Central Florida College Winter Park Statistics Worksheet

Learning Goal: I’m working on a statistics multi-part question and need a sample draft to help me learn.

Chapter 7 problems begin on pg. 280 – No’s 1, 2, 6, 7, 10, 13, 17, 24 and RMC Excel Solver

Chapter 8 problems begin on pg. 332: 1, 2, 5, 6, 12, 13, and 14

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

5
10

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…

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