How to check on Euler progress?

[vaughan]

I e-mailed the Euler admin and got this reply. Hope it helps:

Ok, here is how to read it:

Euler 2000 Resta 6: Power 6, from 76001 to 76002

Means that you compute the value A=76001.
Where A is part of the equation:
A^6 + B^6 = C^6+D^6+E^6+F^6+G^6

<6,2,5> 76001+38173-45442 (407435 seconds)

This means that you reached A=76001, B=38173, C=45442 in 407435 seconds.
B is computed from B=1 to B=76001, C is computed from 1 to the highest value
where A^6+B^6-C^6 >= 0

Since B=38173 is almost A/2, this means that you finished 50% of its
computation.

[AMDave]

so an expected runtime of 10 days
a bit more hefty than expected

what CPU was that reading taken from?

[vaughan]

The cpu is an AMD Athlon XP3200+

[NeoGen]

It depends on the workunit you get. Some are as fast as just a few hours, others can drag through days and nights.
And the application supports suspending and resuming so there is no problem for people that can't leave the computer on too long.
The only thing is that it's really boring to look at. Just one line of text with a few numbers that change a bit every few seconds...

[vaughan]

Resta 6 76001 took my XP3200+ 677028 seconds or 7 days 20 hours.

[ltd]

To get a fairly good impression how long a WU will run do the following:
Let the WU run for at least 20 minutes.
Then you see something like this:
74645+39077.......(12345 seconds)
Now make the following calculation:

74645/39077*12345= Runtime for the WU in Seconds
The value is not exact as the long runing WU tend to get slower at the end of the calculations but it gives you at least a rough estimate.

Lars

Edit: corrected at least some of the typos :oops:

[vaughan]

Lars,

What do I do when I see 75967+57863-61134-53071 (613106 seconds)?

This task has been running for several days on a P4 2.66GHz notebook. My guess is about 10 days total run-time.

Vaughan

[ltd]

Lars,

What do I do when I see 75967+57863-61134-53071 (613106 seconds)?

This task has been running for several days on a P4 2.66GHz notebook. My guess is about 10 days total run-time.

Vaughan

Your guess with the 10 days is not bad. My "formula" gives:
75967/57863*613106=804932seconds
Its always "first number" divided by "second number" multiply with runtime.
804932seconds/86400seconds per day=9.31 days With the calculations getting slower to the end i would guess about 9.5 days.

My longest runing WU so far was about 15 days.

Lars

[NeoGen]

I got one that will take me almost a week too, based on these calculations.

So, the bigger the difference between the first two numbers, the longer the workunit takes. And since the first number tends to be bigger from one workunit to the next, does it mean that the workunits will take more and more everytime?

[ltd]

In principle yes. but the maximum number to be tested in this cycle is around 100000 as value of the first nuimber.
One more opservation: If the first number is even the tests are typicaly very fast. from seconds to maximum a day. (At least i never had a very long runing even number)
The odd numbers tend to run between 3 days and ? ( as said before my maximum was 15 days) with an average of 6 days.(personal guess) There are sometimes odd numbers that also run only a few hours but thats not typical.

Lars

[AMD-USR_JL]

I got one of those lucky odd numbers! :D 79161 finished in 2553 seconds.