Solution Manual For Introduction To Management Science, 11th Edition

Name: Solution Manual For Introduction To Management Science, 11th Edition
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Master complex problems with Solution Manual For Introduction To Management Science, 11th Edition, your go-to guide for step-by-step solutions.

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1-1
Chapter One: Management Science
PROBLEM SUMMARY
1. Total cost, revenue, profit, and
break-even
2. Total cost, revenue, profit, and
break-even
3. Total cost, revenue, profit, and
break-even
4. Break-even volume
5. Graphical analysis (1−2)
6. Graphical analysis (1−4)
7. Break-even sales volume
8. Break-even volume as a percentage
of capacity (1−2)
9. Break-even volume as a percentage
of capacity (1−3)
10. Break-even volume as a percentage
of capacity (1−4)
11. Effect of price change (1−2)
12. Effect of price change (1−4)
13. Effect of variable cost change (1−12)
14. Effect of fixed cost change (1−13)
15. Break-even analysis
16. Effect of fixed cost change (1−7)
17. Effect of variable cost change (1−7)
18. Break-even analysis
19. Break-even analysis
20. Break-even analysis
21. Break-even analysis; volume and
price analysis
22. Break-even analysis; profit analysis
23. Break-even analysis
24. Break-even analysis; profit analysis
25. Break-even analysis; price and volume analysis
26. Break-even analysis; profit analysis
27. Break-even analysis; profit analysis
28. Break-even analysis; profit analysis
29. Linear programming
30. Linear programming
31. Linear programming
32. Linear programming
33. Forecasting/statistics
34. Linear programming
35. Waiting lines
36. Shortest route
PROBLEM SOLUTIONS
1. a)= =
= =
= + = + =
= = =
= − =
f
v
f v
300, $8,000,
$65 per table, $180;
TC $8,000 (300)(65) $27,500;
TR (300)(180) $54,000;
$54,000 27,500 $26,500 per month
v c
c p
c vc
vp
Z
b)f
v
8,000 69.56 tables per month
180 65
c
v p c
= = =
− −
2. a)= = =
= = +
= +
=
= = =
= − =
f v
f v
12,000, $60,000, $9,
$25; TC
60,000 (12,000)(9)
$168,000;
TR (12,000)($25) $300,000;
$300,000 168,000 $132,000 per year
v c c
p c vc
vp
Z
b)= = =
− −
f

1-2
5.
6.
7.−
−
f
v
$25,000
= = = 1,250 dolls
30 10
c
v p c
8.= = = =
Break-even volume as percentage of capacity
3,750 .469 46.9%
8,000
v
k
9.= = = =
Break-even volume as percentage of capacity
24,750.88 .988 98.8%
25,000
v
k
10.= = = =
Break-even volume as percentage of
100,000
capacity .833 83.3%
120,000
v
k
11.f
v
60,000 2,727.3tires
31 9
per year; it reduces the break-even
volume from 3,750 tires to 2,727.3
tires per year.
c
v p c
= = =
− −
12.= = =
− −
f
v
25,000 55,555.55 lb
.60 .15
per month; it reduces the break-even
volume from 100,000 lb per month
to 55,555.55 lb.
c
v p c
13.f
v
25,000 65,789.47 lb
.60 .22
per month; it increases the break-even
volume from 55,555.55 lb per month
to 65,789.47 lb per month.
c
v p c
= = =
− −
14.= = =
− −
f
v
39,000 102,613.57 lb
.60 .22
per month; it increases the break-even
volume from 65,789.47 lb per month
to 102,631.57 lb per month.
c
v p c
15.f v
f
v
Initial profit: (9,000)(.75)
4,000 (9,000)(.21) 6,750 4,000 1,890
$860 per month; increase in price:
(5,700)(.95) 4,000 (5,700)(.21) 5, 415
4,000 1,197 $218 per month; the dair
Z vp c vc
Z vp c
vc
= − − = −
− = − − =
= − −
= − − = −
− = y should not
raise its price.
16.−
f
v
35,000
= = = 1,750
30–10
c
v p c
The increase in fixed cost from $25,000 to
$35,000 will increase the break-even point from
1,250 to 1,750 or 500 dolls; thus, he should not
spend the extra $10,000 for advertising.
17. Original break-even point (from problem 7) = 1,250
New break-even point:= = =
− −
f
v
17,000 1,062.5
30 14
c
v p c
Reduces BE point by 187.5 dolls.
18. a)= = =
− −
f
v
$27,000 5,192.30 pizzas
8.95 3.75
c
v p c
b)=
5,192.3 259.6 days
20
c) Revenue for the first 30 days = 30(pv

1-35,685.7
Total time to break-even 30 20
314.3 days
= +
=
19. a) Cost of Regular plan = $55 + (.33)(260 minutes)
= $140.80
Cost of Executive plan = $100 + (.25)(60 minutes)
= $115
Select Executive plan.
b) 55 + (x − 1,000)(.33) = 100 + (x − 1,200)(.25)
− 275 + .33x = .25x − 200
x = 937.50 minutes per
month or 15.63 hrs.
20. a)= −
7,500
14,000 .35p
p = $0.89 to break even
b) If the team did not perform as well as expected
the crowds could be smaller; bad weather could
reduce crowds and/or affect what fans eat at the
game; the price she charges could affect demand.
c) This will be a subjective answer, but $1.25 seems
to be a reasonable price.
Z = vp − cf − vcv
Z = (14,000)(1.25) − 7,500 − (14,000)(0.35)
= 17,500 − 12,400
= $5,100
21. a) cf = $1,700
cv = $12 per pupil
p = $75= −
1,700
75 12
v
= 26.98 or 27 pupils
b) Z = vp − cf − vcv
$5,000 = v(75) − $1,700 − v(12)
63v = 6,700
v = 106.3 pupils
c) Z = vp − cf − vcv
$5,000 = 60p − $1,700 − 60(12)
60p = 7,420
p = $123.67
22. a) cf = $350,000
cv = $12,000
p = $18,000= −
f
v
c
v p c= −
350,000
18,000 12,000
= 58.33 or 59 students
b) Z = (75)(18,000) − 350,000 − (75)(12,000)
= $100,000
c) Z = (35)(25,000) − 350,000 − (35)(12,000)
= 105,000
This is approximately the same as the profit for
75 students and a lower tuition in part (b).
23. p = $400
cf = $8,000
cv = $75
Z = $60,000+
= −
f
v
Z c
v p c+
= −
60,000 8,000
400 75
v
v = 209.23 teams
24. Fixed cost (cf) = 875,000
Variable cost (cv) = $200
Price (p) = (225)(12) = $2,700
v = cf

1-4
The volume to achieve the goal of $150,000 is,
v = (Z + cf)/(p − cv)
= (150,000 + 100,000)/(6 − .695)
= 47,125
26. a) Monthly fixed cost (cf) = cost of van/60 months
+ labor (driver)/month
= (21,500/60) + (30.42
days/month)($8/hr)
(5 hr/day)
= 358.33 + 1,216.80
= $1,575.13
Variable cost (cv) = $1.35 + 15.00
= $16.35
Price (p) = $34
v = cf/(p − vc)
= (1,575.13)/(34 − 16.35)
v = 89.24 orders/month
b) 89.24/30.42 = 2.93 orders/day − Monday through
Thursday
Double for weekend = 5.86 orders/day − Friday
through Sunday
Orders per month = approximately (18 days)
(2.93 orders) + (12.4 days)(5.86 orders)
= 125.4 delivery orders per month
Profit = total revenue − total cost
= vp – (cf + vcv)
= (125.4)(34) − 1,575.13 – (125.4)(16.35)
= 638.18
27. a)= =
− −
f
v
500
30 14
c
v p c
v = 31.25 jobs
b) (8 weeks)(6 days/week)(3 lawns/day) = 144
lawns
Z = (144)(30) − 500 − (144)(14)
Z = $1,804
c) (8 weeks)(6 days/week)(4 lawns/day) = 192 lawns
Z = (192)(25) − 500 − (192)(14)
Z = $1,612
No, she would make less money than (b)
28. a)= =
− −
f
v
700
35 3
c
v p c
v = 21.88 jobs
b) (6 snows)(2 days/snow)(10 jobs/day) = 120 jobs
Z = (120)(35) − 700 − (120)(3)
Z = $3,140
c) (6 snows)(2 days/snow)(4 jobs/day) = 48 jobs
Z = (48)(150) − 1800 − (48)(28)
Z = $4,056
Yes, better than (b)
d) Z = (120)(35) − 700 − (120)(18)
Z = $1,340
Yes, still a profit with one more person
29. There are two possible answers, or solution points:
x = 25, y = 0 or x = 0, y = 50
Substituting these values in the objective function:
Z = 15(25) + 10(0) = 375
Z = 15(0) + 10(50) = 500
Thus, the solution is x = 0 and y = 50
This is a simple linear programming model, the
subject of the next several chapters. The student
should recognize that there are only two possible
solutions, which are the corner points of the
feasible solution space, only one of which is
optimal.
30. The solution is computed by solving
simultaneous equations,
x

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2-13
b) Solution:
x1 = 87.3 cups
x2 = 130.9 cups
Z = $421.09
61. a) The only binding constraint is for
Colombian; the constraints for Kenyan and
Indonesian are nonbinding and there are
already extra, or slack, pounds of these
coffees available. Thus, only getting more
Colombian would affect the solution.
One more pound of Colombian would
increase sales from $421.09 to $463.20.
Increasing the brewing capacity to 40

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2-15
Increasing the amount available to invest
(i.e., $120,000 to $120,001) will increase
profit from Z = $29,691.37 to
Z = $29,691.62 or approximately $0.25.
Increasing by another dollar will increase
profit by another $0.25, and increasing the
amount available by one more dollar
will again increase profit by $0.25. This
indicates that for each extra dollar invested a
return of $0.25 might be expected with this
investment strategy.
Thus, the marginal value of an extra dollar
to invest is $0.25, which is also referred to
as the “shadow” or “dual” price as described
in Chapter 3.

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