Converting "% faster" to "X times speedup"

Question

I'm confusing myself over a rather pedantic math question. I'm comparing two simulation's runtimes (measured in seconds). I want to know how many times faster is simulation A compared to simulation B.

Normally, I report the result as a percentage, e.g., simulation A is 50% faster than simulation B. To get this percentage, I can simply divide $$t_A / t_B$$. 50% faster is equivalent to "1 times speedup". I'm wondering what is the equation to get this speedup number on less nice numbers?

It's subtle. Does this answer your question? What does "X% faster" mean? — Ethan Bolker, Oct 04 '22 at 18:39
I would say $A$ needs $\left(\frac{t_a}{t_b}-1\right)\cdot 100%$ more time than $B$. I am cautious about using velocity/speed here. — callculus42, Oct 04 '22 at 18:45
According to Wikipedia, A using $50%$ less time than B would be a $2$x speedup, not "1 times". Calculated from dividing the two run times, and assuming the same amount of workload. — peterwhy, Oct 04 '22 at 21:12

score 1 · Answer 1 · answered Oct 05 '22 at 16:50

To say a value $x$ is $p\%$ greater than a value $y$ means that $\dfrac {x-y}y \times 100\% = p\%$. That fraction can also be expressed as $\left(\dfrac xy - 1\right)$. In this case, you are comparing the speeds to accomplish the task, which will be $x = \dfrac 1{t_A}$ and $y = \dfrac 1{t_b}$, so $$\dfrac xy = \dfrac{\dfrac 1{t_A}}{\dfrac 1{t_B}} = \frac {t_B}{t_A}$$ and the answer to your question is

Simulation A is $\left(\dfrac {t_B}{t_A} - 1\right)\times 100\%$ faster than Simulation B.

In particular A will be $50\%$ faster than B if $t_B = (1.5)t_A$

Converting "% faster" to "X times speedup"

1 Answers1