Run Nested FOR-loop Parallelly for Multivariable Function Optimization
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F_max = 0; % Temp var for max of F
F_curr = 0; % Temp var for current F
for x = -0.02:0.001:0.02
for x_1 = 20:100
for x_2 = 20:100
for x_3 = 20:100
for x_4 = 20:100
for x_5 = -15:15
for x_6 = -15:15
for x_7 = -15:15
for x_8 = -15:15
F_curr = double(F(x,x_1,x_2,x_3,x_4,x_5,x_6,x_7,x_8));
if F_curr>F_max
F_max = F_curr;
end
end
end
end
end
end
end
end
end
end
F is a (symbolic) function of 9 (symbolic) variables x, x_1, x_2, x_3, x_4, x_5, x_6, x_7, x_8. The formula of F is given in the following Link1.
How do I edit this code so that it can run parallelly? The reason I can't apply the usual solution is because the variables aren't looping by integer iterations starting from 0 (e.g. i = 1:n (some integer)).
1 个评论
Sam Marshalik
2021-12-17
Hey Joshua, I do not currently have an opportunity to play around with the code, but you will want to employ parfor to speed this up. The issue is that F_max is a temporary variable on all of the workers and they will not be able to exchange information to determine who has the highest F_max (parfor workers are not able to communicate with one another).
I think something you can try is making F_max a sliced output variable (https://www.mathworks.com/help/parallel-computing/sliced-variable.html#bq_tiga) - this will give you a large array with all of the values from your loops. You can then determine the highest value (max(F_max)) from the entire list.
You can also do the following to deal with the outermost parfor-loop, since the non-integers will be an issue:
x = 0:0.1:1;
parfor idx = 1:length(x)
disp(x(idx))
end
采纳的回答
Matt J
2021-12-18
编辑:Matt J
2021-12-18
I don't think a loop over all 9 variables is going to be practical (10^15 combinations).
An important observation, though, is that your function F() is linear with respect to x5,x6,x7,x8,x9. This means that the maximum will be achieved at one of the extreme values of these variables. Consequently, you don't have to search -15:15. You only have to search the two end points -15 and 15 for each of these variables for a total of 16 combinations. For each of the 16 combinations, you need to do a grid search over X, X_1,X_2,X_3,X_4 but the dimension of that search (EDIT:) can be done with increased vectorization as follows.
Xgrid ={ -0.02:.001:0.02;
[-15,15]};
Xgrid= Xgrid([1,2,2,2,2]);
sizeT=repelem(numel(20:100),1,4);
[X_1,X_2,X_3,X_4]=ndgridVecs(20:100); % ndgridVecs available at https://www.mathworks.com/matlabcentral/fileexchange/74956-ndgridvecs
[X_0, X_5,X_6,X_7,X_8]=ndgrid(Xgrid{:});
Fdouble=@(x1,x2,x3,x4,x5,x6,x7,x8,x9) -(x1.^2+x2.^2+x3.^2+x4.^2+x5.^2+x6.^2+x7.^2+x8.^2+x9.^2); %example function
%Fdouble=matlabFunction(F); %true function
N=numel(X_0);
F_max=nan(1,N);
loc=cell(1,N);
tic
parfor n=1:numel(X_0)
[x_0, x_5,x_6,x_7,x_8] = deal(X_0(n), X_5(n),X_6(n),X_7(n), X_8(n));
T=Fdouble(x_0,X_1,X_2,X_3,X_4, x_5,x_6,x_7,x_8);
[F_max(n),I]=max(T,[],'all','linear');
loc{n}=[x_0,I, x_5,x_6,x_7,x_8]; %location of maximum
end
[F_max,nmax]=max(F_max);
loc=loc{nmax};
[j,k,l,m]=ind2sub(sizeT,loc(2));
loc=[loc(1), X_1(j),X_2(k),X_3(l),X_4(m) ,loc(3:end)];
toc%Elapsed time is 46.687547 seconds.
3 个评论
Matt J
2021-12-18
编辑:Matt J
2021-12-18
The reason for stating that is because, the coefficients of [x5,x6,x7,x8] in the function is [-1,1,1,-1]
The coefficients that I see are complicated functions of x,x1..x4. I don't see how you can anticipate their signs.
I've modified my answer above, however, and tested that it runs with more modest memory consumption.
更多回答(1 个)
Matt J
2021-12-17
编辑:Matt J
2021-12-17
Xgrid ={ -0.02:0.001:0.02;
20:100;
20:100;
20:100;
20:100;
-15:15;
-15:15;
-15:15;
-15:15};
sz=cellfun(@numel,Xgrid);
N=prod(sz);
J=numel(sz);
Fdouble=matlabFunction(F);
F_max=-inf;
parfor n=1:N
sub=cell(J,1);
[sub{1:J}]=ind2sub(sz,n); %convert to subscripts
X=cellfun(@(A,B) A(B), Xgrid,sub,'uni',0); %lookup grid values
F_max=max(F_max, Fdouble(X{:}) ); %reduction
end
3 个评论
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