I was able to find the theoretical DP peak performance 371 GFlop/s for the Xeon E5-2690 in this Processor Comparison (interesting that it is easier to find this information in Intel's competitor than Intel support pages itself). However, when I try to derive that peak performance my derivation doesn't match:
add
and mul
operation per cycle so we get: 3.8 x 2 = 7.6Thus, the peak performance in Gflop/s would be 243.2 GFlop/s and not 371 GFlop/s?
here the formula: performance in GFlops = (CPU speed in GHz) x (number of CPU cores) x (CPU instruction per cycle) x (number of CPUs per node).
Intel Xeon E5-2690 2.9GHz/20M/1600MHz Eight Cores 135W (SR0L0) (Renewed)
CPU: Intel XEON Dual processor with minimum of 4 cores each, 3GHz or higher, with Turbo support. 16GB of RAM. To avoid occasional memory reallocation, provide 24GB of RAM. Dedicated hard-drive for the operating system, minimum 500GB.
Turbo Mode
is not used to calculate Theoretical Peak Performance
, you have to consider something like:
CPU speed = 2.9 GHz
CPU Cores = 8
CPU instruction per cycle = 8 (considering AVX-256 -> 256 bits unit, can hold 8 single precision values) x 2 (add and mul operations like you said) = 16
Putting all together:
2.9x8x16 = 371 GFlops/s
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With