Suppose Alice wants to borrow $1000 from a bank for one year. The bank agrees to lend her the money provided that she agrees to pay the original amount plus 10% interest next year. This means she will have to pay $1000 + (10%)×($1000) = $1000 + $100 = $1100 next year.

The formula for the future value $S$ of some investment with simple interest is:

$S\=P \left(1 \+ r t\right)$$$

where $P$ is the principal amount, $r$ is the interest rate, and $t$ is the time period of the investment.

For this example it means: $P \= \$1000$, $r\=0.10$ per year, and $t\=1$ year. Notice that $r$ is in decimal form rather than percent form.

$$\begin{gathered} S \= \$1000\cdot \left(1\+0.10\cdot 1\right) \\ S \= \$1000\cdot \left(1.10\right) \= \$1100 \end{gathered}$$

Suppose Alice wants to borrow $1000 for 2 years at the same simple interest rate of 10% per year. Here $P \= \$1000$, $r\=0.10$per year, and now $t\=2$ years. Therefore, at the end of the 2 years she will owe:

$$\begin{gathered} S \= \$1000\cdot \left(1\+\frac{0.10}{\mathrm{year}}\cdot 2 \mathrm{years}\right) \\ S \= \$1000\cdot \left(1.20\right)\= \$1200 \end{gathered}$$

Consider the same problem of Alice wanting to borrow $1000 from the bank for 2 years at 10% interest per year. Rather than charging simple interest on the loan, the bank can use a more widely used form of interest calculation, compound interest.

Compound interest is interest that is added to the principal of a loan such that the added interest also earns interest. The addition of interest to the principal amount is referred to as compounding.

This means that for the first year, you can borrow $1000, and adding the interest for the first year, means that you owe the bank $1000*(1+0.10) = $1100 at the end of the year.

$\mathrm{S\_\_1}\= \$1000 \+ 0.10\cdot \left(1000\right) \= \$1100$$$$$

Then for the second year, the principal owed is $1100, and subsequently you owe the bank interest on that amount at the end of the term.

$\mathrm{S\_\_2}\= \$1100\+0.10\cdot \left(1100\right)\= \$1210$$$

Using compounded interest, the bank receives $10 more than with simple interest.

Compound interest can also be used to your advantage. Buying guaranteed investment certificates (GICs) or government bonds, can make the bank pay you interest. GICs pay compound interest, which as you will see, is much better than simple interest for investments.

Suppose Bob buys a GIC from the bank worth $1000 at 2.5% interest compounded yearly for a period of 12 years. After the first year his investment will be worth:

$S_1\= \$1000 \left(1 \+ 0.025\right)\= \$1025$

The next year he will make 2.5% interest on $S_1$, so:

$S_2\= {\mathrm{S}}_1 \left(1 \+ 0.025\right)\= \$1050.63$

The difference between compound and simple interest is only $0.63 after 2 years. After the ${12}^{\mathrm{th}}$ year however, his initial investment has grown to:

$S_{12}\= \$ 1000 {\left(1\+0.025\right)}^{12} \= \$1 344.89$

The general formula for compound interest after $n$ compounding periods is:

$S_n\=P{\left(1\+r\right)}^n$

where $P$ is the principal value of the money (initial investment) and $r$ is the interest rate for each compounding period in decimal form. This can be further generalized for any time periods paying $r$ interest rate per year as follows:

$\mathrm{S\_\_n}\=P{\left(1\+\frac{r}{m}\right)}^{\mathrm{mt}}$

where $m$ is the number of times the interest is compounded per year.

Suppose Bob buys two GICs from the bank worth $5000 each. Both will last 3 years and have an interest rate of 2.5% per year; however, one pays simple interest and the other is compounded monthly. After 3 years, the simple interest GIC will pay Bob:$$

$$\begin{gathered} S \= \$5000\left(1\+0.025\times 3 \mathrm{years}\right) \\ \= \$5000\left(1\+0.025\times 3\right) \\ \= \$5375.00 \end{gathered}$$

In contrast, the compound interest GIC will pay

$$\begin{gathered} S \= \$5000{\left(1\+\frac{0.025}{12}\right)}^{3\cdot 12} \\ \= \$5389.00 \end{gathered}$$

which is more than the simple interest GIC.

It is clear that the more frequent the compounding periods, the faster the investment will grow. If you take the limit as the frequency goes to infinity (or, equivalently as the duration of the compounding period goes to zero), you arrive at continuous interest. The return of continuously compounding interest is given by the formula:

$S \= P e^{ r t}\,$

where $t$ is the duration of the investment, $P$ is the principal value, and $r$ is the interest rate.

Now, compare continuously compounded interest with biannually (twice a year) compounded interest. Suppose the annual interest rate is 5% and the principal value is $5000. Over 10 years, the compounded interest will give a return of:

$$\begin{gathered} S \= \$5000{\left(1\+\frac{0.05}{2}\right)}^{2\cdot 10} \\ \= \$8193.08 \end{gathered}$$                  

whereas the continuously compounded interest will make:

$$\begin{gathered} S \= \$5000 e^{0.05\cdot 10} \\ \= \$ \end{gathered}$$$8243.61$

Continuous compounding always generates more interest than discrete compounding. Some loans demand continuous interest, which makes them especially difficult to pay back if they are left to grow for too long.