parfor using

Hi everybody, i have a question about parfor using. I tried a classical example for parfor on my PC but i couldn't get gainly results. How ?

nmax=1000;
y = zeros(nmax,1);
tic
parfor n = 1:nmax
    z = randn(n);
    y(n) = max(svd(z));
end
toc

Results:

(nmax, pool size, time)

(500, 0, 12.51) (500, 2, 8.71) (500, 4, 5.16) That's OK.

(750, 0, 46.03) (750, 2, 38.65) (750, 4, 30.32) Maybe OK.

(1000, 0, 141.40) (1000, 2, 156.19) (1000, 4, 174.81) Why ?

Thanks for replies.

Answer 1

[Laurens Bakker]

Hi Yildirim,

There is considerable overhead in gathering the data back into a single variable. I removed the assignments and that improves the situation.

nmax=1000;
y = zeros(nmax,1);
tic
parfor n = 1:nmax
    max( svd( randn(n) ));
end
toc

Cheers,

Laurens

Answer 2

[N. Yildirim]

Hi Laurens,

thanks a lot for your reply. It's also very logical to me. But I havent still time saving. Nothing has changed :(

Cheers, Yildirim