Question

Consider the following three mutually exclusive investment project alternatives (A, B, C). Use the annual worth analysis (AW) to determine the best investment alternative assuming MARR=15%.

Project name:                                        A                    B                    C

Cash Flow (Year 0):                                    -200,000        -300,000        -400,000

Cash Flow (Year 1-N)                     +150,000       +180,000       +150,000

Salvage value at year N:                +30,000         +50,000          +60,000

Project Life (N):                                    2 years         3 years          6 years

Answer 1

AW of alternative A = -200,000(A/P, 15%, 2) + 150,000 + 30,000(A/F, 15%, 2)

                               = -200,000(0.6151) + 150,000 + 30,000(0.4651)

                               = -123,020 + 150,000 + 13,953

                               = $40,933

AW of alternative B = -300,000(A/P, 15%, 3) + 180,000 + 50,000(A/F, 15%, 3)

                               = -300,000(0.4380) + 180,000 + 50,000(0.2880)

                               = -131,400 + 180,000 + 14,400

                               = $63,000

AW of alternative C = -400,000(A/P, 15%, 6) + 150,000 + 60,000(A/F, 15%, 6)

                               = -400,000(0.2642) + 150,000 + 60,000(0.1142)

                               = -105,680 + 150,000 + 6,852

                               = $51,172

Since the AW of alternative B is highest, therefore, alternative B is the best investment alternative.