Cash flows related to three mutually exclusive capital equipment projects are given in table below

For a given MARR, the project with the highest Net Present Worth (NPW) should be selected.

5.1). Using r (MARR) = 4.13% and n (project life) = 10 years, NPW for each Alternative is:

(P/A, r, n) = 1/r*[1-(1+r)^-n]; (P/F, r, n) = 1/(1+r)^n

Alternative A NPW = -initial cost + Uniform annual*(P/A, r, n) + Salvage value*(P/F, r, n)

= -1,000 + 125*(P/A, 4.13%, 10) + 750*(P/F, 4.13%, 10)

= -1,000 + 125*[1-(1+4.13%)^-10]/4.13% + 750/(1+4.13%)^10

=-1,000 + 1,007.33 + 500.38 = 507.72

Alternative B NPW = -800 + 120*(P/A, 4.13%, 10) + 500*(P/F, 4.13%, 10)

= -800 + 120*[1-(1+4.13%)^-10]/4.13% + 500/(1+4.13%)^10

= -800 + 967.04 + 333.59 = 500.63

Alternative C NPW = -600 + 100*(P/A, 4.13%, 10) + 250*(P/F, 4.13%, 10)

= -600 + 100*[1-(1+4.13%)^-10]/4.13% + 250/(1+4.13%)^10

= -600 + 805.87 + 166.79 = 372.66

Alternative D NPW = -500 + 125*(P/A, 4.13%, 10) + 0*(P/F, 4.13%, 10)

= -500 + 125*[1-(1+4.13%)^-10]/4.13% + 0

= -500 + 1,007.33 = 507.33

At the MARR of 4.13%, Alternative A has the highest NPW so Alternative A shoule be chosen (Option A)

5.2). Again, we calculate NPW using a MARR of 21.4%:

Alternative A NPW =  -1,000 + 125*(P/A, 21.4%, 10) + 750*(P/F, 21.4%, 10)

= -1,000 + 125*[1-(1+21.4%)^-10]/21.4% + 750/(1+21.4%)^10

= -1,000 + 500.11 + 107.86 = -392.03

Alternative B NPW = -800 + 120*(P/A, 21.4%, 10) + 500*(P/F, 21.4%, 10)

= -800 + 120*[1-(1+21.4%)^-10]/21.4% + 500/(1+21.4%)^10

= -800 + 480.10 + 71.91 = -247.99

Alternative C NPW = -600 + 100*(P/A, 21.4%, 10) + 250*(P/F, 21.4%, 10)

= -600 + 100*[1-(1+21.4%)^-10]/21.4% + 250/(1+21.4%)^10

= -600 + 400.09 + 35.95 = -163.96

Alternative D NPW = -500 + 125*(P/A, 21.4%, 10) + 0*(P/F, 21.4%, 10)

= -500 + 125*[1-(1+21.4%)^-10]/21.4% + 0

= -500 + 500.11 = 0.11

At the MARR of 21.4%, Alternative D should be chosen as it has the highest NPW. (Option D)