Cores vs Frequency

snakyjake

Distinguished
Mar 17, 2009
28
0
18,530
I'm contemplating building a system with either the Intel E5-2620 or E3-1275. It mainly comes down to total processing power that I need to take advantage of multi-core statisical calculation (floating point) applications and large datasets (memory and drive storage).

Is the method to estimate total CPU processing power simply base frequency multiplied by cores?

E5-2620 2.1 GHz * 8 Cores = 16.8 GHz
E3-1275 3.6 GHz * 4 Cores = 14.4 GHz
Diff = 4.4 GHz
And perhaps more of a difference if I use Max Frequency instead of the Base, though i don't think I can rely on this number for sustained workloads that can take hours to complete.

Thank you.
 
one also should factor in instructions per clock, but as these are pretty much the same chips in terms of technology, it does not matter. You mentioned this is a multi threaded application, and thus hit it with as many threads as possible. I would get that six core. It might be slower per core, but it has 4 extra threads to throw at tasks. also it is possible to get a small OC by raising the base clock a few MHz
 


Hi,

Modern Intel microprocessors can perform one FPADD and one FPMUL per core per cycle. These can be either IEEE-754 compliant or fused for added precision. The same hardware is used for both scalar and vector operations. FPDIV is slower and shares a port with FPMUL.

Sandybridge and Ivybridge based CPUs support the AVX instructions, which extend floating point operations to 256 bit vectors. Integer vectors are still limited to 128 bits.

Haswell and Broadwell based CPUs support the AVX2 instructions, which extends integer vector operations to 256 bits. Fused Multiply Add is also supported for floating point operations.

Skylake-E will support AVX3/AVX-512 on its Xeon models, which extend floating point vector operations to 512 bits.

If you're looking for a good number cruncher, Skylake based E5 CPUs will be the way to go. If you can't wait that long, then the peak double precision FLOPS can be calculated by simply multiplying 2*8*<# cores>*<clock>