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#1 |
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Oct 2006
103 Posts |
Is there a way to predict/calculate the time a LLRnet-Test with a higher n-value would take?
I know that it depenends on the cpu, but is there a way to calculate it in relation to a lower n? Like: Code:
n=300.000 : n=1.500.000
5h : xh
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#2 |
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Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
102538 Posts |
Maybe this is different with base 5 numbers than base 2 numbers, but a doubling of n produces roughly a quadrupling of testing time, so e.g. if n=300K is 5 hours (from your example), n=600K would be 20 hours, n=1200K would be 80 hours, so I'd guesstimate that n=1500K would be around 100 hours.
Of course, I could just be completely wrong. A better way would be to run some iterations and multiply appropriately.
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#3 | |
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Jul 2003
wear a mask
67A16 Posts |
Quote:
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#4 |
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May 2007
Kansas; USA
33×5×7×11 Posts |
Ditto here also from my experience with the numerous different bases on the CRUS project.
To be more exact, if an n=300K test takes 5 hours, an n=1.5M test, since the exponent is 5 times as high, would take 25 times longer, i.e. 125 hours. |
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#5 |
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Oct 2006
11001112 Posts |
I started the following three pairs in llrnet, and noted the time and the percentage after restarting llrnet and calculated the time a certain pair would need to finish.
Code:
Pairs: 64258 328758 64258 657558 64258 1972518 (I got the first pair from the server after starting llrnet. The other two n-values are generated with sr5sieve [sr5sieve -a 64258 657516 2000000] and are about 2 times // 6 times as big as the n-value from the server) If you increase the n-value with the factor 2, the needed time is increased by factor ~4 (2^2) Increasing the n-value with the factor 6 causes the needed time to be increased by the factor ~36 (6^2) So, your answers were correct. Thanks. Last fiddled with by Rincewind on 2009-06-11 at 13:08 Reason: wrong parameter (1972548 -> 2000000) |
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