Machine X has an initial cost of $10,000 and annual maintenance of $500 per year

Question

Machine X has an initial cost of $10,000 and annual maintenance of $500 per year. It has a useful life of four years and no salvage value at the end of that time. Machine Y costs $20,000 initially and has no maintenance costs during the first year. Maintenance is $100 at the end of the second year and increases by S100 per year thereafter. Machine Y has a useful life of eight years and an anticipated salvage value of $5,000 at the end of its useful life. If the MARR is 8%, which machine should be recommended? Use the present worth method.

Explanation / Answer

Since the life of both the machine is unequal, we have to take LCM of two numbers ,i.e., 4 and 8. The LCM is 8.

In 8 years Machine X will be purchased two times.

PW of Machine X = 10,000 + 10,000(P/F, 8%, 4) + 500(P/A, 8%, 8)

                             = 10,000 + 10,000(0.7350) + 500(5.747)

                             = 10,000 + 7,350 + 2,873.5

                            = $20,223.5

PW of machine Y = 20,000 + [100 + 100(A/G, 8%, 7)] (P/A, 8%, 7) (P/F, 8%, 1) - 5,000(P/F, 8%, 8)

                            = 20,000 + [100 + 100(2.694)] (5.206) (0.9259) - 5,000(0.5403)

                           = 20,000 + 1780.6 - 2701.5

                           = $19,079.1

Since it is cost dominated project, and the PW of machine Y is less than machine X, therefore, Machine Y should be recomended.

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