Question

Your company needs a machine for the next seven years, and you have two choices.

Machine A: costs $113000 and has an annual operating cost of $31000. Machine A has a useful life of seven years and a salvage value of $15600 at the end of 7 years.

Machine B: costs $233000 and has an annual operating cost of $5000. Machine B has a useful life of five years and no salvage value. However, the life of machine B can be extended by two years with a major overhaul. If machine B’s life is extended, it will still cost $5000 annually to operate and still have no salvage value.

What is the maximum amount that you would pay at the end of year 5 to extend the life of machine B by two years? Use AEC method. Assume an annual interest rate of 8.3%.

SHOW ALL WORK

Answer 1

Equivalent annual cost of machine A = 113000*(A/P,8.3%,7) + 31000 - 15600*(A/F,8.3%,7)

= 113000*(0.083 *((1 + 0.083)^7)/((1 + 0.083)^7-1)) + 31000 - 15600*(0.083/((1 + 0.083)^7-1))

= 113000*(0.083 *((1.083)^7)/((1.083)^7-1)) + 31000 - 15600*(0.083/((1.083)^7-1))

= 113000*0.194048 + 31000 - 15600*0.111048

= 51195.075

Equivalent annual cost of machine B = 233000*(A/P,8.3%,5) + 5000

= 233000*(0.083 *((1 + 0.083)^5)/((1 + 0.083)^5-1)) + 5000

= 233000*(0.083 *((1.083)^5)/((1.083)^5-1)) + 5000

= 233000*0.252440 + 5000

= 63818.52

As equivalent annual cost of machine A is less, it should be selected

Let amount paid at EOY 5 to extend life be S, then,

Equivalent annual cost of machine B = 233000*(A/P,8.3%,7) + 5000 + S *(P/F,8.3%,5)*(A/P,8.3%,7)

= 233000*(0.083 *((1 + 0.083)^7)/((1 + 0.083)^7-1)) + 5000 + S *((1 + 0.083)^-5)*(0.083 *((1 + 0.083)^7)/((1 + 0.083)^7-1))

= 233000*(0.083 *((1.083)^7)/((1.083)^7-1)) + 5000 + S *((1.083)^-5)*(0.083 *((1.083)^7)/((1.083)^7-1))

= 233000*0.194048 + 5000 + S *0.671209* 0.194048

As per given condition

233000*0.194048 + 5000 + S *0.671209* 0.194048 = 51195.075

S = (51195.075 - 233000*0.194048 - 5000) / (0.671209* 0.194048) = 7538.698