Question

3. 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

Cashflow/yr. $1450 $1600

Useful Life. 3. 5

Salvage Value $500. $500

Answer 1

Given,

MARR = 10% per year

Project A:

Investment = -$5,000

Cashflow per year = $1,450

Useful life = 3 years

Salvage value = $500

From the compound interest table, we obtain

(A/P, 10%, 3) = 0.4021

(A/F, 10%, 3) = 0.3021

Equivalent annual worth of project A = AW of the Investment + Annual Cashflows + AW of the Salvage Value = -$5,000(A/P, 10%, 3) + $1,450 + $500(A/F, 10%, 3) = -$5,000*0.4021 + $1,450 + $500 * 0.3021 = -$409.45

Project B:

Investment = -$6,000

Cashflow per year = $1,600

Useful life = 5 years

Salvage value = $500

From the compound interest table, we obtain

(A/P, 10%, 5) = 0.2638

(A/F, 10%, 5) = 0.1638

Equivalent annual worth of project B = AW of the Investment + Annual Cashflows + AW of the Salvage Value = -$6,000(A/P, 10%, 5) + $1,600 + $500(A/F, 10%, 5) = -$6000*0.2638 + $1,600 + $500*0.1638 = $99.1

From the above analysis, it can be observed that equivalent annual worth of project B is higher than that of project A. (Also note that the equivalent AW of project A is negative)

Hence, project B is more desirable.