Compound Interest Rate - My Risk Book

Compound Interest Rates

Let's re-state our previous example: if $1,000 is invested for a year at the interest rate of 10% per annum with annual compounding, you are going to receive at the end of the year a return of

$\$1,000 \times 0.10 = \$100$

Your investment earns $100 at the end of the year. The value of your investment at the end of the first year is $1,100. We can write the value of your investment as

$\$1,000 + \$1,000 \times 0.10 = \$1,000(1+0.10) = \$1,100$

If $1,000 is invested for two years at the interest rate of 10% per annum with annual compounding, you are going to receive at the end of the first year a return of

$\$1,000 \times 0.10 = \$100$

The value of your investment is

$\$1,000 + \$1,000 \times 0.10 = \$1,100$

$\$1,000(1+0.10) = \$1,100$

In the second year, you invest $1,100 at 10% for another year. The value of your investment would be

$\$1,100 + \$1,100 \times 0.10 = \$1,210$

$\$1,100(1+0.10) = \$1,210$

Or we can write as follows:

$\$1,100(1+0.10) = \$1,210$

$\$1,000(1+0.10)(1+0.10) = \$1,210$

$\$1,000(1+0.10)^2 = \$1,210$

Now, if you invest $1,210 for the third year at 10%, the value of your investment would be

$\$1,210(1+0.10) = \$1,331$

We can also write the above equation as

$\$1,000(1+0.10)(1+0.10)(1+0.10) = \$1,331$

⇒

$\$1,000(1+0.10)^3 = \$1,331$

We can continue this way for years 4 and 5 with outcomes as follows:

$\$1,000(1+0.10)^4 = \$1,464.10$

$\$1,000(1+0.10)^5 = \$1,610.51$

In this case, your investment of $1,000 earns $610.51 at the end of five years. The value of your investment grows to $1,610.51 rather than $1,500.

In the above example, (1+0.10), (1+0.10)², (1+0.10)³, (1+0.10)⁴, (1+0.10)⁵ show how $1 investment grows every year:

$(1+0.10) = 1.10$

$(1+0.10)^2= 1.21$

$(1+0.10)^3= 1.331$

$(1+0.10)^4= 1.4641$

$(1+0.10)^5= 1.61051$

If you subtract $1 in each equation above, we get the effective interest rate:

$(1+0.10) - 1= 0.10$

$(1+0.10)^2 - 1 = 0.21$

$(1+0.10)^3 - 1 = 0.331$

$(1+0.10)^4 - 1 = 0.4641$

$(1+0.10)^5 -1 = 0.61051$

The effective interest rate when the interest rate compounds annually,

$\left(1+r\right)^t - 1$

Compare the above calculations with the simple interest rate, where interest rates do not compound. It remains 10% without compounding annually. When 10% compounds annually, the effective rate increases from 10% in the first year to 61.051% in the fifth year.

As you can see now that a $1,000 investment at a 10% continuously compounded interest rate can $610.51 as opposed to $500 with a simple interest rate.

Let's $$1,000 = M₀, and (1+0.10)^t = (1+r)^t, then we can write, Future Value, FV_t:

$M_t = M_0(1+r)^t$

In other words, the future value of $1 can be written as (1+r)^t, which is the compound factor.

Let's now consider one year:

When we compound r semi-annually in one year the amount $1 grows to
$\left(1+\frac{r}{2}\right)^2$
When we compound r quarterly in one year the amount $1 grows to
$\left(1+\frac{r}{4}\right)^4$
When we compound r monthly in one year the amount $1 grows to
$\left(1+\frac{r}{12}\right)^{12}$
When we compound r weekly in one year the amount $1 grows to
$\left(1+\frac{r}{52}\right)^{52}$
When we compound r daily in one year the amount $1 grows to
$\left(1+\frac{r}{365}\right)^{365}$