Weighted Adjustment of a Closed Level Network: 18-Geom-A2 Adjustment of Observations, May 2014, Q6

Question 6: Weighted Adjustment of a Closed Level Network (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 among four stations, each with its own standard deviation; $H_A=437.596$ m fixed. Here the weights come from stated $\sigma$'s rather than distances.

Find. Most-probable elevations of B, C, D (and, as a bonus, their standard deviations).

Figure 6.1 — Closed level network: fixed benchmark A (triangle) with observed height differences to B, C, D including the interior chords B–D and A–C. Arrows give the observation sense; labels are ΔElev in metres.

Approach. Identical machinery to Q1, but weights are $w_i=1/\sigma_i^2$: three unknowns, six observations, redundancy three.

Observation equations and weights. $H_{\text{to}}-H_{\text{from}}=\Delta_i+v_i$ with $x=[H_B,H_C,H_D]^{\mathsf T}$; $W=\operatorname{diag}(1/\sigma_i^2)$, so the sharpest line D–A ($\sigma=3$ mm, $w=1.11\times10^{5}$) dominates and the weakest A–C ($\sigma=12$ mm) contributes least.
Solve the weighted normal equations. $\hat x=(A^{\mathsf T}WA)^{-1}A^{\mathsf T}W\ell$: $$\boxed{H_B=448.109,\quad H_C=453.469,\quad H_D=444.944\ \text{m}}$$
Reference standard deviation. With $n-u=6-3=3$, $$s_0=\sqrt{\frac{v^{\mathsf T}Wv}{3}}=\boxed{0.65}$$ Being well below unity, it confirms the observations agree better than their stated $\sigma$'s — a well-conditioned, consistent network (the largest residual, $-8.5$ mm, sits on the weak A–C chord).
Precision of the unknowns. From the cofactor matrix $Q_{xx}=(A^{\mathsf T}WA)^{-1}$, $\sigma_{H}=s_0\sqrt{\operatorname{diag}Q_{xx}}$: $$\sigma_{H_B}=\pm2.3,\;\sigma_{H_C}=\pm2.6,\;\sigma_{H_D}=\pm1.8\ \text{mm}$$

Station	Most-probable elev (m)	Std dev (mm)
A	437.596 (fixed)	—
B	448.109	±2.3
C	453.469	±2.6
D	444.944	±1.8

Concept & Theory

Topic. Weighted least-squares adjustment of a level network with individually stated observation precisions.

Key relations. $w_i=\sigma_0^2/\sigma_i^2$ (here referred to unit weight $\sigma_0=1$); $\hat x=(A^{\mathsf T}WA)^{-1}A^{\mathsf T}W\ell$; $s_0^2=v^{\mathsf T}Wv/(n-u)$; $\Sigma_{xx}=s_0^2\,(A^{\mathsf T}WA)^{-1}$.

Why this works. When each observation carries its own variance, the maximum-likelihood (and minimum-variance) estimate weights by the inverse variance; the redundancy of three independent loops lets the adjustment both fix the heights and estimate their precision from the residuals.

Common pitfall. Using $1/\sigma$ instead of $1/\sigma^2$ as the weight, or reporting $s_0$ as if it had units — it is a dimensionless ratio to the a-priori precision and its departure from 1 is the diagnostic.

Source. Ghilani, Adjustment Computations §12 (level-net adjustment) and §13 (precision from the cofactor matrix).

Question 6 of 7: Weighted Adjustment of a Closed Level Network

Question 6: Weighted Adjustment of a Closed Level Network (20 marks)