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Dependant linear equations

If there are not enough independent condition equations, the normal matrix A cannot be inverted, and a call to WNMLTN will fail with a .false. return value.

The equations could still be solved if some additional `constraint' equations would be introduced. In the more complex cases the precise, let alone the best, form for these additional equations is difficult to determine (e.g. the redundancy situation in Westerbork).

A method known as `Singular value decomposition' (SVD) can be used to obtain the minimal set of orthogonal equations that have to be added to solve the LSQ problem. Several implementations exist in the literature.

In general we can distinguish three types of constrained equations:

the minimum number of sufficient constraint equations are known to be able to solve a system of equations
constraints are used to add additional information (e.g. the sum of angles of a triangle)
no actual constraint equations are known

All three cases are handled in the LSQ package, the first two in the same way.

The general constraint situation arises from the use of Lagrange multiplicators. Assume that in addition to the condition equations, with measured values, we have a set of p rigorous equations:

$\displaystyle \phi_{i}^{}$ (x) = 0 i = 0,..., p - 1

(31)

We must therefore make $\chi^{{2}}_{}$ minimal, subject to the set of (31), or:

$\displaystyle \sum_{{i=0}}^{{n-1}}$ $\displaystyle {\frac{{\partial \chi^{2}}}{{\partial x_{i}}}}$ dx_i = 0

(32)

subject to the conditions:

$\displaystyle \sum_{{i=0}}^{{n-1}}$ $\displaystyle {\frac{{\partial \phi_{k}}}{{\partial x_{i}}}}$ dx_i = 0 k = 0,..., p - 1

(33)

which leads to a set of n + p equations:

$\displaystyle {\frac{{\partial \chi^{2}}}{{\partial x_{i}}}}$ + $\displaystyle \sum_{{k=0}}^{{p-1}}$ $\displaystyle \lambda_{{k}}^{}$ $\displaystyle {\frac{{\partial \phi_{k}}}{{\partial x_{i}}}}$ = 0 i = 0,..., n - 1

(34)

together with the (31).

Note that I have chosen for having constraint equations linear in the unknowns, with a zero value. In cases where this is not adequate (e.g. x + y + z = 360) a simple linear transformation will suffice to make it e.g. x' + y' + z' = 0.

Defining the second term in (34) as B_ik, we can write our expanded set of normal equations as:

$\displaystyle \left(\vphantom{\begin{array}{cc} \mathbf{A} & \mathbf{B} \\ \mathbf{B}^{T} & 0 \end{array} }\right.$ $\displaystyle \begin{array}{cc} \mathbf{A} & \mathbf{B} \\ \mathbf{B}^{T} & 0 \end{array}$ $\displaystyle \left.\vphantom{\begin{array}{cc} \mathbf{A} & \mathbf{B} \\ \mathbf{B}^{T} & 0 \end{array} }\right)$ $\displaystyle \left(\vphantom{\begin{array}{c} \mathbf{x} \\ \mathbf{\lambda} \end{array} }\right.$ $\displaystyle \begin{array}{c} \mathbf{x} \\ \mathbf{\lambda} \end{array}$ $\displaystyle \left.\vphantom{\begin{array}{c} \mathbf{x} \\ \mathbf{\lambda} \end{array} }\right)$ = $\displaystyle \left(\vphantom{ \begin{array}{c} \mathbf{L} \\ 0 \end{array}}\right.$ $\displaystyle \begin{array}{c} \mathbf{L} \\ 0 \end{array}$ $\displaystyle \left.\vphantom{ \begin{array}{c} \mathbf{L} \\ 0 \end{array}}\right)$

(35)

Subsections

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