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In every month, we put ex.: 20 000 "money" to the bank = 240 000 yearly.

In every year the goverment gives 72 000 "money" to our bank account.

This goes for 4 years = (4 * 12 * 20 000) + (4 * 72 000) = 960 000 + 288 000 = 1 248 000

In the end, we will get: 1 248 000

The big questionHow can I count that how much "profit percentage" did I get from this per year?

If I try (BAD!), then:

  • In 4 year I put in 4 * 12 * 20 000 = 960 000
  • At the end I got back = 1 248 000
  • So the profit for 4 years look like: 1 248 000 - 960 000 = 288 000
  • So 288 000 is the 30% of 960 000 for 4 years. But I want it yearly.
  • So 30% / 4 = 7,5% yearly profit?

Lets reverse count to make sure:

  • First year, I put in 12 * 20 000 = 240 000, so 240 000 * 1,075 = 258 000
  • Second year, I already have 258 000 + current year put in 240 000, so 498 000 * 1,075 = 535 350
  • Third year, I already have 535 350 + current year put in 240 000, so 775 350 * 1,075 = 833 501
  • Fourth year, I already have 833 501 + current year put in 240 000, so 1 073 501 * 1,075 = 1 154 013 .. this is not 1 248 000... what am I missing?
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7.5 is not so far off.

If you want to calculate percent per year if we pretend it is a fixed interest rate then it is

$$\sqrt[4]{\frac{(240000+72000)\times 4}{240000\times 4}}=1.0678$$

or 6.78%, quite close to 7.5%

It's gonna be lower every year since it is fixed amount of money.


That your approximation almost works can be explained by Taylor expanding for example $(1+k)^x$ to linear equation (first order polynomial) for small values of $k$.

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  • $\begingroup$ wow, thx, but I counted 4 years with 7.5% and got 1 154 013, less than 1 248 000. How can less percentage be good if 7.5% is smaller already than 1 248 000? what am I missing? $\endgroup$ – KollarA Jun 8 '18 at 8:06
  • $\begingroup$ can you please correct me if I am wrong? :) or anybody else? Many thanks $\endgroup$ – KollarA Jun 12 '18 at 9:30
  • $\begingroup$ Sorry I have forgotten how to explain things like this. You want to check out en.wikipedia.org/wiki/Geometric_series and geometric sum. I can maybe write a better answer later after work. $\endgroup$ – mathreadler Jun 12 '18 at 11:01

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