Question

Problem 6 The plant engineer is considering purchase of a solar water heating system. The firm’s MARR is 8%. The solar water heating system would save the company $900 annually in electricity costs, and would be expected to last 20 years. Maintenance will cost $150 per year. The expected salvage value for the system at the end of 20 years is $500.

a) In order for this solar water heating system to pay for itself (that is, break even), what is the maximum capital cost that the plant engineer should be willing to invest for this system?

b) If the plant engineer is able to purchase the solar water heating system for the break-even cost calculated in part (a), what is the internal rate of return for the solar water heating system?

c) If the solar water heating system can be obtained for $2,000, what is the annual equivalent worth of the solar water heating system?

d) How much additional annual worth in savings would be gained if the solar water heater could be purchased for $2,000 instead of the break-even cost calculated in part (a)? (Showing work not required)

Answer 1

a)

Maximum Capital Cost=PW of future cash flows=(900-150)*(P/A,0.08,20)+500*(P/F,0.08,20)

Let us calculate the interest factors

(P/F,0.08,20)=1/(1+0.08)^20=0.214548

So,

Maximum Capital Cost=(900-150)*9.818147+500*0.214548=$7470.88

b)

At the price calculated in part (b), NPW will be zero. So

IRR=MARR=8%

c)

If P=$2000

NPW=-Initial Cost+PV of future cash flows

NPW=-2000+7470.88=$5470.88

Annual equivalent worth=NPW/(P/A,0.08,20)=5470.88/9.818147=$557.22

d)

If solar system is purchased at break even, AW is zero.

If solar system is purchased at $2000,

Gain in Annual saving=AW of solar system=$557.22