Compound interest

We compound interest when it has to be applied more than once. If we have a debt that is charged interest once a year and we owe it for five years, the debt will have interest added five times – we then talk about compound interest.

Compound interest

If the amount owed is charged interest several times, we speak of compound interest. In the previous formula we used percentages; in the next one it will be easier to turn the percentage into an ordinary decimal. We do this simply by dividing the percentage by a hundred. If the interest was 50%, we get the value 0.5, that is, one half. If it was 7%, we get the number 0.07. I will denote this decimal by the letter “r”. The formula could look like this:

$r = \frac{interest rate}{100}$

If we multiply the amount owed by this number, we get the interest directly. Suppose Uncle Vic lent us £20,000 at an annual interest rate of 10%. As a decimal, that is 0.1. After one year the interest is 20,000 · 0.1 = 2,000. So we pay £2,000 more than we borrowed. What will the interest be in the next year? We have to take the original amount, add the first interest and multiply by 0.1 again:

$(20,000 + 2,000) \cdot 0.1 = 2,200$

A year later we would pay £2,200 in interest, which is more than in the first year.

If we want the amount owed itself, not just the interest, it is enough to add one to the interest rate written as a decimal and multiply the amount owed by that number:

$20,000 \cdot 1.1 = 22,000$

So after the first year we would owe £22,000. If we want the amount owed after three years, we multiply by 1.1 several times in a row:

$((20,000 \cdot 1.1) \cdot 1.1) \cdot 1.1 = 26,620$

A formula is slowly taking shape here, because in multiplication the order and the brackets don't matter, so we can rewrite the expression like this:

$20,000 \cdot (1.1 \cdot 1.1 \cdot 1.1)$

And we simply rewrite that using powers:

$20,000 \cdot {1.1}^{3}$

From this we can easily derive the general formula.

The final compound interest formula

Here is the formula for compound interest, which I will explain below:

$P_{t} = P_{0} \cdot (1 + r)^{t}$

The letter “P” stands for the principal, that is, the amount owed. The subscript tells us how many times interest has already been added. So the symbol P₀ denotes the amount that has not yet had any interest added – it is the amount we received at the start (the £20,000 from Uncle Vic). The letter “t” denotes the number of times we want to add interest. If we want to look ten years ahead, then t = 10. The symbol P_t then denotes the resulting amount owed after t compounding periods. The letter “r” denotes the interest rate as a decimal (see above).

Let's work out how much interest we would pay if we gave the whole amount back to Uncle Vic only after 15 years:

$P_{t} = P_{0} \cdot (1 + r)^{t} = 20,000 \cdot (1 + 0.1)^{15} = 83,544.96$

Note that this is not the interest itself but the final amount owed. If we wanted only the interest, we would have to subtract the original amount owed, that is, take away another £20,000: £63,544.96.

Example

Oliver wanted to buy a new car. Of course he couldn't afford it, so he borrowed £5,000. The interest rate was 8% and interest was added once every year. Oliver was a big spender, so in the first year he repaid only £200. The amount owed fell to £4,800. But at the turn of the year the interest was added, and it turned out that the interest came to 4,800 · 0.08 = 384. On New Year's Day Oliver therefore already owed 4,800 + 384 = 5,184. That is more than at the start.

This didn't worry Oliver at all, and he again repaid only £200. So he owed £4,984. After the interest was added, it shot up to £5,382.72. The amount owed was higher than last year and the year before. It didn't dawn on Oliver, and he carried on repaying only £200 a year. He kept paying it for the rest of his life and still died owing far more than he had borrowed. Better not to do it like this.

Compound interest