20071111, 04:41  #1 
Jun 2004
2^{2}×3×5 Posts 
Wrong credit in P90 years?
Here is my concern:
Recently I finished 2 exponents, one is 391xxxxx (FFT size 2048k), another is 402xxxxx (FFT size 2560k). Because they are different in FFT size, I expected the 2nd one add much more p90 years than 1st one, but the actual p90 years i earned are 8.36 p90 years for 1st one and 8.60 years for 2nd one! 
20071111, 06:19  #2 
Dec 2004
The Land of Lost Content
3×7×13 Posts 
If it helps at all, you can cross check your credits here:
http://www.teamprimerib.com/rr1/bin/calc_p90.php 
20071111, 07:15  #3  
Jun 2004
2^{2}×3×5 Posts 
Quote:


20071111, 08:57  #4 
Sep 2006
Brussels, Belgium
2^{2}·3^{2}·47 Posts 
The credit FFT sizes are based on non SSE2 processors, they are a bit higher : cross over to 2560KB at 40250000, while in the program you find the cross over at 39500000.
Jacob 
20071111, 10:31  #5 
"Lucan"
Dec 2006
England
2·3·13·83 Posts 
Putting pentium 90MHz into the calculator on the "benchmarks"
page gives over 12 years for an exponent of 40000000. So the credit given seems a bit stingy all round. 
20071111, 12:45  #6  
"Lucan"
Dec 2006
England
2×3×13×83 Posts 
Quote:
on Primenet stats page is being underestimated while the bulk of the work is coming fromm LL tests on exponents between 39.5M and 40.25M? Can SSE2 processors run non SSE2 code? If so would they be better off running this code with a 2048K FFT? David Last fiddled with by davieddy on 20071111 at 13:02 

20071111, 15:08  #7  
Sep 2006
Brussels, Belgium
2^{2}·3^{2}·47 Posts 
Quote:
Quote:
Yes you can force prime95 not to use SSE2, see the help or the readme and undoc files. No, it would be slower to use the bigger FFT, even if you got a higher credit. I think that in the end the result in credit / time spent would be lower. You have to benchmark your processor in both situations to have an exact result. When I say benchmark, I do not mean the benchmark option of the program because that gives a BEST time over a small number of iterations, but real crunching and having an average time over a large number of iterations. Jacob 

20071111, 17:07  #8  
"Lucan"
Dec 2006
England
14512_{8} Posts 
Quote:
non SSE2 code) than the 2560K FFT". David PS I would test this, but don't have an SSE2 machine. Last fiddled with by davieddy on 20071111 at 17:52 

20071111, 18:47  #9 
Jun 2003
3·19·89 Posts 
In general, for the same FFT size (machines that support) SSE2 code will be faster than nonSSE2. However, for the crossover exponents, we are comparing smaller nonSSE2 FFT sizes and larger SSE2 FFT size. Depending on the CPU type and the actual FFT sizes, you may find one or the other faster.
A noticeable exception is the P4 which plain sucks at nonSSE2 code 
20071111, 18:59  #10  
Sep 2006
Brussels, Belgium
2^{2}×3^{2}×47 Posts 
Quote:
Jacob 

20071111, 19:24  #11 
"Lucan"
Dec 2006
England
194A_{16} Posts 
Thx. I was really enquiring on behalf of the 15000 or so computers
currently testing this "crossover" range of exponents. Since the majority are P4s(?) I shall accept AXN's assurance that nonSSE2 code is a nonstarter. (Although there is a 30% increase in iteration time going from 2048K to 2560K FFT size). 
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