Compound interest does not only apply to loans and investments. Concepts such as appreciation, depreciation, inflation, population growth and substance decay are examples of practical applications of compound interest.
The compound formula is just one of a variety of methods for calculating the value of appreciating or depreciating assets.
When an object or investment is said to appreciate, this means it increases in value by a given average percentage.
When an object or investment is said to depreciate, this means it decreases in value by a given average percentage. For the purposes of this course we will consider appreciating assets only.
For example, property prices in a given suburb might have appreciated by $2.1%$2.1% in the last quarter, or perhaps $5.8%$5.8% over the last year.
$A=P\times(1+i)^n$A=P×(1+i)n
where:
$A$A is the final amount of money (principal and interest together)
$P$P is the principal (the initial amount of money invested)
$i$i is the interest rate per compounding period, expressed as a decimal
$n$n is the total number of compounding periods
If a house is valued by a real estate agent to be worth $\$580000$$580000, what would the house be worth three years later, if house prices are expected to appreciate by $0.8%$0.8% per year?
Think: The expected value of the house can be calculated using the compound interest formula. We must change $0.8%$0.8% to the decimal $0.008$0.008 by dividing by $100$100.
Do:
Value after three years: | $A$A | $=$= | $P\left(1+i\right)^n$P(1+i)n |
$=$= | $580000\times\left(1+0.008\right)^3$580000×(1+0.008)3 | ||
$=$= | $580000\times1.008^3$580000×1.0083 | ||
$=$= | $\$594031.66$$594031.66 |
If a piece of land appreciates at an average rate of $3.7%$3.7% per annum and its current value is $\$430000$$430000, calculate its value in $3$3 years. Give your answer to the nearest dollar.
A vintage collectors item that costs $\$6000$$6000, appreciates at approximately $6.6%$6.6% p.a.
After how many full years, $n$n, will the value of the vintage collectors item be over $\$15000$$15000?
Inflation is a very similar concept to appreciation, but instead of looking at the increase in the value of an investment, we instead examine the increase in the prices of goods and services in an economy over time.
The rate of inflation is expressed as a percentage. As inflation causes an increase in prices, the compound interest formula can be used again.
Often, the rate of inflation in a particular country is reported as the average annual inflation rate.
A one year sports club membership currently costs $\$332$$332. Calculate the cost in $6$6 years’ time if the inflation rate is on average $2.6%$2.6% per annum. Give your answer correct to the nearest dollar.
Compound interest is an example of exponential growth because it increases by a constant multiplier each period. There are many other real-life applications to exponential growth including growth of bacteria, population growth for humans or animals, and spread of diseases or rumours. Such examples will have a common form of the equation that models the growth and a graph which looks like:
To find the amount, $A$A, of a population or similar after a given time, we can use our compound interest formula $A=P\left(1+i\right)^n$A=P(1+i)n, where $P$P is the initial amount, $i$i is the growth rate per period and $n$n is the number of periods.
A country's population of $18$18 million people is expected to increase by $2.1%$2.1% p.a. over the next $38$38 years .
Calculate the expected population in $38$38 years time.
Round your answer to the nearest whole number.
A scientist observed that a certain species of bacteria grew by $13%$13% each hour. How many bacteria will be present after $17$17 hours if there were $79$79 initially? Give your answer to the nearest whole number.