That is helpful, but my basic objection still stands: you're computing
with *times* while the claim in the article was about *speeds*, I think.
He says "50% more work using the same amount of CPU cycles", which I
read as work/time. So you need to take the reciprocal of all your
values. Don't you?
On 2015-04-13 08:15AM, George Neuner wrote:
> So:
> percent change = 100% * ( (new - old) / old )
> Dropping the common multiplier to keep the numbers simple:
>
> 3.7 -> 3.8
> percent change = 100% * ( (95 - 143) / 143 )
> = 100% * ( -48 / 143 )
> = 100% * ( -0.34 )
> = -34%
Should be:
3.7 -> 3.8
percent change = 100% * ( ( (1/95) - (1/143) ) / (1/143) )
= 100% * ( ( 0.0105 - 0.00699 ) / (0.00699) )
= 100% * ( 0.0351 / .00699 )
= 100% * ( .50 )
= 50%
Yes? No? Maybe I'm reading the article wrong...
--Josh
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