When interest is always calculated only on the principal or initial amount, it is called simple interest. The amount of interest, when calculating simple interest, 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. We multiply the total amount by the interest rate and then add it to the total.
Suppose $\$500$$500 is invested in a compound interest account with an interest rate of $10%$10% p.a. compounded annually (that is, with a compounding period of one year) for $3$3 years. Then after one year, the interest is calculated:
Interest$=500\times10%=500\times0.1=\$50$=500×10%=500×0.1=$50
This interest is then added to the account
Balance after $1$1 year$=500+50=\$550$=500+50=$550.
After the second year interest is calculated again, but this time the interest rate is applied to the balance from the previous year:
Interest | $=550\times0.1=\$55$=550×0.1=$55 |
Balance after $2$2 years | $=550+55=\$605$=550+55=$605 |
Finally, after the third year we have
Interest | $=605\times0.1=\$60.50$=605×0.1=$60.50 |
Balance after $3$3 years | $=605+60.5=\$665.50$=605+60.5=$665.50 |
Hence, the interest is calculated on the previous year's amount including the interest.
$\$8000$$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.
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.
$A=P\left(1+r\right)^t$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 compounding period, expressed as a decimal or fraction
$t$t is the number of compounding periods
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=A−P
John's investment of $\$3000$$3000 earns interest at a rate of $3%$3% p.a, compounded annually over $4$4 years.
What is the value of the investment in dollars at the end of the $4$4 years?
Round your answer to the nearest cent.
Bob borrows $\$5000$$5000 at a rate of $5.2%$5.2% p.a. compounded annually. If he pays off the loan in a lump sum at the end of $6$6 years, how much interest does he pay?
Give your answer in dollars.
Round your answer to the nearest cent.
What do we do if the interest is being compounded more frequently; perhaps daily, weekly, monthly, quarterly or semi-annually?
$A=P\left(1+\frac{r}{n}\right)^{nt}$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.
Suppose $\$500$$500 is invested in a compound interest account with an interest rate of $10%$10% p.a. compounded semi-annually (that is, with a compounding period of $6$6 months) for $3$3 years.
Since the interest is being compounded semi-annually, the number of compounding periods in a year is $n=2.$n=2. The interest rate is $10%$10% per year and so the interest rate per compounding period as a decimal is $\frac{r}{n}=\frac{0.01}{2}=0.005$rn=0.012=0.005 . Moreover, in $3$3 years, there are a total of $nt=2\times3=6$nt=2×3=6 compounding periods. Now we can substitute into the formula:
$A$A | $=$= | $P\left(1+\frac{r}{n}\right)^{nt}$P(1+rn)nt |
$=$= | $500\times\left(1+\frac{0.01}{2}\right)^{2\times3}$500×(1+0.012)2×3 | |
$=$= | $500\times\left(1+0.005\right)^6$500×(1+0.005)6 | |
$\approx$≈ | $\$670.05$$670.05 |
For comparison, if the $\$500$$500 is invested in a compound interest account with an interest rate of $10%$10% p.a. compounded annually for $3$3 years, then
$A$A | $=$= | $P\left(1+r\right)^t$P(1+r)t |
$=$= | $500\times\left(1+0.1\right)^3$500×(1+0.1)3 | |
$\approx$≈ | $\$665.50$$665.50 |
A $\$3400$$3400 investment earns interest at $3%$3% p.a. compounded quarterly over $19$19 years.
Use the compound interest formula to calculate the value of this investment in dollars.
Round your answer to the nearest cent.
Katrina borrows $\$4000$$4000 at a rate of $6.6%$6.6% p.a. compounded semi-annually. If she pays off the loan in a lump sum at the end of $6$6 years, find how much interest she pays in dollars.
Round your answer to the nearest cent.
A $\$8920$$8920 investment earns interest at $3.3%$3.3% p.a. compounded monthly over $5$5 years.
Use the compound interest formula to calculate the value of this investment in dollars.
Round your answer to the nearest cent.
It is common to use tables to view the progression of an investment or loan. In the first investigation of this chapter we explore further how to use spreadsheets to automate calculating the values for such a table. This can be particularly useful when there are frequent compounds and/or a long investment/loan term. Using a spreadsheet we can quickly explore the impact of changes such as interest rates, compounding periods, or regular payments.
Let's explore this interactive compound interest spreadsheet.
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Click on the coloured cells to explore the effect of:
The following spreadsheet shows the balance (in dollars) in a savings account where interest is compounded quarterly.
A | B | C | D | |
1 | Quarter | Balance at beginning of quarter | Interest | Balance at end of quarter |
2 | $1$1 | $Z$Z | $100$100 | $5100$5100 |
3 | $2$2 | $5100$5100 | $Y$Y | $5202.00$5202.00 |
4 | $3$3 | $5202.00$5202.00 | $104.04$104.04 | $X$X |
5 | $4$4 | $5306.04$5306.04 | $106.12$106.12 | $5412.16$5412.16 |
Calculate the value of $X$X.
Use the numbers for quarter $3$3 to calculate the quarterly interest rate, to three decimal places.
Calculate the value of $Y$Y.
Calculate the value of $Z$Z.