You can solve a least-squares problem of the form
such that lb ≤ x ≤ ub,
for problems where C is very large, perhaps too
large to be stored, by using a Jacobian multiply function. For this
technique, use the 'trust-region-reflective'
algorithm.
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
circularly shifted version of v
times Y
. Use
circshift
to circularly shift
v
.
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),fliplr(v)];
temp = [circshift(temp,1,2),circshift(temp,1,2)]; % Now temp = C'(1,:)
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 dolsqJac3
function in the following code sets up the vector
v
and calls the solver lsqlin
:
function [x,resnorm,residual,exitflag,output] = dolsqJac3(n) % r = 1:n-1; % index for making vectors 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 vector v so that it does not need to be % computed in the Jacobian multiply function options = optimoptions('lsqlin','Algorithm','trust-region-reflective',... 'JacobianMultiplyFcn',@(Jinfo,Y,flag)lsqcirculant3(Jinfo,Y,flag,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 lsqcirculant3
is as follows:
function w = lsqcirculant3(Jinfo,Y,flag,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 = circshift(temp,1,2); % shift the circulant end end function a = Jnegative(q) % Calculate C'*q temp = fliplr(v); temp = circshift(temp,1,2); % shift the circulant% 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 = circshift(temp,1,2); % 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] = dolsqJac3(3000);
Local minimum possible. lsqlin stopped because the relative change in function value is less than the function tolerance.
x(1)
ans = 5.0000
x(1500)
ans = -0.5201
x(3000)
ans = -5.0000
output
output = struct with fields: iterations: 16 algorithm: 'trust-region-reflective' firstorderopt: 5.9351e-05 cgiterations: 36 constrviolation: [] linearsolver: [] message: 'Local minimum possible.↵↵lsqlin stopped because the relative change in function value is less than the function tolerance.'