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Groups > comp.programming.threads > #1884
| From | aminer <aminer@toto.net> |
|---|---|
| Newsgroups | comp.programming.threads, comp.programming |
| Subject | Re: Here is again my Proof |
| Date | 2013-10-09 20:56 -0700 |
| Organization | albasani.net |
| Message-ID | <l358hg$3cf$1@news.albasani.net> (permalink) |
| References | <l356ou$lo5$1@news.albasani.net> |
Cross-posted to 2 groups.
each thread is running on a separate core in my example. Hope you have understood my ideas. Thank you, Amine Moulay Ramdane. On 10/9/2013 8:26 PM, aminer wrote: > > Hello, > > I have noticed that Robert Wessel didn't understood my ideas.. > > So i will prove it to you right now , so follow with me carefully please... > > Let say we have 4 threads, and 4 cores, and let say that each > thread is running the same parallel code, and let say we have also a > serial code inside some critical section, and let say that the parallel > part is > 0.1% and the serial part is 0.9%, so let say that the serial part > takes 1 second(it means 0.1%) and the parallel part takes 9 seconds(it > means 0.9%), what i have tried to explain to you is that the Amadhl's > law or equation is not a correct law and it doesn't give a correct > results, > here is why: so if the 4 threads are all looping and looping again > for a number of times running the same parallel code and the same serial > code and let say that they are contending at the same time > for the critical section, this is why i have called an ideal contention > scenario, so if they are contending AT THE SAME TIME for the critical > section(the serial part) they will run 4 serial parts in 4 seconds in > one loop and they will run 4 parallel parts in 9 seconds in one loop, > hence it will take 13 seconds in one loop , but if you run the parallel > part and the serial part serially they will take 40 seconds, so the > scalability will equal 40 seconds divide by 13 seconds = 3.07X, this is > the result that gives us the Amdahl's law, it's 1 /(0.1 + 0.9/4) = > 3.07X, but this is not the end of the story , so follow now carefully > with me, so let divide the parallel part into 9 small parts each equal to > 1 seconds, and the serial part is equal to 1 second, so if the threads > are looping and not contending at the same time for the critical section > and let say that when the threads are looping the first thread will be > on the first small part equal to 1 seconds of the 9 small parts of the 9 > seconds of the parallel part , the second thread will be on the second > small part equal to 1 seconds of the 9 parts of the 9 seconds of the > parallel part, and the third threads will be on the third small part > equal to 1 second of the 9 parts of the 9 seconds of the parallel part , > and the fourth thread will be on the third small part equal to 1 second > of the 9 parts of the 9 seconds of the parallel part , so imagine the 4 > threads > looping again and again and not changing there places like that , so > there will be no serial part at all, cause when each thread will be on > the 1 second of the serial part the other threads will not be on the the > same serial part, so this is a none contention scenario , so since > there is no serial part so the scalability will be perfect and > equal to 4X , so as you have noticed the Amdahl equation gives > the scalability of the ideal contention scenario all the threads > are contending at the same time so the scalability will equal 3.07X, but > if they are not contending at all the scalability will be perfect > and equal to 4X, and if you have less contention this will scale > better than 3.07X, so now i have proved to you that the Amdahl's law > doesn't give you a correct result. > > > > Hope you have understood my arguments and my ideas against the > Amdahl's law. > > > > > Thank you, > Amine Moulay Ramdane. > > > > > > > >
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Here is again my Proof aminer <aminer@toto.net> - 2013-10-09 20:26 -0700
Re: Here is again my Proof aminer <aminer@toto.net> - 2013-10-09 20:40 -0700
Re: Here is again my Proof aminer <aminer@toto.net> - 2013-10-09 20:45 -0700
Re: Here is again my Proof aminer <aminer@toto.net> - 2013-10-09 20:56 -0700
Re: Here is again my Proof Robert Wessel <robertwessel2@yahoo.com> - 2013-10-10 00:03 -0500
Re: Here is again my Proof aminer <aminer@toto.net> - 2013-10-10 11:51 -0700
Re: Here is again my Proof aminer <aminer@toto.net> - 2013-10-10 12:32 -0700
Re: Here is again my Proof blmblm@myrealbox.com <blmblm.myrealbox@gmail.com> - 2013-10-18 18:58 +0000
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