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Let's say I make an investment with monthly deposits of 100€ (starts from 0 and add 100 every month).
After 10 years my account have 16.000€ (120 deposits of 100€).
How could I calculate the yearly average rate of this investment over the 10 years?
The interests are payed at the end of the year, and I would like to take into account the compound interests.
I have the following formula but I don't manage to isolate the rate:
$Ac = De * ( \frac{(1 + Ra)^{n+1} - 1}{Ra} - 1 )$
Ac: amount of money at the end (after the 10 years = 16.000)
De: amount of a deposit (100)
Ra: monthly rate
n: number of deposits (120)

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  • $\begingroup$ Do you make the deposits at the beginning of the month and withdraw at the end of month $120$ so the last deposit earns one month interest? $\endgroup$ – Ross Millikan Jul 31 '18 at 15:24
  • $\begingroup$ How did you get this formula? $\endgroup$ – yW0K5o Jul 31 '18 at 15:34
  • $\begingroup$ @yW0K5o I found the formula on this web site : cbanque.com/placement/versement_periodique.php $\endgroup$ – Nicolas Aug 1 '18 at 16:24
  • $\begingroup$ @Nicolas, I take a look at the website. They explain in reverse, i.e. yearly return rate is known. Here's what I understand about your question. You need to find the yearly return rate base on the total value after 10 years, do you? $\endgroup$ – yW0K5o Aug 1 '18 at 17:10
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    $\begingroup$ Sorry, it was typo in my previous comment $161*Ra+1=(1+Ra)^{121}$. WolframAlpha result is $0.00456063$ monthly or $=5.472756\%$ yearly interest. $\endgroup$ – yW0K5o Aug 1 '18 at 17:59
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For your specific case, as yW0K5o wrote in comment, you need to find the zero of function $$f(x)=161\,x+1-(1+x)^{121}\qquad \text{where} \qquad x=Ra$$ and for that, as already said, you need some numerical method.

You can have reasonable estimates considering that $x \ll1$. For example, using the binomial expansion or Taylor series, you would get $$f(x)=40 x-7260 x^2-287980 x^3+O\left(x^4\right)$$ which reduces to a quadratic equation; its positive solution is $$x=\frac{\sqrt{2041}-33}{2618}\approx 0.00465142$$ corresponding to $5.582$% which is not too bad compared to the exact solution.

Sooner or later, you will learn than, better than with Taylor series, functions can be approximated using Padé approximants. A rather simple one could be $$f(x)=\frac{40 x-\frac{11642740 }{1327} x^2} {1-\frac{50218 }{1327}x+\frac{878339 }{2654}x^2 }$$ leading to $$x=\frac{2654}{582137}\approx 0.00455906$$ corresponding to $5.471$% which is much better.

For an exact solution, Newton method would be easy. Considering the first two terms of the Taylor expansion, let us start using $x_0=\frac{40}{7260}=\frac{2}{363}$.

The iterates would be $$\left( \begin{array}{cc} n & x_n \\ 0 & 0.0055096419 \\ 1 & 0.0047266909 \\ 2 & 0.0045673223 \\ 3 & 0.0045606399 \\ 4 & 0.0045606282 \end{array} \right)$$

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  • $\begingroup$ Thank you for mentioning my idea in your post. Could you tell in more details how to use Taylor series and Newton method? $\endgroup$ – yW0K5o Aug 2 '18 at 13:53
  • $\begingroup$ @yW0K5o. To me, quoting you is normal since I started from your comment. Could you clarify your request ? Do you already about Taylor series or not ? What about numerical methods for finding the zero of a function ? I am totally ready for any discussion around these topics. $\endgroup$ – Claude Leibovici Aug 2 '18 at 13:59
  • $\begingroup$ How you get to $f(x)=40x−7260x^2−287980x^3+O(x^4)$ by using Taylor series. I am not familiar with Taylor series. $\endgroup$ – yW0K5o Aug 2 '18 at 14:04
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    $\begingroup$ @yW0K5o. In this case, it is just the binomial expansion limited to third order. Don't worry : you will learn very soon about them and I am sure that you will enjoy them. I felt in love with them 61 years ago and I use them almost every day. $\endgroup$ – Claude Leibovici Aug 2 '18 at 14:11
  • $\begingroup$ @ClaudeLeibovici, thanks for your interesting solutions! Could you explain how you calculated the iterances $x^n$ with the Newton method? $\endgroup$ – Nicolas Aug 6 '18 at 8:49
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You can't isolate the rate algebraically. You need a numerical approximation. Spreadsheets have an IRR (internal rate of return) function, or you can make a table that assumes an interest rate and computes the final balance. Then use Goal Seek to adjust the interest rate to make the final balance correct. If you have all positive inflows and one withdrawal at the end there will be only one interest rate that works. If you have a mix of inflows and outflows along the way there can be more than one.

Assuming deposits at the beginning of the month and withdrawal at the end, so the last deposit earns one month interest, I find an annual interest rate of about $5.473\%$ works.

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  • $\begingroup$ Thanks for your answer, I hoped there was a formula to calculate it. I will try as you suggest with a table, but that seems fastidious and not very accurate. $\endgroup$ – Nicolas Aug 1 '18 at 16:26
  • $\begingroup$ If you are using a numerical method that needs a starting value, you can just do simple interest. Here you have on average $6000$ on deposit for $10$ years and earned $4000$. The rate would be $\frac {4000}{6000\cdot 10}\frac 1{15}\approx 6.6667\%$. The actual rate will be somewhat smaller due to compounding $\endgroup$ – Ross Millikan Aug 1 '18 at 16:32
  • $\begingroup$ @RossMillikan, it look like I get the same result: 5.472756% $\endgroup$ – yW0K5o Aug 1 '18 at 18:46

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