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Suppose I have a loan of M dollars. At the end of each year, I am charged interest at rate R and make a repayment of P. The loan is repaid after n years.

  1. How long (n) does it take to repay the loan if I am given the other variables?
  2. How much are the repayments of P if I am given the other variable?
  3. Suppose that the payments were at the start of the year. How would this change the problem?
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I asked this question since a much more specific question was asked here and so I think that it is worthwhile answering the general question rather than just the specific –  Casebash Jul 24 '10 at 1:10
    
This question is seeded –  Casebash Jul 24 '10 at 4:08

2 Answers 2

up vote 3 down vote accepted

Basic Theory

The way to solve this problem is to calculate how much each payment reduces your debt after you have been repaying your loan for $n$ years. Let $r=1+R/100$, ie. this converts the interest rate from a percentage to a value you can multiply your debt by to calculate how much you owe after adding one time period's interest.

If I make a payment of $P$ at the end of the $k$th year, then we avoid paying interest on this money $n-k$ times and so we reduce our debt by $Pr^{n-k}$. We sum up the future values of all our payments:

$\sum\limits_{k=1}^n Pr^{n-k}$

If we reverse this, it is equivalent to:

$\sum\limits_{k=0}^{n-1} Pr^k$

This is a geometric series, which can be solved using the formula $\frac{ar^{n-1}}{r-1}$ where $a$ is the first term, $r$ is the factor and $n$ is the number of terms being summed. We then attempt to equate this with the debt owed after $n$ years, which is $Mr^n$.

We now compare the two equations:

$\frac{Pr^{n-1}}{r-1} = Mr^n$

Calculating $n$

We group the $r^n$ terms:

$\frac{P}{r-1} = r^n\frac{M-P}{r-1}$

$r^n = \frac{P}{M(r-1)-P}$

So we just take the $n$th log of the right hand side.

Calculating repayments

Given the principal ($M$) and the interest rate ($r$), what will my payment-per-term ($P$) be over $n$ accruation terms?

$P=\frac{Mr^n(r-1)}{r^{n-1}}$

Payments made at the start of the year

In this case, the future values of our interest payment simply become:

$\sum\limits_{k=1}^n Pr^k$

We proceed as we did before.

Notes

We could also solve this problem using present value instead of future value.

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Suggest you look into amortization calculators.

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Thanks, thats a good link –  Casebash Jul 24 '10 at 2:31

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