Chapter 11 Answers

158

TEACHING SUGGESTIONS

Teaching Suggestion 11.1: Topics in This Chapter.The overall purpose of this chapter is to provide a framework forthe topics of integer programming, branch and bound, nonlinearprogramming and goal programming. These are fairly advancedtopics in a mathematical sense, and the chapter’s intention issolely to introduce them through a series of simple graphical problems. Some of the topics are on the cutting edge of QA. Forexample, in integer and nonlinear programming, no one solutionprocedure exists to handle all problems.

Teaching Suggestion 11.2: Using the Computer to Solve Mixed-Integer Programming Problems.Note that the Excel printout in Program 11.2 allows users to spec-ify which variables are integers and which, by default, can be fractional.

Teaching Suggestion 11.3: How the Branch and Bound MethodCan Help.In this section we illustrate how branch and bound is used to solvesmall assignment and integer programming problems. But its realstrength is in dealing with huge problems (for example, thousandsof variables/constraints). Branch and bound allows us to divide alarge problem into smaller parts, thereby eliminating one-half ortwo-thirds of the options and reducing the problem to a more man-ageable level.

Teaching Suggestion 11.4: Multiple Goals.Ask studdents what other goals a company might have beyondmaximizing profit. Socially conscious firms need to state as theirmission a whole series of objectives. Encourage students to re-search an article showing a goal programming application. Thereis a wealth of research in journals. One interesting application is inthe box later in this section that deals with budgeting for prisons.

Teaching Suggestion 11.5: Deviational Variables Are the Key inGoal Programming.The concept of deviational variables requires careful explanationto the class. Students are accustomed to the decision variables ofX1 and X2. Now they need to concentrate on goal achievement.The minus and plus signs on deviational variables need a thought-ful classroom discussion.

Teaching Suggestion 11.6: Difficulty of Graphical Goal Programming.Solving goal programming problems graphically can be a confus-ing concept relative to graphical LP. Students often have difficultywith the direction of deviational variables.

Teaching Suggestion 11.7: Using the Goal Programming Simplex Method.Point out the similarities and differences between the simplexmethod and the modified goal programming tableau. You canshow that the structure is almost the same. The big change is theaddition of two rows for each new goal. Surprisingly, the compu-tation is not as difficult as it looks.

ALTERNATIVE EXAMPLESAlternative Example 11.1: 0–1 Integer Programming.

Indiana’s prison budget allows it to consider four new installationsnext year. They are

X1 � 1 if maximum security prison in Ft. Wayne, � 0 otherwise

X2 � 1 if minimum security prison in Bloomington, � 0 otherwise

X3 � 1 if halfway house in Indianapolis, 0 otherwiseX4 � 1 if expanded tricounty jail in South Bend, 0 otherwise

The state wants to maximize the number of people that can be“served,” while only building one of the two prisons (X1 or X2) andobserving cost and space limitations. Here is the formulation:

maximize number served � 3,000X1 � 900X2

� 4,000X3 � 1,500X4

subject to

X1 � X2 � 1 prison

4X1 � 2X2 � 7X3 � 3X4 � 12 acres available

3.5X1 � 1X2 � 2.5X3 � 9X4 � 12 million dollars budgeted

Solution: Using software, we find that X1 � 1, X2 � 0, X3 � 1, X4 � 0, number served � 7,000.

Alternative Example 11.2: The Quality University (QU) is aprivate noncredit training firm that specializes in total qualitymanagement (TQM) courses. QU wants to determine how many

11C H A P T E R

Integer Programming, Goal Programming, NonlinearProgramming and the Branch and Bound Methods

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CHAPTER 11 INTEGER PROGRAMMING, GOAL PROGRAMMING, NONLINEAR PROGRAMMING 159

of each of two programs to offer in order to maximize profit. Theirinteger program can be formulated as follows:

maximize profit � $8,500X1 � $6,000X2

subject to

X1 � X2 � courses max. of 10

$1,000X1 � $700X2 � instructor’s pay of $7,200

X1, X2 � 0 and are integers

Using LP, the solution is: X1 � X\c, X2 � 9Z\c, profit � $61,667.

SOLUTION

Initial upper bound (UB) � $61,667 (X1 � X\c, X2 � 9Z\c)

Initial lower bound (LB) � $54,000 (X1 � 0, X2 � 9)(See graph below for this example.)

All nodes are either integer or infeasible, so the solution isseen to be X1 � 3, X2 � 6, profit � $61,500.

X1 = 0X2 = 10P = $60,000

X1 = 2/3X2 = 9 1/3P = $61,667

X1 = 1X2 = 8 7/8P = $61,643

X1 = 1 3/5X2 = 8P = $61,000

X1 = 1X2 = 8P = $56,500

X1 = 2X2 = 7 3/7P = $61,571

X1 = 2 3/10X2 = 7P = $61,550

X1 = 2X2 = 7P = $59,000

X1 = 3X2 = 6P = $61,500

NotFeasible

NotFeasible

UB = $61,667LB = $54,000

UB = $61,600LB = $54,000[X1 = 0, X2 = 9]

X 1 ≤ 0

X2 ≥ 9

X1 ≤ 1

X2 ≥ 8

X1 ≤ 2

X1 ≥ 1

X2 ≤ 8

X1 ≥ 2

X2 ≤ 7

X1 ≥ 3

Optimal Solution

Figure for Alternative Example 11.2

2X1 + 1 X2 = 60

5 10 15 20 25 30 35 40 450

5

10

15

20

25

30

35

40

45

2X1 + 2 X2 = 80

X1

X2

Goal Constraints

2X1 + 4X2 = 80

/21

/21

Alternative Example 11.3:Minimize P1d1

� � P2d2� � P3d3

� � P4d1�

subject to

2x1 � 4x2 � d1� � d1

� � 80

2x1 � 2.5x2 � d2� � d2

� � 80

2x1 � 1.5x2 � d3� � d3

� � 60

All variables � 0

See the graph to the right:

Graph for Alternative Example 11.3

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Third Priority

Goal: Minimize d3�

The area above the constraint line 2X1 � 1Z\xX2 � 60 is eliminated.

Fourth Priority


Cannot minimize d1� totally without violating first two priority

goals.

SOLUTION

X1 � 15

X2 � 20

d1� � 30

2X1 + 1 X2 = 60

5 10 15 20 25 30 35 40 450

5

10

15

20

25

30

35

40

45

X1

X2

d3

d3

d2

d2

d1

d1

–

–

–

+

+

+

B

A

C

/21

2X1 + 1 X2 = 60

5 10 15 20 25 30 35 40 450

5

10

15

20

25

30

35

40

45

X1

X2

d3

d3

d2

d2

d1

d1

–

–

–

+

+

+

B

A

C

D

X1 = 15X2 = 20

/21

First Priority


2X1 + 4X2 = 80

5 10 15 20 25 30 35 40 450

5

10

15

20

25

30

35

40

45

X1

X2

d1

d1

–

+

2X1 + 2 X2 = 80

5 10 15 20 25 30 35 40 450

5

10

15

20

25

30

35

40

45

X1

X2

d2

d2

d1

d1

–

–

+

+ /21

The area below the constraint line d1� is eliminated.

Second Priority


The area below the constraint line d2� is eliminated.

160 CHAPTER 11 INTEGER PROGRAMMING, GOAL PROGRAMMING, NONLINEAR PROGRAMMING



⁄5

a

l

⁄2

a

l

a

l

Alternative Example 11.4: Here is the simplex solution to thegoal programming problem in Alternative Example 11.3.

Initial Goal Programming Tableau

Cj l 0 0 P1 P2 0 P4 0 P3

Solutionb Mix X1 X2 d1

� d2� d3

� d1� d2

� d3� Quantity

P1 d1� 2 41⁄2 1 0 0 �1 0 0 80

P2 d2� 2 21⁄2 0 1 0 0 �1 0 80

0 d3� 2 11⁄2 0 0 1 0 0 �1 60

Cj � Zj P4 0 01⁄2 0 0 0 1 0 0 0P3 0 01⁄2 0 0 0 0 0 1 0P2 �2 �21⁄2 0 0 0 0 1 0 80P1 �2 �41⁄2 0 0 0 1 0 0 80

Pivot column

Pivot row

The Second Goal Programming Tableau

Cj l 0 0 P1 P2 0 P4 0 P3


� d2� d3

� d1� d2

� d3� Quantity

0 X21⁄2 1 1⁄2 0 0 �1⁄2 0 0 20

P2 d2� 3⁄2 0 �5⁄2 1 0 5⁄2 �1 0 60

0 d3� 5⁄2 0 �3⁄2 0 1 3⁄2 0 �1 60

Cj � Zj P4 0 0 05⁄2 0 0 1 0 0 0P3 0 0 05⁄2 0 0 0 0 1 0P2 �3⁄2 0 �5⁄2 0 0 �5⁄2 1 0 60P1 0 0 15⁄2 0 0 0 0 0 0

Pivot column

Pivot row

The Third Goal Programming Tableau

Cj l 0 0 P1 P2 0 P4 0 P3


� d2� d3

� d1� d2

� d3� Quantity

0 X24⁄5 1 0 1⁄5 0 0 �1⁄5 0 32

P4 d1� 3⁄5 0 �1 2⁄5 0 1 �2⁄5 0 24

0 d3� 8⁄5 0 0 �3⁄5 1 0 3⁄5 �1 24

Cj � Zj P4 �3⁄5 0 1 �2⁄5 0 0 2⁄5 0 24P3 02⁄5 0 0 02⁄5 0 0 02⁄5 1 0P2 02⁄5 0 0 12⁄5 0 0 02⁄5 0 0P1 02⁄5 0 1 02⁄5 0 0 02⁄5 0 0

Pivot column

Pivot row



SOLUTIONS TO DISCUSSION QUESTIONS

AND PROBLEMS

11-1. a. Linear programming allows only one goal (forexample, profit maximization) whereas goal programmingpermits multiple goals.b. LP always optimizes; goal programming sometimesonly “satisfies.”c. In goal programming, we deal with “deviationalvariables” as well as real variables.

11-2. When a non-integer solution to an LP problem is found andbranching is performed on one of the variables, an additional con-straint is added to each of two subproblems. These subproblemshave all of the previous constraints plus one new one. Therefore,the feasible region for the subproblem must be smaller than thefeasible region for the original LP problem. No new points areadded to the feasible region. Consequently, it is impossible for thesubproblem to have a better objective function value than the pre-vious LP problem.

11-3. a. Rounding off is the easiest way to solve an integerprogram, but it can give an infeasible or nonoptimalsolution.b. Enumeration is simple in concept, but it can be verytime consuming in large problems.c. The branch and bound method, which can becomputerized, is especially useful when solving largeproblems where enumeration is impractical. It does notalways reach an optimal solution in large problems,however.

11-4. The three types of integer programs are (1) pure integerprogramming, where all variables are integer; (2) mixed-integerprogramming, where some but not all variables are integer; and(3) zero–one integer programming, where all variables are either 0or 1 in value.

11-5. The upper and lower bounds are limits set at each branchand bound stage on the highest and lowest possible costs of a pos-sible assignment. The process is described in Section 11.2. Thebounds help us decide which branches can be discarded.

11-6. Satisficing is a term used in goal programming because itis often not possible to “optimize” a multi-goal problem. We comeas close as possible to reaching goals.

11-7. Deviational variables, similar to slack variables in LP, arethe difference between set goals and the current solution. In LP

problems, only “real” variables are used, representing physicalquantities. This is discussed in Section 11.3.

11-8. A college president’s goals might be to (1) increase en-rollments by 1,000 students; (2) stay within budget; (3) keep classsizes down to an average of 25 students; (4) increase facultysalaries; (5) develop 10 new off-campus courses; (6) reduce aver-age teaching loads to three courses per semester, and so on. Therewill be financial, space, tenure, and many other constraints.

11-9. Ranking goals just means more weight can be placed onone goal over another. The higher-ranked goals must be achievedcompletely before goal programming moves on to meet lower-ranked goals.

11-10. There are four differences between the LP and GP sim-plex methods.

1. GP has negative and positive deviational variables, eachwith a priority.2. The negative deviational variables provide the initialbasic feasible solutions and are analogous to slack variablesin LP.3. There is a separate Zj and Cj – Zj row for each of the pri-ority goals.4. The highest-priority row and the most negative Cj – Zj

value determine the variable to enter the solution next.

11-11. a. Linearb. Nonlinear because of 8X1X2 in objectivec. Goal programmingd. Nonlinear because X1

2 in first constrainte. Nonlinear and quadratic objective function

11-12. a. Let X � number of prime time ads per weekY � number of off-peak ads per week

Maximize audience exposure � 8200X � 5100YSubject to:

390X � 240Y � 1800X � 2Y � 6X, Y � 0

Solution: X � 2; Y � 4.25; audience � 38,075b. X � 2, Y � 4; audience � 36,800

There are other good solutions.c. Optimal integer solution:

X � 4, Y � 1; audience � 37,900

The Final Goal Programming Tableau

Cj l 0 0 P1 P2 0 P4 0 P3


� d2� d3

� d1� d2

� d3� Quantity

0 X2 0 1 �0 �1⁄2 �1⁄2 0 �1⁄2 1⁄2 20P4 d1

� 0 0 �1 �5⁄8 �3⁄8 1 �5⁄8 3⁄8 150 X1 1 0 �0 �3⁄8 �5⁄8 0 3⁄8 �5⁄8 15

Cj � Zj P4 0 0 �1 �5⁄8 �3⁄8 0 5⁄8 �3⁄8 15P3 0 0 �0 �0 �0 0 0 1 0P2 0 0 �0 �1 �0 0 0 0 0P1 0 0 �1 �0 �0 0 0 0 0



2 e X1 � 1 e X2 � 0 (one-third 757s)

800,000X1 � 500,000X2 � $8,000,000 (maintenance)

X1 � X2 � 17 (planes)

X1, X2 to be integers � 0

This is a pure integer programming problem.

The QM for Windows integer programming solution is:

X1 � 5

X2 � 8

Passengers carried � 1,273,000

11-16. a. X1 � X2 � X3 � X4 � X5 � X6 � 3b. X1 � X4 � 1c. X4 � X6

d. 2X5 � X2 � X3

e. X5 � 1 � X2 + X3

11-17. Let Xi � 1 if location i is selected and 0 otherwise, for i � 1 to 6.Minimize X1 � X2 � X3 � X4 � X5 � X6

Subject to: X1 � X6 � 1X1 � X2 � 1X2 � X3 � 1X1 � X3 � X6 � 1X3 � X4 � X5 � 1 X3 � X4 � X6 � 1X2 � X5 � 1X4 � X5 � 1

Xi � 0, 1 for i � 1 to 6.

Solution using QM for Windows Mixed-Integer ProgrammingModule:

X1 � X2 � X4 � 1; all other variables � 0. Objective functionvalue � 3This means only locations 1, 2, and 4 will be used.

11-18.

a.

Maximize profit �$5,000X1 � 6,000X2 � 10,000X3

� 12,000X4 � 8,000X5 � 3,000X6

� 9,000X7 � 10,000X8

subject to $60,000X1 � 50,000X2 � 82,000X3

� 103,000X4 � 50,000X5

� 41,000X6 � 80,000X7

� 69,000X8 � $300,000

b. X1 � 0, X2 � 1, X3 � 1, X4 � 0, X5 � 1, X6 � 1, X7 � 0,X8 � 1, Profit � 37,000

11-19. a. Let X1 � 1 if apartment project is undertaken; 0otherwise

Let X2 � 1 if shopping center project is undertaken; 0 otherwise

Let X3 � 1 if mini-warehouse project is undertaken; 0 otherwise

Maximize NPV � 18X1 � 15X2 � 14X3

Subject to:

40X1 � 30X2 � 20X3 � 80

30�1 � 20X2 � 20X3 � 50

X1, X2, X3 � 1 or 0

Let:1 if location is selected

0 ifX

ii =

location is not selectedi

⎧⎨⎩

2X1 + X2 ≤ 10

X2 ≤ 5

X1 ≤ 3

X1

X2a

b

c

0 1 2 3 4 50

1

2

3

4

5

6

7

8

9

10

OptimalSolution

Figure for Problem 11-14

11-13. Let Xi � 1 if item i is selected and 0 otherwise, for i � 1 to 8.Maximize 80X1 � 20X2 � 50X3 � 55X4 � 50X5 � 75X6

� 30X7 � 70X8

Subject to: 8X1 � X2 � 7X3 � 6X4 � 3X5 � 12X6

� 5X7 � 14X8 � 35Xi � 0, 1

Solution using QM for Windows Mixed-Integer ProgrammingModule:

X1 � X2 � X4 � X5 � X6 � X7 � 1X3 � X8 � 0

Objective function � 310

11-14. X1 � number of larger posters

X2 � number of smaller posters

Maximize profit � 3X1 � 2X2

subject to X1 � 3

X2 � 5

2X1 � X2 � 10

X1, X2 � 0

See graph below.

Step 1. Optimal LP solution at a is (X1 � 21⁄2, X2 � 5, profit �$17.50). Step 2. Integer solution at b is (X1 � 3, X2 � 4, profit �$17). Integer solution at c is (X1 � 2, X2 � 5, profit � $16).Hence the optimal integer solution is X1 � 3 large posters and X2 � 4 small posters (seen at point b).

11-15. X1 � number of Boeing 757s purchased

X2 � number of Boeing 767s purchased

Maximize passenger carrying capability � 125,000X1 � 81,000X2

subject to

80X1 � 110X2 � 1,600 ($ million available)

[X1 � 1 e (X1 � X2)] or



b. The optimal solution is X1 � 1, X2 � 1, X3 � 0. NPV � 33.This means that both the apartment project and the shopping cen-ter project will be undertaken. The amount of money spent in year1 would be $70 (thousand) and in year 2 would be $50 (thousand).

11-20. a. X1 � X2 This means that if the apartment is not built(X1 � 0), the shopping center cannot be built (X2 must equal 0).

b. X1 � X2 � X3 � 2

11-21. a. Let Xij � 1 if generator i is functioning during timeperiod j, and 0 otherwise; where i � 1, 2, 3 and j � 1 for 6–2 timeperiod; j � 2 for 2–10 time period; j � 3 for 6�10 time period.

Let Yij � megawatts produced by generator i in time period j,where i � 1, 2, 3 and j � 1 for 6–2 time period; j � 2 for 2–10time period.

Minimize cost � 6,000(X11 � X12 � X13) � 5,000(X21 � X22 � X23)� 4,000(X31 � X32 � X33) � 8(Y11 � Y12) � 9(Y21 � Y22)� 7(Y31 � Y32)

Subject to:

Y11 � Y21 � Y31 � 3,200 megawatts requirements from 6–2

Y12 � Y22 � Y32 � 5,700 megawatts requirements from 2–10

Y11 � 2,400(X11 � X13) maximum megawatts from #1 from 6–2






X11 � X12 � X13 � 1 generator #1 starts up at most once

�21 � X22 � X23 � 1 generator #2 starts up at most once

�31 � X32 � X33 � 1 generator #3 starts up at most once

�ij � 0 or 1 for all i, j

Yij � 0 for all i, j

b. The solution is: X12 � 1, X33 � 1, Y12 � 2,400, Y31 � 3,200, Y31 � 3,300, total cost � $74,700.

Thus, generator #1 will be utilized in the period 2–10 and willgenerate 2,400 megawatts of electricity. Generator #3 will bestarted at 6 and utilized for the entire 16 hours. It will generate3,200 megawatts during the 6–2 time period, and 3,300 megawattsduring the 2–10 time period.

11-22. Let T � number of TV ads, R � number of newspaper ads,B � number of billboard ads, and N � number of newspaper ads.

Minimize P1d1� � P2d2

� � P3d3� � P4d4

� � P4d5� � P4d6

� � P4d7�

Subject to:

(1) number of people reached

40,000T � 32,000R � 34,000B � 17,000N � d1� � d1

� �1,500,000

(2) budget

900T � 500R � 600B � 180N � d2� � d2

� � 16,000budget constraint

(3) number of TV or radio ads

T � R � d 3� � d 3

� � 6

(4) restriction on number of each individual type of ad

T � d4� � d4

� � 10

R � d5� � d5

� � 10

B � d6� � d6

� � 10

N � d7� � d7

� � 10

All variables � 0

b. T � 0, R � 0.73, B � 0, N � 88.86

c. Goal 1 (number of people reached) and goal 2 (budget) aremet completely. The number of TV, radio, and billboard ads areeach less than 10. The other goals are not met.

11-23. Maximize profit � 2X1 � 3X2

subject to X1 � 3X2 � 9

3X1 � X2 � 7

X1 � X2 � 1

X1, X2 � 0

1. Solve graphically as an LP problem:

X1 � 1.5X2 � 2.5

profit � $10.50

This provides an upper bound value.

2. Round down to X1 � 1, X2 � 2, profit � $8.00 for afeasible solution. The lower bound is $8.00.

3. Branch on X2 to begin:

Subproblem A Subproblem B

New constraint: X2 � 2 New constraint: X2 � 3Optimal solution: X2 � 2 Optimal solution: X2 � 3

X1 � 1.6 X1 � 0profit � $9.33 profit � $9.00

(new upper bound) (new lower bound)

Subproblem C Subproblem D

New constraints: X2 � 2 New constraints: X2 � 2X1 � 1 X1 � 2

Optimal solution: X2 � 2 Optimal solution: X2 � 1X1 � 1 X1 � 2

profit � $8.00 profit � $7.00

5. Both of these subproblems yield all-integer solutions.Comparing them to the lower bound of $9.00, we seethey are both smaller (see the graph in the next col-umn). The solution to the problem (see subproblemB) is

X1 � 0, X2 � 3, profit � $9.00.

4. Branch on X1 now from subproblem A:



11-26. X1 � number of 64MB chips produced

X2 � number of 256MB chips produced

X3 � number of 512MB chips produced

d1� � underfilling customers’ orders of 64MB chips

d2� � underfilling customers’ orders of 256MB chips

d3� � underachievement of sales quotas for 64MB chips

d4� � underachievement of sales quotas for 256MB

chips

d5� � underachievement of sales quotas for 512MB

chips

d6� � underutilization of plant capacity

Minimize deviations

� P1d1� � P1d2

� � P2d3� � P2d4

� � P2d5� � P3d6

�

subject to

X1 � d1� � d1

� � 30 (64MB chips order)

X2 � d2� � d2

� � 35 (256MB chips order)

X1 � d3� � d3

� � 40 (64MB sales goal)

X2 � d4� � d4


X3 � d5� � d5


8X1 � 13X2 � 16X3 � d6� � 1,200 (hours capacity)

All variables � 0


X1

X2

A

B

200 400 600 800

Production Limit

1,0001,300

1,2001,100

100

200

300

400

500

600650700733

800

C

Two-DrawerSales Limit

Profit Target

Three-DrawerSales Limit

d4

d4

d2

d2d1

d1

–

d3–

––

+

+

d3+ +

11-24. Let: X1 � number of two-drawer cabinets produced eachweek

X2 � number of three-drawer cabinets producedeach week

d1� � underachievement of profit goal

d1� � overachievement of profit goal

d2� � idle time in production capacity

d3� � underachievement of sales goal for two-

drawer files

d4� � underachievement of sales goal for three-

drawer files

Minimize deviations

� P1d1� � P1d1

� � P2d2� � P3d3

� � P3d4�

subject to

10X1 � 15X2 � d1� � d1

� � $11,000 (profit target)

1X1 � 2X2 � d2� � 1,300 hours (production limit)

1X1 � d3� � 600 (two-drawer sales limit)

X2 � d4� � 400 (three-drawer sales limit)

All Xi, di variables � 0

11-25. Because we want to achieve the profit goal as closely aspossible (minimize both d1

� and d1�), the line ABC becomes the

feasible region. When the P2 priority is included, the feasible re-gion is reduced to the segment AB. P3 priority applies to both d3

�

and d4�. The three-drawer goal (d4

�) is fully attained at point Band the two-drawer goal (d3

�) is almost reached.


X1 = 1.6X2 = 2P = 9.33

X1 = 0X2 = 3P = 9.00

X1 = 1X2 = 2P = 8.00

X1 = 1X2 = 1P = 7.00

X1 = 1.5X2 = 2.5P = 10.50

Upper Bound = $10.50Lower Bound = $8.00

Upper Bound = $9.33Lower Bound = $9.00

Subproblem A

Infeasible, noninteger

Subproblem C

Subproblem D

Subproblem B

Feasible, integer solution

Optimal Solution

X 2 ≤ 2

X1 ≤ 1

X2 ≥ 3

X1 ≥ 2

The best solution is X1 � 500, X2 � 400. The value of d3� �

100, meaning the two-drawer sales goal is underachieved by 100cabinets. See the graph below.


a


Cj l 0 0 P1 P2 0 P4 0 0 P3 0Solution

b Mix X1 X2 d1� d2

� d3� d4

� d1� d2

� d3� d4

� Quantity

0 X1 1 0 1–3 – 2–3 0 0 – 1–32–3 0 0 2

0 X2 0 1 – 2–97–9 0 0 2–9 – 7–9 0 0 8–3

0 d3� 0 0 – 8–9

1–9 1 0 8–9 – 1–9 �1 0 14–3P4 d4

� 0 0 2–9 – 7–9 0 1 – 2–97–9 0 �1 13–3

0 0 2–9 – 7–9 0 1 – 2–97–9 0 �1 13–3

P4 0 0 – 2–97–9 0 0 2–9 – 7–9 0 �1

0 0 0 0 0 0 0 0 0 0 0

P3 0 0 0 0 0 0 0 0 1 0

0 0 0 0 0 0 0 0 0 0 0

P2 0 0 0 1 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0

P1 0 0 1 0 0 0 0 0 0 0

a Pivot rowPivot

column

Z

C Zj

j j–

⎧⎨⎪

⎩⎪

Z

C Zj

j j–

⎧⎨⎪

⎩⎪

Z

C Zj

j j–

⎧⎨⎪

⎩⎪

Z

C Zj

j j–

⎧⎨⎪

⎩⎪

11-27. Third tableau for Harrison Electric:

Fourth tableau for Harrison Electric:

Cj l 0 0 P1 P2 0 P4 0 0 P3 0Solution

b Mix X1 X2 d1� d2

� d3� d4

� d1� d2

� d3� d4

� Quantity

0 d2� 3–2 0 1–2 �1 0 0 – 1–2 1 0 0 3

0 X27–6 1 1–6 0 0 0 – 1–6 0 0 0 5

0 d3� 1–6 0 – 5–6 0 1 0 5–6 0 �1 0 5

P4 d4� – 7–6 0 – 1–6 0 0 1 1–6 0 0 �1 2

– 7–6 0 – 1–6 0 0 1 1–6 0 0 �1 2

P47–6 0 1–6 0 0 0 – 1–6 0 0 �1

0 0 0 0 0 0 0 0 0 0 0

P3 0 0 0 0 0 0 0 0 1 0

0 0 0 0 0 0 0 0 0 0 0

P2 0 0 0 1 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0

P1 0 0 1 0 0 0 0 0 0 0Z

C Zj

j j–

⎧⎨⎪

⎩⎪

Z

C Zj

j j–

⎧⎨⎪

⎩⎪

Z

C Zj

j j–

⎧⎨⎪

⎩⎪

Z

C Zj

j j–

⎧⎨⎪

⎩⎪

The third tableau corresponds to point C on the graph. The fourth tableau corresponds to point D on the graph.



Cj l 0 0 P1 P2 0 P4 0 P3


� d2� d3

� d1� d2

� d3� Quantity

0 X2 0 1 0 1–4 – 1–4 0 – 1–41–4 20

P4 d1� 0 0 �1 5–8 – 3–8 1 – 5–8

3–8 300 X1 1 0 0 – 3–16

5–16 0 3–16 – 5–16 15

0 0 �1 5–8 – 3–8 1 – 5–83–8 30

P4 0 0 1 – 5–83–8 0 5–8 – 3–8

0 0 0 0 0 0 0 0 0

P3 0 0 0 0 0 0 0 1

0 0 0 0 0 0 0 0 0

P2 0 0 0 1 0 0 0 0

0 0 0 0 0 0 0 0 0

P1 0 0 1 0 0 0 0 0Z

C Zj

j j–

⎧⎨⎪

⎩⎪

Z

C Zj

j j–

⎧⎨⎪

⎩⎪

Z

C Zj

j j–

⎧⎨⎪

⎩⎪

Z

C Zj

j j–

⎧⎨⎪

⎩⎪

11-28. a.

11-28. b.

Cj l 0 0 P1 P2 0 P4 0 P3


� d2� d3

� d1� d2

� d3� Quantity

P1 d1� 2 4 1 0 0 �1 0 0 80

P2 d2� 8 10 0 1 0 0 �1 0 320

0 d3� 8 6 0 0 1 0 0 �1 240

0 0 0 0 0 0 0 0 0

P4 0 0 0 0 0 1 0 0

0 0 0 0 0 0 0 0 0

P3 0 0 0 0 0 0 0 1

8 10 0 1 0 0 �1 0 320

P2 �8 �10 0 0 0 0 1 0

2 4 1 0 0 �1 0 0 80

P1 �2 �4 0 0 0 1 0 0

a

Pivotcolumn

Z

C Zj

j j–

⎧⎨⎪

⎩⎪

Z

C Zj

j j–

⎧⎨⎪

⎩⎪

Z

C Zj

j j–

⎧⎨⎪

⎩⎪

Z

C Zj

j j–

⎧⎨⎪

⎩⎪

k Pivot row

The best solution is

X1 � 15

X2 � 20

d1� � 30



11-29. a. d1� � underachievement of class and study goal

d1� � overachievement of class and study goal

d2� � overachievement of sleeping goal

d3� � underachievement of social time goal

Major Bligh’s objective function becomes

minimize � d1� � d1

� � d2� � d3

�

subject to constraints (per week)

1X1 � 1X2 � 1X3 � 1X4 � 168

1X3 � d1� � d1

� � 30

1X1 � d2� � 49

1X4 � d3� � 20

All variables � 0

Since the goals have priority, they can be rewritten in this order,yielding to the absolute completion of each goal before attemptingto achieve the next goal. The objective function would become

minimize � P1d1� � P1d1

� � P2d2� � P3d3

�

where P1 � meet class and study goal

P2 � meet sleeping goal

P3 � meet socializing goal

b. X1 � 49X2 � 69X3 � 30X4 � 20All goals are fully met.

11-30. a. Let S � dollars invested in stocks; B � dollarsinvested in bonds;

R � dollars invested in real estateMinimize d1

� � d2� � d3

�

Subject to

0.13S + 0.08B + 0.10R � Return is at least 10%d1

� � d1� � 25,000

B � d2� � d2

� = 75,000 Amount in bonds is at least 30%

R � d3� � d3

� = 0.50(S + B) Real estate is less than half of stocks and bonds

S � B � R � 250,000

S �150,000

� 150,000

� 150,000

S, , � 0

b. S � $50,000 invested in stocks

B � 75,000 invested in bonds;

R � $125,000 invested in real estate

The total return is $25,000 (10%). The amount invested in real es-tate is not less than half than the amount invested in stocks andbonds. This is the only goal that is not met.

11-31. Maximize profit � X1(1,800 � 50X1)

� X2(2,400 � 70X2)

� 1,800X1 � 50X21

� 2,400X2 � 70X22

X1, X2 � 0

subject to 100X1 � 130X2 � 5,000 hours

X1, X2 � 0

11-32. Let X1 � no. of XJ6’s and X2 � no. of XJ8’s

a. Maximize Z � 4X1 � .1X21 � 5X2 � .2X2

2

subject to X1 � 2X2 � 40

X1, X2 � 0

b. X1 � 18.3; X2 � 10.8; revenue � $70,420

11-33. The optimal solution found using Solver in Excel is X � 62.73, Y � 8.64, Profit � 720.41.

11-34. The optimal solution found using Solver in Excel is X � 0.333, Y � 0.667, with a variance of 0.102 and a return of 0.09.

11-35. a. Total profit � (P1�6)X1 � (P2�8)X2

b. The optimal solution found using Solver in Excel is X1 � 260, X2 � 140, P1 � 20, P2 � 17.33,

profit � $4,946.67.

11-36. a. Z � $665,000

SOLUTIONS TO INTERNET HOMEWORK PROBLEMS

11-37. Maximize return � 50X1 � 100X2 � 30X3

� 45X4 � 65X5 � 20X6

� 90X7 � 35X8

subject to 500X1 � 1,000X2 � 350X3 � 490X4

� 700X5 � 270X6 � 800X7 � 400X8 � 3,000

X1 � X2 � X3 � X4 � X5 � X6 � X7 � X8 � 5

X1 � X2 � 1

X3 � X4 � X5 � 2

X6 � X7 � X8 � 2

All Xi � 0 or 1

Variable Value

X1 0X2 0X3 0X4 0X5 (South Orlando) 1X6 0X7 0X8 (Apopka) 1X9 (Lake Mary) 1X10 (Cocoa Beach) 1

Variable Value Variable Value

X1 0 X6 0X2 0 X7 0X3 0 X8 1X4 0 X9 1X5 1 X10 1

b. The expected return drops to $625,000. Osceola opensand Cocoa Beach closes.c. As seen below, with Apopka corrected, the new solutionhas a return of $635,000 but the same locations as part a.

Solution:

Z � $635,000



11-38. Lower bound set on rows with assignment A4 ($10), B1($6), C3 ($5), D4 ($25): Total cost $46.

Two optimal solutions (see the figure above) with a total costof $105:

Optimal Solutions

A2B1C3D4

$106

A4B2C3D4

$55

A4B1C3D4

$46

A1B2C3D4

$130

A4B2C1D4

$100

A4B2C3D1

$105

A3B2C1D4

$150

A4B2C1D3

$150

A4B1C2D4

$131

A4B1C3D2

$105

LowerBound = $46

Feasible

Feasible

Not Feasible

Feasible

Feasible

Feasible

Feasible

Not Feasible

Not Feasible

Lower Bound = $55Upper Bound = $105

Lower Bound = $100Upper Bound = $105

A2B2

C2

D2

C1

A1

D1

A3

D3

Upper Bound = –

Branch and Bound Solution for Problem 11-38.


X1

X2

B

A

C

ExposureConstraint

BudgetConstraint

TV Spots Constraint

Newspaper AdsConstraint

d4

d4

d2d2

d1

d1

–

d3–

–

–

+

+

d3+

+

0 5 10 15 20 25 30

TV Spots

10

20

30

40

50

60

70

Ads

11-40. The first two priorities, P1 and P2, are fully satisfied by theregion ABC. But the P3 priority requires that we select a solutionabove the exposure constraint line (minimize d4

�). Point A comesclosest to reaching the P3 goal. The best solution is

X1 � 10 TV spots

X2 � 35 newspaper ads

Total exposure here is 8,250,000 people, so d4� � 750,000 people.

In other words, the exposure goal was underachieved by 3–4 millionpeople.

Notice that in this problem d2� and d3

� are of equal (P2) priorityand hence are equally important. See the graph below.

Assignment Cost Assignment Cost

A4 $ 10 A4 $ 10B1 6 B2 15C3 5 C3 5D2 $184 D1 $175

$105 $105

11-39. X1 � number of TV spots

X2 � number of newspaper ads

d1�� deviation above budget funds of $120,000

d2�� number of TV spots below 10

d3�� number of newspaper ads below 20

d4�� deviation below exposure of 9 million persons

desired

Minimize deviations � P1d1� � P2d2

� � P2d3� � P3d4

�

subject to

5,000X1 � 2,000X2 � d1�� $120,000 (budget

constraint)

X1 � d2� � 10 (TV spots)

X2 � d3� � 20 (newspaper ads)

300,000X1 � 150,000X2 � d4�� 9,000,000 (exposures)

All variables � 0



11-41. Let S � number of Standard blenders produced eachweek

D � number of Deluxe blenders produced each week

C � number of Chef’s Delight blenders produced each week

Minimize d1� � d1

� � d2� � d3

� � d4� � d5

�

Subject to:

(1) use 240 hours per week

1.5S � 2D � 2.5C � d1� � d1

� � 240

(2) produce 60 of the Chef’s Delight blenders

C � d2� � d2

� � 60

(3) produce 60 of the Deluxe blenders

D � d3� � d3

� � 60

(4) produce 60 of the Standard blenders

S � d4� � d4

� � 60

(5) generate profit of at least $3,500

28S � 32D � 35C � d5� � d5

� � 3,500

All variables � 0

11-42. The constraints are the same as in Problem 11-41. The objective function changes to:

Minimize d1� � d1

� � 0.5d2� � 0.5d3

� � 0.5d4� � 0.333d5

�

SOLUTION TO SCHANK MARKETING RESEARCH CASE

1. The first part of this case is an assignment problem that can beformulated with LP. A dummy project manager can be added tocreate a balanced 4 � 4 cost matrix.

Minimize �

where Xij �

i � 1, 2, 3, 4 for Gardener, Ruth, Hardgraves, Dummy

j � 1, 2, 3, 4 for Hines, NASA, General, CBT

2. This part is a goal programming formulation with five goals,ranked from P1 (highest) to P5 (lowest):

P1: assign a manager to the NASA account.

P2: do not assign Gardener to CBT Television account.

P3: meet demands of Hines; they are twice as important asthose of General Foundry.

P4: place Ruth on a project that will earn him $3,000 ormore.

P5: minimize the total cost of all assignments.

Constraints

For client’s demand:

X11 � X21 � X31 � d1� � 1 Hines

X12 � X22 � X32 � d2� � 1 NASA

X13 � X23 � X33 � d3� � 1 General

X14 � X24 � X34 � d4� � 1 CBT

1 if project leader i is assignedto client j{ 0 if otherwise

C Xij iji

m

j

n

==∑∑

11

These constraints assume no more than one assignment per manager.

For project managers:

X11 � X12 � X13 � X14 � d5� � d5

� � 1 Gardener

X21 � X22 � X23 � X24 � d6� � d6

� � 1 Ruth

X31 � X32 � X33 � X34 � d7� � d7

� � 1 Hardgraves

These constraints permit assigning three managers to four clientswhile minimizing positive and negative deviational variables (d5, d6, d7).

Gardener to CBT restriction:

X14 � d8� � 0

This constraint looks at the deviation of d8� from 0. In other

words, the closer d8� is to 0 (not assigning Gardener to CBT), the

closer it comes to meeting the restriction.Ruth earns $3,000 or more:

2,700X21 � 3,200X22 � 3,000X23

� 3,100X24 � d9� � d9

� � $3,000

Here d9� represents underachievement of the goal, while d9

� isoverachievement. The coefficients are the costs per assignment.

Total costs:

This attempts to minimize total cost, bringing it as close to zero aspossible; d10

� is the deviation from the goal.Objective function:

minimize Z � P1d2� � P2d8

� � P3(2d1� � d3

�) � P4d9

� � P5d10�

SOLUTION TO THE OAKTON RIVER BRIDGE CASE

For a given set of requirements, the smallest number of toll collec-tors that will meet them can be obtained from the following inte-ger linear programming problem:

minimize Z � X1 � X2 � X3 � X4 � X5 � X6 � X7

subject to

X1 � X2 � X3 � X4 � X5 � X6 � X7 � R5

X1 � X2 � X3 � X4 � X5 � R5

X2 � X3 � X4 � X5 � X6 � R6

X3 � X4 � X5 � X6 � X7 � R7

X1 X3 � X4 � X5 � X6 � X7 � R1

X1 � X2 � X5 � X6 � X7 � R2

X1 � X2 � X3 � X6 � X7 � R3

X1 � X2 � X3 � X4 � X7 � R4

All variables � 0

where Xj is the number of collectors starting on day j ( j � 1 isSunday) and Rj is the number required on day j.

1. The following table summarizes the requirements for shifts A,B, and C for each of the three days of the week along with the allo-cations that yield the minimum numbers of collectors startingeach: 18 for shift A, 16 for shift B, and 18 for shift C.

( ) –ij

ij ijC X d==

+∑∑ =1

3

1

4

10 0



2. If mixing of shifts is allowed, the daily requirements becomethe sum of the shift requirements, as shown in the second part ofthe table. The minimum number of collectors starting each day isshown in the last day. The total 50 is a reduction of two from thetotal required without allowing for the mixing of shifts.

SOLUTION TO PUYALLUP MALL CASE

The problem can be expressed as the following integer linear pro-gramming problem with Xi being a 0–1 variable, 1 if store i is tobe included and 0 if not:

Maximize

28.1X1 � 34.6X2 � 50.0X3 � 162.0X4 � 77.8X5

� 100.4X6 � 45.2X7 � 80.2X8 � 51.4X9 � 62.5X10

� 18.0X11 � 11.6X12 � 50.4X13 � 73.6X14

� 51.2X15

subject to the space constraint

1.0X1 � 1.6X2 � 2.0X3 � 3.2X4 � 1.8X5 � 2.1X6

� 1.2X7 � 2.4X8 � 1.6X9 � 2.0X10 � 0.6X11

� 0.5X12 � 1.4X13 � 2.0X14 � 1.0X15 � 16

the annual rent constraint

4.4X 1 � 6.1X2 � 8.3X3 � 24.0X4 � 19.5X5 � 20.7X6

� 7.7X7 � 19.4X8 � 11.7X9 � 15.2X10 � 3.9X11

� 3.2X12 � 11.3X13 � 16.0X14 � 9.6X15 � 130

the construction cost constraint

24.6 X1 � 32.0X2 � 41.4X3 � 124.4X4 � 64.8X5

� 79.8X6 � 38.6X7 � 66.8X8 � 45.1X9 � 54.3X10

� 15.0X11 � 13.4X12 � 42.0X13 � 63.7X14

� 40.0X15 � 700

at least one clothing store

X1 � X2 � X3 � 1

at least one hard goods store

X8 � X9 � X10 � 1

at least one miscellaneous-type store

X11 � X12 � X13 � X14 � X15 � 1

at least two restaurants

X4 � X5 � X6 � X7 � 2

no more than two clothing stores

X1 � X2 � X3 � 2

miscellaneous types cannot exceed total of clothing and hardgoods

X1 � X2 � X3 � X8 � X9 � X10 � X11 � X12 � X13

� X14 � X15 � 0

The optimum solution is to include stores 1, 4, 5, 6, 8, 10, 12,14, and 15. The present value is $647,400, all 16,000 square feetof space will be used, the annual rent is $132,000, and the con-struction cost is $531,800.

Toll Collector Requirements for Oakton River Case

SHIFT A B C Mix

DAY Req. Start Req. Start Req. Start Req. Start

Sun. 8 0 10 0 15 5 33 3Mon. 13 3 10 1 13 2 36 9Tue. 12 5 10 5 13 1 35 8Wed. 12 0 10 1 12 4 34 6Thu. 13 5 10 5 12 1 35 9Fri. 13 1 13 1 13 5 39 7Sat. 15 14 15 13 8 10 38 18Total 18 16 18 50

Note: Alternative optimal solutions for each shift may be possible.


Chapter 11 Answers

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Transcript of Chapter 11 Answers