FE Practice Problems w/ Solutions
6-48
You just won the sweepstakes! You have the option of either receiving a lump sum now
for $125,000 or receiving a check for $50,000 each year for three years. You would
receive the first check for $50,000 immediately. At what interest rate (MARR) would you
have to invest your winnings to be indifferent as to how you receive your winnings? [6.5]
Choose the closest answer.
(a) 15.5%
(b) 20.0%
(c) 21.6%
(d) 23.3%
Solution:
Set PWall now (i') = PW3 checks (i') and solve for i'
(
125, 000 = 50, 000 + 50, 000 P , i ', 2
A
P , i ', 2 = 125, 000 - 50, 000 = 1.5
A
50, 000
i ' » 21.6%
(
)
)
Select (c)
If the MARR > 21.6%, select $125,000 now;
if MARR < 21.6%, select $50,000 at times 0, 1, and 2, and
if MARR = 21.6%, you would be indifferent.
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FE Practice Problems w/ Solutions
6-49
TTA is a small feeder freight air line that started with very limited capital to serve the
independent petroleum operators in the arid Southwest. All of its planes are identical,
although they are painted different colors. TTA has been contracting its overhaul work to
Alamo Airmotive for $40,000 per plane per year. TTA estimates that , by building a
$500,000 maintenance facility with a life of 15 years and a residual (market) value of
$100,000 at the end of its life, they could handle their own overhaul at a cost of only
$30,000 per plane per year. What is the minimum number of planes they must operate to
make it economically feasible to build this facility? The MARR is 10% per year. [6.4]
(a) 7
(b) 4
(c) 5
(d) 3
(e) 8
Solution:
Savings per plan = $40,000 - $30,000 = $10,000 per year
Let X = number of planes operated per year.
(
)
(
500, 000 = 10, 000 ( X ) P ,10%,15 + 100, 000 P ,10%,15
A
F
500, 000 - 100, 000 P ,10%,15
F
X=
10, 000 P ,10%,15
A
$500, 000 - $100, 000(0.2394)
X=
= 6.27
$10, 000(7.6061)
(
(
)
)
)
X = 6.26 or 7 planes per year
Select (a)
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FE Practice Problems w/ Solutions
6-50
Complet the following analysis of investment alternatives and select the preferred
alternative. The study period is three years and the MARR = 15% per year. [6.4]
Capital Investment
Annual Revenues
Annual Costs
Market Value EOY 3
PW (15%)
Alternative A
$11,000
4,000
250
5,000
850
(a) do nothing
(b) Alternative A
Alternative B
$16,000
6,000
300
6,150
???
(c) Alternative B
Alternative C
$13,000
5,540
400
2,800
577
(d) Alternative C
Solution:
(
)
(
)
PWB (15%) = - 16, 000 + (6, 000 - 300) P ,15%,3 + 6,150 P ,15%,3
A
F
2.2832
0.6575
= 1, 058
Select (c) - Alternative B to maximize PW
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FE Practice Problems w/ Solutions
6-51
Complet the following analysis of cost alternatives and select the preferred alternative.
The study period is 10 years and the MARR = 12% per year. “Do Nothing” is not an
alternative. [6.4]
Capital
Investment
Annual Costs
Market Value
EOY 10
FW (12%)
Alternative A
$11,000
Alternative B
$16,000
250
1,000
Alternative C
$13,000
300
1,300
Alternative D
$18,000
400
1,750
-$37,551
-$53,658
???
(a) Alt. A
(b) Alt B
(c) Alt C
100
2,000
-$55,660
(d) Alt D
Solution:
(
FWC (12%) = - 13, 000 F
)
(
)
,12%,10 - (400) F ,12%,10 + 1, 750
P
A
3.1058
17.5487
= - 45, 645
Select (a) – Alternative A to minimize costs.
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FE Practice Problems w/ Solutions
6-52
For the following table, assume a MARR of 10% per year, and a useful life for each
alternative of six years, which equals the study period. The rank order of alternatives from
least capital investment to greatest capital investment is DO NOTHING  A  C  B.
Conplete the IRR analysis by selecting the preferred alternative. [6.4]
DO NOTHING
A
AC
(A-C)
CB
(B-C)
-$15,000
-$2,000
-$3,000
 Capital
Investment
 Annual
Revenues
 Annual Costs
 Market Value
 IRR
4,000
900
460
-1,000
6,000
12.7%
-150
-2,220
10.9%
100
3,350
???
(a) DO NOTHING
(b) Alt A
(c) Alt B
(d) Alt C
Solution:
IRR on (B-C)
(
)
(
0 = - 3, 000 + (460 - 100) P , i ', 6 + 3,350 P , i ', 6
A
F
i ' = 13.4% > 10%
)
Select (c) – Alternative B
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FE Practice Problems w/ Solutions
6-53
For the following table, assume a MARR of 15% per year, and a useful life for each
alternative of eight years, which equals the study period. The rank order of alternatives
from least capital investment to greatest capital investment is Z  Y  W  X.
Conplete the incremental analysis by selecting the preferred alternative. “Do Nothing” is
not an option. [6.4]
 Capital
Investment
 Annual Cost
Savings
 Market Value
 PW(15%)
(a) Alt. W
ZY
YW
WX
-$250
-$400
-$550
70
90
15
100
97
50
20
200
???
(b) Alt. X
(c) Alt. Y
(d) Alt. Z
Solution:
(
)
(
)
D PWW ® X (15%) = - 550 + 15 P ,15%,8 + 200 P ,15%,8
A
F
= - $ 417.31 < 0
Select (a) – Alternative W
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FE Practice Problems w/ Solutions
The following mutially exclusive investment alternatives have been presented to you. The
life of all alternatives is 10 years. Use this information to solve problems 6-54 through 657. [6.4]
A
Capital
Investment
Annual
Expenses
Annual
Revenues
Market Value
at EOY 10
IRR
B
C
D
E
$60,000
$90,000
$40,000
$30,000
#70,000
30,000
40,000
25,000
15,000
35,000
50,000
52,000
38,000
28,000
45,000
10,000
15,000
10,000
10,000
15,000
???
42.5%
31.5%
7.4%
9.2%
6-54
After the base alternative has been identified, the first comparison to be made in an
incremental analysis should be which of the following?
(a) C  B
(b) A  B
(c) D  E
(d) C  D
(e) D  C
Solution:
Rank Order: Do Nothing→D→C→A→E→B
Assuming the MARR ≤ 42.5%, Alternative D is the base alternative. The first
Comparison to be made based on the rank ordering would be D→C.
Select (e)
6-55
Using a MARR of 15%, the present worth of the investment in B when compaired
incremently to A is most nearly:
(a) -$69,000
(b) -$21,000
(c) $80,000
(d) $31,000
(e) $53,000
Solution:
PWA→B(15%) = [-$90,000 - (-$60,000)] + ($12,000 - $20,000) (P/A,15%,10) + ($15,000 $10,000) (P/F,15%,10) = -$68,914
Select (a)
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FE Practice Problems w/ Solutions
6-56
The IRR for alternative C is most nearly:
(a) 30%
(b) 15%
(c) 36%
(d) 10%
(e) 20%
Solution:
PWC (i ') = 0
(
= - 4, 000 + 13, 000 P , i ',10
A
P , i ',10 = 4, 000 = 0.3077
A
13, 000
i ' = 30.8%
(
)
)
Select (a)
6-57
Using a MARR of 15%, the preferred alternative is:
(a) Do Nothing
(e) Alt. D
(b) Alt. A
(f) Alt. E
(c) Alt. B
(d) Alt. C
Solution:
Eliminate Alt. B and Alt. E (IRR < 15%)
(
)
(
)
(
)
(
)
(
)
(
)
PWA (15%) = - 60, 000 + 20, 000 P ,15%,10 + 10, 000 P ,15%,10
A
F
= $ 42,848
PWC (15%) = - 40, 000 + 13, 000 P ,15%,10 + 10, 000 P ,15%,10
A
F
= $ 27, 716
PWD (15%) = - 30, 000 + 13, 000 P ,15%,10 + 10, 000 P ,15%,10
A
F
= $37, 716
Select (b) – Alternative A to maximize PW.
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