A piece of machinery costs $7500 and has no salvage value after it is installed.

Question

A piece of machinery costs $7500 and has no salvage value after it is installed. The manufacturer's warranty will pay the first year's maintenance and repair costs. In the second year, maintenance costs will be $900, and this item will increase on a $900 arithmetic gradient in subsequent years. Also, operating expenses for the machinery will be $500 the first year and will increase on a $400 arithmetic gradient in the following years. If interest is 8%, compute the useful life of the machinery that results in a minimum EAC. (find its minimum cost life).

Explanation / Answer

Ans So minimum EAC is at end of fourth years $4590 (rounded off) thereafter cost is increasing

Minium cost life= 4 years

Year EAC of Machiney cost EAC of Repair & Maintainenece EAC for Operating cost Total Formula $7500*(A/P,8%,n) $900*(A/G,8%,n) $500+$400*(A/G,8%,n) 1 8100 0 500 8600 2 4206 433 692 5331 3 2910 854 880 4644 4 2264 1264 1062 4590 5 1878 1661 1238 4777 6 1622 2048 1410 5080 7 1440 2425 1578 5443

Ans So minimum EAC is at end of fourth years $4590 (rounded off) thereafter cost is increasing

Minium cost life= 4 years

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