# Calculate “x % slower/faster”?

Ok, this might sound a bit stupid, but I find these kind of statements "x % slower/faster" pretty confusing.

Let's say I have algorithm A and algorithm B. Algorithm A takes 50 seconds to complete a task and algorithm B takes 25 seconds.

Now I could make the following statements:

1. "B is twice as fast as A"
2. "B is 100 % faster than A" (speedA - speedB)/speedB = 100%
3. "A is 50 % slower than B" (speedA - speedB)/speedA = 50%

• I'm not sure that the tag logic is the appropriate one for this question. It looks more like a question about language than about mathematics. – Fabian Aug 25 '12 at 12:41
• Yes, you are right...as long as you explain yourself clearly. – DonAntonio Aug 25 '12 at 12:42
• @Fabian I wasn't very happy with the tag "Logic". Thanks to Asaf for retagging! – tsauerwein Aug 25 '12 at 12:56
• Talk about confusing? Out here in the real world we see things like "ten times less" and so on. – GEdgar Aug 25 '12 at 13:11

In an ideal world, possibly, we would always use logarithms rather than ratios (percentages) when measuring relative differences. Then, supposing we standardized on the base-2 logarithm we could say

B is $1$ doubling faster than A.

A is $1$ doubling slower than B.

or if we used base-10 logarithms:

B is $0.3$ decades faster than A. ($\log_{10}\frac{50}{25}=0.3$)

A is $0.3$ decades slower than B. ($\log_{10}\frac{25}{50}=-0.3$)

The problem with that "ideal" world is that everyone who needed to speak about relative differences (a rather large fraction of the population) would have to know how logarithms work, which may or may not be a realistic goal.

• "may or may not" hahahahaha – Almo May 10 at 19:20