e04hy:: Minimizing or Maximizing a Function (NAG Toolbox)

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:

Minimize F (x) = \sum_{i = 1}^{m} {[f_{i} (x)]}^{2}

where

x = {(x_{1}, x_{2}, \dots, x_{n})}^{T}

and

m \geq n

. (The functions

f_{i} (x)

are often referred to as ‘residuals’.)

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)

References

Parameters

Compulsory Input Parameters

Optional Input Parameters

Output Parameters

Error Indicators and Warnings

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

If you are not satisfied with the result (e.g., because ifail lies between

3

and

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

decimals, one would expect to get about

t / 2 - 1

decimals accuracy in the components of

x

and between

t - 1

(if

F (x)

is of order

1

at the minimum) and

2 t - 2

(if

F (x)

is close to zero at the minimum) decimals accuracy in

F (x)

Further Comments

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

m ≫ n

is approximately

n \times m^{2} + O (n^{3})

. 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)

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

This example finds least squares estimates of

x_{1}

x_{2}

and

x_{3}

in the model

y = x_{1} + \frac{t_{1}}{x_{2} t_{2} + x_{3} t_{3}}

using the

15

sets of data given in the following table.

\begin{array}{r} y & t_{1} & t_{2} & t_{3} \\ 0.14 & 1.0 & 15.0 & 1.0 \\ 0.18 & 2.0 & 14.0 & 2.0 \\ 0.22 & 3.0 & 13.0 & 3.0 \\ 0.25 & 4.0 & 12.0 & 4.0 \\ 0.29 & 5.0 & 11.0 & 5.0 \\ 0.32 & 6.0 & 10.0 & 6.0 \\ 0.35 & 7.0 & 9.0 & 7.0 \\ 0.39 & 8.0 & 8.0 & 8.0 \\ 0.37 & 9.0 & 7.0 & 7.0 \\ 0.58 & 10.0 & 6.0 & 6.0 \\ 0.73 & 11.0 & 5.0 & 5.0 \\ 0.96 & 12.0 & 4.0 & 4.0 \\ 1.34 & 13.0 & 3.0 & 3.0 \\ 2.10 & 14.0 & 2.0 & 2.0 \\ 4.39 & 15.0 & 1.0 & 1.0 \end{array}

The program uses

(0.5, 1.0, 1.5)

as the initial guess at the position of the minimum.

function e04hy_example


fprintf('e04hy example results\n\n');

global y t;

% Model fitting data.
m = int64(15);
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];

% Initial guess
n = 3;
x = [0.5;  1;  1.5];

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

fprintf('Best fit model parameters are:\n');
for i = 1:n
  fprintf('        x_%d = %10.3f\n',i,x(i));
end
fprintf('\nSum of squares of residuals = %7.4f\n',fsumsq);



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

  global y t;

  fvecc = zeros(m, 1);
  fjacc = zeros(ljc, n);

  for i = 1:m
   denom      = xc(2)*t(i,2) + xc(3)*t(i,3);
   fvecc(i)   = xc(1) + t(i,1)/denom - y(i);
   fjacc(i,1) = 1;
   dummy      = -1/(denom*denom);
   fjacc(i,2) = t(i,1)*t(i,2)*dummy;
   fjacc(i,3) = t(i,1)*t(i,3)*dummy;
  end

function [b, user] = lshes2(m, n, fvecc, xc, lb, user)

  global y t;

  b = zeros(lb, 1);

  sum22 = 0;
  sum32 = 0;
  sum33 = 0;
  for i = 1:m
    dummy = 2*t(i,1)/(xc(2)*t(i,2)+xc(3)*t(i,3))^3;
    sum22 = sum22 + fvecc(i)*dummy*t(i,2)^2;
    sum32 = sum32 + fvecc(i)*dummy*t(i,2)*t(i,3);
    sum33 = sum33 + fvecc(i)*dummy*t(i,3)^2;
  end
  b(3) = sum22;
  b(5) = sum32;
  b(6) = sum33;

On entry,	$n < 1$ ,
or	$m < n$ ,
or	$lw < 8 \times n + 2 \times n \times n + 2 \times m \times n + 3 \times m$ , when $n > 1$ ,
or	$lw < 11 + 5 \times m$ , when $n = 1$ .

NAG Toolbox: nag_opt_lsq_uncon_mod_deriv2_easy (e04hy)

▸▿ Contents

Purpose

Syntax

Description