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Old 2007-11-11, 04:41   #1
nngs
 
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Default 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!
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Old 2007-11-11, 06:19   #2
99.94
 
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If it helps at all, you can cross check your credits here:
http://www.teamprimerib.com/rr1/bin/calc_p90.php
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Old 2007-11-11, 07:15   #3
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Quote:
Originally Posted by 99.94 View Post
If it helps at all, you can cross check your credits here:
http://www.teamprimerib.com/rr1/bin/calc_p90.php
This link gave the same results which puzzles me too. Because these two exponents use different FFT size (2048k versus 2560k), their P90 years should have much larger difference (based on the list at http://www.mersenne.org/bench.htm)
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Old 2007-11-11, 08:57   #4
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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
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Old 2007-11-11, 10:31   #5
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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.
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Old 2007-11-11, 12:45   #6
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Quote:
Originally Posted by S485122 View Post
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
Does this mean that the "sustained throughput" as quoted
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 2007-11-11 at 13:02
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Old 2007-11-11, 15:08   #7
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Quote:
Originally Posted by davieddy View Post
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.
Please refer to several threads like this one Number of P90 years per exponent If results are diverging the calculator is not giving correct results...

Quote:
Originally Posted by davieddy View Post
Can SSE2 processors run non SSE2 code? If so would they be better off running this code with a 2048K FFT?
Yes, No

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
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Old 2007-11-11, 17:07   #8
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Quote:
Originally Posted by S485122 View Post
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.
So you mean "Yes, it would be faster to use the 2048K (with
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 2007-11-11 at 17:52
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Old 2007-11-11, 18:47   #9
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In general, for the same FFT size (machines that support) SSE2 code will be faster than non-SSE2. However, for the cross-over exponents, we are comparing smaller non-SSE2 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 non-SSE2 code
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Old 2007-11-11, 18:59   #10
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Quote:
Originally Posted by davieddy View Post
So you mean "Yes, it would be faster to use the 2048K (with
non SSE2 code) than the 2560K FFT".
Oups ! I mixed up in my answer : non SSE2 code is slower than SSE2 code, I think the iteration times with SSE2 (and thus bigger FFT size) would be smaller than without SSE2 (even with the smaller FFT size.) Like AXN1 said and I suggested to be sure you have to benchmark it if you want a definitive answer for your computer (memory speeds and chipset have some influence as well.)

Jacob
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Old 2007-11-11, 19:24   #11
davieddy
 
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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 non-SSE2
code is a non-starter. (Although there is a 30% increase in iteration
time going from 2048K to 2560K FFT size).
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