Question

Two methods can be used to produce expansion anchors. Method A costs $80,000 initially and will have a $14,000 salvage value after 3 years. The operating cost with this method will be $24,000 in year 1, increasing by $2800 each year. Method B will have a first cost of $104,000, an operating cost of $10000 in year 1, increasing by $10000 each year, and a $34,000 salvage value after its 3-year life. At an interest rate of 12% per year, which method should be used on the basis of a present worth analysis? The present worth for method A is $ . The present worth for method B is $ .

Answer 1

Working note:

In year 3 (terminal year), Net cost = Operating cost in year 3 - Salvage value (for both methods)

PW of costs, Method A = 80,000 + 24,000 x P/F(12%, 1) + 26,800 x P/F(12%, 2) + (26,800 + 2,800 - 14,000) x P/F(12%, 3)

= 80,000 + 24,000 x 0.8929** + 26,800 x 0.7972** + 15,600 x 0.7118**

= 80,000 + 21,429.6 + 21,364.96 + 11,104.08

= 133,898.64

PW of costs, Method B = 104,000 + 10,000 x P/F(12%, 1) + 20,000 x P/F(12%, 2) + (20,000 + 10,000 - 34,000) x P/F(12%, 3)

= 104,000 + 10,000 x 0.8929** + 20,000 x 0.7972** - 4,000 x 0.7118**

= 104,000 + 8,929 + 15,944 - 2,847.2

= 126,025.80

Since Method B has lower PW of costs, method B should be selected.

NOTE: If negative value has to be entered (as PW of costs), enter above values with a minus sign.

**From P/F Factor table