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You can solve a least-squares problem of the form
$$\underset{x}{\mathrm{min}}\frac{1}{2}{\Vert C\cdot x-d\Vert}_{2}^{2}$$
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:
$$C=\left[\begin{array}{ccccc}1& -1/2& 1/3& \mathrm{...}& -1/n\\ -1/n& 1& -1/2& \mathrm{...}& 1/(n-1)\\ 1/(n-1)& -1/n& 1& \mathrm{...}& -1/(n-2)\\ \vdots & \vdots & \vdots & \ddots & \vdots \\ -1/2& 1/3& -1/4& \mathrm{...}& 1\\ 1& -1/2& 1/3& \mathrm{...}& -1/n\\ -1/n& 1& -1/2& \mathrm{...}& 1/(n-1)\\ 1/(n-1)& -1/n& 1& \mathrm{...}& -1/(n-2)\\ \vdots & \vdots & \vdots & \ddots & \vdots \\ -1/2& 1/3& -1/4& \mathrm{...}& 1\end{array}\right].$$
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
$$T=\left[\begin{array}{ccccc}0& 1& 0& \mathrm{...}& 0\\ 0& 0& 1& \mathrm{...}& 0\\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0& 0& 0& \mathrm{...}& 1\\ 1& 0& 0& \mathrm{...}& 0\end{array}\right].$$
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: 'trust-region-reflective' firstorderopt: 7.5143e-05 cgiterations: 36 message: 'Optimization terminated: relative function value changing by less...'