Here is again my Proof

[aminer]

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.

[aminer]

And of course in my example each thread have the same number of loops
and the same number of work, so that you understand more my example.

Thank you,
Amine Moulay Ramdane.

[aminer]

On 10/9/2013 8:26 PM, aminer wrote:

> 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

i mean the fourth thread will be on the fourth small part equal to 1
second of the 9 parts of the 9 seconds of the parallel part.


Hope you have understood now.

[aminer]

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:
>

[Robert Wessel]

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.

[aminer]

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.

[aminer]

I wrote:
> 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.

But, Amine, how can you say that the serial part of the Amdahl's law is
not
the critical section ?


In my example if you don't have context switches , and you have the same
number of threads than the number of cores, and the time inside the
critical
section is constant and the time inside the parallel part is constant,
so the Amdahl's law can predict the worst case contention scenario ,
that means when all the threads are contending for the same critical
section, hence you can predict the worst case contention scenario by
just calculating the time inside the critical section and the time
inside the parallel part and doing the Amdahl calculation, this will
give you the
exact result for the worst case contention scenario, so as you have
noticed the serial part of the Amdahl equation is not only the critical
section, it's the critical section with a context, so you have to take
into consideration also the context, that means the context of the worst
case contention scenario.

Thank you,
Amine Moulay Ramdane.

[blmblm.m...@gmail.com]

In article <l36kri$3t3$1...@news.albasani.net>, aminer <ami...@toto.net> wrote:
>
> I wrote:
> > 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.
>

Exactly. The serial part is the part that must be done exactly once
and cannot be somehow split up among multiple threads or processes.

>
> But, Amine, how can you say that the serial part of the Amdahl's law is
> not
> the critical section ?

Having got it right once (above), why now go off the rails again
(below) ....

(And why start a new thread for each new thought, rather than keeping
the whole discussion in one thread?)

> In my example if you don't have context switches , and you have the same
> number of threads than the number of cores, and the time inside the
> critical
> section is constant and the time inside the parallel part is constant,
> so the Amdahl's law can predict the worst case contention scenario ,
> that means when all the threads are contending for the same critical
> section, hence you can predict the worst case contention scenario by
> just calculating the time inside the critical section and the time
> inside the parallel part and doing the Amdahl calculation, this will
> give you the
> exact result for the worst case contention scenario, so as you have
> noticed the serial part of the Amdahl equation is not only the critical
> section, it's the critical section with a context, so you have to take
> into consideration also the context, that means the context of the worst
> case contention scenario.
>
>
> Thank you,
> Amine Moulay Ramdane.
>
>
>

[ snip ]

--
B. L. Massingill
ObDisclaimer: I don't speak for my employers; they return the favor.