With simple interest, the interest earned each time period is the same because it is always based on the principal amount invested. For example, if you invest \$ 1000 at 2\% per annum simple interest, you will earn 2\% of \$ 1000 each year. The interest of \$ 20 earned each year does not change over the term of the investment.
With compound interest, the interest earned in each time period is added to the principal. So the interest earned in each time period increases because it is calculated on a growing balance. For example, if you invest \$ 1000 at 2\% per annum compounded yearly, you will earn \$ 20 interest in the first year, which is added to the principal. In the second year, interest of 2\% is calculated on a new balance of \$ 1020, not \$ 1000.

This means compound interest usually leads to greater returns than simple interest because the principal is updated each period after interest is earned.

Let's compare the two with an example of earning 5\% interest on an initial investment principal of \$1000.

Year	Principal used	Simple interest earned	New total
1	\qquad \$1000	\quad \quad \$50	\$1050
2	\qquad \$1000	\quad \quad \$50	\$1100
3	\qquad \$1000	\quad \quad \$50	\$1150

Year	Principal used	Compound interest earned	New total
1	\qquad \$1000	\quad \quad \$50	\$1050
2	\qquad \$1050	\quad \quad \$52.50	\$1102.50
3	\qquad \$1102.50	\quad \quad \$55.13	\$1157.63

We can see that after just two years, there is already a financial benefit in earning compound interest over simple interest.

Of course, compound interest can also apply to loans, where the amount we need to repay increases in the same way.

In the previous example on compound interest, the new total after 1 year was 1000 + 1000 \times 0.05. If we factor out the principal amount, we can see that our new total after 1 year is equal to 1000 \times 1.05.

Notice that this is the same way we calculate simple interest over one period.

However, since compound interest calculates interest based on the new total, the principal amount for the second year's interest will be equal to the total amount after the first year. This means that the new total after two years will be equal to 1000 \times 1.05 \times 1.05.

Since the total after each year will be equal to 1.05 times the previous year's total, we can calculate the total after any number of years by multiplying by 1.05 the required number of times.

For example: if John takes out a loan of \$2000 at an interest rate of 4\%, his total amount to repay at the end of each year will be equal to 1.04 times the total of the previous year. After 5 years, his total amount to repay will be equal to \$2000 \times 1.04 \times 1.04 \times 1.04 \times 1.04 \times 1.04 .

Since we are multiplying by the same value multiple times, we can condense the expression using indices. For the case above, we can rewrite John's total amount after 5 years as \$2000 \times 1.04^{5}.

Writing the total in this form leads us to the compound interest formula.

If a principal amount P is invested at an interest rate of r for n periods, the future value A of the investment after n periods will be:A=P(1+r)^{n}

Note that r is a decimal value corresponding to some percentage interest rate. For example, if the interest rate is 4\% then r=0.04.

If we only want to find the amount of interest earned, we can simply subtract the principal amount from the future value of the investment.

\text{Interest} =\text{Future value} - \text{Principal amount}

Now that we have a formula for calculating compound interest, as long as we know the principal amount P, the interest rate r and the number of periods n, we can calculate the future value A,

In addition to this, since we have a formula relating the different variables of an investment, if we know any three of the four variables, we can use them to calculate the last unknown variable.

We can do this by substituting our known values into the formula and then solving the equation for the unknown variable.

Examples

Example 1

Valerie borrows \$1250 at a rate of 7.6\% p.a, compounding annually.

Which of the following expressions represents the amount Valerie must repay after 15 years, assuming that she hasn't paid anything back?

1250 \times (1+0.076) \times 15

1250 \times (1+0.076) \times \dfrac{15}{12}

1250 \times (1+0.076)^{\frac{15}{12}}

1250 \times (1+0.076)^{15}

Worked Solution

Create a strategy

Use the compound interest formula: A=P(1+r)^{n}

Apply the idea

From the problem, we have P=1250, \, r=0.076, \, n=15.

\displaystyle A	\displaystyle =	\displaystyle P(1+r)^{n}	Write the formula
	\displaystyle =	\displaystyle 1250 \left(1+0.076\right)^{15}	Substitute the known values

The correct answer is D.

How much interest was generated on the loan over the fifteen years? Round your answer to two decimal places.

Worked Solution

Create a strategy

Subtract the principal from the future value.

Apply the idea

In part (a), we have the future value, 1250 \left(1+0.076\right)^{15}. So we have:

\displaystyle \text{Interest}	\displaystyle =	\displaystyle 1250 \left(1+0.076\right) ^{15}-1250	Substitute the values
	\displaystyle =	\displaystyle 3750.54-1250	Subtract the principal
	\displaystyle =	\displaystyle \$2500.54	Evaluate

The interest generated on the loan over the fifteen years was \$2500.54.

Idea summary

Compound interest formula

If a principal amount P is invested at an interest rate of r for n periods, the future value A of the investment after n periods will be:A=P(1+r)^{n}

Note that r is a decimal value corresponding to some percentage interest rate. For example, if the interest rate is 4\% then r=0.04.

Calculating interest\text{Interest} =\text{Future value} - \text{Principal amount}

Different periods

So far, the examples we have looked at have all had interest calculated annually. However, interest can be calculated over any period and the interest rate per period will need to match it.

To match the interest rate to the period duration, we can use interest rate conversions.

It is also important to match the total number of periods to the duration of the investment. To do this, we can simply divide the duration of the investment by the duration of the period.

Examples

Example 2

A \$9450 investment earns interest at 2.6\% p.a., compounded monthly over 14 years.

What is the future value of the investment?

Worked Solution

Create a strategy

Use the compound interest formula: A=P(1+r)^{n}

Apply the idea

We are given that P=9450,\, n=14 \text{ years}= 14\times 12 \text{ months}=168\text{ months} and r=\dfrac{0.026}{12}=0.002\,166\,667.

\displaystyle A	\displaystyle =	\displaystyle P(1+r)^{n}
	\displaystyle =	\displaystyle 9450 \left(1+0.002\,166\,667\right) ^{168}	Substitute the values
	\displaystyle =	\displaystyle \$13\,593.90	Evaluate

The future value is \$13\,593.90.

Idea summary

If interest is compounded daily, weekly, monthly, quarterly, or half-yearly, then we need to divide the annual rate of interest, r, and multiply the number of years, n, to get the rate and time periods in the same units.

For example, if interested is compounded quarterly we multiply n by 4 and divide r by 4.

We usually use the following values when converting the rate and periods:

There are 365 days in a year.
There are 52 weeks in a year.
There are 12 months in a year.
There are 4 quarters in a year.
There are 2 half-years in a year.

Visualise compound interest

As we saw when calculating compound interest rate using a table, the interest earned each period was greater than the interest earned last period. This tells us that an investment earning compound interest will be increasing at an increasing rate.

To see what this looks like on a graph, we can compare an investment with simple interest to an investment with compound interest.

\text{Time}

\text{Investment value }

The curve representing compound interest is increasing at an increasing rate, while the simple interest curve is linear.

These relationships are reflected in the formulas for simple and compound interest.

The value of an investment earning simple interest is calculated using the formula A=P+Prn which is a linear equation in terms of n, the number of periods. Meanwhile, compound interest uses the formula A=P(1+r)^{n} which is non-linear equation in terms of n.

Examples

Example 3

The graph below shows both an investment with simple interest, and one with compound interest, labelled Investment A and Investment B respectively.

\text{Time (years)}

\text{Investment } \left(\$1000\right)

Which investment has a higher principal amount?

Worked Solution

Create a strategy

Choose the graph with the higher y-intercept.

Apply the idea

The graph for Investment A has a higher y-intercept than the graph for Investment B. So Investment A has a higher principal amount.

Which investment has a higher final amount after 10 years?

Worked Solution

Create a strategy

Find the point on each graph that lines up with x=10.

Apply the idea

\text{Time (years)}

\text{Investment } \left(\$1000\right)

From the coordinate plane we can see that when x=10 the value of Investment B is higher that the value of a Investment A.

So Investment B has a higher final amount after 10 years.

After how many years are the investments equal in value?

Worked Solution

Create a strategy

Find the intersection point of the graphs.

Apply the idea

\text{Time (years)}

\text{Investment } \left(\$1000\right)

From the the coordinate plane we can see that the two graphs intersect at a point where x=9.

So this means that after 9 years the investments are equal in value.

Idea summary

\text{Time}

\text{Investment value }

The curve representing compound interest is increasing at an increasing rate, while the simple interest curve is linear.

These relationships are reflected in the formulas for simple and compound interest. The simple interest formula is linear: \\ I=Prn while the compound interest formula is non-linear: A=P(1+r)^{n} in terms of n.

Applications of the compound interest formula

While the compound interest formula is used for calculating the result of compound interest, it can also be used for situations that follow the same non-linear pattern.

For example: appreciation or depreciation over one year is a single percentage change, but we can model them over multiple years using the compound interest formula.

In fact, any percentage change that happens multiple times can be modelled using the compound interest formula.

Another way to think of this is that compound interest is simply the percentage increase of money happening multiple times.

Examples

Example 4

Bart is planning to invest \$110\,000 into a saving scheme which accumulates interest weekly. If he aims to have his investment reach \$120\,000 after 2 years, what interest rate per annum, r, will he need? Round your answer to four decimal places.

Worked Solution

Create a strategy

Use the compound interest formula and solve for the rate.

Apply the idea

We are given P=110\,000, A=120\,000, and n=2\times52=104.

The weekly interest rate is represented by \dfrac{r}{52}.

\displaystyle A	\displaystyle =	\displaystyle P(1+r)^{n}
\displaystyle 120\,000	\displaystyle =	\displaystyle 110\,000 \left(1 + \dfrac{r}{52}\right)^{104}	Substitute the values
\displaystyle \dfrac{120\,000}{110\,000}	\displaystyle =	\displaystyle \dfrac{110\,000 \left(1 + \dfrac{r}{52}\right)^{104}}{110\,000}	Divide both sides by 110\,000
\displaystyle \dfrac{12}{11}	\displaystyle =	\displaystyle \left(1 + \dfrac{r}{52}\right)^{104}	Simplify
\displaystyle \left(\dfrac{12}{11}\right)^{\frac{1}{104}}	\displaystyle =	\displaystyle \left(\left(1 + \dfrac{r}{52}\right)^{104}\right)^{\frac{1}{104}}	Take both sides to the power of \dfrac{1}{104}
\displaystyle 1.000837	\displaystyle =	\displaystyle 1 + \dfrac{r}{52}	Evaluate
\displaystyle 1.000837-1	\displaystyle =	\displaystyle 1-1 + \dfrac{r}{52}	Subtract 1 from both sides
\displaystyle 0.000837	\displaystyle =	\displaystyle \dfrac{r}{52}	Evaluate
\displaystyle 0.000837 \times 52	\displaystyle =	\displaystyle \dfrac{r}{52} \times 52	Multiply both sides by 52
\displaystyle 0.0435	\displaystyle =	\displaystyle r	Evaluate

Bart will need an interest rate per annum of 0.0435.

Example 5

A swimming pool is losing water at the rate of 2\% per week due to evaporation. The pool currently holds 850 kilolitres of water.

How much water will be lost in the next 27 days? Round your answer correct to two decimal places.

Worked Solution

Create a strategy

Use the depreciation formula: A=P(1-r)^{n}.

Apply the idea

We have P=850,\,r=2\%=0.02, and n=\dfrac{27}{7} because there are 7 days in a week.

\displaystyle A	\displaystyle =	\displaystyle P(1-r)^{n}
	\displaystyle =	\displaystyle 850(1-0.02)^{\frac{27}{7}}	Substitute the values
	\displaystyle =	\displaystyle 786.2789	Evaluate

To get the amount of water lost, we subtract the amount remaining after 27 days from current amount of water.

\displaystyle \text{Water lost}	\displaystyle =	\displaystyle 850-786.2789	Substitute the values
	\displaystyle =	\displaystyle 63.72 \text{ kL}	Evaluate

How many weeks will it take for the pool to lose at least half its water?

Worked Solution

Create a strategy

Use the depreciation formula: A=P(1-r)^{n}

Apply the idea

We have A=425,\,P=850,\,r=2\%=0.02:

\displaystyle A	\displaystyle =	\displaystyle P(1-r)^{n}	Write the formula
\displaystyle 425	\displaystyle =	\displaystyle 850(1-0.02)^{n}	Substitute the values
\displaystyle \dfrac{425}{850}	\displaystyle =	\displaystyle (1-0.02)^{n}	Divide both sides by 850
\displaystyle (1-0.02)^{n}	\displaystyle =	\displaystyle 0.5	Simplify

Let us try different values for n until we find the closest value to 0.5.

If n=20, we have n=(1-0.02)^{20}=0.6676

If n=30, we have n=(1-0.02)^{30}=0.5455

We are now closer to 0.5 so we can use smaller values.

If n=34, we have n=(1-0.02)^{34}=0.5031

If n=35, we have n=(1-0.02)^{35}=0.4931

So we have: n=35

Reflect and check

You can use calculator to solve for the n:

\displaystyle \log(1-0.02)^{n}	\displaystyle =	\displaystyle \log0.5	Take the logarithm of both sides
\displaystyle n\log(1-0.02)	\displaystyle =	\displaystyle \log0.5	Use the power rule for logarithms
\displaystyle n	\displaystyle =	\displaystyle \dfrac{\log0.5}{\log(1-0.02)}	Divide both sides by \log(1-0.02)
\displaystyle n	\displaystyle =	\displaystyle 34.31	Evaluate

We need to lose at least half its water, so we must go by the end of that week. So:n=35

Idea summary

Any percentage change that happens multiple times can be modelled using the compound interest formula.

If the value is appreciating we can use the formula: A=P(1+r)^n

If the value is depreciating we can use the formula: A=P(1-r)^n

Outcomes

VCMNA328

Connect the compound interest formula to repeated applications of simple interest using appropriate digital technologies

VCMNA359 (10a)

Describe, interpret and sketch parabolas, hyperbolas, circles and exponential functions and their transformations.

5.01 Compound interest

Ideas

The compound interest formula

Examples

Example 1

Different periods

Examples

Example 2

Visualise compound interest

Examples

Example 3

Applications of the compound interest formula

Examples

Example 4

Example 5

Outcomes

VCMNA328

VCMNA359 (10a)

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