Tricks to increase computational speed for D = (A + A*B).*C

Hello

I am working on computing the time evolution of a distribution. The distribution evolves as follows

g(t + dt) = (g(t) + dt*MAT*g(t)).*exp(-k*dt)

here g and k are column vector of size m*1, MAT is a square matrix of size m*m.

Say i have to evolve it "n" times, the way I have attempted my solution is

g = zeros(m,n)

g(:,1) = g0 % initalising

MAT, k = something pre calculated

for i = 1:1:n

g(:,i+1) = (g(:,i) + dt*MAT*g(:,i) ).*exp(-k*dt)

end

This is the conventional basic way I guess but it takes a long time. How else can I implement the evolution to make it faster.

For perspective the matrices are of the order of 10,000*10,000 and I would like to evolve it for another 5000 ish points

MAT is a mostly a sparse matrix. It has significant non zero diagonal elements and in a given row maybe 50-60 elements are non zero.

Grateful for any suggestions!

Answer 1

[Animesh]

To optimize the computation of the large matrix sizes and the sparsity of MAT, you should consider the following approaches:

1. Use Sparse Matrices: If you haven't already, convert MAT to a sparse matrix format in MATLAB using sparse(MAT). This will significantly speed up matrix-vector multiplications for large, sparse matrices.
2. Precompute Static Values: If exp(-k*dt) does not change over iterations, compute it once before the loop and reuse the result.
3. Avoid Growing Arrays in a Loop: Preallocate the full size of g before the loop to avoid dynamically resizing it during each iteration, which is costly.

Given these points, your revised code snippet would look something like this:

% Assuming MAT is already defined and is sparse

% Convert MAT to sparse format if not already

% MAT = sparse(MAT); % Uncomment if MAT is not already sparse

% Precompute constants

exp_neg_kdt = exp(-k*dt);

% Preallocate g

g = zeros(m, n+1);

g(:,1) = g0; % Initial condition

% Time evolution loop

for i = 1:n

g(:,i+1) = (g(:,i) + dt*MAT*g(:,i)) .* exp_neg_kdt;

end

I hope this helps!

Animesh

[Arun S]

Yes, it does help. thank you!