Question

A university spent $1.8 million to install solar panels atop a parking garage. These panels will have a capacity of 400 kilowatts (kW) and have a life expectancy of 20 years. Suppose that the discount rate is 20%, 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. a.3,696.48

b.14,785.92

c.9,241.20

If the solar panels can operate only for 8,317 hours a year at maximum, the project (would/would not)break even?

Continue to assume that the solar panels can operate only for 8,317 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.126,010.32

b.270,022.11

c.180,014.774

d.144,011.79

Answer 1

Discount rate = 20% = 0.2

Looking at one hour of operation in each year = 400 kW x 0.10 $ Kw/hr = 40$ value of electricity per year

Present value = [40/(1+0.2)] + [40/(1+0.2)²] +  [40/(1+0.2)³] +  [40/(1+0.2)⁴] + ------------------------- [40/(1+0.2)²⁰]

PV = 194.76$

a) For break even it would need to run = (1.8 million)/ 194.76 = 9242.145 hours per year

Answer option C

b) So from the above data the break even is 9241 hours per year so 8317 hours effecient system is not productive.