Question

A piece of equipment has a first cost of $150,000, a maximum useful life of 7 years, and a market (salvage) value described by the relation S = 120,000 – 17,000k, where k is the number of years since it was purchased. The salvage value cannot go below zero. The AOC series is estimated using AOC = 60,000 + 7,000k. The interest rate is 14% per year. Determine the economic service life and the respective AW.

The economic service life is ...... year(s) and the AW value is ........

Answer 1

Economic service life is 4 years and AW is -117,273.11

Find EUAC for n = 1 to 7 years

EUAC (n = 1) = (150,000 + 67000(P/F, 14%, 1) - 103000(P/F, 14%, 1))(A/P, 14%, 1) = 135,000

EUAC (n = 2) = (150,000 + 67000(P/F, 14%, 1) + 74000(P/F,14%, 2) - 103000(P/F, 14%, 2))(A/P, 14%, 2) = 121,177.57

Similarly, EUAC for next 5 years is computed below

Year	Annual cost	Salvage value	PV of the annual cost	PV of the market value	Net present value	EUAC
1	67000	103000	58771.93	90350.88	118421.05	135000.00
2	74000	86000	115712.53	66174.21	199538.32	121177.57
3	81000	69000	170385.22	46573.03	273812.19	117939.53
4	88000	52000	222488.28	30788.17	341700.11	117273.11
5	95000	35000	271828.31	18177.90	403650.40	117576.72
6	102000	18000	318298.14	8200.56	460097.58	118317.54
7	109000	1000	361858.60	399.64	511458.97	119268.33