Question

Machine A lasts 5 years and Machine B lasts 10 years. Machine A costs $6,000 and Machine B costs $9,000. The salvage value of Machine A is $4,000 and the salvage value of Machine B is $2,200. Annual operation and maintenance costs are $5,000 for Machine A and $3,900 for Machine B. Both machines can be purchased in the future at the same price as today, and their salvage values and annual costs will remain as they are now. Your MARR is 13.3%.Find the Annual Equivalent Cost (AEC) of both and decide which to select.

Answer 1

AEC for both machines is computed as follows.

AEC, Machine A ($) = 6,000 x A/P(13.3%, 10) + 5,000 - 4,000^# x P/F(13.3%, 10) x A/P(13.3%, 10)

= 6,000 x 0.1865** + 5,000 - 4,000 x 0.2869** x 0.1865** = 1,119 + 5,000 - 214 = 5,905

AEC, Machine B ($) = 9,000 x A/P(13.3%, 10) + 3,900 - 2,200^# x P/F(13.3%, 10) x A/P(13.3%, 10)

= 9,000 x 0.1865** + 3,900 - 2,200 x 0.2869** x 0.1865** = 1,679 + 3,900 - 118 = 5,461

Since Machine B has lower EAC, Machine B should be selected.

**A/P(r%, N) = r / [1 - (1 + r)^-N]

A/P(13.3%, 10) = 0.133 / [1 - (1.133)^-10] = 0.133 / (1 - 0.2869) = 0.133 / 0.7131 = 0.1865

P/F(r%, N) = (1 + r)^-N

P/F(13.3%, 10) = (1.133)^-10 = 0.2869

#Since salvage value is a cash inflow, it is deducted for purpose of computing AEC.