Author: Vincent Diepeveen
Date: 03:52:43 07/08/03
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On July 07, 2003 at 17:28:06, Tom Kerrigan wrote: >On July 07, 2003 at 15:20:15, Vincent Diepeveen wrote: > >>sign in now for 15% speedup i would sign blindfolded. 15% from 250Ghz = >>37.5Ghz. > >A 15% speedup from 250GHz = 287.5GHz, not 37.5GHz. > >You mean "efficiency" every time you say speedup. > >-Tom Speedup as it is in all the ICCA journals from the 80s and onwards is the number of times you are faster out of n processors when compared to 1 processor of such a system. So if i would get the blessed efficiency of 15% out of 500 processors that is 0.15 * 500 = 75 times speedup However to compare it to the speed of a K7 1Ghz it is interesting to express it in Ghz in this case: 37.5 Ghz So it effectively then 37.5 times faster than a K7 at 1 Ghz, assuming that K7 has a hashtable of 250GB. Most programmers talk always about speedup, but that's for a fixed amount of cpu's they always have. Like 2. However with many cpu's i cannot always test with the same amount of CPUs so speedup efficiency is a better form then to measure than speedup. Best regards, Vincent
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