A
B
Capital investment
-3,500
-5,000
Annual revenue
1,900
2,500
Annual expenses
-645
-1,020
56
Useful life (years)
4
6
Market value at end of useful life
0
0
Solution:
LCM = 12 years
Alt. A.
PW
(10%)
= -3,500
– 3,500 [(P/F, 10%, 4) + (P/F, 10%, 8)] + (1,900 – 645) (P/A, 10%, 12)
= $1,028
Alt. B.
PW
(10%)
= -5,000
– 5,000 (P/F, 10%, 6) + (2,500 – 1,020) (P/A, 10%, 12)
= $2,262
Based on the PW method we would select Alternative B because it has the larger value.
Suppose that the above problem is modified such that an analysis period of 6 years is used (co-
terminated assumption) instead of 12 years.
Solution: Use the FW to analyze the situation.
Alt. A.
FW
(10%)
= [-3,500 (F/P, 10%, 4) + (1,900
– 645) (F/A, 10%, 4)] (F/P, 10%, 2)
= $847
Alt. B.
FW
(10%)
= -5,000 (F/P, 10%, 6) + (2,500
– 1,020) (F/A, 10%, 6)
= $2,561
Based on the FW of each alternative at the end of the six-year stuffy period, we would select Alt. B
because it has the larger value.
Imputed Market Value Technique
Obtaining a current estimate from the marketplace for a piece of equipment or another type of
asset is the preferred procedure in engineering practice when a market value at time T < usefu l
life is required.
The imputed market value technique is sometimes called the implied market value.
It is calculated based on the sum of two parts as follows:
MV
T
= [PW at the end of year T of remaining capital recovery amounts] + [PW at the end of
year T of original market value at the end of useful life}
57
Sample Problem. A piece of production equipment is to be replaced immediately because it no
longer meets quality requirements for the end product. The two best alternatives are a used piece of
equipment (E1) and a new automated model (E2). Select the preferred alternative with a 5 year
study period using the imputed market value technique. The MARR is 15% per year. Select the best
alternative using n= 5 years.
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