# Compound Interest

In compound interest, the interest earned by the principal at the end of each interest period (compounding period) is added to the principal. The sum (principal + interest) will earn another interest in the next compounding period.

Consider \$1000 invested in an account of 10% per year for 3 years. The figures below shows the contrast between simple interest and compound interest.

At 10% simple interest, the \$1000 investment amounted to \$1300 after 3 years. Only the principal earns interest which is \$100 per year.

At 10% compounded yearly, the \$1000 initial investment amounted to \$1331 after 3 years. The interest also earns an interest.

**Elements of Compound Interest**

$P$ = principal, present amount

$F$ = future amount, compound amount

$i$ = interest rate per compounding period

$r$ = nominal annual interest rate

$n$ = total number of compounding in t years

$t$ = number of years

$m$ = number of compounding per year

Future amount,

The factor $(1 + i)^n$ is called single-payment compound-amount factor and is denoted by $(F/P, \, i, \, n)$.

Present amount,

The factor $\dfrac{1}{(1 + i)^n}$ is called single-payment present-worth factor and is denoted by $(P/F, \, i, \, n)$.

Number of compounding periods,

Interest rate per compounding period,

**Values of $i$ and $n$**

In most problems, the number of years $t$ and the number of compounding periods per year $m$ are given. The example below shows the value of $i$ and $n$.

*Example*

Number of years, $t = 5 \text{ years}$

Nominal rate, $r = 18\%$

- Compounded annually ($m = 1$)
$n = 1(5) = 5$

$i = 0.18 / 1 = 0.18$

- Compounded semi-annually ($m = 2$)
$n = 2(5) = 10$

$i = 0.18 / 2 = 0.09$

- Compounded quarterly ($m = 4$)
$n = 4(5) = 20$

$i = 0.18 / 4 = 0.045$

- Compounded semi-quarterly ($m = 8$)
$n = 8(5) = 40$

$i = 0.18 / 4 = 0.0225$

- Compounded monthly ($m = 12$)
$n = 12(5) = 60$

$i = 0.18 / 12 = 0.015$

- Compounded bi-monthly ($m = 6$)
$n = 6(5) = 30$

$i = 0.18 / 6 = 0.03$

- Compounded daily ($m = 360$)
$n = 360(5) = 1800$

$i = 0.18 / 360 = 0.0005$

**Continuous Compounding (m → ∞)**

In continuous compounding, the number of interest periods per year approaches infinity. From the equation

$F = \left( 1 + \dfrac{r}{m} \right)^{mt}$

when $m \to \infty$, $mt = \infty$, and $\dfrac{r}{m} \to 0$. Hence,

$\displaystyle F = P \lim_{m \to \infty}\left( 1 + \dfrac{r}{m} \right)^{mt}$

Let $x = \dfrac{r}{m}$. When $m \to \infty$, $x \to 0$, and $m = \dfrac{r}{x}$.

$\displaystyle F = P \lim_{x \to 0}(1 + x)^{\frac{r}{x}t}$

$\displaystyle F = P \lim_{x \to 0}(1 + x)^{\frac{1}{x}rt}$

From Calculus, $\displaystyle \lim_{x \to \infty}(1 + x)^{1/x} = e$, thus,

- Log in to post comments