Percentage Change versus Percentage Difference

Question

I have two processors running the same benchmark. Processor A completes the benchmark in 42.5 seconds, and Processor B completes it in 32.9 seconds. The goal is to determine how much faster Processor B is than Processor A.

With benchmark results indicating "higher is better", that's simple: you divide the larger number by the smaller number. With benchmark results indicating "lower is better", however, do I do the same thing? Am I to find the percentage change between the two numbers, or their percentage difference? Up until now, I've been using computation #1:

1. $$(42.5 / 32.9) - 1 = 29.179\%$$
2. $$1 - (32.9 / 42.5) = 22.588\%$$

Could someone help explain the difference between these two calculations, and which one I need?

Answer 1

It's useful to think of it this way: A processor's speed is inversely proportional to the time it takes to complete the task. Therefore, for processor A we can give a "non-dimensional speed" of $\frac{1}{42.5}$. For processor B, this value is $\frac{1}{32.9}$. It's easy to see that B is faster in this case, because the speed is greater.

Now you want to compare their speeds, or in particular answer the question "How many per cent is processor B faster than A?". So you can just calculate the ratio of their speeds like this: $$ \frac{\left( \frac{1}{32.9}\right)}{\left( \frac{1}{42.5}\right)} = \frac{42.5}{32.9}\approx 1.2917 $$ Therefore, our answer is: Processor B is faster than processor A by $29.2~\%$.

Should you calculate it the other way around, you'd be asking the question "How much slower is processor A compared to processor B?".

Answer 2

Processor A completes the benchmark in 42.5 seconds, and Processor B completes it in 32.9 seconds. The goal is to determine how much faster Processor B is against Processor A.

Let $p$ denote te number of floating-point operations executed by the benchmark; then Processors A and B have speeds $\dfrac p{42.5}$ FLOPS and $\dfrac p{32.9}$ FLOPS, respectively.

So, B is $$\dfrac{\frac p{32.9}-\frac p{42.5}}{\left|\frac p{42.5}\right|}= 29.179\%$$ faster than A, as required.

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1. $$(42.5 / \boldsymbol{32.9}) - 1 = 29.179\%$$

This means that A takes $29\%$ more time than B to complete the benchmark. (Here, B is the comparison baseline or reference point.)

2. $$1 - (32.9 / \boldsymbol{42.5}) = 22.588\%$$

This means that B takes $23\%$ less time than A to complete the benchmark. (Here, A is the comparison baseline or reference point.)

the percentage change between the two numbers, or their percentage difference?

"Percentage change" is not really applicable here. On the other hand, the percentage difference between speeds A and B is commonly computed as $$\dfrac{\frac p{32.9}-\frac p{42.5}}{\frac12\left|\frac p{42.5}+\frac p{32.9}\right|}= 25.464\%.$$