Question

Two methods can be used to produce expansion anchors. Method A costs $90,000 initially and will have a $16,000 salvage value after 3 years. The operating cost with this method will be $26,000 in year 1, increasing by $3200 each year. Method B will have a first cost of $106,000, an operating cost of $6000 in year 1, increasing by $6000 each year, and a $36,000 salvage value after its 3-year life. At an interest rate of 14% per year, which method should be used on the basis of a present worth analysis? The present worth for method A is $ 152027 The present worth for method B is $ is used to produce expansion anchors. Method B

Answer 1

Answer to blank 1: $-146,355.65

Answer to blank 1: $-108,351.92

Answer to blank 1: B

Explanation:

PW of method A = - P - [A1 + G(A/G, i, n)](P/A, i, n) + F(P/F, i, n)

                           = - 90,000 - [26,000 + 3200(A/G, 14%, 3)](P/A, 14%, 3) + 16,000(P/F, 14%, 3)

                           = - 90,000 - [26,000 + 3200(0.913)](2.322) + 16,000(0.6750)

                           = -90,000 - 67,155.95 + 10,800

                           = $-146,355.65

PW of method B = - P - [A1 + G(A/G, i, n)](P/A, i, n) + F(P/F, i, n)

                           = - 106,000 - [6,000 + 6000(A/G, 14%, 3)](P/A, 14%, 3) + 36,000(P/F, 14%, 3)

                           = - 106,000 - [6,000 + 6000(0.913)](2.322) + 36,000(0.6750)

                           = -106,000 - 26,651.92 + 24,300

                           = $-108,351.92

Since the PW of method B is less than method A, therefore, Method B is used to produce expansion anchors