A university spent $1.3 million to install solar panels atop a parking garage. These panels will have a capacity of 200 kilowatts (kW) and have a life expectancy of 20 years. Suppose that the discount rate is 30%, that electricity can be purchased at $0.30 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?

It will take 6,534.31 hours per year for the solar panels to operate to enable this project to break even

Explanation:

Discount rate = 30% = 0.3

Looking at one hour of operation in each year = 200 kW x $0.30 Kw/hr

= $60 value of electricity per year

Compound interest factor for a discount rate of 30% = 3.3158

(taken from compound interest factor table or computed using formula ∑1/(1+r)^t , where r = 30%, and t = 1 to 30)

Present value of operating the solar panels for 1 hour per year = 60 × 3.3158 = $ 198.95

For break even it would need to run = 1.3 million ÷ 198.95

= 6,534.31 hours per year

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