Chapter 14: Replacement Decisions
Sunk Costs, Opportunity Costs, and Cash Flows
14.1
(a) Sunk cost = $12,000- $2,000 = $10,000.
(b) Opportunity cost = $2,000.
14.2
(a) Purchase cost = $15,000, market value = $6,000,
sunk cost = $15,000 – $6,000 = $9,000
14.3
(a) Opportunity cost = $30,000
(b) Assume that the old machine’s operating cost is $30,000 per year. Then the n
ew machine’s operating cost is zero per year. The cash flows associated with r
etaining the defender for two more years are
14.4
(a) Initial cash outlay for the new machine = $120,000
PW(15%) $7,500 $3, 000( / ,15%,1) $3,500( / ,15%, 2)
PF PF
=−− −
(c)
14.5
(a) Cash flows
Year:
0
1
2
3
4
5
Defender
-$10K
0
0
0
0
$5K
Challenger
-$75K
14.6
(a) and (b) Cash flows:
Year:
0
1
2
3
4
5
Defender
-$5,500
$21,000
$21,000
$22,200
Challenger
-$36,500
$24,000
$24,000
$24,000
$24,000
$30,300
Economic Service Life
14.7
14.8
Note: Missing Table P14.8 in the first printing.
Year (n)
OC
MV
0
7,700
1
3,200
4,300
2
3,700
3,300
3
4,800
1,100
4
5,850
(a) Interest Defender:
2
$
3
,
7
0
0
$
3
,
3
0
0
$
3
,
4
3
6
$
2
,
9
9
9
$
6
,
4
3
5
3
$
4
,
8
0
0
$
1
,
1
0
0
$
3
,
8
4
0
$
2
,
8
8
0
$
6
,
7
2
0
4
$
5
,
8
5
0
0
$
4
,
2
6
1
$
2
,
5
3
5
$
6
,
7
9
6
Annual changes in MV 20%
Interest rate 12%
nMarket Value O&M Costs CR(12%) OC(12%) AEC(12%)
0 $200,000
1 $130,000 $20,000 $94,000 $20,000 $114,000
12%i=
Y
e
a
r
O
C
M
V
C
R
(
1
2
%
)
Total
A
EC
(12%)
0
$
7
,
7
0
0
1
$
3
,
2
0
0
$
4
,
3
0
0
$
3
,
2
0
0
$
4
,
3
2
4
$
7
,
5
2
4
OC
(b)
(c) n* = 0.
14.9
(a) Economic service life = 2 years:
Interest rate
12%
n
Market Value
O&M Costs
CR(12%)
OC(12%)
AEC(12%)
(b) Replacement decision:
(c) When to replace?
10 years
C
N=
14.10
(a) Interest
(b) Interest
Replacement Decision with an Infinite Planning Horizon and No
Technological Change
14.11
It is assumed that the required service period is very long.
10%i=
nOC MV AEOC CR(10%) AEC(10%)
0 $15,000
1 $2,500 $12,800 $2,500 $3,700 $6,200
15%i=
nOC MV AEOC CR(10%) AEC(10%)
0 $15,000
1 $2,500 $12,800 $2,500 $4,450 $6,950
14.12 (a) and (b)
14.13
(a) Opportunity cost = $0
(b) The cash flows are:
Defender Challanger
0 -$4,000 -$6,000
1 -$3,000 -$2,000
n
Year: 0 1 2 3 4 5
14.14
(a)
Yes, the new machine should be purchased now.
(b)
Let P as the current market value of the old machine
14.15
Assume that the old system has a current market value of P.
−P(A/P,15%,5) +$10,000 −$7,000 =$3,216.84
14.16
14.17
14.18
(a) Economic service life:
(b) Rate of return calculation:
nOR MV
AEOR CR(10%) AE(10%)
020,000
135,550 17,000 35,550 5,000 30,550
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14.19
For the challenger, we have:
For the defender, since salvage value at year 10 is $1,000 and the problem is
stated that if sold at the end of first year, it will bring $1,500. We assume the
market values will be declined same amount (($1,500-$1,000)/4 = $125) for the
next years.
Since the new machine should not be purchased.
14.20
Option 1
AEC(15%) $15,000 ($6,000 0)( / ,15%,10)
$12,000 ($48,000 $5,000)( / ,15%,10) $5,000(0.15)
AP
AP
= +−
++ − +
Year OC MV AE
OC
CR(14%) AEC(14%)
0 $2,000
1 $3,800 $1,500 $3,800 $780 $4,580
Replacement Problem with a Finite Planning Horizon
14.21
(a)
(b)
Defender
Annual changes in MV 20%
Annual increases in A/T O&M 35%
Interest rate 12%
nMarket Value O&M Costs CR(12%) OC(12%) AEC(12%)
0$12,000
AEC
Defender
Challenger
1
$7,590
$9,960
defender for 3 years in order to optimize the replacement strategy.
14.22
There are several plausible scenarios. Some of the most feasible scenarios are;
14.23
There are several plausible scenarios. Some of the most feasible scenarios are;
2
$7,847
$9,642
3
$8,275
$9,547
14.24
14.25
Replacement Analysis with Tax considerations
14.26
(a) Comments: Sunk cost can be defined as either the difference between book
value and market value or the cost that has already been expended ($).
(c) Equivalent annual cost of operating the truck for two more years:
Cash Flow
Elements
0
End of Period
1
2
(d) Equivalent annual cost of owning and operating the truck for 5 years:
Cash Flow
Elements
0
1
2
3
4
5
14.27
(a)
(b) , (c) and (d): Replace the defender now with the challenger.
Book value = 0
Option 1: Keep the defender
n0 1 2
Depreciation $0 $0 $0
Book value 0 0 0
Market value $30,000 $12,000
Option 2: Replace the defender
n0 1 2 3 4 5 6 7 8
Depreciation $23,579 $40,409 $28,859 $20,609 $14,735 $14,718 $14,735 $7,359
Book value $165,000 $141,422 $101,013 $72,155 $51,546 $36,812 $22,094 $7,359 $0
14.28
(a) Based on the opportunity cost approach:
Cost basis = $120,000
Gains or losses at the time of disposal:
• Old machine:
• New machine:
(b) Cash flow for the old machine:
Cash Flow
Elements
0
1
2
3
4
5
Investment Net
-$26,000
(c) Replacement analysis: Replace the old machine now.
Total depreciation = $109,272
14.29 (a) & (b): Decision – Replace the defender now.
Financial Data n0 1 2 3 4 5 6 7
Depreciation $17,148 $29,388 $20,988 $14,988 $10,716 $10,716 $5,352
Book value ($120,000) ($137,148) ($166,536) ($187,524) ($202,512) ($213,228) ($223,944) ($229,296)
Salvage value $30,000
(1) Keep the defender
n0 1 2 3 4 5
Depreciation $9,600 $5,760 $5,760 $2,880 $0 $0
Book value 14,400 8,640 2,880 0 0 0
Market value $10,000 5,000
Cas h Flow S tatement
(2) Replace the defender
n0 1 2 3 4 5
Depreciation $15,000 $24,000 $14,400 $8,640 $4,320
Book value $75,000 $60,000 $36,000 $21,600 $12,960 $8,640
14.30 (a), (b) & (c) Decision: Do not replace the defender now.
(1) Keep the defender
n0 1 2 3
Depreciation
Book value
Market value $5,500 $1,200
PW (10%) = $28,575 AE (10%) = $11,491
(2) Replace the defender
n0 1 2 3 4 5
Depreciation $7,300 $11,680 $7,008 $4,205 $2,102
Book value $36,500 $29,200 $17,520 $10,512 $6,307 $4,205
Market value $6,300
14.31
(a) Interest Defender:
10%i=
MARR 10%
Permitted Annual Depreciation Amounts over the
Holding Holding Period Total Book
Period 1 2 3 4 5 6 7 8
Depreciation
Value
0$7,810
1$1,116 $1,116 $6,694
2$2,233 $1,115 $3,348 $4,463
3$2,233 $2,230 $1,117 $5,579 $2,231
4$2,233 $2,230 $2,233 $1,115 $7,810 $0
Annual O&M Costs over the Holding Period Total PW
Holding
Total PW of
of A/T
Period 1 2 3 4 5 6 7 8
O&M Costs
O&M Costs
0`
1$3,200 $2,909 $1,891