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I would love if someone were to clear up my confusion. Who is right and what are these things telling me?

I'm trying to show the percent change in some software performance times here at work, and I'm having trouble figuring out the math. As I Google "percent change", I find several sources who confuse this with "percent difference", and I find 3 different equations to calculate it. I have a stats major friend trying to help me and either we're not communicating and they aren't understanding what I'm asking for, I'm wrong/stupid and showing it/crazy, or they've been the recipient of a poor education.

So most sources, including the stats major, tell me percent change is $\frac{old - new}{old} * 100$

Ok, let's play with some numbers. I use Google as my calculator. My numbers are going to be 200k as the old value and 4 as the new value. The units are in clock ticks, if anyone cares, but I don't think it should matter.

According to Google, $\frac{200000 - 4}{200000} * 100 = 99.998\%$ faster.

What this is telling me that increasing $4$ by $99.998\%$, $99.998\% * 4 = 200000$

How am I interpreting this entirely wrong? What does this percentage mean? How do I change $4$ by $99.998\%$ and get to the old value? I can't believe that this is only a $\approx 100\%$ change...

RandomUser
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Matthew Reddington
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  • There's already an answer for this, but I'll point out that percent change is $\frac{new-old}{old}\cdot 100$. This serves to change the sign, and a decrease an in amount is negative while an increase is positive. – RandomUser Apr 29 '14 at 18:02

2 Answers2

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When you raise a value by some percentage and then lower the result by the same percentage, you don't get back where you started. When the percentages are small, you almost come back. Let's use smaller numbers for an example. If you start with $200$ and reduce it $50\%$, you get $100$. If you raise that by $50\%$ you get $100 \cdot 1.5=150$ and you are down $50$ at the end. The general case is if you increase by $x$, you multiply by $1+x$. If you decrease by $x$ you multiply by $1-x$ The product of these if $1-x^2$, so you will always end up lower. If $x$ is small, $x^2$ is even smaller, so might be negligible.

Ross Millikan
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  • I have no idea what you just told me. – Matthew Reddington Apr 29 '14 at 18:23
  • Your expectation that if you lower something by a certain percentage, then raise the result by the same percentage you will get back where you started is incorrect. Many people have the same expectation. I was trying to show how close it would be. – Ross Millikan Apr 29 '14 at 18:26
  • I have no idea what you just told me.

    But, I think I understand what you're trying to say. And that is something I surprisingly understand: the percents aren't reversible. Percent down is not the same as percent up. It dawned on me that putting the percent in word form, "The process formerly taking 200k clock ticks has been reduced by 99.998%, to 4 clock ticks." I was expecting some sort of sense of magnitude would be conveyed in the percent, like 4 is 5m times smaller than 200k, and it is, but not in a way that was intuitive to me. Thanks for your help, from you, I got it.

     – Matthew Reddington Apr 29 '14 at 18:34
  • Sorry, hitting enter doesn't mean "new line" it means "submit." – Matthew Reddington Apr 29 '14 at 18:34
  • That is correct. You reduced it by almost $100%$, so it went almost to zero. Increasing it by the same percentage will (almost) double it, but twice a small number is still a small number. The difference comes about because the denominator is so much smaller for the second step. – Ross Millikan Apr 29 '14 at 18:36
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It's not that it's 99.998% faster, 4 is 100-99.998 = 0.002% of 200,000. Because speed is defined as x/time, the change in time is 1/0.002 = 500x as fast

Kloppie5
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