How do I emphasize a speed increase between two sets of data? I've run a test on a program, upgraded the program and then run the same tests again.  The first test showed that the median time for the program to respond was approximately 2.0 seconds.  The second test showed that, after the upgrade, the median time for the program to respond was approximately 0.5 seconds.  
While writing a paper, I really want to emphasize the speed increase of the program after the upgrade.  I recognize that I can multiply 0.5 seconds by 4 to get 2 seconds, but that really means, "the pre-upgraded program was 4 times slower than the upgraded program".  I want to say something like, "After the upgrade, the program ran x times faster than before the upgrade."
What is the proper way to get x?
 A: Often phrases related to "x times faster" is potentially going to lead to difficulty if combined with percentages (especially if phrased in terms of percentages in the form of "less time"). People regularly say things using these terms that are nonsensical (and use the terms in an inconsistent way) - indeed it happens so often that using such expressions phrased correctly may be misinterpreted.
[Edit: to clarify, what I am particularly getting at is that phrases akin to "4 times less" and "400% less", while common, are especially to be avoided ... as is anything that might be interpreted as implying those.]
I suggest that where easy, it may even be better avoiding use of percentages in this context for that reason, and instead say things like "based on a comparison of median times, the program ran four times faster." --- which is similar to the expression you started with.
But if you want to particularly emphasize it, you might also display either times or speeds visually (not necessarily just the median, if you have the data).
A: In this case, I was measuring the amount of time a request took.  So:
$Speed = Requests / Time Taken$
For the first run, 1 request took 2 seconds.
$ 1R / 2s = 0.5 RPs$
For the second run, 1 request took 0.5 seconds.
$ 1R / 0.5s = 2 RPs$
X is calculated by: 
$X = Slower Rate / Faster Rate$
So:
$ 2 RPs / 0.5 RPs = 4$
Thereby, "After the upgrade, the program ran 4 times faster than before the upgrade."
Additionally, "After the upgrade, requests ran in 1/4 of the time as requests before the upgrade." is also correct. 
