Speed up sparse matrix calculations

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John Doe
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!
  4 Comments
Jan
Jan on 13 May 2013
Edited: Jan on 13 May 2013
@Robert P: It would be much more convenient and useful to post the Matlab code instead of a mathematical notation. 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 mathematical notation is even not clear:
start loop # 2 (10 iterations)
dVa = - (U \ (L \ (P * P)));
dVaComp = (U \ (L \ (P * M * c * M' * dVa)));
end loop # 2
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
John Doe on 13 May 2013
Edited: John Doe on 13 May 2013
Thanks for the hints Jan! I have updated the question, please let me know if I should do further changes, or provide additional information.

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