Successive percentage changes

Date of the original text: 25 October 2009.

In general, the final change after a sequence of increases, represented by positive values, or decreases, represented by negative values, is the product of the corresponding multipliers, minus 1:

[(100 + a) / 100] · [(100 + b) / 100] · … − 1

The case of two changes

Let us formalise the case of two percentage changes, a and b. The combined change, initially expressed as a decimal number, is:

[(100 + a) / 100] · [(100 + b) / 100] − 1

= [(100 + a)(100 + b) − 10000] / 10000

= [10000 + 100b + 100a + ab − 10000] / 10000

= (100b + 100a + ab) / 10000

Multiply by 100 to express the result as a percentage:

Final change = [(100b + 100a + ab) / 100] %
             = [a + b + (ab / 100)] %

This gives the formula for solving a pair of successive percentage changes.

Example: a 10% increase followed by a 4% increase

If a price increases by 10% and then by 4%, set a = 10 and b = 4:

[(100 · 10 + 100 · 4 + 10 · 4) / 100] %
= (1000 + 400 + 40) / 100 %
= 14.4%

The overall change is therefore a 14.4% increase, not merely 14%: the second increase also applies to the increase already produced by the first.

Special case: equal and opposite changes

An interesting special case occurs when the two changes are opposites, that is, b = −a. For example, a price rises by 10% and then falls by 10%. Substituting b = −a into the formula gives:

[(100(−a) + 100a + a(−a)) / 100] %
= [−a² / 100] %

If a change is followed by an equal and opposite change, the final change is always negative whenever a ≠ 0.

For example, if a price rises by 30% and then decreases by 30%, the final change is:

[−a² / 100] % = [−30² / 100] % = −9%

The final value is therefore 9% lower than the initial value.

Prepared by Salvatore Mosaico.