20190329, 12:36  #1 
Random Account
Aug 2009
3625_{8} Posts 
Gigahertz Day
Gigahertz day is a concept that still seems a bit muddy to me after all these years working on this project.
One gigahertz day is a CPU running at 1 GHz for 24 hours. So, if a person has a 3.8 GHz processor, for example, this would be 3.8 GHz/Days in 24 hours. I know it is probably much more complex than this when applying it to assignments. Is what I have the basics of this? 
20190329, 12:52  #2 
"Composite as Heck"
Oct 2017
1100011001_{2} Posts 
It's 1GHz for 24 hours of a specific CPU that's ancient by today's standards, an early core 2 duo or something near there. As a yard stick for throughput it's not the best especially when you start trying to compare work of different types, but it's only meant as a rough guide.

20190329, 13:04  #3 
"Brian"
Jul 2007
The Netherlands
110011000101_{2} Posts 
@storm5510 I think what you write sums it up exactly. At least that's my understanding too. I compare it in my mind with the "kilowatt hour" as a measure of consumption of electricity.
@M344587487 Indeed. But I was running a machine on GIMPS with a processor speed of about 1 GHz right up until it packed up in November 2017. By the end it was taking about 56 months (running about a third of the time  I didn't leave it on 24 hours a day) to do a double check in the 4045 million exponent range which the project had reached by then. My machine was occasionally the subject of a certain amount of hilarity here. 
20190329, 13:36  #4 
6809 > 6502
"""""""""""""""""""
Aug 2003
101×103 Posts
2·3·37·43 Posts 
AMD CPU's never lived up to the 1GHz days per day.
The oldsters will remember the P90 years as a measurement of work done. 
20190329, 15:00  #5 
Einyen
Dec 2003
Denmark
2·7·223 Posts 
You can calculate what 1 Ghzday 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 GHzdays. 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 Ghzdays is 172.8*10^{12} Floating Point operations. But all the Ghzdays calculations are approximate, I think it is hard to determine the exact number of FLOP that LL, PRP, P1, TF and ECM tests requires. Last fiddled with by ATH on 20190329 at 15:01 
20190329, 18:53  #6  
"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest
2×2,543 Posts 
Quote:
The GhzD/day unit of throughput rate is particularly ugly. My engineering professors would never have tolerated the uncanceled units. Ghzequivalent might have flown; Gheq. 

20190329, 18:56  #7 
"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest
2·2,543 Posts 
Indeed. The number of cycles required is not determinate. It varies depending on the results of roundoff checks, Jacobi checks, Gerbicz checks, etc. for the same exponent and computation type. (continuing from the last save file...)

20190329, 22:43  #8 
Random Account
Aug 2009
3625_{8} Posts 
I forgot about the cores.
I remember the big deal over Y2K. I had a 1 GHz. P3 at the time in the place I worked. I thought I was in hogheaven. It seems like the FLOP value would be determined by the they type of work. Anything pure integer would be zero. Thank you all for your replies. Perhaps somebody will come along in the future and find this useful to them. 
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