Question

use present worth (PW) to select the best alternative among A, B, C and D, Alternative A : Capital investment = $8000, Salvage value= $5500, Total annual revenues = $4850, total annual expenses $ 2000. Alternative B: Capital investment = $10,000, Salvage value= $3700, Total annual revenues = $5000,total annual expenses $1500 , Alternative C: Capital investment = $12,000, Salvage value= $4500, Total annual revenues = $6150, total annual expenses $1150, Alternative D: Capital investment = $13,000, Salvage value= $4000, Total annual revenues = $6750, total annual expenses $1000 , Assume that MARR is 10% per year compounded annually and useful life for all alternatives is 8 years .

Answer 1

NPW of A = -8000 + 5500*(P/F,10%,8) + (4850-2000)*(P/A,10%,8)

= -8000 + 5500*0.466507 + (4850-2000)*5.334926

= 9770.33

NPW of B = -10000 + 3700*(P/F,10%,8) + (5000-1500)*(P/A,10%,8)

= -10000 + 3700*0.466507 + (5000-1500)*5.334926

= 10398.32

NPW of C = -12000 + 4500*(P/F,10%,8) + (6150-1150)*(P/A,10%,8)

= -12000 + 4500*0.466507 + (6150-1150)*5.334926

= 16773.91

NPW of D = -13000 + 4000*(P/F,10%,8) + (6750-1000)*(P/A,10%,8)

= -13000 + 4000*0.466507 + (6750-1000)*5.334926

= 19541.83

As NPW of D is highest, it should be selected