8350rocks
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juanrga :
gamerk316 :
juanrga :
A more general equation used in HPC is
Performance = (CPU speed in GHz) x (number of CPU cores) x (CPU instruction per cycle) x (number of CPUs per node)
http://www.novatte.com/our-blog/197-how-to-calculate-peak-theoretical-performance-of-a-cpu-based-hpc-system
For desktops, the "number of CPUs per node" = 1 (unless some of you have a dual socket mobo or a cluster in the home ) and the general HPC formula reduces to that I gave above:
Performance = GHz x cores x instruction per cycle.
You are essentially trying to ignore SMT concerns. While you can get away with this for HTT, other forms of SMT, such as AMD's CMT, start causing issues, due to the overhead involved in using the second core of a BD module. Hence why I tried to keep the comparisons between just quads, because you start making the formula a LOT more complicated.
And if you REALLY want to be technical, all this math assumes the processor is doing 100% work on the program in question, and if any CPU core isn't stuck at 100% load, you grossly overestimate IPC, so you need to account for core loading on a Per-Core basis, then figure out how much each core affected performance, hence why Physical and Logicial cores have to be considered separately, factoring in the "average" performance benefit/loss of various core loading profiles into the formula, which goes well outside what we are trying to discuss at this point.
Hence, the IPC numbers given by my formulas are grossly inflated. But for comparing two processors where the app in question is the only major program running, the numbers are good enough to be able to calculate the relative difference in performance between two chips, since the numbers would be inflated equally for both.
Also, Juan, learn some math:
http://mathforum.org/library/drmath/view/58083.html
http://www.mathsisfun.com/percentage-difference.html
You are trying to use % Change, which is not valid when comparing two significantly different items. You have to use % difference instead, which is defined as:
| (V1 - V2) / (V1 + V2) / 2 | * 100%
Hence my numbers.
First, SMT != CMT.
Second, the above formula for performance also works for SMT. If you check the above HPC link you can found the application of the formula to Intel Xeon processors. Of course, one uses physical cores, not virtual cores.
Third, yes the processors have to be loaded in order to measure performance. If the processors are at iddle, then we aren't measuring their potential. I believed this class of stuff was self-evident. I try to avoid self-evident stuff in my posts. Would I mention that the computer has to be turned on?
Fourth, you continue making the same mistake about percentages.
If Phenom II scores 13.51 and Kaveri scores 16.77 then Kaveri is 24% faster than Phenom II. It is not 21% faster as you pretend. I already gave you the formula. I repeat it:
(( 16.77 - 13.51 ) / 13.51 ) * 100 = 24.13
You can find this formula in the same site that you mention
http://www.mathsisfun.com/numbers/percentage-change.html
SMT can be CMT