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This is used for calculating monthly costs that increase by a certain percentage month-over-month when given a forecasted annual budget.

An annual increase of 15% annually or 1.28% monthly (equalling 15% annually) is expected.

Given that I know the target value (a sum of all the months) to be, $100,000, how would I calculate the initial value?

Final value: $100,000
Period growth rate: 1.28%
Periods: 11
Initial value: X

Is there a formula for this use case?

Artem Yevtushenko
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  • How do get that that $1.28%$ monthly is equivalent to $15%$ yearly? Give a reply if you are really interested in the question. – callculus42 Oct 21 '22 at 05:16
  • @callculus42 I used 15^(1/11) as this is how many growth periods I can account for - not quite annual. 15^(1/12) may also be used. I hope the solution should be able to generalize regardless. What I'm really curious about is 1. the mathematical notation/form used to describe the scenario and 2. how to describe the form in mathematical terms. I'll look into the response from John Forkosh and your suggestion of finite geometric series. ty kindly. – Artem Yevtushenko Oct 25 '22 at 03:43

1 Answers1

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Not seeing a general formula in Gradshteyn and Ryzhik, etc. But it's easy enough to formulate generally: let $V_f=\$100,000$ be your final value, $p=0.0128$ your $1.28$% growth rate, $n=11$ your number of periods, and $V_0$ your initial value to be determined. Then $$V_f = \sum_{i=1}^n V_0(1+p)^{i-1} = V_0\sum_{i=1}^n (1+p)^{i-1}$$ so you just need a general formula for $\sum\limits_{i=0}^{n-1} x^i$ (where $x$ is our $1+p$), which is what I'm not seeing. But it's trivial to program, and given your postings on stackoverflow, I'd guess extremely trivial for you.

John Forkosh
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  • It does exist a closed form for the sum. – callculus42 Oct 21 '22 at 05:39
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    @callculus42 So you think you might want to actually post it (as an answer), and preferably along with the source you're referencing??? And please "Give a reply if you are really interested in the question". – John Forkosh Oct 21 '22 at 05:41
  • At the moment, I don't want to post an answer, since I don't know the OP comes from 15% yearly to 1.28% monthly. – callculus42 Oct 21 '22 at 06:09
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    @callculus42 He presumably made a small numerical miscalculation of $1.15^{1/12}$. And are you sure you actually have an answer? I pretty carefully checked, and there doesn't seem to be a general closed-form representation of $\sum\limits_{i=1}^n x^i$. Personally, that's all I want to see, if it exists. Couldn't care less about the monthly/yearly thing. And that's totally irrelevant with respect to the general $\sum\limits_{i=1}^n x^i$ question. – John Forkosh Oct 21 '22 at 06:26
  • That's why I asked the OP for a reply. // The key word is finite geometric series. – callculus42 Oct 21 '22 at 06:31
  • @callculus42 Okay, thanks. How'd I miss that??? :) I'll leave you to post the answer, rather than editing mine. – John Forkosh Oct 21 '22 at 06:33