Compound Interest and Depreciation

Why repeated percentage change compounds

A percentage change is applied to the current amount, not repeatedly to the original amount. This makes the changes compound.

For example, suppose money in an account earns interest at the end of every year. After the first year, the interest is added to the balance. The second year's interest is then calculated from this larger balance, so interest is earned on both the original investment and the earlier interest.

Depreciation follows the same mechanism in reverse. If a car loses a percentage of its value each year, the second decrease is calculated from its value after the first decrease. Equal percentage decreases therefore do not usually mean equal cash decreases.

This is different from simple interest, where the interest added in every period is calculated from the original investment.

Turning a percentage into a multiplier

For an increase of $r\%$, the amount left after one interval is $(100+r)\%$ of the previous amount. The multiplier is

$1+\frac{r}{100}.$

For example, an increase of $3\%$ has multiplier $1.03$ because the new amount is $103\%$ of the old amount.

For a decrease of $r\%$, the amount remaining is $(100-r)\%$ of the previous amount. The multiplier is

$1-\frac{r}{100}.$

A depreciation of $12\%$ therefore has multiplier $0.88$: the item retains $88\%$ of its previous value.

The multiplier represents the whole new amount. Multiplying by $0.03$ finds only $3\%$ of an amount; multiplying by $1.03$ increases the whole amount by $3\%$.

Working through compound interest period by period

Suppose £2,500 is invested at $3\%$ compound interest per year for four years. The annual multiplier is $1.03$, and it is applied once for each complete year:

• after year 1: $2500\times1.03=2575$;
• after year 2: $2575\times1.03=2652.25$;
• after year 3: $2652.25\times1.03=2731.8175$;

The final balance is therefore £2,813.77, rounded to the nearest penny. If the question asks for the interest earned rather than the balance, subtract the original investment:

$£2813.77-£2500=£313.77.$

Keeping the unrounded calculator value until the end avoids small rounding errors building up from year to year.

Working through depreciation with changing rates

A rate need not remain constant. If different percentages apply in different intervals, find and apply the appropriate multiplier at each stage.

Consider a new car worth £15,000 which depreciates by $30\%$, $20\%$ and $15\%$ respectively over three successive years. The word respectively fixes the order of the rates.

• Year 1: a $30\%$ decrease leaves $70\%$, so $15000\times0.70=10500$.
• Year 2: a $20\%$ decrease leaves $80\%$, so .

The car's value after three years is £7,140. Its total depreciation is the original value minus the final value:

$£15000-£7140=£7860.$

It would be incorrect to add the three rates and reduce the original value by $65\%$. Each percentage acts on a different current value.

Matching multipliers to intervals

First identify the interval attached to the rate. An annual rate is applied once per year; a monthly rate is applied once per month. Thus a monthly change lasting two years involves 24 monthly intervals, not two.

A useful written chain is

$\text{starting amount}\ \longrightarrow\ \text{after interval 1}\ \longrightarrow\ \text{after interval 2}\ \longrightarrow\cdots$

Place the correct multiplier above each step. This makes changing rates and incomplete reading of the time period easier to spot.

Interpreting the result

Finish by connecting the number to the context. State whether it is a bank balance, an item's remaining value, the interest earned or the value lost. A final amount after growth should normally exceed the starting amount, while a remaining value after depreciation should be smaller. This simple check can expose a reversed multiplier or an incorrect number of intervals.