# 5.07 Compound interest

Lesson

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$$500 for 33 years at a rate of 6%6% per year using compound interest, then: • After the first year, the value of the investment would be 500\times1.06500×1.06. • After the second year, the value of the investment would be \left(500\times1.06\right)\times1.06(500×1.06)×1.06, which simplifies to 500\times1.06^2500×1.062 • After the third year, the investment would be \left(500\times1.06^2\right)\times1.06(500×1.062)×1.06, which we can simplify to 500\times1.06^3500×1.063 #### Worked example ##### Example 1 Han's investment of \6000$$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:

 Year 1: $A_1$A1​ $=$= $6000\times1.02$6000×1.02 $=$= $\$6120$$6120 Year 2: A_2A2​ == 6120\times1.026120×1.02 == \6240.40$$6240.40 Year 3: $A_3$A3​ $=$= $6240.40\times1.02$6240.40×1.02 $=$= $\$6367.25$$6367.25 (to the nearest cent) (b) What is the amount of interest earned? Think: The interest is the difference between the total amount of the investment and the principal (the initial amount invested). Do: 6367.25-6000=\367.256367.256000=367.25 Notice that when we calculate compound interest, we are calculating the total value of an investment or loan. To find the amount of interest earned, II, we subtract the principal PP from the total value AA. That is, I=P-AI=PA. This is different to simple interest calculations, where we calculated the interest earned. To find the total value of the investment, we needed to add the interest and the principal amount. That is, A=P+IA=P+I. Notice that this is the same relation between AA, II and PP as above, just rearranged. #### Practice question ##### Question 1 Luke's investment of \5000$$5000 earns interest at $3%$3% per year, compounded annually over $4$4 years.

Answer the following questions by repeated multiplication.

1. What is the value of the investment after $4$4 years?

2. What is the amount of interest earned?

## 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.

Compound interest formula (annual compounding)

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

where:

$A$A is the final amount of money (principal and interest together)

$P$P is the principal (the initial amount of money invested)

$r$r is the interest rate per year, expressed as a decimal or fraction

$t$t is the number of years

This formula gives us the total amount (ie. the principal and interest together). If we just want to know the value of the interest, we can work it out by subtracting the principal from the total amount of the investment. In symbols:

$I=A-P$I=AP

#### Practice questions

##### Question 4

Kathleen has just won $\$20000$$20000. When she retires in 2121 years, she wants to have \52000$$52000 in her fund which earns $8%$8% interest per annum.

How much of her winnings, to the nearest cent, does she need to invest now to achieve this?

## Other compounding periods

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

Compound interest formula (other compounding periods)

$A=P\left(1+\frac{r}{n}\right)^{nt}$A=P(1+rn)nt

where:

$A$A is the final amount of money (principal and interest together)

$P$P is the principal (the initial amount of money invested)

$r$r is the interest rate per year, expressed as a decimal or fraction

$n$n is the number of compounding periods in a year

$t$t is the number of years

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$$500 is invested in a compound interest account with an interest rate of 10%10% per year compounded semi-annually (that is, with a compounding period of 66 months) for 33 years. Since the interest is being compounded semi-annually, the number of compounding periods in a year is n=2.n=2. The interest rate is 10%10% per year and so the interest rate per compounding period as a decimal is \frac{r}{n}=\frac{0.01}{2}=0.005rn=0.012=0.005 . Moreover, in 33 years, there are a total of nt=2\times3=6nt=2×3=6 compounding periods. Now we can substitute into the formula:  AA == P\left(1+\frac{r}{n}\right)^{nt}P(1+rn​)nt == 500\times\left(1+\frac{0.01}{2}\right)^{2\times3}500×(1+0.012​)2×3 == 500\times\left(1+0.005\right)^6500×(1+0.005)6 \approx≈ \670.05$$670.05

For comparison, if the $\$500$$500 is invested in a compound interest account with an interest rate of 10%10% per year compounded annually for 33 years, then  AA == P\left(1+r\right)^tP(1+r)t == 500\times\left(1+0.1\right)^3500×(1+0.1)3 \approx≈ \665.50$$665.50
Compounding more frequently produces more interest!

Did you know?

The two versions of the compound interest formula that we have - one for annual compounding and one for other compounding periods - actually come from the same formula:

$A=P(1+R)^N$A=P(1+R)N

where:

$R$R is the interest rate per compounding period, expressed as a decimal or fraction

$N$N is the number of compounding periods.

#### Practice questions

##### Question 7

Charlie is expecting a Christmas bonus of $\$20002000 in $6$6 months time. What is the most he can borrow now, $x$x in dollars, at a rate of $3.9%$3.9% per year compounded daily, and still be able to pay off the loan with his bonus?

1. Assume there are $365$365 days in a year

### Outcomes

#### A.CED.A.2'

Create equations in two or more variables to represent relationships between quantities; graph equations on coordinate axes with labels and scales. 'Linear and exponential (integer inputs only)

#### F.BF.A.1'

Write a function that describes a relationship between two quantities. 'Linear and exponential (integer inputs)

#### F.BF.A.1.A'

Determine an explicit expression, a recursive process, or steps for calculation from a context. 'Linear and exponential (integer inputs)

#### F.LE.A.1

Distinguish between situations that can be modeled with linear functions and with exponential functions.

#### F.LE.A.1.C

Recognize situations in which a quantity grows or decays by a constant percent rate per unit interval relative to another.

#### F.LE.A.3'

Observe using graphs and tables that a quantity increasing exponentially eventually exceeds a quantity increasing linearly, quadratically, or (more generally) as a polynomial function. 'Linear and exponential