7.02 Compound interest

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.

 

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

 

Hence, the interest is calculated on the previous year's amount including the interest.

 

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. 

As with simple interest we need to ensure that the time period for the rate and $n$n match. For example: the annual rate and the duration both given in 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:

 

Practice questions

Question 2

 

Question 3

 

Compound interest formula - other compounding periods

What do we do if the interest is being compounded more frequently; perhaps daily, weekly, monthly, quarterly or semi-annually? The same formula applies but we may need to convert the rate and time to match the compounding period.

From an annual rate, the rate per period can be found by dividing the annual rate by the number of compounds per year. For example, $4.5%$4.5% p.a. compounded monthly gives us a rate of $i=\frac{4.5%}{12}=0.375%$i=4.5%12​=0.375% per month.

 

Worked example

Example 2

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.

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

Do: Now we can substitute into the formula:

$A$A	$=$=	$P\left(1+i\right)^n$P(1+i)n
	$=$=	$500\times\left(1+0.005\right)^6$500×(1+0.005)6
	$\approx$≈	$\$670.05$$670.05

Reflect: 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	$=$=	$P\left(1+i\right)^n$P(1+i)n
	$=$=	$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 4

Question 5

Question 6

 

Modelling compound interest using a spreadsheet

It is common to use tables to view the progression of an investment or loan. In first the 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.

Exploration

Let's explore this interactive compound interest spreadsheet.

Created with Geogebra

Click on the coloured cells to explore the effect of:

• Changing the amount invested (the blue cell).
• Changing the annual interest rate (the green cell). 
• Changing the number of compounding periods (the pink cell) to quarterly ($4$4), monthly ($12$12), weekly ($52$52) or perhaps daily ($365$365).

Investigate

• What happens as you increase the number of compounding periods?
• What happens as you increase the annual interest rate?
• How has the value in cell C10 been calculated?
• How has the value in D12 been calculated?

 

Worked example

Example 3

David invests $\$8000$$8000 in the bank with an interest rate of $3%$3% p.a. compounded monthly.  He creates the following spreadsheet to help him do "what if analysis" to examine the problem:

	A	B	C	D
1	Principal	$\$8000$$8000
2	Annual interest rate	$3%$3%
3	Compounds per year	$12$12
4
5	Month	Balance start of month	Interest	Balance end of month
6	1	$\$8000$$8000	$\$20$$20	$\$8020$$8020
7	2	$\$8020$$8020	$\$20.05$$20.05	$\$8040.05$$8040.05

Some of the formulae David used to create this spreadsheet are shown in the table below:

	A	B	C	D
5	Month	Balance start of month	Interest	Balance end of month
6	1	=B1	=$B$2/$B$3*B6	=B6+C6
7	=A6+1	=D6
8

Note:

• The spreadsheet is designed so that if the values in cells B1, B2 and B3 are changed, the whole page instantly updates. This makes it quick and easy to investigate different investment options. 
• The $ signs in the cell references make the reference absolute. That means the cell name will not change as the formula is copied down the column. 
• The month numbers here have been calculated using a formula in cell A7. 

(a) What is the purpose of the formula in cell C6?

Think: Ignore the $ signs. The formula is =B2/B3 * B6, which is calculating $3%\div12\times\$8000$3%÷12×$8000.

Do: This calculates the amount of monthly interest by taking the yearly interest rate (cell B2) and dividing by the number of compounds per year (cell B3) then multiplying by the amount at the start of the month (cell B6). 

(b) What is the purpose of the formula in cell D6?

Think: The formula is =B6+C6 so it is added two values together.

Do: Write that it calculates the balance at the end of the month by calculating start balance + interest.

(c) Use a spreadsheet program to recreate this spreadsheet. How long does it take David to save the $\$9000$$9000?

Reflect: The spreadsheet should show that it takes him $48$48 months for his balance to first exceed $\$9000$$9000 and hence, it will take David approximately $4$4 years if he doesn't make any additional payments.

 

Practice question

Question 7