## How can I multiply two big matrices, avoiding out of memory?

### Xin Liu (view profile)

on 25 Aug 2019
Latest activity Commented on by Bruno Luong

on 26 Aug 2019

### Bruno Luong (view profile)

for example, I = A*B, in which size(A) = [1024^2, 3], size(B) = [3, 1001^2]. So, size(I) = [1024^2, 1001^2] which could cause out of memory.
I have tried using tall arrays like the following code:
AA = tall(A);
II = AA*B;
I = gather(II);
but the command line still shows error: out of memory.
Sincerely thanks!
%% update %%
Thanks for your answer very much! Originally, I want to calculate this equation: where so I write the following matrix equation: m and n are variable, for example,m = 1024, n=1001^2 whatever. It is worth noting that the two matrices in exp(...) are not too big, but the result of their multiplication is too large to be storaged in memory. I've tried to split the two matrices into several small patches, but it need to input the number of the patches manually rather than automatically, moreover, the for loop could not be avoided in this case.
I have no good ideal currently.
Best regards!

Bruno Luong

### Bruno Luong (view profile)

on 25 Aug 2019
Storage of your matrix requires 8 Terabytes. Do you have a HD that big?
Xin Liu

### Xin Liu (view profile)

on 26 Aug 2019
I don't have so big HD - -、, so I'm searching for a better method to avoid big matrix :)

### Bruno Luong (view profile)

on 26 Aug 2019
Edited by Bruno Luong

### Bruno Luong (view profile)

on 26 Aug 2019

You can process by chunk of smaller (100 here);
% Generate small test data
m = 100;
n = 100^2;
C = rand(m^2,3); % your [alpha,beta,gamma]
XYZ = rand(3,n); % your [x; y; z];
A = rand(1,m^2);
psize = 100; % chunk size
E = zeros(1,n);
count = 0;
while count < n
p = min(psize,n-count);
j = count+(1:p);
E(j) = A*exp(C*XYZ(:,j));
count = count + p;
end

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Bruno Luong

### Bruno Luong (view profile)

on 26 Aug 2019
There is nothing wrong with using loop. This is the most often miss-prejudice that average MATLAB users might have about using for-loop. The for-loop by itself is OK, what you do and how you do inside the for-loop matters.
In in the case like that, it avoid you to buy a HD of 10 Tb to do your calculation.
To prove the point run this by yourself and see how much you lost.
% Generate small test data
m = 100;
n = 100^2;
C = rand(m^2,3); % your [alpha,beta,gamma]
XYZ = rand(3,n); % your [x; y; z];
A = rand(1,m^2);
psize = 100; % chunk size
tic
E = zeros(1,n);
count = 0;
while count < n
p = min(psize,n-count);
j = count+(1:p);
E(j) = A*exp(C*XYZ(:,j));
count = count + p;
end
toc % Elapsed time is 0.546385 seconds.
tic
E = A*exp(C*XYZ);
toc % Elapsed time is 0.545368 seconds.
Xin Liu

### Xin Liu (view profile)

on 26 Aug 2019
Well, I think I should accept your answer, :), haha~thank you very much.
Bruno Luong

### Bruno Luong (view profile)

on 26 Aug 2019
Well if you have considered this option, then it is your answer too.