5.07 Compound interest

We've now learned about simple interest, where interest is calculated at a fixed rate on just the principal amount, so the amount of interest earned in each time period remains constant. This type of interest is very rare to encounter in most situations, however. Instead, banks and other financial institutions that calculate interest do so using compound interest instead.

Compound interest as repeated multiplication

Compound interest occurs when we earn interest on the principal amount and the interest earned so far, rather than just on the principal amount. That is, each time period we calculate interest earned by using the total amount from the previous year. This means that the interest increases at an exponential rate, as we earn interest on interest that has already been earned. 

For example, if we were to invest $\$500$$500 for $3$3 years at a rate of $6%$6% per year using compound interest, then:

• After the first year, the value of the investment would be $500\times1.06$500×1.06.
• After the second year, the value of the investment would be $\left(500\times1.06\right)\times1.06$(500×1.06)×1.06, which simplifies to $500\times1.06^2$500×1.062
• After the third year, the investment would be $\left(500\times1.06^2\right)\times1.06$(500×1.062)×1.06, which we can simplify to $500\times1.06^3$500×1.063

 

Worked example

Example 1

Han's investment of $\$6000$$6000 earns interest at $2%$2% per year, compounded annually over $3$3 years.

Answer the following questions by using repeated multiplication.

(a) What is the value of the investment after $3$3 years? Round your answer to the nearest cent.

Think: We should apply the interest rate to the new amount each year.

Do:

(b) What is the amount of interest earned?

Think: The interest is the difference between the total amount of the investment and the principal (the initial amount invested).

Do: $6367.25-6000=\$367.25$6367.25−6000=$367.25

 

Notice that when we calculate compound interest, we are calculating the total value of an investment or loan. To find the amount of interest earned, $I$I, we subtract the principal $P$P from the total value $A$A. That is, $I=P-A$I=P−A.

This is different to simple interest calculations, where we calculated the interest earned. To find the total value of the investment, we needed to add the interest and the principal amount. That is, $A=P+I$A=P+I. Notice that this is the same relation between $A$A, $I$I and $P$P as above, just rearranged.

 

Practice question

Question 1

 

 

The compound interest formula

Notice that in the above example, at the end of each compounding period there is a two step process: calculate the interest and then add it to the account balance. We could have combined these two steps as follows:

Balance after $1$1 year $=500+500\times0.1=500\times(1+0.1)=550$=500+500×0.1=500×(1+0.1)=550

This suggests a rule: 

New balance $=$= Previous balance $\times(1+0.1)$×(1+0.1)

In other words, we can find the balance at the end of each year by repeatedly multiplying by $(1+0.1)$(1+0.1)

Balance after $1$1 years	$=500\times(1+0.1)$=500×(1+0.1)
Balance after $2$2 years	$=500\times(1+0.1)\times(1+0.1)=500\times(1+0.1)^2$=500×(1+0.1)×(1+0.1)=500×(1+0.1)2
Balance after $3$3 years	$=500\times(1+0.1)\times(1+0.1)\times(1+0.1)=500\times(1+0.1)^3$=500×(1+0.1)×(1+0.1)×(1+0.1)=500×(1+0.1)3

This leads us to the compound interest formula. 

 

 

Practice questions

Question 2

 

Question 3

Question 4

 

Other compounding periods

What do we do if the interest is being compounded more frequently; perhaps daily, weekly, monthly, quarterly or semi-annually?

 

Notice that since $n$n is the number of compounding periods in a year, $\frac{r}{n}$rn​ is the interest rate per compounding period, and $nt$nt is the total number of compounding periods.

 

Worked example

Suppose $\$500$$500 is invested in a compound interest account with an interest rate of $10%$10% per year compounded semi-annually (that is, with a compounding period of $6$6 months) for $3$3 years.

Since the interest is being compounded semi-annually, the number of compounding periods in a year is $n=2.$n=2. The interest rate is $10%$10% per year and so the interest rate per compounding period as a decimal is $\frac{r}{n}=\frac{0.01}{2}=0.005$rn​=0.012​=0.005 . Moreover, in $3$3 years, there are a total of $nt=2\times3=6$nt=2×3=6 compounding periods. Now we can substitute into the formula:

$A$A	$=$=	$P\left(1+\frac{r}{n}\right)^{nt}$P(1+rn​)nt
	$=$=	$500\times\left(1+\frac{0.01}{2}\right)^{2\times3}$500×(1+0.012​)2×3
	$=$=	$500\times\left(1+0.005\right)^6$500×(1+0.005)6
	$\approx$≈	$\$670.05$$670.05

For comparison, if the $\$500$$500 is invested in a compound interest account with an interest rate of $10%$10% per year compounded annually for $3$3 years, then

$A$A	$=$=	$P\left(1+r\right)^t$P(1+r)t
	$=$=	$500\times\left(1+0.1\right)^3$500×(1+0.1)3
	$\approx$≈	$\$665.50$$665.50

Compounding more frequently produces more interest!

 

 

Practice questions

Question 5

Question 6

Question 7