# 5.01 Compound interest

Lesson

### Comparing simple and compound interest

When you invest money, you usually earn interest. There are two main ways you can earn interest: simple and compound.

• 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$$1000 at 2%2% per annum simple interest, you will earn 2%2% of \1000$$1000 each year. The interest of $\$20$$20 earned each year does not change over the term of the investment. • With compound interest, the interest earned in each time period is added to the principal. So the interest earned in each time period increases because it is calculated on a growing balance. For example, if you invest \1000$$1000 at $2%$2% per annum compounded yearly, you will earn $\$20$$20 interest in the first year, which is added to the principal. In the second year, interest of 2%2% is calculated on a new balance of \1020$$1020, not $\$1000$$1000. This means compound interest usually leads to greater returns than simple interest because the principal is updated each period after interest is earned. Let's compare the two with an example of earning 5%5% interest on an initial investment principal of \1000$$1000.

Year Principal Used Simple interest earned New total
1 $\$1000$$1000 \50$$50 $\$1050$$1050 2 \1000$$1000 $\$50$$50 \1100$$1100
3 $\$1000$$1000 \50$$50 $\$1150$$1150 Year Principal Used Compound interest earned New total 1 \1000$$1000 $\$50$$50 \1050$$1050
2 $\$1050$$1050 \52.50$$52.50 $\$1102.50$$1102.50 3 \1102.50$$1102.50 $\$55.13$$55.13 \1157.63$$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.

### Repeated applications of simple interest

In the previous example on compound interest, the new total after $1$1 year was $1000+1000\times0.05$1000+1000×0.05. If we factor out the principal amount, we can see that our new total after $1$1 year is equal to $1000\times1.05$1000×1.05.

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\times1.05\times1.05$1000×1.05×1.05.

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

For example: if John takes out a loan of $\$2000$$2000 at an interest rate of 4%4%, his total amount to repay at the end of each year will be equal to 1.041.04 times the total of the previous year. After 55 years, his total amount to repay will be equal to \2000\times1.04\times1.04\times1.04\times1.04\times1.04$$2000×1.04×1.04×1.04×1.04×1.04.

### 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.

#### Worked example

Cam is investing $\$1000$$1000 into an account that accumulates interest at a rate of 2.4%2.4% p.a., compounding monthly. What will the account's total be after 33 years? Think: Since interest is compounding monthly, interest will be calculated 1212 times a year. Since the interest rate of 2.4%2.4% is an annual interest rate, we need to divide it by 1212. This will give us a monthly interest rate of \frac{2.4%}{12}2.4%12. Since there are 1212 months in a year, after 33 years Cam's investment will have experienced 3\times123×12 periods. Do: Substituting our values into the compound interest formula gives us: A=1000\left(1+\frac{2.4%}{12}\right)^{3\times12}A=1000(1+2.4%12)3×12 Simplifying our values gives us: A=1000\times1.005^{36}A=1000×1.00536 Evaluating this and rounding to the nearest cent tells us that, after 33 years, Cam's investment will reach a total of: A=\1196.68A=1196.68 ### Visualising 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. #### Practice question ##### Question 1 The graph below shows both an investment with simple interest, and one with compound interest, labelled Investment A and Investment B respectively. Loading Graph... 1. Which investment has a higher principal amount? Investment A A Investment B B Investment A A Investment B B 2. Which investment has a higher final amount after 1010 years? Investment A A Investment B B Investment A A Investment B B 3. After how many years are the investments equal in value? 99 years A 66 years B 1010 years C 1212 years D 99 years A 66 years B 1010 years C 1212 years D As we can see, 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. Simple interest is calculated using the formula A=P+PrnA=P+Prn which is a linear equation in terms of nn, the number of periods. Meanwhile, compound interest uses the formula A=P\left(1+r\right)^nA=P(1+r)n which is non-linear equation in terms of nn. ### 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. #### Practice questions ##### Question 2 Victoria borrows \35000$$35000 at a rate of $4.8%$4.8% p.a, compounding monthly.

1. After $4$4 months, Victoria repays the loan all at once. How much money does she pay back in total?

Round your answer to the nearest cent.

2. How much interest was generated on the loan over the four months?

Valerie borrows $\$1250$$1250 at a rate of 7.6%7.6% p.a, compounding annually. 1. Which of the following expressions represents the amount Valerie must repay after 1515 years, assuming that she hasn't paid anything back? 1250\times\left(1+0.076\right)\times151250×(1+0.076)×15 A 1250\times\left(1+0.076\right)\times\frac{15}{12}1250×(1+0.076)×1512 B 1250\times\left(1+0.076\right)^{\frac{15}{12}}1250×(1+0.076)1512 C 1250\times\left(1+0.076\right)^{15}1250×(1+0.076)15 D 1250\times\left(1+0.076\right)\times151250×(1+0.076)×15 A 1250\times\left(1+0.076\right)\times\frac{15}{12}1250×(1+0.076)×1512 B 1250\times\left(1+0.076\right)^{\frac{15}{12}}1250×(1+0.076)1512 C 1250\times\left(1+0.076\right)^{15}1250×(1+0.076)15 D 2. How much interest was generated on the loan over the fifteen years? Round your answer to two decimal places. ##### Question 4 Bart is planning to invest \110000$$110000 into a saving scheme which accumulates interest weekly. If he aims to have his investment reach $\$120000120000 after $2$2 years, what interest rate per annum, $r$r, will he need?

Enter each line of working as an equation.

Round your answer to four decimal places.

##### Question 5

A swimming pool is losing water at the rate of $2%$2% per week due to evaporation.

The pool currently holds $850$850 kilolitres of water.

1. How much water will be lost in the next $27$27 days?

Give your answer in kilolitres and correct to two decimal places .

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

### 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.