Question

A university spent $1.3 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 $0.20 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.

1. Approximately how many hours per year will the solar panels need to operate to enable this project to break even?

3,003.31

3,337.01

4,338.11

1,668.51

2. If the solar panels can operate only for 3,003 hours a year at maximum, the project would or would not break even.

Continue to assume that the solar panels can operate only for 3,003 hours a year at maximum.

3. In order for the project to be worthwhile (i.e., at least break even), the university would need a grant of at least 130,121.29 or 182,169.81 or 65,060.65

Please show your work!

Answer 1

1)

AW of costs of project=1300000*(A/P,0.20,20)

Let us calculate the annual worth of cost of project

AW of costs of project=1300000*0.20535653=$266963.49

Let plant operates x hours per year

Annual savings =x*400*0.20=80x

In break even

Annual savings=Annual worth of costs

80x=266963.49

x=266963.49/80=3337.04

Closest option is

3337.01

2)

Since maximum capacity is less than breakeven quantity.

Project would not break even

3)

Present value of savings=3003*400*0.2/(A/P,0.20,20)

Present value of savings=3003*400*0.2/0.20535653=$1169867.84

Subsidy required=1300000-1169867.84=$130132.16

Closest option is

130,121.29