parfor using

2012 3 3
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回答已采纳
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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.

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 采纳的回答

Laurens Bakker
Laurens Bakker 2012-3-7

0 个投票

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

更多回答（1 个）

N. Yildirim
N. Yildirim 2012-3-17

0 个投票

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

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