Author: Robert Hyatt
Date: 14:25:13 08/25/03
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On August 25, 2003 at 15:11:57, Sune Fischer wrote: >On August 25, 2003 at 14:53:46, Robert Hyatt wrote: > > >>>Your calculation was based on 2.4 Mnps and you compared that to 1 Mnps (nps >>>being the _only_ factor changing in the equation) but you neglected to include >>>the double bandwith _for the cache copying_ which would unify it back down. >> >>I'm not sure what you mean, exactly. I always referred to my 2.4M nps as >>"the benchmark number". > >Yes, but note this doesn't mean 1 node every 400 ns, it means 2 nodes every 800 >ns (ball park figures as always). > >I believe this distinction is important when you talk about memory trafic, >because using two cpus does actually double the _cache_ bandwidth (for things in >cache), and that is an important factor. > >-S. This depends on what you believe. IE I don't believe that it is only cache bandwidth. Copy/make means you copy a big block of stuff that is probably in L2 cache, to another couple of lines, and then you use the new stuff. But you just (a) destroyed something in L2 already and (b) you now have something in L2 that even if it is never used again, it all has to be written back to memory. So you get memory bandwidth thrown in no matter what. And with two processors, you are doing just as much of anything as you would be with a single cpu that is twice as fast. IE if I do 2.4M nodes per second at 2.8ghx X 2, I will do 2.4M nodes per second at 5.6ghz X 1, which means that the cache bandwidth and memory bandwidth will be the same on both of those machines...
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