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Old 2009-06-07, 19:45   #1
Rincewind
 
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Default 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
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Old 2009-06-07, 20:27   #2
TimSorbet
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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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Old 2009-06-07, 20:53   #3
masser
 
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Quote:
Originally Posted by Mini-Geek View Post
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.
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Old 2009-06-09, 03:19   #4
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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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Old 2009-06-11, 12:24   #5
Rincewind
 
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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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