# CFCWP Statistics Excel Spreadsheet

i have 2 chapters of work for quantitative methods that have multiple parts to be answered. Chapter 9: 1,2,3,9 and 15

Please use Excel Solver to solve all Chapter 9 problems. Each problem requires its tab.

Your workbook needs to have a worksheet with Excel Solver inputs, an Answer Report, a Sensitivity Report, and a Limits report for each problem. Also, problem 9-2 part c is a separate Excel Solver solution.

Chapter 10: 1,2,6,8, and 11

Submit your solutions as an Excel workbook with a separate spreadsheet for each problem.

Appendix G Self-Test Solutions and Answers

to Even-Numbered Problems

Chapter 1

12. a. If x represents the number of pairs of shoes produced,

a mathematical model for the total cost of producing x

pairs of shoes is TC 2000 60x. The two components of total cost in this model are fixed cost ($2,000)

and variable cost (60x).

b. If P represents the total profit, the total revenue (TR) is

80x and a mathematical model for the total profit

realized from an order for x pairs of shoes is P TR

TC 80x (2000 60x) 20x 2000.

c. The breakeven point is the number of shoes produced

(x) at the point of no profit (P 0).

Thus the breakeven point is the value of x when P

20x 2000 0. This occurs when 20x 2000 or x

100 (i.e., the breakeven point is 100 pairs of shoes).

2. Define the problem; identify the alternatives; determine the

criteria; evaluate the alternatives; choose an alternative.

4. A quantitative approach should be considered because the

problem is large, complex, important, new, and repetitive.

6. Quicker to formulate, easier to solve, and/or more easily

understood

8. a. Max

s.t.

b.

c.

d.

e.

10x 5y

5x 2y 40

x 0, y 0

Controllable inputs: x and y

Uncontrollable inputs: profit (10, 5), labor-hours (5, 2),

and labor-hour availability (40)

See Figure G1.8c.

x 0, y 20; Profit $100 (solution by trial and error)

Deterministic

Total units received x y

Total cost 0.20x 0.25y

x y 5000

x 4000 Kansas City

y 3000 Minneapolis

e. Min 0.20x 0.25y

s.t.

x

y 5000

x

4000

y 3000

x, y 0

10. a.

b.

c.

d.

FIGURE G1.8c SOLUTION

Profit:

$10/unit for x

$5/unit for y

Labor-hours:

5/unit for x

2/unit for y

40 labor-hour capacity

Production quantities

x and y

Controllable

Input

Max 10x + 5y

s.t.

5x + 2y ≤ 40

x

≥0

y≥0

Mathematical

Model

Projected profit and

check on production

time constraint

Output

© Cengage Learning 2013

Uncontrollable Inputs

14. a. If x represents the number of copies of the book that

are sold, total revenue (TR) 46x and total cost

(TC) 160,000 6x, so Profit TR TC 46x

(160,000 6x) 40x 160,000. The breakeven point

is the number of books produced (x) at the point of

no profit (P 0). Thus the breakeven point is the value

of x when P 40x 160,000 0. This occurs when

40x 160,00 or x 4000 (i.e., the breakeven point is

4000 copies of the book).

b. At a demand of 3800 copies, the publisher can expect a

profit of 40(3800) 160,000 152,000 160,000

8000 (i.e., a loss of $8,000).

c. Here we know demand (d 3800) and want to determine the price p at which we will breakeven (the point

at which profit is 0). The minimum price per copy

that the publisher must charge to break even is

Profit p(3800) (160,000 6(3800)) 3800p

182,800. This occurs where 3800p 182,800 or p

48.10526316 or a price of approximately $48.

d. If the publisher believes demand will remain at 4000

copies if the price per copy is increased to $50.95, then

the publisher could anticipate a profit of TR TC

50.95(4000) (160,000 6(4000)) 203,800

184,000 19,800 or a profit of $19,800. This is a

return of p兾TC 10.8% on the total cost of $184,000,

and the publisher should proceed if this return is

sufficient.

16. a. The annual return per share of Oil Alaska is $6.00 and

the annual return per share of Southwest Petroleum is

$4.00, so the objective function that maximizes the

total annual return is Max 6x 4y.

b. The price per share of Oil Alaska is $50.00 and the

price per share of Southwest Petroleum is $30.00, so

846

Appendix G

Self-Test Solutions and Answers to Even-Numbered Problems

(1) the mathematical expression for the constraint

that limits total investment funds to $800,000 is 50x

30y 800000,

(2) the mathematical expression for the constraint that

limits investment in Oil Alaska to $500,000 is 50x

500000, and

(3) the mathematical expression for the constraint that

limits investment in Southwest Petroleum to $450,000

is 30x 450000.

c.

Chapter 2

1. a. Record the number of persons waiting at the X-ray

department at 9:00 A.M.

b. The experimental outcomes (sample points) are the

number of people waiting: 0, 1, 2, 3, and 4. (Note:

Although it is theoretically possible for more than four

people to be waiting, we use what has actually been

observed to define the experimental outcomes.)

c.

Number Waiting

Probability

0

1

2

3

4

0.10

0.25

0.30

0.20

0.15

Total

d.

e.

1.00

d. The relative frequency method

2. a. Choose a person at random, and have him/her taste the

four blends of coffee and state a preference.

b. Assign a probability of 1/4 to each blend, using the classical method of equally likely outcomes.

c.

Blend

Probability

1

2

3

4

0.20

0.30

0.35

0.15

Total

1.00

The relative frequency method was used.

4. a. Of the 132,275,830 individual tax returns received by

the IRS in 2006, 31,675,935 were in the 1040A,

Income Under $25,000 category. Using the relative frequency approach, the probability a return from the

1040A, Income Under $25,000 category would be chosen at random is 31675935兾132275830 0.239.

b. Of the 132,275,830 individual tax returns received by

the IRS in 2006, 3,376,943 were in the Schedule C, Reciepts Under $25,000 category; 3,867,743 were in the

Schedule C, Reciepts $25,000–$100,000 category; and

were 2,288,550 in the Schedule C, Reciepts $100,000

6. a.

b.

7. a.

b.

c.

d.

e.

8. a.

& Over category. Therefore, 9,533,236 Schedule Cs

were filed in 2006, and the remaining 132,275,830

9,533,236 122,742,594 individual returns did not

use Schedule C. By the relative frequency approach,

the probability that the chosen return did not use

Schedule C is 122742594兾132275830 0.928.

Of the 132,275,830 individual tax returns received by

the IRS in 2006, 12,893,802 were in the Non 1040A,

Income $100,000 & Over category; 2,288,550 were in

the Schedule C, Reciepts $100,000 & Over category;

and 265,612 were in the Schedule F, Reciepts

$100,000 & Over category. By the relative frequency

approach, the probability that the chosen return

reported income/reciepts of $100,000 and over is

(12893802 2288550 265612)兾132275830

15447964兾132275830 0.117.

26,463,973 Non 1040A, Income $50,000–$100,000

returns were filed in 2006, so assuming examined returns were evenly distributed across the 10 categories

(i.e., the IRS examined 1% of individual returns in each

category), the number of returns from the Non 1040A,

Income $50,000–$100,000 category that were examined is 0.01(26463973) 264,639.73 (or 264,640).

The proportion of total 2006 returns in the Schedule C,

reciepts $100,000 & Over is 2,288,550兾132,275,830

0.0173. Therefore, if we assume the recommended

additional taxes are evenly distributed across the ten categories, the amount of recommended additional taxes

for the Schedule C, Reciepts $100,000 & Over category

is 0.0173($13,045,221,000.00) $225,699,891.81.

P(A) P(150 199) P(200 and over)

26

5

100

100

0.31

P(B) P(less than 50) P(50 99) P(100 149)

0.13 0.22 0.34

0.69

P(A) 0.40, P(B) 0.40, P(C) 0.60

P(A B) P(E1, E2, E3, E4) 0.80.

Yes, P(A B) P(A) P(B)

Ac {E3, E4, E5}; C c {E1, E4}; P(Ac) 0.60;

P(C c) 0.40

A Bc {E1, E2, E5}; P(A Bc) 0.60

P(B C) P(E2, E3, E4, E5) 0.80

Let P(A) be the probability a hospital had a daily inpatient volume of at least 200 and P(B) be the probability

a hospital had a nurse to patient ratio of at least 3.0.

From the list of 30 hospitals, 16 had a daily inpatient

volume of at least 200, so by the relative frequency approach the probability one of these hospitals had a daily

inpatient volume of at least 200 is P(A) 16兾30

0.533, Similarly, since 10 (one-third) of the hospitals

had a nurse-to-patient ratio of at least 3.0, the probability of a hospital having a nurse-to-patient ratio of at least

Appendix G

847

Self-Test Solutions and Answers to Even-Numbered Problems

3.0 is P(B) 10兾30 0.333. Finally, since seven of the

hospitals had both a daily inpatient volume of at least

200 and a nurse-to-patient ratio of at least 3.0, the probability of a hospital having both a daily inpatient volume of at least 200 and a nurse-to-patient ratio of at

least 3.0 is P(A ¨ B) 7兾30 0.233.

b. The probability that a hospital had a daily inpatient

volume of at least 200 or a nurse-to-patient ratio of at

least 3.0 or both is P(A ´ B) P(A) P(B)

P(A ¨ B) 16兾30 10兾30 7兾30 (16 10 7)兾

30 19兾30 0.633.

c. The probability that a hospital had neither a daily

inpatient volume of at least 200 nor a nurse-to-patient

ratio of at least 3.0 is 1 P(A ´ B) 1 19兾30

11兾30 0.367.

10. P(Defective and Minor) 4兾25

P(Defective and Major) 2兾25

P(Defective) (4兾25) (2兾25) 6兾25

P(Major Defect | Defective) P(Defective and Major)兾

P(Defective) (2兾25)兾(6兾25) 2兾6 1兾3.

P(A傽B)

0.40

0.6667

P(B)

0.60

P(A傽B)

0.40

b. P(B | A)

0.80

P(A)

0.50

c. No, because P(A | B) Z P(A)

12. a. P(A | B)

2 yrs

2 yrs

$0–$499

$500–$999

» $1000

0.12

0.075

0.195

0.24

0.275

0.515

0.09

0.2

0.29

0.45

0.55

1.00

P( 2 yrs) 0.45

P( $1000) 0.29

P(2 accounts have $1000) (0.29)(0.29) 0.0841

P($500 $999 | 2 yrs) P($500 $999 and

2 yrs)兾P( 2yrs) 0.275兾0.55 0.5

e. P( 2 yrs and $1000) 0.09

f. P( 2 yrs | $500 $999) 0.275兾0.515 0.533981

a.

b.

c.

d.

16. a. 0.19

b. 0.71

c. 0.29

18. a. 0.25, 0.40, 0.10

b. 0.25

c. Independent; program does not help

20. a. P(B ¨ A1) P(A1)P(B | A1) (0.20)(0.50) 0.10

P(B ¨ A2) P(A2)P(B | A2) (0.50)(0.40) 0.20

P(B ¨ A3) P(A3)P(B | A3) (0.30)(0.30) 0.09

b. P(A2 | B)

13. a.

0.20

0.51

0.10 0.20 0.09

c.

Reason for Applying

Quality

Cost/

Convenience

Other

Total

Full Time

Part Time

0.218

0.208

0.204

0.307

0.039

0.024

0.461

0.539

Total

0.426

0.511

0.063

1.000

b. A student will most likely cite cost or convenience as

the first reason: probability 0.511; school quality is

the first reason cited by the second largest number of

students: probability 0.426.

c. P(Quality | Full Time) 0.218兾0.461 0.473

d. P(Quality | Part Time) 0.208兾0.539 0.386

e. P(B) 0.426 and P(B | A) 0.473

Because P(B) Z P(B | A), the events are dependent.

Events

P(Ai)

P(B | Ai)

P(Ai 艚 B)

P(Ai | B)

A1

A2

A3

0.20

0.50

0.30

0.50

0.40

0.30

0.10

0.20

0.09

0.26

0.51

0.23

0.39

1.00

1.00

22. a. 0.40

b. 0.67

24. Let S speeding is reported

S C speeding is not reported

F Accident results in fatality for vehicle occupant

We have P(S) 0.129, so P(S C) 0.871. Also P(F | S)

0.196 and P(F | SC ) 0.05. Using the tabular form of

Bayes’ theorem provides:

14.

2 yrs

2 yrs

$0–$499

$500–$999

» $1000

120

75

195

240

275

515

90

200

290

450

550

1000

Events

Prior

Probabilities

Conditional

Probabilities

Joint

Probabilities

Posterior

Probabilities

S

SC

0.129

0.871

0.196

0.050

0.0384

0.0025

0.939

0.061

0.0409

1.000

1.000

848

Appendix G

Self-Test Solutions and Answers to Even-Numbered Problems

3. a.

25. a. P(defective part) 0.0065 (see below)

Events

P(Ai)

P(D | Ai)

Supplier A

Supplier B

Supplier C

0.60

0.30

0.10

0.0025

0.0100

0.020

1.00

P(Ai 僕 D) P(Ai 僕 D)

0.0015

0.0030

0.0020

0.23

0.46

0.31

P(D) 0.0065

1.00

b.

0.2

Male

Applicants

Female

Applicants

70

90

40

80

After combining these two crosstabulations into a single crosstabulation with Accept and Deny as the row

labels and Male and Female as the column labels, we

see that the rate of acceptance for males across the

university is 70兾(70 90) 0.4375 or approximately

44%, while the rate of acceptance for females across

the university is 40兾(40 80) 0.33 or 33%.

b. If we focus solely on the overall data, we would conclude that the university’s admission process is biased

in favor of male applicant. However, this occurs because most females apply to the College of Business

(which has a far lower rate of acceptance that the

College of Engineering). When we look at each college’s acceptance rate by gender, we see that the

acceptance rate of males and females are equal in the

College of Engineering (75%) and the acceptance rate

of males and females are equal in the College of

Business (33%). The data do not support the accusation that the university favors male applicants in its

admissions process.

1. a.

b.

c.

d.

e.

Values: 0, 1, 2, . . . , 20 discrete

Values: 0, 1, 2, . . . discrete

Values: 0, 1, 2, . . . , 50 discrete

Values: 0 x 8 continuous

Values: x 0 continuous

2. a. 0.05; probability of a $200,000 profit

b. 0.70

c. 0.40

f (x)

0.3

28. a.

Chapter 3

3兾20 0.15

5兾20 0.25

8兾20 0.40

4兾20 0.20

0.4

P(D1 | S1) 0.2195, P(D2 | S1) 0.7805

P(D1 | S2) 0.5000, P(D2 | S2) 0.5000

P(D1 | S3) 0.8824, P(D2 | S3) 0.1176

0.1582 and 0.8418

Accept

Deny

f(x)

1

2

3

4

Total 1.00

b. Supplier B (prob. 0.46) is the most likely source.

26. a.

b.

c.

d.

x

0.1

1

2

3

x

4

c. f(x) 0 for x 1, 2, 3, 4

g f(x) 1

4. a.

x

f(x)

xf(x)

3

6

9

0.25

0.50

0.25

0.75

3.00

2.25

1.00

6.00

Totals

b.

E(x) 6.00

x

xⴚ

(x ⴚ )2

f(x)

(x ⴚ )2f(x)

3

6

9

3

0

3

9

0

9

0.25

0.50

0.25

2.25

0.00

2.25

4.50

Var(x) 2 4.50

c. 24.50 2.12

6. a.

x

1

2

3

4

5

f(x)

0.97176

0.026675

0.00140

0.00014

0.00002

If we let x 5 represent quintuplets or more, the probability distribution of the number children born per

pregnancy in 1996 is provided in the first two columns

of the preceding table.

Appendix G

b.

x

f(x)

1

2

3

4

5

0.97176

0.026675

0.00140

0.00014

0.00002

xⴚ

xf(x)

0.97176 0.03000

0.05333

0.97000

0.004218 1.97000

0.00059

2.97000

0.00011

3.97000

(x ⴚ )2

(x )2f(x)

0.00090

0.94090

3.88090

8.82090

15.76089

0.00087

0.02509

0.00544

0.00131

0.00034

1.0300

c.

0.03305

The expected value of the number children born per

pregnancy in 1996 is E[x] 1.030 and the variance of

the number children born per pregnancy in 1996 is

Var[x] 2 0.03305.

y

f( y)

1

2

3

4

5

0.965964

0.0333143

0.0014868

0.0000863

0.0000163

2

2!

c. f (2) a b (0.4)2(0.6)0

(0.16)(1) 0.16

2

2!0!

d. P(x 1) f(1) f(2) 0.48 0.16 0.64

e. E(x) np 2(0.4) 0.8

Var(x) np(1 p) 2(0.4)(0.6) 0.48

20.48 0.6928

10. a. f (0) 0.3487

b. f(2) 0.1937

c. 0.9298

d. 0.6513

e. 1

f. 2 0.9000, 0.9487

12. a. Probability of a defective part being produced must be

0.03 for each trial; trials must be independent.

b. Two outcomes result in exactly one defect.

c. P(no defects) (0.97)(0.97) 0.9409

P(1 defect) 2(0.03)(0.97) 0.0582

P(2 defects) (0.03)(0.03) 0.0009

b.

d.

c.

d.

e.

f(y)

1

2

3

4

5

0.965964

0.0333143

0.0014868

0.0000863

0.0000163

yf(y)

yⴚ

(y ⴚ )2

0.9650964 0.0366118 0.0013404

0.0666286

0.9633882 0.9281168

0.0044604

1.9633882 3.8548932

0.0003451

2.9633882 8.7816695

0.0000814

3.9633882 15.7084459

1.0000000 1.0366118

2xe2

x!

6 for 3 time periods

6xe6

f (x)

x!

2 2

4(0.1353)

2e

0.2706

f (2)

2!

2

66e6

0.1606

f (6)

6!

45e4

f (5)

0.1563

5!

14. a. f (x)

If we let y 5 represent quintuplets or more, the probability distribution of the number children born per

pregnancy in 2006 is provided in the first two columns

of the preceding table.

y

849

Self-Test Solutions and Answers to Even-Numbered Problems

(y )2f(y)

0.001293639

0.030919551

0.005731423

0.000757611

0.000255769

0.038957993

The expected value of the number children born per

pregnancy in 2006 is E[y] 1.030 and the variance of

the number children born per pregnancy in 2006 (after

rounding) is Var[y] 2 0.0390.

e. The number of children born per pregnancy is greater

in 2006 than in 1996, and the variation in the number

of children born per pregnancy is also greater in 2006

than in 1996. However, these data provide no information on which we could base a determination of causes

of this upward trend.

8. a. Medium 145; large 140; prefer medium

b. Medium 2725; large 12,400; prefer medium

2

2!

9. a. f (1) a b (0.4)1(0.6)1

(0.4)(0.6) 0.48

1

1!1!

2

2!

b. f (0) a b (0.4)0(0.6)2

(1)(0.36) 0.36

0

0!2!

f.

16. a.

b.

c.

d.

18. a.

0.0009

0.9927

0.0302

0.8271

f(x)

3

2

1

0.5

1.0

1.5

x

b. P(x 1.25) 0; the probability of any single point is

zero because the area under the curve above any single

point is zero.

c. P(1.0 x 1.25) 2(0.25) 0.50

d. P(1.2 x 1.5) 2(0.30) 0.60

20. a. f(x)

1.0

0.5

0

1

2

x

850

Appendix G

Self-Test Solutions and Answers to Even-Numbered Problems

b.

c.

d.

21. a.

b.

c.

d.

e.

f.

22. a.

b.

c.

d.

e.

f.

23. a.

b.

24.

26.

28.

29.

30.

31.

0.50

0.30

0.40

P(0 z 0.83) 0.7967 0.5000 0.2967

P(1.57 z 0) 0.5000 0.0582 0.4418

P(z 0.44) 1.0000 0.6700 0.2300

P(z 0.23) 1.0000 0.4090 0.5910

P(z 1.20) 0.8849

P(z 0.71) 0.2389

1.96

1.96

0.61

1.12

0.44

0.44

Area 0.2119 z 0.80

Area outside the interval 0.0970 must be split between

the two tails.

Cumulative probability 0.5(0.0970) 0.9030

0.9515 z 1.66

c. Area outside the interval 0.7948 must be split between

the two tails.

Cumulative probability 0.5(0.7948) 0.2052

0.6026 z 0.26

d. Area 0.9948 z 2.56

e. Area 1.0000 0.6915 0.3085 z 0.50

a. 0.3830

b. 0.1056

c. 0.0062

d. 0.1603

a. 0.7745

b. 36.32 days

c. 19%

19.23

a. P(x x0) 1 e x 兾3

b. P(x 2) 1 e2兾3 1 0.5134 0.4866

c. P(x 3) 1 P(x 3) 1 (1 e3兾3) e1

0.3679

d. P(x 5) 1 e5兾3 1 0.1889 0.8111

e. P(2 x 5) P(x 5) P(x 2) 0.8111

0.4866 0.3245

a. 0.3935

b. 0.2231

c. 0.3834

a. f (x)

b. P(x 12) 1 e12兾12 0.6321

c. P(x 6) 1 e6兾12 0.3935

d. P(x 30) 1 P(x 30) 1 (1 e30兾12)

0.0821

32. a. 50 hours

b. 0.3935

c. 0.1353

34. a. 0.5130

b. 0.1655

c. 0.3679

Chapter 4

1. a.

s1

d1

s2

2

250

100

s3

1

s1

d2

s2

3

25

100

100

s3

75

b.

Decision

Maximum

Profit

Minimum

Profit

d1

d2

250

100

25

75

0

.09

.08

.07

.06

.05

.04

.03

.02

.01

Optimistic approach: Select d1

Conservative approach: Select d2

Regret or opportunity loss table:

Decision

s1

s2

s3

d1

d2

0

150

0

0

50

0

Maximum regret: 50 for d1 and 150 for d2; select d1

2. a. Optimistic: d1

Conservative: d3

Minimax regret: d3

c. Optimistic: d1

Conservative: d2 or d3

Minimax regret: d2

3. a. Decision: Choose the best plant size from the two

alternatives—a small plant and a large plant.

6

12

18

24

x

Appendix G

851

Self-Test Solutions and Answers to Even-Numbered Problems

Chance event: Market demand for the new product line

with three possible outcomes (states of nature): low,

medium, and high

b. Influence Diagram:

45000 miles (15000 miles for 3 years):

36($299) $0.15(45000 36000) $12,114

54000 miles (18000 miles for 3 years):

36($299) $0.15(54000 36000) $13,464

For the Midtown Motors lease option:

Market

Demand

Plant

Size

36000 miles (12000 miles for 3 years):

36($310) $0.20*max(36000 45000,0) $11,160.00

45000 miles (15000 miles for 3 years):

36($310) $0.20*max(45000 45000,0) $11,160.00

54000 miles (18000 miles for 3 years):

36($310) $0.20*max(54000 45000,0) $12,960.00

Profit

c.

For the Hopkins Automotive lease option:

Low

Small

150

Medium

200

36000 miles (12000 miles for 3 years):

36($325) $0.15*max(36000 54000,0) $11,700

45000 miles (15000 miles for 3 years):

36($325) $0.15*max(45000 54000,0) $11,700

54000 miles (18000 miles for 3 years):

36($325) $0.15*max(54000 54000,0) $11,700

So the payoff table for Amy’s problem is:

High

200

Low

50

Medium

Large

High

Actual Miles Driven Annually

Dealer

12,000

15,000

18,000

Hepburn Honda

Midtown Motors

Hopkins Automotive

$10,764

$11,160

$11,700

$12,114

$11,160

$11,700

$13,464

$12,960

$11,700

200

500

c. The minimum and maximum payoffs for each of

Amy’s three alternatives are:

d.

Decision

Maximum

Profit

Small

Large

200

500

Minimum Maximum

Profit

Regret

150

50

300

100

Optimistic Approach: Large plant

Conservative Approach: Small plant

Minimax Regret: Large plant

4. a. The decision faced by Amy is to select the best lease

option from three alternatives (Hepburn Honda, Midtown Motors, and Hopkins Automotive). The chance

event is the number of miles Amy will drive.

b. The payoff for any combination of alternative and

chance event is the sum of the total monthly charges

and total additional mileage cost; that is,

For the Hepburn Honda lease option:

36000 miles (12000 miles for 3 years):

36($299) $0.15(36000 36000) $10,764

Dealer

Hepburn Honda

Midtown Motors

Hopkins Automotive

Minimum

Cost

Maximum

Cost

$10,764

$11,160

$11,700

$13,464

$12,960

$11,700

Thus:

The optimistic approach results in selection of the

Hepburn Automotive lease option (which has the

smallest minimum cost of the three alternatives—

$10,764).

The conservative approach results in selection of the

Hopkins Automotive lease option (which has the smallest maximum cost of the three alternatives—$11,700).

To find the lease option to select under the minimax

regret approach, we must first construct an opportunity

loss (or regret) table. For each of the three chance

events (driving 12,000 miles, driving 15,000 miles,

driving 18,000 miles), subtract the minimum payoff

from the payoff for each decision alternative.

852

Appendix G

Self-Test Solutions and Answers to Even-Numbered Problems

Regret Table

State of Nature

(Actual Miles Driven Annually)

Maximum Regret

Decision Alternative

Hepburn Honda

Midtown Motors

Hopkins Automotive

12,000

$0

$396

$936

15,000

$954

$0

$540

18,000

$1,764

$1,260

$0

The maximum regret associated with each of the three

decision alternatives is as follows:

Decision Alternative

Maximum Regret

Hepburn Honda

Midtown Motors

Hopkins Automotive

$1764

$1260

$ 936

The minimax regret approach results in selection of the

Hopkins Automotive lease option (which has the

smallest regret of the three alternatives: $936).

d. We first find the expected value for the payoffs associated with each of Amy’s three alternatives:

EV(Hepburn Honda) 0.5($10,764) 0.4($12,114)

0.1($13,464) $11,574

EV(Midtown Motors) 0.5($11,160) 0.4($11,160)

0.1($12,960) $11,340

EV(Hopkins Automotive) 0.5($11,700) 0.4($11,700)

0.1($11,700) $11,700

The expected value approach results in selection of the

Midtown Motors lease option (which has the minimum

expected value of the three alternatives—$11,340).

e. The risk profile for the decision to lease from Midtown

Motors is as follows:

Probability

1.0

0.8

the chance outcomes (whether Amy drives 12,000

miles or 15,000 miles annually, her payoff is $11,160).

f. We first find the expected value for the payoffs associated with each of Amy’s three alternatives:

EV(Hepburn Honda) 0.3($10,764) 0.4($12,114)

0.3($13,464) $12,114

EV(Midtown Motors) 0.3($11,160) 0.4($11,160)

0.3($12,960) $11,700

EV(Hopkins Automotive) 0.3($11,700) 0.4($11,700)

0.3($11,700) $11,700

The expected value approach results in selection of either

the Midtown Motors lease option or the Hopkins Automotive lease option (both of which have the minimum

expected value of the three alternatives—$11,700).

5. a. EV(d1) 0.65(250) 0.15(100) 0.20(25) 182.5

EV(d2) 0.65(100) 0.15(100) 0.20(75) 95

The optimal decision is d1.

6. a. Pharmaceuticals; 3.4%

b. Financial; 4.6%

7. a. EV(own staff) 0.2(650) 0.5(650) 0.3(600) 635

EV(outside vendor) 0.2(900) 0.5(600)

0.3(300) 570

EV(combination) 0.2(800) 0.5(650) 0.3(500)

635

Optimal decision: Hire an outside vendor with an

expected cost of $570,000

b.

Cost

Probability

Own staff

Outside vendor

Combination

300

600

900

0.3

0.5

0.2

1.0

8. a. EV(d1) p(10) (1 p)(1) 9p 1

EV(d2) p(4) (1 p)(3) 1p 3

10

0.6

0.4

0.2

10

11

12

Cost ($ 1000s)

13

Note that although we have three chance outcomes

(drive 12,000 miles annually, drive 15,000 miles annually, and drive 18,000 miles annually), we only have

two unique costs on this graph. This is because for this

decision alternative (lease from Midtown Motors)

there are only two unique payoffs associated with the

three chance outcomes—the payoff (cost) associated

with the Midtown Motors lease is the same for two of

0

0.25

1

p

Value of p for

which EVs are equal

9p 1 1p 3 and hence p 0.25

d2 is optimal for p 0.25, d1 is optimal for p 0.25

Appendix G

b. d2

c. As long as the payoff for s1 2, then d2 is optimal.

EV (node 9) 0.18(400) 0.82(200) 236

EV (node 10) 0.40(100) 0.60(300) 220

EV (node 11) 0.40(400) 0.60(200) 280

EV (node 3) Max(186,314) 314d2

EV (node 4) Max(264,236) 264d1

EV (node 5) Max(220,280) 280d2

EV (node 2) 0.56(314) 0.44(264) 292

EV (node 1) Max(292,280) 292

‹ Market Research

If favorable, decision d2

If unfavorable, decision d1

10. b. Space Pirates

EV $724,000

$84,000 better than Battle Pacific

c. $200

0.18

$400

0.32

$800

0.30

$1600 0.20

d. P(Competition) 0.7273

12. a. Decision: Whether to lengthen the runway

Chance event: The location decisions of Air Express

and DRI

Consequence: Annual revenue

b. $255,000

c. $270,000

d. No

e. Lengthen the runway.

14. a. If s1, then d1; if s2, then d1 or d2; if s3, then d2

b. EvwPI 0.65(250) 0.15(100) 0.20(75) 192.5

c. From the solution to Problem 5, we know that

EV(d1) 182.5 and EV(d2) 95; thus, recommended

decision is d1; hence, EvwoPI 182.5.

d. EVPI EvwPI EvwoPI 192.5 182.5 10

16. a.

d1

F

Market

Research

6

7

2

d1

U

1

Profit

Payoff

100

s2

300

s1

400

s2

200

s1

8

s2

100

300

4

d2

d1

No Market

Research

s1

3

d2

s1

9

10

400

s2

200

s1

100

s2

300

s1

400

s2

200

5

d2

853

Self-Test Solutions and Answers to Even-Numbered Problems

11

b. EV (node 6) 0.57(100) 0.43(300) 186

EV (node 7) 0.57(400) 0.43(200) 314

EV (node 8) 0.18(100) 0.82(300) 264

18. a. 5000 200 2000 150 2650

3000 200 2000 150 650

b. Expected values at nodes:

8: 2350

5: 2350

9: 1100

6: 1150

10: 2000

7: 2000

4: 1870

3: 2000

2: 1560

1: 1560

c. Cost would have to decrease by at least $130,000.

d.

Payoff (in millions)

Probability

$200

800

2800

0.20

0.32

0.48

1.00

20. b. If Do Not Review, Accept

If Review and F, Accept

If Review and U, Accept

Always Accept

c. Do not review; EVSI $0

d. $87,500; better method of predicting success

22. a. Order two lots; $60,000

b. If E, order two lots

If V, order one lot

EV $60,500

c. EVPI $14,000

EVSI $500

Efficiency 3.6%

Yes, use consultant.

23.

State

of Nature

P(sj)

P(I |sj)

P(I 艚 sj)

P(sj |I)

s1

s2

s3

0.2

0.5

0.3

0.10

0.05

0.20

0.020

0.025

0.060

0.1905

0.2381

0.5714

P(I) 0.105

1.0000

1.0

24. a. 0.695, 0.215, 0.090

0.98, 0.02

0.79, 0.21

0.00, 1.00

854

Appendix G

Self-Test Solutions and Answers to Even-Numbered Problems

b. d2

c. Risk takers

d. Between 0 and 0.26

c. If C, Expressway

If O, Expressway

If R, Queen City

26.6 minutes

Chapter 5

1. a. EV(d1) 0.40(100) 0.30(25) 0.30(0) 47.5

EV(d2) 0.40(75) 0.30(50) 0.30(25) 52.5

EV(d3) 0.40(50) 0.30(50) 0.30(50) 50.0

The optimal solution is d2.

b. Using utilities

Decision Maker A

Decision Maker B

EU(d1) 4.9

EU(d2) 5.9

EU(d1) 6.0 Best

EU(d1) 4.45 Best

EU(d2) 3.75

EU(d1) 3.00

10. a. EV(Comedy) 0.30(30%) 0.60(25%)

0.10(20%) 26.0%

and

EV(Reality Show) 0.30(40%) 0.40(20%)

0.30(15%) 24.5%

Using the expected value approach, the manager

should choose the Comedy.

b. p probability of a 40% percentage of viewing

audience

1 p probability of a 15% percentage of viewing

audience

c. Arbitrarily using a utility of 10 for the best payoff and

a utility of 0 for the worst payoff, the utility table is as

follows:

c. Difference in attitude toward risk; decision maker A

tends to avoid risk, whereas decision maker B tends to

take a risk for the opportunity of a large payoff.

2. a. d2; EV(d2) $5000

b. p probability of a $0 cost

1 p probability of a $200,000 cost

c. d1; EV(d1) 9.9

d. Expected utility approach; it avoids risk of large loss.

Probability

4. a. Route B; EV 58.5

b. p probability of a 45-minute travel time

1 p probability of a 90-minute travel time

c. Route A; EV 7.6; risk avoider

5. a.

1.0

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0.0

A

100

350

0

Utility

Value

40%

30%

25%

20%

15%

Does not apply

0.40

0.30

0.10

Does not apply

10

4

3

1

0

EV(Comedy) 0.30(4) 0.60(3) 0.10(1) 3.1

and

EV(Reality Show) 0.30(10) 0.40(1)

0.30(0) 3.4

Using the expected utility approach, the manager

should choose the Reality Show.

Although the Comedy has the higher expected payoff

in terms of percentage of viewing audience, the Reality

Show has the higher expected utility. This suggests the

manager is a risk taker.

11.

b. A—risk avoider

B—risk taker

C—risk neutral

c. Risk avoider A, at $20 payoff p 0.70

EV(Lottery) 0.70(100) 0.30(100) $40

Therefore, will pay 40 20 $20

Risk taker B, at $20 payoff p 0.45

EV(Lottery) 0.45(100) 0.55(100) $10

Therefore, will pay 20 (10) $30

6. A: d1; B: d2; C: d2

8. a.

Win

Lose

Bet

Do not bet

Indifference

Value of p

and so the expected payoffs in terms of utilities are as

follows:

C B

-100 -50 0

50

Payoff

Percentage

of Viewing

Audience

10

0

Player B

Player A

b1

b2

b3

Minimum

a1

a2

8

2

5

4

7

10

2

Maximum

8

嘷

5

10

嘷

5

The maximum of the row minimums is 5 and the minimum

of the column maximums is 5. The game has a pure strategy. Player A should take strategy a1 and Player B should

take strategy b2. The value of the game is 5.

Appendix G

12. a. The payoff table is as follows:

If a2, EV 4q 3(1 q)

1q 2(1 q) 4q 3(1 q)

1q 2 2q 4q 3 3q

10q 5

q 0.50

P(b2) q 0.50

P(b3) 1 0.50 0.50

c. 1p 4(1 p) (0.70) 4(0.30) 0.50

Blue Army

Attack Defend Minimum

Red Army

Attack

Defend

30

40

50

0

Maximum

40

50

30

0

The maximum of the row minimums is 30 and the minimum of the column maximums is 40. Because these

values are not equal, a mixed strategy is optimal.

Therefore, we must determine the best probability, p,

for which the Red Army should choose the Attack

strategy. Assume the Red Army chooses Attack with

probability p and Defend with probability 1 p. If the

Blue Army chooses Attack, the expected payoff is

30p 40(1 p). If the Blue Army chooses Defend,

the expected payoff is 50p 0(1 p).

Setting these equations equal to each other and solving for p, we get p 2兾3.

Red Army should choose to Attack with probability

2兾3 and Defend with probability 1兾3.

b. Assume the Blue Army chooses Attack with probability q and Defend with probability 1 q. If the Red

Army chooses Attack, the expected payoff for the Blue

Army is 30q 50(1 q). If the Red Army chooses

Defend, the expected payoff for the Blue Army is

40q 0(1 q). Setting these equations equal to each

other and solving for q, we get q 0.833. Therefore,

the Blue Army should choose to Attack with probability 0.833 and Defend with probability 1 0.833

0.167.

14. a. Strategy a3 dominated by a2

Strategy b1 dominated by b2

Player B

Player A

a1

a2

855

Self-Test Solutions and Answers to Even-Numbered Problems

b2

b3

1

4

2

3

b. Let p probability of a1 and (1 p) probability of a2

If b1, EV 1p 4(1 p)

If b2, EV 2p 3(1 p)

1p 4(1 p) 2p 3(1 p)

1p 4 4p 2p 3 3p

10p 7

p 0.70

p(a1) p 0.70

p(a2) 1 0.70 0.30

Let q probability of b2 and (1 q) probability of b3

If a1, EV 1q 2(1 q)

16. A: P(a3) 0.80, P(a4) 0.20

B: P(b1) 0.40, P(b2) 0.60

Value 2.8

Chapter 6

1. The following table shows the calculations for parts (a),

(b), and (c).

Absolute

Absolute

Time

Value of Squared Percent- Value of

Series

Forecast Forecast Forecast

age

Percentage

Week Value Forecast Error

Error

Error

Error

Error

1

2

3

4

5

6

18

13

16

11

17

14

a.

b.

c.

d.

18

13

16

11

17

5

3

5

6

3

5

3

5

6

3

25

9

25

36

9

38.46

18.75

45.45

35.29

21.43

38.46

18.75

45.45

35.29

21.43

Totals

22

104

51.30

159.38

MAE 22兾5 4.4

MSE 104兾5 20.8

MAPE 159.38兾5 31.88

The forecast for week 7 is F7 Y7 14.

2. The following table shows the calculations for parts (a),

(b), and (c).

Absolute

Absolute

Time

Value of Squared Percent- Value of

Series

Forecast Forecast Forecast

age Percentage

Week Value Forecast Error

Error

Error

Error

Error

1

2

3

4

5

6

18

13

16

11

17

14

18.00

15.50

15.67

14.50

15.00

5.00

0.50

4.67

2.50

1.00

5.00

0.50

4.67

2.50

1.00

25.00

0.25

21.81

6.25

1.00

38.46

3.13

42.45

14.71

7.14

38.46

3.13

42.45

14.71

7.14

Totals

13.67

54.31

70.21

105.86

a. MAE 13.67兾5 2.73

b. MSE 54.31兾5 10.86

c. MAPE 105.89兾5 21.18

Appendix G

Self-Test Solutions and Answers to Even-Numbered Problems

d. The forecast for week 7 is F7 (Y1 Y2 Y3 Y4

Y5 Y6)兾6 (18 13 16 11 17 14)兾6

14.83.

3. The following table shows the measures of forecast error

for both methods.

MAE

MSE

MAPE

Exercise 1

4.40

20.80

31.88

Exercise 2

2.73

10.86

21.18

For each measure of forecast accuracy, the average of all

the historical data provided more accurate forecasts than

simply using the most recent value.

5. a.

20

Time Series Value

856

18

16

14

12

10

8

6

4

2

0

1

2

Month

1

2

3

4

5

6

7

Forecast

Forecast

Error

24

13

20

12

19

23

11

7

8

7

4

8

Squared

Forecast

Error

121

49

64

49

16

64

Total 363

The forecast for month 8 is F8 Y8 15.

b.

Week

1

2

3

4

5

6

7

Forecast

Forecast Error

24.00

18.50

19.00

17.25

17.60

18.50

5

6

Week

1

2

3

4

5

6

Time

Series

Value

18

13

16

11

17

14

Forecast

15.67

13.33

14.67

Forecast

Error

Squared

Forecast

Error

4.67

3.67

0.67

Total

21.78

13.44

0.44

35.67

MSE 35.67兾3 11.89.

The forecast for week 7 is F7 (Y4 Y5 Y6)兾3

(11 17 14)兾3 14.

c. Smoothing constant ␣ 0.2

MSE 363/6 60.5

Time

Series

Value

24

13

20

12

19

23

15

4

Week (t)

The data appear to follow a horizontal pattern.

b. Three-week moving average

4. a.

Time

Series

Value

24

13

20

12

19

23

15

3

Squared

Forecast

Error

11.00

121.00

1.50

2.25

7.00

49.00

1.75

3.06

5.40

29.16

3.50

12.25

Total 216.72

MSE 216.72兾6 36.12

Forecast for month 8 is F8 (Y1 Y2 Y3 Y4 Y5

Y6 Y7)兾7 (24 13 20 12 19 23 15)兾

7 18.

c. The average of all the previous values is better because

MSE is smaller.

Week

1

2

3

4

5

6

Time

Series

Value

18

13

16

11

17

14

Forecast

Forecast

Error

18.00

17.00

16.80

15.64

15.91

5.00

1.00

5.80

1.36

1.91

Squared

Forecast

Error

25.00

1.00

33.64

1.85

3.66

Total 65.15

MSE 65.15兾5 13.03

The forecast for week 7 is F7 ␣Y6 (1 ␣)F6

0.2(14) (1 0.2)15.91 15.53.

d. The three-week moving average provides a better forecast since it has a smaller MSE.

e. Several values of ␣ will yield an MSE smaller than the

MSE associated with ␣ 0.2. The value of ␣ that

yields the minimum MSE is ␣ 0.367694922, which

yields an MSE of 12.060999.

Appendix G

␣ 0.367694922

Week

1

2

3

4

5

6

Time

Series

Value

18

13

16

11

17

14

Forecast

18

16.16

16.10

14.23

15.25

Forecast

Error

Squared

Forecast

Error

5.00

25.00

0.16

0.03

5.10

26.03

2.77

7.69

1.25

1.55

Total 60.30

MSE 252.87兾6 42.15

The forecast for week 8 is F8 ␣Y7 (1 ␣)F7

0.2(15) (1 0.2)20.15 19.12.

d. The three-week moving average provides a better forecast since it has a smaller MSE.

e. Several values of ␣ will yield an MSE smaller than the

MSE associated with ␣ 0.2. The value of ␣ that

yields the minimum MSE is ␣ 0.351404848, which

yields an MSE of 39.61428577.

␣ 0.351404848

MSE 60.30兾5 12.060999

6. a.

Week

1

2

3

4

5

6

7

30

Time Series Value

25

20

15

10

Time

Series

Value

24

13

20

12

19

23

15

Forecast

24.00

20.13

20.09

17.25

17.86

19.67

5

0

1

2

3

4

Week (t)

5

6

7

Week

1

2

3

4

5

6

7

Forecast

19.00

15.00

17.00

18.00

Forecast

Error

Squared

Forecast

Error

7.00

49.00

4.00

16.00

6.00

36.00

3.00

9.00

Total 110.00

MSE 110兾4 27.5.

The forecast for week 8 is F8 (Y5 Y6 Y7)兾3

(19 23 15)兾3 19.

c. Smoothing constant ␣ 0.2

Week

1

2

3

4

5

6

7

Time

Series

Value

24

13

20

12

19

23

15

Forecast

24.00

21.80

21.44

19.55

19.44

20.15

Forecast

Error

Squared

Forecast

Error

11.00

121.00

1.80

3.24

9.44

89.11

0.55

0.30

3.56

12.66

5.15

26.56

Total 252.87

Forecast

Error

Squared

Forecast

Error

11.00

121.00

0.13

0.02

8.09

65.40

1.75

3.08

5.14

26.40

4.67

21.79

Total 237.69

MSE 237.69兾6 39.61428577

8. a.

The data appear to follow a horizontal pattern.

b. Three-week moving average

Time

Series

Value

24

13

20

12

19

23

15

857

Self-Test Solutions and Answers to Even-Numbered Problems

Week

1

2

3

4

5

6

7

8

9

10

11

12

Time

Series

Value

17

21

19

23

18

16

20

18

22

20

15

22

Weighted

Moving

Average

Forecast

19.33

21.33

19.83

17.83

18.33

18.33

20.33

20.33

17.83

Forecast

Error

Squared

Forecast

Error

3.67

13.47

3.33

11.09

3.83

14.67

2.17

4.71

0.33

0.11

3.67

13.47

0.33

0.11

5.33

28.41

4.17

17.39

Total 103.43

b. MSE 103.43兾9 11.49

Prefer the unweighted moving average here; it has a

smaller MSE.

c. You could always find a weighted moving average at

least as good as the unweighted moving average.

Actually, the unweighted moving average is a special

case of the weighted average for which the weights are

equal.

10. a. F13 0.2Y12 0.16Y11 0.64(0.2Y10 0.8F10)

0.2Y12 0.16Y11 0.128Y10 0.512F10

858

Appendix G

Self-Test Solutions and Answers to Even-Numbered Problems

The MSE for the 3-month moving average is smaller,

so use the 3-month moving average.

F13 0.2Y12 0.16Y11 0.128Y10 0.512(0.2Y9

0.8F9) 0.2Y12 0.16Y11 0.128Y10 0.1024Y9

0.4096F9

b. The more recent data receive the greater weight or

importance in determining the forecast. The moving

averages method weights the last n data values equally

in determining the forecast.

12. a.

11.0

10.5

Interest Rate (%)

10.0

13. a.

400

Value (millions of dollars)

F13 0.2Y12 0.16Y11 0.128Y10 0.1024Y9

0.4096(0.2Y8 0.8F8) 0.2Y12 0.16Y11

0.128Y10 0.1024Y9 0.08192Y8 0.32768F8

c. The forecast for month 13 is F13 (Y10 Y11 Y12)兾

3 (9.7 9.6 9.6)兾3 9.63.

350

300

250

200

150

100

50

0

9.5

1

2

3

4

9.0

5

6

7

Month (t)

8

9

10

11

12

The data appear to follow a horizontal pattern.

8.5

b.

8.0

␣ 0.2

7.5

7.0

1

2

3

4

5

6

7

Month (t)

8

9

10

11

12

The data appear to follow a horizontal pattern.

b.

Month

1

2

3

4

5

6

7

8

9

10

11

12

Time

Series

Value

3-Month

Moving

Average

Forecast

9.5

9.3

9.4

9.6

9.8

9.7

9.8

10.5

9.9

9.7

9.6

9.6

9.40

9.43

9.60

9.70

9.77

10.00

10.07

10.03

9.73

(Error)2

0.04

0.14

0.01

0.01

0.53

0.01

0.14

0.18

0.02

1.08

4-Month

Moving

Average

Forecast

9.45

9.53

9.63

9.73

9.95

9.98

9.97

9.92

MSE(3-Month) 1.08兾9 0.12

MSE(4-Month) 1.09兾8 0.14

(Error)2

0.12

0.03

0.03

0.59

0.00

0.08

0.14

0.10

1.09

Month

1

2

3

4

5

6

7

8

9

10

11

12

3-Month

Time Moving

Series Average

Value Forecast (Error)2

240

350

230

260

280

320

220

310

240

310

240

230

273.33

280.00

256.67

286.67

273.33

283.33

256.67

286.67

263.33

177.69

0.00

4010.69

4444.89

1344.69

1877.49

2844.09

2178.09

1110.89

17,988.52

Average

Forecast

(Error)2

240.00

262.00

255.60

256.48

261.18

272.95

262.36

271.89

265.51

274.41

267.53

12100.00

1024.00

19.36

553.19

3459.79

2803.70

2269.57

1016.97

1979.36

1184.05

1408.50

27,818.49

MSE(3-Month) 17,988.52兾9 1998.72

MSE(␣ 0.2) 27,818.49兾11 2528.95

Based on the above MSE values, the 3-month moving average appears better. However, exponential

smoothing was penalized by including month 2,

which was difficult for any method to forecast. Using only the errors for months 4 to 12, the MSE for

exponential smoothing is as follows:

MSE(␣ 0.2) 14,694.49兾9 1632.72

Appendix G

Thus, exponential smoothing was better considering

months 4 to 12.

c. Using exponential smoothing,

F13 ␣Y12 (1 ⫺ ␣)F12 0.20(230) 0.80(267.53)

260.

Month

1

2

3

4

5

6

7

8

9

10

11

12

14. a.

160

140

Sales

120

100

80

60

40

20

0

1

2

3

4

5

6

7

Month (t)

859

Self-Test Solutions and Answers to Even-Numbered Problems

8

9

10

11

Time

Series

Value

Forecast

Forecast

Error

Squared

Error

105

135

120

105

90

120

145

140

100

80

100

110

105

105.98

106.43

106.39

105.85

106.31

107.57

108.63

108.35

107.43

107.18

30.00

14.02

1.43

16.39

14.15

38.69

32.43

8.63

28.35

7.43

2.82

900.00

196.65

2.06

268.53

200.13

1496.61

1051.46

74.47

803.65

55.14

7.93

Total

5056.62

12

MSE 5056.62兾11 459.6929489

16. a.

The data appear to follow a horizontal pattern.

12

b. Smoothing constant ␣ 0.3.

10

1

2

3

4

5

6

7

8

9

10

11

12

105

135

120

105

90

120

145

140

100

80

100

110

Squared

Error

(Yt ⴚ Ft)2

8

Rating

Time

Forecast

Series

Error

Month t Value Yt Forecast Ft Yt ⴚ Ft

6

4

105.00

114.00

115.80

112.56

105.79

110.05

120.54

126.38

118.46

106.92

104.85

30.00

6.00

10.80

22.56

14.21

34.95

19.46

26.38

38.46

6.92

5.15

900.00

36.00

116.64

508.95

201.92

1221.50

378.69

695.90

1479.17

47.89

26.52

Total

5613.18

MSE 5613.18兾11 510.29

The forecast for month 13 is F13 ␣Y12

(1 ␣)F12 0.3(110) 0.7(104.85) 106.4.

c. The value of ␣ that yields the smallest possible MSE

is ␣ 0.032564518, which yields an MSE of

459.6929489.

␣ 0.032564518

2

0

1997 1998 1999 2000 2001 2002 2003 2004 2005 2006 2007 2008

Year (t)

The time series plot indicates a possible linear trend in

the data. This could be due to decreasing viewer interest in watching the Masters. But closer inspection of

the data indicates that the two highest ratings correspond to years 1997 and 2001, years in which Tiger

Woods won the tournament. In fact, four of the five

highest ratings occurred when Tiger Woods won the

tournament. So, instead of an underlying linear trend in

the time series, the pattern observed may be simply due

to the effect Tiger Woods has on ratings and not necessarily on any long-term decrease in viewer interest.

b. The methods discussed in this section are only applicable for a time series that has a horizontal pattern.

So, if there is really a long-term linear trend in the

data, the methods discussed in this section are not

appropriate.

860

Appendix G

Self-Test Solutions and Answers to Even-Numbered Problems

c. The following time series plot shows the ratings for

years 2002–2008.

which results in the following forecasts, errors, and MSE:

9

8

Year

1

2

3

4

5

6

7

Rating

6

5

4

3

2

Sales

6.00

11.00

9.00

14.00

15.00

Forecast

6.80

8.90

11.00

13.10

15.20

17.30

Squared

Forecast

Forecast

Error

Error

0.80

0.64

2.10

4.41

2.00

4.00

0.90

0.81

0.20

0.04

Total 9.9

1

2003

2004

2005

Year (t)

2006

2007

2008

The time series plot for the data for years 2002–2008

exhibits a horizontal pattern. It seems reasonable to

conclude that the extreme values observed in 1997 and

2001 are more attributable to viewer interest in the performance of Tiger Woods. Basing the forecast on years

2002–2008 does seem reasonable. But because of the

injury that Tiger Woods experienced in the 2008 season, if he is able to play in the 2009 Masters then the

rating for 2009 may be significantly higher than suggested by the data for years 2002–2008. These types of

issues are what make forecasting in practice so difficult. For the methods to work, we have to be able to

assume that the pattern in the past is appropriate for the

future. But, because of the great influence Tiger Woods

has on viewer interest, making this assumption for this

time series may not be appropriate.

17. a.

16

Time Series Value

14

MSE 9.9兾5 1.982.475

c. F6 b0 b1t 4.7 2.1(6) 17.3

18. a.

Percentage of Stocks in Portfolio

0

2002

33

32

31

30

29

28

27

26

25

0

1

2

3

4

5

Period (t)

6

7

8

9

b. The value of the MSE will vary depending on the ultimate value of ␣ that you select. The value of ␣ that

yields the smallest possible MSE is ␣ 0.467307293,

which yields an MSE of 1.222838367.

␣ 0.467307293

12

10

8

6

4

2

0

1

2

3

Time Period (t)

4

5

The time series plot shows a linear trend.

b. The regression estimates for the slope and y-intercept

are as follows:

b1

n

n

n

t1

n

t1

a tYt a t a Ytnn

t1

n

2

2

a t a a t b nn

t1

186 (15)(55)兾5

55 (15)2兾5

t1

55

15

b0 Y b1t

2.10 a b 4.70

5

3

2.10

Period

1st-2007

2nd-2007

3rd-2007

4th-2007

1st-2008

2nd-2008

3rd-2008

4th-2008

1st-2009

2nd-2009

Stock%

29.8

31.0

29.9

30.1

32.2

31.5

32.0

31.9

30.0

Forecast

29.80

30.36

30.15

30.12

31.09

31.28

31.62

31.75

30.93

MSE 1.222838367

Forecast

Error

Squared

Forecast

Error

1.20

1.44

0.46

0.21

0.05

0.00

2.08

4.31

0.41

0.16

0.72

0.51

0.28

0.08

1.75

3.06

Total 9.78

c. The forecast for second quarter 2009 will vary depending on the ultimate value of ␣ that you selected in

part (b). Using an exponential smoothing model with

␣ 0.467307293, the forecast for second quarter

2009 30.93.

Enrollment (1000s)

20. a.

20

18

16

14

12

10

8

6

4

2

0

22. a.

$40

$35

$30

$25

$20

$15

$10

$5

$0

1

2

3

4

5

6

Year (t)

7

8

9

n

n

t1 t1

2

n

n

t1

t1

2

a t a a tb nn

b0 Y b1t

627.4 (45)(108)兾9

285 (45)2兾9

2

3

4

5

Year (t)

6

4.7167

which results in the following forecasts, errors, and

MSE:

Period

Year

1

2

3

4

5

6

7

8

9

10

2001

2002

2003

2004

2005

2006

2007

2008

2009

2010

6.50

8.10

8.40

10.20

12.50

13.30

13.70

17.20

18.10

Forecast

6.17

7.63

9.09

10.54

12.00

13.46

14.91

16.37

17.83

19.28

Forecast

Error

8

The time series plot shows an upward linear trend.

b. The regression estimates for the slope and y-intercept

are as follows:

b1

n

n

a tYt a t a Ytnn

t1

t1 t1

n

n

t1

t1

2

2

a t a a tb nn

1081.6 (36)(223.8)兾8

204 (36)2兾8

1.7738

which results in the following forecasts, errors, and

MSE:

108

45

4.7167 a b 1.4567

9

9

Enrollment

7

36

223.8

b0 Y b1t

1.774 a b 19.9928

8

8

n

a tYt a t a Ytnn

t1

1

n

The time series plot shows a linear trend.

b. The regression estimates for the slope and y-intercept

are as follows:

b1

861

Self-Test Solutions and Answers to Even-Numbered Problems

Cost/Unit ($)

Appendix G

Squared

Forecast

Error

0.33

0.47

0.69

0.34

0.50

0.16

1.21

0.83

0.27

Total

0.11

0.22

0.47

0.12

0.25

0.02

1.47

0.69

0.07

3.427

Year

1

2

3

4

5

6

7

8

9

Cost/

Unit($)

20.00

24.50

28.20

27.50

26.60

30.00

31.00

36.00

Forecast

21.77

23.54

25.31

27.09

28.86

30.64

32.41

34.18

35.96

Forecast

Error

1.77

0.96

2.89

0.41

2.26

0.64

1.41

1.82

Total

Squared

Forecast

Error

3.12

0.92

8.33

0.17

5.12

0.40

1.99

3.30

23.34619

MSE 2.9183

c. The average cost/unit has been increasing by approximately $1.77 per year.

d. F9 b0 b1t 19.9928 1.7738(9) 35.96

MSE 0.3808

c. F10 b0 b1t 4.7167 1.4567(10) 19.28

862

Appendix G

Self-Test Solutions and Answers to Even-Numbered Problems

26. a.

90

3500

80

3000

70

Time Series Value

Time Series Value

24. a.

60

50

40

30

20

2000

1500

1000

500

10

0

2500

0

1

2

3

4

5

6

7

Period (t)

8

9

10

11

12

The time series plot shows a horizontal pattern. But

there is a seasonal pattern in the data. For instance, in

each year the lowest value occurs in quarter 2 and the

highest value occurs in quarter 4.

b. After putting the data into the following format:

Dummy Variables

Year Quarter Quarter 1 Quarter 2 Quarter 3 Yt

1

1

1

1

2

2

2

2

3

3

3

3

1

2

3

4

1

2

3

4

1

2

3

4

1

0

0

0

1

0

0

0

1

0

0

0

0

1

0

0

0

1

0

0

0

1

0

0

0

0

1

0

0

0

1

0

0

0

1

0

71

48

58

78

68

41

60

81

62

51

53

72

we can use the LINEST function to find the regression

model:

Value 77.00 10.00 Qtr1 30.33 Qtr2 20.00 Qtr3

c. The quarterly forecasts for next year are as follows:

Quarter 1 forecast 77.0 10.0(1) 30.33(0) 20.0(0)

67.00

Quarter 2 forecast 77.0 10.0(0) 30.33(1) 20.0(0)

46.67

Quarter 3 forecast 77.0 10.0(0) 30.33(0) 20.0(1)

57.00

Quarter 4 forecast 77.0 10.0(0) 30.33(0) 20.0(0)

77.00

1

2

3

4

5

6

7

Period (t)

8

9

10

11

12

There appears to be a seasonal pattern in the data and

perhaps a moderate upward linear trend.

b. After putting the data into the following format:

Dummy Variables

Year Quarter Quarter 1 Quarter 2 Quarter 3

1

1

1

1

2

2

2

2

3

3

3

3

1

2

3

4

1

2

3

4

1

2

3

4

1

0

0

0

1

0

0

0

1

0

0

0

0

1

0

0

0

1

0

0

0

1

0

0

0

0

1

0

0

0

1

0

0

0

1

0

Yt

1690

940

2625

2500

1800

900

2900

2360

1850

1100

2930

2615

we can use the LINEST function to find the regression

model:

Value 2491.67 711.67 Qtr1 1511.67 Qtr2

326.67 Qtr3

c. The quarterly forecasts for next year are as follows:

Quarter 1 forecast 2491.67 711.67(1) 1511.67(0)

326.67(0) 1780.00

Quarter 2 forecast 2491.67 711.67(0) 1511.67(1)

326.67(0) 980.00

Quarter 3 forecast 2491.67 711.67(0) 1511.67(0)

326.67(1) 2818.33

Quarter 4 forecast 2491.67 711.67(0) 1511.67(0)

326.67(0) 2491.67

Appendix G

863

Self-Test Solutions and Answers to Even-Numbered Problems

d. After putting the data into the following format:

b. After putting the data into the following format:

Dummy Variables

Year

Quarter

Quarter 1

Quarter 2

Quarter 3

t

Yt

1

1

1

1

2

2

2

2

3

3

3

3

1

2

3

4

1

2

3

4

1

2

3

4

1

0

0

0

1

0

0

0

1

0

0

0

0

1

0

0

0

1

0

0

0

1

0

0

0

0

1

0

0

0

1

0

0

0

1

0

1

2

3

4

5

6

7

8

9

10

11

12

1690

940

2625

2500

1800

900

2900

2360

1850

1100

2930

2615

we can use the LINEST function to find the regression

model:

Value 2306.67 642.29 Qtr1 1465.42 Qtr2

349.79 Qtr3 23.13t

Dummy Variables

Year Quarter Quarter 1 Quarter 2 Quarter 3

1

1

1

0

0

1

2

0

1

0

1

3

0

0

1

1

4

0

0

0

2

1

1

0

0

2

2

0

1

0

2

3

0

0

1

2

4

0

0

0

3

1

1

0

0

3

2

0

1

0

3

3

0

0

1

3

4

0

0

0

4

1

1

0

0

4

2

0

1

0

4

3

0

0

1

4

4

0

0

0

5

1

1

0

0

5

2

0

1

0

5

3

0

0

1

5

4

0

0

0

The quarterly forecasts for next year are as follows:

Quarter 1 forecast 2306.67 642.29(1) 1465.42(0)

349.79(0) 23.13(13) 1965.00

Quarter 2 forecast 2306.67 642.29(0) 1465.42(1)

349.79(0) 23.13(14) 1165.00

Quarter 3 forecast 2306.67 642.29(0) 1465.42(0)

349.79(1) 23.13(15) 2011.33

Quarter 4 forecast 2306.67 642.29(0) 1465.42(0)

349.79(0) 23.13(16) 2676.67

Yt

20

100

175

13

37

136

245

26

75

155

326

48

92

202

384

82

176

282

445

181

we can use the LINEST function to find the regression

model:

Revenue 70.0 10.0 Qtr1 105 Qtr2 245 Qtr3

Quarter 1 forecast 70.0 10.0(1) 105(0) 245(0) 80

Quarter 2 forecast 70.0 10.0(0) 105(1) 245(0) 175

Quarter 3 forecast 70.0 10.0(0) 105(0) 245(1) 315

Quarter 4 forecast 70.0 10.0(0) 105(0) 245(0) 70

c. After putting the data into the following format:

Sales ($1000s)

28. a.

500

450

400

350

300

250

200

150

100

50

0

Dummy Variables

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Period (t)

The time series plot shows both a linear trend and

seasonal effects.

Year

Quarter

Quarter 1

Quarter 2

Quarter 3

t

Yt

1

1

1

1

2

2

2

2

3

3

3

3

4

4

4

4

5

5

5

5

1

2

3

4

1

2

3

4

1

2

3

4

1

2

3

4

1

2

3

4

1

0

0

0

1

0

0

0

1

0

0

0

1

0

0

0

1

0

0

0

0

1

0

0

0

1

0

0

0

1

0

0

0

1

0

0

0

1

0

0

0

0

1

0

0

0

1

0

0

0

1

0

0

0

1

0

0

0

1

0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

20

100

175

13

37

136

245

26

75

155

326

48

92

202

384

82

176

282

445

181

864

Appendix G

Self-Test Solutions and Answers to Even-Numbered Problems

we can use the LINEST function to find the regression

model:

Revenue 70.10 45.03 Qtr1 128.35 Qtr2

256.68 Qtr3 11.68t

Quarter 1 forecast 70.10 45.03(1) 128.35(0)

256.68(0) 11.68(21) 221

Quarter 2 forecast 70.10 45.03(0) 128.35(1)

256.68(0) 11.68(22) 315

Quarter 3 forecast 70.10 45.03(0) 128.35(0)

256.68(1) 11.68(23) 456

Quarter 4 forecast 70.10 45.03(0) 128.35(0)

256.68(0) 11.68(24) 211

Chapter 7

B

100

80

60

(b)

(c)

40

20

–100 –80 –60 –40 –20

1. Parts (a), (b), and (e) are acceptable linear programming

relationships.

Part (c) is not acceptable because of 2×22.

Part (d) is not acceptable because of 3 2×1.

Part (f) is not acceptable because of 1x1x2.

Parts (c), (d), and (f) could not be found in a linear programming model because they contain nonlinear terms.

2. a.

6. 7A 10B 420

6A 4B 420

4A 7B 420

(a)

0

40

50

60

80

100

A

7.

B

100

B

(0,8)

8

50

4

0

50

100

150

200

A

250

(4,0)

0

b.

4

8

A

10.

B

B

6

8

5

4

4

0

c.

4

8

A

Optimal solution

A = 12/7, B = 15/7

3

Value of Objective Function =

2(12/7) + 3(15/7) = 69/7

2

B

A+

2

1

8

Points on

line are only

feasible points

1

2

3

4

5

5A

4

0

B =

6

B=

+3

4

8

A

15

0

6

A

Appendix G

A 2B

6 (1)

5A 3B 15 (2)

Equation (1) times 5:

5A 10B 30 (3)

Equation (2) minus equation (3):

7B 15

B 15兾7

From equation (1):

A 6 2(15兾7)

6 30兾7 12兾7

18. b. A 18兾7, B 15兾7

c. 0, 0, 4兾7

20. b. A 3.43, B 3.43

c. 2.86, 0, 1.43, 0

22. b.

Extreme Point

Coordinates

Profit ($)

1

2

3

4

5

(0, 0)

(1700, 0)

(1400, 600)

(800, 1200)

(0, 1680)

0

8500

9400

8800

6720

12. a. A 3, B 1.5; Value of optimal solution 13.5

b. A 0, B 3; Value of optimal solution 18

c. Four: (0, 0), (4, 0), (3, 1.5), and (0.3)

13. a.

B

Extreme point 3 generates the highest profit.

c. A 1400, C 600

d. Cutting and dyeing constraint and the packaging

constraint

e. A 800, C 1200; profit $9200

8

6

Feasible region

consists of this

line segment only

4

24. a. Let R number of units of regular model

C number of units of catcher’s model

Max 5R 8C

1R C 3/2C 900 Cutting and sewing

1/ R

1/ C 300 Finishing

2

3

1/ R

1/ C 100 Packaging and

8

4

shipping

R, C 0

b.

C

2

0

2

4

6

8

A

b. The extreme points are (5, 1) and (2, 4).

c. B

6

900

Optimal solution

A = 2, B = 4

700

2B

Catcher’s model

A+

800

=1

0

2

0

2

4

6

540 standard bags, 252 deluxe bags

7668

630, 480, 708, 117

0, 120, 0, 18

16. a. 3S 9D

b. (0,540)

c. 90, 150, 348, 0

17. Max 5A 2B 0s1 0s2 0s3

s.t.

1A 2B 1s1

420

2A 3B

1s2

610

6A 1B

1s3 125

A, B, s1, s2, s3 0

8

A

F

4

14. a.

b.

c.

d.

865

Self-Test Solutions and Answers to Even-Numbered Problems

600

C&

500

400

S

P&

300

S

Optimal solution

R = 500, C = 150

200

100

0

100 200 300 400 500 600 700 800 900

Regular model

c. 5(500) 8(150) $3700

d. C & S

1(500) 3/2(150) 725

1

F

/2(500) 1/3(150) 300

1/ (500) 1/ (150) 100

P&S

8

4

e.

Department

Capacity

Cutting and sewing

900

Finishing

300

Packaging and shipping 100

Usage

Slack

725 175 hours

300

0 hours

100

0 hours

R

866

Appendix G

26. a. Max

s.t.

Self-Test Solutions and Answers to Even-Numbered Problems

34. a.

50N 80R

N R 1000

N

250

R 250

N 2R

0

N, R 0

b. N 666.67, R 333.33; Audience exposure 60,000

28. a. Max 1W 1.25M

s.t.

5W

7M 4480

3W

1M 2080

2W

2M 1600

W, M 0

b. W 560, M 240; Profit 860

30. a. Max 15E 18C

s.t.

40E 25C 50,000

40E

15,000

25C 10,000

25C 25,000

E, C 0

31.

B

4

Feasible

region

3

(21/4, 9/4)

2

1

(4, 1)

0

1

2

3

4

5

(A 4, B 1) and (A 21兾4, B 9兾4)

c. The optimal solution [see part (a)] is A 4, B 1.

35. a. Min

s.t.

B

6A 4B 0s1 0s2 0s3

2A 1B s1

12

1A 1B

s2

10

1B

s3 4

A, B, s1, s2, s3 0

b. The optimal solution is A 6, B 4

c. s1 4, s2 0, s3 0

36. a. Min

s.t.

10,000T 8000P

8

P 10

T

P 25

3T

2P 84

c. (15, 10); (21.33, 10); (8, 30); (8, 17)

d. T 8, P 17

Total cost $216,000

6

T

Feasible

region

4

2

4

6

Optimal solution

A = 3, B = 1

8

A

3A + 4B = 13

Objective function value 13

32.

Objective

Extreme

Points

Function

Value

Surplus

Demand

(250, 100)

(125, 225)

(125, 350)

800

925

1300

125

—

—

Stock

Total

Processing

Production

Time

—

—

125

A

b. There are two extreme points:

c. (375, 400); (1000, 400); (625, 1000); (375, 1000)

d. E 625, C 1000

Total return $27,375

2

6

—

125

—

38. a. Min

s.t.

7.50S 9.00P

0.10S 0.30P 6

0.06S 0.12P 3

S

P 30

S, P 0

c. The optimal solution is S 15, P 15.

d. No

e. Yes

40. P1 30, P2 25, Cost $55

Appendix G

42.

867

Self-Test Solutions and Answers to Even-Numbered Problems

Chapter 8

B

1. a.

Satisfies constraint #2

10

B

10

A = 4, B = 6

8

7)

3(

8

Infeasibility

)=

(3

+2

6

Satisfies constraint #1

4

2

2

43.

4

B

6

4

Unbounded

3

Feasible

region

8

A

10

1

1

2

3

4

5

A

44. a. A 30兾16, B 30兾16; Value of optimal solution 60/16

b. A 0, B 3; Value of optimal solution 6

46. a. 180, 20

b. Alternative optimal solutions

c. 120, 80

48. No feasible solution

50. M 65.45, R 261.82; Profit $45,818

52. S 384, O 80

54. a. Max

s.t.

160M1 345M2

15

M2 10

M1

5

M2

5

40M1 50M2 1000

M1, M2 0

M1

b. M1 12.5, M2 10

2

0

2

0

Optimal Solution

A = 7, B = 3

27

6

4

2

4

6

8

10

A

b. The same extreme point, A 7 and B 3, remains

optimal; Value of the objective function becomes

5(7) 2(3) 41.

c. A new extreme point, A 4 and B 6, becomes

optimal; Value of the objective function becomes

3(4) 4(6) 36.

d. The objective coefficient range for variable A is 2 to 6;

the optimal solution, A 7 and B 3, does not change.

The objective coefficient range for variable B is 1 to 3;

resolve the problem to find the new optimal solution.

2. a. The feasible region becomes larger with the new optimal solution of A 6.5 and B 4.5.

b. Value of the optimal solution to the revised problem is

3(6.5) 2(4.5) 28.5; the one-unit increase in the

right-hand side of constraint 1 increases the value of

the optimal solution by 28.5 27 1.5; therefore, the

shadow price for constraint 1 is 1.5.

c. The right-hand-side range for constraint 1 is 8 to 11.2;

as long as the right-hand side stays within this range,

the shadow price of 1.5 is applicable.

d. The value of the optimal solution will increase by 0.5

for every unit increase in the right-hand side of constraint 2 as long as the right-hand side is between 18

and 30.

4. a.

b.

c.

d.

X 2.5, Y 2.5

2

5 to 11

The value of the optimal solution will increase by 3 for

every unit increase in the right-hand side of constraint

2 as long as the right-hand side is between 9 and 18.

868

Appendix G

Self-Test Solutions and Answers to Even-Numbered Problems

5. a. Regular glove 500; Catcher’s mitt 150;

Value 3700

b. The finishing, packaging, and shipping constraints are

binding; there is no slack

c. Cutting and sewing 0

Finishing 3

Packaging and shipping 28

Additional finishing time is worth $3 per unit, and additional packaging and shipping time is worth $28 per unit.

d. In the packaging and shipping department, each additional hour is worth $28.

6. a. The optimal value for the Regular Glove variable is 5,

the Allowable Decrease is 1, and the Allowable

Increase is 7. The optimal value for the Catcher’s Mitt

variable is 8, the Allowable Decrease is 4.667, and the

Allowable Increase is 2. Therefore, we can express the

Objective Coefficient Ranges as follows:

Variable

Objective Coefficient Range

Regular Glove

5 1 4 to 5 7 12

Catcher’s Mitt 8 4.667 3.333 to 8 2 10

b. As long as the profit contribution for the regular glove

is between $4.00 and $12.00, the current solution is

optimal; as long as the profit contribution for the

catcher’s mitt stays between $3.33 and $10.00, the current solution is optimal; the optimal solution is not

sensitive to small changes in the profit contributions for

the gloves.

c. The shadow prices for the resources are applicable over

the following ranges:

Right-HandSide Range

Constraint

Cutting and sewing 900 175 725 to No Upper Limit

Finishing

300 166.667 133.333 to

300 100 400

Packaging

100 25 75 to 100

35 135

d. The shadow price of packaging and shipping constraint

is 28, so the amount of increase (28) (20) $560.

8. a. More than $7.00

b. More than $3.50

c. None

10. a. S 4000

M 10,000

Total risk 8(4000) 3(10,000) 62,000

b.

Variable

Objective Coefficient Range

S

M

8.000 4.250 3.750 to No Upper Limit

No Upper Limit to 3.000 3.400 6.400

c.

d.

e.

f.

5(4000) 4(10,000) $60,000

60,000兾1,200,000 0.05 or 5%

0.057 risk units

0.057(100) 5.7%

12. a. E 80, S 120, D 0

Profit 63(80) 95(120) 135(0) $16,440

b. Fan motors and cooling coils

c. The manufacturing time constraint has slack;

2400 2080 320 hours are available.

d. This represents an increase in the objective function

coefficient for D of $150 $135 $15. Because this

is less than the allowable increase of $24 for the objective function coefficient for D, there is no change in the

optimal solution.

13. a. The range of optimality for each objective function

coefficient is as follows:

E 63.000 15.5000 47.500 to 63.000 12.000 75

S 95.000 8.000 87.000 to 95.000 31.000 126

D No lower limit to 135.000 24.000 159.000

b. Because more than one objective function coefficient

value is changing at the same time here, we must resolve the problem to answer this question. Re-solving

the problem with the new profit values shows that the

optimal solution will not change. However, the change

in total profit will be 69(80) 93(120) 135(0)

$16,680.

c. The range of feasibility for the right-hand side values

for each constraint is as follows:

Fan motors constraint

Cooling coils constraint

Manufacturing time

constraint

200.000 40.000 160.000 to

200.000 80.000 280.000

320.000 120.000 200.000 to

320.000 80.000 400.000

2400.000 320.000 2080.000

to No Upper Limit

d. Yes, 100 is greater than the allowable increase for the

fan motors constraint (80.000).

The shadow price will change.

14. a. The optimal solution is to manufacture 100 cases of

model A and 60 cases of model B and purchase 90

cases of model B.

Total Cost 10(100) 6(60) 14(0) 9(90) $2170

b. Demand for A, demand for B, assembly time

c.

Constraint

Shadow Price

1

2

3

4

12.25

9.0

0

0.375

Appendix G

If demand for model A increases by 1 unit, total cost

will increase by $12.25.

If demand for model B increases by 1 unit, total cost

will increase by $9.00.

If an additional minute of assembly time is available,

total cost will decrease by $.375.

d. Assembly time constraint

16. a. 100 suits, 150 sport coats

Profit $40,900

40 hours of cutting overtime

b. Optimal solution will not change.

c. Consider ordering additional material.

$34.50 is the maximum price.

d. Profit will improve by $875.

20. a. Max

b. H $400,000, P $225,000, A $375,000

Total annual return $88,750

Annual percentage return 8.875%

c. No change

d. Increase of $890

e. Increase of $312.50, or 0.031%

AO

BN

AN

BN

AO

AN, AO, BN, BO 0

50,000

BO 70,000

80,000

BO 60,000

b. Optimal solution

Model A

Model B

30L

22. a. Min

New Line

Old Line

50,000

30,000

0

40,000

Total cost: $3,850,000

c. The first three constraints are binding.

d. The shadow price for the new production line capacity

constraint is 15. Because the shadow price is negative, increasing the right-hand side of constraint 3 will

cause the objective function to decrease. Thus, every

1-unit increase in the right hand side of this constraint

will actually reduce the total production cost by $15.

In other words, an increase in capacity for the new

production line is desirable.

e. Because constraint 4 is not a binding constraint, any

increase in the production line capacity of the old production line will have no effect on the optimal solution;

thus, increasing the capacity of the old production line

results in no benefit.

f. The reduced cost for model A made on the old production line is 5; thus, the cost would have to decrease by

at least $5 before any units of model A would be produced on the old production line.

g. The right-hand-side range for constraint 2 shows an

allowable decrease of 40,000. Thus, if the minimum production requirement is reduced 10,000 units to 60,000,

the shadow price of 40 is applicable. Thus, total cost

would decrease by 10,000(40) $400,000.

25D

18S

L

D

S 100

0.6L 0.4D

0

0.15L 0.15D 0.85S 0

0.25L 0.25D

S 0

L

50

L, D, S 0

30AN 50AO 25BN 40BO

AN

0.07H 0.12P 0.09A

H

P

A 1,000,000

0.6H 0.4P 0.4A 0

P 0.6A 0

H, P, A 0

18. a. The linear programming model is as follows:

Min

869

Self-Test Solutions and Answers to Even-Numbered Problems

b. L 48, D 72, S 30

Total cost $3780

c. No change

d. No change

24. a. 333.3, 0, 833.3; Risk 14,666.7; Return 18,000,

or 9%

b. 1000, 0, 0, 2500; Risk 18,000; Return 22,000, or

11%

c. $4000

26. a. Let M1 units of component 1 manufactured

M2 units of component 2 manufactured

M3 units of component 3 manufactured

P1 units of component 1 purchased

P2 units of component 2 purchased

P3 units of component 3 purchased

Min 4.50M1 5.00M2 2.75M3 6.50P1 8.80P2 7.00P3

2M1 3M2 4M3

21,600 Production

15,000 Assembly

1M1 1.5M2 3M3

18,000 Testing/Packaging

1.5M1 2M2 5M3

1P1

6,000 Component 1

1M1

1P2

4,000 Component 2

1M2

1P3 3,500 Component 3

1M3

M1, M2, M3, P1, P2, P3 0

b.

Source

Manufacture

Purchase

Component Component Component

1

2

3

2000

4000

4000

1400

2100

Total Cost $73,550

c. Production: $54.36 per hour

Testing & Packaging: $7.50 per hour

d. Shadow prices $7.969; it would cost Benson $7.969

to add a unit of component 2.

870

Appendix G

28. a. Let

b.

c.

d.

e.

f.

30. a.

b.

c.

d.

e.

Self-Test Solutions and Answers to Even-Numbered Problems

G amount invested in growth stock fund

S amount invested in income stock fund

M amount invested in money market fund

Max 0.20G 0.10S 0.06M

s.t.

0.10G 0.05S 0.01M (0.05)(300,000)

G

(0.10)(300,000)

S

(0.10)(300,000)

M (0.20)(300,000)

G

S

M 300,000

G, S, M 0

G 120,000; S 30,000; M 150,000

0.15 to 0.60; No Lower Limit to 0.122; 0.02 to 0.20

4668

G 48,000; S 192,000; M 60,000

The client’s risk index and the amount of funds available

L 3, N 7, W 5, S 5

Each additional minute of broadcast time increases

cost by $100.

If local coverage is increased by 1 minute, total cost

will increase by $100.

If the time devoted to local and national news is increased by 1 minute, total cost will increase by $100.

Increasing the sports by 1 minute will have no effect

because the shadow price is 0.

32. a. Let P1 number of PT-100 battery packs produced at

the Philippines plant

P2 number of PT-200 battery packs produced at

the Philippines plant

P3 number of PT-300 battery packs produced at

the Philippines plant

M1 number of PT-100 battery packs produced at

the Mexico plant

M2 number of PT-200 battery packs produced at

the Mexico plant

M3 number of PT-300 battery packs produced at

the Mexico plant

Min 1.13P1 1.16P2 1.52P3 1.08M1 1.16M2 1.25M3

P1

M1

200,000

M2

100,000

P2

M3 150,000

P3

P2

175,000

P1

M2

160,000

M1

75,000

P3

M3 100,000

P1, P2, P3, M1, M2, M3 0

b. The optimal solution is as follows:

PT-100

PT-200

PT-300

Philippines

40,000

100,000

50,000

Mexico

160,000

0

100,000

Total production and transportation cost is $535,000.

c. The range of optimality for the objective function

coefficient for P1 shows a lower limit of $1.08; thus, the

production and/or shipping cost would have to

decrease by at least 5 cents per unit.

d. The range of optimality for the objective function

coefficient for M1 shows a lower limit of $1.11; thus,

the production and/or shipping cost would have to

decrease by at least 5 cents per unit.

Chapter 9

1. a. Let

T number of television spot advertisements

R number of radio advertisements

N number of newspaper advertisements

Max 100,000T 18,000R 40,000N

s.t.

2,000T 300R 600N 18,200 Budget

T

10 Max TV

R

20 Max Radio

N

10 Max News

0.5T 0.5R 0.5N

0 Max 50% Radio

0.9T 0.1R 0.1N

0 Min 10% TV

T, R, N, 0

Budget $

Solution: T 4

R 14

N 10

$ 8000

4200

6000

$18,200

Audience 1,052,000

b. The shadow price for the budget constraint is 51.30.

Thus, a $100 increase in budget should provide an

increase in audience coverage of approximately 5130.

The right-hand-side range for the budget constraint

will show this interpretation is correct.

2. a. Let

Max

s.t.

x1 units of product 1 produced

x2 units of product 2 produced

30×1 15×2

x1 0.35×2 100 Dept. A

0.30×1 0.20×2 36 Dept. B

0.20×1 0.50×2 50 Dept. C

x1, x2 0

Solution: x1 77.89, x2 63.16; Profit $3284.21

b. The shadow price for Department A is $15.79; for

Department B it is $47.37; and for Department C it is

$0.00. Therefore, we would attempt to schedule overtime in Departments A and B. Assuming the current

labor available is a sunk cost, we should be willing to

pay up to $15.79 per hour in Department A and up to

$47.37 in Department B.

c. Let

xA hours of overtime in Department A

xB hours of overtime in Department B

xC hours of overtime in Department C

Appendix G

Max

s.t.

30×1

15×2 18xA 22.5xB 12xC

x1 0.35×2

0.30×1 0.20×2

0.20×1 0.50×2

xA

xA

xB

xB

x1, x2, xA, xB, xC 0

100

36

xC 50

10

6

xC 8

x1 87.21

x2 65.12

Profit $3341.34

Overtime

Department A

10 hours

Department B

3.186 hours

Department C

0 hours

Increase in profit from overtime $3341.34 3284.21

$57.13

4. Let X1 the number of pounds of Party Nuts to produce

X2 the number of pounds of Mixed Nuts to produce

X3 the number of pounds of Premium Nuts to

produce

Max 2(1.00)X1 2(2.10)X2 2(3.63)X3 1.5(X1

0.55X2) 5.35(0.25X2 0.40X3) 6.25(0.1X2 0.2X3)

s.t.

X1 0.55X2

500 (Peanuts)

0.25X2 0.40X2 180 (Cashews)

0.1X2 0.2X2

100 (Brazil Nuts)

0.1X2 0.4X2

80

(Hazelnuts)

X1, X2, X2

0

The optimal solution is as follows:

133 1/3 pounds of Party Nuts (or 266 2/3 bags)

666 2/3 pounds of Mixed Nuts (or 1333 1/3 bags)

33 1/3 pounds of Premium Nuts (or 66 2/3 bags)

Profit of $537.33

The binding constraints are Peanuts, Cashews, and

Hazelnuts.

Brazil Nuts are not binding (only 73 1/3 pounds are used

resulting in slack of 26 2/3 pounds).

6. Let

Max

s.t.

871

Self-Test Solutions and Answers to Even-Numbered Problems

x1 units of product 1

x2 units of product 2

b1 labor-hours Department A

b2 labor-hours Department B

25×1 20×2 0b1 0b2

6×1 8×2 1b1

0

12×1 10×2

1b2 0

1b1 1b2 900

x1, x2, b1, b2 0

Solution: x1 50, x2 0, b1 300, b2 600; Profit:

$1250

8. Let x1 the number of officers scheduled to begin at

8:00 A.M.

x2 the number of officers scheduled to begin at noon

x3 the number of officers scheduled to begin at

4:00 P.M.

x4 the number of officers scheduled to begin at

8:00 P.M.

x5 the number of officers scheduled to begin at

midnight

x6 the number of officers scheduled to begin at

4:00 A.M.

The objective function to minimize the number of officers

required is as follows:

Min

x1 x2 x3 x4 x5 x6

The constraints require the total number of officers on

duty each of the six 4-hour periods to be at least equal to

the minimum officer requirements. The constraints for the

six 4-hour periods are as follows:

Time of Day

8:00 A.M.–Noon

x1

x6 5

Noon-4:00 P.M.

x1 x2

6

4:00 P.M.–8:00 P.M.

x2 x3

10

8:00 P.M.–Midnight

x3 x4

7

Midnight–4:00 A.M.

x4 x5

4

4:00 A.M.–8:00 A.M.

x5 x6 6

x1, x2, x3, x4, x5, x6 0

Schedule 19 officers as follows:

x1 3 begin at 8:00 A.M.

x2 3 begin at noon

x3 7 begin at 4:00 P.M.

x4 0 begin at 8:00 P.M.

x5 4 begin at midnight

x6 2 begin at 4:00 A.M.

9. a. Let each decision variable, A, P, M, H, and G, represent the fraction or proportion of the total investment

placed in each investment alternative.

Max

s.t.

0.073A 0.103P 0.064M 0.075H 0.045G

A

0.5A

0.5A

0.6A

P

M

H

0.5P 0.5M 0.5H

0.5P 0.5M 0.5H

0.25M 0.25H

0.4P

A, P, M, H, G 0

Solution: Objective function 0.079 with

Atlantic Oil 0.178

Pacific Oil 0.267

Midwest Oil 0.000

Huber Steel 0.444

Government Bonds 0.111

b. For a total investment of $100,000, we show

Atlantic Oil $ 17,800

Pacific Oil

26,700

Midwest Oil

0

Huber Steel

44,400

Government Bonds

11,100

Total

$100,000

G1

0

0

G0

0

872

Appendix G

Self-Test Solutions and Answers to Even-Numbered Problems

c. Total earnings $100,000 (0.079) $7,900

d. Marginal rate of return 0.079

S the proportion of funds invested in stocks

B the proportion of funds invested in bonds

M the proportion of funds invested in mutual

funds

C the proportion of funds invested in cash

The linear program and optimal solution are as follows:

Max 0.1S 0.03B 0.04M 0.01C

s.t.

(1) 1S 1B 1M 1C 1

(2) 0.8S 0.2B 0.3M 0.4

(3) 1S (0.75

(4) 1B 1M 0

(5) 1C 0.1

(6) 1C 0.3

10. a. Let

The optimal allocation among the four investment alternatives:

Stocks

Bonds

Mutual Funds

Cash

40.9%

14.5%

14.5%

30.0%

The annual return associated with the optimal portfolio

is 5.4%.

Total risk 0.409(0.8) 0.145(0.2) 0.145(0.3)

0.300(0.0) 0.4

b. Changing the right-hand-side value for constraint 2 to

0.18 and re-solving, we obtain the following optimal

solution:

Stocks

Bonds

Mutual Funds

Cash

0.0%

36.0%

36.0%

28.0%

The annual return associated with the optimal portfolio

is 2.52%.

Total risk 0.0(0.8) 0.36(0.2) 0.36(0.3)

0.28(0.0) 0.18

c. Changing the right-hand-side value for constraint 2

to 0.7 and re-solving, we obtain the following optimal

allocation among the four investment alternatives:

Stocks

Bonds

Mutual Funds

Cash

75.0%

0.0%

15.0%

10.0%

The annual return associated with the optimal

portfolio is 8.2%.

Total risk 0.75(0.8) 0.0(0.2) 0.15(0.3)

0.10(0.0) 0.65

d. Note that a maximum risk of 0.7 was specified for

this aggressive investor, but that the risk index for the

portfolio is only 0.65. Thus, this investor is willing to

take more risk than the solution shown above provides.

There are only two ways the investor can become even

more aggressive: by increasing the proportion invested

in stocks to more than 75% or reducing the cash

requirement of at least 10% so that additional cash

could be put into stocks. For the data given here, the

investor should ask the investment advisor to relax

either or both of these constraints.

e. Defining the decision variables as proportions means

the investment advisor can use the linear programming

model for any investor, regardless of the amount of the

investment. All the investor advisor must do is to

establish the maximum total risk for the investor and

resolve the problem using the new value for maximum

total risk.

12. Let Bi pounds of shrimp bought in week i, i 1, 2, 3, 4

Si pounds of shrimp sold in week i, i 1, 2, 3, 4

Ii pounds of shrimp held in storage (inventory) in

week i

Total purchase cost 6.00B1 6.20B2 6.65B3 5.55B4

Total sales revenue 6.00S1 6.20S2 6.65S3 5.55S4

Total storage cost 0.15I1 0.15I2 0.15I3 0.15I4

Total profit contribution (total sales revenue) (total

purchase cost) (total storage cost)

Objective: Maximize total profit contribution subject to

balance equations for each week, storage capacity for

each week, and ending inventory requirement for week 4.

Max 6.00S1 6.20S2 6.65S3 5.55S4 6.00B1

6.20B2 6.65B3 5.55B4 0.15I1 0.15I2

0.15I3 0.15I4

s.t.

20,000 B1 S1 I1 Balance equation—week 1

I1

B2 S2 I2 Balance equation—week 2

I2

B3 S3 I3 Balance equation—week 3

I3

B4 S4 I4 Balance equation—week 4

I1 100,000 Storage capacity—week 1

I2 100,000 Storage capacity—week 2

I3 100,000 Storage capacity—week 3

I4 100,000 Storage capacity—week 4

I4 25,000 Required inventory—week 4

all variables 0

Note that the first four constraints can be written as follows:

I1 B1 S1

20,000

I1 I2 B2 S2

0

I2 I3 B3 S3

0

I3 I4 B4 S4

0

The optimal solution follows:

Week (i)

1

2

3

4

Bi

Si

Ii

80,000

0

0

25,000

0

0

100,000

0

100,000

100,000

0

25,000

Appendix G

Self-Test Solutions and Answers to Even-Numbered Problems

s.t.

R1 S1

110,000 Input 1 Capacity

R2 S2

350,000 Input 2 Capaicty

R3 S3

300,000 Input 3 Capacity

R1 R2 R3 350,000 Max Demand for Regular

S1 S2 S3 500,000 Max Demand for Super

100R1 87R2 110R2 90 (R1 R2 R3)

Required Octane Level

Regular

100S1 87S2 110S2 100 (S1 S2 S3)

Required Octane Level

Super

R1, R2, R3, S1, S2, SS3 0

Total profit contribution $12,500

Note, however, that ASC started week 1 with 20,000

pounds of shrimp and ended week 4 with 25,000 pounds

of shrimp. During the 4-week period, ASC has taken

profits to reinvest and build inventory by 5000 pounds in

anticipation of future higher prices. The amount of profit

reinvested in inventory is ($5.55 $0.15)(5000) $28,500.

Thus, total profit for the 4-week period including reinvested profit is $12,500 $28,500 $41,000.

xi number of Classic 21 boats produced in

Quarter i; i 1, 2, 3, 4

si ending inventory of Classic 21 boats in

Quarter i; i 1, 2, 3, 4

Min 10,000×1 11,000×2 12,100×3 13,310×4

250s1 250s2 300s3 300s4

s.t.

x1 s1 1900 Quarter 1 demand

s1 x2 s2 4000 Quarter 2 demand

s2 x3 s3 3000 Quarter 3 demand

s3 x4 s4 1500 Quarter 4 demand

s4 500 Ending Inventory

x1 4000 Quarter 1 capacity

x2 3000 Quarter 2 capacity

x3 2000 Quarter 3 capacity

x4 4000 Quarter 4 capacity

b.

14. a. Let

Quarter

1

2

3

4

Production

4000

3000

2000

1900

Ending

Inventory

2100

1100

100

500

Cost

($)

40,525,000

33,275,000

24,230,000

25,439,000

$123,469,000

c. The shadow prices tell us how much it would cost if

demand were to increase by one additional unit. For

example, in Quarter 2 the shadow price is $12,760;

thus, demand for one more boat in Quarter 2 will

increase costs by $12,760.

d. The shadow price of 0 for Quarter 4 tells us we have

excess capacity in Quarter 4. The negative shadow

prices in Quarters 1–3 tell us how much increasing the

production capacity will decrease costs. For example,

the shadow price of $2510 for Quarter 1 tells us that

if capacity were increased by 1 unit for this quarter,

costs would go down $2510.

15. Let Ri the number of barrels of input i to use to

produce Regular, i 1, 2, 3

Si the number of barrels of input i to use to

produce Super, i 1, 2, 3

Max {18.5 (Ri1 Ri2 Ri3) 20(Ri1 Ri2 Ri3)

16.5(Ri1 Si1) 14(Ri2 Si2) 17.5(Ri3 Si3)}

873

Maximum Profit $2,845,000 by making 260,000 barrels of Regular and 500,000 barrels of Super. The 260,000

barrels of Regular are produced by mixing 110,000 barrels

of Input 1, 132,608.7 barrels of Input 2, and 17,391.3 barrels of Input 3. The 500,000 barrels of Super are produced

by mixing 217,391.3 barrels of Input 2, and 282,608.7 barrels of Input 3.

All available inputs are used, so each input capacity

constraint is binding. The limit on maximum amount of

Super we can sell is binding, as is the minimum octane requirement for Super.

16. Let

Min

s.t.

xi number of 10-inch rolls of paper processed by

cutting alternative i; i 1, 2, . . . , 7

x1 x2 x3 x4 x5 x6 x7

6×1

x5 x6 4×7 1000

x4 3×5 2×6

2000

x6 x7 4000

2×3 2×4

x1, x2, x3, x4, x5, x6, x7 0

4×2

2×3

11/2″ production

21/2″ production

31/2″ production

x1

0

x2 125

x3 500

x4 1500

x5

0

x6

0

x7

0

Total Rolls 125 500 1500 2125 Rolls

Production:

11/2″ 1000

21/2″ 2000

31/2″ 4000

Waste: Cut alternative 4 (1/2″ per roll)

Therefore, waste 1/2(1500) 750 inches

b. Only the objective function needs to be changed. An

objective function minimizing waste production and

the new optimal solution are given.

Min x1 0x2 0x3 0.5×4 x5 0x6 0.5×7

x1 0

x2 500

x3 2000

x4 0

874

Appendix G

Self-Test Solutions and Answers to Even-Numbered Problems

x5 0

x6 0

x7 0

Total Rolls 2500 Rolls

Production:

11/2″ 4000

21/2″ 2000

31/2″ 4000

Waste is 0; however, we have overproduced the 11/2″

size by 3000 units. Perhaps these can be inventoried for

future use.

c. Minimizing waste may cause you to overproduce. In

this case, we used 375 more rolls to generate a 3000

surplus of the 11/2″ product. Alternative b might be preferred on the basis that the 3000 surplus could be held

in inventory for later demand. However, in some trim

problems, excess production cannot be used and must

be scrapped. If this were the case, the 3000 unit 11/2″

size would result in 4500 inches of waste, and thus

alternative 1 would be the preferred solution.

18. a. Let x1 number of Super Tankers purchased

x2 number of Regular Line Tankers purchased

x3 number of Econo-Tankers purchased

Min

s.t.

or

550×1

350×3

4600×3 600,000 Budget

6700×1 55000×2

15(5000)x1 20(2500)x2 25(1000)x3 550,000

75000×1

x1

or

425×2

50000×2

x2

x1 1/2 (x1 x2 x3)

1/ x

2 1

25000×3 550,000 Meet Demand

x3 15 Max. Total Vehicles

x3 3 Min. Econo-Tankers

1/2×2 1/2×3 0 No more than 50% Super Tankers

x1, x2, x3 0

Solution: 5 Super Tankers, 2 Regular Tankers, 3 EconoTankers

Total Cost: $583,000

Monthly Operating Cost: $4650

b. The last two constraints in the preceding formulation

must be deleted and the problem re-solved.

The optimal solution calls for 71/3 Super Tankers at

an annual operating cost of $4033. However, because a

partial Super Tanker can’t be purchased, we must

round up to find a feasible solution of 8 Super Tankers

with a monthly operating cost of $4400.

Actually, this is an integer programming problem,

because partial tankers can’t be purchased. We were

fortunate in part (a) that the optimal solution turned out

integer.

The true optimal integer solution to part (b) is

x1 6 and x2 2, with a monthly operating cost of

$4150. This is 6 Super Tankers and 2 Regular Line

Tankers.

19. a. Let x11 amount of men’s model in month 1

x21 amount of women’s model in month 1

x12 amount of men’s model in month 2

x22 amount of women’s model in month 2

s11 inventory of men’s model at end of month 1

s21 inventory of women’s model at end of

month 1

s12 inventory of men’s model at end of month 2

s22 inventory of women’s model at end of

month 2

The model formulation for part (a) is given.

Min

s.t.

or

or

120×11 90×21 120×12 90×22 2.4s11 1.8s21 2.4s12 1.8s22

20 x11 s11 150

x11 s11 130

30 x21 s21 125

Satisfy Demand

(1)

x21 s21 95

s11 x12 s12 200

s21 x22 s22 150

s12 25

s22 25

Labor-hours: Men’s

2.0 1.5 3.5

Women’s 1.6 1.0 2.6

3.5 x11 2.6 x21 900

3.5 x11 2.6 x21 1100

3.5 x11 2.6 x21 3.5 x12 2.6 x22 100

3.5 x11 2.6 x21 3.5 x12 2.6 x22 100

x11, x12, x21, x22, s11, s12, s21, s22 0

Satisfy Demand

Satisfy Demand

Satisfy Demand

Ending Inventory

Ending Inventory

(2)

(3)

(4)

(5)

(6)

Labor Smoothing for

Month 1

Labor Smoothing for

Month 2

(7)

(8)

(9)

(10)

The optimal solution is to produce 193 of the men’s

model in month 1, 162 of the men’s model in month 2,

95 units of the women’s model in month 1, and 175 of

the women’s model in month 2. Total Cost $67,156.

Inventory Schedule

Month 1

Month 2

63 Men’s

25 Men’s

0 Women’s

25 Women’s

Labor Levels

Previous month

Month 1

Month 2

1000.00 hours

922.25 hours

1022.25 hours

b. To accommodate this new policy, the right-hand sides of

constraints 7–10 must be changed to 950, 1050, 50, and

50, respectively. The revised optimal solution is given.

x11 201

x21 95

x12 154

x22 175 Total Cost $67,175

We produce more men’s models in the first month and

carry a larger men’s model inventory; the added cost,

however, is only $19. This seems to be a small expense

to have less drastic labor force fluctuations. The new labor levels are 1000, 950, and 994.5 hours each month.

Because the added cost is only $19, management might

want to experiment with the labor force smoothing restrictions to enforce even less fluctuations. You may

want to experiment yourself to see what happens.

Appendix G

875

Self-Test Solutions and Answers to Even-Numbered Problems

LM1 No. of large on machine M1

LM2 No. of large on machine M2

LM3 No. of large on machine M3

MM2 No. of meal on machine M2

MM3 No. of meal on machine M3

20. Let xm number of units produced in month m

Im increase in the total production level in month m

Dm decrease in the total production level in month m

sm inventory level at the end of month m

where

m 1 refers to March

m 2 refers to April

m 3 refers to May

Min 1.25 I1 1.25 I2 1.25 I3 1.00 D1 1.00 D2 1.00 D3

s.t.

Change in production level in March:

The formulation and solution follows. Note that constraints 1–3 guarantee that next week’s schedule will be

met and constraints 4–6 enforce machine capacities.

MIN

20SM124SM232SM315LM128LM235LM318MM236MM3

x1 10,000 I1 D1

S.T.

or

x1 I1 D1 10,000

Change in production level in April:

x2 x1 I2 D2

or

x2 x1 I2 D2 0

1)

2)

3)

4)

5)

6)

1SM11SM21SM380000

1LM11LM21LM380000

1MM21MM365000

0.03333SM10.04LM12100

0.02222SM20.025LM20.03333MM22100

0.01667SM30.01923LM30.02273MM32400

Change in production level in May:

x3 x2 I3 D3

or

x3 x2 I3 D3 0

Demand in March:

2500 x1 s1 12,000

or

x1 s1 9500

Demand in April:

Optimal Solution

Objective Function Value

Variable

———–SM2

SM3

LM1

LM2

LM3

MM2

MM3

Value

———–0.00000

80000.00000

52500.00000

0.00000

27500.00000

63006.30063

1993.69937

5515886.58866

SM1

0.00000

s1 x2 s2 8000

Demand in May:

s2 x3 15,000

Inventory capacity in March:

s1 3000

Inventory capacity in April:

s2 3000

Optimal Solution:

Total cost of monthly production increases and decreases

$2500

D1 0

x1 10,250 I1 250

x2 10,250 I2 0

D2 0

x3 12,000 I3 1750 D3 0

s1 750

s2 3000

22. Let SM1 No. of small on machine M1

SM2 No. of small on machine M2

SM3 No. of small on machine M3

Constraint

———–1

2

3

4

5

6

Slack/Surplus

————-0.00000

0.00000

0.00000

0.00000

0.00000

492.25821

Note that 5,515,887 square inches of waste are generated.

Machine 3 has 492 minutes of idle capacity.

24. Let

x1 proportion of investment A undertaken

x2 proportion of investment B undertaken

s1 funds placed in savings for period 1

s2 funds placed in savings for period 2

s3 funds placed in savings for period 3

s4 funds placed in savings for period 4

L1 funds received from loan in period 1

L2 funds received from loan in period 2

L3 funds received from loan in period 3

L4 funds received from loan in period 4

876

Appendix G

Self-Test Solutions and Answers to Even-Numbered Problems

Objective Function:

In order to maximize the cash value at the end of the four

periods, we must consider the value of investment A, the

value of investment B, savings income from period 4, and

loan expenses for period 4.

Max

Chapter 10

1. The network model is shown:

Atlanta

1400

Dallas

3200

Columbus

2000

Boston

1400

3200×1 2500×2 1.1s4 1.18L4

Constraints require the use of funds to equal the source of

funds for each period.

2

6

6

Phila.

5000

2

Period 1:

1000×1 800×2 s1 1500 L1

or

1

1000×1 800×2 s1 L1 1500

Period 2:

3000

800×1 500×2 s2 1.18L1 400 1.1s1 L2

New

Orleans

2

5

7

or

800×1 500×2 1.1s1 s2 1.18L1 L2 400

Period 3:

200×1 300×2 s3 1.18L2 500 1.1s2 L3

or

200×1 300×2 1.1s2 s3 1.18L2 L3 500

Period 4:

s4 1.18L3 100 200×1 300×2 1.1s3 L4

or

200×1 300×2 1.1s3 s4 1.18L3 L4 100

Limits on Loan Funds Available:

L1 200

L2 200

L3 200

L4 200

Min 14×11 9×12 7×13 8×21 10×22 5×23

s.t.

x11 x12 x13

30

x21 x22 x23 20

x11

x21

25

x12

x22

15

x13

x23 10

x11, x12, x13, x21, x22, x23 0

Jefferson City–Des Moines

Jefferson City–Kansas City

Jefferson City–St. Louis

Omaha–Des Moines

x1 1

x2 1

Optimal Solution: $4340.40

x1 0.458

x2 1.0

Amount

Cost

5

15

10

20

70

135

70

160

Total

or 45.8%

or 100.0%

Savings/Loan Schedule:

Savings

Loan

x11 amount shipped from Jefferson City to

Des Moines

x12 amount shipped from Jefferson City to

Kansas City

b. Optimal Solution:

Proportion of Investment Undertaken:

Investment A

Investment B

2. a. Let

Period 1

Period 2

Period 3

Period 4

242.11

—

—

200.00

—

127.58

341.04

—

435

4. The optimization model can be written as

xij Red GloFish shipped from i to j i M for Michigan, T

for Texas; j 1, 2, 3.

yij Blue GloFish shipped from i to j, i M for

Michigan, T for Texas; j 1, 2, 3.

Appendix G

877

Self-Test Solutions and Answers to Even-Numbered Problems

Min xM1 2.50xM2 0.50xM3 yM1 2.50yM2 0.50yM3 2.00yT1 1.50yT2 2.80yT3

subject to

xM2

xM3

1,000,000

xM1

yM1

yM2

1,000,000

yM3

yT1

yT2

yT3

600,000

320,000

300,000

160,000

380,000

450,000

290,000

xM1

xM2

xM3

yM1

yT1

yM2

yT2

yM3

yT3

xij 0

Using this new objective function and constraint the

optimal solution is $2.2 million, so the savings are

$150,000.

6. The network model, the linear programming formulation,

and the optimal solution are shown. Note that the third

constraint corresponds to the dummy origin. The variables

x31, x32, x33, and x34 are the amounts shipped out of the

Demand

Supply

5000

C.S.

D1

2000

D2

5000

D3

3000

D4

2000

32

34

3000

34

30

28

D.

0

38

And we change the constraints

320,000

xM1

xM2

300,000

xM3

160,000

to

xM1 xT1 320,000

xM2 xT2 300,000

xM3 xT3 160,000

dummy origin; they do not appear in the objective function because they are given a coefficient of zero.

32 40

Solving this linear program using Solver, we find that

we should produce 780,000 red GloFish in Michigan,

670,000 blue GloFish in Michigan, and 450,000 blue

GloFish in Texas.

Using the notation in the model, the number of GloFish

shipped from each farm to each retailer can be expressed

as follows:

xM1 320,000

xM2 300,000

xM3 160,000

yM1 380,000

yM2 0

yM3 290,000

yT1 0

yT2 450,000

yT3 0

a. From Solver, the minimum transportation cost is

$2.35 million.

b. We have to add variables xT1, xT2, and xT3 for Red

GloFish shipped between Texas and Retailers 1, 2 and

3. The revised objective function is

Minimize xM1 2.50xM2 0.50xM3 yM1 2.50yM2

0.50yM3 2.00yT1 1.50yT2 2.80yT3 xT1 2.50xT2

0.50xT3

We replace the third constraint above with

xT1 xT2 xT3 yT1 yT2 yT3 600,000

4000

Dum

0

0

0

Note: Dummy origin has supply of 4000.

Max 32×11 34×12 32×13 40×14 34×21 30×22 28×23 38×24

s.t.

5000

x11 x12 x13 x14

3000

x21 x22 x23 x24

4000 Dummy

x31 x32 x33 x34

x21

x31

2000

x11

x22

x32

5000

x12

x23

x33

3000

x13

x24

x34

2000

x14

xij 0 for all i, j

Optimal Solution

Units

Cost

Clifton Springs–D2

Clifton Springs–D4

Danville–D1

Danville–D4

4000

1000

2000

1000

$136,000

40,000

68,000

38,000

Total Cost $282,000

878

Appendix G

Self-Test Solutions and Answers to Even-Numbered Problems

Cost Matrix ($1000s)

Customer 2 demand has a shortfall of 1000.

Customer 3 demand of 3000 is not satisfied.

Supplier

8. a.

1

Boston

7

3

4

5

6

660

639

830

553

648

534

702

775

511

684

680

693

850

581

693

590

693

900

595

657

630

630

930

553

747

11

1

Denver

b. Optimal Solution:

13

2

Dallas

20

Supplier 1–Division 2

Supplier 2–Division 5

Supplier 3–Division 3

Supplier 5–Division 1

Supplier 6–Division 4

70

17

2

Atlanta

12

8

3

Chicago

$ 603

648

775

590

553

Total $3169

10

150

2

614

603

865

532

720

50

8

100

1

1

2

3

4

5

3

Los

Angeles

18

60

11. a. Network Model

Demand

13

Supply

1

P1

450

16

4

St. Paul

80

b. There are alternative optimal solutions.

Solution 1

Denver to St. Paul:

Atlanta to Boston:

Atlanta to Dallas:

Chicago to Dallas:

Chicago to Los Angeles:

Chicago to St. Paul:

Total Profit: $4240

10

50

50

20

60

70

4

W1

7

8

5

5

Solution 2

Denver to St. Paul:

Atlanta to Boston:

Atlanta to Los Angeles:

Chicago to Dallas:

Chicago to Los Angeles:

Chicago to St. Paul:

3

P3

380

10

50

50

70

10

70

6

4

8

6

C1

300

7

C2

300

8

C3

300

9

C4

400

4

2

P2

600

4

3

100

Division

W25

W2

6

6

7

7

b. & c. The linear programming formulation and solution

is shown below:

LINEAR PROGRAMMING PROBLEM

If solution 1 is used, Forbelt should produce 10 motors at

Denver, 100 motors at Atlanta, and 150 motors at Chicago.

There will be idle capacity for 90 motors at Denver.

If solution 2 is used, Forbelt should adopt the same

production schedule but a modified shipping schedule.

10. a. The total cost is the sum of the purchase cost and the

transportation cost. We show the calculation for Division

1–Supplier 1 and present the result for the other DivisionSupplier combinations.

Division 1–Supplier 1

MIN 4X14 7X15 8X24 5X25 5X34 6X35

6X46 4X47 8X48 4X49 3X56 6X57

7X58 7X59

S.T.

(1)

(2)

(3)

(4)

(5)

Purchase cost (40,000 $12.60)

$504,000

Transportation Cost (40,000 $2.75) 110,000

Total Cost:

$614,000

(6)

(7)

(8)

(9)

X14

X24

X34

X46

X34

X56

X35

X46

X47

X48

X49

X15

X25

X35

X47

0

X57

0

X56

X57

X58

X59

450

600

380

X48 X49 X14 X24

X58 X59 X15 X25

300

300

300

400

Appendix G

OPTIMAL SOLUTION

OPTIMAL SOLUTION

Value

———–450.000

0.000

0.000

600.000

250.000

0.000

0.000

300.000

0.000

400.000

300.000

0.000

300.000

0.000

Objective Function Value 11220.000

11850.000

Variable

———–X14

X15

X24

X25

X34

X35

X46

X47

X48

X49

X56

X57

X58

X59

X39

X45

X54

Reduced Costs

————–0.000

3.000

3.000

0.000

0.000

1.000

3.000

0.000

1.000

0.000

0.000

2.000

0.000

3.000

There is an excess capacity of 130 units at plant 3.

12. a. Three arcs must be added to the network model in

Problem 11a. The new network is shown:

Supply

1

P1

4

7

8

600

5

5

380

6

7

14.

7

C2

300

3

2

W25

W2

6

7

7

8

C3

8

6

300

6

9

C4

1

Muncie

400

b. & c. The linear programming formulation and optimal

solution is shown below:

9

5

3

Xenia

3

34

34

5

Cincinnati

S.T.

(1)

(2)

(3)

(4)

(5)

(6)

(7)

(8)

(9)

X14

X24

X34

X45

X34

X54

X35

X46

X47

X48

X39

X15

X25

X35

X46

X54

X56

X45

X56

X57

X58

X49

450

600

X39

X47

0

X57

0

300

300

300

X59

380

X48 X49 X14 X24

X58 X59 X15 X25

400

7

Greenwood

4

8

Concord

3

9

Chatham

3

35

28

24

LINEAR PROGRAMMING PROBLEM

MIN 4X14 7X15 8X24 5X25 5X34

6X35 6X46 4X47 8X48 4X49 3X56

6X57 7X58 7X59 7X39 2X45 2X54

2

44

4

Louisville

3

8

2

Brazil

6

Macon

32

3

P3

2

300

4

2

P2

6

4

8

4

W1

3

450

Reduced Costs

————–0.000

2.000

4.000

0.000

2.000

2.000

2.000

0.000

0.000

0.000

0.000

3.000

0.000

4.000

0.000

1.000

3.000

The value of the solution here is $630 less than the value

of the solution for Problem 23. The new shipping route

from plant 3 to customer 4 has helped (x39 380). There

is now excess capacity of 130 units at plant 1.

Demand

6

C1

Value

———–320.000

0.000

0.000

600.000

0.000

0.000

0.000

300.000

0.000

20.000

300.000

0.000

300.000

0.000

380.000

0.000

0.000

57

Objective Function Value

Variable

———–X14

X15

X24

X25

X34

X35

X46

X47

X48

X49

X56

X57

X58

X59

879

Self-Test Solutions and Answers to Even-Numbered Problems

A linear programming model is

Min 8×146×153×248×259×343×3544×4634×4734×4832×4957×5635×5728×5824×59

s.t.

3

x14 x15

6

x24 x25

5

x34 x35

x24

x34

x46 x47 x48 x49

0

x14

x25

x35

x56 x57 x58 x59 0

x15

x56

2

x46

x57

4

x47

x58

3

x48

x59 3

x49

xij 0 for all i, j

880

Appendix G

Self-Test Solutions and Answers to Even-Numbered Problems

b.

Optimal Solution

Muncie–Cincinnati

Cincinnati–Concord

Brazil–Louisville

Louisville–Macon

Louisville–Greenwood

Xenia–Cincinnati

Cincinnati–Chatham

Units

Shipped

Cost

1

3

6

2

4

5

3

6

84

18

88

136

15

72

Min 10×11 16×12 32×13 14×21 22×22 40×23 22×31 24×32 34×33

s.t.

1

x11 x12 x13

1

x21 x22 x23

x31 x32 x33 1

x21

x31

1

x11

x22

x32

1

x12

x23

x33 1

x13

xij 0 for all i, j

Solution: x12 1, x21 1, x33 1

Total completion time 64

419

18. a.

Crews

Jobs

Two rail cars must be held at Muncie until a buyer is found.

16. a.

1

8

5

3

3

2

5

6

1

2

White

1

3

Blue

x53 5

x54 0

x56 5

x67 0

x74 6

x56 5

1

4

Green

2

34 6

x12 0

x15 0

x25 8

x27 0

x31 8

x36 0

x42 3

25

44

1

1

2

1

3

1

4

1

5

1

38

xij 0 for all i, j

b.

30

47 1

3

Min 20×12 25×15 30×25 45×27 20×31 35×36

30×42 25×53 15×54 28×56 12×67 27×74

s.t.

x31 x12 x15

x25 x27 x12 x42

x31 x36 x53

x54 x74 x42

x53 x54 x56 x15 x25

x36 x56 x67

x74 x27 x67

1

Red

44

43

1

Total cost of redistributing cars $917

5

Brown

28

17. a.

b.

1

Jackson

1

2

Ellis

10

16

1

Client 1

1

2

Client 2

1

32

1

14

22

40

Min 30×11 44×12 38×13 47×14 31×15 25×21 . . . 28×55

s.t.

x11 x12 x13 x14 x15

1

x21 x22 x23 x24 x25

1

x31 x32 x33 x34 x35

1

1

x41 x42 x43 x44 x45

x51 x52 x53 x54 x55

1

x11 x21 x31 x41 x51

1

x12 x22 x32 x42 x52

1

x13 x23 x33 x43 x53

1

x14 x24 x34 x44 x54

1

x15 x25 x35 x45 x55

1

xij 0, i 1, 2, . . . , 5; j 1, 2, . . . , 5

22

Optimal Solution:

1

3

Smith

24

34

3

Client 3

1

Green to Job 1

Brown to Job 2

Red to Job 3

Blue to Job 4

White to Job 5

$ 26

34

38

39

25

$162

Appendix G

Self-Test Solutions and Answers to Even-Numbered Problems

Because the data are in hundreds of dollars, the total installation cost for the five contracts is $16,200.

20. a. This is the variation of the assignment problem in

which multiple assignments are possible. Each distribution center may be assigned up to three customer

zones.

The linear programming model of this problem has

40 variables (one for each combination of distribution

center and customer zone). It has 13 constraints. There

are five supply (3) constraints and eight demand

(1) constraints.

The optimal solution is as follows:

Cost

($1000s)

Assignments

Plano

Flagstaff

Springfield

Boulder

Kansas City, Dallas

Los Angeles

Chicago, Columbus, Atlanta

Newark, Denver

34

15

70

97

Total Cost $216

b. The Nashville distribution center is not used.

c. All the distribution centers are used. Columbus is

switched from Springfield to Nashville. Total cost increases by $11,000 to $227,000.

22. A linear programming formulation of this problem can be

developed as follows. Let the first letter of each variable

name represent the professor and the second two the

course. Note that a DPH variable is not created because

the assignment is unacceptable.

Max 2.8AUG 2.2AMB 3.3AMS 3.0APH 3.2BUG · · · 2.5DMS

s.t.

AUG AMB AMS APH

1

BUG BMB BMS BPH

1

CUG CMB CMS CPH 1

DUG DMB DMS 1

AUG BUG CUG DUG

1

AMB BMB CMB DMB

1

AMS BMS CMS DMS 1

APH BPH CPH 1

All Variables 0

Optimal Solution

Rating

A to MS course

B to Ph.D. course

C to MBA course

D to Undergraduate course

3.3

3.6

3.2

3.2

Max Total Rating 13.3

23. Origin—Node 1

Transshipment—Nodes 2–5

Destination—Node 7

881

The linear program will have 14 variables for the arcs

and 7 constraints for the nodes.

Let

xij e

1

0

if the arc from node i to node j is on the shortest route

otherwise

Min 7×12 9×13 18×14 3×23 5×25 3×32 4×35

3×46 5×52 4×53 2×56 6×57 2×65 3×67

s.t.

Flow Out

Flow In

Node 1 x12 x13 x14

1

Node 2 x23 x25

x12 x32 x52 0

Node 3 x32 x35

x13 x23 x53 0

Node 4 x46

x14

0

Node 5 x52 x53 x56 x57 x25 x35 x65 0

Node 6 x65 x67

x46 x56

0

Node 7

x57 x67

1

xij 0 for all i and j

Optimal Solution: x12 1, x25 1, x56 1, and x67 1

Shortest Route: 1–2–5–6–7

Length 17

24. The linear program h…