mersenneforum.org  

Go Back   mersenneforum.org > Great Internet Mersenne Prime Search > PrimeNet

Reply
 
Thread Tools
Old 2017-11-30, 22:34   #1
preda
 
preda's Avatar
 
"Mihai Preda"
Apr 2015

101001100002 Posts
Default 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?
preda is offline   Reply With Quote
Old 2017-12-01, 00:03   #2
science_man_88
 
science_man_88's Avatar
 
"Forget I exist"
Jul 2009
Dumbassville

202618 Posts
Default

Quote:
Originally Posted by preda View Post
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
science_man_88 is offline   Reply With Quote
Old 2017-12-01, 00:08   #3
sdbardwick
 
sdbardwick's Avatar
 
Aug 2002
North San Diego County

2A816 Posts
Default

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

Last fiddled with by sdbardwick on 2017-12-01 at 00:23
sdbardwick is offline   Reply With Quote
Old 2017-12-01, 00:41   #4
science_man_88
 
science_man_88's Avatar
 
"Forget I exist"
Jul 2009
Dumbassville

8,369 Posts
Default

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
science_man_88 is offline   Reply With Quote
Old 2017-12-01, 00:45   #5
ATH
Einyen
 
ATH's Avatar
 
Dec 2003
Denmark

297910 Posts
Default

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
ATH is offline   Reply With Quote
Old 2017-12-01, 01:23   #6
Dubslow
Basketry That Evening!
 
Dubslow's Avatar
 
"Bunslow the Bold"
Jun 2011
40<A<43 -89<O<-88

11100000111012 Posts
Default

Quote:
Originally Posted by ATH View Post
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.

Last fiddled with by Dubslow on 2017-12-01 at 01:25
Dubslow is offline   Reply With Quote
Old 2017-12-01, 01:52   #7
preda
 
preda's Avatar
 
"Mihai Preda"
Apr 2015

132810 Posts
Default

Quote:
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.
preda is offline   Reply With Quote
Old 2017-12-01, 19:03   #8
VictordeHolland
 
VictordeHolland's Avatar
 
"Victor de Hollander"
Aug 2011
the Netherlands

100100101102 Posts
Default

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 is offline   Reply With Quote
Old 2017-12-02, 01:23   #9
VictordeHolland
 
VictordeHolland's Avatar
 
"Victor de Hollander"
Aug 2011
the Netherlands

2×587 Posts
Default GFLOPS of a Core i5 2500k @4GHz

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.
VictordeHolland is offline   Reply With Quote
Old 2017-12-02, 05:34   #10
LaurV
Romulan Interpreter
 
LaurV's Avatar
 
Jun 2011
Thailand

22·7·11·29 Posts
Default

Quote:
Originally Posted by Dubslow View Post
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)
LaurV is offline   Reply With Quote
Old 2017-12-02, 18:38   #11
vasyannyasha
 
vasyannyasha's Avatar
 
"Vasiliy"
Apr 2017
Ukraine

22×3×5 Posts
Default

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
vasyannyasha is offline   Reply With Quote
Reply

Thread Tools


Similar Threads
Thread Thread Starter Forum Replies Last Post
"Nehalem" quad-cores faster than 100 GFLOPS? ixfd64 Hardware 11 2009-03-09 18:17
v5 Server Conversion compusion Software 3 2008-11-14 19:22
Units Conversion Puzzle JHagerson Lounge 19 2005-11-24 05:38
conversion to GF(2) bigbud Math 9 2005-04-16 01:13
Date conversion with Python leifbk Programming 2 2005-01-26 23:00

All times are UTC. The time now is 10:57.

Sat Nov 28 10:57:13 UTC 2020 up 79 days, 8:08, 3 users, load averages: 1.49, 1.16, 1.10

Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2020, Jelsoft Enterprises Ltd.

This forum has received and complied with 0 (zero) government requests for information.

Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation.
A copy of the license is included in the FAQ.