the Zen of Communication.

Running some performance tuning, the app needed to know how many flops (floating point operations) per cycle the system could handle.

I used SiSoftware’s Sandra benchmarking app. It told me that my Intel Pentium D Dual Core 1.8ghz proc was producing 10.86gflops, but not the flops per clock cycle.

From this we know: a) the total gigaflops (10.86), b) the number of cores (2), and c) the number of clock cycles per second (1.8ghz)

Example of the standard formula:
The formula to determine total gigaflops is:
Flops per cycle x # of Cores x Clock speed.

This involves four values:

a = flop per clock cycle
b = clock speed (ghz)
c = cores
n = gflops

For a dual core 3ghz system with 4 flops per cycle, we can deduce 24gflops (a x c x b = n, or 4 x 2 x 3 = 24) . But I only have the total gflops, clock speed, and number of cores.

Reverse algebra:

a = n / b / c

Or in my case:
10.86 gflops / 1.8ghz / 2 cores = 3.01 flops per cycle (per core). So the E2610 chip at 1.8ghz produces 3 flops per cycle per core, or 6 flops total. Ta da.

Note: It’s worth mentioning that in this case, 10.86 gflops and 1.8ghz seem like closely related numbers, and that it would be quick to figure out how many gflops a system can handle by its clock speed (i.e. 1.8ghz equals 10.86gflops).  This is not the case.  In the first example of a dual core 3ghz proc producing 24gflops, you can’t deduce the one from the other.  It was just a coincidence in my case, so don’t do that.

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