Where does the loan calculation formula come from?

By MUNGAI KIHANYA

The Sunday Nation

Nairobi,

11 December 2016

Last week I illustrated how the monthly balances of a loan are calculated as the borrower makes payments. It turned out that a Sh100,000-loan at 14 per cent interest can be cleared in just 11 months by paying Sh10,000 monthly. However, the final payment is only Sh6,962 and so the question that followed was: “is it possible to calculate equal monthly payments that clear the loan in eleven months?”.

To find the answer, we follow a process similar to the one used last week, but this time, with unknown quantities – I told you to cut out and keep the article, didn’t I?

Suppose you take out a Loan of L shillings at an Interest rate I per annum and you wish to clear it in N months by Paying P shillings monthly. The first step is to convert the interest from per annum to per month, thus we find that monthly rate, i = I/12.

At the end of the first month, the loan earns interest and the Balance, B, becomes; B= (L + L x i%) = L x (100% + i%). At first sight, that looks complicated; and we are just getting started! It can be simplified by defining a new factor r = (100% + i%).

The equation changes to: B = L x r. Mathematicians like to remove the “x” sign to make formulae look neat and tidy, so we write it as: B = Lr – nice and simple, isn’t it?

When you make the first payment of P shillings, the new balance becomes B(1) = Lr – P. At the end of the second month, this unpaid amount also earns interest at the same factor r and thus increases to (Lr – P)r = Lrr – Pr.

You pay another P shillings at this point and the balance the end of the second month comes to B(2) = Lrr – Pr – P. This rises again to (Lrr – Pr – P)r = Lrrr – Prr – Pr at the end of the third month. When you pay the third instalment you will be left with B(3) =  Lrrr – Prr – Pr – P to pay. This process continues until the loan is cleared.

A patter appears immediately: The balance at the end of month one begins with “Lr”; that at month 2 has “Lrr”; and at month 3 it has “Lrrr”.

Now “rrr” is r raised to the power of three. So, we should expect that the balance after making the final payment at the end of the last month – month N – should start with r raised to the power of N. That is: B(N) = Lr^N – P{r^(N-1) + r^(N-2) + … + r² + r + 1}.

But since this is the final payment, this balance must be equal tozero shillings; in other words: Lr^N – P{r^(N-1) + r^(N-2) + … + r² + r + 1} = 0. Making P the subject of this formula is an easy step and it yields: P = Lr^N/{r^(N-1) + r^(N-2) + … + r² + r + 1}.

This looks nice and tidy, but there is a challenge: suppose you wish to pay the loan in 10 years (N = 120 months). It would be quite tedious to calculate r¹¹⁹ = r¹¹⁸ + r¹¹⁷ + … + r² + r + 1. There are over 100 values to be evaluated and the added! Luckily, there is a mathematical short-cut that simplifies that calculation – it is called the sum of a geometric series.

I would be glad to show how the short-cut comes about, but I think we’ve done enough maths for a lazy Sunday! It turns out that {r^(N-1) + r^(N-2) + … + r² + r + 1} = {r^N – 1}/{r – 1}.

Applying this short-cut, the equation for the monthly instalment changes to: P = Lr^N{r – 1}/(r^N – 1}. All that remains now is to enter the numbers. In the example from last week, L = Sh100,000; r = 1.01167 (from I = 14%); and N = 11 months. Putting these numbers in the above formula yields a monthly instalment of Sh9,739.76. Try it out and see.