Parfor and data copy to workers

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Giovanni De Luca 2013-12-11

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https://ww2.mathworks.cn/matlabcentral/answers/109445-parfor-and-data-copy-to-workers

编辑： Matt J 2013-12-11

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Hello,

I read this brief explanation about when MATLAB does a copy of data when working with the PCT in

http://www.mathworks.com/matlabcentral/newsreader/view_thread/321502

but I would need a more clear answer, if possible. Assuming I have 2 available parallel workers, and I want to perform a time-expensive function fnc on a large sparse matrix A with dimension [10^5x10^5] and B with dimension [10^5x10^2], having a vector vec with scalar values and dimensione [1x1000], and the function return a positive scalar scal:

function scal=fnc(A,B,elem)
    X=(elem*A)\B;  % most expensive routine
    % do something on X
    scal=...
end

Then,

parfor i=1:size(vec,2)  % 1000 cycles
  scal(1,i)=fnc(A,B,vec(1,i));
end
s_max=max(scal(1,:));

My question is: how many times the data-copy is done, having 2 available workers and the loop index=1000? The point is that, if I split vec in 2 parts (as the number of available workers), i.e.

vec_new=[vec(1,1:500);vec(1,501:1000)]; % [2x500] matrix

and I slightly modify the function fnc, creating a new function fnc_new:

function scal=fnc_new(A,B,vec_elem)
    scal=0;
    for i=1:size(vec_elem,2)
        X=(elem*A)\B;
        % do something on X
        scal_temp=...
    end
    scal=max(scal,scal_temp);
end

and

parfor i=1:size(vec_new,1)  % 2 cycles
  scal(1,i)=fnc_new(A,vec_new(i,:));
end
s_max=max(scal(1,:));

the two approach ( fnc plus first parfor, and fnc_new plus second parfor ) provide the same final results s_max , but I experienced a further speedup on the second one, I wonder if it's a data-copy issue. I hope I was clear. Thank you in advance.

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Matt J 2013-12-11

编辑：Matt J 2013-12-11

在 MATLAB Online 中打开

It looks quite sub-optimal to be doing

X=(elem*A\B)

repeatedly in a loop for the same A and B. You should really be pre-computing X=A\B once. Then inside fnc() do

function scal=fnc(X,elem)
  X=elem*X;  % most expensive routine
  % do something on X
  scal=...
end

Once you do this, the rest of your question might be irrelevant, since you noted that X=(elem*A)\B is your bottleneck anyway. I can't see why the second version would be faster, though, not from the detail that your code provides.

Other than that, you appear to have a typo where you call

scal(1,i)=fnc_new(A,vec_new(i,:));

with only two arguments. Shouldn't it be

scal(1,i)=fnc_new(A,B,vec_new(i,:));

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