Author: Vincent Diepeveen
Date: 11:47:30 09/03/02
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On September 03, 2002 at 14:33:52, Uri Blass wrote: On Tue, 03 Sep 2002 18:52:38 +0100, Simon Waters <Simon@wretched.demon.co.uk> wrote: Page 13 from icca from the DTS article from bob says: 4. DTS performance results The DTS algorithm was tested using a Cray C916/1024. this machine has 16 processors with a cycle time of 4.166 nanoseconds and also has 1024 million words of memory (8 gigabytes). 4.1 testing methodology In producing these results, all testing used the machine in a dedicated mode so that all of the machines memory was used, regardless of the number of processors utilized in each test (except the one processor test, which is explained later). [snip] ==> regarding the 1 processor result that he had to share the machine As a result, memory conflicts were much higher (bank conflicts to those that are Cray-savvy) as well as swapping overhead which gets charged to the user. While this is well below the .1% level of noise, it is noticeable, and should be remembered. The whole math i did was based upon bob's search times table. table 3 in icca march 1997. I'll give a few numbers for you to verify it. procs: 1 , 2 , 4 , 8 , 16 pos 1 2830 1415 832 435 311 2 2849 1424 791 438 274 3 3274 1637 884 467 239 4 2308 1154 591 349 208 5 1584 792 440 243 178 6 4294 2147 1160 670 452 7 1888 993 524 273 187 8 7275 3637 1966 1039 680 9 3940 1790 1094 635 398 10 2431 1215 639 333 187 11 3062 1531 827 425 247 and so on. see the article for all other numbers. you'll see i didn't make mistakes in my table. the fact is that if you get a speedup at a position of 3.7 then that means your speedup in reality calculates when dividing search time of 1 processor by 2,4,8,16 that it is somewhere in the range of 3.65 - 3.74 However if it is 3.70 your speedup, then the chance that happens next time again and again and again is not so big. Because speedups are pretty relevant and only to base upon search times, only this table matters IMHO. The other tables are interesting to see, but this is THE important table. If you want to let your 16 processor run (which is the tournament run) look better, you obviously have to modify your results for 1..8 processors in order to let your 16 processor version look better. That's exactly what Bob did. If you divide the above numbers in next manner you'll get the table i posted which shows he is a big fraud. >On September 03, 2002 at 14:22:05, Vincent Diepeveen wrote: > >>On September 03, 2002 at 13:51:52, Uri Blass wrote: >> >>It is about the second digit being round, because >>that makes the chance you have such a speedup 1/10 of >>a chance. >> >>Bob claims a 2.0 speedup which bob claims according to >>his paper based upon counting up all times then dividing >>by total times. >> >>However if we look at every speedup individually then >>if you get a 2.0 speedup that's in a range of 1.95-2.04 >>RIGHT? >> >>Please answer that question. >> >>A speedup of 2.0 lies in a range of about 1.95-2.04 >> >>So i hope you realize that chance it is 2.00 is exactly 1/10 >>of a chance. >> >>How big is the chance that 50 numbers in a row have >>a speedup in parallel search of 2.00 then? > >I see. >I agree that the chances are very small. > >> >>How big is the chance that all numbers end at a 0? > >I agree that the chances are very small and the data is enough to convince me >that I do not believe that the data is correct. > >Uri
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