Weighted Least-Squares Adjustment of a Level Net: 18-Geom-A2 Adjustment of Observations, December 2018, Q1

Question 1: Weighted Least-Squares Adjustment of a Level Net (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Six observed height differences on a braced quadrilateral level net (the four sides A–B–C–D plus the two diagonals A–C and B–D), each with its own standard deviation, and one fixed benchmark $H_A=437.596$ m.

Find. The most-probable elevations of the three unknown stations B, C and D.

Figure 1.1 — Braced quadrilateral level net. A (triangle) is the fixed benchmark; arrows give the sense of each observed height difference (m).

Approach. Because each observation carries a different standard deviation, weight by $w_i=1/\sigma_i^2$, write one linear observation equation per course in the three unknowns $[H_B,H_C,H_D]$, and solve the weighted normal equations.

Form the observation equations. Every course models $H_{\text{to}}-H_{\text{from}}=\Delta_i+v_i$. The four courses that touch the fixed station A carry $H_A$ to the right-hand side, e.g. $H_B-H_A=10.509\Rightarrow H_B=448.105$ (provisional) and $H_A-H_D=-7.348\Rightarrow -H_D=-444.944$. The design matrix rows are $[+1,0,0],[-1,1,0],[0,-1,1],[0,0,-1],[-1,0,1],[0,1,0]$ for $x=[H_B,H_C,H_D]^{\mathsf T}$.
Build the weight matrix. $W=\operatorname{diag}(1/\sigma_i^2)$. The sharp 3 mm course D–A carries weight $1.11\times10^{5}$ while the weak 12 mm diagonal A–C carries only $6.9\times10^{3}$ — the diagonal is trusted about sixteen times less.
Solve the weighted normal equations. With $N=A^{\mathsf T}WA$ and $t=A^{\mathsf T}W\ell$, $$\hat{x}=N^{-1}t\;\Rightarrow\;\boxed{\,H_B=448.109,\ H_C=453.468,\ H_D=444.944\ \text{m}\,}$$
Check the fit. Back-computing each course from the adjusted heights gives residuals of $+3.7,-0.2,-1.9,+0.4,+1.9,-8.5$ mm; the largest lands on the weakly-weighted diagonal A–C, exactly where the network absorbs its misclosure. With $n=6$, $u=3$ (redundancy $r=3$), the reference standard deviation is $s_0=\pm0.65$ (unit weight), and the station precisions are $\sigma_{H_B}=\pm2.3$ mm, $\sigma_{H_C}=\pm2.6$ mm, $\sigma_{H_D}=\pm1.8$ mm.

Station	A	B	C	D
Elevation (m)	437.596*	448.109	453.468	444.944
Std. dev. (mm)	—	±2.3	±2.6	±1.8

*held fixed.

Concept & Theory

Topic. Weighted parametric (observation-equation) least-squares adjustment of a differential level network.

Key relations. Observation equation $A\hat{x}=\ell+v$; weighted normal equations $A^{\mathsf T}WA\,\hat{x}=A^{\mathsf T}W\ell$; weight $w_i=\sigma_0^2/\sigma_i^2$ (unit $\sigma_0=1$); reference variance $s_0^2=v^{\mathsf T}Wv/(n-u)$; covariance $\Sigma_{xx}=s_0^2(A^{\mathsf T}WA)^{-1}$.

Why this works. When the observations have individually known standard deviations, minimising $v^{\mathsf T}Wv$ with $w_i=1/\sigma_i^2$ is the Best Linear Unbiased Estimator under the Gauss–Markov model. Fixing $H_A$ removes the net’s single datum defect so the normal-equation matrix is invertible; the two diagonals supply the redundancy that makes an adjustment (rather than a mere computation) possible.

Common pitfall. Using $w_i=1/\sigma_i$ instead of $1/\sigma_i^2$, or mixing the two levelling weight rules ($1/\sigma^2$ vs $1/\text{distance}$); both distort the relative trust and bias every height.

Source. Ghilani, Adjustment Computations §12 (least-squares adjustment of level nets), §10–11 (weights and the stochastic model).

Question 1 of 7: Weighted Least-Squares Adjustment of a Level Net

Question 1: Weighted Least-Squares Adjustment of a Level Net (20 marks)