Percentage Change versus Percentage Difference

algebra-precalculuspercentagesterminology

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?

Best Answer

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?".

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