Question

The following data is available for three different alternatives. Assume an interest rate of 9% per year, compounded annually. Initial Cost Annual Benefit Useful Life (yrs) Alternative A 7,000 1,275 infinite Alternative B 9,400 1,327 4 Alternative 14,000 4,867 17 1 Alternatives B and Care replaced at the end of their useful lives with identical replacements. Using present worth analysis, find the best alternative. Choose Alternative A because it lasts the longest Choose Alternative A because its net present worth is postive Choose Alterntive C because it has the highest annual benefit Choose Alternative C because its net present worth is $15,999.99 more than its nearest competitor

Answer 1

Alternative A

Initial Cost = 7000

Annual benefits = 1275

Net present worth = -Initial Cost + (Annual benefits ÷ Interest)

= -7000 + (1275 ÷ 0.09) = 7166.67

Alternative B

Initial Cost = 9400

Annual benefits = 1327

Life = 14 years

Net present worth = - (1327 - 9400 )(A/P, 9%, 14)(1/9%) = -11517.48

Alternative C

Initial Cost = 14000

Annual benefits = 4867

Life = 5 years

It is to be repeated two times

Calculate PW

PW = (4867 - 14000) (A/P,9%,7)(1/9%)

PW = -9133(0.1987)(1/9%) = -20163.63

So, choose alternative A as it is having a positive net present worth.

Option B.