Parfor performance are too good to be true
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Code:
function X
z=ones(1000);
tic
parfor j = 1:200
for k = 1:200
z = z+1;
end
end;
toc
zz=ones(1000);
tic
for j = 1:200
for k = 1:200
zz = zz+1;
end
end;
toc
isequal(z,zz)
Output:
Elapsed time is 0.091983 seconds.
Elapsed time is 12.810199 seconds.
ans =
1
I have two questions:
1. I have only 4 cores, how does parfor run more than 100 times faster?
2. Is the code legal? z = z+1 from 4 cores simultaneously can cause unwanted result (two +1 operations in the same time can result only one increment).
Thanks
0 Comments
Accepted Answer
Thorsten
on 4 Nov 2015
Edited: Thorsten
on 4 Nov 2015
1. I guess the code is optimized, such that, e.g., the for loop is replaced by z = z + 200;
2. That's valid, because the computation of z can be done in any order. It's a simplified version of the last example in http://de.mathworks.com/help/distcomp/parfor.html
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