Question

With the estimates shown below, Sarah needs to determine the trade-in (replacement) value of machine X that will render its AW equal to that of machine Y at an interest rate of 9% per year. Determine the replacement value.

	Machine X	Machine Y
Market Value, $	?	89,000
Annual Cost, $ per Year	−59,000	−40,000 for year 1,increasing by 2000 per year thereafter.
Salvage Value	18,500	23,000
Life, Years	3	5

The replacement value is $ ...........?

Answer 1

ANS ....

First, we compute Present Worth (PW) of costs for Machine Y, as follows. Note that PV factor in year N = (1.09)^-N

Annual worth (AW) for Machine Y ($) = PW / P/A(9%, 5) = - 243,860 / 3.8897** = - 62,694

If replacement value for Machine X be $R, then

AW of Machine X = AW of Machine Y

- R x A/P(9%, 3) - 59,000 + 18,500 x P/F(9%, 3) x A/P(9%, 3) = - 62,694

- R x 0.3951 - 59,000 + 18,500 x 0.7722 x 0.3951 = - 62,694

- R x 0.3951 - 59,000 + 5,644 = - 62,694

- R x 0.3951 = - 62,694 + 59,000 - 5,644

- R x 0.3951 = - 9,338

R = $23,634

Thus we solved.........