Use simple and compound interest formulas to find future values, present values, interest rates and time periods

What this dot point is asking

NESA wants you to apply the simple and compound interest formulas to investments and debts, switch between an annual interest rate and a per-period rate, and solve for any one of $A$, $P$, $r$ or $n$ given the others. You also need to compare scenarios with different compounding frequencies.

The answer

The whole topic turns on a single contrast. Simple interest pays a fixed dollar amount every period, always a percentage of the original deposit, so the balance climbs in equal steps and its graph is a straight line. Compound interest pays interest on the interest already earned, so each period's growth is a little larger than the last and the graph bends upward as an exponential curve. Over one period the two are identical; over a few decades the gap is enormous. The chart below makes that gap concrete before we write down the formulas.

Simple interest

Simple interest is paid only on the original principal. After $n$ time periods at per-period rate $r$,

$P$ is the principal, $I$ is total interest, $A$ is the total amount. Because the same fixed interest $Pr$ is added every period, the balance is a linear (arithmetic) sequence and the graph of $A$ against $n$ is a straight line of gradient $Pr$.

Compound interest

Compound interest is added to the principal at the end of each period and itself earns interest in every subsequent period. After $n$ compounding periods at per-period rate $r$,

The factor $(1 + r)$ is the per-period multiplier: multiplying by it once advances the balance one period, and raising it to the power $n$ advances $n$ periods in one step. That is the whole reason the formula is a power rather than a product: $n$ repeated multiplications by $(1 + r)$ collapse into $(1 + r)^n$. The graph of $A$ against $n$ is exponential, and from the second period onward compound interest beats simple interest for the same nominal rate, because the interest credited in period one is itself earning in period two while the simple-interest balance ignores it.

Per-period rate and number of periods

Rates are usually quoted as a nominal annual rate, but interest may compound more often than annually. Convert before applying the formula.

• Annual compounding: $r = \frac{R}{1}$, $n =$ years.
• Monthly: $r = \frac{R}{12}$, $n = 12 \times$ years.
• Quarterly: $r = \frac{R}{4}$, $n = 4 \times$ years.
• Daily: $r = \frac{R}{365}$, $n = 365 \times$ years.

Where $R$ is the nominal annual rate expressed as a decimal.

Present value

The present value $P$ is the amount you must invest today to grow to a target $A$ in $n$ periods. Rearranging the compound interest formula,

Discounting a future amount back to the present is exactly compounding run in reverse. A dollar promised in the future is worth less than a dollar today, because today's dollar could be invested and would grow; dividing by $(1 + r)^n$ is precisely that discount. A useful sanity check: the present value is always smaller than the future amount whenever $r > 0$.

Discounting back to today, stage by stage

To make the discount tangible, take the present-value calculation from the worked example below: how much must you invest today to have $20000 in $5$ years at $6\%$ per annum compounded annually? Rather than divide by $(1.06)^5$ in one move, walk the future amount back one year at a time. Each year you step toward today, you divide by $1.06$ once.

Stage 1, start at maturity. The $20000 is fixed at year $5$, the moment the money is needed. Nothing has been discounted yet.

Stage 2, discount back one year. Its value at year $4$ is $\dfrac{20000}{1.06} \approx 18867.92$, i.e. $18867.92: the amount you would need a year earlier to reach $20000.

Stage 3, keep discounting. Dividing by $1.06$ again gives the value at year $3$, $\dfrac{20000}{(1.06)^2} \approx 17799.93$, i.e. $17799.93. Each step back shaves off another year's growth.

Stage 4, arrive at today. Five divisions by $1.06$ land on the present value, $\dfrac{20000}{(1.06)^5} \approx 14945.16$, i.e. $14945.16. That single figure is what the closed form $A(1 + r)^{-n}$ computes in one step, and the five arcs are why the exponent is $5$.

Solving for the rate or time

Solving $A = P(1 + r)^n$ for $r$ or $n$:

The time formula needs logarithms, which is fair game in Maths Advanced.

Effective annual rate

The effective annual rate makes different compounding frequencies comparable. If the nominal annual rate is $R$ compounded $m$ times per year,

A nominal $6\%$ compounded monthly is an effective $6.17\%$, slightly more than $6\%$ compounded annually.

The link to geometric series and exponential growth

Compound interest is geometric growth: each period the balance is multiplied by the fixed factor $1 + r$, so the successive balances $P, P(1+r), P(1+r)^2, \dots$ form a geometric sequence with first term $P$ and common ratio $1 + r$. This is why the compound-interest formula and the $n$th term of a geometric sequence are the same equation. It also explains the shape of the graph: an exponential curve that climbs ever more steeply, in contrast to the straight line of simple interest. When interest is paid into a series of regular deposits rather than a single lump, the geometric-series sum formula gives the future value of an annuity, the next step in the financial mathematics topic.

How exam questions ask about simple and compound interest

The wording varies, but each version maps to one of the four rearrangements. Decide which quantity is unknown first, then pick the form:

• "Find the value / future value / amount after $n$ years." Straight $A = P(1 + r)^n$. Convert the rate to a per-period rate first, then substitute.
• "How much should be invested now / deposited today to have $X in $n$ years?" A present-value question: divide, $P = A(1 + r)^{-n}$.
• "How long until the investment reaches / first exceeds $X?" Solve for $n$ with logarithms, then round up to the next whole period, because the balance only clears the target at the end of that period.
• "At what (annual) interest rate...?" Solve for $r$ with the $n$th-root form $r = (A/P)^{1/n} - 1$, then multiply by $k$ if the question wants the nominal annual rate.
• "How much more does compound interest earn than simple interest?" Compute both, $A_{\text{compound}} = P(1 + r)^n$ and $A_{\text{simple}} = P(1 + rn)$, and subtract.
• "Which investment is the better deal?" Compute both future values at the same horizon, or both effective annual rates, and state which is larger with the dollar gap.

Edge cases worth knowing

• Simple interest can beat compound in the first period only if the rate is quoted differently. For the same per-period $r$ they tie after one period and compound wins forever after. They never cross again.
• Rounding the per-period rate too early skews the answer. Carry $r$ to at least six decimal places (or as a fraction); rounding $0.06/12$ to $0.005$ is fine, but rounding $0.052/12$ to $0.004$ instead of $0.004333$ shifts a multi-year answer by dollars.
• "First exceeds" rounds up; "after $n$ years" does not. A doubling time of $10.24$ years means the balance has not yet doubled at the end of year $10$, so $11$ full years are needed; but "the value after $10$ years" is just the year-$10$ figure.
• A present value can never exceed the future amount while $r > 0$. If your $P$ comes out larger than $A$, you multiplied instead of divided.