Cray Unleashes 50-Petaflop XK6 Supercomputer
Since it runs on Nvidia you are most probably better off with your 17GHash/s farm.maybe ur right
thank u for getting me back up from under my bed ^^ 
Topic: Mining power of Bitcoin vs other networks (Read 12767 times)
maybe ur right
thank u for getting me back up from under my bed ^^ 
Cray Unleashes 50-Petaflop XK6 Supercomputer
Since it runs on Nvidia you are most probably better off with your 17GHash/s farm.
Cray Unleashes 50-Petaflop XK6 Supercomputer
http://www.pcmag.com/article2/0,2817,2385857,00.asp
i want that one my 17ghash are sooooo ugly in comparison lol
http://www.pcmag.com/article2/0,2817,2385857,00.asp
i want that one
my 17ghash are sooooo ugly in comparison lol
The FPGA is interesting. The ASIC numbers, including cost, would be even more interesting.
Yes, that is the most efficient one.(The 5970)
Though nVidia pulls about 0.7 mhashes/joule
AMD pulls 0.6 mhashes/joule
ARM 0.11 mhashes/joule
Intel 0.24 mhashes/joule
And this weird board: Altera EP4CE115C7 18.80 Mhashes/joule.($300 academic version) It is quite weird look at what it is.
https://en.bitcoin.it/wiki/Mining_hardware_comparison
Though nVidia pulls about 0.7 mhashes/joule
AMD pulls 0.6 mhashes/joule
ARM 0.11 mhashes/joule
Intel 0.24 mhashes/joule
And this weird board: Altera EP4CE115C7 18.80 Mhashes/joule.($300 academic version) It is quite weird look at what it is.
https://en.bitcoin.it/wiki/Mining_hardware_comparison
I gathered from elsewhere that the average number of hashes needed to get one block is given by the difficulty times 2^32. If this is correct then:
The average number of hashes per block right now is 1,564,057 * 2^32 = 6,717,573,664,079,872 = 6,717,573,664 Megahashes/Block
Using your number of 2.3 Megahashes/Joule yields 6,717,573,664 / 2.3 = 2,920,684,201 Joules/Block
Finally 2,920,684,201 Joules/Block is 811 KWh/Block
Using your number of 0.15 $/KWh yields $121.65 per block or $2.43 per Bitcoin.
I have checked and rechecked these numbers and given the assumptions I believe they are correct.
Of course the 2.3 Megahashes/Joule is only the energy used by the video card and does not take other power consumption and cooling costs into consideration.
So the actual cost is more, this is just the lower bound.
The average number of hashes per block right now is 1,564,057 * 2^32 = 6,717,573,664,079,872 = 6,717,573,664 Megahashes/Block
Using your number of 2.3 Megahashes/Joule yields 6,717,573,664 / 2.3 = 2,920,684,201 Joules/Block
Finally 2,920,684,201 Joules/Block is 811 KWh/Block
Using your number of 0.15 $/KWh yields $121.65 per block or $2.43 per Bitcoin.
I have checked and rechecked these numbers and given the assumptions I believe they are correct.
Of course the 2.3 Megahashes/Joule is only the energy used by the video card and does not take other power consumption and cooling costs into consideration.
So the actual cost is more, this is just the lower bound.
5.8 petaflops, or 5800 teraflops...
on pretty much all recent archs with vector units, one 32-bit INTOP = 2 single precision FLOP
one bitcoinhash is ~6.35k x86 INTOP
450G hash/s * 6.35k INTOP/hash = 2858T INTOP/s
2858T INTOP/s * 2 FLOP/INTOP = ~6700T FLOP/s
on pretty much all recent archs with vector units, one 32-bit INTOP = 2 single precision FLOP
one bitcoinhash is ~6.35k x86 INTOP
450G hash/s * 6.35k INTOP/hash = 2858T INTOP/s
2858T INTOP/s * 2 FLOP/INTOP = ~6700T FLOP/s
I asked the admin and he responded that the conversion used on the http://bitcoinwatch.com/ page is from this very thread! The site uses the formula from above (after correction for the typo).
The page simply uses the following assumptions/estimates:
1 INTOP = 2 FLOP
1 hash = 6.35K INTOP
1 hash = 12.7K FLOP
So the hashrate in TeraFLOP/s is simply 12.7 times the hashrate in Gigahashes/s.
For example as I wrote this: 11,558.55 Gigahashs/s * 12.7 TeraFLOP/Gigahash = 146,794 TeraFLOP/s = 146 PetaFLOP/s
So it is true.
I wonder how much electricity it is. Wait a second for the edit.
EDIT: So, assuming everyone is mining with 5970s, the most efficient, it would mean that we got a 2.3 Mhashes/joule -> 13,300,000 Mhashes/x -> 5,782,608 Joules
3.6 mJ = 1 kWh -> ~1.6
Assuming 15 cents per kWh that is an expensive bill for just one block. D:
5.8 petaflops, or 5800 teraflops...
on pretty much all recent archs with vector units, one 32-bit INTOP = 2 single precision FLOP
one bitcoinhash is ~6.35k x86 INTOP
450G hash/s * 6.35k INTOP/hash = 2858T INTOP/s
2858T INTOP/s * 2 FLOP/INTOP = ~6700T FLOP/s
on pretty much all recent archs with vector units, one 32-bit INTOP = 2 single precision FLOP
one bitcoinhash is ~6.35k x86 INTOP
450G hash/s * 6.35k INTOP/hash = 2858T INTOP/s
2858T INTOP/s * 2 FLOP/INTOP = ~6700T FLOP/s
I asked the admin and he responded that the conversion used on the http://bitcoinwatch.com/ page is from this very thread! The site uses the formula from above (after correction for the typo).
The page simply uses the following assumptions/estimates:
1 INTOP = 2 FLOP
1 hash = 6.35K INTOP
1 hash = 12.7K FLOP
So the hashrate in TeraFLOP/s is simply 12.7 times the hashrate in Gigahashes/s.
For example as I wrote this: 11,558.55 Gigahashs/s * 12.7 TeraFLOP/Gigahash = 146,794 TeraFLOP/s = 146 PetaFLOP/s
I am wondering about the number of TeraFLOP/s reported on the http://bitcoinwatch.com/ page. For example the number I see right now is 145906.
If this number is really in TeraFLOP/s then it is claiming 146 PetaFLOP/s!
That can't be right. What do you think? This number is larger than all the supercomuters and all other distributed computing projects put together.
I have an email into the page admin and will let you know how they respond.
If this number is really in TeraFLOP/s then it is claiming 146 PetaFLOP/s!
That can't be right. What do you think? This number is larger than all the supercomuters and all other distributed computing projects put together.
I have an email into the page admin and will let you know how they respond.
There is no way 5970 can achieve the rated 4.64 TFLOPS in any benchmark. Don't delude yourself that Bitcoin universe is that powerful (although it is still impressive). Supercomputers are benchmarked with Linpack and for 5970, it will be 0.5 TFLOPS tops in double precision. In single precision, it'll be less than 2 TFLOPS. In real life calculations (say molecular dynamics from Folding@Home) it's going to be even less.
5970 is extremely efficient in sha256 hashing, an extremely simple, predictable and low bandwidth application. No so much for anything more practical.
I calculated Tianhe-1A hashspeed by counting its number of Teslas and Xeons and it is about 845 GH/s. It is rated 2.5 PFLOPS.
https://bitcointalksearch.org/topic/m.66931
5970 is extremely efficient in sha256 hashing, an extremely simple, predictable and low bandwidth application. No so much for anything more practical.
I calculated Tianhe-1A hashspeed by counting its number of Teslas and Xeons and it is about 845 GH/s. It is rated 2.5 PFLOPS.
https://bitcointalksearch.org/topic/m.66931
from wiki
I think they wouldn't be too far from that
Quote
Operation costs
In energy cost, according to the Green500 list, as of June 2010 the most efficient TOP500 supercomputer runs at 773.38 MFLOPS per watt. This translates to an energy requirement of 1.29 watts per GFLOPS, however this energy requirement will be much greater for less efficient supercomputers.
In energy cost, according to the Green500 list, as of June 2010 the most efficient TOP500 supercomputer runs at 773.38 MFLOPS per watt. This translates to an energy requirement of 1.29 watts per GFLOPS, however this energy requirement will be much greater for less efficient supercomputers.
I think they wouldn't be too far from that
Interesting number 773 MFLOPS per watt. This means that 5970 wattage equivalent (300W) would be 2.3 TFLOPS, while 5970 is rated twice that. This means that bitcoin miners run their state of the art bitcoin rigs twice more efficiently than "June 2010 the most efficient TOP500 supercomputer". Moreover for INTOPS 5970's are than arguably 6-10 times more efficient than your typical CPU or NVIDIA GPU based supercomputer.
I'm so tired I didn't even notice
from wiki
I think they wouldn't be too far from that
Quote
Operation costs
In energy cost, according to the Green500 list, as of June 2010 the most efficient TOP500 supercomputer runs at 773.38 MFLOPS per watt. This translates to an energy requirement of 1.29 watts per GFLOPS, however this energy requirement will be much greater for less efficient supercomputers.
In energy cost, according to the Green500 list, as of June 2010 the most efficient TOP500 supercomputer runs at 773.38 MFLOPS per watt. This translates to an energy requirement of 1.29 watts per GFLOPS, however this energy requirement will be much greater for less efficient supercomputers.
I think they wouldn't be too far from that
Order of magnitude is about right ... 30,000 GPU cores ... maybe knock some off for auxiliary device power consumption, cooling, lights, coffee maker.
I'd say 20-25K GPU cores puts them double anything else out there in total hardware cores which is a good feel for where the nuts 'n bolts bleeding edge is at, so yeah maybe 6-7 THash/sec.... sounds like a good factor of safety target for bitcoin network.
If you could get a peek at their electricity bill
Around 40 million a year last I heard. Who knows though really.
The NSA has without a doubt the biggest integer machine(s) on the planet but their hash power is classified.
If you could get a peek at their electricity bill it maybe possible to get a conservative order of magnitude idea of hash power using 5970 type Hash/Watt ratios.
Very nice. Thanks for all the replies & outside info.
I could be wrong, however, we are already at the level of SETI and F@H.
Yeah, ours pays you
Seriously though folding and all the others are pretty awesome.
Now a 5870 does 320 MHashes with latest poclbm miner.
A 5870 does roughly 2.72 TeraFLOPS
That means that if we divide the 568000 MHashes/sec the network is doing between 320 we get 1175 5870s.
We then take that number and multiply it by its computational power of 2.72 TFLOPS and we get 4828 TeraFlops or almost 5 Petaflops.
I could be wrong, however, we are already at the level of SETI and F@H.
A 5870 does roughly 2.72 TeraFLOPS
That means that if we divide the 568000 MHashes/sec the network is doing between 320 we get 1175 5870s.
We then take that number and multiply it by its computational power of 2.72 TFLOPS and we get 4828 TeraFlops or almost 5 Petaflops.
I could be wrong, however, we are already at the level of SETI and F@H.
How about another way to measure, based on the total capacity of the hardware even if we're not using all of its floating point capabilities for mining. Assuming entire network is 5970s, even though in reality, it's probably much or most of the network.
Published FLOPS for 5970: 4640G (FLOPS/5970)
Hashrate of 5970 (stock): 580M ((hash/s)/5970)
Network hashrate: 450G (hash/s)
Total 5970s: 450G (hash/s) / (580M ((hash/s)/5970) = 775.86 (5970)
Total network FLOPS based on total 5970s: 4640G (FLOPS/5970) * 775.86 (5970) = 3.6PFLOPS
The number might not mean a whole lot - other than how many FLOPS are available if suddenly the supercomputing power of all miners could be bought and redirected to some other purpose.
It might be a sanity check for the higher figure - if my figures are right, the idea that the Bitcoin network has more total FLOPS than the sum of the published FLOPS of all the individual GPU's that make it up might garner some doubt.
Published FLOPS for 5970: 4640G (FLOPS/5970)
Hashrate of 5970 (stock): 580M ((hash/s)/5970)
Network hashrate: 450G (hash/s)
Total 5970s: 450G (hash/s) / (580M ((hash/s)/5970) = 775.86 (5970)
Total network FLOPS based on total 5970s: 4640G (FLOPS/5970) * 775.86 (5970) = 3.6PFLOPS
The number might not mean a whole lot - other than how many FLOPS are available if suddenly the supercomputing power of all miners could be bought and redirected to some other purpose.
It might be a sanity check for the higher figure - if my figures are right, the idea that the Bitcoin network has more total FLOPS than the sum of the published FLOPS of all the individual GPU's that make it up might garner some doubt.
Yeah, typo, meant 2*2858T = ~ 5700TFLOP/s
current estimated network speed is all over the place depending on if you use difficulty, 1-day avg, 1-week avg or whatever
so "somewhere around 5-6 PFLOP/s"
[edit]
argh, 2*2858T = ~ 5700TFLOP/s, not G
current estimated network speed is all over the place depending on if you use difficulty, 1-day avg, 1-week avg or whatever
so "somewhere around 5-6 PFLOP/s"
[edit]
argh, 2*2858T = ~ 5700TFLOP/s, not G
Jump to: