Question

The following data have been estimated for two mutually exclusive investment alternatives, A and B,
associated with a small engineering project for which revenues as well as expenses are involved. They
have useful lives of 3 and 5 years, respectively. If MARR = 10% per year, show which alternative is more
desirable by using equivalent-worth methods. Use the repeatability assumption.

Project	A	B
Investment	$5,000	$6,000
Cash flow/yr	$1,450	$1,600
Usual Life	3	5
Salvage value	$500	$500

Answer 1

Alternative A

PW of salvage value= 500/(1+10%)^3=500/1.1^3=500/1.331=$375.66

Net investment=5000-375.66=$4624.34

Annual cost of $4624.34 for 3 years at 10% will be given by

PW=A*(1-(1+r)^-n)/r

Or, 4624.34=A*(1-(1+10%)^-3)/10%

Or, 4624.34=A*(1-1.1^-3)/0.1

Or, 4624.34=A*(1-0.7513)/0.1

Or, A=4624.34*0.1/0.2487

Or, A=1859.52

Hence Net annual worth = 1450-1859.52=-$409.52

Alternative B

PW of salvage value= 500/(1+10%)^5=500/1.1^5=500/1.6105=$310.46

Net investment=6000-310.46=$5689.54

Annual cost of $5689.54 for 5 years at 10% will be given by

PW=A*(1-(1+r)^-n)/r

Or, 5689.54=A*(1-(1+10%)^-5)/10%

Or, 5689.54=A*(1-1.1^-5)/0.1

Or, 5689.54=A*(1-0.6209)/0.1

Or, A=5689.54*0.1/0.3791

Or, A=1500.89

Hence Net annual worth = 1600-1500.89=$99.11

Hence alternative B annual worth is more and is more desirable