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Chapter Contents
Chapter Introduction
NAG Toolbox

# NAG Toolbox: nag_lapack_dgglse (f08za)

## Purpose

nag_lapack_dgglse (f08za) solves a real linear equality-constrained least squares problem.

## Syntax

[a, b, c, d, x, info] = f08za(a, b, c, d, 'm', m, 'n', n, 'p', p)
[a, b, c, d, x, info] = nag_lapack_dgglse(a, b, c, d, 'm', m, 'n', n, 'p', p)

## Description

nag_lapack_dgglse (f08za) solves the real linear equality-constrained least squares (LSE) problem
 $minimize x c-Ax2 subject to Bx=d$
where $A$ is an $m$ by $n$ matrix, $B$ is a $p$ by $n$ matrix, $c$ is an $m$ element vector and $d$ is a $p$ element vector. It is assumed that $p\le n\le m+p$, $\mathrm{rank}\left(B\right)=p$ and $\mathrm{rank}\left(E\right)=n$, where $E=\left(\begin{array}{c}A\\ B\end{array}\right)$. These conditions ensure that the LSE problem has a unique solution, which is obtained using a generalized $RQ$ factorization of the matrices $B$ and $A$.

## References

Anderson E, Bai Z, Bischof C, Blackford S, Demmel J, Dongarra J J, Du Croz J J, Greenbaum A, Hammarling S, McKenney A and Sorensen D (1999) LAPACK Users' Guide (3rd Edition) SIAM, Philadelphia
Anderson E, Bai Z and Dongarra J (1992) Generalized QR factorization and its applications Linear Algebra Appl. (Volume 162–164) 243–271
Eldèn L (1980) Perturbation theory for the least squares problem with linear equality constraints SIAM J. Numer. Anal. 17 338–350

## Parameters

### Compulsory Input Parameters

1:     $\mathrm{a}\left(\mathit{lda},:\right)$ – double array
The first dimension of the array a must be at least $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{m}}\right)$.
The second dimension of the array a must be at least $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$.
The $m$ by $n$ matrix $A$.
2:     $\mathrm{b}\left(\mathit{ldb},:\right)$ – double array
The first dimension of the array b must be at least $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{p}}\right)$.
The second dimension of the array b must be at least $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$.
The $p$ by $n$ matrix $B$.
3:     $\mathrm{c}\left({\mathbf{m}}\right)$ – double array
The right-hand side vector $c$ for the least squares part of the LSE problem.
4:     $\mathrm{d}\left({\mathbf{p}}\right)$ – double array
The right-hand side vector $d$ for the equality constraints.

### Optional Input Parameters

1:     $\mathrm{m}$int64int32nag_int scalar
Default: the dimension of the array c and the first dimension of the array a. (An error is raised if these dimensions are not equal.)
$m$, the number of rows of the matrix $A$.
Constraint: ${\mathbf{m}}\ge 0$.
2:     $\mathrm{n}$int64int32nag_int scalar
Default: the second dimension of the arrays a, b.
$n$, the number of columns of the matrices $A$ and $B$.
Constraint: ${\mathbf{n}}\ge 0$.
3:     $\mathrm{p}$int64int32nag_int scalar
Default: the dimension of the array d and the first dimension of the array b. (An error is raised if these dimensions are not equal.)
$p$, the number of rows of the matrix $B$.
Constraint: $0\le {\mathbf{p}}\le {\mathbf{n}}\le {\mathbf{m}}+{\mathbf{p}}$.

### Output Parameters

1:     $\mathrm{a}\left(\mathit{lda},:\right)$ – double array
The first dimension of the array a will be $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{m}}\right)$.
The second dimension of the array a will be $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$.
2:     $\mathrm{b}\left(\mathit{ldb},:\right)$ – double array
The first dimension of the array b will be $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{p}}\right)$.
The second dimension of the array b will be $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$.
3:     $\mathrm{c}\left({\mathbf{m}}\right)$ – double array
The residual sum of squares for the solution vector $x$ is given by the sum of squares of elements ${\mathbf{c}}\left({\mathbf{n}}-{\mathbf{p}}+1\right),{\mathbf{c}}\left({\mathbf{n}}-{\mathbf{p}}+2\right),\dots ,{\mathbf{c}}\left({\mathbf{m}}\right)$; the remaining elements are overwritten.
4:     $\mathrm{d}\left({\mathbf{p}}\right)$ – double array
5:     $\mathrm{x}\left({\mathbf{n}}\right)$ – double array
The solution vector $x$ of the LSE problem.
6:     $\mathrm{info}$int64int32nag_int scalar
${\mathbf{info}}=0$ unless the function detects an error (see Error Indicators and Warnings).

## Error Indicators and Warnings

${\mathbf{info}}=-i$
If ${\mathbf{info}}=-i$, parameter $i$ had an illegal value on entry. The parameters are numbered as follows:
1: m, 2: n, 3: p, 4: a, 5: lda, 6: b, 7: ldb, 8: c, 9: d, 10: x, 11: work, 12: lwork, 13: info.
It is possible that info refers to a parameter that is omitted from the MATLAB interface. This usually indicates that an error in one of the other input parameters has caused an incorrect value to be inferred.
${\mathbf{info}}=1$
The upper triangular factor $R$ associated with $B$ in the generalized $RQ$ factorization of the pair $\left(B,A\right)$ is singular, so that $\mathrm{rank}\left(B\right); the least squares solution could not be computed.
${\mathbf{info}}=2$
The $\left(N-P\right)$ by $\left(N-P\right)$ part of the upper trapezoidal factor $T$ associated with $A$ in the generalized $RQ$ factorization of the pair $\left(B,A\right)$ is singular, so that the rank of the matrix ($E$) comprising the rows of $A$ and $B$ is less than $n$; the least squares solutions could not be computed.

## Accuracy

For an error analysis, see Anderson et al. (1992) and Eldèn (1980). See also Section 4.6 of Anderson et al. (1999).

When $m\ge n=p$, the total number of floating-point operations is approximately $\frac{2}{3}{n}^{2}\left(6m+n\right)$; if $p\ll n$, the number reduces to approximately $\frac{2}{3}{n}^{2}\left(3m-n\right)$.
nag_opt_lsq_lincon_solve (e04nc) may also be used to solve LSE problems. It differs from nag_lapack_dgglse (f08za) in that it uses an iterative (rather than direct) method, and that it allows general upper and lower bounds to be specified for the variables $x$ and the linear constraints $Bx$.

## Example

This example solves the least squares problem
 $minimize x c-Ax2 subject to Bx=d$
where
 $c = -1.50 -2.14 1.23 -0.54 -1.68 0.82 ,$
 $A = -0.57 -1.28 -0.39 0.25 -1.93 1.08 -0.31 -2.14 2.30 0.24 0.40 -0.35 -1.93 0.64 -0.66 0.08 0.15 0.30 0.15 -2.13 -0.02 1.03 -1.43 0.50 ,$
 $B = 1.0 0 -1.0 0 0 1.0 0 -1.0$
and
 $d = 0 0 .$
The constraints $Bx=d$ correspond to ${x}_{1}={x}_{3}$ and ${x}_{2}={x}_{4}$.
```function f08za_example

fprintf('f08za example results\n\n');

% Minimize ||c - Ax|| given Bx=d
a = [-0.57, -1.28, -0.39,  0.25;
-1.93,  1.08, -0.31, -2.14;
2.30,  0.24,  0.40, -0.35;
-1.93,  0.64, -0.66,  0.08;
0.15,  0.30,  0.15, -2.13;
-0.02,  1.03, -1.43,  0.50];
c = [-1.50; -2.14;  1.23; -0.54;  -1.68;  0.82];
b = [ 1,     0,    -1,     0;
0,     1,     0,    -1];
d = [ 0;
0];

%
[~, ~, resid, ~, x, info] = f08za( ...
a, b, c, d);

sqres = norm(resid(3:6),2);
disp('Constrained least-squares solution');
disp(x);
disp('Square root of the residual sum of squares');
disp(sqres);

```
```f08za example results

Constrained least-squares solution
0.4890
0.9975
0.4890
0.9975

Square root of the residual sum of squares
0.0251

```