Question

.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?

Answer 1

EUAC for 1 yr of operation = 200000 * (A/P,10%,1) + 50000 - 100000 * (A/F,10%,1)

= 200000 *1.1 + 50000 - 100000

= 170000

EUAC for 2 yr of operation = 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

EUAC for 3 yr of operation = 200000 * (A/P,10%,3) + 50000 + 50000*(A/G,10%,3)

= 200000 *0.402115 + 50000 + 50000*0.936556

= 177250.8

EUAC for 4 yr of operation = 200000 * (A/P,10%,4) + 50000 + 50000*(A/G,10%,4)

= 200000 *0.315471 + 50000 + 50000*1.381168

= 182152.6

As EUAC has started increasing, it will keep on increasing as years of operation are increased

Minimum EUAC = 165238.1, ESL = 2 yrs