Chapter 11 Answers

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11 H

A

P

T

E

R

Integer Programming, Goal Programming, Nonlinear Programming and the Branch and Bound Methods TEACHING SUGGESTIONS Teaching Suggestion 11.1: Topics in This Chapter.

The overall purpose of this chapter is to provide a framework for the topics of integer programming, branch and bound, nonlinear programming and goal programming. These are fairly advanced topics in a mathematical sense, and the chapter’s intention is solely to introduce them through a series of simple graphical problems. Some of the topics are on the cutting edge of QA. For example, in integer and nonlinear programming, no one solution procedure exists to handle all problems. Teaching Suggestion 11.2: Using the Computer to Solve Mixed-Integer Programming Programming Problems. Problems.

Note that the Excel printout in Program 11.2 allows users to specify which variables are integers and which, by default, can be fractional. Teaching Suggestion 11.3: How the Branch and Bound Method Can Help.

In this section we illustrate how branch and bound is used to solve small assignment and integer programming problems. But its real strength is in dealing with huge problems (for example, thousands of variables/constraints). Branch and bound allows us to divide a large problem into smaller parts, thereby eliminating one-half or two-thirds of the options and reducing the problem to a more manageable level. Teaching Suggestion 11.4: Multiple Goals.

Ask studdents what other goals a company might have beyond maximizing profit. Socially conscious firms need to state as their mission a whole series of objectives. Encourage students to research an article showing a goal programming application. There is a wealth of research in journals. One interesting application application is in the box later in this section that deals with budgeting for prisons. Teaching Suggestion 11.5: Deviational Variables Variables Are the Key in Goal Programming.

The concept of deviational variables requires careful explanation to the class. Students are accustomed to the decision variables of X 1 and X 2. Now they need to concentrate on goal achievement. The minus and plus signs on deviational variables need a thoughtful classroom discussion.

158

Teaching Suggestion 11.6: Difficulty 11.6: Difficulty of Graphical Graphical Goal Programming.

Solving goal programming problems graphically can be a confusing concept relative to graphical LP. Students often have difficulty with the direction of deviational variables. Teaching Suggestion 11.7: Using the Goal Programming Simplex Method.

Point out the similarities and differences between the simplex method and the modified goal programming tableau. You can show that the structure is almost the same. The big change is the addition of two rows for each new goal. Surprisingly, the computation is not as difficult as it looks.

ALTERNATIVE EXAMPLES Alternative Example 11.1:

0–1 Integer Programming.

Indiana’s prison budget allows it to consider four new installations next year. They are X 1 ϭ 1 if maximum security prison in Ft. Wayne, ϭ

0 otherwise

X 2 ϭ 1 if minimum security prison in Bloomington, ϭ

0 otherwise

X 3 ϭ 1 if halfway house in Indianapolis, 0 otherwise X 4 ϭ 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 ( X 1 or X 2) and observing cost and space limitations. Here is the formulation: maximize number served ϭ 3,000 X 1 ϩ 900 X 2 ϩ 4,000 X 3 ϩ 1,500 X 4 subject to X 1 ϩ X 2 р 1 prison

4 X 1 ϩ 2 X 2 ϩ 7 X 3 ϩ 3 X 4

р

12 acres available

3.5 X 1 ϩ 1 X 2 ϩ 2.5 X 3 ϩ 9 X 4 р 12 million dollars budgeted Solution: Using software, we find that X 1 X 4 ϭ 0, number served ϭ 7,000.

ϭ

1, X 2

ϭ

0, X 3

ϭ

1,

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


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I N T E G E R P R O G R A M M I N G , G O AL P R O G R A M M I N G , N O N L I N E A R P R O G R A M M I N G

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

S OLUTION

Initial upper bound (UB)

maximize profit ϭ $8,500 X 1 ϩ $6,000 X 2

ϭ

$61,667 ( X 1 ϭ X\c, X 2 ϭ 9Z\c)

Initial lower bound (LB) ϭ $54,000 ( X 1 ϭ 0, X 2 ϭ 9) (See graph below for this example.)

subject to X 1 ϩ X 2 р courses max. of 10

All nodes are either integer or infeasible, so the solution is seen to be X 1 ϭ 3, X 2 ϭ 6, profit ϭ $61,500.

$1,000 X 1 ϩ $700 X 2 р instructor’s pay of $7,200 X 1, X 2 ജ 0 and are integers

Using LP, the solution is: X 1 ϭ X\c, X 2 ϭ 9Z\c, profit ϭ $61,667.

=0 = 10 P = $60,000 X 1 X 2

0 ≤

X 1 = 2/3 = 9 1/3 P = $61,667 X 1 X 2

UB = $61,667 LB = $54,000 X 1 ≥

1

Not Feasible

≥ 9

=1 = 8 7/8 P = $61,643 X 1 X 2

X 2

X

≤ 1

2 ≤

8

= 1 3/5 =8 P = $61,000 X 1 X 2

Figure for Alternative Example 11.2

=1 =8 P = $56,500 X 1 X 2

X 1

UB = $61,600 LB = $54,000 [X 1 = 0, X 2 = 9]

X

1 ≥

Not Feasible

≥ 8

2

=2 = 7 3/7 P = $61,571 X 1 X 2

X 2

X

2 ≤

X 1

7

= 2 3/10 =7 P = $61,550 X 1 X 2

≤ 2

X

1 ≥

=2 =7 P = $59,000 X 1 X 2

3

=3 =6 P = $61,500 X 1 X 2

Optimal Solution

Alternative Example 11.3: Minimize P1d 1Ϫ ϩ P2d 2Ϫ ϩ P3d 3ϩ ϩ P4d 1ϩ

Graph for Alternative Example 11.3 X 2

subject to 2 x 1 ϩ 4 x 2 ϩ d 1Ϫ Ϫ d 1ϩ ϭ 80 Ϫ

Ϫ

d 2 ϭ 80

Ϫ

Ϫ

d 3ϩ ϭ 60

2 x 1 ϩ 2.5 x 2 ϩ d 2 2 x 1 ϩ 1.5 x 2 ϩ d 3 All variables у 0

See the graph to the right:

ϩ

45 40 35

2X 1 + 11 / 2 X 2 = 60

30 2X 1 + 4X 2 = 80

25 20

2X 1 + 21 / 2X 2 = 80

15 10 5 0

5

10

15

20

25

30

35

Goal Constraints

40

45

X 1


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First Priority

Third Priority

Goal: Minimize d 1Ϫ

Goal: Minimize d 3ϩ

X 2

X 2

45

45

40

40

35

35

30

30

25

25

20

2X 1 + 4X 2 = 80

d 1+

5

10

15

20

25

30

35

40

45

X 1

C

d 1–

0

5

10

15

20

25

30

35

40

45

X 1

Fourth Priority Goal: Minimize d 1ϩ

X 2

X 2

45

45

40

B

40

35

35

d 2+ d 2–

d 3+

2X 1 + 11 / 2X 2 = 60

d 3–

30

20

A

d 2+

25 2X 1 + 21 / 2X 2 = 80

d 1+

d 2–

20

15

d 1+

15

d 1–

C

X 1 X 2

= 15 = 20

d 1–

10

5 0

d 1+

The area above the constraint line 2 X 1 ϩ 1Z\x X 2 ϭ 60 is eliminated.

Goal: Minimize d 2Ϫ

10

d 2 –

5

Second Priority

25

d 2+

A

10

The area below the constraint line d 1Ϫ is eliminated.

30

2X 1 + 11 / 2 X 2 = 60

d 3–

15 d 1 –

5 0

d 3+

20

15 10

B

5 5

10

15

20

25

30

35

40

The area below the constraint line d 2Ϫ is eliminated.

45

X 1

D

0

5

10

15

20

25

30

35

40

45

X 1

Cannot minimize d 1ϩ totally without violating first two priority goals. S OLUTION X 1

ϭ

15

X 2

ϭ

20

ϭ

30

ϩ

d 1


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

Initial Goal Programming Tableau C j l

b a

P 1 P 2

0 C j Ϫ Z j

Solution Mix

0

0

P 1

P 2

0

P 4

0

P 3

X 1

X 2

d 1Ϫ

d 2Ϫ

d 3Ϫ

d 1ϩ

d 2ϩ

d 3ϩ

Quantity

1 0 0

0 1 0

0 0 1

Ϫ1

0

0 0

Ϫ1

0

0 0 Ϫ1

80 80 60

0 0 0 0 Pivot column

0 0 0 0

0 0 0 0

1 0 0 1

0 0 1 0

0 1 0 0

0 0 80 80

Ϫ

2 2 2

d 1 d 2Ϫ d 3Ϫ

0 0 Ϫ2 Ϫ2

P 4 P 3 P 2 P 1

4 ⁄ 2 1 2 2 ⁄ 1 2 1 ⁄ 1

0 ⁄ 2 1 2 0 ⁄ 1 Ϫ2 ⁄ 2 1 l 2 Ϫ4 ⁄ 1

Pivot row

The Second Goal Programming Tableau C j l

b 0 a

P 2

0 C j Ϫ Z j

0

0

P 1

P 2

0

P 4

0

P 3

Solution Mix

X 1

X 2

d 1Ϫ

d 2Ϫ

d 3Ϫ

d 1ϩ

d 2ϩ

d 3ϩ

Quantity

X 2 d 2Ϫ d 3Ϫ

1 ⁄ 2 3 ⁄ 2 5 ⁄ 2

1 0 0

1 ⁄ 2 Ϫ5 ⁄ 2 Ϫ3 ⁄ 2

0 1 0

0 0 1

1 Ϫ ⁄ 2

0 Ϫ1 0

0 0 Ϫ1

20 60 60

0 0 3 Ϫ ⁄ 2 0

0 0 0 0

05 ⁄ 2 05 ⁄ 2 5 ϩ ⁄ 2 15 ⁄ 2

0 0 0 0

0 0 0 0

1 0 5 2 Ϫ ⁄ ⁄ 0 l

P 4 P 3 P 2 P 1

⁄ 2 3 ⁄ 2 5

2

0 0 1 0 Pivot column

0 1 0 0

0 0 60 0

Pivot row

The Third Goal Programming Tableau C j l

b 0 P 4 a

0 C j Ϫ Z j

Pivot row

0

0

P 1

P 2

0

P 4

0

P 3

Solution Mix

X 1

X 2

d 1Ϫ

d 2Ϫ

d 3Ϫ

d 1ϩ

d 2ϩ

d 3ϩ

Quantity

X 2 d 1ϩ d 3Ϫ

4 5 ⁄ 3 ⁄ 5 8 5 ⁄

1 0 0

0

⁄ 5 2 ⁄ 5 Ϫ3 ⁄ 5

0 0 1

0 1 0

5 Ϫ1 ⁄

Ϫ1

⁄ 5

0 0 Ϫ1

32 24 24

Ϫ ⁄ 5

0 0 0 0

1 0 0 1

Ϫ ⁄ 5

0 0 0 0

0 0 0 0

⁄ 5 02 ⁄ 5 02 ⁄ 5 02 ⁄ 5

0 1 0 0

24 0 0 0

P 4 P 3 P 2 P 1

3

0 ⁄ 5 02 ⁄ 5 2 0 l ⁄ 5 ⁄ Pivot column 2

5

0

1

2

0 ⁄ 5 12 ⁄ 5 02 ⁄ 5 2

Ϫ2 ⁄ 5 3 2


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The Final Goal Programming Tableau C j l

b 0 P 4

0 C j Ϫ Z j

0

0

P 1

P 2

0

P 4

0

P 3

Solution Mix

X 1

X 2

d 1Ϫ

d 2Ϫ

d 3Ϫ

d 1ϩ

d 2ϩ

d 3ϩ

Quantity

X 2 d 1ϩ X 1

0 0 1

1 0 0

Ϫ0

Ϫ ⁄ 2

2 Ϫ ⁄

5 Ϫ ⁄ 8

3 Ϫ ⁄ 8

Ϫ0

3 8 Ϫ ⁄

5 8 Ϫ ⁄

0 1 0

Ϫ ⁄ 2

Ϫ1

⁄ 2 3 8 ⁄ 5 8 Ϫ ⁄

20 15 15

P 4 P 3 P 2 P 1

0 0 0 0

0 0 0 0

Ϫ1

5 8 Ϫ ⁄

3 8 Ϫ ⁄

Ϫ0

Ϫ0

Ϫ0

Ϫ1

Ϫ0

Ϫ1

Ϫ0

Ϫ0

0 0 0 0

5 8 ⁄ 0 0 0

3 8 Ϫ ⁄

Ϫ0

15 0 0 0

SOLUTIONS TO DISCUSSION QUESTIONS AND PROBLEMS 11-1.

a. Linear programming allows only one goal (for example, profit maximization) whereas goal programming permits multiple goals. b. LP always optimizes; goal programming sometimes only “satisfies.” c. In goal programming, we deal with “deviational variables” as well as real variables.

11-2. When a non-integer solution to an LP problem is found and

branching is performed on one of the variables, an additional constraint is added to each of two subproblems. These subproblems have all of the previous constraints plus one new one. Therefore, the feasible region for the subproblem must be smaller than the feasible region for the original LP problem. No new points are added to the feasible region. Consequently, it is impossible for the subproblem to have a better objective function value than the previous LP problem. 11-3.

a. Rounding off is the easiest way to solve an integer program, but it can give an infeasible or nonoptimal solution. b. Enumeration is simple in concept, but it can be very time consuming in large problems. c. The branch and bound method, which can be computerized, is especially useful when solving large problems where enumeration is impractical. It does not always reach an optimal solution in large problems, however.

11-4. The three types of integer programs are (1) pure integer programming, where all variables are integer; (2) mixed-integer programming, where some but not all variables are integer; and (3) zero–one integer programming, where all variables are either 0 or 1 in value. 11-5. The upper and lower bounds are limits set at each branch and bound stage on the highest and lowest possible costs of a possible assignment. The process is described in Section 11.2. The bounds help us decide which branches can be discarded. 11-6. Satisficing is a term used in goal programming because it is often not possible to “optimize” a multi-goal problem. We come as close as possible to reaching goals. 11-7. Deviational variables, similar to slack variables in LP, are the difference between set goals and the current solution. In LP

1

1

1

5 Ϫ ⁄ 8 3 8 ⁄

1

1 0 0

problems, only “real” variables are used, representing physical quantities. This is discussed in Section 11.3. 11-8. A college president’s goals might be to (1) increase enrollments by 1,000 students; (2) stay within budget; (3) keep class sizes down to an average of 25 students; (4) increase faculty salaries; (5) develop 10 new off-campus courses; (6) reduce average teaching loads to three courses per semester, and so on. There will be financial, space, tenure, and many other constraints. 11-9. Ranking goals just means more weight can be placed on one goal over another. The higher-ranked goals must be achieved completely before goal programming moves on to meet lowerranked goals.

There are four differences between the LP and GP simplex methods. 1. GP has negative and positive deviational variables, each with a priority. 2. The negative deviational variables provide the initial basic feasible solutions and are analogous to slack variables in LP. 3. There is a separate Z j and C j – Z j row for each of the priority goals. 4. The highest-priority row and the most negative C j – Z j value determine the variable to enter the solution next. 11-10.

11-11.

a. b. c. d. e.

Let X ϭ number of prime time ads per week Y ϭ number of off-peak ads per week Maximize audience exposure ϭ 8200 X ϩ 5100 Y Subject to: 390 X ϩ 240Y р 1800 X у 2 Y р 6 X , Y у 0 Solution: X ϭ 2; Y ϭ 4.25; audience ϭ 38,075 b. X ϭ 2, Y ϭ 4; audience ϭ 36,800 There are other good solutions. c. Optimal integer solution: X ϭ 4, Y ϭ 1; audience ϭ 37,900 11-12.

a.

Linear Nonlinear because of 8 X 1 X 2 in objective Goal programming Nonlinear because X 12 in first constraint Nonlinear and quadratic objective function


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11-13. Let X i ϭ 1 if item i is selected and 0 otherwise, for i ϭ 1 to 8. Maximize 80 X 1 ϩ 20 X 2 ϩ 50 X 3 ϩ 55 X 4 ϩ 50 X 5 ϩ 75 X 6 ϩ 30 X 7 ϩ 70 X 8 Subject to: 8 X 1 ϩ X 2 ϩ 7 X 3 ϩ 6 X 4 ϩ 3 X 5 ϩ 12 X 6 ϩ 5 X 7 ϩ 14 X 8 р 35 X i ϭ 0, 1

Solution using QM for Windows Mixed-Integer Programming Module: X 1 ϭ X 2 ϭ X 4 ϭ X 5 ϭ X 6 ϭ X 7 ϭ 1 X 3 ϭ X 8 ϭ 0 Objective function ϭ 310

2 e X 1

800,000 X 1 ϩ 500,000 X 2

This is a pure integer programming problem. The QM for Windows integer programming solution is: X 1 ϭ 5 X 2 ϭ 8

Passengers carried ϭ 1,273,000 11-16.

Maximize profit ϭ 3 X 1 ϩ 2 X 2 р

X 1

3

р

10

Subject to:

X 1, X 2 у 0

X 1 ϩ X 6 у 1 X 1 ϩ X 2 у 1 X 2 ϩ X 3 у 1 X 1 ϩ X 3 ϩ X 6 у 1 X 3 ϩ X 4 ϩ X 5 у 1 X 3 ϩ X 4 ϩ X 6 у 1 X 2 ϩ X 5 у 1 X 4 ϩ X 5 у 1 X i ϭ 0, 1 for i ϭ 1 to 6.

See graph below. Figure for Problem 11-14 10 2X 1 + X 2 ≤ 10

9

X 1

≤3

8

Solution using QM for Windows Mixed-Integer Programming Module: X 1 ϭ X 2 ϭ X 4 ϭ 1; all other variables ϭ 0. Objective function value ϭ 3 This means only locations 1, 2, and 4 will be used.

7

≤5

X 2

6 X 2

c

5

a

11-18.

b

4

a.

3 2

0

0

1

2

3

4

Let: X i

⎧1 =⎨ ⎩0

5

subject to

X 1

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

11-15.

if location i is selected if location i is not selected

Maximize profit ϭ $5,000 X 1 ϩ 6,000 X 2 ϩ 10,000 X 3

Optimal Solution

1

a. X 1 ϩ X 2 ϩ X 3 ϩ X 4 ϩ X 5 ϩ X 6 у 3 b. X 1 ϩ X 4 ϭ 1 c. X 4 р X 6 d. 2 X 5 р X 2 ϩ X 3 e. X 5 ϩ 1 у X 2 + X 3

11-17. Let X i ϭ 1 if location i is selected and 0 otherwise, for i ϭ 1 to 6. Minimize X 1 ϩ X 2 ϩ X 3 ϩ X 4 ϩ X 5 ϩ X 6

X 2 р 5

2 X 1 ϩ X 2

$8,000,000 (maintenance)

X 1, X 2 to be integers у 0

X 2 ϭ number of smaller posters

subject to

р

0 (one-third 757s)

X 2 р 17 (planes)

X 1 ϩ

X 1 ϭ number of larger posters

11-14.

Ϫ 1e X 2 у

163

ϩ

12,000 X 4 ϩ 8,000 X 5 ϩ 3,000 X 6

ϩ

9,000 X 7 ϩ 10,000 X 8

$60,000 X 1 ϩ 50,000 X 2 ϩ 82,000 X 3 ϩ

103,000 X 4 ϩ 50,000 X 5

ϩ

41,000 X 6 ϩ 80,000 X 7

ϩ

69,000 X 8 р $300,000

b. X 1 ϭ 0, X 2 ϭ 1, X 3 ϭ 1, X 4 ϭ 0, X 5 ϭ 1, X 6 ϭ 1, X 7 ϭ 0, X 8 ϭ 1, Profit ϭ 37,000 a. otherwise 11-19.

Let X 1 ϭ 1 if apartment project is undertaken; 0

X 1 ϭ number of Boeing 757s purchased

Let X 2 ϭ 1 if shopping center project is undertaken; 0 otherwise

X 2 ϭ number of Boeing 767s purchased

Let X 3 ϭ 1 if mini-warehouse project is undertaken; 0 otherwise

Maximize passenger carrying capability ϭ 125,000 X 1 ϩ 81,000 X 2

Maximize NPV ϭ 18 X 1 ϩ 15 X 2 ϩ 14 X 3

subject to 80 X 1

ϩ

110 X 2 [ X 1

р

1,600 ($ million available)

у 1e ( X 1 ϩ X 2)]

or

Subject to: 40 X 1 ϩ 30 X 2 ϩ 20 X 3 р 80 30 ⌾ 1 ϩ 20 X 2 ϩ 20 X 3 р 50 X 1, X 2, X 3 ϭ 1 or 0


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b. The optimal solution is X 1 ϭ 1, X 2 ϭ 1, X 3 ϭ 0. NPV ϭ 33. This means that both the apartment project and the shopping center project will be undertaken. The amount of money spent in year 1 would be $70 (thousand) and in year 2 would be $50 (thousand). 11-20. a. X 1 у X 2 This means that if the apartment is not built ( X 1 ϭ 0), the shopping center cannot be built ( X 2 must equal 0).

b.

X 1 ϩ X 2 ϩ X 3 ϭ 2

11-21. a. Let X ij ϭ 1 if generator i is functioning during time period j, and 0 otherwise; where i ϭ 1, 2, 3 and j ϭ 1 for 6–2 time period; j ϭ 2 for 2–10 time period; j ϭ 3 for 6Ϫ10 time period.

Let Y ij ϭ megawatts produced by generator i in time period j, where i ϭ 1, 2, 3 and j ϭ 1 for 6–2 time period; j ϭ 2 for 2–10 time period. Minimize cost ϭ 6,000( X 11 ϩ X 12 ϩ X 13) ϩ 5,000( X 21 ϩ X 22 ϩ X 23) ϩ 4,000( X 31 ϩ X 32 ϩ X 33 ) ϩ 8( Y 11 ϩ Y 12 ) ϩ 9( Y 21 ϩ Y 22 ) ϩ 7(Y 31 ϩ Y 32)

(3) number of TV or radio ads Ϫ T ϩ R Ϫ d ϩ 3 ϩ d 3 ϭ 6

(4) restriction on number of each individual type of ad T Ϫ d 4ϩ ϩ d Ϫ 4 ϭ 10 Ϫ R Ϫ d ϩ 5 ϩ d 5 ϭ 10 Ϫ B Ϫ d ϩ 6 ϩ d 6 ϭ 10 Ϫ N Ϫ d ϩ 7 ϩ d 7 ϭ 10

All variables у 0 b.

T ϭ 0, R ϭ 0.73, B ϭ 0, N ϭ 88.86

c. Goal 1 (number of people reached) and goal 2 (budget) are met completely. The number of TV, radio, and billboard ads are each less than 10. The other goals are not met. 11-23.

Maximize profit ϭ 2 X 1 ϩ 3 X 2 subject to

X 1 ϩ 3 X 2 р 9

3 X 1 ϩ X 2 р 7

Subject to:

X 1 Ϫ X 2 р 1

Y 11 ϩ Y 21 ϩ Y 31 у 3,200

megawatts requirements from 6–2

Y 12 ϩ Y 22 ϩ Y 32 у 5,700

megawatts requirements from 2–10

Y 11 р 2,400( X 11 ϩ X 13)

maximum megawatts from #1 from 6–2

Y 12 р 2,400( X 12 ϩ X 13)


Y 21 р 2,100( X 21 ϩ X 23)


profit ϭ $10.50

Y 22 р 2,100( X 22 ϩ X 23)


This provides an upper bound value.

Y 31 р 3,300( X 31 ϩ X 33)


Y 32 р 3,300( X 32 ϩ X 33)


2. Round down to X 1 ϭ 1, X 2 ϭ 2, profit ϭ $8.00 for a feasible solution. The lower bound is $8.00.

X 11 ϩ X 12 ϩ X 13 р 1

generator #1 starts up at most once

3. Branch on X 2 to begin:

⌾ 21 ϩ X 22 ϩ X 23 р

1


⌾ 31 ϩ X 32 ϩ X 33 р

1


⌾ ij ϭ

0 or 1 for all i, j

Y ij у 0 for all i, j

b. The solution is: X 12 ϭ 1, X 33 ϭ 1, Y 12 ϭ 2,400, Y 31 ϭ 3,200, Y 31 ϭ 3,300, total cost ϭ $74,700. Thus, generator #1 will be utilized in the period 2–10 and will generate 2,400 megawatts of electricity. Generator #3 will be started at 6 and utilized for the entire 16 hours. It will generate 3,200 megawatts during the 6–2 time period, and 3,300 megawatts during the 2–10 time period. 11-22. Let T ϭ number of TV ads, R ϭ number of newspaper ads,

B ϭ number of billboard ads, and N Ϫ

Minimize P1d 1

ϩ

ϩ

P2d 2

ϩ

Ϫ

P3d 3

ϩ

ϭ number ϩ

P4d 4

ϩ

of newspaper ads.

ϩ

P4d 5

ϩ

P4d 6ϩ ϩ P4d ϩ 7

Subject to: (1) number of people reached 40,000T ϩ 32,000R 1,500,000

34,000B

ϩ

180N Ϫ d 2ϩ ϩ d 2Ϫ ϭ 16,000

ϩ

17,000N

Ϫ d ϩ 1 ϩ d 1 ϭ

ϩ

Ϫ

(2) budget 900T ϩ 500R ϩ 600B budget constraint

X 1, X 2 у 0

1. Solve graphically as an LP problem: X 1 ϭ 1.5 X 2 ϭ 2.5

Subproblem A

New constraint: X 2 р 2 Optimal solution: X 2 ϭ 2 X 1 ϭ 1.6 profit ϭ $9.33 (new upper bound)

Subproblem B

New constraint: X 2 у 3 Optimal solution: X 2 ϭ 3 X 1 ϭ 0 profit ϭ $9.00 (new lower bound)

4. Branch on X 1 now from subproblem A: Subproblem C

New constraints: X 2 р 2 X 1 р 1 Optimal solution: X 2 ϭ 2 X 1 ϭ 1 profit ϭ $8.00

Subproblem D

New constraints: X 2 р 2 X 1 у 2 Optimal solution: X 2 ϭ 1 X 1 ϭ 2 profit ϭ $7.00

5. Both of these subproblems yield all-integer solutions. Comparing them to the lower bound of $9.00, we see they are both smaller (see the graph in the next column). The solution to the problem (see subproblem B) is X 1 ϭ 0, X 2 ϭ 3, profit ϭ $9.00.


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Figure for Problem 11-23 Subproblem C =1 =2 P = 8.00 X 1 X 2

≤ 1

Subproblem A X 1 X 2

= 1.6 =2 9.33

The best solution is X 1 ϭ 500, X 2 ϭ 400. The value of d 3Ϫ ϭ 100, meaning the two-drawer sales goal is underachieved by 100 cabinets. See the graph below. Figure for Problem 11-25

X 1

800

Infeasible, noninteger

733 700 650 600

P =

2 ≤

Upper Bound = $9.33 Lower Bound = $9.00

X 2

X

1 ≥

2

Subproblem D

= 1.5 = 2.5 P = 10.50

=1 =1 P = 7.00

X 1 X 2

X 1 X 2

Upper Bound = $10.50 Lower Bound = $8.00

A

Profit Target Three-Drawer Sales Limit

500 X 2

+ d 4

B

400

X

– d 4

2 ≥

3

Subproblem B =0 =3 P = 9.00 X 1 X 2

300 Feasible, integer solution

Production Limit

Two-Drawer Sales Limit

200

– d 3

100

+ d 3 – d 1

Optimal Solution

200

11-24.

each week

11-26.

d 3Ϫ ϭ underachievement of sales quotas for 64MB chips d 4Ϫ ϭ underachievement of sales quotas for 256MB

d 4Ϫ ϭ underachievement of sales goal for three-

chips

drawer files

d 5Ϫ ϭ underachievement of sales quotas for 512MB

Minimize deviations

chips

P1d 1Ϫ ϩ P1d 1ϩ ϩ P2d 2Ϫ ϩ P3d 3Ϫ ϩ P3d 4Ϫ

d 2Ϫ ϭ 1,300 hours (production limit)

1 X 1

ϩ

d 3Ϫ ϭ 600 (two-drawer sales limit)

ϩ

d 4Ϫ ϭ 400 (three-drawer sales limit)

X 2

All X i, d i variables у 0 Because we want to achieve the profit goal as closely as possible (minimize both d 1Ϫ and d 1ϩ), the line ABC becomes the feasible region. When the P2 priority is included, the feasible region is reduced to the segment AB. P3 priority applies to both d 3Ϫ and d 4Ϫ. The three-drawer goal ( d 4Ϫ) is fully attained at point B and the two-drawer goal (d 3Ϫ) is almost reached.

X 1 ϭ number of 64MB chips produced

d 2Ϫ ϭ underfilling customers’ orders of 256MB chips

drawer files

ϩ

C

1,000 1,200 1,100 1,300

d 1Ϫ ϭ underfilling customers’ orders of 64MB chips

d 3Ϫ ϭ underachievement of sales goal for two-

1 X 1 ϩ 2 X 2

+ d 2 – d 2


d 2Ϫ ϭ idle time in production capacity

10 X 1 ϩ 15 X 2 ϩ d 1Ϫ Ϫ d 1ϩ ϭ $11,000 (profit target)

800

+ d 1


d 1ϩ ϭ overachievement of profit goal

subject to

600 X 1

d 1Ϫ ϭ underachievement of profit goal

11-25.

400

Let: X 1 ϭ number of two-drawer cabinets produced each week X 2 ϭ number of three-drawer cabinets produced

ϭ

165


CHAPTER 11

Ϫ

d 6 ϭ underutilization of plant capacity

Minimize deviations ϭ

P1d 1Ϫ ϩ P1d 2Ϫ ϩ P2 d 3Ϫ ϩ P2 d 4Ϫ ϩ P2 d 5Ϫ ϩ P3 d 6Ϫ

subject to X 1 ϩ d 1Ϫ Ϫ d 1ϩ ϭ 30

(64MB chips order)

Ϫ

ϩ

(256MB chips order)

Ϫ

ϩ

X 1 ϩ d 3 Ϫ d 3 ϭ 40

(64MB sales goal)

X 2 ϩ d 4Ϫ Ϫ d 4ϩ ϭ 50

(256MB sales goal)

X 2 ϩ d 2 Ϫ d 2 ϭ 35

Ϫ

ϩ

X 3 ϩ d 5 Ϫ d 5 ϭ 60

(512MB sales goal)

8 X 1 ϩ 13 X 2 ϩ 16 X 3 ϩ d 6Ϫ ϭ 1,200 (hours capacity) All variables у 0


166

11-27.

12:04 PM

Page 166


CHAPTER 11

Third tableau for Harrison Electric:

C j l

b

5/12/08

Solution Mix

0

0

P 1

P 2

0

P 4

0

0

P 3

0

X 1

X 2

d 1Ϫ

d 2Ϫ

d 3Ϫ

d 4Ϫ

d 1ϩ

d 2ϩ

d 3ϩ

d 4ϩ

– –23 –79 –19 7 –– 9

0

0

–23 –79 –19 –79

0

0

2

0

0

X 2

0

1

0

d 3Ϫ

0

0

P 4

d 4Ϫ

0

0

1 –3 – –29 8 –– 9 –29

0

0

–29

P 4

⎧⎪ Z j ⎩⎪C j – Z j

0

0

⎧⎪ Z j ⎩⎪C j – Z j

0

P 3

0

P 2

⎪ Z j ⎩⎪C j – Z j

P 1

⎧⎪ Z j ⎪⎩C j – Z j

0

X 1

0

0

1

0

0

1

1 –– 3 –29 –89 – –29

7 –– 9

0

1

– –29

– –29

–79

0

0

–29

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

1

0

0

0

0

0

0

0

0

0

0

1

0

0

0

1

0

Quantity

Ϫ1

0

0

Ϫ1

–83 14 –3 13 –3

–79

0

Ϫ1

13 –3

7 –– 9

0

ϩ1

0

0

0

0

0

0

1

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

– –

0

0

0

0

Pivot row

a Pivot column

Fourth tableau for Harrison Electric: C j l

b

Solution Mix

0

d 2ϩ

0

X 2

0

d 3Ϫ

P 4

d 4Ϫ

0

0

P 1

P 2

0

P 4

0

0

P 3

0

X 1

X 2

d 1Ϫ

d 2Ϫ

d 3Ϫ

d 4Ϫ

d 1ϩ

d 2ϩ

d 3ϩ

d 4ϩ

–32 –76 –16 – –76

0

–12 1 –6 5 –– 6 – 1–6

Ϫ1

0

0

– –12

1

0

0

–16 –56 –16

0

0

0

5

0

Ϫ1

0

5

0

0

Ϫ1

2

0

0

Ϫ1

2

0

0

ϩ1

1 0 0

– –76 –76

0

0

0

0

0

1

0

0

0

1

–

0

0

1

0

1 –6 1 –6

0

0

0

–16 1 –– 6

–

P 4

⎧⎪ Z j ⎩⎪C j – Z j

0

0

0

0

0

0

0

0

0

0

P 3

⎪ Z j ⎩⎪C j – Z j

0

0

0

0

0

0

0

0

1

0

⎧⎪ Z j ⎩⎪C j – Z j

0

0

0

0

0

0

0

0

0

0

P 2

0

0

0

1

0

0

0

0

0

0

⎪ Z j ⎩⎪C j – Z j

0

0

0

0

0

0

0

0

0

0

P 1

0

0

1

0

0

0

0

0

0

0

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

Quantity

3

0

0

0

a


5/12/08

CHAPTER 11

11-28.

12:04 PM

Page 167


a.

C j l

0

0

P 1

P 2

0

P 4

0

P 3

X 1

X 2

d 1Ϫ

d 2Ϫ

d 3Ϫ

d 1ϩ

d 2ϩ

d 3ϩ

b

Solution Mix

P 1

d 1Ϫ

2

4

1

0

0

Ϫ1

0

0

80

P 2

d 2Ϫ

8

10

0

1

0

0

Ϫ1

0

320

0

d 3Ϫ

8

6

0

0

1

0

0

Ϫ1

240

0

0

0

0

0

0

0

0

0

P 4

⎪ Z j ⎩⎪C j – Z j

0

0

0

0

0

1

0

0

⎧⎪ Z j ⎩⎪C j – Z j

0

0

0

0

0

0

0

0

P 3

0

0

0

0

0

0

0

1

P 2

⎪ Z j ⎩⎪C j – Z j

Ϫ8

P 1

⎧⎪ Z j ⎩⎪C j – Z j

Ϫ2

8

2

10

0

1

0

0

Ϫ1

0

Ϫ10

0

0

0

0

1

0

4

1

0

0

Ϫ1

0

0

Ϫ4

0

0

0

1

0

0

Quantity

0

320

80

a Pivot column 11-28.

b.

C j l

b 0 P 4

0

Solution Mix X 2 d 1ϩ X 1

P 4

⎪ Z j ⎪⎩C j – Z j

P 3

0

0

P 1

P 2

0

P 4

0

P 3

X 1

X 2

d 1Ϫ

d 2Ϫ

d 3Ϫ

d 1ϩ

d 2ϩ

d 3ϩ

Quantity

–14 –38 5 – 16 –

20 30 15 30

0 0 1

1 0 0

0 Ϫ1 0

–14 –58 3 –16 –

0

0

Ϫ1

–58

– –38

–58

–38

0

–58

–

– –14 – –38 –5 16

0 1 0

– –14 – –58 –3 16

1

– –58

–38 –38

–

0

0

1

⎧⎪ Z j ⎪⎩C j – Z j

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

1

⎧⎪ Z j ⎪⎩C j – Z j

0

0

0

0

0

0

0

0

P 2

0

0

0

1

0

0

0

0

⎧⎪ Z j – ⎩⎪C j Z j

0

0

0

0

0

0

0

0

P 1

0

0

1

0

0

0

0

0

The best solution is X 1 ϭ 15 X 2 ϭ 20 d 1ϩ ϭ 30

0

0

0

k Pivot row

167


168

11-29.

5/12/08

12:04 PM

CHAPTER 11

Page 168


a. d 1Ϫ ϭ underachievement of class and study goal d 1ϩ ϭ overachievement of class and study goal d 2ϩ ϭ overachievement of sleeping goal d 3Ϫ ϭ underachievement of social time goal

subject to 100 X 1 ϩ 130 X 2 р 5,000 hours X 1, X 2 у 0 11-32.

a. Maximize Z ϭ 4 X 1 Ϫ .1 X 12 ϩ 5 X 2 Ϫ .2 X 22 subject to X 1 ϩ 2 X 2 ϭ 40

Major Bligh’s objective function becomes minimize ϭ d 1Ϫ ϩ d 1ϩ ϩ d 2ϩ ϩ d 3Ϫ subject to constraints (per week)

X 1, X 2 у 0

b. X 1 ϭ 18.3; X 2 ϭ 10.8; revenue ϭ $70,420

1 X 1 ϩ 1 X 2 ϩ 1 X 3 ϩ 1 X 4 р 168

11-33.

1 X 3 ϩ d 1Ϫ Ϫ d 1ϩ ϭ 30

11-34.

1 X 4 ϩ d 3Ϫ ϭ 20

The optimal solution found using Solver in Excel is X ϭ 0.333, Y ϭ 0.667, with a variance of 0.102 and

All variables у 0

a return of 0.09.

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

11-35.

P1 ϭ meet class and study goal

11-36.

P2 ϭ meet sleeping goal P3 ϭ meet socializing goal

b. X 1 ϭ 49 X 2 ϭ 69 X 3 ϭ 30 X 4 ϭ 20 All goals are fully met. a. Let S ϭ dollars invested in stocks; B invested in bonds;

ϭ

dollars

R ϭ dollars invested in real estate Minimize d1Ϫ ϩ d2Ϫ ϩ d3ϩ Subject to

a. Total profit ϭ (P1Ϫ6) X 1 ϩ (P2Ϫ8) X 2 b. The optimal solution found using Solver in Excel is X 1 ϭ 260, X 2 ϭ 140, P1 ϭ 20, P2 ϭ 17.33, profit ϭ $4,946.67.

minimize ϭ P1d 1Ϫ ϩ P1d 1ϩ ϩ P2 d 2ϩ ϩ P3d 3Ϫ

11-30.

The optimal solution found using Solver in Excel is X ϭ 62.73, Y ϭ 8.64, Profit ϭ 720.41.

1 X 1 Ϫ d 2ϩ ϭ 49

where

Let X 1 ϭ no. of XJ6’s and X 2 ϭ no. of XJ8’s

a.

Z ϭ $665,000

Variable

Value

X 1 X 2 X 3 X 4 X 5 (South Orlando) X 6 X 7 X 8 (Apopka) X 9 (Lake Mary) X 10 (Cocoa Beach)

0 0 0 0 1 0 0 1 1 1

Return is at least 10%

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

B ϩ d2Ϫ Ϫ d2ϩ = 75,000

Amount in bonds is at least 30%

Solution:

R ϩ d3Ϫ Ϫ d3ϩ = 0.50(S + B)

Real estate is less than half of stocks and bonds

0.13S + 0.08B + 0.10R d1Ϫ Ϫ d1ϩ ϭ 25,000

ϩ

S ϩ B ϩ R ϭ 250,000 S Յ150,000 〉 Յ 150,000 ⌹ Յ 150,000

S, 〉, ⌹ Ն 0

Z ϭ $635,000 Variable

Value

Variable

Value

X 1 X 2 X 3 X 4 X 5

0 0 0 0 1

X 6 X 7 X 8 X 9 X 10

0 0 1 1 1

SOLUTIONS TO INTERNET HOMEWORK PROBLEMS

b. S ϭ $50,000 invested in stocks

11-37.

B ϭ 75,000 invested in bonds;

Maximize return ϭ 50 X 1 ϩ 100 X 2 ϩ 30 X 3

R ϭ $125,000 invested in real estate The total return is $25,000 (10%). The amount invested in real estate is not less than half than the amount invested in stocks and bonds. This is the only goal that is not met. 11-31.

Maximize profit ϭ X 1(1,800

Ϫ

50 X 1)

ϩ X 2(2,400 Ϫ

70 X 2)

ϩ

90 X 7 ϩ 35 X 8

X 1 ϩ X 2 ϩ X 3 ϩ X 4 ϩ X 5 ϩ X 6 ϩ X 7 ϩ X 8 у 5 X 1 ϩ X 2 X 3 ϩ X 4 ϩ X 5

2,400 X 2 Ϫ 70 X 22 X 1, X 2 у 0

45 X 4 ϩ 65 X 5 ϩ 20 X 6

subject to 500 X 1 ϩ 1,000 X 2 ϩ 350 X 3 ϩ 490 X 4 ϩ 700 X 5 ϩ 270 X 6 ϩ 800 X 7 ϩ 400 X 8 р 3,000

2 ϭ 1,800 X 1 Ϫ 50 X 1

ϩ

ϩ

р

1

у

2

X 6 ϩ X 7 ϩ X 8 у 2

All X i ϭ 0 or 1


5/12/08

12:04 PM

Page 169

169


CHAPTER 11

A2 B 1 C 3

A1 B 2

$106

C 3

D 4

C 3

A4 B 1 C 3

$55

C 2

D 4

D 2

C 2

Lower Bound = $100 Upper Bound = $105

3

1

$131

Not Feasible

A4 B 2 C 3

Upper Bound = –

A4 B 2

$105

C 1

D 1

A4 C 3

D

D

D 4

B 1

$100

D 4

A4 B 1

Lower Bound = $46

B 2 C 1

Lower Bound = $55 Upper Bound = $105

$46

Feasible

A 3

A4 C 1

D 4

B 2

$150

D 4

Not Feasible

A4 B 2

C 1

Feasible

A 1

Not Feasible A

B 2

D 4

Feasible 2

A3

$130

$105

D 3

Feasible

Feasible

$150

Feasible

D 2

Optimal Solutions

Branch and Bound Solution for Problem 11-38.

11-38. Lower bound set on rows with assignment A4 ($10), B1 ($6), C 3 ($5), D4 ($25): Total cost $46.

Two optimal solutions (see the figure above) with a total cost of $105: Assignment

Cost

Assignment

Cost

A4 B1 C 3 D2

$ 10 6 5 $184 $105

A4 B2 C 3 D1

$ 10 15 5 $175 $105

11-39.

X 1 ϭ number of TV spots X 2 ϭ number of newspaper ads d 1ϩ ϭ deviation above budget funds of $120,000

The first two priorities, P1 and P2, are fully satisfied by the region ABC. But the P3 priority requires that we select a solution above the exposure constraint line (minimize d 4Ϫ). Point A comes closest to reaching the P3 goal. The best solution is 11-40.

X 1 ϭ 10 TV spots X 2 ϭ 35 newspaper ads

Total exposure here is 8,250,000 people, so d 4Ϫ ϭ 750,000 people. In other words, the exposure goal was underachieved by –43 million people. Notice that in this problem d 2Ϫ and d 3Ϫ are of equal (P2) priority and hence are equally important. See the graph below. Figure for Problem 11-40 Ads

d 2Ϫ ϭ number of TV spots below 10

70

d 3Ϫ ϭ number of newspaper ads below 20

60

TV Spots Constraint

d 4Ϫ ϭ deviation below exposure of 9 million persons

desired Minimize deviations ϭ P1d 1ϩ ϩ P2d 2Ϫ ϩ P2d 3Ϫ ϩ P3d 4Ϫ 2

subject to ϩ

5,000 X 1 ϩ 2,000 X 2 Ϫ d 1 X 1

d 2 –

50

ϩ

ϭ

$120,000 (budget constraint)

d 2Ϫ ϭ 10 (TV spots)

X 2 ϩ d 3Ϫ ϭ 20 (newspaper ads) Ϫ

300,000 X 1 ϩ 150,000 X 2 ϩ d 4

ϭ

40

Budget Constraint

30

Newspaper Ads Constraint

A

C

20

+ d 2

+ d 3

B + d 1 – d 1

10

9,000,000 (exposures)

All variables у 0

0

5

10

15

20 X 1

+ d 4 – d 4

25

d 3–

Exposure Constraint

30 TV Spots


170

5/12/08

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CHAPTER 11

week

These constraints assume no more than one assignment per manager.

D ϭ number of Deluxe blenders produced each week

For project managers:

11-41.

Let S ϭ number of Standard blenders produced each

C ϭ number of Chef’s Delight blenders produced each week Ϫ

Minimize d 1

ϩ

ϩ

Ϫ

Ϫ

Ϫ

Ϫ

d 1 ϩ d 2 ϩ d 3 ϩ d 4 ϩ d 5

Subject to: (1) use 240 hours per week ϩ

1.5S ϩ 2D ϩ 2.5C Ϫ d 1

ϩ

Ϫ

d 1 ϭ 240

(2) produce 60 of the Chef’s Delight blenders C Ϫ d 2ϩ ϩ d 2Ϫ ϭ 60 D Ϫ d 3ϩ ϩ d 3Ϫ ϭ 60 (4) produce 60 of the Standard blenders S Ϫ d 4ϩ ϩ d 4Ϫ ϭ 60 ϩ

ϩ Ϫ Ϫ Ϫ Ϫ Minimize dϪ 1 ϩ d1 ϩ 0.5d 2 ϩ 0.5d 3 ϩ 0.5d 4 ϩ 0.333d 5

SOLUTION TO SCHANK MARKETING RESEARCH CASE 1. The first part of this case is an assignment problem that can be formulated with LP. A dummy project manager can be added to create a balanced 4 ϫ 4 cost matrix.

Xij ϭ

where

Hardgraves

These constraints permit assigning three managers to four clients while minimizing positive and negative deviational variables (d 5, d 6, d 7). Gardener to CBT restriction: This constraint looks at the deviation of d 8ϩ from 0. In other words, the closer d 8ϩ is to 0 (not assigning Gardener to CBT), the closer it comes to meeting the restriction. Ruth earns $3,000 or more: ϩ

3,100 X 24 ϩ d 9Ϫ Ϫ d 9ϩ ϭ $3,000

Ϫ

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

ϭ

Ruth

X 31 ϩ X 32 ϩ X 33 ϩ X 34 ϩ d 7Ϫ Ϫ d 7ϩ ϭ 1

d 5 ϭ 3,500

11-42.

Minimize

Gardener

X 21 ϩ X 22 ϩ X 23 ϩ X 24 ϩ d 6 Ϫ d 6 ϭ 1

Ϫ

All variables у 0

n

ϩ

2,700 X 21 ϩ 3,200 X 22 ϩ 3,000 X 23

(5) generate profit of at least $3,500 28S ϩ 32D ϩ 35C Ϫ d 5

Ϫ

X 14 Ϫ d 8ϩ ϭ 0

(3) produce 60 of the Deluxe blenders

ϩ

X 11 ϩ X 12 ϩ X 13 ϩ X 14 ϩ d 5Ϫ Ϫ d 5ϩ ϭ 1

m

Here d 9 represents underachievement of the goal, while d 9ϩ is overachievement. The coefficients are the costs per assignment. Total costs: 4

3

∑ ∑ (C ij X ij ) – d 10+ = 0

j =1 i =1

This attempts to minimize total cost, bringing it as close to zero as possible; d 10ϩ is the deviation from the goal. Objective function: minimize Z ϭ P1d 2Ϫ ϩ P2d 8ϩ ϩ P3(2d 1Ϫ ϩ d 3Ϫ) ϩ

∑ ∑ Cij X ij

P4d 9Ϫ ϩ P5d 10ϩ

j = 1 i = 1

{

1

if project leader i is assigned to client j if otherwise

0

i ϭ 1, 2, 3, 4 for Gardener, Ruth,

Hardgraves, Dummy j ϭ 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):

SOLUTION TO THE OAKTON RIVER BRIDGE CASE For a given set of requirements, the smallest number of toll collectors that will meet them can be obtained from the following integer linear programming problem: minimize Z ϭ X 1 ϩ X 2 ϩ X 3 ϩ X 4 ϩ X 5 ϩ X 6 ϩ X 7 subject to X 1 ϩ X 2 ϩ X 3 ϩ X 4 ϩ X 5 ϩ X 6 ϩ X 7 у R5 X 1 ϩ X 2 ϩ X 3 ϩ X 4 ϩ X 5

P1: assign a manager to the NASA account.

X 2 ϩ X 3 ϩ X 4 ϩ X 5 ϩ X 6

P2: do not assign Gardener to CBT Television account. P3: meet demands of Hines; they are twice as important as

those of General Foundry. P4: place Ruth on a project that will earn him $3,000 or

у R5 у R6

X 3 ϩ X 4 ϩ X 5 ϩ X 6 ϩ X 7 у R7 X 3 ϩ X 4 ϩ X 5 ϩ X 6 ϩ X 7 у R1

X 1 X 1 ϩ X 2

more.

X 1 ϩ X 2 ϩ X 3

P5: minimize the total cost of all assignments.

X 1 ϩ X 2 ϩ X 3 ϩ X 4

ϩ X 5 ϩ X 6 ϩ X 7 у R2 ϩ X 6 ϩ X 7 у R3 ϩ X 7 у R4

All variables у 0

Constraints

For client’s demand: X 11 ϩ X 21 ϩ X 31 ϩ d 1Ϫ ϭ 1 Ϫ

Hines

X 12 ϩ X 22 ϩ X 32 ϩ d 2 ϭ 1

NASA

X 13 ϩ X 23 ϩ X 33 ϩ d 3Ϫ ϭ 1

General

Ϫ

X 14 ϩ X 24 ϩ X 34 ϩ d 4 ϭ 1

CBT

where X j is the number of collectors starting on day j ( j Sunday) and R j is the number required on day j.

ϭ

1 is

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 allocations that yield the minimum numbers of collectors starting each: 18 for shift A, 16 for shift B, and 18 for shift C.


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CHAPTER 11

the annual rent constraint

Toll Collector Requirements for Oakton River Case SHIFT DAY

A

B

C

Mix

Req. Start

Req. Start

Req. Start

Req. Start

Sun. Mon. Tue. Wed. Thu. Fri. Sat. Total

8 13 12 12 13 13 15

0 3 5 0 5 1 14 18

10 10 10 10 10 13 15

0 1 5 1 5 1 13 16

15 13 13 12 12 13 8

5 2 1 4 1 5 10 18

33 36 35 34 35 39 38

171

3 9 8 6 9 7 18 50

4.4 X 1 ϩ 6.1 X 2 ϩ 8.3 X 3 ϩ 24.0 X 4 ϩ 19.5 X 5 ϩ 20.7 X 6 ϩ

7.7 X 7 ϩ 19.4 X 8 ϩ 11.7 X 9 ϩ 15.2 X 10 ϩ 3.9 X 11

ϩ

3.2 X 12 ϩ 11.3 X 13 ϩ 16.0 X 14 ϩ 9.6 X 15 у 130

the construction cost constraint 24.6 X 1 ϩ 32.0 X 2 ϩ 41.4 X 3 ϩ 124.4 X 4 ϩ 64.8 X 5 ϩ

79.8 X 6 ϩ 38.6 X 7 ϩ 66.8 X 8 ϩ 45.1 X 9 ϩ 54.3 X 10

ϩ

15.0 X 11 ϩ 13.4 X 12 ϩ 42.0 X 13 ϩ 63.7 X 14

ϩ

40.0 X 15 р 700

at least one clothing store Note: Alternative optimal solutions for each shift may be possible.

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

SOLUTION TO PUYALLUP MALL CASE The problem can be expressed as the following integer linear programming problem with X i being a 0–1 variable, 1 if store i is to be included and 0 if not: Maximize 28.1 X 1 ϩ 34.6 X 2 ϩ 50.0 X 3 ϩ 162.0 X 4 ϩ 77.8 X 5 ϩ

100.4 X 6 ϩ 45.2 X 7 ϩ 80.2 X 8 ϩ 51.4 X 9 ϩ 62.5 X 10

ϩ

18.0 X 11 ϩ 11.6 X 12 ϩ 50.4 X 13 ϩ 73.6 X 14

ϩ

51.2 X 15

subject to the space constraint 1.0 X 1 ϩ 1.6 X 2 ϩ 2.0 X 3 ϩ 3.2 X 4 ϩ 1.8 X 5 ϩ 2.1 X 6 ϩ

1.2 X 7 ϩ 2.4 X 8 ϩ 1.6 X 9 ϩ 2.0 X 10 ϩ 0.6 X 11

ϩ

0.5 X 12 ϩ 1.4 X 13 ϩ 2.0 X 14 ϩ 1.0 X 15 р 16

X 1 ϩ X 2 ϩ X 3 у 1

at least one hard goods store X 8 ϩ X 9 ϩ X 10 у 1

at least one miscellaneous-type store X 11 ϩ X 12 ϩ X 13 ϩ X 14 ϩ X 15 у 1

at least two restaurants X 4 ϩ X 5 ϩ X 6 ϩ X 7 у 2

no more than two clothing stores X 1 ϩ X 2 ϩ X 3 р 2

miscellaneous types cannot exceed total of clothing and hard goods X 1 ϩ X 2 ϩ X 3 ϩ X 8 ϩ X 9 ϩ X 10 Ϫ X 11 Ϫ X 12 Ϫ X 13 Ϫ X 14 Ϫ X 15 у

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 feet of space will be used, the annual rent is $132,000, and the construction cost is $531,800.

Chapter 11 Answers

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