A university spent $2 million to install solar panels atop a parking garage. These panels will have a capacity of 300 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. Approximately how many hours per year will the solar panels need to operate to enable this project to break even?

a. 6,371.11
b. 4,900.85
c. 2,450.43
d. 5,881.02

10,053 Hours

Explanation:

The total capacity of Solar panels = 300 KW

Electricity cost per KWh = $0.20

Savings per hour of operation per year = 300KW * $0.2 = $60 per year

Now we will compound this benefit for 20 years:

Present Value of Savings = [$60/(1+0.3)] + [$60/(1+0.3)^2] +  [$60/(1+0.3)^3] +  [$60/(1+0.3)^4] + [$60/(1+0.3)^20]

Present Value of Savings for 1 hour = $198.948

Now

Break-even = Fixed Cost / PV of operating 1 hour per year = $2,000,000 / $198.948 per hour = 4,900.85 hours

The number of Hours per year the solar panels must operate to breakeven the situation is as under:

Break-even = Fixed cost / PV of operating 1 hour per year

= 1,300,000 / 265.64 = 10,053 Hours

answer; ///(c) i believe its it ignores cash flows following the payback period;

start by isolating the variable on one side by adding 43w^2 to both sides.   the equation will read:

25w^2+5=0

*subtract 5 from both sides*

25w^2=-5

*divide both sides by 25*

w^2=-1/5

*take the square root of both sides*

w=√(-1/5)

the square root of -1 is i.   the square root of 5 remains √5

the answer is i/√5.

hope this