Required information An electric switch manufacturing company is trying to decid
ID: 1138525 • Letter: R
Question
Required information An electric switch manufacturing company is trying to decide between three different assembly methods. Method A has an estimated first cost of $42,000, an annual operating cost (AOC) of $9,000, and a service life of 2 years. Method B will cost $83,000 to buy and will have an AOC of $5,000 over its 4-year service life. Method C costs $119,000 initially with an AOC of $5,000 over its 8-year life. Methods A and B will have no salvage value, but Method C will have equipment worth 10% of its first cost. Perform a present worth analysis to select the method at-9% per year. The present worth of method A is $ The present worth of method B is $ The present worth of method C is $ Metho (Click to select) S selected.Explanation / Answer
Solution:-
Salvage value, Method C = $119,000 x 10% = $11,900
(a) FW of methods is computed as follows.
FW, Method A ($) = 42,000 x F/P (9%, 4) + 9,000 x F/A (9%, 4)
= 42,000 x 1.412 + 9,000 x 4.573
= 59304 + 41157
= 1, 00,461
FW, Method B ($) = 83,000 x F/P (9%, 4) + 5,000 x F/A (9%, 4)
= 83,000 x 1.412 + 5,000 x 4.573
= 117,196 + 22,865
= 1, 40,061
FW, Method C ($) = 119,000 x F/P (9%, 8) + 5,000 x F/A (9%, 8) - 11,900
= 119,000 x 1.993 + 5,000 x 11.028 - 11,900
= 237,167 + 55,140 - 11,900
= 2, 80,407
Since Method A has lowest FW of costs, this is preferred.
(b) PW is computed as follows.
PW, Method A ($) = 42,000 + 9,000 x P/A (9%, 4)
= 42,000 + 9,000 x 3.240
= 42,000 + 24,298
= 71,160
PW, Method B ($) = 80,000 + 5,000 x P/A (9%, 4)
= 83,000 + 5,000 x 3.0373
= 83,000 + 15,186.5
= 98,186.5
PW, Method C ($) = 119,000 + 5,000 x P/A (9%, 8) - 11,900 x P/F (9%, 8)
= 119,000 + 5,000 x 5.535 - 11,900 x 0.5019
= 119,000 + 27,675 – 5,972.61
= 140,702.39
Since Method A has lowest PW of costs, this is preferred.
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