So far we have explored how the initial value of an investment can increase to some larger future value by earning simple or compound interest over time. In these cases we knew the principal or present value ($PV$PV), the interest rate ($r$r), and the investment duration ($n$n), and we wanted to find the future value ($FV$FV) of the investment.
But what if we already know the future value, or if we have a future value in mind that we want our investment to reach after a certain amount of time? What present value do we need to begin with in order to reach this future value?
For example, if you would like to have saved $\$50000$$50000 to pay a deposit on an apartment in $5$5 years’ time, you can calculate how much you’d need to put into a savings account now so that it grow to $\$50000$$50000 in $5$5 years, at the current interest rate.
This new approach involves a rearrangement of the relevant interest formula to make $PV$PV the subject.
$FV=PV(1+r)^n$FV=PV(1+r)n
To make $PV$PV the subject, divide both sides by $\left(1+r\right)^n$(1+r)n
$PV=\frac{FV}{(1+r)^n}$PV=FV(1+r)n
Saving for a purchase
Say we want to purchase a used car in $3$3 years’ time with a budget of $\$6000$$6000, and we know that the bank can offer an interest rate of $5.3%$5.3% p.a., compounded annually. If we open a new savings account with the bank, how much will we need to deposit so that we can purchase the car after $3$3 years?
This scenario involves compound interest, so we will be using the formula $FV=PV\left(1+r\right)^n$FV=PV(1+r)n. We know the final value, the interest rate, and the investment duration. The variable we want to find is the present value, $PV$PV.
| $FV$FV | $=$= | $PV\left(1+r\right)^n$PV(1+r)n | |
| $\frac{FV}{\left(1+r\right)^n}$FV(1+r)n | $=$= | $PV$PV | (Rearrange to make $PV$PV the subject) |
| $PV$PV | $=$= | $\frac{6000}{\left(1+\frac{5.3}{100}\right)^3}$6000(1+5.3100)3 | (Substitute the known values) |
| $=$= | $\frac{6000}{\left(1+0.053\right)^3}$6000(1+0.053)3 | ||
| $=$= | $\frac{6000}{1.053^3}$60001.0533 | ||
| $=$= | $5138.85$5138.85 (2 d.p.) |
So we will need to deposit at least $\$5138.85$$5138.85, to the nearest cent.
We can double check that this present value will give us a sufficient future value by substituting $PV=5138.85$PV=5138.85 into the formula $FV=PV\left(1+r\right)^n$FV=PV(1+r)n to get $FV=5138.85\left(1+\frac{5.3}{100}\right)^3=\$6000$FV=5138.85(1+5.3100)3=$6000, to the nearest cent.
Practice questions
Question 1
A $\$4080$$4080 investment earns interest at $6.2%$6.2% p.a., compounded annually over $7$7 years.
What is the future value of the investment to the nearest cent?
Another investment in the same account will grow to $\$6000$$6000 in $7$7 years. What is the present value of the investment, to the nearest cent?
Question 2
An investment earns interest at $6.53%$6.53% p.a., compounded monthly. What is the present value of the investment if it will be worth $\$5840$$5840 in $14$14 years?
Question 3
The sports stands at a local playing field will need to be replaced in $16$16 years. The construction costs for new sports stands are estimated at $\$25000$$25000. If the mayor opens an account with $3.41%$3.41% p.a. interest, compounded monthly, what is the minimum initial deposit that they can make in order to be able to purchase the new stands in $16$16 years?
Finding other unknown values
We can also use the future value formula to calculate the interest rate if we know the present value, the future value and the term of the investment. To find the interest rate, you can solve the equation using technology or use trial and error.
For example, if we want a $\$2000$$2000 investment to double in value over $10$10 years, what annual compound interest rate would we need to invest at?
Let's start with the compound interest formula, substitute the known values, and rearrange the equation to find the value of $r$r.
| $FV$FV | $=$= | $PV\left(1+r\right)^n$PV(1+r)n | |
| $4000$4000 | $=$= | $2000\left(1+r\right)^{10}$2000(1+r)10 | (The future value is double the principal investment) |
| $2$2 | $=$= | $\left(1+r\right)^{10}$(1+r)10 | (Divide both sides by $2000$2000) |
| $\sqrt[10]{2}$10√2 | $=$= | $1+r$1+r | (Take the tenth root of both sides) |
| $r$r | $=$= | $\sqrt[10]{2}-1$10√2−1 | (Isolate $r$r) |
| $r$r | $=$= | $7.18%$7.18% (2 d.p.) |
The interest rate is $7%$7% p.a. correct to the nearest percent.