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 : [b]x[/b]h
[/code]

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. :smile: A better way would be to run some iterations and multiply appropriately.

[QUOTE=Mini-Geek;176423]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. :smile: A better way would be to run some iterations and multiply appropriately.[/QUOTE]

Ditto.

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.

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)
[/code]

And the results are:

[b]
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)
[/b]

So, your answers were correct. Thanks.