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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?

amoeba
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GamerJosh
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  • And here is where I admit that this is my first time on this SE and I am by no means an expert in statistics. I'm open to suggestions on how to improve this question, including better tags or moving it to a different SE (if there is one that is more appropriate). – GamerJosh Apr 07 '14 at 18:22
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    You might find this question and this question, located on the EL&U site, to be helpful. – Kit Z. Fox Apr 09 '14 at 13:45
  • @KitFox Thank you. Those are helpful, and I recognize I may have my wording incorrect in a few places here; if those inaccuracies have confused what the question is, I apologize. – GamerJosh Apr 09 '14 at 13:50
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    The language is fine, but the math is off. If Y is 400% faster than X, then X is not 400% slower than Y. It is 75% slower. – RegDwight Apr 09 '14 at 13:53
  • @RegDwight Agree. – GamerJosh Apr 09 '14 at 14:11
  • @RegDwight Actually, I'm not sure I entirely agree, because I don't believe that's what I said. I do think it's the language. In my example, the pre-upgraded program finished in 2 seconds and the upgraded program finished in 0.5 seconds. The pre-upgraded program was 4 times slower than the upgraded program. – GamerJosh Apr 09 '14 at 14:24
  • Aye, "four times slower" is fine, but that is not what you were saying, and not what my comment was about. But Glen_b has actually explained it all in his answer, and now that you have edited the question accordingly, it is of course a moot point. – RegDwight Apr 09 '14 at 15:39

2 Answers2

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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).

Glen_b
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  • I see what you're saying (with regards to phrasing), and even agree with it. However, I don't feel like it answers my question. Is "four times faster" actually accurate? Is it more accurate to say "1/4 times faster"? - I'd most prefer to be accurate, but I'm not convinced that has the same impact. And, as an aside, I am displaying a graph (which really drives home the performance improvement), but I need to summarize the results in a way that could be used as a high-level summary while still emphasizing the improvement. – GamerJosh Apr 08 '14 at 13:41
  • "Accurate" is probably a red herring; it's no good saying "well by the dictionary it's the right way to say it" if few people understand you. What's really important is to convey your message - it matters whether you're correctly understood. And that's why my answer emphasizes that and discusses what might be misunderstood (most especially, I was trying to get you to avoid the common-but-bad "takes 300% less time" or "takes 3 times less time" type of expression, but avoided actually saying them). In this case, "4 times faster" is both actually correct and likely to be understood correctly. – Glen_b Apr 09 '14 at 00:13
  • Since the question, ultimately, is: 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? If you can show how to get "4" and explain why that is correct, then I can accept this answer. – GamerJosh Apr 09 '14 at 16:34
  • The value 4 came from your question. Earlier in my comments I pointed out that it's important to qualify such a statement by explaining at some point how it was calculated (in this case it was a ratio of sample medians). If your phrasing is accompanied by an explanation of how you got a ratio of "4" (i.e. has 'median time' in there somewhere as the things in the ratio), you're doing nothing more than explaining exactly what you computed. The question of what you should compute ... depends on your needs or the needs of your audience. (ctd) – Glen_b Apr 09 '14 at 21:17
  • (ctd) ... If these times would be experienced regularly some people might care more about average time, but that doesn't of itself make either choice 'right'. I can't tell you your own preference function, so as long as you work out what you want/need and explain that's what was calculated, how could it be wrong? – Glen_b Apr 09 '14 at 21:21
  • Note that "1/4 times faster" is not a correct description of the half-second vs "2 second" comparison. You'd say "ran in 1/4 of the time" (and again, include how that is computed - i.e. based on 'median time') – Glen_b Apr 09 '14 at 22:49
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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.

GamerJosh
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    That last statement--"3/4 less time"--is so cumbersome, unusual, and prone to misinterpretation that it should be avoided. The program now runs four times faster than it used to and requests can be processed in one quarter of the time. That's a simple and straightforward characterization. – whuber Apr 08 '14 at 20:36
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    I'm afraid I don't follow any of this because I don't see where the twos and eights come from. According to the information set forth in your question, the new program is 2/0.5 = four times faster. If that's wrong, then please edit your question. To say "25% faster" is inconsistent with the other information you have offered: a program that is 25% faster does 5 calculations in the time that it used to do 4. I hope you see how confusing the use of percents can be in this context. – whuber Apr 08 '14 at 21:21
  • I think you would benefit by consulting a dictionary definition of "speed." If you wish to discuss this issue further, a better site for it would be [english.se]. – whuber Apr 09 '14 at 13:33
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    @whuber Please don't refer people to our site if what they need is a dictionary. That said, this question and this question might be helpful. I'll comment under the question as well. – Kit Z. Fox Apr 09 '14 at 13:45
  • @Kit Thank you for the links--they look relevant and helpful. I believe the issue discussed in these comments, although closely related to those links, may yet be subtly different: it concerns usage about comparing rates, not just quantities. My recommendation to consult a dictionary was merely a starting point to help focus the dialog. – whuber Apr 09 '14 at 13:49