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NAG Toolbox: nag_opt_lsq_uncon_mod_deriv_easy (e04gz)

Purpose

nag_opt_lsq_uncon_mod_deriv_easy (e04gz) is an easy-to-use modified Gauss–Newton algorithm for finding an unconstrained minimum of a sum of squares of mm nonlinear functions in nn variables (mn)(mn). First 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] = e04gz(m, lsfun2, x, 'n', n, 'user', user)
[x, fsumsq, user, ifail] = nag_opt_lsq_uncon_mod_deriv_easy(m, lsfun2, x, 'n', n, 'user', user)

Description

nag_opt_lsq_uncon_mod_deriv_easy (e04gz) is similar to the function LSFDN2 in the NPL Algorithms Library. It is applicable to problems of the form
m
MinimizeF(x) = [fi(x)]2
i = 1
MinimizeF(x)=i=1m[fi(x)]2
where x = (x1,x2,,xn)Tx=(x1,x2,,xn)T and mnmn. (The functions fi(x)fi(x) are often referred to as ‘residuals’.)
You must supply a function to evaluate the residuals and their first derivatives at any point xx.
Before attempting to minimize the sum of squares, the algorithm checks the function 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(x).

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 mm of residuals, fi(x)fi(x), and the number nn of variables, xjxj.
Constraint: 1nm1nm.
2:     lsfun2 – function handle or string containing name of m-file
You must supply this function to calculate the vector of values fi(x)fi(x) and the Jacobian matrix of first derivatives (fi)/(xj) fi xj  at any point xx. It should be tested separately before being used in conjunction with nag_opt_lsq_uncon_mod_deriv_easy (e04gz).
[fvec, fjac, user] = lsfun2(m, n, xc, ldfjac, user)

Input Parameters

1:     m – int64int32nag_int scalar
mm, the numbers of residuals.
2:     n – int64int32nag_int scalar
nn, the numbers of variables.
3:     xc(n) – double array
The point xx at which the values of the fifi and the (fi)/(xj) fi xj  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_deriv_easy (e04gz) is called.
5:     user – Any MATLAB object
lsfun2 is called from nag_opt_lsq_uncon_mod_deriv_easy (e04gz) with the object supplied to nag_opt_lsq_uncon_mod_deriv_easy (e04gz).

Output Parameters

1:     fvec(m) – double array
fvec(i)fveci must be set to the value of fifi at the point xx, for i = 1,2,,mi=1,2,,m.
2:     fjac(ldfjac,n) – double array
fjac(i,j)fjacij must be set to the value of (fi)/(xj) fi xj at the point xx, for i = 1,2,,mi=1,2,,m and j = 1,2,,nj=1,2,,n.
3:     user – Any MATLAB object
3:     x(n) – double array
n, the dimension of the array, must satisfy the constraint 1nm1nm.
x(j)xj must be set to a guess at the jjth component of the position of the minimum, for j = 1,2,,nj=1,2,,n. The function checks the first derivatives calculated by lsfun2 at the starting point and so is more likely to detect any error in your functions if the initial x(j)xj are nonzero and mutually distinct.

Optional Input Parameters

1:     n – int64int32nag_int scalar
Default: The dimension of the array x.
The number mm of residuals, fi(x)fi(x), and the number nn of variables, xjxj.
Constraint: 1nm1nm.
2:     user – Any MATLAB object
user is not used by nag_opt_lsq_uncon_mod_deriv_easy (e04gz), but is passed to lsfun2. 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.

Input Parameters Omitted from the MATLAB Interface

w lw iuser ruser

Output Parameters

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

Error Indicators and Warnings

Note: nag_opt_lsq_uncon_mod_deriv_easy (e04gz) 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 = 1ifail=1
On entry,n < 1n<1,
orm < nm<n,
orlw < 8 × n + 2 × n × n + 2 × m × n + 3 × mlw<8×n+2×n×n+2×m×n+3×m, when n > 1n>1,
orlw < 11 + 5 × mlw<11+5×m, when n = 1n=1.
  ifail = 2ifail=2
There have been 50 × n50×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_deriv_easy (e04gz) from the final point held in x.
W ifail = 3ifail=3
The final point does not satisfy the conditions for acceptance as a minimum, but no lower point could be found.
  ifail = 4ifail=4
An auxiliary function has been unable to complete a singular value decomposition in a reasonable number of sub-iterations.
W ifail = 5ifail=5
W ifail = 6ifail=6
W ifail = 7ifail=7
W ifail = 8ifail=8
There is some doubt about whether the point xx found by nag_opt_lsq_uncon_mod_deriv_easy (e04gz) is a minimum of F(x)F(x). The degree of confidence in the result decreases as ifail increases. Thus, when ifail = 5ifail=5 it is probable that the final xx gives a good estimate of the position of a minimum, but when ifail = 8ifail=8 it is very unlikely that the function has found a minimum.
  ifail = 9ifail=9
It is very likely that you have made an error in forming the derivatives (fi)/(xj) fi xj  in lsfun2.
If you are not satisfied with the result (e.g., because ifail lies between 33 and 88), 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 tt decimals, one would expect to get about t / 21t/2-1 decimals accuracy in the components of xx and between t1t-1 (if F(x)F(x) is of order 11 at the minimum) and 2t22t-2 (if F(x)F(x) is close to zero at the minimum) decimals accuracy in F(x)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_deriv_easy (e04gz) varies, but for mnmn is approximately n × m2 + O(n3)n×m2+O(n3). In addition, each iteration makes at least one call of lsfun2. So, unless the residuals and their derivatives can be evaluated very quickly, the run time will be dominated by the time spent in lsfun2.
Ideally, the problem should be scaled so that the minimum value of the sum of squares is in the range (0, + 1)(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_deriv_easy (e04gz) 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_deriv_easy_example
m = int64(15);
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_deriv_easy(m, @lsfun2, 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
 

xOut =

    0.0824
    1.1330
    2.3437


fsumsq =

    0.0082


user = 

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


ifail =

                    0


function e04gz_example
m = int64(15);
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] = e04gz(m, @lsfun2, 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
 

xOut =

    0.0824
    1.1330
    2.3437


fsumsq =

    0.0082


user = 

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


ifail =

                    0



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Chapter Introduction
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