Answer:
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.
