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Asked by John Doe
on 12 May 2013

Is it possible to speed up large sparse matrix calculations by e.g. placing parantheses optimally?

What I'm asking is: Can I speed up the following code by forcing Matlab to do the operations in a specified order (for instance "from right to left" or something similar)?

I have a sparse square symmetric matrix H (4000x4000), that previously has been factorized, and a sparse vector M with length equal to the dimension of H. What I want to do is the following:

k = 0; cList = 1:3500 % List of contingecies to study nc = length(cList); nb = size(bus,1); % bus is a matrix with information about buses in the system resVa = zeros(nb,nc); [L U P] = lu(H); % H is sparse (thus also L, U and P). Dimension = nbxnb

while k < nc k = k + 1; cont = cList(k);

bf = branch(cont,1); % branch is a matrix with information about branches bt = branch(cont,2); % Col 1 = from-bus, col 2 = to-bus

dy = -H(bf,bt); % M = sparse([bf,bt],1,[1,-1],nb,1);

z = -M'*(U \ (L \ (P * M))); c = (1/dy + z)^-1;

% V = Vm*exp(j*Va); - A complex vector of dimension nbx1 % Sbus is a complex vector of dimension nbx1 % Ybus is a complec matrix of dimension nbxnb mis = (V .*conj((Ybus)*V) - Sbus) ./ Vm; P_mis = real(mis);

converged = 0; i = 0;

while (~converged && i < max_it) i = i + 1; %% The lines I hope to optimize: dVa = - (U \ (L \ (P * P_mis))); dVaComp = (U \ (L \ (P * M * c * M' * dVa))); %%

Va = Va + dVa + dVaComp;

V = Vm .* exp(1j * Va); mis = (V .*conj((Ybus)*V) - Sbus) ./ Vm; P_mis = real(mis);

normP = norm(P, inf); if normP < tol converged = 1; break; end end resVa(:,k) = Va * 180 / pi; end

Some additional information regarding the matrices:

All diagonal elements of H are non-zeros (it's still square, sparse and symmetrical).

Ybus and H have the same structure, but Ybus is complex and H is real.

Vm is updated using the imaginary part of `mis`, and some other matrices, but I've left it out for simplicity.

**EDIT:**

I achieved almost 75% reduction in computation time by using the extended syntax for `lu` when factorizing H.

[L U P Q] = lu(H);

Please let me know if any additional savings are possible to achieve.

Thanks!

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## 4 Comments

## Cedric Wannaz

Direct link to this comment:http://www.mathworks.com/matlabcentral/answers/75506#comment_148534

What size is

Htypically?## John Doe

Direct link to this comment:http://www.mathworks.com/matlabcentral/answers/75506#comment_148573

Typically around 4000x4000. Each column contains about 4 non-zero elements.

The code above (except the lu-factorization) is in a loop that runs 4000 times. The two lines in the end (

dVa, anddVaComp) run ten times for each of those 4000 rounds, thus a total of 40000 times (P is updated within that loop). I added some information in the question now.## Jan Simon

Direct link to this comment:http://www.mathworks.com/matlabcentral/answers/75506#comment_148586

@Robert P: It would be much more convenient and useful to post the Matlab code instead of a

mathematicalnotation. Instead of optimizing your code, we have to write it from scratch. On one hand this is more work for us, on the other hand the created code will not necessarily match your one sufficiently. So please post the code, which you want to get optimized.You

mathematicalnotation is even not clear:It is not specified, what is changed in the iterations. Then it is not meaningful to run the loop 10 times at all.

Calculating the inverse explicitly is not useful in general: it is slower and less accurate compared to solving the corresponding linear equation system.

## John Doe

Direct link to this comment:http://www.mathworks.com/matlabcentral/answers/75506#comment_148589

Thanks for the hints Jan! I have updated the question, please let me know if I should do further changes, or provide additional information.