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#12 | ||
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"Richard B. Woods"
Aug 2002
Wisconsin USA
22×3×641 Posts |
Quote:
Perhaps someone can put together a table on mersennewiki (I presume there's none there now). Quote:
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#13 |
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Dec 2003
Hopefully Near M48
175810 Posts |
For some reason, it looks like Mr. Woltman doesn't want to answer these questions; I don't know why.
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#14 |
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P90 years forever!
Aug 2002
Yeehaw, FL
1D6616 Posts |
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#15 |
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Dec 2003
Hopefully Near M48
2×3×293 Posts |
Thank you for the reply Mr. Woltman. So how will the amount of credit be calculated?
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#16 | |
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"Richard B. Woods"
Aug 2002
Wisconsin USA
11110000011002 Posts |
Quote:
Some raw excerpts: xjmptable DD 743, 32, 0.00000111, 896 ... allfft DD 68130000, 3670016, 0.323, 9456000 ... allfft DD 77910000, 4194304, 0.382, 5724160 ... allfft DD 96830000, 5242880, 0.485, 7045376 ... allfft DD 115300000, 6291456, 0.668, 8364416 ... allfft DD 134200000, 7340032, 0.886, 9683456 ... xjmptablep DD 739, 32, 0.00000111, 1152 ... allfft DD 58410000, 3145728, 0.289, 4143104 ... allfft DD 77700000, 4194304, 0.425, 5462016 ... DD 115000000, 6291456, 0.668, 9650176 ... DD 153100000, 8388608, 1.042, 12828672 ... I'll leave those to your interpretation. :-) Last fiddled with by cheesehead on 2006-11-12 at 10:26 |
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#17 |
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Jan 2006
Tampa, Florida
2×97 Posts |
I have a fair way of calculating how much credit I should recieve for M100,000,007. Since I am testing two exponents (on a two processor computer so they recieve equal resources), M36,234,713 (known credit after completion) and M100,000,007 (unknown credit), we can simply compare the two:
Fastest time per iteration possible: M36,234,713: 0.074 seconds M100,000,007: 0.254 seconds Fastest possible time to test: M36,234,713: (0.074 sec / iter)(36,234,713 iter) = 2,681,368 sec = 31.034 days M100,000,007: (0.254 sec / iter)(100,000,007 iter) = 25,400,002 sec = 293.982 days Obviously, these tests are taking slightly longer than this as the computer is slowed by other applications. However, this is the ideal time. Now we take the ratio of the two: (293.982 days / 31.034 days) = 9.4729 Note that (100,000,007 / 36,234,713)^2 = 7.6164, so the time actually increases slightly more than quadratic. However, once M36,234,713 finishes, I can check and see how much credit I recieved for that test and multiply it by 9.4729 to find out how much credit I will recieve for M100,000,007. It should end up being ~65 - 70 P90 CPU years. |
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#18 |
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Jan 2006
Tampa, Florida
2·97 Posts |
To George Woltman:
When are you going to update your site to include work on exponents larger than 79.3 million, not just M100,000,007, but also the 100M digits prefactor project, MasterPrime2006, and Operation Billion Digits? There is significant factoring work going on in this range, and LL tests are coming soon. Also, there are some banners on http://www.mersenne.org/primenet/status.shtml that need to be updated due to the incredible improvement and growth of GIMPS over the last few years: Some banners (that need to be updated) say: GIMPS / Primenet = 38 Cray T916 Supercomputers. Whoa. GIMPS does 79 years of computing- every day. Help out. GIMPS / Primenet. Join 15,000 fellow computer users. 38 of them are known in the universe. Find one more. 27,000 computers. Working as one. GIMPS / Primenet. It exists. Hiding. Cloaked. Find Mersenne Prime #39. GIMPS. Because 2,098,960 digits just aren't enough. GIMPS. Because 2^6,972,593-1 just isn't big enough. 950,000,000,000 floating point operations / sec. Nifty. These banners should say instead: GIMPS / Primenet = 400 Cray T932 Supercomputers. Whoa. GIMPS does 1900 years of computing- every day. Help out. GIMPS / Primenet. Join 50,000 fellow computer users. 44 of them are known in the universe. Find one more. 76,000 computers. Working as one. GIMPS / Primenet. It exists. Hiding. Cloaked. Find Mersenne Prime #45. GIMPS. Because 9,808,358 digits just aren't enough. GIMPS. Because 2^32,582,657-1 just isn't big enough. 22,800,000,000,000 floating point operations / sec. Nifty. |
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#19 | |
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Sep 2006
Brussels, Belgium
2·3·281 Posts |
Quote:
According to YOUR formula 100 000 007 should get you 73.3932 P90 CPU years. |
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#20 | |
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Dec 2003
Hopefully Near M48
6DE16 Posts |
You can use this: http://www.teamprimerib.com/rr1/bin/...onent=36234713
But is it really correct to give the smallest iteration times? Wouldn't it be more accurate to give the average iteration time? Quote:
Last fiddled with by jinydu on 2006-11-12 at 19:09 |
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#21 |
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Jan 2006
Tampa, Florida
2·97 Posts |
Sorry. I thought M36,234,713 would be about 7.2 CPU years. But if M100,000,007 is 73.3932 CPU years, then that is even better. I'll know for sure in about a week (when M36,234,713 finishes). If this is so, then it will mean that M100,000,007 requires 7.34 x 10^16 calculations:surprised to prove whether or not it is prime!
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#22 | |
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"Richard B. Woods"
Aug 2002
Wisconsin USA
22×3×641 Posts |
My data? Are you referring to the entries in the "[x]jmptable[p]" tables in Prime95 source module mult.asm?
Quote:
Last fiddled with by cheesehead on 2006-11-13 at 04:56 |
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