Question

A company is considering 3 mutually exclusive alternatives along with "Do-Nothing" as part of a new quality improvement initiative. The alternatives are in the table below. For each alternative, the salvage value at the end of the useful life is zero. At the end of 10 years, Alternative Z can be replaced by another Z with identical costs and benefits. If the MARR is 6.5 %, and the analysis period is 20 years, which alternative should be selected?

	X	Y	Z
Installed Cost	$15,000	$20,000	$10,000
Uniform Annual Benefit	$1,625	$1,890	$1,625
Useful Life, in years	20	20	10
MARR	6.50%	6.50%	6.50%

Answer 1

NPW of X = -15000 + 1625*(P/A,6.5%,20)

= -15000 + 1625*(((1 + 0.065)^20-1)/(0.065*(1 + 0.065)^20))

= -15000 + 1625*(((1.065)^20-1)/(0.065*(1.065)^20))

= -15000 + 1625*11.018507

= 2905.07

NPW of Y = -20000 + 1890*(P/A,6.5%,20)

= -20000 + 1890*(((1 + 0.065)^20-1)/(0.065*(1 + 0.065)^20))

= -20000 + 1890*(((1.065)^20-1)/(0.065*(1.065)^20))

= -20000 + 1890*11.018507

= 824.98

NPW of Z = -10000 + 1625*(P/A,6.5%,20) - 10000*(P/F,6.5%,10)

= -10000 + 1625*(((1 + 0.065)^20-1)/(0.065*(1 + 0.065)^20)) - 10000*((1 + 0.065)^-10)

= -10000 + 1625*(((1.065)^20-1)/(0.065*(1.065)^20)) - 10000*((1.065)^-10)

= -10000 + 1625*11.018507 - 10000*0.532726

= 2577.81

As NPW of X is highest, it should be selected