The answers are already there but I do not know how they got the answers they di

Question

The answers are already there but I do not know how they got the answers they did. How did they get the minimum EUAC and the marginal cost?

A high-speed electronic assembly machine was purchased two years ago for $50,000. At the present time, it can be sold for $23,000 and replaced by a newer model having a purchase price of $45,000; or it can be kept in service for a maximum of one more year. The new assembly machine, if purchased, has a useful life of not more than two years. If the before-tax MARR is 15%, when should the old assembly machine be replaced? Use the following data table for your analysis. Challenger Defender Market Value 0&M; Costs Year Market Value 0&M; Costs $45,000 31,000 23,000 $9,000 13,500 $23,000 16,500 $14,500 Click the icon to view the interest and annuity table for discrete compounding when the MARR is 15% per year The minimum EUAC value of the chalenger is $ 28,076. (Round to the nearest dollar.) The marginal cost of keeping the defender in service for one more year is $ 24,450. (Round to the nearest dollar.) The old assembly machine should be replaced in one year

Explanation / Answer

Find the marginal cost as the annual equivalent cost of defender = 23000 (A/P, 15%, 1) + 14500(A/F, 15%, 1) - 16500

= 23000 x 1.1500 + 14,500 x 1 - 16,500 = 24450

Find the EUACs for 1 year and 2 year for the challenger

EUAC for 1 year = (45000 + 9000(P/F, 15%, 1) - 31000(P/F, 15%, 1))(A/P, 15%, 1) = (45000 - 22000*0.8696)*(1.15) = 29749

EUAC for 2 year = (45000 + 9000(P/F, 15%, 1) + 13500(P/F, 15%, 2) - 23000(P/F, 15%, 2))(A/P, 15%, 2) = (45000 + 9000*0.8696 + 13500*0.75614 - 23000*0.75614)*(0.6151) = 28076.

Since EUAC for 2 years is more than that of the defender, defender should be kept for 1 year and replaced after it

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