Transcript TM 661 Chapter 6 Solutions 1

TM 661
Replacement
Solutions 1
A firm is considering an optimal replacement policy for a truck. The initial cost of a
new truck is $100,000 and is expected to remain so for the forseeable future.
Maintenance costs are expected to be $10,000 per year and are expected to escalate by
10% per year. The following table estimates the salvage value of a new truck after n
years of use. Using a minimum attractive rate of return of 20%, determine the optimal
replacement interval for this truck.
Years
1
2
3
4
5
>5
Salvage
$50,000
$30,000
$20,000
$10,000
$ 5,000
$0
Soln: Costs are represented as positive. For example, the cash flow 1 Yr replacement
for t=1 is computed as 10,000 in maintenance less 50,000 for salvage. The cash flow
for 2 Yr replacement, t=2, is computed as 10,000 x 1.1 less 30,000 salvage. Cash flows,
NPV and EUAC for replacement intervals 1-5 are shown below.
Optimal Replacement
t
0
1
2
3
4
5
6
7
NPV =
EUAC =
1 Yr
100,000
(40,000)
2 Yr
100,000
10,000
(19,000)
MARR =
3 Yr
100,000
10,000
11,000
(7,900)
66,667
(80,000)
95,139
(62,273)
111,400
(52,885)
20%
4 Yr
100,000
10,000
11,000
12,100
3,310
124,571
(48,120)
5 Yr
100,000
10,000
11,000
12,100
13,310
9,641
6 Yr
100,000
10,000
11,000
12,100
13,310
14,641
16,105
133,268
(44,562)
135,277
(45,234)
TM 661
Replacement
Solutions 2
A company owns a 5-year old turret lathe that has a book value of $28,000. The present
market value for the lathe is $20,000. The company has just finished closing out the books
for the current year where maintenance costs totaled $4,000. Costs for next year are
estimated to be $4,500 and are expected to increase by $500 per year. The current lathe is
expected to have a remaining life of 5 years. A new lathe can be purchased for $75,000.
Annual operating and maintenance costs are expected to average $1,000 over the first 5
years with a salvage value of $40,000 at that time. Determine if the company should retain
the current machine or purchase a new lathe. MARR is 15%.
Soln: I used the cash flow approach.
NPWK
= -4,500(P/A,15,5) - 500(P/G,15,5)
= -4,500(3.3522) - 500(5.7751)
= -17,972
Keep
1
2
3
4
5
4,500
6,500
Replace
NPWR = -55,000 - 1,000(P/A,15,5)
+ 40,000(P/F,15,5)
= -55,000 - 1,000(3.3522)
+ 40,000(.4972)
= -38,464
40,000
20,000
1
1,000
75,000
Keep the current Lathe
2
3
4
5