e04gz is an easy-to-use modified Gauss–Newton algorithm for finding an unconstrained minimum of a sum of squares of $m$ nonlinear functions in $n$ variables $\left(m\ge n\right)$. 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

C# |
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public static void e04gz( int m, int n, E04..::..E04GZ_LSFUN2 lsfun2, double[] x, out double fsumsq, out int ifail ) |

Visual Basic |
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Public Shared Sub e04gz ( _ m As Integer, _ n As Integer, _ lsfun2 As E04..::..E04GZ_LSFUN2, _ x As Double(), _ <OutAttribute> ByRef fsumsq As Double, _ <OutAttribute> ByRef ifail As Integer _ ) |

Visual C++ |
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public: static void e04gz( int m, int n, E04..::..E04GZ_LSFUN2^ lsfun2, array<double>^ x, [OutAttribute] double% fsumsq, [OutAttribute] int% ifail ) |

F# |
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static member e04gz : m : int * n : int * lsfun2 : E04..::..E04GZ_LSFUN2 * x : float[] * fsumsq : float byref * ifail : int byref -> unit |

#### Parameters

- m
- Type: System..::..Int32
*On entry*: the number $m$ of residuals, ${f}_{i}\left(x\right)$, and the number $n$ of variables, ${x}_{j}$.*Constraint*: $1\le {\mathbf{n}}\le {\mathbf{m}}$.

- n
- Type: System..::..Int32
*On entry*: the number $m$ of residuals, ${f}_{i}\left(x\right)$, and the number $n$ of variables, ${x}_{j}$.*Constraint*: $1\le {\mathbf{n}}\le {\mathbf{m}}$.

- lsfun2
- Type: NagLibrary..::..E04..::..E04GZ_LSFUN2You must supply this method to calculate the vector of values ${f}_{i}\left(x\right)$ and the Jacobian matrix of first derivatives $\frac{\partial {f}_{i}}{\partial {x}_{j}}$ at any point $x$. It should be tested separately before being used in conjunction with e04gz.
A delegate of type E04GZ_LSFUN2.

- x
- Type: array<System..::..Double>[]()[][]An array of size [n]
*On entry*: ${\mathbf{x}}\left[\mathit{j}-1\right]$ must be set to a guess at the $\mathit{j}$th component of the position of the minimum, for $\mathit{j}=1,2,\dots ,n$. The method checks the first derivatives calculated by lsfun2 at the starting point and so is more likely to detect any error in your methods if the initial ${\mathbf{x}}\left[j-1\right]$ are nonzero and mutually distinct.*On exit*: the lowest point found during the calculations. Thus, if ${\mathbf{ifail}}={0}$ on exit, ${\mathbf{x}}\left[j-1\right]$ is the $j$th component of the position of the minimum.

- fsumsq
- Type: System..::..Double%
*On exit*: the value of the sum of squares, $F\left(x\right)$, corresponding to the final point stored in x.

- ifail
- Type: System..::..Int32%
*On exit*: ${\mathbf{ifail}}={0}$ unless the method detects an error or a warning has been flagged (see [Error Indicators and Warnings]).

# Description

e04gz is similar to the method LSFDN2 in the NPL Algorithms Library. It is applicable to problems of the form

where $x={\left({x}_{1},{x}_{2},\dots ,{x}_{n}\right)}^{\mathrm{T}}$ and $m\ge n$. (The functions ${f}_{i}\left(x\right)$ are often referred to as ‘residuals’.)

$$\mathrm{Minimize}\u200aF\left(x\right)=\sum _{i=1}^{m}{\left[{f}_{i}\left(x\right)\right]}^{2}$$ |

You must supply a method to evaluate the residuals and their first derivatives at any point $x$.

Before attempting to minimize the sum of squares, the algorithm checks the method 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 $F\left(x\right)$.

# 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# Error Indicators and Warnings

**Note:**e04gz may return useful information for one or more of the following detected errors or warnings.

Errors or warnings detected by the method:

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

- ${\mathbf{ifail}}=2$

- ${\mathbf{ifail}}=3$
- The final point does not satisfy the conditions for acceptance as a minimum, but no lower point could be found.

- ${\mathbf{ifail}}=4$
- An auxiliary method has been unable to complete a singular value decomposition in a reasonable number of sub-iterations.

- ${\mathbf{ifail}}=5$
- ${\mathbf{ifail}}=6$
- ${\mathbf{ifail}}=7$
- ${\mathbf{ifail}}=8$
- There is some doubt about whether the point x$x$ found by e04gz is a minimum of $F\left(x\right)$. The degree of confidence in the result decreases as ifail increases. Thus, when ${\mathbf{ifail}}={5}$, it is probable that the final $x$ gives a good estimate of the position of a minimum, but when ${\mathbf{ifail}}={8}$ it is very unlikely that the method has found a minimum.

- ${\mathbf{ifail}}=9$
- It is very likely that you have made an error in forming the derivatives $\frac{\partial {f}_{i}}{\partial {x}_{j}}$ in lsfun2.

- ${\mathbf{ifail}}=-9000$
- An error occured, see message report.
- ${\mathbf{ifail}}=-6000$
- Invalid Parameters $\u2329\mathit{\text{value}}\u232a$
- ${\mathbf{ifail}}=-8000$
- Negative dimension for array $\u2329\mathit{\text{value}}\u232a$
- ${\mathbf{ifail}}=-6000$
- Invalid Parameters $\u2329\mathit{\text{value}}\u232a$

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 $F\left(x\right)$ is of order $1$ at the minimum) and $2t-2$ (if $F\left(x\right)$ is close to zero at the minimum) decimals accuracy in $F\left(x\right)$.

# Parallelism and Performance

None.

# 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 e04gz varies, but for $m\gg n$ is approximately $n\times {m}^{2}+\mathit{O}\left({n}^{3}\right)$. 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 $\left(0,+1\right)$ 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 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 (E04YCF not in this release), using information returned in segments of the workspace array w. See (E04YCF not in this release) for further details.

# Example

This example finds least squares estimates of ${x}_{1}$, ${x}_{2}$ and ${x}_{3}$ in the model

using the $15$ sets of data given in the following table.

The program uses $\left(0.5,1.0,1.5\right)$ as the initial guess at the position of the minimum.

$$y={x}_{1}+\frac{{t}_{1}}{{x}_{2}{t}_{2}+{x}_{3}{t}_{3}}$$ |

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

Example program (C#): e04gze.cs