Question

A university spent $1.7 million to install solar panels atop a parking garage. These panels will have a capacity of 900 kilowatts (kW) and have a life expectancy of 20 years. Suppose that the discount rate is 30%, that electricity can be purchased at $0.10 per kilowatt-hour (kWh), and that the marginal cost of electricity production using the solar panels is zero. Hint: It may be easier to think of the present value of operating the solar panels for 1 hour per year first. Approximately how many hours per year will the solar panels need to operate to enable this project to break even?

A. 5,696.67 B.2,278.67 C.7,405.67

If the solar panels can operate only for 5,127 hours a year at maximum, A. would B. would not  the project break even

. Continue to assume that the solar panels can operate only for 5,127 hours a year at maximum.

In order for the project to be worthwhile (i.e., at least break even), the university would need a grant of at least A. 170,00.66 B. 68,00.26 C. 221,000.86

Answer 1

Solution :

Given :

Interest rate (i) = 30%,

Time (t) = 20 years

Initial cost = 1.7milliom or 1,700,000

(1) :- Let no. of hour of operation = h per year

Total savings per year = 900 × 0.1 × h

For breakeven,

1,700,000 = 90× h × (P/A, 30%, 20)

1,700,000 = 90 × h × 3.315794

h = 1,700,000 / (90 × 3.315794) = 5696.67 hours

So,.Option A is correct.

(b) :- As Breakeven hours (5696.67) > 5,127 hours, so the project will not breakeven

(c) :- Total saving per year for 5127 hours of operation = 900 × 0.1 × 5127 = 461,430

Present value of savings = 461,430 × (P/A, 30%,20) = 461,430 × 3.315794 = 1,530,006.85

Grant required = 1,700,000 - 1,530,006.85 = 170,000.66

So, Option A is correct.

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