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ATH · 2019-03-29, 15:00

You can calculate what 1 Ghz-day is defined as in terms of FLOP (Floating Point operations):

https://www.mersenne.org/primenet/

Right now in the "Aggregate Computing Power" last 24 hours:
498.551 TFLOP/sec = 498,551 GFLOP/sec but it is also equal to: 249275 GHz-days.

So 1 Ghzdays is defined as 498551/249275 = 2 GFLOP/sec for 24 hours which is:

2 GFLOP/s * 86400 sec = 172,800 GFLOP = 172.8 TFLOP

So 1 Ghz-days is 172.8*10¹² Floating Point operations.

But all the Ghz-days calculations are approximate, I think it is hard to determine the exact number of FLOP that LL, PRP, P-1, TF and ECM tests requires.

2019-03-29, 15:00	#5
ATH Einyen Dec 2003 Denmark 2×11×137 Posts	You can calculate what 1 Ghz-day is defined as in terms of FLOP (Floating Point operations): https://www.mersenne.org/primenet/ Right now in the "Aggregate Computing Power" last 24 hours: 498.551 TFLOP/sec = 498,551 GFLOP/sec but it is also equal to: 249275 GHz-days. So 1 Ghzdays is defined as 498551/249275 = 2 GFLOP/sec for 24 hours which is: 2 GFLOP/s * 86400 sec = 172,800 GFLOP = 172.8 TFLOP So *1 Ghz-days is 172.810¹² Floating Point operations*. But all the Ghz-days calculations are approximate, I think it is hard to determine the exact number of FLOP that LL, PRP, P-1, TF and ECM tests requires. Last fiddled with by ATH on 2019-03-29 at 15:01*