5-1. MARR = 12% [Note: the text is in error; the interest rate should be 12%, not 15%]
EOY A B C D E
0 -$8,000.00 -$10,000.00 -$12,000.00 -$15,000.00 -$16,000.00
5-2. MARR = 12% [Note: the text is in error; the interest rate should be 12%, not 15%]
EOY A B C D E
5-3. MARR = 10%
EOY A B C D E F
0 -$5,000.00 -$5,000.00 -$8,000.00 -$12,000.00 $0.00 -$15,000.00
1 -$5,000.00 $800.00 $1,000.00 $1,500.00 $0.00 $1,000.00
2 $1,000.00 $1,200.00 $1,000.00 $1,500.00 $0.00 $1,500.00
Chapter 5
Inde
p
endent Investment O
pp
ortunit
y
Mutuall
y
Exclusive Investment Alternative
Inde
p
endent Non-re
p
eatin
g
Investment O
pp
ortunit
y
38
5-4. MARR = 10%
EOY A B C D E F
0 -$5,000.00 -$5,000.00 -$8,000.00 -$12,000.00 $0.00 -$15,000.00
1 -$5,000.00 $800.00 $1,000.00 $1,500.00 $0.00 $1,000.00
5-5. The answer, of course, depends on the magnitude of the investment. For example, if an
investment of no more than $11,365.14 is required then all three should be chosen. If more
than $11,365.14, but less than $11,978.13, is required then B and C should be chosen. If
more than $11,978.13, but less than $12,591.12 is required, then only B should be chosen.
5-6. Each alternative involves receipt of $15,000 over a 5-year period. The only differences are
5-7. With 3
p
ro
p
osals, there are 23, or 8,
p
ossible alternatives. The
y
are:
Alternative Opportunity Feasible? Alternative Opportunity Feasible?
Mutuall
y
Exclusive Investment Alternative
39
5-8. With 4
p
ro
p
osals, there are 24, or 16,
p
ossible alternatives. The
y
are:
Alternative Opportunity Feasible? Alternative Opportunity Feasible?
1 none Yes 9 B,C Yes
5-9. With 5
p
ro
p
osals, there are 25, or 32,
p
ossible alternatives. The
y
are:
Alternative Opportunity Feasible? Alternative Opportunity Feasible?
1 none No 17 A,B,C No
2 A Yes 18 A,B,D No
3 B Yes 19 A,B,E No
5-10. With 5
p
ro
p
osals, there are 25, or 32,
p
ossible alternatives. The
y
are:
Alternative Opportunity Feasible? Alternative Opportunity Feasible?
1 none Yes 17 A,B,C No
2 A Yes 18 A,B,D No
40
5-11. MARR = 10%
Geometric rate = 10% 15%
EOY CF( Q ) CF( R )
5-12. MARR = 12%
EOY CF[A] CF[B] CF[C]
0 $0.00 -$100,000 -$175,000
5-13. MARR = 12%
EOY CF[A] CF[B] CF[C] CF[D]
0 -$50,000 -$125,000 -$200,000 -$100,000
1 -$100,000 -$75,000 $50,000 -$150,000
2 $50,000 $70,000 $50,000 $75,000
3 $50,000 $70,000 $50,000 $75,000
41
5-14. MARR = 10%
EOY CF[A] CF[B] CF[C] CF[D]
0 $0 -$100,000 -$180,000 -$125,000
1 $0 -$100,000 -$20,000 -$75,000
5-15. MARR = 15%
EOY CF[A] CF[B] CF[C] CF[D]
0 -$50,000 -$100,000 -$250,000 -$225,000
1 -$100,000 -$100,000 $75,000 -$75,000
5-16. MARR = 12%
EOY CF[A] CF[B] CF[C]
0 -$1,150,000 -$1,250,000 -$2,000,000
1 -$425,000 -$400,000 -$275,000
5-17. The difference in the annual worth-costs for B and C is $615. Therefore, for C to be chosen,
42
5-18. MARR = 0%
EOY CF[A] CF[B] CF[C]
0 -$1,150,000 -$1,250,000 -$2,000,000
1 -$425,000 -$400,000 -$275,000
2 -$425,000 -$400,000 -$275,000
MARR = 11%
EOY CF[A] CF[B] CF[C]
0 -$1,150,000 -$1,250,000 -$2,000,000
1 -$425,000 -$400,000 -$275,000
2 -$425,000 -$400,000 -$275,000
MARR = 22%
EOY CF[A] CF[B] CF[C]
0 -$1,150,000 -$1,250,000 -$2,000,000
1 -$425,000 -$400,000 -$275,000
2 -$425,000 -$400,000 -$275,000
43
5-19. MARR = 10%
EOY CF[A]
0 -$7,500.00
5-20. MARR = 12%
EOY CF[A] CF[B] CF[C]
0 -$4,000 -$12,000 -$24,000
1 -$7,000 -$2,000 -$500
2 -$7,000 -$2,000 -$500
5-21. MARR = 15%
Annual Output = 20,000
Alternative
GP machines SP machine
First Cost $35,000 $55,000
Op Hrs/Yr 1,000 1,000
44
5-22. Annual cost = -$250t
[(
A/P,10%,15
)
+0.03
]
-10t
(
A/F,10%,15
)
-4,500
(
0.82
)
t-3
(
0.01
)(
365
)(
24
)
/1,000
Annual cost = -$40.69t-$394.2
(
0.82
)
t-3
Insulation
Thickness
(
t
)
Annual Cost
(
$
)
Insulation
Thickness
(
t
)
Annual Cost
$
3.00 516.2700 5.00 468.5101
3.25 507.3624 5.10 467.3708
3.50 499.3783 5.20 466.3338
3.75 492.2730 5.30 465.3972
4.00 486.0040 5.40 464.5590
5-23. AW(I) = -($10,000-$1,000)(A/P,12%,5)-$1,000(0.12)-$14,150
5-24. MARR = 10%
EOY CF[A] CF[B]
0 -$20,000 -$25,000
1 -$5,000 -$1,500
2 -$5,000 -$1,500
45
5-25. AW(A) = -$6,800(A/P,12%,8)
5-26. MARR = 10%
EOY CF[Buy] CF[Contract]
0 -$4,250.00 -$2,300
5-27. AW(A) = -$3,500(A/P,12%,15)-$600
5-28. Annual land cost is the same for all alternatives: $500,000(0.12) = $60,000.
AW(2) = -$200,000(A/P,12%,30)+$30,000
5-29. PW(A) = -$20,000+$12,000(P/A,12%,10)
5-30. MARR = 10% MARR = 8%
EOY CF(A) CF(B) EOY CF(A) CF(B)
0 -$10,000.00 -$15,000 0 -$10,000.00 -$15,000
46
5-31. MARR = 10%
AW(D1) = -$50,000(A/P,10%,20) -$9,000 + $10,000(A/F,10%,20)
AW(D1) = -$14,698.38
5-32. MARR = 6%
AW(A) = –
{[
$90/ft+2
(
$3/ft
)](
5,280ft/mi
)
+3
(
$9,000/mi
)}(
A/P,6%,20
)
– $1,800/mi
– 3($450/mi)(A/F,6%,5)
AW(A) = -$48,585.58
EOY CF[A] CF[B]
0 -$533,880 -$260,190 =-[$45+(2)($1.5)](5280)-(3)($2,250)
1 -$1,800 -$19,215 =-$2,700-(3)($225)-(2)($1.5)(5280)
2 -$1,800 -$19,215
3 -$1,800 -$19,215
4 -$1,800 -$19,215
5 -$3,150 -$19,215
6 -$1,800 -$19,215
7 -$1,800 -$19,215
8 -$1,800 -$19,215
9 -$1,800 -$19,215
10 -$3,150 -$279,855 =-$19,215-[$45+(2)($1.5)](5280)-(3)($2,400)
11 -$1,800 -$19,215
12 -$1,800 -$19,215
13 -$1,800 -$19,215
14 -$1,800 -$19,215
15 -$3,150 -$19,215
16 -$1,800 -$19,215
17 -$1,800 -$19,215
18 -$1,800 -$19,215
19 -$1,800 -$19,215
20 -$3,150 -$19,215
AW = -$48,585.58 -$54,588.39 Select Design A
5-33. PW(A) = -$80,000+$21,750(P/A,10%,5) = $2,449.61
PW(B) = (P/F,10%,2)[$-80,000+$33,333(P/A,10%,3)] = $2,393.12. Choose Alternative A
47
5-34. MARR = 12%
AW(A) = -$75,000(A/P,12%,5)+$20,435+$15,000(A/F,12%,5)
AW(A) = $1,990.42
5-35. (a) repeatability assumption holds
AW(X) = -$6,000(A/P,12%,12)-$2,500
AW(X) = -$3,468.40 or from Excel: -$3,468.62
5-36. (a) annual worth comparison
AW(A) = -$1,200(A/P,8%,3)-$160-$0.07/kW(0.746kW/hp)(60hp)(800hr/yr)/.90
5-37. (a) annual worth comparison
AW(C) = -$1,100(A/P,8%,5)-$150-$0.085/kW(0.746kW/hp)(50hp)(1000hr/yr)/.82
5-38. AW(A) = -$100,000(A/P,10%,6)+$20,000+$30,000(A/F,10%,6)
5-39. (a) solve for preferred alternative using annual worth
AW(A) = -$20,000(A/P,20%,5)-$5,500+$1,000(A/F,20%,5)
AW(A) = -$12,053.21
5-40. AW(A) = -$15,000(A/P,8%,10)+$3,500-(0.07)($15,000)+$2,500(A/F,8%,10)
48