Coaching Centre
Future value and present value
The compound interest formula FV = PV(1 + r)^n with any compounding period, finding the rate and number of periods, the effect of compounding more often, and working backwards to find present value.
Banks don't pay interest once a year any more. Most savings accounts add interest every month, and some calculate it daily. The formula is the same one you used in Year 9. What changes is how you cut the year into pieces, and that is exactly where HSC marks go missing.
One formula, two values
- $PV$ is the present value: the amount invested or borrowed now.
- $FV$ is the future value: what it grows to.
- $r$ is the interest rate per compounding period, as a decimal.
- $n$ is the number of compounding periods.
The interest earned is $FV - PV$. Each period, the balance is multiplied by $(1 + r)$, so after $n$ periods it has been multiplied by $(1 + r)^n$. That is why the growth speeds up: each period's interest is worked out on a bigger balance than the last.
Rate per period and number of periods
Interest rates are always quoted per year ("p.a." means per annum). If interest is added more often, split the rate and multiply the time by the same number.
| Compounded | Periods per year | $r$ for $6\%$ p.a. | $n$ for $3$ years |
|---|---|---|---|
| yearly (annually) | $1$ | $0.06$ | $3$ |
| half-yearly | $2$ | $0.06 \div 2 = 0.03$ | $3 \times 2 = 6$ |
| quarterly | $4$ | $0.06 \div 4 = 0.015$ | $3 \times 4 = 12$ |
| monthly | $12$ | $0.06 \div 12 = 0.005$ | $3 \times 12 = 36$ |
| daily | $365$ | $0.06 \div 365$ | $3 \times 365 = 1095$ |
Priya puts $\$8000$ from her gap-year job into an account paying $4.2\%$ p.a., compounded monthly. Find the value after $5$ years, and the interest earned.
Does compounding more often help?
Yes, but less than most people expect. With more frequent compounding, interest starts earning its own interest sooner.
Compare $\$5000$ invested for $3$ years at $6\%$ p.a. compounded yearly, quarterly and monthly.
So when two accounts are compared, the rate matters far more than the compounding period. A slightly higher rate compounded yearly can still beat a lower rate compounded monthly, and the only way to be sure is to calculate both.
Set $\$5000$ at $6\%$ for $3$ years and check Example 2. Now stretch the time to $30$ years. How big is the gap between yearly and daily compounding now? Is it bigger as a dollar amount, or as a percentage?
Switch to Work backwards and set a goal of $\$15\,000$ in $4$ years at $4.8\%$. Check Example 3.
Working backwards: present value
Often you know the goal and need the starting amount: "How much do I need to put away now?" Rearrange the formula by dividing both sides by $(1 + r)^n$.
Josh wants $\$15\,000$ for a car deposit in $4$ years. His account pays $4.8\%$ p.a., compounded monthly. How much does he need to deposit now?
Your turn
$\$12\,000$ is invested at $5\%$ p.a., compounded quarterly, for $6$ years. Find the future value and the interest earned.
Show the working
$r = 0.05 \div 4 = 0.0125$ and $n = 6 \times 4 = 24$. $FV = 12\,000(1.0125)^{24} \approx \$16\,168.21$. Interest $= 16\,168.21 - 12\,000 = \$4168.21$.
$\$3000$ is invested at $6\%$ p.a., compounded monthly, for $3$ years. In $FV = PV(1 + r)^n$, the values of $r$ and $n$ are:
Monthly: $r = 0.06 \div 12 = 0.005$ and $n = 3 \times 12 = 36$.
Find the future value of $\$4000$ invested at $5\%$ p.a., compounded annually, for $6$ years. Give your answer to the nearest cent.
$FV = 4000(1.05)^6 \approx \$5360.38$.
Find the future value of $\$2500$ invested at $6\%$ p.a., compounded monthly, for $2$ years, to the nearest cent.
$r = 0.005$, $n = 24$. $FV = 2500(1.005)^{24} \approx \$2817.90$.
Find the interest earned when $\$10\,000$ is invested at $4\%$ p.a., compounded quarterly, for $5$ years, to the nearest cent.
$r = 0.01$, $n = 20$. $FV = 10\,000(1.01)^{20} \approx \$12\,201.90$. Interest $= \$2201.90$.
The same amount is invested at the same annual rate for $10$ years. Which compounding period gives the largest future value?
The more often interest is added, the sooner it starts earning interest itself, so daily gives the most, though only slightly more than monthly.
How much must be invested now at $5\%$ p.a., compounded annually, to have $\$20\,000$ in $5$ years? Answer to the nearest cent.
$PV = \dfrac{20\,000}{1.05^5} \approx \$15\,670.52$.
An account pays $7.3\%$ p.a., compounded daily. Find the interest rate per day, as a percentage.
$7.3 \div 365 = 0.02\%$ per day.
Which expression gives the value of $\$6000$ after $4$ years at $3.6\%$ p.a., compounded monthly?
$r = 0.036 \div 12 = 0.003$ and $n = 4 \times 12 = 48$.
Exam style: Mia needs $\$8000$ in $3$ years for a car. Her savings account pays $4.5\%$ p.a., compounded monthly. How much must she deposit now? Answer to the nearest cent.
$r = 0.045 \div 12 = 0.00375$, $n = 36$. $PV = \dfrac{8000}{1.00375^{36}} \approx \$6991.49$.
Exam style: $\$10\,000$ is invested for $4$ years. Account A pays $5\%$ p.a. compounded annually. Account B pays $4.9\%$ p.a. compounded monthly. How much more does the better account pay, to the nearest cent?
A: $10\,000(1.05)^4 = \$12\,155.06$. B: $10\,000\left(1 + \dfrac{0.049}{12}\right)^{48} \approx \$12\,160.42$. B is better by $12\,160.42 - 12\,155.06 = \$5.35$ (using unrounded values). The lower rate wins here because it compounds monthly, but only just.