How to compute the theoretical peak performance of CPU

Question

Here is my cat /proc/cpuinfo output:

...

processor           : 15
vendor_id           : GenuineIntel
cpu family          : 6
model               : 26
model name          : Intel(R) Xeon(R) CPU           E5520  @ 2.27GHz
stepping            : 5
cpu MHz             : 1600.000
cache size          : 8192 KB
physical id         : 1
siblings            : 8
core id             : 3
cpu cores           : 4
apicid              : 23
fpu                 : yes
fpu_exception       : yes
cpuid level         : 11
wp                  : yes
flags               : fpu vme de pse tsc msr pae mce cx8 apic ...
bogomips            : 4533.56
clflush size        : 64
cache_alignment     : 64
address sizes       : 40 bits physical, 48 bits virtual
power management    :

This machine has two CPUs, each with 4 cores with hyperthreading capability, so the total processor number is 16(2 CPU * 4 core * 2 hyperthreading). These processors have same output, to keep clean, I just show the last one's info and omit part of flags in the flags line.

So how do I calculate the peak performance of this machine in terms of GFlops? Let me know if more info should be supplied.

Thanks.

Solution 1

You can check the Intel export spec. The GFLOP in the chart is usually referred as the peak of a single chip. It shows 36.256 Gflop/s for E5520.

This single chip has 4 physical cores with SSE. So this GFLOP can also be calculated as: 2.26GHz*2(mul,add)*2(SIMD double precision)*4(physical core) = 36.2.

You system has two CPUs, so your peak is 36.2*2 = 72.4 GFLOP/S.

Solution 2

you can find a formula in this website:

http://www.novatte.com/our-blog/197-how-to-calculate-peak-theoretical-performance-of-a-cpu-based-hpc-system

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).

so in your case: 2.27x4x4x2=72.64 GFLOP/s see here for the configuration of your CPU http://ark.intel.com/products/40200