Question

A piece of equipment has a first cost of $160,000, a maximum useful life of 7 years, and a market (salvage) value described by the relation S = 120,000 – 23,000k, 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 + 12,000k. 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

The economic service life is 3 years and the AW value is -137044.49

This is found by first finding the PV of annual cost and then finding the PV of salvage value. Finally the initial cost is added and net present value for all 7 years is computed. Using (A/P) factor, the NPV is converted in EUAC. Minimum EUAC occurs in year 3 and so AW = -EUAC = -137044.49

Year	Annual cost	Salvage value	PV of the annual cost	PV of the market value	Net present value	EUAC
1	72000	97000	63157.89	85087.72	138070.18	157400.00
2	84000	74000	127793.17	56940.60	230852.57	140194.39
3	96000	51000	192590.43	34423.55	318166.89	137044.49
4	108000	28000	256535.10	16578.25	399956.85	137267.11
5	120000	5000	318859.34	2596.84	476262.50	138727.43
6	132000	0	378996.77	0.00	538996.77	138607.06
7	144000	0	436544.54	0.00	596544.54	139109.64