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Groups > comp.programming.threads > #1884

Re: Here is again my Proof

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.

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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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Thread

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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