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Ratio, proportion & rates of change · Percentages & growth

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Compound growth and decay

Why repeated percentage change is a power of the multiplier, how compound interest differs from simple interest, and how to build growth and decay models.

Ratio, proportion & rates of change · Percentages & growth

Compound growth and decay

Why repeated percentage change is a power of the multiplier, how compound interest differs from simple interest, and how to build growth and decay models.

Why it works

Compound growth is what happens when the same percentage change is applied over and over, each time to the new total. Because each step is a multiplication by the same multiplier, doing it nn times is that multiplier raised to the power nn.

Say £2000\pounds 2000 earns 3%3\% interest a year, compounded. Each year the balance is multiplied by 1.031.03:

2000→×1.032060→×1.032121.80→×1.03…2000 \xrightarrow{\times 1.03} 2060 \xrightarrow{\times 1.03} 2121.80 \xrightarrow{\times 1.03} \dots

After nn years you have multiplied by 1.031.03 a total of nn times, so

amount=2000×(1.03)n.\text{amount} = 2000 \times (1.03)^n.

That single formula is the heart of the topic:

final=initial×(1±r100)n,\text{final} = \text{initial} \times \left(1 \pm \frac{r}{100}\right)^{n},

with ++ for growth and −- for decay, rr the percentage rate per period, and nn the number of periods.

Why it isn't simple interest. Simple interest pays the same fixed amount each year — 3%3\% of the original £2000\pounds 2000, i.e. £60\pounds 60 every year, £180\pounds 180 over three years, giving £2180\pounds 2180. Compound interest pays interest on the interest already earned, so year 2 earns 3%3\% of £2060\pounds 2060, not of £2000\pounds 2000. Over three years compound gives 2000×1.033=£2185.452000\times 1.03^3 = \pounds 2185.45 — a little more than simple, and the gap widens fast over longer periods. The tell-tale error is doing simple interest (n×n \times one year's interest) when the question says "compound".

Decay is the same machine with a multiplier below 1. A car worth £15 000\pounds 15\,000 losing 18%18\% of its value each year is multiplied by 0.820.82 each year (it keeps 82%82\%), so after nn years it is worth 15 000×0.82n15\,000 \times 0.82^{n}. Depreciation, radioactive-style decay, and population decline all use this. The common slip is to use ×1.18\times 1.18 or ×0.18\times 0.18 instead of ×0.82\times 0.82 — build the multiplier from 11 as always: 18%18\% lost means 82%82\% kept.

Reading the power tells you the time. Because nn is an exponent, questions like "how many years until the investment first exceeds £2500\pounds 2500?" are answered by trying values of nn until the amount passes the target — the balance grows faster and faster, so it crosses once and stays above.