ch05_ism
TRANSCRIPT
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
Des Moines to Job 1 – 2,000 tons
Des Moines to Job 2 – 3,000 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
6634 CH05 UG 8/23/02 2:10 PM Page 28
CHAPTER 5 TRANSPORTAT ION ASSIGNMENT AND NETWORK MODELS 29
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:
Decatur to Des Moines – 300 units
Minneapolis to Blue Earth – 200 units
Carbondale to Ciro – 100 units
Carbondale to Des Moines – 50 units
St. Louis to Blue Earth – 50 units
St. Louis to Ciro – 100 units
Total cost = $17,250
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
Decatur to Des Moines – 250 units
Minneapolis to Blue Earth – 200 units
Carbondale to Ciro – 200 units
Carbondale to Des Moines – 100 units
East St. Louis to Carbondale – 150 units
Total cost = $16,850
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
30 CHAPTER 5 TRANSPORTAT ION ASSIGNMENT AND NETWORK MODELS
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
CHAPTER 5 TRANSPORTAT ION ASSIGNMENT AND NETWORK MODELS 31
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
Total maximum flow: 17.
6634 CH05 UG 8/23/02 2:10 PM Page 31
32 CHAPTER 5 TRANSPORTAT ION ASSIGNMENT AND NETWORK MODELS
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
Total maximum flow: 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
2
3
4
5
6
6
1
2
4
7
5
7
5
6
6
5
7 9
8
1
2
3
4
5
6
7
8 93
3
3
34
44
4
7
1
8
5
5
2 2
TimePerson Job (Minutes)
Tom Preparation 100Cathy Assembly 70George Finishing 60Leon Packaging 210
Total time 240
TimePerson Job (Minutes)
George Preparation 80Tom Assembly 60Leon Finishing 80Randy Packaging 210
Total time 230
6634 CH05 UG 8/23/02 2:10 PM Page 32
CHAPTER 5 TRANSPORTAT ION ASSIGNMENT AND NETWORK MODELS 33
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.
TimePerson Job (Minutes)
Cathy Preparation 120Tom Assembly 60George Finishing 60Leon or Randy Packaging 210
Total time 250
TimePerson Job (Minutes)
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
34 CHAPTER 5 TRANSPORTAT ION ASSIGNMENT AND NETWORK MODELS
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
PLANT Chicago Milwaukee Minneapolis Detroit Capacity
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
PLANT Chicago Milwaukee Minneapolis Detroit Capacity
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
PLANT Chicago Milwaukee Minneapolis Detroit Capacity
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
CHAPTER 5 TRANSPORTAT ION ASSIGNMENT AND NETWORK MODELS 35
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
PLANT Chicago Milwaukee Minneapolis Detroit Capacity
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
36 CHAPTER 5 TRANSPORTAT ION ASSIGNMENT AND NETWORK MODELS
(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
CHAPTER 5 TRANSPORTAT ION ASSIGNMENT AND NETWORK MODELS 37
1
4
3
2
5
6
7
10
11
12
8
9
13
18
19
20
21
22
23
26
25
24
27
14
15
16
17
Pond
Bold lines indicatethe selected branches
Stream
Stream
5
55
55
1
1
1
6
6
6
8
9
4
4
4
44
4
3
3
2
2
2
2
2
2
2
7
7
7
11
10
8
8
8
11
Figure for RDPA Case (Part 1)
6634 CH05 UG 8/23/02 2:10 PM Page 37