Compound interest

We now apply our knowledge of the simple interest model to build the compound interest model.

Half-yearly compound interest

Let us first consider the scenario of interest compounded half-yearly.

A year comprises of \(2\) half-years (i.e., \(2\) periods of six months). Assume that the bank pays out a simple interest over each of these half-years at a rate of \(R\)% per annum. Then on each half-year, the (simple) interest would be pay out at \(\frac{R}{2}\)% per half-yearly.

At the end of the first half-year, the amount accumulated would be \(P \left(1+ \frac{R}{200}\right)\) dollars.

This amount is viewed as the principal sum at the start of the second half-year. So, the amount generated over the second half-year would be \(P \left(1+ \frac{R}{200}\right)^2\) dollars.

Half-yearly compound interest

Principal = \(\$ P\)
Interest rate = \(R \%\) per annum
Compounded half-yearly
Time = \(n\) years
Compounded amount = \(\$A\)
Then

\[A=P \left(1+ \frac{R}{2\times 100}\right)^{2n}\]

Total compound interest \(I\) compounded half-yearly over \(n\) years is given by

\[I=A-P=P\left[\left(1+ \frac{R}{2(100)}\right)^{2n} -1\right]\]

Quarterly compound interest

A year comprises of \(4\) quarters (i.e., \(4\) periods of three months). Assume that the bank pays out a simple interest over each of these quarters at a rate of \(R\)% per annum. Then on each quarter, the (simple) interest would be pay out at \(\frac{R}{4}\)% per quarter. At the end of the first quarter, the amount accumulated would be \(P \left( 1 + \frac{\big(\frac{R}{4}\big)}{100} \right)\) dollars. This amount is viewed as the principal sum at the start of the second quarter. So, the amount generated over the second quarter would be \(P \left( 1 + \frac{\big(\frac{R}{4}\big)}{100} \right)^2\) dollars.

Repeating the argument twice, the amount generated over the fourth quarter is \(P \left( 1 + \frac{\frac{R}{4}}{100} \right)^4\).

Quarterly compound interest

Principal = \(\(P\)
Interest rate = \(R\)% per annum
Compounded quarterly
Time = \(n\) years
Compounded amount = \)\(A\)
Then

\[A = P \left( 1 + \frac{\big(\frac{R}{4}\big)}{100} \right)^{4n}\]

Total compound interest \(I\) compounded quarterly over \(n\) years is given by

\[I = A - P = P \left[ \left( 1 + \frac{R}{4(100)} \right)^{4n} - 1 \right]\]

In general, if a bank pays out interest compounded per \(\frac{1}{k}\) of a year over \(n\) years, then the compound interest paid out can be shown to be

\[I = P \left[ \left( 1 + \frac{R}{100k} \right)^{kn} - 1 \right]\]

Remarks. Note that over each frame of \(\frac{1}{k}\)-year, the amount accumulated has dilated by a factor of \(\left( 1 + \frac{R}{100k} \right)\), which can be viewed as a common ratio. This common ratio is multiplied to the current amount to yield the next amount. In this way, what lies in the heart of a compound interest model is the geometric progression.