Solutions to Chapter 9 Problems
9-1 Defender (old lift truck):
Using the outsider viewpoint, the investment value of the old lift truck is its current market value.
9-2 Option1: Keep old car
The investment value of the old car is its current market value (outsider viewpoint).
Capital Investment= $400
Annual Expenses: Maintenance= $800
Gasoline=
year
mi000,15
mi
gal
10
1
gal
50.1$ = $2,250/yr.
Total= $3,050
104
9-3 Old Crane (Defender):
Using the outsider viewpoint, the investment value of the defender is its current market value plus the cost of
the overhaul required to keep it in service.
Capital investment= $8,000 + $4,000= $12,000
9-4 (a)
EOY Market Loss Loss in Value Cost of
Capital
Annual
Expenses
Total Cost EUAC (0%)
1
2
$0
0
$5,000
0
$0
0
$3,000
4,000
$8,000
4,000
$8,000
6,000
105
EUAC 3= ($3,750+ $4,145+ $4,413)(A/P, 12%, 3)= $5,124
(c)
EOY Market Value Loss in Value Cost of
Capital
Annual
Expenses
Total Cost PW (12%)
1
2
$0
0
$10,000
0
$1,200
0
$3,000
4,000
$14,200
4,000
$12,679
3,189
9-5
EOY Market Value Loss in Value Cost of
Capital
Annual
Expenses
Total Cost EUAC (12%)
1
$2,250
$750
$360
$950
$2,060
$2,060
106
9-6 Defender:
EOY Market Value Loss in Value Cost of
Capital
Annual
Expenses a
Total Cost EUAC
0
$87, 000
—
—
—
—
—
9-7
(a) Present Machine (Defender)
Year Capital Recovery Amount Expense for Year Total EUAC
1
2
3
$4,000(0.15)+(4,000-3,000)=$1,600
$3,000(0.15)+(3,000-2,500)= $950
$2,500(0.15)+(2,500-2,000)= $875
$20,000
$25,000
$30,000
$21,600
$25,950
$30,875
$21,600*
$23,622
$25,713
*The economic life of the “defender” is 1 year, and the EUAC during this year is $21,600.
Improved Machine (Challenger)
107
9-8 The older tractor’s (defender’s) marginal costs and economic life are calculated when MARR= 0% using the format
below.
EOY Market Value Loss in Value Cost of
Capital
O&M
Expenses
Total Cost EUAC a
0
1
2
$80,000
70,000
60,000
—
$10,000
10,000
—
0
0
—
$20,000
25,000
—
$30,000
35,000
—
$30,000*
32,500
108
9-9 The repeatability assumption and the AW method (over one useful life cycle) are used in the comparison of the two
robots. The use of repeatability as a simplified modeling approach can be supported in this case.
9-10 Challenger:
AW(15%)= -$9,000- [$65,000 (A/P, 15%, 20) – $13,000 (A/F, 15%, 20)]= -$19,260
Assuming an infinitely long study period: PW (15%)= 15.0
260,19$−= -$128,400
Defender:
109
9-11 The overpass can be reinforced to extend its life for 5 years and then be replaced by a new concrete overpass. An
alternative is to build the new concrete overpass immediately. Coterminate the study period at 40 years.
Cash Flow Analysis
EOY Reinforce Now, Replace Later Replace with New Overpass
Now (∆Replace Now-
Reinforce)
0
1
-$36,000 a
0
-$140,000
3,200 b
-$104,000
3,200
a -$22,000 (reinforcement) -$14,000 (MV of scrap foregone)= -$36,000
b Savings in annual expenses
c $16,000 (sharp value) -$140,000(investment in new overpass)= -$124,000
110
9-12 Abandonment Interval:
Keep for N= 1 year:
PW(10%)= -$7,500 + ($6,200 + $2,000)(P/F, 10%, 1) = -$45
Keep for N= 2 years:
9-13
EOY, k Market Value Loss in Value Cost of Capital Annual Expenses Approximate
After-Tax Total
(Marginal) Cost a
0
1
$8,000
4,700
—
$3,300
—
$560
—
$3,000
—
$4,116
111
EOY, k MACRS BV Interest on Tax
Credit b
Adjusted After-
Tax Total
(Marginal) Cost c
PW (7%) EUAC
0
1
2
$8,000
6,400
3,840
—
$224
179
—
$4,340
3,076
—
$4,056
2,687
—
$4,340
3,729
9-14
BV 0= $62,000(1- 0.2 – 0.32 – 0.192 – 0.1152) = $10, 714
MV 0= $12, 000
Expense for year 0 repair work = $4,000
9-15
(a)
Year, k
1
$15,702*
$20,866
9-16 Keep the Defender:
EOY BTCF Depr TI T (40%) ATCF
0
1
-$3,200
– 439 b
—
$1,000
-($3,200 – $5,000 a)= $2,300
– 1, 439
– $920
576
-$4, 120
137
Defender
EUAC throu
g
h
y
ear k
Challenger
EUAC through year k
113
aBV 0= $9,000 (1- 0.5/9 – 3/9) = $5,500
Loss on disposal = $3,200 – $5,500 = $2,300
1
2
3
4
5
– 100
– 110
– 121
– 133
– 146
$2,600
4,160
2,496
1,498
749 b
-$2,700
– 4,270
– 2, 617
– 1,631
– 895
$1, 080
1, 708
1, 047
652
358
980
1,598
926
519
212