Question

Thank you for your assistance in advanced. Please show your work so I can learn/understand how it was done.

Your company needs a machine for the next seven years, and you have two choices (assume an annual interest rate of 12%) Machine A costs $120,000 and has an annual operating cost of $45,000. Machine A has a useful life of seven years and a salvage value of $13,000 Machine B costs $190,000 and has an annual operating cost of $27,000. 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 $27,000 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? Click the icon to view the interest factors for discrete compounding when i: 12% per year. You should pay SRound to the nearest dollar.)

Answer 1

First of all determine the present value of machine A. The present value of cash flow can be written as follows

PW_A = - $ 120,000 - $ 45,000(P/A, 12%, 7) + $ 13,000(P/F,12%,7)

PW_A = - $ 120,000 - 45,000*4.5638 + 13,000*0.4523

PW_A = - 120,000 - 205,371 + 5,879.90

PW_A = - $ 319,491.10

Similarly we can calculate the present worth of machine B for 7 years. Let us assume the firm pays an amount A after 5 years to extend life of machine B.

PW_B = - 190,000 - 27,000(P/A,12%,7) -A(P/F,12%,5)

PW_B = - 190,000 - 27,000*4.5638 - A*0.5674

PW_B = - 190,000 - 123,222.60 - 0.5674A

PW_B = - 313,222.60 - 0.5674A

The firm will be indifferent two machine A and B when their present worth is equal.

-319,491.10 = - 313,222.60 - 0.5674A

-319,491.10 + 313,222.6 = - 0.5674A

- 6268.5 = - 0.5674A

0.5674A = 6268.50

A = 6268.5 / 0.5674

A = $ 11,047.76

Amount = $ 11,048 (Approximately)

The firm will be ready to pay an additional amount of $ 11,048 in order to extend the life of machine B by two years.

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