Groups | Search | Server Info | Keyboard shortcuts | Login | Register [http] [https] [nntp] [nntps]

Groups > comp.programming.threads > #1887

Re: Here is again my Proof

Cross-posted to 2 groups.

Show all headers | View raw

Hello,

I think i know what is my mistake: the serial part in
the Amdahl's law is not the critical section, you must not
confuse the two.



Amine Moulay Ramdane.


On 10/9/2013 10:03 PM, Robert Wessel wrote:
> On Wed, 09 Oct 2013 20:26:18 -0700, aminer <aminer@toto.net> 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.
>
>
> You've redefined the serial work as parallel work, leading to an
> absurd result.
>
> If you have a bit of work, with serial (Sn) and parallel (Pn) parts,
> but you run multiple instances of that, so the S1, S2, S3... are
> serial with respect to themselves and their respective parallel parts,
> but parallel with each other (IOW, S1, S2, S3... can run at the same
> time), S1, S2, S3... as a group are not serial.
>
> Assuming for the sake of brevity that each Sn is the same size, the
> actual unit of serial work in the system is *one* of the Sn units, not
> all of them.  Thus if you had 100 iterations (each with .1 S and .9 P
> work), Amdahl's law would imply that the maximum theoretical speedup
> is about *1000*, since all 100 Sn's can run in parallel.
>
> Amdahl's law is just fine.
>
> And please stop misusing the percent sign.  Your writing is convoluted
> enough when half your numbers are not off two orders of magnitude.
>

Back to comp.programming.threads | Previous | Next — Previous in thread | Next in thread | Find similar

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

csiph-web