Team drive #1: k=400-1001 n=333.2K-600K

[em99010pepe]

It's going to be very close counting with the increase of n size.
G4000 is doing 4200 pairs/day, IB400 8814 pair/day and C443 1800 pairs/day. I'm doing 1000 pairs/day, Mini-Geek 236 pairs/day, what about you Ian?

I'll leave a message on XS forum to see if Glenn can add a few more cores for a week.

[mdettweiler]

Originally Posted by em99010pepe

567-570 done only waiting for Max confirmation. Please see results from 30/11/2008 and 01/12/2008. Thank you.

Carlos

LLRnet C443 has officially completed 567K-570K, lresults emailed to Gary.

[mdettweiler]

Originally Posted by em99010pepe

It's going to be very close counting with the increase of n size.
G4000 is doing 4200 pairs/day, IB400 8814 pair/day and C443 1800 pairs/day. I'm doing 1000 pairs/day, Mini-Geek 236 pairs/day, what about you Ian?

I'll leave a message on XS forum to see if Glenn can add a few more cores for a week.

Hmm...if I add up all those numbers, I get 16050 k/n pairs per day (not counting Ian's manual work, of course, which we don't yet have figures for). Based on an average of 1100 k/n pairs per n=100 range, and about n=25K left in Drive #1, there are ~275000 k/n pairs left to do.

If I divide 275000 by 16050, I get approximately 17 days to complete Drive #1. Of course, that is not factoring in any increases in the amount of time it takes to run each k/n pair, nor is it factoring in the amount of manual work that Ian does each day.

I'll see if I can come up with some more accurate calculations later, factoring in FFT jumps and Ian's manual work.

[mdettweiler]

Originally Posted by mdettweiler

Hmm...if I add up all those numbers, I get 16050 k/n pairs per day (not counting Ian's manual work, of course, which we don't yet have figures for). Based on an average of 1100 k/n pairs per n=100 range, and about n=25K left in Drive #1, there are ~275000 k/n pairs left to do.

If I divide 275000 by 16050, I get approximately 17 days to complete Drive #1. Of course, that is not factoring in any increases in the amount of time it takes to run each k/n pair, nor is it factoring in the amount of manual work that Ian does each day.

I'll see if I can come up with some more accurate calculations later, factoring in FFT jumps and Ian's manual work.

Okay...I just checked and it turns out that there are no more FFT jumps (based on a median k of k=701) in the 1st Drive! All the remaining work will be at the 40960 FFT length.

On my 2.2Ghz Core 2 Duo, I get about 0.82 ms per iteration at n=600K, again for k=701. For n=570K, I get almost the exact same time per iteration - 0.82 ms!

Based on this, and presuming that my assumption of 11000 k/n pairs per n=1K is correct, then we're on course to finish Drive #1 by December 17!

If you guys see any glaring errors in these calculations, feel free to point them out.

[Mini-Geek]

Originally Posted by mdettweiler

Based on this, and presuming that my assumption of 11000 k/n pairs per n=1K is correct, then we're on course to finish Drive #1 by December 17!

If you guys see any glaring errors in these calculations, feel free to point them out.

I don't see any glaring errors, but something you're not taking in to account is that not everything will finish at the same moment. I'm currently on course to finish on December 18th, so I shouldn't hold us up too much, but still...It should probably only be 1-5 days longer and still well within the end of the year, but it is something to consider.

[MyDogBuster]

I should easily finish my ranges by the 17th. Don't want to hold anything up. We certainly picked up some speed somewhere.

I also think I can sieve 1005-2000K to 1T by then. Should make for an interesting last 2 weeks this year in NPLB land.

Manually I'm doing about 800/day.

[Lennart]

578.0-578.1
578.1-578.2
578.2-578.3
578.3-578.4
578.4-578.5

Taking those.

/Lennart

[gd_barnes]

Originally Posted by mdettweiler

Okay...I just checked and it turns out that there are no more FFT jumps (based on a median k of k=701) in the 1st Drive! All the remaining work will be at the 40960 FFT length.

On my 2.2Ghz Core 2 Duo, I get about 0.82 ms per iteration at n=600K, again for k=701. For n=570K, I get almost the exact same time per iteration - 0.82 ms!

Based on this, and presuming that my assumption of 11000 k/n pairs per n=1K is correct, then we're on course to finish Drive #1 by December 17!

If you guys see any glaring errors in these calculations, feel free to point them out.

HOLY SMOLY!! surprised:surprised

Something in my gut tells me this is incorrect. It's taking a good couple of weeks for us to process the only n=~11K range that is currently loaded in
port 400. I still can't see how we'll do the entire n=578K-600K range that is left in just 2.5 weeks. Perhaps BlisteringSheep adding the equivalent of ~2 quads has helped more than I expected. I'm going to have to do a very detailed check on this since I'll quickly be able to get the exact k/n pairs remaining for unreserved ranges.

But if it is correct, heck we don't need rallies or anything else. That said, we'll still have at least 1 rally because they're fun! Regardless, to finish 2 weeks early would be very exciting!

Edit: BTW, if some people's reservations are still being worked on a few days after the servers are dry, that's no big deal as long as they're done before year end! :-)


Gary

[IronBits]

In the past 5 days, we have completed (now located at the top of the stats page) http://nplb.ironbits.net/progress_400.html
Total knpairs for all days shown: 41408

[gd_barnes]

Originally Posted by mdettweiler

Hmm...if I add up all those numbers, I get 16050 k/n pairs per day (not counting Ian's manual work, of course, which we don't yet have figures for). Based on an average of 1100 k/n pairs per n=100 range, and about n=25K left in Drive #1, there are ~275000 k/n pairs left to do.

If I divide 275000 by 16050, I get approximately 17 days to complete Drive #1. Of course, that is not factoring in any increases in the amount of time it takes to run each k/n pair, nor is it factoring in the amount of manual work that Ian does each day.

I'll see if I can come up with some more accurate calculations later, factoring in FFT jumps and Ian's manual work.

I couldn't resist going into more detail now that we have all of the figures. There were no glaring errors although there was a "moderate-sized" error in that the amount that we have remaining was underestimated by a fair amount. Thanks to everyone's info., we can now get a very detailed estimate:

First some assumptions:

Port 400:
1. Apparently BlisteringSheep dropped his machines off of port 400 sometime yesterday so I'm assuming that he won't have any on for the rest of the time.
2. I added an 8th quad to port 400 about the same time yesterday.

The approximate average on port 400 the last few days without Blis is 7150. Adding my additional quad adds another 750. This results in an average of 7900 / day for port 400; moderately lower than Carlos's 8800 / day figure.

Manual ranges:
I will assume that everyone is approximately 1/3rd done with their currently reserved manual ranges. I will exclude Lennart's range altogether because I'll assume that he'll be done well before Dec. 17th with it.

Misc.:
The port 4000 figure looked close and I'll use everyone else's estimates for port 443 and their manual ranges.

Although we're losing about 900 pairs / day on port 400 with this estimate, my estimate WILL include the 800 pairs / day done manually by Ian. So the 2 will almost offset one another.

Now to the remaining pairs. There are a lot more remaining than originally calculated. Calculations as of ~8 AM GMT on Dec. 2nd:

n=578.5K-600K  250,170
port 400        39,569
port 4000       16,237
port 443         8,384
n=547.3K-547.7K  4,798 * 2/3 = 3,200  (assumes 1/3rd done)
n=575.2K-576K    9,278 * 2/3 = 6,185
n=576K-577K     11,613 * 2/3 = 7,742
 
Total remaining:  331,487
 
Process per day:
port 400   7,900
port 4000  4,200
port 443   1,800
Carlos     1,000
Ian          800
Mini         236
 
Total processing per day:  15,936

Days remaining: 331,487 / 15,936 = 20.8

Now, even though there are no fftlen increases, the # of iterations per test still goes up. Example, let's say n=100K takes 100 secs. to test. For the same fftlen at n=120K, it would take 120 secs. to test.

The approximate average current test is n=572K. If you split the difference on what is left, you have (600K+572K)/2=586K. So the added time that it will take for the average remaining test will be 586K / 572K or 2.45% longer.

Adding 2.45% to 20.8 gives 21.3 days. Argh! That was a worthless adjustment. lol

So we're looking at finishing 21 days after Dec. 2nd or Dec. 23rd. Alas, Santa Claus arrives a bit early!

Hopefully Blis will be back, which will help. Regardless, let's see if we can get a rally in there and finish by the 20th or 21st!

It looks like I was way off base in thinking that there must be a serious error in the original calculations. I really thought it would be shortly after the end of the year when we would finish. Before Ian came on board, I was figuring on mid-late Jan.

Thanks to everyone for calculating your processing rates and for contributing all of your mean machines to our effort!


Gary

[MyDogBuster]

Hopefully Blis will be back, which will help. Regardless, let's see if we can get a rally in there and finish by the 20th or 21st!

So everyone has been challenged to finish this BEFORE the fat guy gets stuck in my chimmney again. I read Gary's analysis post and got a headache halfway thru, but all seems plausible barring any unforseen disasters.