Difference between sandy bridge and ivy bridge

Difference between sandy bridge and ivy bridge how to#

Yet if i apply the same math like for the nvidia then: 316.8 gflops / (45GB/s * 80%) = 8.8 gflops per byte on the Haswell Xeons. As it appears the results i could only reproduce at P4's with overclocked memory controller (FSB), years years later. Regrettably cheapest motherboard i saw that i would find acceptable is a supermicro X10dri, though i do not know the memory bandwidth it can achieve.ДЄs those intel testmachines with 100% sureness are always overclocked major league everywhere - i remember there how Johan de Gelas start this century also had new P4's with Diep on it and it took me many years until i could reproduce his results. In diepeveen-double precision throughput as compared to the nvidia chip that's giving then: 3 * 4 = 12 doubles per core per clocktick throughput.ДЎ2 cores * 2.2Ghz * 12 = 316.8 gflops double precision for a 12 core chip at 2.2Ghz. Now i didn't check how many uops a haswell AVX2.0 instruction eats - i'm assuming somehow you can get to 3 instructions per clock AVX2.0. The 12 core cpu's are sometimes bit cheaper on ebay, as opposed to the 4k dollar a chip for the 18 core things. At Johan de Gelas test you can see putting in 1 rdimm ddr4 per channel is fastest, provided you do not need overly much RAM, for maximum bandwidth.Д©0GB/s roughly is very impressive there from a single socket CPU. Seems like that you need 4 rdimm 2133Mhz (well that's how it's called actually clocked of course a lot lower) for the 4 channels.

I was looking at building a 2 socket Haswell. If we compare with a CPU, it's very impressive what latest generation Xeons deliver there - they have made up much terrain compared to GPU's last years. That gives an efficiency need of 2.5 gflops per byte at this nvidia titan-z gpu. Maybe 80% is usable of that practical which is roughly 0.8 * 336.5 = 270GB/s

If i ignore the manufacturer cheating and use diepeveen double precision gflops a single chip in the titan-z (it has 2 gpu's on each card) delivers roughly: 0.7Ghz * 15 SM's * 64 doubles a clock = 672 Gflops (without cheating, with cheating multiply it by 2 or so)

Difference between sandy bridge and ivy bridge how to#

With those 512 doubles which you have to share with 32 cuda cores, it's all about how to efficiently use the GDDR5 bandwidth. Practical with power of 2 that's just 512 doubles and 4KB (allows 12 warps). Which is just a meager 48KB or so divided by say at least 8 warps (number of simultaneously running threads at the same SM-core) you have at most 6 KB for a vector of 32 cudacores. That's one of the major problems to solve at a GPU for transforms that do not fit inside the tiny L1 datacache of a gpu. What's at limbnumber binary 0b0000001010001 for example has to go then to 0b1000101000000 and vice versa. First problem that cpu's do not share because of caches is rewriting the data.

Though most of it will be testing to see if it's possible to get the bandwidth. Yeah i had bought a Titan-Z from nvidia to toy here to get FFT to work for k * 2^n +- c transforms.