Compound Interest — Mathslive.ie

Section 1 of 9

The Setup

Compound interest is just interest stacking on top of interest. The pile grows each period, then the next period's interest is worked out on the new bigger pile.

Before any formula, learn the letters. You will meet these in every money question on the paper.

The letters:
$P$ = present value = principal (what you start with)
$F$ = future value = amount (what it grows to)
$t$ = time
$i$ = rate of interest (as a decimal, not a %)

2% rate means $i = 0.02$. Always convert before plugging in.

Three rate names — learn the difference:
AER = annual equivalent rate → save
APR = annual percentage rate → borrow
Nominal rate = used when the rate's period and the question's period are different (e.g. "12% nominal compounded monthly")

Must learn — the formula:
$F = P(1+i)^{t}$

Rearranged for $P$:
$P = \dfrac{F}{(1+i)^{t}}$

Both forms are in the Tables.

TRY THIS

Match each value below to $P$, $F$, $t$ or $i$:
"€500 is invested for 4 years at 3% AER. It grows to €562.75."

Write out P, F, t, i on paper first. Watch the % — it has to become a decimal.

"is invested" → starting amount → $P = 500$

"It grows to" → final amount → $F = 562.75$

"for 4 years" → $t = 4$

"3%" → $i = 0.03$

$P = 500$, $F = 562.75$, $t = 4$, $i = 0.03$

Section 2 of 9

Basic Compound Interest

Before we just plug into the formula — let's see why it works. Watch the pile grow year by year.

(i) Building it up year by year

Example: €200 is invested for 2 years at 2% AER. Find its value at the end of year 2.

After year 1: $0.02 \times 200 = 4$ → €204

i.e. $200 \times 1.02 = 204$

After year 2: $204 \times 1.02$

$= 200 \times 1.02 \times 1.02$

$= 200 \times (1.02)^{2}$

Each year multiplies by $(1+i)$. After $t$ years, multiply $t$ times.

(ii) Now use the formula

$P = 200, \quad i = 0.02, \quad t = 2$

$F = P(1+i)^{t}$

$F = 200(1.02)^{2}$

F = €208.08

TRY THIS

€1,500 is invested for 4 years at 3.5% AER. Find its value at the end of the 4 years.

Write down $P$, $i$, $t$. Then $F = P(1+i)^{t}$.

$P = 1500, \quad i = 0.035, \quad t = 4$

$F = 1500(1.035)^{4}$

F = €1721.28

TRY THIS

€8,000 is invested for 10 years at 4.2% AER. What is it worth at the end?

Same formula. Just careful with the decimal.

$P = 8000, \quad i = 0.042, \quad t = 10$

$F = 8000(1.042)^{10}$

F = €12,071.67

TRY THIS

€600 is invested for 3 years at 2% AER. Find its value at the end of the 2nd year.

Careful — the money sits for 3 years total, but they're asking about the end of year 2.

$P = 600, \quad i = 0.02$

$t = 2$ (we want value at end of year 2, not end of investment)

$F = 600(1.02)^{2}$

F = €624.24

Section 3 of 9

Finding Present Value

Sometimes you're given the future value and asked what was originally invested. Same formula, rearranged.

Must learn:
$P = \dfrac{F}{(1+i)^{t}}$

(i) Worked example

A sum of money is invested for 3 years at 2.5% AER. It is worth €6,250 at the end of year 3 compounded annually. How much was invested?

$F = 6250, \quad i = 0.025, \quad t = 3$

$P = \dfrac{F}{(1+i)^{t}}$

$P = \dfrac{6250}{(1.025)^{3}}$

P = €5803.75

TRY THIS

An investment is worth €10,000 after 5 years at 3% AER. How much was originally invested?

Future value is given. Use $P = \dfrac{F}{(1+i)^{t}}$.

$F = 10000, \quad i = 0.03, \quad t = 5$

$P = \dfrac{10000}{(1.03)^{5}}$

P = €8626.09

TRY THIS

A bond matures at €25,000 in 8 years. The AER is 4.5%. What should it cost today?

"What should it cost today" = the present value $P$.

$F = 25000, \quad i = 0.045, \quad t = 8$

$P = \dfrac{25000}{(1.045)^{8}}$

P = €17,579.63

Section 4 of 9

Continuous Growth

Some investments are described as compounded continuously — meaning the interest is being added every instant, not just once a year.

Different scenario, different formula. $e$ shows up because that's what continuous growth looks like.

Must learn:
$F = P\, e^{rt}$

Rearranged for $P$:
$P = \dfrac{F}{e^{rt}}$

$e \approx 2.71$ — your calculator has an $e^x$ button.

(i) Worked example

€200 is invested for 2 years at 2% nominal interest per annum, compounded continuously. Find its value at the end of year 2.

$P = 200, \quad r = 0.02, \quad t = 2$

$F = P\, e^{rt}$

$F = 200\, e^{2 \times 0.02}$

$F = 200\, e^{0.04}$

F = €208.16

Tiny bit bigger than the €208.08 we got compounding annually at 2%. Makes sense — continuous compounding stacks more often, so more interest.

TRY THIS

€3,000 is invested at 4% nominal per annum, compounded continuously, for 5 years. Find its value at the end.

$F = P e^{rt}$. Punch $e^{rt}$ in first, then multiply by $P$.

$P = 3000, \quad r = 0.04, \quad t = 5$

$F = 3000\, e^{5 \times 0.04} = 3000\, e^{0.2}$

F = €3664.21

TRY THIS

A bacteria culture grows continuously. The current mass is 0.5 g. After 6 hours, find its mass if the rate is 25% per hour, compounded continuously.

Same formula even though it's not money. Continuous growth is continuous growth.

$P = 0.5, \quad r = 0.25, \quad t = 6$

$F = 0.5\, e^{6 \times 0.25} = 0.5\, e^{1.5}$

$F = 2.241$ g

Section 5 of 9

Nominal Rate vs AER

This is where most students lose marks. Take it slowly.

AER means "the rate that, applied once a year, gives you the actual growth". So if AER is 3%, your money actually grows by 3% over the year — full stop.

But what if interest is added every month? You can't just divide by 12. Let's see why.

(i) The trap

An annual rate of 12% is charged on a loan. Find the equivalent monthly compounded rate.

Tempting (and wrong): just do $12 \div 12 = 1\%$.

Why is it wrong? Because if you charged 1% every month and compounded it, the actual yearly growth would be more than 12%. The rate has to be a bit smaller to land on exactly 12% per year.

(ii) The right way

Let $P$ stay general. After a year of growing by 12%: $F = 1.12 P$.
Let's pick nice numbers: $P = 100$, $F = 112$.

After 12 months of monthly compounding at rate $i$:

$112 = 100(1+i)^{12}$

$1.12 = (1+i)^{12}$

$1+i = \sqrt[12]{1.12}$

$1+i = 1.009$

$i = 0.009 = 0.9\%$ per month

So the monthly rate is 0.9%, not 1%. Tiny difference per month, big difference after years.

Must learn — the conversion:
If AER (or APR) is $i_{\text{annual}}$, then the monthly rate $i_{\text{m}}$ satisfies:
$(1+i_{\text{m}})^{12} = 1+i_{\text{annual}}$
$1+i_{\text{m}} = \sqrt[12]{\,1+i_{\text{annual}}\,}$

(iii) Proof it works — annually = monthly

€5,000 is invested for 2 years at 3% AER. Find its value when compounded (a) annually, (b) monthly.

(a) Annually

$P = 5000, \quad i = 0.03, \quad t = 2$

$F = 5000(1.03)^{2}$

F = €5304.50

(b) Monthly

First find the monthly rate:

$(1+i)^{12} = 1.03$

$1+i = \sqrt[12]{1.03} = 1.002$

Now $t$ is in months: $t = 2 \times 12 = 24$

$F = 5000(1.002)^{24}$

F = €5304.50

Same answer! That's because AER is by definition the equivalent yearly rate — so however often you compound (using the correct equivalent rate), you must land on the same place.

TRY THIS

The AER on a savings account is 4%. Find the equivalent monthly compounded rate (to 4 d.p.).

$(1+i)^{12} = 1.04$. Then take 12th root.

$(1+i)^{12} = 1.04$

$1+i = \sqrt[12]{1.04} = 1.00327$

$i = 0.00327 = 0.3274\%$ per month

$i \approx 0.3274\%$ per month

TRY THIS

€2,000 is invested for 3 years at 5% AER, compounded monthly. Find its value at the end of year 3.

Two-stage problem: first find the monthly rate from AER, then plug into $F = P(1+i)^{t}$ with $t$ in months.

Monthly rate: $(1+i)^{12} = 1.05$

$1+i = \sqrt[12]{1.05} = 1.00407$

$t = 3 \times 12 = 36$

$F = 2000(1.00407)^{36}$

F = €2315.25

TRY THIS

A credit card charges interest at 1.5% per month. Find the AER.

Going the other way now. AER = the yearly equivalent. Compound 1.015 twelve times.

$(1.015)^{12} = 1 + i_{\text{annual}}$

$1.1956 = 1 + i_{\text{annual}}$

AER $= 0.1956 = 19.56\%$

AER $\approx 19.56\%$

Section 6 of 9

Solving for time t

Now flip it. You know what you start with, you know what you want it to grow to — how long will it take?

$t$ is stuck up in an exponent, so we need logs to pull it down. Easier with continuous growth — use $\ln$. With annual compounding, use $\log$.

(i) Worked example — continuous

How long will it take €500 to grow to €600 compounded continuously at a nominal rate of 3%?

$P = 500, \quad F = 600, \quad r = 0.03$

$F = P\, e^{rt}$

$600 = 500\, e^{0.03 t}$

$\dfrac{6}{5} = e^{0.03 t}$

$\ln\!\left(\dfrac{6}{5}\right) = 0.03 t$

$t = \dfrac{\ln(6/5)}{0.03} = 6.08$

$t = 7$ years (always round up for "how long")

Round-up rule: if you say "6 years" the money hasn't quite reached €600 yet. So 7 full years is the smallest integer that works.

(ii) Worked example — annual

How long will it take €1,000 to double at 4% AER, compounded annually?

$P = 1000, \quad F = 2000, \quad i = 0.04$

$F = P(1+i)^{t}$

$2000 = 1000(1.04)^{t}$

$2 = (1.04)^{t}$

$\log 2 = t \log 1.04$

$t = \dfrac{\log 2}{\log 1.04} = 17.67$

$t = 18$ years

TRY THIS

How long will it take €800 to grow to €1,200 at 5% AER, compounded annually?

Annual compounding → use $\log$. Round up the answer.

$1200 = 800(1.05)^{t}$

$1.5 = (1.05)^{t}$

$\log 1.5 = t \log 1.05$

$t = \dfrac{\log 1.5}{\log 1.05} = 8.31$

$t = 9$ years

TRY THIS

How long does it take €1,500 to grow to €2,000 at 4% nominal interest, compounded continuously?

Continuous → use $\ln$.

$2000 = 1500\, e^{0.04 t}$

$\dfrac{4}{3} = e^{0.04 t}$

$\ln(4/3) = 0.04 t$

$t = \dfrac{\ln(4/3)}{0.04} = 7.19$

$t = 8$ years

Section 7 of 9

Inflation

Inflation is just compound interest dressed up differently — prices growing instead of an investment growing. Same formula.

$F = P(1+i)^{t}$ where now:
$P$ = today's price · $F$ = future price · $i$ = inflation rate

(i) Forward — what will it cost?

A car costs €7,500. If inflation runs at an average of 3.5%, find the cost of the same car in 3 years' time.

$P = 7500, \quad i = 0.035, \quad t = 3$

$F = 7500(1.035)^{3}$

F = €8315.38

(ii) Backward — what did it cost?

The average annual inflation rate over the last 3 years is 5.6%. An item at present costs €3,500 this year. How much should it have cost 3 years ago?

$F = 3500, \quad i = 0.056, \quad t = 3$

$P = \dfrac{F}{(1+i)^{t}} = \dfrac{3500}{(1.056)^{3}}$

P = €2972.19

TRY THIS

A house costs €280,000 today. If inflation runs at 2.4%, find its price in 8 years' time.

Same formula as basic CI.

$P = 280000, \quad i = 0.024, \quad t = 8$

$F = 280000(1.024)^{8}$

F = €338,499.23

F ≈ €338,499.23

TRY THIS

A pair of shoes now costs €120. Inflation has averaged 3% over the last 5 years. What did the same pair cost 5 years ago?

Backwards — $F$ is now, $P$ is 5 years ago. Use $P = F/(1+i)^{t}$.

$F = 120, \quad i = 0.03, \quad t = 5$

$P = \dfrac{120}{(1.03)^{5}}$

P = €103.51

P ≈ €103.51

Section 8 of 9

Depreciation — Reducing Balance

Depreciation is compound interest in reverse — value going down, not up. A car, a tractor, a laptop — all lose value each year.

Must learn:
$F = P(1-i)^{t}$ (reducing balance)

Notice the minus. Each year you keep $(1-i)$ of the previous value.
$F$ here is called the net book value (NBV).

(i) Year-by-year build-up

A tractor costs €250,000. It reduces in value at a rate of 12% per annum. Find its value at the end of each year for 3 years.

$P = 250000, \quad i = 0.12, \quad (1-i) = 0.88$

Year 1: F = 250000 × 0.88 = €220,000

Year 2: F = 220000 × 0.88 = €193,600

Year 3: F = 193600 × 0.88 = €170,368

Or in one shot using the formula: F = 250000(0.88)³ = €170,368. Same answer.

(ii) Backward — finding the original price

A car is worth €12,000 after 3 years with depreciation of 15% per annum. Find its value when new.

$F = 12000, \quad i = 0.15, \quad t = 3$

$P = \dfrac{F}{(1-i)^{t}} = \dfrac{12000}{(0.85)^{3}}$

P = €19,540

TRY THIS

A laptop costs €1,400 new. It depreciates at 20% per annum. Find its value after 4 years.

$(1 - 0.20) = 0.80$ each year.

$P = 1400, \quad i = 0.20, \quad t = 4$

$F = 1400(0.80)^{4}$

F = €573.44

TRY THIS

A van is worth €8,500 after 5 years. It depreciates at 18% per annum. Find its price when new.

Going backwards: $P = F/(1-i)^{t}$.

$F = 8500, \quad i = 0.18, \quad t = 5$

$P = \dfrac{8500}{(0.82)^{5}}$

P = €22,927.13

P ≈ €22,927.13

Section 9 of 9

Continuous Depreciation

Same idea as continuous growth, but the rate is negative — value falls continuously.

Must learn:
$F = P\, e^{-rt}$

Note the minus sign in the exponent — that's what makes it shrink.

(i) Worked example — finding the rate

A car cost €21,000 new. It has a value of €16,300 after 3 years of continuous depreciation. Find the rate of depreciation.

$P = 21000, \quad F = 16300, \quad t = 3$

$F = P\, e^{-rt}$

$16300 = 21000\, e^{-3r}$

$\dfrac{16300}{21000} = e^{-3r}$

$\ln\!\left(\dfrac{16300}{21000}\right) = -3r$

$-0.2534 = -3r$

$r = 0.0845$

$r \approx 8.4\%$ per annum

TRY THIS

A machine cost €40,000 new. After 5 years of continuous depreciation it is worth €25,000. Find the rate of depreciation.

$F = Pe^{-rt}$. Isolate $e^{-rt}$, then $\ln$ both sides.

$25000 = 40000\, e^{-5r}$

$\dfrac{25000}{40000} = e^{-5r}$

$\ln(0.625) = -5r$

$-0.4700 = -5r$

$r = 0.0940 = 9.4\%$

$r \approx 9.4\%$ per annum

TRY THIS

A laptop costs €1,800 new. It depreciates continuously at 22% per annum. Find its value after 4 years.

Continuous depreciation forward: $F = Pe^{-rt}$.

$P = 1800, \quad r = 0.22, \quad t = 4$

$F = 1800\, e^{-0.22 \times 4} = 1800\, e^{-0.88}$

F = €746.61

F ≈ €746.61

End of Compound Interest.

Coming next: Annuities & Series — savings instalments, loans, pensions, NPV and bonds.