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Old 2017-12-01, 01:52   #7
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"Mihai Preda"
Apr 2015

44E16 Posts

Originally Posted by Dubslow View Post
That number, 172.8, is not a coincidence: it is in fact exactly (24*60*60)/500, which is another way of saying that 1 EGHz (1 [Core 2] Equivalent GHz, = 1GHz-day/day) is exactly 2 GFLOPS. Not sure which unit is defining the other, but that looks like the definition used by PrimeNet. Does anyone have an old Core 2 lying around?

So a 100 EGHz CPU (?? GPU?) is doing 200 GFLOPS according to the PrimeNet definition of EGHz and FLOP. I'm not sure how these line up to an actual Core 2 or to the "standard" definitions of FLOP.
OK, thanks, so 1 EGHz == 2 GFLOPS.

Now, an LL / PRP test at 76M exponent is evaluated to about 216 GHz-Days, which comes to 37.3 PFLOP (i.e. 216 * 24 * 60 * 60 * 2 GFLOP).

I would be curious to make a comparison with the actual number of FP64 operations done by the FFTs at that size.
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