If produced by Method A, a product's initial capital cost will be $100,000, its annual operating cost will be $20,000, and its salvage value after 3 years will be $20,000. With Method B there is a first cost of $150,000, an annnual operation cost of $100,000, and a $50,000 salvage value after its 3-year life. Baed on a present worth analysis at a 15% interest rate, which method should be used?

The correct answer is Method A should be selected.

Explanation:

According to the scenario, computation of the given data are as follow:-

                                                   Method A                  Method B

Initial capital cost=                   $1,00,000       $1,50,000

Operating cost=          $20,000       $1,00,000

Salvage value=          $20,000       $50,000  

Present worth = -Initial capital cost - Operating cost × [( 1 + i)^n - i÷1 (1+i)^n] + Salvage value × 1÷(1+i)^n

Method A = -$100,000 - $20,000 × [(1 + 0.15)^3 - 1÷0.15 (1 + 0.15)^3] + $20,000 *(1 ÷ (1+0.15)^3

= -$100,000 - $20,000 × [1.520875 - 1 ÷ 0.228131] + $20,000 × (1 ÷ 1.520875)

= -$100,000 - $20,000 × 2.283225 + $20,000 × 0.6575

= -$100,000 - $45,664.5 + $13,150.324

= -$132,513.68

Method B = -$150,000 - $100,000 × [(1 + 0.15)^3 - 1 ÷ 0.15 (1 + 0.15)^3] + $50,000 × (1 ÷ (1 + 0.15)^3

= - $150,000 - $100,000 × [1.520875 - 1 ÷ 0.228131 ] + $50,000 × ( 1 ÷ 1.520875 )

= - $150,000 - $100,000 × 2.283225 + $50,000 × 0.6575

= - $150,000 - $228,322.5 + $32,875

= - $345,447

According to the analysis Method A will be selected because it’s show low negativity.

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