MARR = 20%
Present worth of Gate 1 = -15000 - 6500 * (P/A, 20%,5)
= -15000 - 6500 * 2.990612
= -34438.98
Present worth of Gate 2 = -19000 - 5600 * (P/A, 20%,5) + 2000 * (P/F, 20%,5)
= -19000 - 5600 * 2.990612 + 2000 * 0.401878
= -34943.67
Present worth of Gate 3 = -24000 - 4000 * (P/A, 20%,5) + 5000 * (P/F, 20%,5)
= -24000 - 4000 * 2.990612 + 5000 * 0.401878
= -33953.06
As the present cost of gate 3 is lowest, it should be selected
The management of Brawn Engineering is considering three alternatives to satisfy an OSHA requirement for safety...
A firm is considering three mutually exclusive alternatives as part of an upgrade to an existing transportation network. At EOY 10, alternative III would be replaced with another alternative Ill having the same installed cost and net annual revenues. If MARR is 10% per year, which alternative (if any) should be chosen? Use the incremental IRR procedure. $40,000 $6,500 $20,000 $5,200 Installed cost Net annual revenue Salvage value Useful life Calculated IRR $30,000 $5,600 0 20 years 18.0% 20 years...