This adjustment uses the Observation Model =F(X)
The adjusted observations are modeled in terms of the estimated adjusted parameters.
In this case the coordinates of the individiual OPUS solutions are treated as observations and the estimated parameters are the XYZ coordinates of the combined solution

Here is the observation model for the specific case of combining OPUS solutions for days 095, 096 and 097. One OPUS solution would provide a unique answer. Three OPUS solutions provide an over determined condition, which can be adjusted.

The design matrix for this adjustment is derived from the partial differentials with respect to each parameter. This works out to be a 3x9 matrix consisting of 1s and 0s.

The degrees of freedom of this adjustment is 6. One OPUS solution provides all the information needed for a unique solution. Three OPUS solutions provide six more measurements than are needed. The degree of freedom of the adjustment is also equal to the number of rows in the design matrix minus the number of columns.

The Raw Data:
Matrices are the individual OPUS solutions
Covariance matrices for the individual solutions are from the OPUS extended output

Using the raw data above, build the vector, the vector of observations

Set up the Observation equations as a function of the estimated parameters:
is a vector consisting of the estimated station coordinates for the unknown station
is the vector of initial coordinates for the solution

Use an intial estimate of 0 for all three parameters. Since the design matrix is linear the first iteration will compute the entire differential correction.

The vector of misclosures is formed by difference between and the vector

Using the individual covariance matrices for the OPUS positions, build the overall covariance matrix
for the least squares adjustment of the combined solution.

The OPUS solutions were independently determined on different days so assume that there are no correlations between the individual day's solutions. In this case the matrix is block diagonal.

The weight matrix is the inverse of the covariance matix.

The actual solution to the system of equations follows:

N is the normal matrix

X is the vector of corrections to the initial estimate. Because the design matrix is linear it is also the final solution.

Compute the residuals

Compute the Variance of Unit Weight

Standard Deviation of Unit Weight:

If the original covariance matrices are multiplied by the Variance of Unit weight and the Standard Deviation of Unit weight would by 1

Compute the covariance matrix of the estimated parameters. The inverse of the normal matrix is the covariance of the estimated parameters. First the normal matrix is properly scaled by the inverse of the variance of unit weight.

This is the covariance of the ECEF UVW coordinates

Compute the misclosures between the adjusted coordinates and the published position of NLIB.

This is the published coordinate of NLIB, transformed to day 097

The misclosures in UVW

and are the published longitude and latitude of NLIB

is the rotation matrix to transform XYZ to NEU

The misclosures as NEU

Transform XYZ covariance matrix to NEU covariance matrix

This was previously scaled by the variance of unit weight.

Compute the standard deviations for N, E and U