Question

Two mutually exclusive alternatives have the estimates shown below. Use annual worth analysis to determine which should be selected at an interest rate of 13% per year.

	Q	R
First Cost	$-40,000	$-80,000
AOC per Year	$-9,000	$-10,000 in year 1, increasing by $1,000 per year thereafter
Salvage Value	$1,000	$8,000
Life	2 years	4 years

Alternative  (Click to select)  Q  R  should be selected.

Answer 1

I = 13%

1) Q:

Pw = first cost + aoc per year(p/a,i,n) + salvage value(p/f,i,n)

pw = -40,000 - 9,000(p/a,13%,2) + 1,000(p/f,13%,2)

pw = -40,000 - 9,000 * 1.668 + 1,000 * 0.7831

pw = -40,000 - 15,012 + 783.1

pw = -54,228.9

aw = pw(a/p,i,n)

aw = -54,228.9(a/p,13%,2)

aw = -54,228.9 * 0.5995

aw = - 32,510.22

2) R:

pw = first cost + aoc per year(p/a,i,n) + increase in aoc per year(p/g,i,n) + salvage value(p/f,i,n)

pw = -80,000 - 10,000(p/a,13%,4) - 1,000(p/g,13%,4) + 8,000(p/f,13%,4)

pw = -80,000 - 10,000 * 2.974 - 1,000 * 4.009 + 8,000 * 0.6133

pw = -80,000 - 29,740 - 4,009 + 4,906.4

pw = -108,842.6

aw = pw(a/p,i,n)

aw = - 108,842.6(a/p,13%,4)

aw = -108,842.6 * 0.3362

aw = -36,592.88

so Q will be selected because it has the highest annual worth that is -32,510.22 > -36,592.88