How to calculate interest rate for given amount?

Question

Mr. X pays 1300 each month for 20yrs.

After 20yrs X getting 550000.

what interest Mr. X got the final amount 550000?

For 20yrs 240 months there.

240 * 1300 = 312000

Full steps solution will be easy for my understanding.

Answer 1

This is an interesting question. First of all, it depends on the time when the interests are added to the account. Let's assume, that we have a monthly interest.

Let's say, the monthly interest rate is $p \in \mathbb{R}$. For example, for an monthly interest of 0.1 % we have $p=0.001$. If we bring the amount of $x \in \mathbb{R}$ to our bank and remain it there for $i \in \mathbb{N}$ months (without adding money), then in the end we have the final amount $s$ with

$ s = x\cdot (1 + p)^i$

So for our question now, let's do the same calculation for each of the 120 payment rates. The sum of the final amounts must be equal to 550000. So we get this equation:

$550000 = \sum_{i = 1}^{120} 1300\cdot (1+p)^i = 1300 \cdot \sum_{i = 1}^{120} (1+p)^i = 1300 \cdot (1+p)\cdot \frac{(1+p)^{120} - 1}{p}$

In the last equation I have used a common formula to simplify such sums. Sadly, there is no standard analytical way to solve this equation now, so that we can find out $p$. What you can do is to solve this numerically. This means, that you try values for $p$. And when the right term of the equation gives you some value over 550000, then you have to decrease $p$ and do the same again, and when it gives you something less then 550000, then you have to increase your $p$. In a few iterations you can approximate $p$ closely.

This is probably not the answer that you've expected, since it doesn't give you a simple exact formula. I guess that such a formula exists, but I don't know it. Another question is what happens when we don't have a monthly interest, but annual interest. This makes it more complicated to find an exact formula. Though a similar numerical approach should be possible to find as well.

Answer 2

Starting from S. M. Roch's answer and keeping his/her notations, you need to solve for $p$ $$s=x\cdot(1+p)\cdot \frac{(1+p)^{n} - 1}{p}$$ which, as already said, does not show analytical solutions and requires numerical methods.

So, let us consider the problem of finding the zero of $$f(p)=(1+p)\cdot \frac{(1+p)^{n} - 1}{p}-a\tag 1$$ using $a=\frac sx$. We know, from the nature of the problem, that the solution is unique, positive an small.

To get a first approximation, using Taylor expansion or the binomial theorem, we can write $$f(p)=(n-a)+\frac{1}{2} n (n+1) p+O\left(p^2\right)$$ Ignoring the higher order terms and solving the linear equation, we can have an estimate $$p=\frac{2 (a-n)}{n (n+1)}$$ Applied to your case $a=\frac{550000}{1300}$, $n=240$, this would give as a first estimate $p=\frac{119}{18798}\approx 0.00633046$.

Now, let us use Newton method which, starting from a reasonable guess $p_0$, will update it according to $$p_{k+1}=p_k-\frac{f(p_k)}{f'(p_k)}$$ So, from $(1)$, $$f'(p)=\frac{(n p-1) (p+1)^n+1}{p^2}\tag 2$$ $$p_{k+1}=\frac{p_k \left(a p_k+((n-1) p_k-2) (p_k+1)^n+p_k+2\right)}{(n p_k-1) (p_k+1)^n+1}$$ Using your numbers, the method will generate the following iterates $$\left( \begin{array}{cc} k & p_k \\ 0 & 0.006330461 \\ 1 & 0.004651664 \\ 2 & 0.004351353 \\ 3 & 0.004343325 \\ 4 & 0.004343319 \end{array} \right)$$ which is the solution for ten significant figures.

Let us check the value of $s$ for $x=1300$ and $n=240$. We should obtain $s=549999.96$ which seems quite good.

Added for your curiosity

Sooner or later, you will learn that it is possible to locally approximate a function using Padé approximants (these are ratios of polynomials of preselected degrees). Using the lowest, we should have $$f(p)\approx \frac{\frac{1}{6} p \left(2 a n-2 a+n^2+5 n\right)-a+n}{\frac{1}{3} (1-n) p+1}$$ Setting the numerator equal to zero would give, as an estimate $$p\approx \frac{6 (a-n)}{n^2+(5+2a)n-2a}$$ Using again your numbers, this would give $p\approx 0.00420817$ and, starting with this initial value, Newton method will probably converge using a single iteration.

Using the next approximant, we should get $$p\approx \frac{2 (a-n) \left(n^2+(2 a+5) n-2 a\right)}{(2 a+3) n^3+\left(a+3\right)^2 n^2-a\left(3 a+8\right) n+2 a^2}$$ which, using your numbers, would give $p\approx 0.00433140$.

By the way, you could be interested by a similar problem.