20090607, 19:45  #1 
Oct 2006
103 Posts 
Predicting the needed time for high nvalues?
Is there a way to predict/calculate the time a LLRnetTest with a higher nvalue 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 
20090607, 20:27  #2 
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. 
20090607, 20:53  #3  
Jul 2003
wear a mask
1,433 Posts 
Quote:


20090609, 03:19  #4 
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. 
20090611, 12:24  #5 
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 nvalues are generated with sr5sieve [sr5sieve a 64258 657516 2000000] and are about 2 times // 6 times as big as the nvalue from the server) If you increase the nvalue with the factor 2, the needed time is increased by factor ~4 (2^2) Increasing the nvalue 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 20090611 at 13:08 Reason: wrong parameter (1972548 > 2000000) 
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