Question 1 of 7: Weighted Least-Squares Adjustment of a Level Network
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — May 2014 — 04-Geom-A2, Adjustment of Observations and Data Analysis. Three-hour, CLOSED-BOOK exam; any non-communicating calculator permitted. Format: seven questions are given and any five (20 marks each) constitute a complete paper — all seven are solved below for completeness. Least-squares adjustments are carried out in the Canadian survey frame (heights on CGVD2013, horizontal on NAD83(CSRS)); US-foot units are retained in Q4 wherever the printed question uses them.
Reference texts: Ghilani, Adjustment Computations: Spatial Data Analysis (6th ed., Wiley, 2017); Wolf & Ghilani, Elementary Surveying: An Introduction to Geomatics (15th ed., Pearson); Mikhail & Gracie, Analysis and Adjustment of Survey Measurements (Van Nostrand, 1981).
Method note. Every weighted least-squares result below was formed from the normal equations $N\hat{x}=A^{\mathsf T}W\,\ell$; the reference (unit-weight) standard deviation is $s_0=\sqrt{v^{\mathsf T}Wv/(n-u)}$ with $n$ observations and $u$ unknowns.
Question 1: Weighted Least-Squares Adjustment of a Level Network (20 marks)
Given. Ten observed height differences around a level net with one fixed benchmark, $H_A=437.596$ m. Levelling variance grows with line length, so each observation is weighted by $w_i=1/d_i$ (distance in km).
Find. (a) most-probable elevations of B–G; (b) the reference standard deviation $s_0$; (c) adjusted observations and residuals.
Figure 1.1 — Level network: triangle A (fixed benchmark) and closing loop B–G, with the three interior chords A–C, C–F, F–D. Arrows show the sense of each observed height difference; labels are line lengths in km.
Approach. Write one linear observation equation per course, $H_{\text{to}}-H_{\text{from}}=\Delta_i+v_i$, in the six unknown elevations (A is held fixed and moved to the right side), then solve the weighted normal equations and back-substitute for residuals.
Form the observation equations. With unknown vector $x=[H_B,H_C,H_D,H_E,H_F,H_G]^{\mathsf T}$, each course contributes a row of $A$ with $+1$ at the "to" station and $-1$ at the "from" station. Courses touching A carry $H_A$ into the constant term, e.g. $H_B-H_A=+5.666\Rightarrow H_B=5.666+437.596$.
Build the weight matrix. $W=\operatorname{diag}(1/d_i)$, so the 1 km course A–B carries weight $1.0$ while the 7.6 km course F–G carries only $0.132$ — long lines are trusted less.
Solve the normal equations. $N=A^{\mathsf T}WA$ (a $6\times6$ system) and $t=A^{\mathsf T}W\ell$; then
$$\hat{x}=N^{-1}t\quad\Rightarrow\quad \boxed{\;H_B=443.263,\ H_C=491.293,\ H_D=494.297\;}$$
$$\boxed{\;H_E=480.294,\ H_F=500.941,\ H_G=468.568\ \text{m}\;}$$
Residuals and reference standard deviation. Each adjusted course difference is recomputed from the fitted elevations, $\bar\Delta_i=H_{\text{to}}-H_{\text{from}}$, and $v_i=\bar\Delta_i-\Delta_i$. With $n=10$ observations and $u=6$ unknowns (redundancy $r=4$),
$$s_0=\sqrt{\frac{v^{\mathsf T}Wv}{n-u}}=\sqrt{\frac{4.30\times10^{-4}}{4}}=\boxed{\pm0.0104\ \text{(unit weight, i.e. }\pm10.4\ \text{mm}/\sqrt{\text{km}})}$$
The largest residuals fall on the long E–F run ($-30.8$ mm) and the C–D and D–E loop lines, with the C–F and F–D chords next — the long, loosely weighted lines, which is exactly where the loosely weighted, redundant observations should absorb the network misclosures.
