execution time with or without parfor

Question

amir 2014-2-3

1
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/114704-execution-time-with-or-without-parfor

评论： Matt J 2014-2-7

I have a simple code for testing parfor in my local profile (with 4 cores)

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%code 1
matlabpool open 4 % 2 or 1
tic;
parfor i = 1:30
   res = 0;
   for n = 1 : 3000000
      res = res + sin(n) + cos(n);
   end
   A(i) = res;
end
toc;
matlabpool close
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%code 2
tic;
for i = 1:30
  res = 0;
  for n = 1 : 3000000
    res = res + sin(n) + cos(n);
  end
  A(i) = res;
end
toc;
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

I have executed code 1 using 4 labs or 2 labs or 1 lab and executed code 2. the results is here:

code-1 - 8 labs(4 core with 4 hypthread) --> 15 sec
code-1 - 4 labs                          --> 22 sec
code-1 - 2 labs                          --> 35 sec
code-1 - 1 labs                          --> 65 sec
code-2 -                                 --> 18 sec

regards the results, it is better to use code-2 and releasing all other cores (you may also consider the time needed to run 'matlabpool open' and 'matlabpool close'). I have read this : http://www.mathworks.co.uk/matlabcentral/answers/44734-there-is-aproblem-in-parfor

but it seems in this case execution time is much longer than setup time of parallel mechanism.

if there is not any thing wrong with my results, main question is when its better to use parfor.

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amir 2014-2-4

在 MATLAB Online 中打开

unfortunately I can not run it on anther system. I have changed the code to generate all times. you can just paste and run it then check your results with my results.

-----------------------------------------------------

clear all;
clc;
max_counter = 3000000;
fprintf(1,'max count %d\n', max_counter );
times = zeros(5,1);
for poolsize = 0:4
  if poolsize > 0
    matlabpool('open',poolsize);
  end
  real_poolsize = matlabpool('size');
  fprintf(1,'pool size %d\nstarted ...\n', real_poolsize );
  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
  tic;
  if poolsize == 0
    for i = 1:30
      res = 0;
      for n = 1 : max_counter
        res = res + sin(n) + cos(n);
      end
      A(i) = res;
      fprintf(1,' %d ',i);
    end
  else
    parfor i = 1:30
      res = 0;
      for n = 1 : max_counter
        res = res + sin(n) + cos(n);
      end
      A(i) = res;
      fprintf(1,' %d ',i);
    end
  end
  t1 = toc;
  fprintf(1,'\npool size %d , time %f\n',poolsize,t1);
  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
  if poolsize > 0
    matlabpool close
  end
  times(poolsize + 1) = t1;
end
disp(times);

-----------------------------------------------------

and here is my results:

max count 3000000
pool size 0
started ...
 1  2  3  4  5  6  7  8  9  10  11  12  13  14  15  16  17  18  19  20  21  22  23  24  25  26  27  28  29  30 
pool size 0 , time 18.536493
Starting matlabpool using the 'local' profile ... connected to 1 workers.
pool size 1
started ...
 20  19  18  17  16  15  14  13  12  11  10  9  8  7  6  5  4  3  2  1  30  29  28  27  26  25  24  23  22  21 
pool size 1 , time 68.053388
Sending a stop signal to all the workers ... stopped.
Starting matlabpool using the 'local' profile ... connected to 2 workers.
pool size 2
started ...
 20  19  18  17  16  15  14  13  12  11  25  24  23  22  21 
 10  9  8  7  6  5  4  3  2  1  28  27  26  30  29 
pool size 2 , time 40.965749
Sending a stop signal to all the workers ... stopped.
Starting matlabpool using the 'local' profile ... connected to 3 workers.
pool size 3
started ...
 14  13  12  11  10  9  8  26  25  30 
 20  19  18  17  16  15  24  23  22  21 
 7  6  5  4  3  2  1  28  27  29 
pool size 3 , time 28.554977
Sending a stop signal to all the workers ... stopped.
Starting matlabpool using the 'local' profile ... connected to 4 workers.
pool size 4
started ...
 10  9  8  7  6  27  28 
 5  4  3  2  1  24  23 
 15  14  13  12  11  26  25  30 
 19  18  17  16  22  21  20  29 
pool size 4 , time 23.066573
Sending a stop signal to all the workers ... stopped.
   18.5365
   68.0534
   40.9657
   28.5550
   23.0666

as you can see the best answer is without using parfor.

Matt J 2014-2-6

You're not doing any of this over a network are you? This is all on a local CPU?

amir 2014-2-6

编辑：amir 2014-2-6

在 MATLAB Online 中打开

I think I have found the reason:

If I run the code as a function, I will get your results and If I run it as a script (without deceleration of function and name) I get bad results. and my new results (poolnum = 4) :

Please tell me how did you run this code (as a function in a function file or as a script) ?

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Answer 1

Matt J 2014-2-7

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/114704-execution-time-with-or-without-parfor#answer_123549

编辑：Matt J 2014-2-7

在 MATLAB Online 中打开

If I run the code as a function, I will get your results and If I run it as a script (without deceleration of function and name) I get bad results. and my new results (poolnum = 4) :

I think you're on to something! I was indeed running inside a function. When I repeated the test, but running it as a script instead, I get bad behavior similar to what you were reporting.

As can be seen here, PARFOR eventually does outperform a plain for-loop, but it takes a very large worker pool, and with very marginal benefits.

I'm just wondering now whether this is known/documented behavior, or a bug...

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amir 2014-2-7

maybe if I had 12 pools, using 'parfor' was better than for. I think using 'parfor' in script makes some overhead. you can see it in your results too (20.7096 -->'for' 67.3600-->'parfor').

I think its not a known behavior because at least it must shows some warning or something.

Thanks for your help.