Question

Your company needs a machine for the next seven​ years, and you have two choices​ (assume an annual interest rate of 6​%).

Machine A costs ​$140,000 and has an annual operating cost of ​$46,000. Machine A has a useful life of seven years and a salvage value of ​$8,000.

Machine B costs ​$180,000 and has an annual operating cost of ​$24,000. 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 certain amount of investment. If Machine​ B's life is​ extended, it will still cost ​$24,000 annually to operate and still have no salvage value.

What would you pay at the end of year 5 to extend the life of Machine B by two​ years?

You should pay ​$________.​(Round to the nearest​ dollar.)

Answer 1

Let the amount to be paid at the end of year 5 is X for machine B so that their present values are equal

140,000 + 46,000(P/A, 6%, 7) - 8,000(P/F, 6%, 7) = 180,000 + 24,000(P/A, 6%, 7) + X(P/F, 6%, 5)

140,000 + 46,000*5.5824 - 8,000*0.66506 = 180,000 + 24,000*5.5824 + X*0.74726

391,469.09 = 313,977.6 + X*0.74726

This gives X = 103,702

You should pay 103,702 to remain indifferent between A and B.