Hello there i have the following problem and i would be very thankful if someone could help me out a little:
So i have to calculate the value of an investment in t years. I also want to be able to add money inbetween in intervals (every month, every 2 months, every 4 months and once a year) The parameters i have are:
Initial investment $Inv$ ... values like 100.000
interest $i$ ... values like 1,03
reinvestment money $R$ ... values like 100
factor by how much i want to raise my reinvested money every year $f$ ... values like 1,02 etc.
How often to invest $R$ again per year $u$ ... possible values:(12,3,2,1)
Number of years i want to invest for $t$ ... values like 10
You have to be careful because for example if $ u = 12$ and $t = 1$ then the first payment gets interest for 12 months the second for 11 and so on. I figured out for 1 year you can say $(u+1)/2$ gets full interest and $(u-1)/2$ gets none at all. I have figured out an explicit formula for the payment of the current year with the geometric series. But my formula for the final value ist still recursive and i want to get it explicitly. So far i have:
$value_n= (value_{n-1} + R*(1-f^n)/(1-f)*(u+1)/2)*i + R*(1-f^n)/(1-f)*(u-1)/2$
and i hope it is alright so far.
Wolfram alpha didnt help either. I would be super happy if someone helped me solve it or at least told me its not possible.
Thank you!
Without the increase of the re-investment R we have for the one year the formula
$$R\cdot \frac{\left(1+\frac{i}m \right)^m-1}{\frac{i}m}$$
This is the value after one year if we invest an amount of $R$ monetary units $m$ times a year. To get the future value (in $n$ years) we have to compound it $m\cdot (n-1)$ times with the interest rate of $\frac{i}m$.
Then the total future value is
$S_n=R\cdot \frac{\left(1+\frac{i}m \right)^m-1}{\frac{i}m}\cdot \left(1+\frac{i}m \right)^{m\cdot (n-1)}+R\cdot 1.02\cdot \frac{\left(1+\frac{i}m \right)^m-1}{\frac{i}m}\cdot \left(1+\frac{i}m \right)^{m\cdot (n-2)}+R\cdot 1.02^2\cdot \frac{\left(1+\frac{i}m \right)^m-1}{\frac{i}m}\cdot \left(1+\frac{i}m \right)^{m\cdot (n-3)}+\ldots$
Let´s substitute some constant terms. $T=\frac{\left(1+\frac{i}m \right)^m-1}{\frac{i}m},Z=\left(1+\frac{i}m \right)^{m}$ and $q=1.02$. Then it becomes
$$S_n=R\cdot T\left[q^0\cdot Z^{n-1}+q^1\cdot Z^{n-2}+q^2\cdot Z^{n-3}+\ldots +q^{n-2}\cdot Z^1+q^{n-1}\cdot Z^0 \right]$$
In the brackets we have $n$ summands. Dividing both sides by $q^{n-1}$
$\frac{S_n}{q^{n-1}}=R\cdot T\left[ \qquad \qquad \ \ \quad\left(\frac{Z}{q}\right)^{n-1}+ \left(\frac{Z}{q}\right)^{n-2}+\left(\frac{Z}{q}\right)^{n-3}+\ldots +\left(\frac{Z}{q}\right)^{1}+\color{red}{\underbrace{\left(\frac{Z}{q}\right)^{0}}_{=1}}\right] \quad (1)$
Now we multiply the equation by $\frac{Z}q$
$\frac{Z}q\cdot \frac{S_n}{q^{n-1}}=R\cdot T\left[\color{red}{\left(\frac{Z}{q}\right)^{n}}+\left(\frac{Z}{q}\right)^{n-1}+ \left(\frac{Z}{q}\right)^{n-2}+\left(\frac{Z}{q}\right)^{n-3}+\ldots +\left(\frac{Z}{q}\right)^{1}\right] \quad (2)$
Next we subtract (2) from (1). Only the two red summands are remaining.
$\left(1-\frac{Z}q\right)\cdot \frac{S_n}{q^{n-1}}=R\cdot T\cdot \left[1- \left(\frac{Z}{q}\right)^{n}\right]$ And finally we have
$$S_n=R\cdot T\cdot \frac{1- \left(\frac{Z}{q}\right)^n}{1-\frac{Z}q}\cdot q^{n-1}$$
In terms of structure it looks somehow familiar.