Question

4.A machine cost $200,000 and has a salvage value of $100,000 if kept for one year. The salvage value will decrease by $50,000 in years 2 and 3 and remain zero after year 3. The operating costs are $50,000 the first year and increase by $50,000 per year. So operating costs in year two will be $100,000, and in year three $150,000 and so on. How long should the equipment be kept so that annual cost is minimized if the MARR is 10%/year compounded annually or stated another way what is the economic service life (ESL) and what is the associated annual cost for this service life?

Please solve with no table or excel

Answer 1

Determining EUAC for all the years as under:

First Year EUAC:
= 200000 * (A/P,10%,1) + 50000 - 100000 * (A/F,10%,1)
= 200000 *1.1 + 50000 - 100000
= $170000

Second Year EUAC:
= 200000 * (A/P,10%,2) + 50000 + 50000*(A/G,10%,2) - 50000 * (A/F,10%,2)
= 200000 *0.576190 + 50000 + 50000*0.476190 - 50000 *0.476190
= $165238.10

Third Year EUAC:
= 200000 * (A/P,10%,3) + 50000 + 50000*(A/G,10%,3)
= 200000 *0.402115 + 50000 + 50000*0.936556
= $177250.8

Fourth Year EUAC:
= 200000 * (A/P,10%,4) + 50000 + 50000*(A/G,10%,4)
= 200000 *0.315471 + 50000 + 50000*1.381168
= $182152.6

As per above calculations, as the EUAC has started increasing year on year, it will keep on increasing as years of operations are also increased.

Thus, Minimum EUAC = $165238.10
Economic Service Life (ESL) = 2 years.