Jacobian Multiply Function with Linear Least Squares

You can solve a least-squares problem of the form


such that A·x ≤ b, Aeq·x = beq, lb ≤ x ≤ ub, for problems where C is very large, perhaps too large to be stored, by using a Jacobian multiply function.

For example, consider the case where C is a 2n-by-n matrix based on a circulant matrix. This means the rows of C are shifts of a row vector v. This example has the row vector v with elements of the form (–1)k+1/k:

v = [1, –1/2, 1/3, –1/4, ... , –1/n],

cyclically shifted:


This least-squares example considers the problem where

d = [n – 1; n – 2; ...; –n],

and the constraints are –5 ≤ x(i) ≤ 5 for i = 1, ..., n.

For large enough n, the dense matrix C does not fit into computer memory. (n = 10,000 is too large on one tested system.)

A Jacobian multiply function has the following syntax:

w = jmfcn(Jinfo,Y,flag)

Jinfo is a matrix the same size as C, used as a preconditioner. If C is too large to fit into memory, Jinfo should be sparse. Y is a vector or matrix sized so that C*Y or C'*Y makes sense. flag tells jmfcn which product to form:

  • flag > 0 ⇒  w = C*Y

  • flag < 0 ⇒  w = C'*Y

  • flag = 0 ⇒  w = C'*C*Y

Since C is such a simply structured matrix, it is easy to write a Jacobian multiply function in terms of the vector v; i.e., without forming C. Each row of C*Y is the product of a shifted version of v times Y. The following matrix performs one step of the shift: v shifts to v*T, where


To compute C*Y, compute v*Y to find the first row, then shift v and compute the second row, and so on.

To compute C'*Y, perform the same computation, but use a shifted version of temp, the vector formed from the first row of C':

temp = [fliplr(v)*T,fliplr(v)*T];

To compute C'*C*Y, simply compute C*Y using shifts of v, and then compute C' times the result using shifts of fliplr(v).

The dolsqJac function in the following code sets up the vector v and matrix T, and calls the solver lsqlin:

function [x,resnorm,residual,exitflag,output] = dolsqJac(n)
r = 1:n-1; % index for making vectors

T = spalloc(n,n,n); % making a sparse circulant matrix
for m = r
T(n,1) = 1;

v(n) = (-1)^(n+1)/n; % allocating the vector v
v(r) =( -1).^(r+1)./r;

% Now C should be a 2n-by-n circulant matrix based on v,
% but that might be too large to fit into memory.

r = 1:2*n;
d(r) = n-r;

Jinfo = [speye(n);speye(n)]; % sparse matrix for preconditioning
% This matrix is a required input for the solver;
% preconditioning is not really being used in this example

% Pass the matrix T and vector v so they don't need to be
% computed in the Jacobian multiply function
options = optimoptions('lsqlin','JacobianMultiplyFcn',...

lb = -5*ones(1,n);
ub = 5*ones(1,n);

[x,resnorm,residual,exitflag,output] = ...

The Jacobian multiply function lsqcirculant is as follows:

function w = lsqcirculant(Jinfo,Y,flag,T,v)
% This function computes the Jacobian multiply functions
% for a 2n-by-n circulant matrix example

if flag > 0
    w = Jpositive(Y);
elseif flag < 0
    w = Jnegative(Y);
    w = Jnegative(Jpositive(Y));

    function a = Jpositive(q)
        % Calculate C*q
        temp = v;

        a = zeros(size(q)); % allocating the matrix a
        a = [a;a]; % the result is twice as tall as the input

        for r = 1:size(a,1)
            a(r,:) = temp*q; % compute the rth row
            temp = temp*T; % shift the circulant

    function a = Jnegative(q)
        % Calculate C'*q
        temp = fliplr(v)*T; % the circulant for C'

        len = size(q,1)/2; % the returned vector is half as long
        % as the input vector
        a = zeros(len,size(q,2)); % allocating the matrix a

        for r = 1:len
            a(r,:) = [temp,temp]*q; % compute the rth row
            temp = temp*T; % shift the circulant

When n = 3000, C is an 18,000,000-element dense matrix. Here are the results of the dolsqJac function for n = 3000 at selected values of x, and the output structure:

[x,resnorm,residual,exitflag,output] = dolsqJac(3000);

Optimization terminated: relative function value changing by
 less than OPTIONS.FunctionTolerance.

ans =

ans =

ans =

output = 
       iterations: 16
        algorithm: 'trust-region-reflective'
    firstorderopt: 7.5143e-05
     cgiterations: 36
          message: 'Optimization terminated: relative function value changing by les…'
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