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Chapter Contents
Chapter Introduction
NAG Toolbox

# NAG Toolbox: nag_opt_lsq_uncon_mod_deriv2_easy (e04hy)

## Purpose

nag_opt_lsq_uncon_mod_deriv2_easy (e04hy) is an easy-to-use modified Gauss–Newton algorithm for finding an unconstrained minimum of a sum of squares of m$m$ nonlinear functions in n$n$ variables (mn)$\left(m\ge n\right)$. First and second derivatives are required.
It is intended for functions which are continuous and which have continuous first and second derivatives (although it will usually work even if the derivatives have occasional discontinuities).

## Syntax

[x, fsumsq, user, ifail] = e04hy(m, lsfun2, lshes2, x, 'n', n, 'user', user)
[x, fsumsq, user, ifail] = nag_opt_lsq_uncon_mod_deriv2_easy(m, lsfun2, lshes2, x, 'n', n, 'user', user)

## Description

nag_opt_lsq_uncon_mod_deriv2_easy (e04hy) is similar to the function LSSDN2 in the NPL Algorithms Library. It is applicable to problems of the form:
 m MinimizeF(x) = ∑ [fi(x)]2 i = 1
$Minimize⁡F(x)=∑i=1m[fi(x)]2$
where x = (x1,x2,,xn)T$x={\left({x}_{1},{x}_{2},\dots ,{x}_{n}\right)}^{\mathrm{T}}$ and mn$m\ge n$. (The functions fi(x)${f}_{i}\left(x\right)$ are often referred to as ‘residuals’.)
You must supply a function to evaluate the residuals and their first derivatives at any point x$x$, and a function to evaluate the elements of the second derivative term of the Hessian matrix of F(x)$F\left(x\right)$.
Before attempting to minimize the sum of squares, the algorithm checks the user-supplied functions for consistency. Then, from a starting point supplied by you, a sequence of points is generated which is intended to converge to a local minimum of the sum of squares. These points are generated using estimates of the curvature of F(x)$F\left(x\right)$.

## References

Gill P E and Murray W (1978) Algorithms for the solution of the nonlinear least squares problem SIAM J. Numer. Anal. 15 977–992

## Parameters

### Compulsory Input Parameters

1:     m – int64int32nag_int scalar
The number m$m$ of residuals, fi(x)${f}_{i}\left(x\right)$, and the number n$n$ of variables, xj${x}_{j}$.
Constraint: 1nm$1\le {\mathbf{n}}\le {\mathbf{m}}$.
2:     lsfun2 – function handle or string containing name of m-file
You must supply this function to calculate the vector of values fi(x)${f}_{i}\left(x\right)$ and the Jacobian matrix of first derivatives (fi)/(xj) $\frac{\partial {f}_{i}}{\partial {x}_{j}}$ at any point x$x$. It should be tested separately before being used in conjunction with nag_opt_lsq_uncon_mod_deriv2_easy (e04hy) (see the E04 Chapter Introduction).
[fvec, fjac, user] = lsfun2(m, n, xc, ldfjac, user)

Input Parameters

1:     m – int64int32nag_int scalar
m$m$, the numbers of residuals.
2:     n – int64int32nag_int scalar
n$n$, the numbers of variables.
3:     xc(n) – double array
The point x$x$ at which the values of the fi${f}_{i}$ and the (fi)/(xj) $\frac{\partial {f}_{i}}{\partial {x}_{j}}$ are required.
4:     ldfjac – int64int32nag_int scalar
The first dimension of the array fjac as declared in the (sub)program from which nag_opt_lsq_uncon_mod_deriv2_easy (e04hy) is called.
5:     user – Any MATLAB object
lsfun2 is called from nag_opt_lsq_uncon_mod_deriv2_easy (e04hy) with the object supplied to nag_opt_lsq_uncon_mod_deriv2_easy (e04hy).

Output Parameters

1:     fvec(m) – double array
fvec(i)${\mathbf{fvec}}\left(i\right)$ must be set to the value of fi${f}_{\mathit{i}}$ at the point x$x$, for i = 1,2,,m$\mathit{i}=1,2,\dots ,m$.
2:     fjac(ldfjac,n) – double array
fjac(i,j)${\mathbf{fjac}}\left(\mathit{i},\mathit{j}\right)$ must be set to the value of (fi)/(xj) $\frac{\partial {f}_{\mathit{i}}}{\partial {x}_{\mathit{j}}}$ at the point x$x$, for i = 1,2,,m$\mathit{i}=1,2,\dots ,m$ and j = 1,2,,n$\mathit{j}=1,2,\dots ,n$.
3:     user – Any MATLAB object
3:     lshes2 – function handle or string containing name of m-file
You must supply this function to calculate the elements of the symmetric matrix
 m B(x) = ∑ fi(x)Gi(x), i = 1
$B(x)=∑i=1mfi(x)Gi(x),$
at any point x$x$, where Gi(x)${G}_{i}\left(x\right)$ is the Hessian matrix of fi(x)${f}_{i}\left(x\right)$. It should be tested separately before being used in conjunction with nag_opt_lsq_uncon_mod_deriv2_easy (e04hy) (see the E04 Chapter Introduction).
[b, user] = lshes2(m, n, fvec, xc, lb, user)

Input Parameters

1:     m – int64int32nag_int scalar
m$m$, the number of residuals.
2:     n – int64int32nag_int scalar
n$n$, the number of residuals.
3:     fvec(m) – double array
The value of the residual fi${f}_{\mathit{i}}$ at the point x$x$, for i = 1,2,,m$\mathit{i}=1,2,\dots ,m$, so that the values of the fi${f}_{\mathit{i}}$ can be used in the calculation of the elements of b.
4:     xc(n) – double array
The point x$x$ at which the elements of b are to be evaluated.
5:     lb – int64int32nag_int scalar
The length of the array b.
6:     user – Any MATLAB object
lshes2 is called from nag_opt_lsq_uncon_mod_deriv2_easy (e04hy) with the object supplied to nag_opt_lsq_uncon_mod_deriv2_easy (e04hy).

Output Parameters

1:     b(lb) – double array
Must contain the lower triangle of the matrix B(x)$B\left(x\right)$, evaluated at the point x$x$, stored by rows. (The upper triangle is not required because the matrix is symmetric.) More precisely, b(j(j1) / 2 + k)${\mathbf{b}}\left(\mathit{j}\left(\mathit{j}-1\right)/2+\mathit{k}\right)$ must contain i = 1mfi(2fi)/(xjxk) $\sum _{\mathit{i}=1}^{m}{f}_{\mathit{i}}\frac{{\partial }^{2}{f}_{\mathit{i}}}{\partial {x}_{\mathit{j}}\partial {x}_{\mathit{k}}}$ evaluated at the point x$x$, for j = 1,2,,n$\mathit{j}=1,2,\dots ,n$ and k = 1,2,,j$\mathit{k}=1,2,\dots ,\mathit{j}$.
2:     user – Any MATLAB object
4:     x(n) – double array
n, the dimension of the array, must satisfy the constraint 1nm$1\le {\mathbf{n}}\le {\mathbf{m}}$.
x(j)${\mathbf{x}}\left(\mathit{j}\right)$ must be set to a guess at the j$\mathit{j}$th component of the position of the minimum, for j = 1,2,,n$\mathit{j}=1,2,\dots ,n$. The function checks lsfun2 and lshes2 at the starting point and so is more likely to detect any error in your functions if the initial x(j)${\mathbf{x}}\left(j\right)$ are nonzero and mutually distinct.

### Optional Input Parameters

1:     n – int64int32nag_int scalar
Default: For n, the dimension of the array x.
The number m$m$ of residuals, fi(x)${f}_{i}\left(x\right)$, and the number n$n$ of variables, xj${x}_{j}$.
Constraint: 1nm$1\le {\mathbf{n}}\le {\mathbf{m}}$.
2:     user – Any MATLAB object
user is not used by nag_opt_lsq_uncon_mod_deriv2_easy (e04hy), but is passed to lsfun2 and lshes2. Note that for large objects it may be more efficient to use a global variable which is accessible from the m-files than to use user.

w lw iuser ruser

### Output Parameters

1:     x(n) – double array
The lowest point found during the calculations. Thus, if ${\mathbf{ifail}}={\mathbf{0}}$ on exit, x(j)${\mathbf{x}}\left(j\right)$ is the j$j$th component of the position of the minimum.
2:     fsumsq – double scalar
The value of the sum of squares, F(x)$F\left(x\right)$, corresponding to the final point stored in x.
3:     user – Any MATLAB object
4:     ifail – int64int32nag_int scalar
${\mathrm{ifail}}={\mathbf{0}}$ unless the function detects an error (see [Error Indicators and Warnings]).

## Error Indicators and Warnings

Note: nag_opt_lsq_uncon_mod_deriv2_easy (e04hy) may return useful information for one or more of the following detected errors or warnings.
Errors or warnings detected by the function:

Cases prefixed with W are classified as warnings and do not generate an error of type NAG:error_n. See nag_issue_warnings.

ifail = 1${\mathbf{ifail}}=1$
 On entry, n < 1${\mathbf{n}}<1$, or m < n${\mathbf{m}}<{\mathbf{n}}$, or lw < 8 × n + 2 × n × n + 2 × m × n + 3 × m$\mathit{lw}<8×{\mathbf{n}}+2×{\mathbf{n}}×{\mathbf{n}}+2×{\mathbf{m}}×{\mathbf{n}}+3×{\mathbf{m}}$, when n > 1${\mathbf{n}}>1$, or lw < 11 + 5 × m$\mathit{lw}<11+5×{\mathbf{m}}$, when n = 1${\mathbf{n}}=1$.
ifail = 2${\mathbf{ifail}}=2$
There have been 50 × n$50×n$ calls of lsfun2, yet the algorithm does not seem to have converged. This may be due to an awkward function or to a poor starting point, so it is worth restarting nag_opt_lsq_uncon_mod_deriv2_easy (e04hy) from the final point held in x.
W ifail = 3${\mathbf{ifail}}=3$
The final point does not satisfy the conditions for acceptance as a minimum, but no lower point could be found.
ifail = 4${\mathbf{ifail}}=4$
An auxiliary function has been unable to complete a singular value decomposition in a reasonable number of sub-iterations.
W ifail = 5${\mathbf{ifail}}=5$
W ifail = 6${\mathbf{ifail}}=6$
W ifail = 7${\mathbf{ifail}}=7$
W ifail = 8${\mathbf{ifail}}=8$
There is some doubt about whether the point x$x$ found by nag_opt_lsq_uncon_mod_deriv2_easy (e04hy) is a minimum of F(x)$F\left(x\right)$. The degree of confidence in the result decreases as ifail increases. Thus, when ${\mathbf{ifail}}={\mathbf{5}}$ it is probable that the final x$x$ gives a good estimate of the position of a minimum, but when ${\mathbf{ifail}}={\mathbf{8}}$ it is very unlikely that the function has found a minimum.
ifail = 9${\mathbf{ifail}}=9$
It is very likely that you have made an error in forming the derivatives (fi)/(xj) $\frac{\partial {f}_{i}}{\partial {x}_{j}}$ in lsfun2.
ifail = 10${\mathbf{ifail}}=10$
It is very likely that you have made an error in forming the quantities Bjk${B}_{jk}$ in lshes2.
If you are not satisfied with the result (e.g., because ifail lies between 3$3$ and 8$8$), it is worth restarting the calculations from a different starting point (not the point at which the failure occurred) in order to avoid the region which caused the failure. Repeated failure may indicate some defect in the formulation of the problem.

## Accuracy

If the problem is reasonably well scaled and a successful exit is made, then, for a computer with a mantissa of t$t$ decimals, one would expect to get about t / 21$t/2-1$ decimals accuracy in the components of x$x$ and between t1$t-1$ (if F(x)$F\left(x\right)$ is of order 1$1$ at the minimum) and 2t2$2t-2$ (if F(x)$F\left(x\right)$ is close to zero at the minimum) decimals accuracy in F(x)$F\left(x\right)$.

The number of iterations required depends on the number of variables, the number of residuals and their behaviour, and the distance of the starting point from the solution. The number of multiplications performed per iteration of nag_opt_lsq_uncon_mod_deriv2_easy (e04hy) varies, but for mn$m\gg n$ is approximately n × m2 + O(n3)$n×{m}^{2}+\mathit{O}\left({n}^{3}\right)$. In addition, each iteration makes at least one call of lsfun2 and some iterations may call lshes2. So, unless the residuals and their derivatives can be evaluated very quickly, the run time will be dominated by the time spent in lsfun2 (and, to a lesser extent, in lshes2).
Ideally, the problem should be scaled so that the minimum value of the sum of squares is in the range (0, + 1)$\left(0,+1\right)$ and so that at points a unit distance away from the solution the sum of squares is approximately a unit value greater than at the minimum. It is unlikely that you will be able to follow these recommendations very closely, but it is worth trying (by guesswork), as sensible scaling will reduce the difficulty of the minimization problem, so that nag_opt_lsq_uncon_mod_deriv2_easy (e04hy) will take less computer time.
When the sum of squares represents the goodness-of-fit of a nonlinear model to observed data, elements of the variance-covariance matrix of the estimated regression coefficients can be computed by a subsequent call to nag_opt_lsq_uncon_covariance (e04yc), using information returned in segments of the workspace array w. See nag_opt_lsq_uncon_covariance (e04yc) for further details.

## Example

```function nag_opt_lsq_uncon_mod_deriv2_easy_example
m = int64(15);
x = [0.5;
1;
1.5];

x = [0.5;
1;
1.5];
y = [0.14,0.18,0.22,0.25,0.29,0.32,0.35,0.39,0.37,0.58,0.73,0.96,1.34,2.10,4.39];
t = [[1.0, 15.0, 1.0],
[2.0, 14.0, 2.0],
[3.0, 13.0, 3.0],
[4.0, 12.0, 4.0],
[5.0, 11.0, 5.0],
[6.0, 10.0, 6.0],
[7.0, 9.0, 7.0],
[8.0, 8.0, 8.0],
[9.0, 7.0, 7.0],
[10.0, 6.0, 6.0],
[11.0, 5.0, 5.0],
[12.0, 4.0, 4.0],
[13.0, 3.0, 3.0],
[14.0, 2.0, 2.0],
[15.0, 1.0, 1.0]];

user = {y, t, 3};

[xOut, fsumsq, user, ifail] = ...
nag_opt_lsq_uncon_mod_deriv2_easy(m, @lsfun2, @lshes2, x, 'user', user)

function [fvecc, fjacc, user] = lsfun2(m, n, xc, ljc, user)
fvecc = zeros(m, 1);
fjacc = zeros(ljc, n);

for i = 1:double(m)
denom = xc(2)*user{2}(i,2) + xc(3)*user{2}(i,3);
fvecc(i) = xc(1) + user{2}(i,1)/denom - user{1}(i);
fjacc(i,1) = 1.0d0;
dummy = -1.0d0/(denom*denom);
fjacc(i,2) = user{2}(i,1)*user{2}(i,2)*dummy;
fjacc(i,3) = user{2}(i,1)*user{2}(i,3)*dummy;
end
function [b, user] = lshes2(m, n, fvecc, xc, lb, user)
b = zeros(lb, 1);

sum22 = 0.0d0;
sum32 = 0.0d0;
sum33 = 0.0d0;
for i = 1:double(m)
dummy = 2.0d0*user{2}(i,1)/(xc(2)*user{2}(i,2)+xc(3)*user{2}(i,3))^3;
sum22 = sum22 + fvecc(i)*dummy*user{2}(i,2)^2;
sum32 = sum32 + fvecc(i)*dummy*user{2}(i,2)*user{2}(i,3);
sum33 = sum33 + fvecc(i)*dummy*user{2}(i,3)^2;
end
b(3) = sum22;
b(5) = sum32;
b(6) = sum33;
```
```

xOut =

0.0824
1.1330
2.3437

fsumsq =

0.0082

user =

[1x15 double]    [15x3 double]    [3]

ifail =

0

```
```function e04hy_example
m = int64(15);
x = [0.5;
1;
1.5];

x = [0.5;
1;
1.5];
y = [0.14,0.18,0.22,0.25,0.29,0.32,0.35,0.39,0.37,0.58,0.73,0.96,1.34,2.10,4.39];
t = [[1.0, 15.0, 1.0],
[2.0, 14.0, 2.0],
[3.0, 13.0, 3.0],
[4.0, 12.0, 4.0],
[5.0, 11.0, 5.0],
[6.0, 10.0, 6.0],
[7.0, 9.0, 7.0],
[8.0, 8.0, 8.0],
[9.0, 7.0, 7.0],
[10.0, 6.0, 6.0],
[11.0, 5.0, 5.0],
[12.0, 4.0, 4.0],
[13.0, 3.0, 3.0],
[14.0, 2.0, 2.0],
[15.0, 1.0, 1.0]];

user = {y, t, 3};

[xOut, fsumsq, user, ifail] = e04hy(m, @lsfun2, @lshes2, x, 'user', user)

function [fvecc, fjacc, user] = lsfun2(m, n, xc, ljc, user)
fvecc = zeros(m, 1);
fjacc = zeros(ljc, n);

for i = 1:double(m)
denom = xc(2)*user{2}(i,2) + xc(3)*user{2}(i,3);
fvecc(i) = xc(1) + user{2}(i,1)/denom - user{1}(i);
fjacc(i,1) = 1.0d0;
dummy = -1.0d0/(denom*denom);
fjacc(i,2) = user{2}(i,1)*user{2}(i,2)*dummy;
fjacc(i,3) = user{2}(i,1)*user{2}(i,3)*dummy;
end
function [b, user] = lshes2(m, n, fvecc, xc, lb, user)
b = zeros(lb, 1);

sum22 = 0.0d0;
sum32 = 0.0d0;
sum33 = 0.0d0;
for i = 1:double(m)
dummy = 2.0d0*user{2}(i,1)/(xc(2)*user{2}(i,2)+xc(3)*user{2}(i,3))^3;
sum22 = sum22 + fvecc(i)*dummy*user{2}(i,2)^2;
sum32 = sum32 + fvecc(i)*dummy*user{2}(i,2)*user{2}(i,3);
sum33 = sum33 + fvecc(i)*dummy*user{2}(i,3)^2;
end
b(3) = sum22;
b(5) = sum32;
b(6) = sum33;
```
```

xOut =

0.0824
1.1330
2.3437

fsumsq =

0.0082

user =

[1x15 double]    [15x3 double]    [3]

ifail =

0

```