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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 2*n*-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(m,m+1)=1; end 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','JacobMult',... @(Jinfo,Y,flag)lsqcirculant(Jinfo,Y,flag,T,v)); lb = -5*ones(1,n); ub = 5*ones(1,n); [x,resnorm,residual,exitflag,output] = ... lsqlin(Jinfo,d,[],[],[],[],lb,ub,[],options);

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); else w = Jnegative(Jpositive(Y)); end 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 end end 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 end end end

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.TolFun. x(1) ans = 5.0000 x(1500) ans = -0.5201 x(3000) ans = -5.0000 output output = iterations: 16 algorithm: 'large-scale: trust-region reflective Newton' firstorderopt: 7.5143e-05 cgiterations: 36 message: 'Optimization terminated: relative function value changing by less than OPT...'

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