Why does MATLAB take a longer time to compute the SVD of matrices whos size is a power of 2?

There is a significant difference in the Singular Value Decomposition computation time when the matrix size is a power of two.

For example, if I compute the SVD of randomly generated square matrices of sizes 511, 512 and 513, note the difference in the computaion times.

A=randn(512);
tic;[u s v]=svd(A);toc

This displays:

 Elapsed time is 24.476422 seconds.

However, the following:

A=randn(511);
tic;[u s v]=svd(A);toc

displays:

 Elapsed time is 5.049256 seconds.

and

A=randn(513);
tic;[u s v]=svd(A);toc

displays:

 Elapsed time is 4.974085 seconds.

I would like to increase the speed and capabilities of SVD computation in MATLAB.

Answer 1

[MathWorks Support Team]

This is expected behavior. This behavior is due to what is called the "cache resonance effect". When the size of a matrix is a power of 2, the cache replacement algorithm is not as effective.

For more information on increasing the speed and capabilities of matrix computation refer to the following URL:

https://www.mathworks.com/company/newsletters/articles/matlab-incorporates-lapack.html

[Philip Borghesani]

Although this effect is expected the times and size of the effect are a bit overstated for modern systems, on my machine (Xeon E5-1650 v3) and R2017b I see:

 A=randn(511);B=randn(512);C=randn(513);
 >> timeit(@()svd(A),3)
 ans =
    0.0470
 >> timeit(@()svd(B),3)
 ans =
    0.0507
 >> timeit(@()svd(C),3)
 ans =
    0.0466