000); W3 (8,000);

W5 (8,000)Plant 3 (O.T.) W3 (3,000); W4(1,000)

Start EndBranch Node Node Cost Include Cost

Branch 1 1 2 3 Y 3Branch 2 1 5 2 Y 2Branch 3 2 3 1 Y 1Branch 4 2 10 6Branch 5 3 4 1 Y 1Branch 6 3 8 5 Y 5Branch 7 4 8 5Branch 8 5 6 2 Y 2Branch 9 5 10 5 Y 5Branch 10 6 7 2 Y 2Branch 11 6 11 4 Y 4Branch 12 7 12 4Branch 13 8 9 2 Y 2Branch 14 9 13 7 Y 7Branch 15 10 11 8Branch 16 10 15 11Branch 17 11 12 2 Y 2Branch 18 11 16 8 Y 8Branch 19 12 17 9Branch 20 13 14 4 Y 4Branch 21 13 18 6 Y 6Branch 22 14 15 4 Y 4Branch 23 15 20 7Branch 24 16 22 8Branch 25 17 23 8 Y 8Branch 26 18 19 2 Y 2Branch 27 19 20 2 Y 2Branch 28 19 24 5 Y 5Branch 29 20 21 4 Y 4Branch 30 21 22 1 Y 1Branch 31 21 25 4 Y 4Branch 32 22 23 6 Y 6Branch 33 22 25 5Branch 34 23 26 7 Y 7Branch 35 24 27 11Branch 36 25 27 3 Y 3Branch 37 26 27 10

Total 100

Start EndBranch Node Node Cost Include Cost

Branch 1 1 2 3 Y 3Branch 2 1 5 2 Y 2Branch 3 2 3 1 Y 1Branch 4 2 10 6Branch 5 3 4 1 Y 1Branch 6 3 8 5 Y 5Branch 7 4 8 5Branch 8 5 6 2 Y 2Branch 9 5 10 5 Y 5Branch 10 6 7 2 Y 2Branch 11 6 11 4 Y 4Branch 12 7 12 4Branch 13 8 9 2 Y 2Branch 14 9 13 7 Y 7Branch 15 10 11 8Branch 16 10 15 11Branch 17 11 12 2 Y 2Branch 18 11 16 9 Y 9Branch 19 12 17 9Branch 20 13 14 4 Y 4Branch 21 13 18 6 Y 6Branch 22 14 15 4 Y 4Branch 23 15 20 7Branch 24 16 22 12Branch 25 17 23 8 Y 8Branch 26 18 19 2 Y 2Branch 27 19 20 2 Y 2Branch 28 19 24 5 Y 5Branch 29 20 21 4 Y 4Branch 30 21 22 1 Y 1Branch 31 21 25 4 Y 4Branch 32 22 23 6 Y 6Branch 33 22 25 5Branch 34 23 26 7 Y 7Branch 35 24 27 11Branch 36 25 27 3 Y 3Branch 37 26 27 10

Total 101

6634 CH05 UG 8/23/02 2:10 PM Page 36


1

4

3

2

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8

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13

18

19

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Pond

Bold lines indicatethe selected branches

Stream

Stream

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1

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6

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Figure for RDPA Case (Part 1)

6634 CH05 UG 8/23/02 2:10 PM Page 37

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