Sighting of the legendary Monster Workunits (630 cr)

John McLeod VII
John McLeod VII
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Joined: 10 Nov 04
Posts: 547
Credit: 632,255
RAC: 0

RE: RE: RE: Based on

Message 69490 in response to message 69470

Quote:
Quote:
Quote:

Based on the client's estimate, yes, the WU should never have been sent. That's granted. But the question still is whether this estimate is realistic or not. To me it looks too pessimistic.

CU

BRM

The part your overlooking here is the estimates from the scheduler log are not client side generated other than they are based on the BM's and other metrics the host has last reported. They are server side calculated, although I'm not 100% sure at the moment (brain fade) about who actually calculates the RDCF.

Anyway, we can compute RDCF ourselves :-)

ETAcpu = Nflop /BMflops * RDCF

where NFlops is a hypothetical floating point operations equivalent of the computational effort for this WU, estimated by the WU-generator

So, for the 400 MHZ box in question, we have a Benchmark of 300 MFlops

ETAcpu ~ 24 d ~ 2 * 10^6 sec
BMFlops ~ 300 *10^6 ops / sec
Nflop ~ 5.5 * 10^14 (taken from one of my own monster results, see the last argument for the client in client_state.xml for the result in question)

==>RDCF would be close to 1 :-(.

Oh-oh, this estimation seems to be realitic after all for that box ...

CU

BRM


The DCF is calculated on the client based on how well the client actually does at completing work in the estimated time.

BTW, I have detached my slowest client as completely hopeless. My second slowest is not going to make it, but may be close enough to get credit as it will take even the fastest machine longer than the difference.

Alinator
Alinator
Joined: 8 May 05
Posts: 927
Credit: 9,352,143
RAC: 0

RE: The DCF is calculated

Message 69491 in response to message 69490

Quote:

The DCF is calculated on the client based on how well the client actually does at completing work in the estimated time.

BTW, I have detached my slowest client as completely hopeless. My second slowest is not going to make it, but may be close enough to get credit as it will take even the fastest machine longer than the difference.

LOL...

So you've been playing 'Russian Roulette' with some of your old timers too! ;-)

My track record for that is 3 successes and 2 flameouts. :-)

Alinator

FJBJClauwens
FJBJClauwens
Joined: 11 Nov 06
Posts: 4
Credit: 2,134,335
RAC: 0

My machine completed 29

My machine completed 29 monster units thus far:
the claimed credit varies between 643.79 and 643.92,
the cpu time between 83595 and 83755 seconds,
ie between 23h 13 min and 23 h 17min, rearkable constant.
Of these 29 there are 14 still pending.
Because of the many pending claims the average credit
oscillates between 2100 and 2400 credits a day.

Frans

Gerry Rough
Gerry Rough
Joined: 1 Mar 05
Posts: 102
Credit: 1,847,066
RAC: 0

Hmmmm. I've been wanting to

Hmmmm. I've been wanting to see the monster WUs on my hosts, but no luck yet. It can't be my hosts, they have plenty of firepower, both from ram and CPU. Am I missing something?


(Click for detailed stats)

Alinator
Alinator
Joined: 8 May 05
Posts: 927
Credit: 9,352,143
RAC: 0

Just give it some time. You

Just give it some time. You have to work through the lower template frequency datapaks you already have first, but you'll get some eventually. ;-)

Alinator

archae86
archae86
Joined: 6 Dec 05
Posts: 2,824
Credit: 3,302,237,374
RAC: 2,558,588

I just built a new Core 2

I just built a new Core 2 Quad (Q6600) in the ashes of my Gallatin whose boot drive died.

Wouldn't you know that the first Einstein units it got issued took about thirty hours and are claiming 655.57? It looks like it may be days before I get my first concrete indication on Einstein that the thing is working right. And if I push for a little overclock it will take a long time to get assurance.

Ah well, better my Q6600 than some new participant walking in with a 1.4 GHz Pentium M Banias and wondering why it is taking so long on a decent machine.

Gary Roberts
Gary Roberts
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Joined: 9 Feb 05
Posts: 5,210
Credit: 43,575,557,131
RAC: 44,317,643

RE: Ah well, better my

Message 69496 in response to message 69495

Quote:
Ah well, better my Q6600 than some new participant walking in with a 1.4 GHz Pentium M Banias and wondering why it is taking so long on a decent machine.

I'd love to have a 1.4GHz Pentium M to do the monsters. Instead I have to make do with this PIII 933MHz machine which has just taken around 5 days to do a 663.58 credit one. At least it didn't have to hang around waiting for validation :).

Cheers,
Gary.

Jim Milks
Jim Milks
Joined: 19 Jun 06
Posts: 116
Credit: 529,852
RAC: 0

I think I may have just

I think I may have just gotten my first monster. The estimated time to completion is 40.5 hours on my 1.83 GHz Macbook whereas the 386 cobblestone workunits only take ~23.75 hours. And my wingman is a Windows box that has a history of client errors.

Bikeman (Heinz-Bernd Eggenstein)
Bikeman (Heinz-...
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Joined: 28 Aug 06
Posts: 3,516
Credit: 455,978,858
RAC: 46,188

RE: I think I may have just

Message 69498 in response to message 69497

Quote:
I think I may have just gotten my first monster. The estimated time to completion is 40.5 hours on my 1.83 GHz Macbook whereas the 386 cobblestone workunits only take ~23.75 hours. And my wingman is a Windows box that has a history of client errors.

Yes, this is definitely a "monster". I'm not sure where exactly the boundary is, but it must be somewhere between 530 Hz and 535 Hz (this is the 3 digit number in the workunit name).

Good Luck!

CU

BRM

Gary Roberts
Gary Roberts
Moderator
Joined: 9 Feb 05
Posts: 5,210
Credit: 43,575,557,131
RAC: 44,317,643

RE: I'm not sure where

Message 69499 in response to message 69498

Quote:
I'm not sure where exactly the boundary is, but it must be somewhere between 530 Hz and 535 Hz (this is the 3 digit number in the workunit name).

For those who are not quite sure whether they have a "monster" or not, I guess the best definition is that a "monster" has a frequency in the very topmost band as shown on Bernd's graph that he announced some time ago in this message. There is a small band of frequencies in the topmost righthand corner of the graph which give by far the longest crunching times. It is difficult to estimate the exact frequency where the transition occurs, however, because of the nature of the graph.

Here are some credit/frequency data for results currently running on my boxes which give a more precise idea of the transition point:-

Frequency .. Credit
520.95 ..... 459.16
530.60 ..... 473.17
532.80 ..... 636.13
533.50 ..... 637.78
537.95 ..... 644.15
545.10 ..... 656.93

ie, it's somewhere between 530.60 and 532.80 :).

Cheers,
Gary.

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