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Trying to make the problem as simple as possible.

I decide to start on January $1^{\text{st}}$ a saving plan for which the bank will give me a constant interest rate $r$ (the interests being paid on December $31^{\text{st}}$).

During a given year $y$, I shall make a constant monthly deposit $d_y$.

But, this deposit will change from year to year; for example $$d_{y+1}=d_y +\delta\qquad \text{or}\qquad d_{y+1}=d_y \,(1+\delta)\qquad \text{with}\qquad \delta>0$$

What could be the formula of the present value after $n$ years ?

Claude Leibovici
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  • Consider the first year. The interest is paid at the end of the year, the rate being $r$. The monthly deposit (paid on the first day of the month) is $d_0$. Is it so that, at the end of the year, the interest takes only into account the amount on the account at the moment of the interest payment? In other words, at the end of the year, is it so that the interest is applied to the amount of $12d$? Or does the interest take into account some kind of "average amount on the account through the year"? If it's the first option, the amount at the end of the first year is $$ 12 r d_0 $$ – Matti P. Feb 05 '19 at 06:30
  • At the end of the first year, the amount on the account is (after applying the interest) $$ 12 rd_0 $$ After two years, the amount is $$ (12rd_0 + 12d_1)r = 12r^2 d_0 + 12r d_1 $$ ... maybe you can continue from here? – Matti P. Feb 05 '19 at 06:35
  • @MattiP. I must confess that I am very bad with finance (and many other areas, for sure). What, in your opinion, would be the simplest and the most complex situations ? I know the solution for $\delta=0$. In any manner, thanks for answering. Cheers ;-) – Claude Leibovici Feb 05 '19 at 06:41

1 Answers1

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I assume you insert the cash at the end of the year.

The result is $$FV=\frac{d_1}{r-\delta}((1+r)^n-(1+\delta)^n)$$

This is done as follows. We have

$$\forall y\geq 1:d_y=(1+\delta)^{y-1}d_1$$

The cash flow is thus $$d_1,d_2=(1+\delta)d_1,d_3=(1+\delta)^2d_1,\dots$$

So, present value of the cash flow would be $$PV=\sum\limits_{y=1}^n\frac{d_y}{(1+r)^{y}}=\sum\limits_{y=1}^n\frac{d_1(1+\delta)^{y-1}}{(1+r)^{y}}=\frac{d_1}{r-\delta}\Big(1-\frac{(1+\delta)^n}{(1+r)^n}\Big)$$

Using sum of geometric sequence. The future value is $$FV=(1+r)^nPV$$

I further explain. For year $y$ the payment $d_y$ gets $n-y$ years of interest. So, if we only look at what happens with this sum of money we get the following $$d_y,(1+r)d_y,(1+r)^2d_y,\dots,(1+r)^{n-y}d_y$$

Don Fanucci
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  • Thanks for answering ! If I am not mistaken, the total amount of my deposits have been $$d_1 \times \frac{(1+\delta) \left((1+\delta)^n-1\right)}{\delta }$$ Is this correct ? – Claude Leibovici Feb 05 '19 at 07:02
  • @ClaudeLeibovici It is almost correct you don't have the $(1+\delta)$ factor at the beginning you sum $\sum\limits_{y=1}^nd_y=d_1\sum\limits_{y=1}^n(1+\delta)^{y-1}=d_1 \times \frac{\left((1+\delta)^n-1\right)}{\delta }$. But, observe that by doing so you forget about the "time value of money", this is, in finance we do not care about the sum of the components of the cash flow, because of the interest (money today is not like money in a year from now), so it is meaningless, in a sense, to consider this sum. – Don Fanucci Feb 05 '19 at 07:31
  • Thanks again. Now, I shall be able to play with this material. To tell the truth, what I want to do is to approximate $r$ knowing $FV$, $\sum_{y=1}^n d_y$ and $\delta$. I am a fanatic lover of function approximations. Cheers. – Claude Leibovici Feb 05 '19 at 07:46
  • I have the feeling that there is something I do not understand. Make $\delta\to r$ and then $FV\to -n (1+r)^{n-1},d_1 < 0$ – Claude Leibovici Feb 05 '19 at 11:21
  • @ClaudeLeibovici you are right. I switched $r-\delta$ with $\delta-r$ in the original solution (a mistake in calculating the geometric sum). Fixed it now. – Don Fanucci Feb 05 '19 at 11:31
  • Thanks to you, I have a lot of fun since, now, knowing the ratio $\frac{FV}{\sum_{y=1}^n d_y}$ and $\delta$ neither $n$ or $r$ can be computed explicitly. – Claude Leibovici Feb 05 '19 at 15:54