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I want to calculate the payoff time for a photovoltaic system. Some constants:

Current electricity price per kWh: 0,122 EUR
Electricity production per yearh: 4427 kWh
Annual electricity price increase: 5%
Feed rate into the power grid: 68%
Costs: 9.080 EUR

How can I calculate the payoff time? I tried it with this formula but failed:

n = log(Kn/Ko)/log(q)

Kn = 9080
Ko = 0,122 * 4427 * 0,68
q = 1,05

The result is ~66 years, but it should be 17. Is the compound interest the wrong way?

testing
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1 Answers1

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The costs are fix costs. They are payed at t=0. Interest rates and the inflation rate are not taken into account.

The profit after n years is

$P=\left( \text{electricity per year [kWh]} \right) \times \left( \text{feed rate}\right) \times \left( \text{price per kWh}\right) \times \left( \text{future value annuity factor}\right)-\text{fix costs}$

$=4427\cdot 0.68\cdot 0.122\cdot \frac{1-1.05^n}{1-1.05}-9080 \geq 0$.

Let $4427\cdot 0.68\cdot 0.122=x$ and $9080=y$. The equation becomes

$x\cdot \frac{1-1.05^n}{1-1.05}-y=0 \Rightarrow x\cdot \frac{1-1.05^n}{1-1.05}=y$

multiplying both sides by $(1-1.05)$

$ x\cdot (1-1.05^n)=y(1-1.05)$

$ (1-1.05^n)=\frac{y(1-1.05)}{x}$

$ 1.05^n=1-\frac{y(1-1.05)}{x} $

Taking logs

$n\cdot log(1.05)=log\left(1-\frac{y(1-1.05)}{x} \right)$

$n=\frac{log\left(1-\frac{y(1-1.05)}{x} \right)}{log(1.05)}$

If you insert the corresponding values, you get $\boxed{n\approx 16.49}$.

Since n is defined for natural numbers and you want to have no loss the result has to be rounded up.

callculus42
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  • Thanks for your answer. I have some questions to your calculation: How do you come to $ \frac{1-1.05^n}{1-1.05} $? When you take the log from $ 1-\frac{y(1-1.05)}{x} $ what happens with the negative sign? If I insert the corresponding values I get a negative result, but the log can't be applied to a negative value. – testing Aug 14 '15 at 20:15
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    You´re are on the right track. You have to take the log of $1-\left( \frac{y(1-1.05)}{x}\right)$ (I had a typo). If you replace the minus sign in front of the fraction by a plus sign, than ONE of the factors of the fraction has to be multiplied by $(-1)$--- in other words the sign of one factor has to be turned around. We take $(1-1.05)$ Multiplied by (-1), we get $-(1-1.05)=-1+1.05=1.05-1=0.05$ The term for log is $1+\frac{y(0.05)}{x}$ It is still positive. – callculus42 Aug 14 '15 at 21:45
  • Thanks, now I get your result! Can I ask you some more questions? How do you come to $ \frac{1-1.05^n}{1-1.05} $? How is this defined? Can you explain that in more detail? I also tried to calculate it with the cummulative method for the calculation of the payoff time. I used $(AW – ZÜ 1) / (ZÜ 2 – ZÜ 1)$ from here and the result is slightly more different. 16,49438602 is yours vs. 16,488288877 from the other method. See here. It's a small difference. – testing Aug 17 '15 at 18:27
  • @testing Sure you can ask more questions. I have to go shopping now, the supermarkets are closing soon. I wil be back in round about one hour. Your linked website is difficult to understand for me, because I´m a german :-) – callculus42 Aug 17 '15 at 18:35
  • No, problem. Thanks! Supermarket open till 9 p.m.? Nice. I wanted to translate first, but then I saw your profile :-D – testing Aug 17 '15 at 18:37
  • @testing Right. Good idea to look at my profile first. – callculus42 Aug 17 '15 at 18:38
  • @testing The formula is for the partial sum of a geometric series (link : https://en.wikipedia.org/wiki/Geometric_series ). And if a constant value is compounded every period on more time, than you have this. Now to the difference of your and my result. It is written on the website, that it is "only" a linear $\texttt{interpolation}$. That is why the difference. But this is ok. The differnce is really small. But be careful: By using the Amortisationsrechnung you do not consider the compounding. – callculus42 Aug 17 '15 at 19:15
  • @testing I meant " And if a constant value is compounded every period $\color{red}{\text{one}}$ more time, than ... – callculus42 Aug 17 '15 at 19:27
  • Now I had the time to read the article. What I haven't fully understood is how the geometric series correspond to something like interest. The geometric series represents a value which is getting smaller and smaller and the interests are "growing". Even if I sum up all fractions this doesn't fully make sense to me, because you would have to take the value before into account. Perhaps you could bring some light into this for me :-) – testing Aug 24 '15 at 21:11