Calculation Procedure:- Compute the required replacement capital
Replacement capital = $56,000(1.07)
3
(1.09)
2
= $56,000(1.225)(1.188) = $81,497. - Compute the annual deposit
From the preceding calculation procedure, annual deposit = $81,497(0.17740) = $14,458.
PRESENT WORTH OF COSTS IN
INFLATIONARY PERIOD
An asset with a first cost of $70,000 is expected to last 6 years and to have the following
additional cost data as based on present costs: salvage value, $5000; annual maintenance,
$8400; major repairs at the end of the fourth year, $9000. The asset will be replaced with
a duplicate when it is retired. Using an interest rate of 12 percent and an inflation rate of 8
percent per year, find the present worth of costs of this asset for the first two lives (i.e., for
12 years).
Calculation Procedure:- Compute the present worth of the capital expenditures
for the first life
The "present" refers to the beginning of the first life. The payment for repairs will be
$9000(1.08)
4
, and the present worth of this payment is $9000(1.08)
4
(SPPW, n = 4, i = 12
percent) = $9000(1.08)
4
/(1.12)
4
= $7780. Similarly, the present worth of the salvage val-
ue is $5000(1.08)
6
/(1.12)
6
= $4020. Thus, the present worth of capital expenditures for
the first life is $70,000 + $7780 - $4020 = $73,760. - Compute the present worth of maintenance for the first life
The annual payments for maintenance constitute a uniform-rate series in which the first
payment R 1 = $8400(1.08) = $9072 and the ratio of one payment to the preceding pay-
ment is r = 1.08. By Eq. 9, the present-worth factor of the series is URSPW =
[(1.08/1.12)^6 - 1]/(1.08 - 1.12) = 4.901. Then present worth of series = ,K 1 (URSPW) =
$9072(4.901) = $44,460. - Compute the present worth of costs for the first life
Summing the results, we see that present worth = $73,760 + $44,460 = $118,220. - Compute the present worth of costs for the second life
Since each payment in the second life is (1.08)^6 times the corresponding payment in the
first life, the value of all payments in the second life, evaluated at the beginning of that
life, is (1.08)^6 times that for the first life, or $118,220(1.08)^6. The present worth of this
amount is $118,220(1.08)^6 /( 1.12)^6 = $95,040. - Compute the present worth of costs for the first two lives
Summing the results yields PW = $118,220 + $95,040 = $213,260.
Related Calculations: Let h = [(I +/)/(! + OF, where N= life of asset, years.
In the standard case, where all annual payments as based on present costs are equal and
no extraordinary intermediate payments occur, the present worth of costs for the first
life is P - Lh + c(\ +/)(/* - !)/(/- 0» where P = initial cost; L = salvage value as based
on present costs; c - annual payment for operation, maintenance, etc., as based on pres-