So in this particular scenario, I have two processors running the same benchmark. One is faster than the other, with the faster result being the one that takes the least amount of time to complete.

As an example, Processor A completes the benchmark in 42.5 seconds, and Processor B completes it in 32.9 seconds.

The goal here is to obtain how much faster Processor B is against Processor A . When the benchmark results in a "higher is better" result, that's simple. You divide the larger number by the smaller number.

With results indicating "lower is better," however, do I do the same thing? Am I finding the percentage change between the two numbers, or the percentage difference?

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

Up until now, I've been using #1.

  1. ((42.5 / 32.9) * 100) - 100 = 29.179%
  2. 100 - ((32.9 / 42.5) * 100) = 22.588%

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?".

  • $\begingroup$ Hi Matti. Thank you so much. You answered everything that I wanted to know. :) $\endgroup$ – DylRicho Feb 6 '19 at 12:37

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