Question

4. (10 points) A firm is considering three mutually exclusive alternatives as part of a production improvement program. The alternatives are: Initial Cost $20,000 $30,000 $50,000 Uniform Annual Benefit $4,000 $5,000 $6,500 Useful Life Salvage Value $2,000 $9,000 The MARR is 10%. Which alternative do you recommend? Be sure to use the proper technique when comparing alternatives with different useful lives.

Answer 1

4.

When alternatives are of different useful lives, then most suitable method is to find Equivalent uniform Annual worth (EUAW) of each alternative. Any alternative with highest positive EUAW will be selected as the best alternative.

R = 10%.

EUAW for alternative A = (-20000 + 4000*(P/A, 10%, 10) )(A/P, 10%, 10)

EUAW for alternative A = (-20000 + 4000*6.1446)*.1627

EUAW for alternative A = $744.91 or $745

==

EUAW for alternative B = (-30000 + 5000*(P/A, 10%, 8) + 2000*(P/F, 10%, 8))*(A/P, 10%, 8)

EUAW for alternative B = (-30000 + 5000*5.3349 + 2000*.4665)*.1874

EUAW for alternative B = -448.35 or -$448

==

EUAW for alternative C = (-50000 + 6500*(P/A, 10%, 20) + 9000*(P/F, 10%, 20))*(A/P, 10%, 20)

EUAW for alternative C = (-50000 + 6500*8.5136 + 9000*.1486)*.1175

EUAW for alternative C = $784.41 or $784

Since EUAW of alternative C is highest among all the three alternatives, then alternative C is best and it should be selected.