38137978-lfch10

130

TEACHING SUGGESTIONS

Teaching Suggestion 10.1: Transportation Models in the Chapter.This is a long chapter, in part, because of the four transportationalgorithms that are discussed. If time is an issue in your course, se-lect one of the two initial solution methods and one of the twofinal solution methods to cover in class. The easiest, but not mostefficient, are the northwest corner and stepping-stone rules.

Teaching Suggestion 10.2: Using the Northwest Corner Rule.This approach is easily understood by students and is appealing toteach for that very reason. Make sure the students understand theweakness of the algorithm (that is, it ignores costs totally). Askthem to come up with their own approaches that could improve onthis. Invariably, a good student will present an approach thatcomes very close to VAM. Name the student’s approach after him(or her) and tell him he could have been famous if he had devisedit 50 years earlier.

Teaching Suggestion 10.3: Using the Stepping-Stone Method.Students usually pick up the concept of a closed path and learn totrace the pluses and minuses fairly quickly. But they run into prob-lems when they have to cross over an empty cell. Stress that the citiesin the tableau are just in random order, so crossing an unoccupiedbox is fine. The big test is Table 10.5. Once students comprehendthis tracing, they are usually ready to move on. Remind students thatthere is only one closed path that can be traced for each unused cell.

Teaching Suggestion 10.4: Dummy Rows and Columns.Another confusing issue to students is whether to add a dummyrow (source) or dummy column (destination) in a transportationproblem. A slow and careful explanation is valuable so that stu-dents can reach an intuitive understanding as to the correct choice.Also note that the software adds these dummies automatically.

Teaching Suggestion 10.5: Handling Degeneracy inTransportation Problems.Just as a warning, be aware that students are often confused by theconcept of where to place the zero so that the closed paths can betraced. Carefully explain why you chose or didn’t choose a certaincell. The choice of cell can affect the number of iterations that fol-low.

Teaching Suggestion 10.6: Facility Location Problems.These are an important application of the transportation model andmake 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 onthese problems.

Teaching Suggestion 10.7: Sensitivity Analysis on the Assignment Problem.This algorithm is easy to use and understand. Tell about solving alarge staffing problem, then discuss the cost implications if oneworker is not available or insists on doing a particular task. It iseasy, with the software, to recompute the answers and conduct asensitivity analysis. This is the basis of Problem 10-37.

Teaching Suggestion 10.8: Maximizing Assignment Problems.This section is needed if students are to solve maximization prob-lems by hand, but QM for Windows and Excel QM softwarenegate the need by handling both types of problems. The sectioncan be skipped if the software is being used.

Teaching Suggestion 10.9: Problem 10-37.In assigning this challenging aggregate planning problem, youmay wish to first provide some background information on how tostructure the plan. Remind students that back ordering is not per-mitted, so very large costs must be inserted in many cells. Notethat Problem 10-23 (Mehta Company) is a warm-up exercise forthis data set problem.

ALTERNATIVE EXAMPLES

Alternative Example 10.1: Let us presume that a product ismade at two of our factories which we wish to ship to three of ourwarehouses. We produce 18 at factory A and 22 at factory B; wewant 10 in warehouse 1, 20 in warehouse 2, and 10 in warehouse3. Per unit transportation costs are A to 1, $4; A to 2, $2; A to 3,$3; B to 1, $3; B to 2, $2; B to 3, $1. The corresponding trans-portation table is

10C H A P T E R

Transportation and Assignment Models

TO WarehousesFROM 1 2 3 Total

4 2 3Factory A 18

3 2 1Factory B 22

Total 10 20 10 40

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CHAPTER 10 TRANSPORTAT ION AND ASSIGNMENT MODELS 131

The northwest corner approach follows:

Let us determine the total cost of transportation with this initialnorthwest corner solution. For each filled cell, simply multiply thenumber of units being shipped by the unit shipping cost and thenadd those transhipment costs. Thus, in the order in which the cellswere filled, we have 10($4) � 8($2) � 12($2) � 10($1) � $90.

Using stepping-stone or MODI, we can find the optimalsolution:

SOLUTION:

Cost � 18($2) � 10($3) � 2($2) � 10($1) � $80.

Alternative Example 10.2: There is often an imbalance be-tween the amounts produced and the amounts desired in the ware-houses. In Alternative Example 10.1, there were 40 units producedand forty units demanded for warehousing. Let us presume that anadditional 4 units are desired at each warehouse, increasing thetotal demand to 14 � 24 � 14 � 52. The supply shortage of 12units prevents a solution of this problem until we create a dummyfactory that produces a fake 12 units. The cost to ship a false unitfrom a dummy factory or to a dummy warehouse is zero. After thefinal optimal solution is computed, the false units and dummy fa-cilities are ignored. Our new example with a dummy factory and anorthwest corner initial solution would look like this:

Alternative Example 10.3: Here is a production application ofthe transportation problem. Set up the following problem in atransportation format and solve for the minimum-cost plan:

See the bottom of the next page for the solution.

Alternative Example 10.4: As an example of an assignmentproblem, let us assume that Susan is a sorority pledge coordinatorwith four jobs and only three pledges. Susan decides that the as-signment problem is appropriate except that she will attempt tominimize total time instead of money (since the pledges aren’tpaid). Susan also realizes that she will have to create a fictitiousfourth pledge and she knows that whatever job gets assigned tothat pledge will not be done (this semester, anyhow). She createsestimates for the respective times and places them in the followingtable:

Zingo is, of course, a fictitious pledge, so her times are all zero.

(a) The first step in this algorithm is to develop the opportu-nity cost table. This is done by subtracting the smallest num-ber in each row from every other value in that row, then,using these newly created figures, by subtracting the smallestnumber in each column from every other value in that col-umn. Whenever these smallest values are zero, the subtrac-tion results in no change. Susan’s resulting matrix is

No change was produced when dealing with the columnssince the smallest values were always the zeros from row four.(b) The next step is to draw lines through all of the zeros.The lines are to be straight and either horizontal or vertical.Furthermore, you are to use as few lines as possible. If it re-quires four of these lines (four because it is a 4 � 4 matrix),an optimal assignment is already possible. If it requires fewerthan four lines, another step is required before optimal as-signments may be made. In our example, draw a line through:row four, column three, either column one or row three. Oneversion of the matrix is


4 2 3Factory A 10 8 18

3 2 1Factory B 12 10 22

Total 10 20 10 40


4 2 3Factory A 18 18

3 2 1Factory B 10 2 10 22

Total 10 20 10 40


4 2 3Factory A 14 4 18

3 2 1Factory B 20 2 22

Dummy 0 0 0Factory C 12 12

Total 14 24 14 52

PERIOD

Feb. Mar. Apr.

Demand 55 70 75Capacity

Regular 50 50 50Overtime 5 5 5Subcontract 12 12 10

Beginning inventory 10Costs

Regular time $60 per unitOvertime 80 per unitSubcontract 90 per unit

Inventory carrying cost $1 per unit per month

Job 1 Job 2 Job 3 Job 4

Barb 4 9 3 8Cindy 7 8 2 6Donna 3 4 5 7Zingo 0 0 0 0



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132 CHAPTER 10 TRANSPORTAT ION AND ASSIGNMENT MODELS

(c) Since the number of lines required was less than thenumber of assignees, a third step is required (as is normallythe case). Looking at the version of the matrix with the linesthrough it, determine the smallest number. Subtract thissmallest number from every number not covered by a lineand add it to every number at the intersection of two lines.Repeat the lining out process, with the following result:

Which is still not an optimum solution.

(d) Since all of the zeros can be lined out with three lines,this is still not optimal. Hence, we repeat the step of findingthe smallest uncovered number and both subtracting thatquantity from uncovered numbers and adding it to thosenumbers at line intersections. The resultant matrix, afterbeing lined again, is

Since this matrix requires four lines to cover all zeros, wehave now reached an optimal solution stage.(e) Although there is more than one sequence in which tomake the assignments, in our example the assignments mustbe: Cindy, job 3; Barb, job 1; Donna, job 2; Zingo, job 4.Since Zingo is a dummy row, the job labeled job 4 does notget completed. The total time is 10.







Table for AlternativeExample 10-3:Transportation Solution

Demand for:

TotalUnused CapacityCapacity Available

Supply from: Feb. Mar. Apr. (Dummy) (Supply)

Beginning 0 1 2 0inventory 10 10

60 61 62 0Regular time 45 5 50

80 81 82 0Overtime 5 5

90 91 92 0Subcontract 3 9 12

999 60 61 0Regular time 50 50

999 80 81 0Overtime 5 5

999 90 91 0Subcontract 2 10 12

999 999 60 0Regular time 50 50

999 999 80 0Overtime 5 5

999 999 90 0Subcontract 10 10

Demand 55 70 75 9 209

April

Mar

chFe

brua

ryPe

riod



SOLUTIONS TO DISCUSSION QUESTIONS

AND PROBLEMS

10-1. The transportation model is an example of decision mak-ing under certainty where a decision maker knows beforehand exactly what state of nature will occur (see Chapter 2). In trans-portation problems, this means that the costs of each shippingroute, the demand at each destination, and the supply at eachsource are all known with certainty.

10-2. Vogel’s approximation method gives a good initial solu-tion because it makes each allocation on the basis of the opportu-nity cost, or penalty, that would be incurred if that allocation is notchosen (see Section 10.6). The northwest corner rule does not takeinto account the shipping costs associated with each route alterna-tive as does VAM. Nevertheless, the northwest corner rule couldprovide as low-cost an initial solution—but only if, by chance, itturned out that the lowest-cost routes happened to be on the ini-tially assigned squares.

10-3. A balanced transportation problem is one in which totaldemand (from all destinations) is exactly equal to total supply(from all sources). If a problem is unbalanced, it is necessary toestablish either a dummy source (if demand is greater than supply)or a dummy destination (if demand is less than supply). Refer toSection 10.7.

10-4. This would cause two filled cells to become empty simul-taneously. This means that the solution in the next table will be de-generate. Placing a 0 in one of these two cells and treating this as afilled cell can resolve this difficulty.

10-5. The total cost will decrease $2 for each unit that is placedin this empty cell. Since the maximum that can be placed in thiscell is 80 units, the total cost will decrease by 2(80) � $160. Thismeans the total cost for the solution in the next table will be $900 � $160 � $740. In general, when moving from one trans-portation table to the next, the total cost will decrease by the im-provement index for the cell to be filled times the minimum num-ber of units in any of the “negative” cells in the steppingstone path.

10-6. When m � n � 1 squares (where m � number of rowsand n � number of columns) are not occupied, the solution is de-generate. Not enough squares are occupied to allow us to draw aclosed path for all unused squares. Hence we would not be able toevaluate all of the unused routes. To handle this problem, we se-lect one empty square, place a zero in it, pretend as if it is occu-pied, and proceed as in a normal, nondegenerate case. (To bringthe number of allocations to m � n � 1, it may be necessary toplace a zero in more than one empty square.)

10-7. The enumeration method is not a practical means of solv-ing 5 � 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.

10-8. The assignment problem is a special case of the trans-portation problem and hence can be solved with the approachshown earlier in this chapter. This is illustrated for the Fix-It Shopproblem. Notice that the column and row requirements will alwaysbe equal to 1.

The northwest corner initial assignment above yields a degeneratesolution (only three squares are filled instead of the required five).This will always be a problem when applying the transportationmethod to assignment problems. The problem will be degeneratebecause there will be only one assignment in a given row orcolumn.

10-9. It is not necessary to rework the assignment solution.Changing each entry in the cost table will not result in differenttotal opportunity cost tables. The optimal cost will, however, beincreased by $25 from $492 to $517 because of the extra $5charge for each of the five workers.

10-10. To exclude any unwanted or unacceptable assignmentfrom occurring, it is necessary only to place a very high artificialcost in the row and column representing that particular assign-ment. If, for example, all of the relocation costs for Simmons’sfirm were in the range $1,000 to $3,000, an artificial cost of$20,000 could be placed on the unwanted assignment. Conversely,if we were dealing with a maximization problem, a very low ratingwould be given to the unacceptable assignment.

10-11. a. Initial solution to modify Executive Furniture Corpo-ration problem using the northwest corner rule:

Total cost of this initial solution

� 200($5) � 100($4) � 100($4) � 50($3)

� 250($5)

� 1,000 � 400 � 400 � 150 � 1,250

� $3,200

b. To see if this initial solution is optimal, we compute im-provement indices for each unused square, namely, D–C,E–A, F–A, and F–B:

TO Albu- FactoryFROM querque Boston Cleveland Capacity

5 4 3Des Moines 200 100 300

8 4 3Evansville 100 50 150

Fort 9 7 5Lauderdale 250 250

Warehouse Requirements 200 200 300 700

TO Project Project Project PersonnelFROM 1 2 3 Available

$11 $14 $6Adams 1 1

8 10 11Brown 1 1

9 12 7Cooper 1 1

Project Needs 1 1 1 3



D–C index path � D–C to E–C to E–B to D–B

� $3 � 3 � 4 � 4 � $0

E–A index path � E–A to E–B to D–B to D–A

� $8 � 4 � 4 � 5 � �$3

F–A index path � F–A to F–C to E–C to E–B to D–B to D–A

� �$9 � 5 � 3 � 4 � 4 � 5 � �$2

F–B index path � F–B to F–C to E–C to E–B

� �$7 � 5 � 3 � 4 � �$1

This solution is optimal, so further stepping–stone computationsare not necessary.

c. The improvement index for square D–C is zero. This im-plies the presence of multiple optimal solutions. Practicallyspeaking, management could close the E–C shipping route andsend 50 units on the D–C route instead. The table below illus-trates the overall changes in this alternative optimal solution.

We eliminate the Albuquerque column from consideration be-cause all 200 units in this column have been allocated. We findnew opportunity costs based on the remaining rows and columns.In the next iteration of this process, the opportunity costs are thesame as in the original table.

Total cost of alternative optimal solution

� 200($5) � 50($4) � 50($3) � 150($4) � 250($5)

� $3,200

10-12. Using VAM, we find the opportunity costs by compar-ing the lowest cost cell in each row (and column) with the secondlowest cost cell in that row (or column). The results are given inthe table below. We avoid the high opportunity cost by putting asmany units as possible in the lowest cost cell for the row or col-umn with the highest opportunity cost.

When the Ft. Lauderdale row is eliminated from further considera-tion, we have the opportunity costs shown below. We assign 50 units from Des Moines to Cleveland. Then the only remainingcolumn is Boston, so the assignments are made where possible.Evaluating the empty cells indicates that this is the optimal solutionwith a cost of $3,200.

10-13. a. Hardrock’s initial solution using the northwest cornerrule is shown below.

Cost � 40($10) � 30($4) � 20($5) � 30($8) � 30($6)

� $1,040

Using the stepping-stone method, the following improvement in-dices are computed:

plant 1–project C � $11 � $4 � $5 � $8 � �$4 (closed path: 1-C to 1-B to 2-B to 2-C)

plant 2–project A � �$12 � $5 � $4 � $10 � $1 (closed path: 2-A to 2-B to 1-B to 1-A)

plant 3–project A � �$9 � $6 � $8 � $5 � $4 � $10 � $0(closed path: 3-A to 3-C to 2-C to 2-B to 1-B to 1-A)

plant 3–project B � �$7 � $6 � $8 � $5 � �$4 (closed path: 3-B to 3-C to 2-C to 2-B)

Since all indices are greater than or equal to zero, this initial solu-tion provides the optimal transportation schedule, namely, 40 units


5 4 3Des Moines 200 50 50 300

8 4 3Evansville 150 0 150




5 4 3Des Moines 200 300 1

8 4 3Evansville 150 1


Warehouse Requirements 200 200 300


5 4 3Des Moines 200 300 1

8 4 3Evansville 150 1

Fort 9 7 5Lauderdale 250 250 2



5 4 3Des Moines 200 50 50 300 1

8 4 3Evansville 150 150 1

Fort 9 7 5Lauderdale 250 250 -


TO PlantFROM A B C Capacity

10 4 111 40 30 70

12 5 82 20 30 50

9 7 63 30 30

Project Requirements 40 50 60 150

3 0 0

0 0

0 0



TO

FROM A B C Dummy Capacity

10 4 11 01 40 30 70

12 5 8 02 20 30 50

9 7 6 03 30 30 60

Requirements 40 50 60 30 180

from 1 to A, 30 units from 1 to B, 20 units from 2 to B, 30 unitsfrom 2 to C, and 30 units from 3 to C.

b. There is an alternative optimal solution to this problem.This fact is seen by the index for plant 3–project A beingequal to zero. The other optimal solution, should you wish forstudents to pursue it, is as follows:

plant 1–project A � 20 units

plant 1–project B � 50 units

plant 2–project C � 50 units

plant 3–project A � 20 units

plant 3–project C � 10 units

Total cost remains unchanged at $1,040.

10-14. Hardrock’s problem now requires the addition of adummy project (destination) because supply exceeds demand. Thenorthwest corner initial solution is as follows:

Cost of initial solution � 40($10) � 30($4) � 20($5) � 30($8) � 30($6) � 30($0)

� $1,040

This is the same initial assignment and cost as that found in Prob-lem 10-13. This coincidence occurs because the change in plantcapacity is at the lower right-hand corner of the table and is unaf-fected by the northwest corner rule.

The second table involves bringing the plant 2–dummy routeinto the solution as follows:

Cost of this iteration � $980.

Testing the unused routes:

plant 1–project C index � $11 � 8 � 5 � 4 � �$4

plant 1–dummy index � �$0 � 0 � 6 � 8 � 5 � 4 � �$1

plant 2–project A index � �$12 � 5 � 4 � 10 � �$1

plant 2–dummy index � �$0 � 0 � 6 � 8 � �$2

plant 3–project A index � �$9 � 6 � 8 � 5 � 4 � 10 � $0

plant 3–project B index � �$7 � 6 � 8 � 5 � �$4

TO


10 4 11 01 40 30 70

12 5 8 02 20 0 30 50

9 7 6 03 60 60


bestimprovement index



Because two squares became zero by opening the plant 2–dummyroute, the current solution is degenerate (fewer than 3 rows � 4columns � 1 square are occupied). We will need to place an artifi-cial zero in an unused square (such as plant 2–project C) to be ableto trace all of the closed paths and evaluate where this solution isoptimal.

We now trace the closed paths for the six unused squares (weassume that the plant 2–project C square has a zero in it). The in-dices are:

plant 1–project C � �$11 � 8 � 5 � 4 � �$4

plant 1–dummy � �$0 � 0 � 5 � 4 � �$1

plant 2–project A � �$12 � 5 � 4 � 10 � �$1

plant 3–project A � �$9 � 6 � 8 � 5 � 4 � 10 � $0

plant 3–project B � �$7 � 6 � 8 � 5 � �$4

plant 3–dummy � �$0 � 0 � 8 � 6 � �$2

Since all indices are zero or positive, an optimal solution has beenreached. Again, note that the plant 3–project A route has an im-provement index of $0, implying that an alternative optimal solu-tion exists. The alternative optimal solution, whose total cost isalso $980, is shown in the following table.

10-15. a. Using the northwest corner rule for the Saussy Lum-ber Company data, the following initial solution isreached:

TO


10 4 11 01 20 50 70

12 5 8 02 20 30 50

9 7 6 03 20 40 60


TO Customer Customer CustomerFROM 1 2 3 Capacity

3 3 2Pineville 25 25

4 2 3Oak Ridge 5 30 5 40

3 2 3Mapletown 30 30

Demand 30 30 35 95

Initial cost � 25($3) � 5($4) � 30($2) � 5($3) � 30($3)

� $260

b. Applying the stepping-stone method, the improvementindices are computed:

The improved solution is shown in the following table. Itscost is $255.

Pineville–customer 2 � �$3 � 2 � 4 � 3 � �$2

Pineville–customer 3 � �$2 � 3 � 4 � 3 � $0

Mapletown–customer 1 � �$3 � 3 � 3 � 4 � �$1

Mapletown–customer 2 � �$2 � 3 � 3 � 2 � $0

bestimprove-ment index


3 3 2Pineville 25 25

4 2 3Oak Ridge 30 10 40

3 2 3Mapletown 5 25 30

Demand 30 30 35 95



K1 �� 3 K2 �� 1 K3 �� 2


3 3 2R1 � 0 Pineville 0 25 25

4 2 3R2 � 1 Oak Ridge 30 10 40

3 2 3R3 � 0 Mapletown 30 30

Demand 30 30 35 95

R1 � 0

R1 � K1 � C11 ⇒ 0 � K1 � 3 or K1 � 3

R2 � K1 � C21 ⇒ R2 � 3 � 4 or R2 � 1

R2 � K2 � C22 ⇒ 1 � K2 � 2 or K2 � 1

R2 � K3 � C23 ⇒ 1 � K3 � 3 or K3 � 2

R3 � K3 � C33 ⇒ R3 � 2 � 3 or R3 � 1

Improvement indices are as follows:

Pineville–customer 2 � I12

� C12 � R1 � K2 � 3 � 0 � 1 � �2


� C13 � R1 � K3 � 2 � 0 � 2 � 0

Mapletown–customer 1 � I31

� C31 � R3 � K1 � 3 � 1 � 3 � �1


� C32 � R3 � K2 � 2 � 1 � 1 � 0

The final solution is also evaluated using MODI below and to theright.

Calculations of the Ri’s, Kj’s, and improvement indices are

R1 � K1 � C11 ⇒ 0 � K1 � 3 or K1 � 3

R3 � K1 � C31 ⇒ R3 � 3 � 3 or R3 � 0

R1 � K3 � C13 ⇒ 0 � K3 � 2 or K3 � 2

R2 � K3 � C23 ⇒ R2 � 2 � 3 or R2 � 1

R2 � K2 � C22 ⇒ 1 � K2 � 2 or K2 � 1

Improvement indices:


� C12 � R1 � K2 � 3 � 0 � 1 � �2

Oak Ridge–customer 1 � I21

� C21 � R2 � K1 � 4 � 1 � 3 � 0


� C32 � R3 � K2 � 2 � 0 � 1 � �1


� C33 � R3 � K3 � 3 � 0 � 2 � �1

Final solution with Ri and Kj values:

K1 K2 K3


3 3 2R1 Pineville 25 25

4 2 3R2 Oak Ridge 5 30 5 40

3 2 3R3 Mapletown 30 30

Demand 30 30 35 95

Checking improvement indices again, we find that this improvedsolution is still not optimal. The improvement index for thePineville–customer 3 route � �$2 � 3 � 3 � 3 � �$1. Henceanother shift is necessary. The third iteration is shown in the fol-lowing table:

The cost of this solution is $230. Since two squares went to zerosimultaneously in this last table, the solution has become degener-ate. However, an examination of improvement indices reveals thatthis current solution is optimal.

10-16. Solving the Saussy Lumber Company problem withMODI, we begin with the same initial solution as found in Prob-lem 10-15:


3 3 2Pineville 0 25 25

4 2 3Oak Ridge 30 10 40

3 2 3Mapletown 30 0 30

Demand 30 30 35 95

bestimprovementindex



K1 �� 50 K2 �� 30 K3 �� 40 K4 �� 90

TO Coal CoalFROM Valley Coaltown Junction Coalsburg Supply

50 30 60 70R1 � 0 Morgantown 30 5 35

20 80 10 90R2 � 50 Youngstown 40 20 60

100 40 80 30R3 � 120 Pittsburgh 5 20 25

Demand 30 45 25 20 120

10-17. Krampf Lines Railway Company’s initial northwest cor-ner solution is shown below.

To test for improvement with MODI, we set up an equation foreach occupied square:

R1 � 0

R1 � K1 � 50 0 � K1 � 50 or K1 � 50

R1 � K2 � 30 0 � K2 � 30 or K2 � 30

R2 � K2 � 80 R2 � 30 � 80 or R2 � 50

R2 � K3 � 10 50 � K3 � 10 or K3 � �40

R3 � K3 � 80 R3 � 40 � 80 or R3 � 120

R3 � K4 � 30 120 � K4 � 30 or K4 � �90

index13 � C13 � R1 � K3 � 60 � 0 � (�40)� �100

index14 � C14 � R1 � K4 � 70 � 0 � (�90)� �160

index21 � C21 � R2 � K1 � 20 � 50 � 50 � �80

index24 � C24 � R2 � K4 � 90 � 50 � (�90)� �130

index31 � C31 � R3 � K1 � 100 � 120 � 50� �70

index32 � C32 � R3 � K2 � 40 � 120 � 30� �110

Second Krampf solution—cost � 5,500 miles:


K1 �� 50 K2 �� 30 K3 �� 40 K4 �� 20


50 30 60 70R1 � 0 Morgantown 30 5 35

20 80 10 90R2 � 50 Youngstown 35 25 60

100 40 80 30R3 � 10 Pittsburgh 5 20 25

Demand 30 45 25 20 120

Initial solution’s total cost

� 30(50 miles) � 5(30 miles) � 40(80 miles) � 20(10 miles) � 5(80 miles) � 20(30 miles)

� 6,050 car-miles



TO FactoryFROM Dallas Atlanta Denver Dummy Capacity

8 12 10Houston 800 50 850

10 14 9Phoenix 250 200 200 650

11 8 12Memphis 300 300

Warehouse 800 600 200 200Requirements

R1 � 0

R1 � K1 � 50 ⇒ K1 � 50

R1 � K2 � 30 ⇒ K2 � 30

R2 � K2 � 80 ⇒ R2 � 50

R2 � K3 � 10 ⇒ K3 � �40

R3 � K2 � 40 ⇒ R3 � 10

R3 � K4 � 30 ⇒ K4 � 20

index13 � C13 � R1 � K3 � 60 � 0 � (�40)� �100

index14 � C14 � R1 � K4 � 70 � 0 � 20 � �50

index21 � C21 � R2 � K1 � 20 � 50 � 50 � �80

index24 � C24 � R2 � K4 � 90 � 50 � 20 � �20

index31 � C31 � R3 � K1 � 100 � 10 � 50� �40

index33 � C33 � R3 � K3 � 80 � 10 � (�40)� �110

Third and optimal Krampf solution � 3,100 miles:

10-18. A dummy destination (column) is added. Using VAM,the initial solution is the optimal solution.

In the optimal solution we ship 800 from Houston to Dallas, 50from Houston to Atlanta, 250 from Phoenix to Atlanta, 200 fromPhoenix to Denver, and 300 from Memphis to Atlanta. The totalcost is $14,700.

10-19. If Vogel’s Approximation is used, the initial solution isthe optimal solution. This is to ship 120 from Reno to Phoenix, 20from Denver to Phoenix, 160 from Pittsburgh to Cleveland, and180 from Denver to Chicago. The total cost is $5,700.

10-20. The problem is unbalanced and a dummy destinationmust be added. The optimal solution is to ship 120 from Reno toPhoenix, 20 from Denver to Phoenix, 160 from Pittsburgh to


50 30 60 70Morgantown 35 35

20 80 10 90Youngstown 30 5 25 60

100 40 80 30Pittsburgh 5 20 25

Demand 30 45 25 20 120


Cleveland, and 130 from Denver to Chicago. There will be 30units left in Denver that are not needed. The total cost is $5,310.

10-21. a. VAM steps are as follows:

1. Assign 30 units to C–W (the W column has thegreatest difference, 7) and place X’s in all otherrow C squares.

2. Assign 20 units to B–X.

3. Assign 10 units to B–W.

4. Assign 20 units to A–Z.

5. Assign 35 units to A–Y and 15 units to B–Y.



TO ExcessFROM W X Y Z Supply

12 4 9 5A X X 35 20 55 4

8 1 6 6B 10 20 15 X 45 0

1 12 4 7C 30 X X X 30

Power Demand 40 20 50 20 130

Total VAM cost � 35(9) � 20(5) � 10(8) � 20(1)� 15(6) � 30(1)

� 635

b. MODI technique to test for optimality:

K1 �� 11 K2 �� 4 K3 �� 9 K4 �� 5

TO ExcessFROM W X Y Z Supply

12 4 9 5R1 � 0 A 35 20 55

8 1 6 6R2 � �3 B 10 20 15 45

1 12 4 7R3 � �10 C 30 30

Power Demand 40 20 50 20 130

R1 � 0

R1 � K3 � 9 K3 � 9

R1 � K4 � 5 K4 � 5

R2 � K3 � 6 R2 � �3

R2 � K1 � 8 K1 � 11

R2 � K2 � 1 K2 � 4

R3 � K1 � 1 R3 � �10

index11 � C11 � R1 � K1 � 12 � 0 � 11 � �1

index12 � C12 � R1 � K2 � 4 � 0 � 4 � 0

index24 � C24 � R2 � K4 � 6 � (�3) � 5 � �4

index32 � C32 � R3 � K2 � 12 � (�10) � 4 � �18

index33 � C33 � R3 � K3 � 4 � (�10) � 9 � �5

index34 � C34 � R3 � K4 � 7 � (�10) � 5 � �12

Since all improvement indices are zero or positive, this solution is optimal. An alternative optimal solution, however, is A–X � 20,A–Y � 15, A–Z � 20, B–W � 10, B–Y � 35, C–W � 30, cost �$635.

10-22. The initial solution using the northwest corner rule showsthat degeneracy exists. The number of rows plus the number ofcolumns minus 1 � 4 � 3 � 1 � 6. But the number of occupiedsquares is only 5. Refer to the numbers not circled. To solve theproblem a zero will have to be placed in a square (such as 2–C).This will enable all unused paths to be closed.



TO Hospital Hospital Hospital HospitalFROM 1 2 3 4 Supply

8 9 11 16Bank 1 50 50

12 7 5 8Bank 2 10 70 80

14 10 6 7Bank 3 30 40 50 120

Demand 90 70 40 50 250

10-23. Using VAM to find an initial solution, we make the fol-lowing assignment:

Cost of VAM � 50($8) � 70($7) � 10($5)

assignment � 40($14) � 30($6) � 50($7)

� $2,030

Application of the MODI or stepping-stone methods willyield the following solution in one more iteration. The optimalcost is $2,020.

TO Hospital Hospital Hospital HospitalFROM 1 2 3 4 Supply

8 9 11 16Bank 1 50 X X X 50 1

12 7 5 8Bank 2 X 70 10 X 80

14 10 6 7Bank 3 40 X 30 50 120 1

Demand 90 70 40 50 250

The optimal solution to Problem 10-22, through the use ofour computer program, is circled. Cost � $1,036.

TO

FROM A B C Supply

72 8 9 41 26 15 31 72

38 5 6 82 38 38

7 9 63 46 34 12 46

5 3 74 19 19 19

Demand 110 34 31 175



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

10-24. The optimal solution to the Hall Real Estate decision isshown in the table below.

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

10-25. Mehta’s production smoothing problem is a good exer-cise in the formulation of transportation problems and applyingthem to real-world issues. The problem may be set up as in thetable on the top of the next page. All squares with X’s representnonfeasible (backorder) solutions. In applying a computer pro-gram to solve such a problem, a very large cost (say about $5,000)would be assigned to each of these squares. This would assure thatthey would not appear in the final solution. The dummy destina-tion (month) is added to balance the problem.

The initial solution has a cost of $65,700.The costs for the beginning inventory in months 1, 2, 3, and 4

could be 0, 10, 20, and 30 respectively if the carrying cost for thebeginning inventory has already been considered. The solution isthe same but the cost would be $65,300.

10-26. 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). The northwest corner method is usedto provide an initial solution. The total cost of this first solution isseen to be $23,600. You should note that the cost of each individ-ual “plant to distribution center” route is found by adding the distribution costs to the respective unit production costs. Thus the total production plus shipping cost of one auto top carrier from Atlanta to Los Angeles is $14 ($8 for shipping plus $6 forproduction).

Table for Problem 10-26

Total cost � (600 units � $14) � (200 units � $9) � (700 units � $12) � (500 units � $10)

� $8,400 � $1,800 � $8,400 � $5,000

� $23,600

Is this initial solution optimal? We once again employ the step-ping-stone method to test it and to compute improvement indicesfor unused routes.

Improvement index for Atlanta to New York route:

�$11 (Atlanta to New York)

�$14 (Atlanta to Los Angeles)

�$9 (Tulsa to Los Angeles)

�$12 (Tulsa to New York)

� �$6

TO ProductionFROM Los Angeles New York Capacity

$14 $11Atlanta 600 600

$9 $12Tulsa 200 700 900

$9 $10New Orleans 500 500

Demand 800 1,200 2,000



10-26 (continued)

Improvement index for New Orleans to Los Angeles route:

�$9 (New Orleans to Los Angeles)

�$10 (New Orleans to New York)

�$12 (Tulsa to New York)

�$9 (Tulsa to Los Angeles)

� �$2

Since the firm can save $6 for every unit it ships from Atlanta toNew York, it will want to improve the initial solution and send asmany as possible (600 in this case) on this currently unused route.

You may want to confirm that the total cost is now $20,000, asavings of $3,600 over the initial solution.

Again, we must test the two unused routes to see if their im-provement indices are negative numbers.

index for Atlanta to Los Angeles� �$14 � $11 � $12 � $9 � �$6

index for New Orleans to Los Angeles� �$9 � $10 � $12 � $9 � �$2

Since both indices are greater than zero, we have reached an opti-mal solution. If Ashley selects to open the New Orleans plant, thefirm’s total distribution system cost will be $20,000. If the Houstonplant site is chosen, the initial solution is as follows:

Total cost of initial solution

� $8,400 � $1,800 � $8,400 � $4,500

� $23,100

Improvement index for Atlanta to New York

� �$11 � $14 � $9 � $12

� �$6

Table for Problem 10-25

Destination (Month)

Sources 1 2 3 4 Dummy Capacity

10 20 30 40 0Beginning inventory 40 40

100 110 120 130 0Regular prod. (month 1) 80 20 100

130 140 150 160 0Overtime (month 1) 50 50

100 110 120 0Regular prod. (month 2) 90 10 100

130 140 150 0Overtime (month 2) 50 50

100 110 0Regular prod. (month 3) 100 100

130 140 0Overtime (month 3) 50 50

100 0Regular prod. (month 4) 100 100

130 0Overtime (month 4) 50 50

150 150 150 150 0Outside purchases 30 420 450

Demand 120 160 240 100 470 1,090


$14 $11Atlanta 600 600

$9 $12Tulsa 800 100 900

$9 $10New Orleans 500 500

Demand 800 1,200 2,000


$14 $11Atlanta 600 600

$9 $12Tulsa 200 700 900

$7 $9Houston 500 500

Demand 800 1,200 2,000

M10_REND6289_10_IM_C10.QXD 5/12/08 11:26 AM 3REVISED


Improvement index for Houston to Los Angeles

� �$7 � $9 � $12 � $9

� �$1

The improved solution by opening Atlanta to New Yorkroute is shown below.

Total cost of improved solution � $19,500.Improvement indices for Atlanta to New York and Houston

to Los Angeles routes are both positive at this point. Hence an op-timal solution has been reached. Upon comparing total costs forthe Houston option ($19,500) to those for the New Orleans option($20,000), we would recommend to Ashley that all factors beingequal, the Houston site should be selected.

10-27. Considering Fontainebleau, we have

Optimal cost � $1,530,000.

Considering Dublin, we have the following initial northwest cornersolution:


$14 $11Atlanta 600 600

$9 $12Tulsa 800 100 900

$7 $9Houston 500 500

Demand 800 1,200 2,000

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

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

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


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

55 55 40 70Pusan 1,000 1,000 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



Optimal cost � $1,535,000.

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.

10-28. Considering East St. Louis, we have:

Initial solution—northwest corner rule:

Final solution

Optimal solution:


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

Decatur Minn. C’dale E. St. L. Demand

20 17 21 29Blue Earth 250 250

25 27 20 30Ciro 50 150 200

22 25 22 30Des Moines 50 150 150 350

Capacity 300 200 150 150 800


20 17 21 29Blue Earth 50 200 250

25 27 20 30Ciro 150 50 200

22 25 22 30Des Moines 250 100 350

Capacity 300 200 150 150 800

Decatur Minn. C’dale St. Louis Demand

20 17 21 27Blue Earth 250 250

25 27 20 28Ciro 50 150 200

22 25 22 31Des Moines 50 150 150 350

Capacity 300 200 150 150 800

Optimal cost using East St. Louis: $17,400.

Considering St. Louis, we have:




Optimal solution:

Optimal cost using St. Louis: $17,250.

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

10-29. Considering East St. Louis, we have:


Optimal solution:

Optimal cost using East St. Louis: $60,900.

Considering St. Louis, we have:



20 17 21 27Blue Earth 200 50 250

25 27 20 28Ciro 100 100 200

22 25 22 31Des Moines 300 50 350

Capacity 300 200 150 150


70 77 91 69Blue Earth 250 250

75 87 90 70Ciro 50 150 200

72 85 92 70Des Moines 50 150 150 350

Capacity 300 200 150 150


70 77 91 69Blue Earth 50 200 250

75 87 90 70Ciro 150 50 200

72 85 92 70Des Moines 250 100 350

Capacity 300 200 150 150


70 77 91 77Blue Earth 250 250

75 87 90 78Ciro 50 150 200

72 85 92 81Des Moines 50 150 150 350

Capacity 300 200 150 150



Optimal solution:

Optimal cost using St. Louis: $62,250.

Therefore, East St. Louis is $1,350 per week less expensive thanSt. Louis.

10-30. Step 1—row subtraction:

Column subtraction:

Step 2—minimum straight lines to cover zeros:

Step 3—subtract the smallest uncovered number from all the un-covered numbers—add it to numbers at intersections of two lines:

Return to step 2—cover all zeros:

Assignment can be made:

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

10-31. The initial table used for the assignment problem is:


Billy 400 90 60 120Taylor 650 120 90 180Mark 480 120 80 180John 500 110 90 150

Solving this using the assignment module in QM for Windows,the following assignments are made:Billy–Job 1; Taylor–Job 2; Mark–Job 3; John–Job 4The total time is 750 minutes.

10-32. For the prohibited route where no assignment may bemade, a very high cost (10,000 miles) used to prevent anythingfrom being assigned here. The initial assignment table is:

Kansas City Chicago Detroit Toronto

Seattle 1500 1730 1940 2070Arlington 460 810 1020 1270Oakland 1500 1850 2080 10000Baltimore 960 610 400 330

The optimal solution found using the QM for Windows assign-ment module is:The Seattle crew will go to Detroit.The Arlington crew will go to Kansas City.The Oakland crew will go to Chicago.The Baltimore crew will go to Toronto.The total distance is 4,580 miles.


70 77 91 77Blue Earth 200 50 250

75 87 90 78Ciro 100 100 200

72 85 92 81Des Moines 300 50 350

Capacity 300 200 150 150

MACHINE

JOB W X Y Z

A12 0 4 6 3

A15 0 1 3 0

B2 0 3 3 2

B9 0 2 4 2

MACHINE

JOB W X Y Z

A12 0 2 3 2

A15 1 0 1 0

B2 0 1 0 1

B9 0 0 1 1

MACHINE

JOB W X Y Z

A12 0 3 3 3

A15 0 0 0 0

B2 0 2 0 2

B9 0 1 1 2

MACHINE

JOB W X Y Z

A12 0 3 3 3

A15 0 0 0 0

B2 0 2 0 2

B9 0 1 1 2

MACHINE

JOB W X Y Z

A12 0 2 3 2

A15 1 0 1 0

B2 0 1 0 1

B9 0 0 1 1



10-33. If the total distance is maximized, we assign a very lowcost (miles) to the prohibited route to prevent this assignment. Acost of 0 is used. The initial table is:

Kansas City Chicago Detroit Toronto

Seattle 1500 1730 1940 2070Arlington 460 810 1020 1270Oakland 1500 1850 2080 0Baltimore 960 610 400 330

With the solution found using QM for Windows, the Seattle crewwill go to Chicago; the Arlington crew will go to Toronto; theOakland crew will go to Detroit; the Baltimore crew will go toKansas City; and the total distance is 6,040 miles. This maximumdistance is 1,460 miles more than the minimum distance (4,580).

10-34. Because this is a maximization problem, each number issubtracted from 95. The problem is then solved using the mini-mization algorithm.

10-35.

10-36. Each rating is subtracted from 27.1 because this is a max-imization problem.

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

Assignment Rating

1–2 P.M. on A 27.12–3 P.M. on C 17.13–4 P.M. on B 18.54–5 P.M. on independent 12.8

Overall rating 75.5

10-37.

Thus, the optimal solution does not change by adding a fourthmember. Davis is assigned to the dummy (nonexistent project).This is because Davis is not the relatively least-cost assignment toany of the first three projects.

10-38. The following optimal assignments can be made:

10-39. Students should note the large numbers used to block in-feasible production plans (see Printout 1 on the nextpage).a. The solution yields a cost of $2,591,200. The plan isshown in Printout 2. There are multiple optimal solutions.b. Yes, the solution now costs $2,640,500 with 275 permonth 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 pro-duction plan remains the same as in Printout 2.

If overtime cost is $1,200 per unit, the total cost is$2,572,100.

Assignment Rating

Adams to project 3 $ 6Brown to project 2 10Cooper to project 1 9Davis to dummy $00

$25

Assignment Cost

Component C53 to plant 6 0.06Component C81 to plant 3 0.04Component D5 to plant 4 0.30Component D44 to plant 5 0.10Component E2 to plant 2 0.07Component E35 to plant 8 0.06Component G99 to plant 1 0.55

Total cost $1.18



Printout 1 for Problem 10-39 (Computer Data Entry. The costs are in $1,000s.)

JAN FEB MARCH APR MAY JUNE JULY AUG Supply

REG 1. 1.1 1.2 1.3 1.4 1.5 1.6 1.7 235OT 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2. 20SUB 1.5 1.6 1.7 1.8 1.9 2. 2.1 2.2 12REG 10. 1. 1.1 1.2 1.3 1.4 1.5 1.6 255OT 10. 1.3 1.4 1.5 1.6 1.7 1.8 1.9 24SUB 10. 1.5 1.6 1.7 1.8 1.9 2. 2.1 15REG 10. 10. 1. 1.1 1.2 1.3 1.4 1.5 290OT 10. 10. 1.3 1.4 1.5 1.6 1.7 1.8 26SUB 10. 10. 1.5 1.6 1.7 1.8 1.9 2. 15REG 10. 10. 10. 1. 1.1 1.2 1.3 1.4 300OT 10. 10. 10. 1.3 1.4 1.5 1.6 1.7 24SUB 10. 10. 10. 1.5 1.6 1.7 1.8 1.9 17REG 10. 10. 10. 10. 1. 1.1 1.2 1.3 300OT 10. 10. 10. 10. 1.3 1.4 1.5 1.6 30SUB 10. 10. 10. 10. 1.5 1.6 1.7 1.8 17REG 10. 10. 10. 10. 10. 1. 1.1 1.2 290OT 10. 10. 10. 10. 10. 1.3 1.4 1.5 28SUB 10. 10. 10. 10. 10. 1.5 1.6 1.7 19REG 10. 10. 10. 10. 10. 10. 1. 1.1 300OT 10. 10. 10. 10. 10. 10. 1.3 1.4 30SUB 10. 10. 10. 10. 10. 10. 1.5 1.6 19REG 10. 10. 10. 10. 10. 10. 10. 1. 290OT 10. 10. 10. 10. 10. 10. 10. 1.3 30SUB 10. 10. 10. 10. 10. 10. 10. 1.5 20

Demand 255. 294. 321. 301. 330. 320. 345. 340.



10-40. a. Here is the first schedule using our software.

RT � regular time; OT � overtime; SUB � subcontracting

b. The revised schedule is

Printout 2 for Problem 10-39 (Computer Solution to HAIFA. Multiple Optimal Solutions)

Optimal Solution: 96.0

1 2 3 4 5 6 7 8 9 10

1 0 0 0 0 1 0 0 0 0 02 0 0 0 0 0 1 0 0 0 03 0 0 0 1 0 0 0 0 0 04 0 0 1 0 0 0 0 0 0 05 1 0 0 0 0 0 0 0 0 06 0 0 0 0 0 0 0 1 0 07 0 0 0 0 0 0 0 0 0 18 0 0 0 0 0 0 1 0 0 09 0 0 0 0 0 0 0 0 1 010 0 1 0 0 0 0 0 0 0 0

Optimal Solution: 92.0

1 2 3 4 5 6 7 8 9 10

1 0 0 0 0 1 0 0 0 0 02 0 0 0 0 0 1 0 0 0 03 0 0 0 0 0 0 1 0 0 04 0 0 1 0 0 0 0 0 0 05 1 0 0 0 0 0 0 0 0 06 0 0 0 0 0 0 0 1 0 07 0 0 0 1 0 0 0 0 0 08 0 0 0 0 0 0 0 0 0 19 0 0 0 0 0 0 0 0 1 010 0 1 0 0 0 0 0 0 0 0

Optimal cost � JAN FEB MARCH APR MAY JUNE JULY AUG Dummy$2,591,200

REG 235.OT 20.SUB 0. 0. 12.REG 255.OT 24.SUB 15.REG 290.OT 26.SUB 5. 10.REG 300.OT 1. 0. 23.SUB 17.REG 300.OT 30.SUB 17.REG 290.OT 28.SUB 2. 17.REG 300.OT 30.SUB 15. 0. 4.REG 290.OT 30.SUB 20.



c. Yes, there is a new schedule:

Optimum Solution: 93.0

1 2 3 4 5 6 7 8 9 10

1 0 0 0 0 1 0 0 0 0 02 0 0 0 0 0 1 0 0 0 03 0 0 0 0 0 0 1 0 0 04 0 0 1 0 0 0 0 0 0 05 0 0 0 1 0 0 0 0 0 06 0 0 0 0 0 0 0 1 0 07 1 0 0 0 0 0 0 0 0 08 0 0 0 0 0 0 0 0 0 19 0 0 0 0 0 0 0 0 1 010 0 1 0 0 0 0 0 0 0 0

2 4 3

TO

FROM A B C Available

4 3 3W 15 35 0

6 7 6Y X 50 0

8 2 5Z X 50 X 50 3

Demand 30 65 40 135

2 4 3

TO


4 3 3W X 15 20 35 1

6 7 6Y X 50 0

8 2 5Z X 50 X 50 3

Demand 30 65 40 135

2 4 3

TO


4 3 3W X 15 20 35 0

6 7 6Y 30 X 20 50 0

8 2 5Z X 50 X 50 3

Demand 30 65 40 135

2 1 2

TO FactoryFROM A B C Availability

4 3 3W 35 0

6 7 6Y 50 0

8 2 5Z X 50 X 50 3

Store Demand 30 65 40 135

SOLUTIONS TO INTERNET HOMEWORK PROBLEMS

10-41. Jessie Cohen Clothing Group’s first VAM assignmenttable:

In the initial assignment table above, we see that the Z row has thegreatest difference (3). We assign the minimum possible numberof units (50) to the least-cost route (Z–B) in that row.

Second VAM assignment with B’s requirement satisfied:

This second VAM table (above) indicates that the greatest differ-ence is now in the B column (4). We may assign up to 15 units tothe W–B square without exceeding the demand at store B.

Third VAM assignment with W’s requirement satisfied:

The third VAM table involves assigning 20 units to the W–Croute. This is done because column C has the highest differenceand square W–C the lowest cost in that column.

Final assignment for Cohen Clothing Group:

The final assignment (above) is made by completing the row andcolumn requirements. This means that 30 units must be assignedto Y–A and 20 units to Y–C.

The total cost of this VAM assignment � (15 units � $3) �(20 units � $3) � (30 units � $6) � (20 units � $6) � (50 units � $2) � $505. A quick check using the stepping-stone indexmethod indicates that this VAM solution is optimal.

10-42.

OFFICE

MAN Omaha Miami Dallas

Jones 800 1,100 1,200

Smith 500 1,600 1,300

Wilson 500 1,000 2,300

Row subtractionis done next.

OFFICE


Jones 0 300 400

Smith 0 1,100 800

Wilson 0 500 1,800

Columnsubtractionis done next.



Optimal assignment:

Jones to DallasSmith to OmahaWilson to MiamiCost � $1,200 � $500 � $1,000

� $2,700

10-43. Original problem:

Optimal assignment:

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.

10-44. Original problem:

OFFICE


Jones 0 0 0

Smith 0 800 400

Wilson 0 200 1,400

Cover zeroswith lines next.

OFFICE


Jones 0 0 0

Smith 0 800 400

Wilson 0 200 1,400

Subtractsmallestnumber next.

OFFICE


Jones 200 0 0

Smith 0 600 200

Wilson 0 0 1,200

Cover zeroswith lines next.

OFFICE


Jones 200 0 0

Smith 0 600 200

Wilson 0 0 1,200

CUSTOMER

SITE A B C D

1 7 3 4 8

2 5 4 6 5

3 6 7 9 6

4 8 6 7 4

Rowsubtractionis donenext.

CUSTOMER

SITE A B C D

1 4 0 1 5

2 1 0 2 1

3 0 1 3 0

4 4 2 3 0

Columnsubtractionis done next.

CUSTOMER

SITE A B C D

1 4 0 0 5

2 1 0 1 1

3 0 1 2 0

4 4 2 2 0

Cover zeroswith lines.

CUSTOMER

SITE A B C D

1 4 0 0 5

2 1 0 0 1

3 0 1 2 0

4 4 2 2 0

CASE

SQUAD A B C D E

1 14 7 3 7 27

2 20 7 12 6 30

3 10 3 4 5 21

4 8 12 7 12 21

5 13 25 24 26 8

Row sub-tractionisdonenext.

CASE

SQUAD A B C D E

1 11 4 0 4 24

2 14 1 6 0 24

3 7 0 1 2 18

4 1 5 0 5 14

5 5 17 16 18 0

Columnsubtrac-tion isdonenext.

CASE

SQUAD A B C D E

1 10 4 0 4 24

2 13 1 6 0 24

3 6 0 1 2 18

4 0 5 0 5 14

5 4 17 16 18 0

Coverzeroswithlines.



Optimal assignment:

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.

10-45.

10-46. The major difference between the MODI and stepping-stone methods is in the procedure used to test for optimality. In thestepping-stone method, we first draw a closed path for each of theempty squares to calculate its improvement index. Then, the mostfavorable square (i.e., the one with the largest negative index) isidentified. In MODI, however, we first identify the most favorablesquare (by using row and column numbers) and then draw a closedpath (only for that path) to direct us in improving the solution.

10-47. A “northeast corner” rule would be directly analogousto the northwest corner rule, but it would simply begin in theupper right-hand corner instead of the upper left-hand corner. Wesee in the table that this initial solution is degenerate because onlyfour squares (instead of the expected five) are occupied. The de-generacy condition, by the way, is just a peculiarity of the Execu-tive Furniture Corporation data.

SOLUTION TO ANDREW–CARTER, INC., CASE

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 solutions, al-though requiring relatively few iterations to optimality, involvedegeneracy if solved manually. The costs are:

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(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.

Getting the solution manually should not be attempted usingthe northwest corner rule. It will take eight tableaux to do the “allplants” configuration, with degeneracy appearing in the seventhtableau; the “1 and 2” configuration takes five tableaux; and so on.It is strongly suggested that software be used.

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.

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

From To (Amount)

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)

CASE

SQUAD A B C D E

1 10 4 0 4 24

2 13 1 6 0 24

3 6 0 1 2 18

4 0 5 0 5 14

5 4 17 16 18 0

Assignment Rating

C53 at plant 1 10 centsC81 at plant 3 4 centsD5 at plant 4 30 centsD44 at plant 2 14 cents

Total manufacturing cost 58 cents


5 4 3Des Moines 100 100

8 4 3Evansville 200 100 300



TimePerson Job (Minutes)

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

Total time 240



2. If Randy is used, the assignment problem becomes unbal-anced and a dummy job must be added. The optimum assignmentwould 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 as-signment with the remaining three workers assigned to the remain-ing three tasks is

If Cathy is assigned to the finishing task, the optimum assign-ment is

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).

INTERNET CASE STUDYNorthwest General Hospital

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.


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

Total time 230


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

Total time 250


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

Total time 250

Optimal Solution

Source Destination Number of Trays

From: Station 5A To: Wing 5 605A 6 805A 3 603G 1 803G 3 903G 4 551S 4 1551S 2 120

Optimal Cost: 4,825 minutes



SHOP

PLANT Chicago Milwaukee Minneapolis Detroit Capacity Ri

10 20 40 25Gary 200 100 300

20 30 50 15Fort Wayne 50 100 150

0 0 0 0Dummy 100 50 50 100 300

Demand 300 100 150 200 750

Kj

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

� $9,000

The costs for the additional plants are shown below.

Cost Table for Custom Vans, Inc.

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.

The total cost is 300($10) � 100($0) � 150($0) � 150($15) �50($0) � $5,250. This is also the optimal solution with no addi-tional plants. The cost of the existing shipping pattern is $9,000and is shown below. Thus the existing shipping pattern can beimproved.

Existing Shipping Pattern

10 20 40 15

SHOP

PLANT Chicago Milwaukee Minneapolis Detroit Capacity

Gary 300 X X X 300

Fort Wayne X X X 150 150

Dummy X 100 150 50 300

750Demand 300 100 150 200

750

The optimal solution is:



Improvement indices (MODI method):

G to Milw: 20 � 20 � 0 � 0G to D: 25 � 5 � 0 � �20FW to Milw: 30 � 20 � 10 � 0FW to Minn: 50 � 40 � 10 � 0D to C: 26 � 10 � (�4) � �20D to Milw: 36 � 20 � (�4) � �20D to Minn: 56 � 40 � (�4) � �20M to C: 7 � 10 � (�18) � �15M to D: 37 � 5 � (�18) � �50


G to Milw: 20 � 5 � 0 � �15G to Minn: 40 � 25 � 0 � �15FW to Milw: 30 � 5 � (�10) � �35FW to C: 20 � 10 � (�10) � �20FW to Minn: 50 � 25 � (�10) � �35

M to D: 37 � 25 � (�3) � �15R to C: 5 � 10 � 5 � �10 � best improvement

(see iteration 2)R to Minn: 10 � 5 � 5 � 0R to D: 35 � 25 � 5 � �5

All solutions are positive; solution is optimal as shown:

G to C: 200 unitsG to Minn: 100 unitsFW to C: 100 unitsFW to D: 50 unitsD to D: 150 unitsM to Milw: 100 unitsM to Minn: 50 units

Total cost � 200(10) � 100(20) � 100(2) � 100(40) � 50(22) � 50(15) � 150(1) � $10,200

SHOP


10 20 40 25Gary 200 100 300 0 (10)

20 30 50 15 (5)Fort Wayne 100 50 150 10 (10)

26 36 56 1Detroit 150 150 �4 (25)

7 2 22 37 (5)Madison 100 50 150 �18 15

Demand 300 100 150 200 750

Kj 10 20 40 5

(3) (18) (13) (14)10 10

Detroit–Madison, Iteration 1 (Vogel’s Approximation Method)

SHOP


10 20 40 25Gary 250 50 300 0 (10)

20 30 50 15Fort Wayne 150 150 �10 (5)

7 2 22 37 (5)Madison 50 100 0* 150 �3 (15)

5 10 30 35Rockford 150 150 5 (1)

Demand 300 100 150 200 750

Kj 10 5 25 25

(2) (8) (8) (10)(10)

*0 supplied to avoid degeneracy.

Madison–Rockford, Iteration 1 (Vogel’s Approximation Method)




G to Milw: 20 � 15 � 0 � �5G to Minn: 40 � 35 � 0 � �5FW to C: 20 � 10 � (�10) � �20FW to Milw: 30 � 15 � (�10) � �25FW to Minn: 50 � 35 � (�10) � �25M to C: 7 � 10 � (�13) � �10M to D: 37 � 25 � (�13) � �25R to Milw: 10 � 15 � (�5) � 0R to D: 35 � 25 � (�5) � �15

Improvement indices (MODI method) for Detroit-Rockford:

G to Minn: 40 � 40 � 0 � 0G to D: 25 � 5 � 0 � �20FW to Milw: 30 � 20 � 10 � 0FW to Minn: 50 � 40 � 10 � 0D to C: 26 � 10 � (�4) � �20D to Milw: 36 � 20 � (�4) � �20D to Minn: 56 � 40 � (�4) � �20R to C: 5 � 10 � (�10) � �5R to D: 35 � 5 � (�10) � �40

Optimal solution:

G to C: 200 unitsG to Milw: 100 unitsFW to C: 100 unitsD to D: 150 unitsFW to D: 50 unitsR to Minn: 150 units

Total costs � 200(10) � 100(20) � 100(20) � 50(15) � 150(1)� 150(30)

� $11,400

Optimal solution:

G to C: 250 unitsG to D: 50 unitsFW to D: 150 unitsM to Milw: 100 unitsM to Minn: 50 unitsR to C: 50 unitsR to Minn: 100 units

Total cost � 250(10) � 50(5) � 100(2) � 50(22) � 100(30) �50(25) � 150(15)

� $10,550

10 15 35 25

SHOP


10 20 40 250 Gary 250 50 300 0

20 30 50 15�10 Fort Wayne 150 150 �10

7 2 22 37�13 Madison 100 50 150 �13

5 10 30 35�5 Rockford 50 100 150 �5

Demand 300 100 150 200 750

Kj 10 15 35 25

Madison–Rockford, Iteration 2

Detroit–Rockford (Vogel’s Approximation Method)

10 30 40 5

SHOP


10 20 40 25 (10)0 Gary 200 100 300 0 (30)

20 30 50 15 (5)10 Fort Wayne 100 50 150 10 (10)

26 36 56 1�4 Detroit 150 150 �4 (25)

5 10 30 35�10 Rockford 0* 150 150 �10 (5)

Demand 300 100 150 200 750

Kj 10 20 40 5

(5) (10) (10) (14)(15) (10)

*0 supplied to avoid degeneracy.


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