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5.07 Modelling with compound interest

Lesson

So far we have used the concept of compound interest to model money investments where an initial amount grows at a certain rate each period. In such cases, the interest rate is applied to a growing balance.

For example, a $\$1000$$1000 principal that earns compound interest at $1%$1% grows to:

  • $1000\times1.01=1010$1000×1.01=1010 after $1$1 year
  • $1010\times1.01=1020.10$1010×1.01=1020.10 after $2$2 years, and so on.

But the underlying idea of compounding can be applied to any situation where an initial amount increases at the same rate each time period.

Population growth is a great example of where compounding occurs.

If a country’s population is currently $10000000$10000000 people, and it is forecast to grow by $3%$3% each year for the next $2$2 years, we can model the population using compound interest:

  • In the first year, the initial population will grow by $3%$3%:
    Population after $1$1 year $=$=$10000000\times1.03=10300000$10000000×1.03=10300000
  • In the second year, the new population of $10300000$10300000 will again grow by $3%$3%:
    Population after $2$2 years $=$=$10300000\times1.03=10609000$10300000×1.03=10609000

Government agencies rely on population growth models to be able to plan economically, socially and environmentally for the future.

As we’ll see below, compounding comes about in many other contexts too.

Appreciation

In general, the price of things like property, jewellery, artwork, and rare collectables will increase over time. This is known as appreciation, and we might say, for example, that the value of a house has appreciated. To determine the value of an asset at some point in the future we use the familiar compound interest formula, provided that we know the rate at which it appreciates.

Exploration

A first edition copy of "A Christmas Carol" by Charles Dickens is worth approximately $\$9500$$9500 today. If this book appreciates in value at a rate of $6.75%$6.75% p.a., we can calculate its value in $15$15 years.

The value of the book increases at a constant rate of $6.75%$6.75% each year, so we can use the compound interest formula to determine its future value. We know the present value of the book, the rate at which it appreciates, and the number of compounding periods.

$FV$FV $=$= $PV\left(1+r\right)^n$PV(1+r)n
  $=$= $9500\left(1+\frac{6.75}{100}\right)^{15}$9500(1+6.75100)15
  $=$= $9500\left(1+0.0675\right)^{15}$9500(1+0.0675)15
  $=$= $9500\times1.0675^{15}$9500×1.067515
  $=$= $25307.07$25307.07 (2 d.p.)

The first edition book will be worth approximately $\$25307.07$$25307.07 in $15$15 years.

Inflation

Many decades ago the price of things like meals, cinema tickets, and clothes were all in the range of a few dollars, or even less than a dollar. These days we pay $10$10 or $20$20 times as much for these same goods. What has happened?

It is not the case that the actual value of these goods has increased! Rather the value, or buying power, of the money we use to buy things has itself decreased. The result is that our money is worth less so we have to use more of it to purchase the same items.

This is known as inflation, and it has been happening throughout all of human history, though it has only become well understood recently in the context of a global economy. The behaviour of inflation is similar to appreciation; some amount of money today will have the same buying power as a larger amount of money in the future. If we know the inflation rate we can use the compound interest formula to model the value of money over time.

Exploration

Amanda has just begun Year 7 at school, and notices that the school canteen sells sandwiches for $\$5.80$$5.80 each. She remembers reading in the paper recently that the inflation rate has been at a consistent $2.31%$2.31% p.a. Amanda wonders how much the same type of sandwich might cost by the time she reaches Year 12.

We are given an inflation rate, and we want to know how much $\$5.80$$5.80 in today's money will be worth in $12-7=5$127=5 years' time.

$FV$FV $=$= $PV\left(1+r\right)^n$PV(1+r)n
  $=$= $5.8\left(1+\frac{2.31}{100}\right)^5$5.8(1+2.31100)5
  $=$= $5.8\left(1+0.0231\right)^5$5.8(1+0.0231)5
  $=$= $5.8\times1.0231^5$5.8×1.02315
  $=$= $6.50$6.50 (2 d.p.)

So we can estimate that Amanda will be paying about $\$6.50$$6.50 for a sandwich from the canteen when she is in Year 12.

In reality both appreciation and inflation will affect the value of an asset over time. Accounting for both factors requires complex models, so in the following questions we will consider only one at a time to determine the future value.

Practice questions

Question 1

What is the value of a wine collection, currently valued at $\$2550$$2550, after $3$3 years if its rate of appreciation is $8%$8% p.a.? Round your answer to the nearest cent.

Question 2

A movie ticket currently costs $\$20$$20 at the local cinema. Calculate the price in $15$15 years’ time if the inflation rate is on average $3.07%$3.07% p.a. Give your answer correct to the nearest dollar.

Question 3

The average price of a new apartment in town is $\$771000$$771000 today. Find the average price of an apartment $23$23 years ago if the inflation rate was an average of $3.74%$3.74% p.a. Give your answer to the nearest dollar.

Outcomes

MS2-12-5

makes informed decisions about financial situations, including annuities and loan repayments

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