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 2009-06-07, 19:45 #1 Rincewind     Oct 2006 103 Posts Predicting the needed time for high n-values? 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
 2009-06-07, 20:27 #2 Mini-Geek Account Deleted     "Tim Sorbera" Aug 2006 San Antonio, TX USA 17·251 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.
2009-06-07, 20:53   #3
masser

Jul 2003

1,433 Posts

Quote:
 Originally Posted by Mini-Geek 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.
Ditto.

 2009-06-09, 03:19 #4 gd_barnes     May 2007 Kansas; USA 5×13×157 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.
 2009-06-11, 12:24 #5 Rincewind     Oct 2006 103 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) And the results are: 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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