An Introduction To Management Science 14th Edition By Anderson – Test Bank

Solutions:

1. a. Profit = (249 – c1 – c2 ) x – 1,000,000
= (249 – 45 – 90) (20,000) – 1,000,000
= $1,280,000 (Engineer’s)

b. Profit = (249 – 45 – 100) (10,000) – 1,000,000
= $40,000 (Financial Analyst)

c. Simulation will provide probability information about the various profit levels possible. What if scenarios show possible profit outcomes but do not provide probability information.

2. a. Let c = variable cost per unit
x = demand

Profit = 50x – cx – 30,000
= (50 – c) x – 30,000

b. Base case: Profit = (50 – 20) 1200 – 30,000 = 6,000
Worst case: Profit = (50 – 24) 300 – 30,000 = -22,200
Best case: Profit = (50 – 16) 2100 – 30,000 = 41,400

c. The possibility of a $41,400 profit is interesting, but the worst case loss of $22,200 is risky. Risk analysis would be helpful in evaluating the probability of a loss.

3. a. Using the random number intervals

Direct Labor Cost Interval of Random Numbers
$43 0.0 but less than 0.1
$44 0.1 but less than 0.3
$45 0.3 but less than 0.7
$46 0.7 but less than 0.9
$47 0.9 but less than 1.0

Random Number Direct Labor Cost
0.3753 $45
0.9218 $47
0.0336 $43
0.5145 $45
0.7000 $46

b. Parts cost = a + r(b – a) = 80 + r(100 – 80) = 80 + r(20)

r = 0.6221 Parts cost = 80 + (0.6221)(20) = $92.44

r = 0.3418 Parts cost = 80 + (0.3418)(20) = $86.84

r = 0.1402 Parts cost = 80 + (0.1402)(20) = $82.80

r = 0.5198 Parts cost = 80 + (0.5198)(20) = $90.40

r = 0.9375 Parts cost = 80 + (0.9375)(20) = $98.75
c. First-year demand = µ + zσ = 15,000 + z(4500)

r = 0.8531 z = 1.05 Demand = 15,000 + 1.05(4500) = 19,725

r = 0.1762 z = -0.93 Demand = 15,000 – 0.93(4500) = 10,815

r = 0.5000 z = 0.00 Demand = 15,000 + 0.00(4500) = 15,000

r = 0.6810 z = 0.47 Demand = 15,000 + 0.47(4500) = 17,115

r = 0.2879 z = -0.56 Demand = 15,000 – 0.56(4500) = 12,480

Random Number Direct Labor Cost
0.3753 $45
0.9218 $47
0.0336 $43
0.5145 $45
0.7000 $46

4. a.
Number of New Accounts Opened
Probability Interval of
Random Numbers
0 .01 .00 but less than .01
1 .04 .01 but less than .05
2 .10 .05 but less than .15
3 .25 .15 but less than .40
4 .40 .40 but less than .80
5 .15 .80 but less than .95
6 .05 .95 but less than 1.00

b.

Trial
Random Number Number of New Accounts Opened
1 0.7169 4
2 0.2186 3
3 0.2871 3
4 0.9155 5
5 0.1167 2
6 0.9800 6
7 0.5029 4
8 0.4154 4
9 0.7872 4
10 0.0702 2

c. For the 10 trials Gustin opened 37 new accounts. With an average first year commission of $5000 per account, the total first year commission is $185,000. The cost to run the 10 seminars is $35,000, so the net contribution to profit for Gustin is $150,000 or $15,000 per seminar. Because the seminars are a very profitable way of generating new business, Gustin should continue running the seminars.

5. a.
Stock Price Change Probability Interval
-2 .05 .00 but less than .05
-1 .10 .05 but less than .15
0 .25 .15 but less than .40
+1 .20 .40 but less than .60
+2 .20 .60 but less than .80
+3 .10 .80 but less than .90
+4 .10 .90 but less than 1.00

b.
Random Number Price Change Ending Price Per Share
0.1091 -1 $38
0.9407 +4 $42
0.1941 0 $42
0.8083 +3 $45

Ending price per share = $45

6. a.

Payment
Probability Interval of
Random Numbers
0 .83 .00 but less than .83
500 .06 .83 but less than .89
1,000 .05 .89 but less than .94
2,000 .02 .94 but less than .96
5,000 .02 .96 but less than .98
8,000 .01 .98 but less than .99
10,000 .01 .99 but less than 1.00

b.

Trial
Random Number
Payment
1 0.9850 8,000
2 0.2122 0
3 0.5590 0
4 0.0004 0
5 0.5177 0
6 0.0064 0
7 0.2044 0
8 0.2390 0
9 0.7305 0
10
11
12
13
14
15
16
17
18
19
20 0.6351
0.4971
0.9569
0.1091
0.9907
0.1900
0.9403
0.3545
0.8083
0.4704
0.5274 0
0
2,000
0
10,000
0
2,000
0
0
0
0

4 claims paid out of 20 for a total of $22,000 in payments to policyholders.

7. Time = µ + zσ = 15 + z(3)

r = 0.1562 z = -1.01 Time = 15 – 1.01(3) = 11.97 minutes

r = 0.9821 z = 2.10 Time = 15 + 2.10(3) = 21.30 minutes

r = 0.3409 z = -0.41 Time = 15 – 0.41(3) = 13.77 minutes

r = 0.5594 z = 0.15 Time = 15 + 0.15(3) = 15.45 minutes

r = 0.7758 z = 0.76 Time = 15 + 0.76(3) = 17.28 minutes

8. a. The following table can be used to simulate a win for Atlanta

Game Interval for Atlanta Win
1 .00 but less than .60
2 .00 but less than .55
3 .00 but less than .48
4 .00 but less than .45
5 .00 but less than .48
6 .00 but less than .55
7 .00 but less than .50
b. Using the random numbers in column 6 beginning with 0.3813, 0.2159 and so on, Atlanta wins games 1 and 2, loses game 3, wins game 4, loses game 5 and wins game 6. Thus, Atlanta wins the 6-game World Series 4 games to 2 games.

c. Repeat the simulation many times. In each case, record who wins the series and the number of games played, 4, 5, 6 or 7. Count the number of times Atlanta wins. Divide this number by the total number of simulation runs to estimate the probability that Atlanta will win the World Series. Count the number of times the series ends in 4 games and divide this number by the total number of simulation runs to estimate the probability of the World Series ending in 4 games. This can be repeated for 5-game, 6-game and 7-game series.

9. a. Base case using most likely completion times.

A 6
B 5
C 14
D 8
33 weeks

Worst case: 8 + 7 + 18 + 10 = 43 weeks
Best case: 5 + 3 + 10 + 8 = 26 weeks

b.

Activity Random Number Completion Time
A 0.1778 5
B 0.9617 7
C 0.6849 14
D 0.4503 8
Total: 34 Weeks

c. Simulation will provide a distribution of project completion time values. Calculating the percentage of simulation trials with completion times of 35 weeks or less can be used to estimate the probability of meeting the completion time target of 35 weeks.

10. a.
Hand Value Probability Interval
17 .1654 .0000 but less than .1654
18 .1063 .1654 but less than .2717
19 .1063 .2717 but less than .3780
20 .1017 .3780 but less than .4797
21 .0972 .4797 but less than .5769
Broke .4231 .5769 but less than 1.000