16.04 Compound interest
Simple interest is calculated only on the principal (that is, the initial amount) so the amount of interest being added to a loan or investment remains constant or fixed. However, most of the time when banks and financial institutions calculate interest, they are using compound interest.
Compound interest is calculated at the end of each compounding period, which is typically a day, month, quarter, or year. At the end of each compounding period, the total amount (principal plus interest) from previous compounding periods is used to calculate the new quantity of interest.
Worked example
Example 1
Suppose $\$500$$500 is invested in a compound interest account with an interest rate of $10%$10% p.a. compounded annually (that is, with a compounding period of one year) for $3$3 years. Then after one year, the interest is calculated:
Interest$=500\times10%=500\times0.1=\$50$=500×10%=500×0.1=$50
This interest is then added to the account
Balance after $1$1 year$=500+50=\$550$=500+50=$550.
After the second year interest is calculated again, but this time the interest rate is applied to the balance from the previous year:
| Interest | $=550\times0.1=\$55$=550×0.1=$55 |
| Balance after $2$2 years | $=550+55=\$605$=550+55=$605 |
Finally, after the third year we have
| Interest | $=605\times0.1=\$60.50$=605×0.1=$60.50 |
| Balance after $3$3 years | $=605+60.5=\$665.50$=605+60.5=$665.50 |
Practice question
Question 1
$\$8000$$8000 is invested for $3$3 years at a rate of $3%$3% p.a. compounded annually.
Complete the table below, rounding to the nearest cent.
Number of periods Interest ($\$$$) Balance ($\$$$) After $0$0 years - $8000$8000 After $1$1 year $\editable{}$ $\editable{}$ After $2$2 years $\editable{}$ $\editable{}$ After $3$3 years $\editable{}$ $\editable{}$ Calculate the total interest accumulated over $3$3 years in dollars.
Round your answer to the nearest cent.
Why does money grow faster with compound interest than simple interest?
As seen before, simple interest is interest earned on the principal invested amount only, whereas compound interest is interest earned on the principal amount plus interest on the interest already earned. So instead of the value of your investment increasing it a straight line as with simple interest, it will exponentially grow something like this:
For example, consider a deposit of $\$1000$$1000 into an online account for $2$2 years that pays $10$10% pa simple interest. The interest earned in the $2$2 years is $\$1000\times10%\times2=\$200$$1000×10%×2=$200.
But suppose that, instead of simple interest, the paid interest compounded annually. In this case, the interest earned in the first year would be $\$1000\times10%\times1=\$100$$1000×10%×1=$100.
The new principal at the end of the first year would be $\$1000+\$100=\$1100$$1000+$100=$1100.
The interest earned in the second year would then be $\$1100\times10%\times1=\$110$$1100×10%×1=$110.
So the total compound interest earned over the two years would be $\$100+\$110=\$210$$100+$110=$210 which is $\$10$$10 more than what was earned with simple interest. Although a $\$10$$10 difference may not seem like much, think of how much the difference would have been if a million dollars was invested instead of a thousand, or if the investment was made for twenty years instead of two.
The compound interest formula
Notice that in the first worked 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. The two steps can be combined 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, the balance at the end of each year can be calculated 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 |
The compound interest formula can then be derived as follows:
$A=P(1+\frac{r}{100})^n$A=P(1+r100)n
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 compounding period, expressed as a whole number
$n$n is the number of compounding time periods
For interest rates $r$r that are expressed as a decimal or a fraction, this formula can be used instead:
$A=P\left(1+r\right)^n$A=P(1+r)n
$I=A-P$I=A−P
Practice questions
Question 2
William's investment of $\$2000$$2000 earns interest at a rate of $6%$6% p.a, compounded annually over $4$4 years.
What is the future value of the investment to the nearest cent?
Question 3
A $\$9450$$9450 investment earns interest at $\frac{13}{5}%$135% p.a. compounded monthly over $14$14 years. What is the future value of the investment to the nearest cent?
Question 4
Tom wants to put a deposit on a house in $4$4 years. In order to finance the $\$12000$$12000 deposit, he decides to put some money into a high interest savings account that pays $5%$5% p.a. interest compounded monthly. If $P$P is the amount of money that he must put into his account now to accumulate enough for the deposit, find $P$P to the nearest cent.
Enter each line of working as an equation.