Conversion GHz to GFLOPS?

preda · 2017-11-30, 22:34

PrimeNet measures the amount of compute work done in GHz-days (e.g. https://www.mersenne.org/report_top_500/ )

Dividing GHz-Days by "days" to get compute throughput (compute per unit of time), would produce "GHz" as a unit of compute power.

E.g. a computer that does 100 GHz-days in 24h, has a compute power of 100GHz.

Another common unit of "compute power" is GFLOPS, e.g. used in https://en.wikipedia.org/wiki/TOP500
https://en.wikipedia.org/wiki/FLOPS

What is the correspondence between the two units of "compute power", "PrimeNet GHz" and GFLOPS?

E.g. for the example computer that does 100GHz-Days per day, how many GFLOPS would that be?

science_man_88 · 2017-12-01, 00:03

Quote:

Originally Posted by preda

PrimeNet measures the amount of compute work done in GHz-days (e.g. https://www.mersenne.org/report_top_500/ )

Dividing GHz-Days by "days" to get compute throughput (compute per unit of time), would produce "GHz" as a unit of compute power.

E.g. a computer that does 100 GHz-days in 24h, has a compute power of 100GHz.

Another common unit of "compute power" is GFLOPS, e.g. used in https://en.wikipedia.org/wiki/TOP500
https://en.wikipedia.org/wiki/FLOPS

What is the correspondence between the two units of "compute power", "PrimeNet GHz" and GFLOPS?

E.g. for the example computer that does 100GHz-Days per day, how many GFLOPS would that be?

even Ghz days isn't consistent https://answers.yahoo.com/question/i...1100447AAeZ0E9

sdbardwick · 2017-12-01, 00:08

IIRC, for Prime95 purposes, 1 GHz/Day = the work done in 24 hrs by a sole 1 Ghz Core2 processor core.
Edit2:

Code:

// In Primenet v4 we used a 90 MHz Pentium CPU as the benchmark machine
// for calculating CPU credit. The official unit of measure became the
// P-90 CPU year. In 2007, not many people own a plain Pentium CPU, so we
// adopted a new benchmark machine - a single core of a 2.4 GHz Core 2 Duo.
// Our official unit of measure became the C2GHD (Core 2 GHz Day). That is,
// the amount of work produced by the single core of a hypothetical
// 1 GHz Core 2 Duo machine. A 2.4 GHz should be able to produce 4.8 C2GHD
// per day.
//
// To compare P-90 CPU years to C2GHDs, we need to factor in both the
// the raw speed improvements of modern chips and the architectural
// improvements of modern chips. Examining prime95 version 24.14 benchmarks
// for 640K to 2048K FFTs from a P100, PII-400, P4-2000, and a C2D-2400
// and compensating for speed differences, we get the following architectural
// multipliers:
//
// One core of a C2D = 1.68 P4.
// A P4 = 3.44 PIIs
// A PII = 1.12 Pentium
//
// Thus, a P-90 CPU year = 365 days * 1 C2GHD *
//	(90MHz / 1000MHz) / 1.68 / 3.44 / 1.12 
//	= 5.075 C2GHDs

Edit: Source of above: Prime95's post from years ago

science_man_88 · 2017-12-01, 00:41

I may be wrong, but I believe Ghz is a measure of clock frequency, flops is a measure of number of a specific type of operation. some operations can be done twice a clock cycle, some more so there's not really a good conversion between the two, hence why the type of operation matters.

ATH · 2017-12-01, 00:45

Long ago I calculated 1 Ghz-days to be ~ 171 TeraFLOP (Not FLOPS)
Post #7 and #25 in this old thread:
http://www.mersenneforum.org/showthread.php?t=10235

Now we can calculate it more easily from here:
https://www.mersenne.org/primenet/

Under "Aggregate Computing Power" - "Today, last 24 hours" we can see that:

123137 Ghz-Days = 246.275 TFLOP/sec * 86400 sec = 21278160 TFLOP =>
1 Ghz-Days = 21278160 TFLOP / 123137 = 172.8 TFLOP

We can test "last 7 days" and "last 30 days" to see if we get the same results:

994150 Ghz-Days = 284.043 TFLOP/s * 86400s/day * 7 days =>
1 Ghz-Days = 172.8 TFLOP

4395208 Ghz-Days = 293.014 TFLOP/s * 86400s/day * 30 days =>
1 Ghz-Days = 172.8 TFLOP

Dubslow · 2017-12-01, 01:23

Quote:

Originally Posted by ATH

Long ago I calculated 1 Ghz-days to be ~ 171 TeraFLOP (Not FLOPS)
Post #7 and #25 in this old thread:
http://www.mersenneforum.org/showthread.php?t=10235

Now we can calculate it more easily from here:
https://www.mersenne.org/primenet/

Under "Aggregate Computing Power" - "Today, last 24 hours" we can see that:

123137 Ghz-Days = 246.275 TFLOP/sec * 86400 sec = 21278160 TFLOP =>
1 Ghz-Days = 21278160 TFLOP / 123137 = 172.8 TFLOP

We can test "last 7 days" and "last 30 days" to see if we get the same results:

994150 Ghz-Days = 284.043 TFLOP/s * 86400s/day * 7 days =>
1 Ghz-Days = 172.8 TFLOP

4395208 Ghz-Days = 293.014 TFLOP/s * 86400s/day * 30 days =>
1 Ghz-Days = 172.8 TFLOP

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.

preda · 2017-12-01, 01:52

Quote:

Originally Posted by Dubslow

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.

VictordeHolland · 2017-12-01, 19:03

A Core2Duo core has a SSE registers of 128bits for SIMD . It can 'split' the SIMDregister to do 2x FP (DoublePrecision, FP64) adds and execute them at the same time in parallel. In addition it can also do 2x FP64 multiplications (mul) at the same time. So in theory a C2D core could do 4 DP FLOPs (2adds+2mul) per clockcycle.

However, in practice getting half that is already extremely difficult, even though Prime95 is very efficient. You hit latencies and memory bandwidth issues pretty fast.

AVX introduced bigger 256bit SIMD registers (Sandy, Ivy Bridge). So theoretically can do 8 DP FLOPs per clockcyle.

AVX2 extends the instruction set introduced with AVX and introduces FMA3 (Haswell, Broadwell, Skylake-S, Kabylake-S). Now it increases to 16 DP FLOPs per cycle if using FMA.

AVX512 extends the SIMD register once again, to 512 bits (and a multitude of subsets, which differ between the Knights Landing co-processor and the normal CPUs). 32 DP FLOPs per cycle.

VictordeHolland · 2017-12-02, 01:23

Intel Core i5 2500k @4GHz
DDR3-2133 (Dual Channel)

Theoretical:
8 DP/clock * 4 GHz = 32 GFLOP/s per core
32 GFLOPS * 4 cores = 128 GFLOP/s total

LINPACK with Intel pre-compiled binaries (read: ultra optimised for highest throughput/ benchmark results).

1 core:

Code:

Performance Summary (GFlops)

Size   LDA    Align.  Average  Maximal
1000   1000   4       23.9988  24.2828 
2000   2000   4       25.9523  26.4666 
5000   5008   4       27.9580  28.1283 
10000  10000  4       28.9317  28.9918 
15000  15000  4       29.4065  29.4075 
18000  18008  4       29.4349  29.4349 
20000  20016  4       29.3201  29.3201 
22000  22008  4       29.4509  29.4509 
25000  25000  4       29.5640  29.5640 
26000  26000  4       29.6370  29.6370 
27000  27000  4       29.4683  29.4683 
30000  30000  1       29.6019  29.6019 
35000  35000  1       29.6650  29.6650 
40000  40000  1       29.5893  29.5893 

Residual checks PASSED

End of tests

4 cores

Code:

Performance Summary (GFlops)

Size   LDA    Align.  Average  Maximal
1000   1000   4       71.0218  73.5734 
2000   2000   4       84.0847  85.3924 
5000   5008   4       88.9042  93.5976 
10000  10000  4       100.6101 100.8921
15000  15000  4       99.3512  104.1565
18000  18008  4       106.5735 106.9429
20000  20016  4       106.7802 107.0437
22000  22008  4       112.7814 112.8765
25000  25000  4       112.9538 112.9650
26000  26000  4       113.1638 113.2873
27000  27000  4       113.0612 113.0612
30000  30000  1       113.7686 113.7686
35000  35000  1       112.7552 112.7552
40000  40000  1       112.7320 112.7320

Residual checks PASSED

End of tests

80-90% of the theoretical throughput (as was to be expected with highly optimised binaries). Even though Linpack is used to benchmark/rank supercomputers, it is a pretty meaningless benchmark as it does not (or barely) take into account how well the cores/sockets/systems are interconnected. It should only be treated as a best practically achievable and nothing more than that.

Prime95 benchmark for this machine (somebody can probably convert these to GHzdays)

Code:

Prime95 64-bit version 29.3, RdtscTiming=1
Timings for 1024K FFT length (4 cores, 4 workers):  6.01,  6.19,  5.98,  6.10 ms.  Throughput: 659.13 iter/sec.
Timings for 2048K FFT length (4 cores, 4 workers): 12.72, 13.06, 12.67, 12.91 ms.  Throughput: 311.53 iter/sec.
Timings for 4096K FFT length (4 cores, 4 workers): 27.12, 27.71, 26.93, 27.32 ms.  Throughput: 146.71 iter/sec.
Timings for 6144K FFT length (4 cores, 4 workers): 40.92, 41.86, 40.62, 41.34 ms.  Throughput: 97.13 iter/sec.
Timings for 8192K FFT length (4 cores, 4 workers): 57.47, 59.16, 56.94, 58.09 ms.  Throughput: 69.08 iter/sec.

LaurV · 2017-12-02, 05:34

Quote:

Originally Posted by Dubslow

Does anyone have an old Core 2 lying around?

Yes, what should I do with it?
(and please don't say "stick it in your... ", hehe)

vasyannyasha · 2017-12-02, 18:38

From post #9
112349 seconds to test exponent M35000011 4 cores
43.75 Ghz-day=12 PFLOP
So 1 Ghz-day=0.27 PFLOP=274 TFLOP

I made measurements on my AMD
760208 seconds to test exponent M35000011 1 core
43.75 Ghz-day=16.7 PFLOP
So 1 GHz-day=0.38 PFLOP=372 TFLOP

Maybe my calculations wrong

2017-11-30, 22:34	#1
preda "Mihai Preda" Apr 2015 2×11×61 Posts	Conversion GHz to GFLOPS? PrimeNet measures the amount of compute work done in GHz-days (e.g. https://www.mersenne.org/report_top_500/ ) Dividing GHz-Days by "days" to get compute throughput (compute per unit of time), would produce "GHz" as a unit of compute power. E.g. a computer that does 100 GHz-days in 24h, has a compute power of 100GHz. Another common unit of "compute power" is GFLOPS, e.g. used in https://en.wikipedia.org/wiki/TOP500 https://en.wikipedia.org/wiki/FLOPS What is the correspondence between the two units of "compute power", "PrimeNet GHz" and GFLOPS? E.g. for the example computer that does 100GHz-Days per day, how many GFLOPS would that be?

2017-12-01, 00:41	#4
science_man_88 "Forget I exist" Jul 2009 Dumbassville 2⁶×131 Posts	I may be wrong, but I believe Ghz is a measure of clock frequency, flops is a measure of number of a specific type of operation. some operations can be done twice a clock cycle, some more so there's not really a good conversion between the two, hence why the type of operation matters. Last fiddled with by science_man_88 on 2017-12-01 at 00:42

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2017-12-01, 00:45	#5
ATH Einyen Dec 2003 Denmark BD7₁₆ Posts	Long ago I calculated 1 Ghz-days to be ~ 171 TeraFLOP (Not FLOPS) Post #7 and #25 in this old thread: http://www.mersenneforum.org/showthread.php?t=10235 Now we can calculate it more easily from here: https://www.mersenne.org/primenet/ Under "Aggregate Computing Power" - "Today, last 24 hours" we can see that: 123137 Ghz-Days = 246.275 TFLOP/sec * 86400 sec = 21278160 TFLOP => 1 Ghz-Days = 21278160 TFLOP / 123137 = 172.8 TFLOP We can test "last 7 days" and "last 30 days" to see if we get the same results: 994150 Ghz-Days = 284.043 TFLOP/s * 86400s/day * 7 days => 1 Ghz-Days = 172.8 TFLOP 4395208 Ghz-Days = 293.014 TFLOP/s * 86400s/day * 30 days => 1 Ghz-Days = 172.8 TFLOP

2017-12-01, 19:03	#8
VictordeHolland "Victor de Hollander" Aug 2011 the Netherlands 2230₈ Posts	A Core2Duo core has a SSE registers of 128bits for SIMD . It can 'split' the SIMDregister to do 2x FP (DoublePrecision, FP64) adds and execute them at the same time in parallel. In addition it can also do 2x FP64 multiplications (mul) at the same time. So in theory a C2D core could do 4 DP FLOPs (2adds+2mul) per clockcycle. However, in practice getting half that is already extremely difficult, even though Prime95 is very efficient. You hit latencies and memory bandwidth issues pretty fast. AVX introduced bigger 256bit SIMD registers (Sandy, Ivy Bridge). So theoretically can do 8 DP FLOPs per clockcyle. AVX2 extends the instruction set introduced with AVX and introduces FMA3 (Haswell, Broadwell, Skylake-S, Kabylake-S). Now it increases to 16 DP FLOPs per cycle if using FMA. AVX512 extends the SIMD register once again, to 512 bits (and a multitude of subsets, which differ between the Knights Landing co-processor and the normal CPUs). 32 DP FLOPs per cycle.

2017-12-02, 18:38	#11
vasyannyasha "Vasiliy" Apr 2017 Ukraine 2²·3·5 Posts	From post #9 112349 seconds to test exponent M35000011 4 cores 43.75 Ghz-day=12 PFLOP So 1 Ghz-day=0.27 PFLOP=274 TFLOP I made measurements on my AMD 760208 seconds to test exponent M35000011 1 core 43.75 Ghz-day=16.7 PFLOP So 1 GHz-day=0.38 PFLOP=372 TFLOP Maybe my calculations wrong