A home-owner is trying to determine which of three devices to purchase to treat

Question

A home-owner is trying to determine which of three devices to purchase to treat water at the tap. Research has shown the following information: If the home-owner does nothing, she can invest her money in a 4.5%/year savings account. In making her determination, she believes that inflation will be 3%/year. Device #1 : Capital Cost ($) = 1000 O&M; Cost ($/yr) = 625 Life Expectancy (years) = 15 Device #2 Capital Cost ($) = 950 O&M; Cost ($/yr) = 550 Life Expectancy (years) = 5 Device #3 Capital Cost ($) = 1010 O&M; Cost ($/yr) = 582 Life Expectancy (years) = 10 (a) If she were to capitalize her costs for a 30-year life, what unit should she choose? The best answer is # 2 but can you demonstrate how to find this?

Explanation / Answer

effective interst rate

(1+r)=((1+0.045)*(1+0.03) =

r= 0.07635

life =30 years

So

present value of Cost incurred in Device 1

= 1000 + 1000/(1+r)^15 + 625*[(1+r)^30 -1]/[r*(1+r)^30] put r=0.076

= 8643.49

present value of Cost incurred in Device 2

= 950 +950/(1+r)^5 +950/(1+r)^10 ...950/(1+r)^25+ 550*[(1+r)^30 -1]/[r*(1+r)^30]

put r=0.076

=6432.98

cost of 3 is defenetely greater than 2

so since cost incurred in 2 is lowest its the best solution

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