GELE Adjustment Computations (Least Squares) — Least Squares — Observation EquationsRevision Notes
Revision notes for GELE Adjustment Computations (Least Squares) Least Squares — Observation Equations — designed for time-pressed reviewers. These notes skip the basics and focus on what Professional Regulation Commission (PRC) — Board of Geodetic Engineering consistently tests, so you spend your revision hours on the content most likely to appear on exam day.
Exam context
On the GELE 2026, the Adjustment Computations (Least Squares) subtest carries a "Core" weight in Professional Regulation Commission (PRC) — Board of Geodetic Engineering's pattern. Least Squares — Observation Equations lands at position 2nd out of 5 in the standard review order. Target score is 70% weighted average, no sub-test below 50%, and roughly a meaningful share of items come from Adjustment Computations (Least Squares) on a typical GELE paper.
Least Squares — Observation Equations - Revision Notes
The method of least squares is the cornerstone of geodetic network adjustment. When more observations are collected than the minimum needed to solve for the unknowns (a condition called redundancy), the system is over-determined and no exact solution exists. Least squares provides the rigorous, statistically optimal solution by finding the set of unknown parameters that minimizes the weighted sum of squared residuals. In Philippine geodetic practice — whether adjusting PRS92-based control networks, PPCS/UTM traverse closures, or differential leveling routes for cadastral surveys under PD 1529 — the observation-equation (parametric) approach is the standard tool. This chapter covers the principle, matrix formulation, the critical special case of the weighted mean, and the post-adjustment variance check.
Sections
Formulas
Example
Three leveling observations with weights 0.5, 1.0, 0.25 and residuals −0.02, +0.02, −0.04 m give: Σwv² = 0.5(0.02²) + 1.0(0.02²) + 0.25(0.04²) = 0.0002 + 0.0004 + 0.0004 = 0.001 m²
Formula
Minimize: Σ wᵢvᵢ² = vᵀPv
Variables
vᵢ = residual of observation i; wᵢ = weight of observation i; P = diagonal weight matrix with Pᵢᵢ = wᵢ
Application
Defines the objective function of least squares adjustment
Example
Route of 2 km → w = 1/2 = 0.50; route of 1 km → w = 1/1 = 1.00; route of 4 km → w = 1/4 = 0.25
Formula
wᵢ = 1 / Kᵢ (leveling by route length)
Variables
Kᵢ = length of level route i in km; wᵢ = weight assigned to that route
Application
Standard weighting formula for differential leveling routes in Philippine cadastral and geodetic surveys
Example
If σ₀ = 5 mm and σ₁ = 5 mm, σ₂ = 10 mm: w₁ = 25/25 = 1.0, w₂ = 25/100 = 0.25
Formula
wᵢ = σ₀² / σᵢ²
Variables
σ₀ = reference (a priori) standard deviation; σᵢ = standard deviation of observation i
Application
General weighting formula for any type of geodetic observation
Exam Tips
- Memorize: for leveling, weight = 1/(route length in km). This appears in almost every board exam leveling problem.
- Check: always verify that r = n − u > 0 before proceeding with adjustment.
- When the problem says 'equal precision' or 'equal weight,' use P = I (identity matrix); the solution becomes the simple arithmetic mean.
- The phrase 'most probable value' in board problems almost always signals a least-squares (weighted mean) computation.
Key Points
- Least squares is applied whenever the number of observations n exceeds the number of unknowns u, creating redundancy r = n − u > 0.
- Each observation contains a small random error; least squares finds the most probable values of the unknowns given those errors.
- The criterion: minimize the weighted sum of squared residuals, Σ wᵢvᵢ².
- A residual vᵢ = (computed value from adjusted parameters) − (observed value); it is the 'leftover' after adjustment.
- When all observations have equal precision, weights are all equal (w = 1) and the method reduces to minimizing Σvᵢ², equivalent to finding the arithmetic mean for a single unknown.
- Weights are inversely proportional to variance: wᵢ = σ₀² / σᵢ², where σ₀ is the reference standard deviation. For leveling, wᵢ = 1/Kᵢ (inverse of route length in km).
- The least-squares solution is unbiased and has minimum variance among all linear unbiased estimators (Gauss-Markov theorem).
Definitions
Term
Residual (v)
Definition
The difference between the value computed from adjusted parameters and the observed value: v = (computed) − (observed). Equivalently written as v = Ax̂ − l in matrix form.
Importance
Residuals quantify the discrepancy between observations and the adjusted solution; their magnitude and sign reveal blunders and systematic errors.
Term
Redundancy (Degrees of Freedom, r)
Definition
r = n − u, where n = number of observations and u = number of unknowns. Redundancy measures how over-determined the system is.
Importance
r must be > 0 for a least-squares adjustment to be possible. r = 0 gives a unique but un-adjustable solution; r < 0 means the system is under-determined.
Term
Weight Matrix (P)
Definition
A diagonal matrix (for independent observations) where each diagonal element Pᵢᵢ = wᵢ = 1/σᵢ². Off-diagonal elements are zero for uncorrelated observations.
Importance
P controls the relative influence of each observation on the adjusted result; heavier observations pull the solution more strongly.
Section Title
The Least-Squares Principle
Common Mistakes
- Using equal weights for observations of clearly different precision (e.g., short and long leveling routes).
- Forgetting that r = n − u must be positive; trying to 'adjust' a system with n = u (no redundancy).
- Confusing 'error' (true, unknown) with 'residual' (computed, observable after adjustment).
- Applying wᵢ = Kᵢ (proportional to length) instead of wᵢ = 1/Kᵢ for leveling — a classic sign error.
Formulas
Example
Two unknowns (x₁, x₂) observed in three combinations: obs₁ = x₁, obs₂ = x₂, obs₃ = x₁ + x₂. Then A = [[1,0],[0,1],[1,1]] and l = [ℓ₁−x₁⁰, ℓ₂−x₂⁰, ℓ₃−(x₁⁰+x₂⁰)]ᵀ
Formula
v = Ax̂ − l
Variables
v = (n×1) residual vector; A = (n×u) design matrix; x̂ = (u×1) vector of parameter corrections; l = (n×1) misclosure vector (observed − computed from x₀)
Application
The fundamental observation equation in matrix form; starting point of every parametric least-squares adjustment
Example
For u = 1 unknown and n = 3 observations: N = Σwᵢaᵢ², AᵀPl = Σwᵢaᵢlᵢ; dividing gives x̂ = Σwᵢaᵢlᵢ / Σwᵢaᵢ²
Formula
Normal equations: (AᵀPA)x̂ = AᵀPl
Variables
N = AᵀPA = (u×u) normal matrix (symmetric, positive definite); AᵀPl = (u×1) right-hand side vector
Application
The system of u equations in u unknowns whose solution gives the least-squares estimates
Example
Single unknown: x̂ = (Σwᵢaᵢ²)⁻¹ (Σwᵢaᵢlᵢ). If aᵢ = 1 for all i, this simplifies to x̂ = Σwᵢlᵢ / Σwᵢ (weighted mean)
Formula
x̂ = (AᵀPA)⁻¹ AᵀPl = N⁻¹ AᵀPl
Variables
N⁻¹ = inverse of the normal matrix; also called the cofactor matrix Qx̂x̂; full covariance matrix of x̂ is Σx̂ = σ₀² N⁻¹
Application
The closed-form least-squares solution; gives the most probable values of parameter corrections
Example
For a distance observation d = √[(X₂−X₁)² + (Y₂−Y₁)²]: ∂d/∂X₂ = (X₂−X₁)/d₀, ∂d/∂Y₂ = (Y₂−Y₁)/d₀
Formula
Design matrix element: Aᵢⱼ = ∂fᵢ/∂xⱼ |x₀
Variables
fᵢ = functional model for observation i; xⱼ = j-th unknown parameter; evaluated at initial approximation x₀
Application
Constructing A for nonlinear observation equations (e.g., GPS baseline components, angles in triangulation)
Exam Tips
- For board exam problems, A is usually given or easily constructed — focus on correctly forming N = AᵀPA and AᵀPl.
- A quick check: N must be square (u×u) and symmetric. If it is not symmetric, recheck your matrix multiplication.
- For a single unknown with aᵢ = 1 for all rows, AᵀPA = Σwᵢ and AᵀPl = Σwᵢlᵢ, giving x̂ = weighted mean directly.
- In Philippine board exams, 'set up the normal equations' typically means write out (AᵀPA)x̂ = AᵀPl in scalar form — show the arithmetic.
Key Points
- Each observation ℓᵢ is modeled as a function of the unknown parameters: ℓᵢ + vᵢ = f(x̂).
- For nonlinear functions, linearize by Taylor expansion about initial approximate values x₀: the result is a linear system.
- The design matrix A (also called the coefficient matrix) contains the partial derivatives ∂f/∂xⱼ evaluated at x₀.
- The misclosure vector l contains (observed value − value computed from x₀); it is NOT the raw observation.
- The system of observation equations in matrix form: v = Ax̂ − l.
- The normal equations are derived by minimizing vᵀPv with respect to x̂: (AᵀPA)x̂ = AᵀPl.
- The solution is: x̂ = (AᵀPA)⁻¹ AᵀPl. The matrix N = AᵀPA is called the normal matrix.
- For linear problems (e.g., leveling, traverse angle adjustment), no iteration is needed — A is exact.
- For nonlinear problems (e.g., geodetic positioning), iterate: update x₀ ← x₀ + x̂ and recompute until x̂ converges to zero.
Definitions
Term
Design Matrix (A)
Definition
The (n×u) matrix of partial derivatives of the observation functions with respect to the unknown parameters, evaluated at the initial approximation. Each row corresponds to one observation; each column to one unknown.
Importance
A completely defines the geometric structure of the network. Its rank must equal u for the normal matrix to be invertible.
Term
Normal Matrix (N)
Definition
N = AᵀPA, a (u×u) symmetric, positive definite matrix formed from the design and weight matrices. The normal equations are Nx̂ = AᵀPl.
Importance
N must be invertible (full rank) for a unique solution to exist. Rank deficiency indicates a datum defect — common in free networks.
Term
Misclosure Vector (l)
Definition
l = ℓ_observed − f(x₀), the difference between observed values and values computed from the initial approximations x₀. Sometimes written as the 'reduced observation' vector.
Importance
l represents the discrepancies that the adjustment must explain through the parameter corrections x̂ and residuals v.
Term
Normal Equations
Definition
The system of u linear equations (AᵀPA)x̂ = AᵀPl obtained by setting ∂(vᵀPv)/∂x̂ = 0. There is exactly one normal equation per unknown.
Importance
Solving the normal equations gives the unique least-squares solution; this is the central computational step of the parametric adjustment.
Section Title
Observation Equations — Matrix Formulation
Common Mistakes
- Confusing l (misclosure = observed − computed from x₀) with the raw observation vector ℓ.
- Setting up A with incorrect signs — always differentiate carefully; a sign error propagates through the entire solution.
- Forgetting to include P in the normal equations when observations have unequal weights (using AᵀA instead of AᵀPA).
- Attempting to invert a singular or near-singular normal matrix without checking the network datum (free networks need constraints).
- For nonlinear problems, stopping after one iteration without checking convergence of x̂.
Formulas
Example
Elevations: 152.34 m (w=0.50), 152.30 m (w=1.00), 152.36 m (w=0.25). x̂ = [0.50(152.34)+1.00(152.30)+0.25(152.36)] / (0.50+1.00+0.25) = [76.17+152.30+38.09] / 1.75 = 266.56 / 1.75 = 152.32 m
Formula
x̂ = Σwᵢℓᵢ / Σwᵢ (Weighted Mean)
Variables
x̂ = adjusted (most probable) value; wᵢ = weight of i-th observation; ℓᵢ = i-th observed value
Application
Least-squares adjustment of a single unknown observed multiple times — leveling, repeated angle measurement, repeated distance measurement
Example
For the leveling example: v₁ = 152.32 − 152.34 = −0.02 m; v₂ = 152.32 − 152.30 = +0.02 m; v₃ = 152.32 − 152.36 = −0.04 m. Check: Σwᵢvᵢ = 0.5(−0.02)+1.0(+0.02)+0.25(−0.04) = −0.01+0.02−0.01 = 0 ✓
Formula
vᵢ = x̂ − ℓᵢ
Variables
vᵢ = residual of observation i; x̂ = adjusted value; ℓᵢ = observed value
Application
Computing residuals after finding the weighted mean; used in variance check
Example
Leveling example: vᵀPv = 0.5(0.02²)+1.0(0.02²)+0.25(0.04²) = 0.0002+0.0004+0.0004 = 0.0010 m²
Formula
vᵀPv = Σwᵢvᵢ²
Variables
Scalar result used as numerator in the reference variance formula
Application
Computing the weighted sum of squared residuals for the variance check
Exam Tips
- Set up a neat tabular solution: columns for ℓᵢ, wᵢ, wᵢℓᵢ, vᵢ, wᵢvᵢ, wᵢvᵢ² — this organized format earns full marks.
- For leveling: immediately write weights as w = 1/K before doing anything else.
- Board exams often ask for the 'most probable value' or 'adjusted elevation' — this is always the weighted mean for single-unknown problems.
- Carry sufficient decimal places (at least 4 significant figures) in wᵢℓᵢ to avoid rounding errors in the final answer.
Key Points
- When a single unknown x is measured directly n times, every row of A equals [1], so A is an (n×1) column vector of ones.
- The normal equation reduces to: (Σwᵢ)x̂ = Σwᵢlᵢ.
- The least-squares solution is simply the weighted mean: x̂ = Σwᵢℓᵢ / Σwᵢ.
- If all weights are equal (wᵢ = 1), the weighted mean becomes the arithmetic mean: x̂ = Σℓᵢ / n.
- This is the most common board-exam scenario for least squares in leveling, angle measurement, and EDM distance measurement.
- After solving, compute residuals: vᵢ = x̂ − ℓᵢ (note: here the 'computed' value is x̂ for each observation).
- The sum of weighted residuals always equals zero: Σwᵢvᵢ = 0 — use this as a numerical check.
Definitions
Term
Weighted Mean
Definition
The least-squares estimate of a single unknown derived from n repeated observations with different weights: x̂ = Σwᵢℓᵢ / Σwᵢ. It is the unique solution to the normal equation for u = 1.
Importance
The most frequently tested least-squares computation in Philippine geodetic engineering board exams; mastery is essential.
Term
Check Condition: Σwᵢvᵢ = 0
Definition
After computing the weighted mean and residuals, the sum of weighted residuals must equal zero. This is a necessary consequence of the normal equation and a reliable numerical check.
Importance
Use this check after every weighted mean computation to catch arithmetic errors before proceeding to the variance calculation.
Section Title
Special Case: Single Unknown — Weighted Mean
Common Mistakes
- Computing the simple arithmetic mean instead of the weighted mean when weights differ.
- Using wᵢ = Kᵢ (length) instead of wᵢ = 1/Kᵢ for leveling route weights.
- Forgetting to check Σwᵢvᵢ = 0 as a numerical verification.
- Rounding intermediate Σwᵢℓᵢ and Σwᵢ values before dividing, causing accumulated rounding error.
Formulas
Example
Leveling example: vᵀPv = 0.0010 m²; n = 3, u = 1, r = 2. σ̂₀² = 0.0010/2 = 0.0005 m². σ̂₀ = √0.0005 = 0.0224 m = 22.4 mm (this represents the precision per unit weight)
Formula
σ̂₀² = vᵀPv / (n − u) = vᵀPv / r
Variables
σ̂₀² = a posteriori reference variance; vᵀPv = weighted sum of squared residuals; n = number of observations; u = number of unknowns; r = n − u = redundancy (degrees of freedom)
Application
Post-adjustment quality check; tests whether the weighting is consistent with the actual observation precision
Example
For single unknown: N = Σwᵢ = 1.75; N⁻¹ = 1/1.75 = 0.5714. Variance of x̂: σ²x̂ = σ̂₀² × 0.5714 = 0.0005 × 0.5714 = 2.857×10⁻⁴ m². Standard deviation: σx̂ = 0.0169 m = 16.9 mm
Formula
Σx̂ = σ̂₀² · N⁻¹ = σ̂₀² · (AᵀPA)⁻¹
Variables
Σx̂ = (u×u) covariance matrix of the adjusted parameters; diagonal elements are the variances of individual adjusted parameters
Application
Propagating the a posteriori variance into precision estimates for each adjusted unknown
Example
12 observations, 8 unknowns: r = 12 − 8 = 4 redundant observations; adjustment is possible and the variance has 4 degrees of freedom
Formula
r = n − u
Variables
r = redundancy = degrees of freedom; n = total observations; u = total unknowns
Application
Determines whether adjustment is possible (r > 0) and how many redundant checks exist in the network
Exam Tips
- Memorize: σ̂₀² = vᵀPv / (n − u). This formula appears directly in board exam questions.
- When asked to 'check the adjustment,' compute σ̂₀² and state whether it is close to 1.0.
- r = n − u is also tested directly as a standalone calculation — a 2-point question in many previous board exams.
- For the single-unknown case, always verify Σwᵢvᵢ = 0 first, then compute σ̂₀² = vᵀPv / (n − 1).
Key Points
- After solving for x̂, compute the adjusted residuals v = Ax̂ − l and the weighted sum of squared residuals vᵀPv.
- The reference variance (a posteriori variance factor) is: σ̂₀² = vᵀPv / (n − u) = vᵀPv / r.
- For a properly weighted network, σ̂₀² should be close to 1.0 (if weights were set using σ₀ = 1 as reference).
- σ̂₀² >> 1 indicates: blunders, incorrect weights, or systematic errors in the observations.
- σ̂₀² << 1 indicates: over-pessimistic weights (observations are better than assumed) or insufficient redundancy.
- The reference standard deviation is σ̂₀ = √(vᵀPv / r).
- The covariance matrix of the adjusted parameters is Σx̂ = σ̂₀² (AᵀPA)⁻¹ = σ̂₀² N⁻¹.
- Individual observation precision after adjustment: σ̂ᵢ = σ̂₀ / √wᵢ.
- A formal chi-squared test can be applied: the statistic r·σ̂₀²/σ₀² follows a χ²(r) distribution.
Definitions
Term
Reference Variance (σ̂₀²)
Definition
The a posteriori estimate of the variance of unit weight, computed as vᵀPv divided by the degrees of freedom r = n − u. It measures how well the adjustment fits the observations relative to their assumed precisions.
Importance
The primary diagnostic tool for assessing adjustment quality; a value significantly different from 1.0 signals problems with the data or the weighting scheme.
Term
Degrees of Freedom (r)
Definition
r = n − u, the number of redundant observations. It equals the number of independent checks available and is the denominator in the reference variance formula.
Importance
r = 0 means no check is possible — the adjustment has zero degrees of freedom and σ̂₀² is undefined. Always strive for r ≥ 3 in practice.
Term
A Posteriori vs. A Priori
Definition
A priori (before adjustment): assumed variances σᵢ² used to form P. A posteriori (after adjustment): estimated variances derived from the actual residuals, characterized by σ̂₀².
Importance
Comparing a priori and a posteriori precision estimates is how geodesists validate their stochastic model — a fundamental quality-control step in all Philippine national geodetic surveys.
Section Title
Post-Adjustment: Reference Variance and Quality Check
Common Mistakes
- Using n (total observations) instead of r = n − u as the denominator in the variance formula.
- Forgetting to compute residuals after adjustment — vᵀPv cannot be calculated without them.
- Interpreting σ̂₀² > 1 as always indicating blunders — it may simply mean the a priori weights were too optimistic.
- Reporting σ̂₀² without specifying the units or the weighting convention, making the value meaningless to others.
Formulas
Example
Three observations: obs₁ = x₁ (A row 1 = [1, 0]); obs₂ = x₂ (A row 2 = [0, 1]); obs₃ = x₁ + x₂ (A row 3 = [1, 1]). A = [[1,0],[0,1],[1,1]]
Formula
For obs i = a·x₁ + b·x₂ + ...: row i of A = [a, b, ...]
Variables
a, b = coefficients of unknowns x₁, x₂ in observation i; these are the partial derivatives for linear models
Application
Constructing A for any linear observation model (leveling, angle loops, direct combinations)
Example
obs₃ = x₁ + x₂; observed value = 10.05 m; initial approximations x₁⁰ = 5.00, x₂⁰ = 5.00; l₃ = 10.05 − (5.00+5.00) = +0.05 m
Formula
lᵢ = ℓᵢ_observed − fᵢ(x₀)
Variables
lᵢ = i-th element of misclosure vector; ℓᵢ_observed = measured value; fᵢ(x₀) = value computed from initial approximations x₀
Application
Building the right-hand side vector for the normal equations
Exam Tips
- Practice writing A from word problems without referring to notes — this matrix setup step is where most time is lost in exams.
- For a three-route leveling problem converging on one point: A = [1; 1; 1] (3×1 column of ones) — instantly recognizable as the weighted mean case.
- Label the A matrix explicitly with observation names on rows and unknown names on columns during the exam to prevent errors.
- The design matrix A for a linear model never contains the numerical values of observations — only coefficients.
Key Points
- For leveling: if each observation directly gives the elevation of unknown Hⱼ, then Aᵢⱼ = +1 if Hⱼ is the 'forward' BM, −1 if 'backward', and 0 otherwise.
- For a traverse: angle observations link adjacent azimuths; distance observations involve coordinate differences.
- For the single-unknown case with all direct observations: A is simply a column vector of 1s.
- For two unknowns (x₁, x₂) observed in combinations: A has two columns; row i contains the coefficients of x₁ and x₂ in the i-th observation equation.
- Always label rows (observations) and columns (unknowns) clearly before writing any entry in A.
- The number of rows of A equals n (observations); the number of columns equals u (unknowns).
- For GPS baseline components in a Philippine CORS network, A contains direction cosines linking baseline vectors to station coordinate corrections.
Definitions
Term
Parametric (Observation Equation) Method
Definition
The approach where each observation is expressed as a function of the unknown parameters, leading to v = Ax̂ − l. The number of equations equals n; the unknowns are x̂.
Importance
The most widely used adjustment method in Philippine geodetic practice; directly applicable to leveling networks, traverse adjustments, and GPS processing.
Term
Condition (Variation of Coordinates) Method
Definition
An alternative approach where geometric conditions among observations are written directly, without explicit unknowns. Less common in modern practice but still tested in board exams.
Importance
Understanding both methods allows candidates to choose the simpler one for a given problem; for most Philippine board exam problems, the parametric method is preferred.
Section Title
Setting Up the Design Matrix A — Practical Examples
Common Mistakes
- Writing A with the wrong number of columns (e.g., u+1 columns due to miscounting unknowns).
- Putting raw observations in l instead of the misclosure (observed − computed from x₀).
- For leveling networks: using elevation differences as entries in A instead of ±1 coefficients.
- Transposing rows and columns — always: rows = observations, columns = unknowns.
Connections
- TRAVERSE ADJUSTMENT (Bowditch/Compass Rule vs. Least Squares): The observation-equation method provides the rigorous alternative to the empirical Bowditch rule for traverse closure distribution. Board exams may ask candidates to compare results — least squares produces smaller residuals but requires more computation.
- DIFFERENTIAL LEVELING NETWORKS (Philippine Vertical Datum): All first-order and second-order leveling networks in the Philippines (referenced to PMSL at Manila or Cebu tide gauges) are adjusted by least squares. Route weights w = 1/K are applied directly using the formulas in this chapter.
- PRS92 HORIZONTAL CONTROL NETWORK: The Philippine Reference System 1992, established under NAMRIA, uses a 3D geodetic network adjusted by least squares with GPS baseline observations. The design matrix A contains direction cosines of baseline vectors.
- WEIGHT MATRIX AND STOCHASTIC MODEL: The weight matrix P is the inverse of the observation covariance matrix Σℓ = σ₀² Q, where Q is the cofactor matrix. Understanding this link connects least squares to error propagation (Law of Propagation of Variances).
- ERROR PROPAGATION: After adjustment, the covariance of adjusted parameters Σx̂ = σ̂₀² N⁻¹ directly applies the law of error propagation — a closely related topic in the PRC Geodetic Engineer board exam syllabus.
- CONDITION EQUATIONS METHOD: The alternative approach to parametric least squares; both methods are covered in the board exam. The parametric method (this chapter) is generally more systematic for networks with many unknowns.
- GPS BASELINE PROCESSING (RA 8560 — Geodetic Engineering Act): Under RA 8560, only licensed Geodetic Engineers may sign and seal survey plans. Network adjustments using least squares are a core professional competency. GPS-derived baselines in PPCS/UTM Zone 51N or 52N are adjusted using observation equations in 3D space.
- CADASTRAL SURVEYS (PD 1529 — Property Registration Decree): Control point coordinates used in cadastral surveys must be derived from properly adjusted networks. The precision standards for cadastral monuments referenced in PD 1529 implicitly require least-squares-quality adjustments.
- CHI-SQUARED TEST: The reference variance ratio r·σ̂₀²/σ₀² follows a χ²(r) distribution, connecting least squares to hypothesis testing and statistical quality control — topics in the Geodetic Engineering board exam under 'Geodetic Statistics'.
Exam Strategy
For PRC Geodetic Engineer board exams on Least Squares — Observation Equations, adopt this systematic approach: (1) READ the problem completely and identify n (observations), u (unknowns), and r = n − u immediately — write these three numbers first. (2) IDENTIFY the type: if u = 1, go directly to the weighted mean formula; if u > 1, set up A, P, and l in matrix form. (3) COMPUTE WEIGHTS first using the appropriate rule (w = 1/K for leveling; w = σ₀²/σᵢ² for general observations) before touching any observation values. (4) USE a tabular format with clearly labeled columns: ℓᵢ | wᵢ | wᵢℓᵢ | vᵢ | wᵢvᵢ | wᵢvᵢ² — this organized layout earns partial credit even if arithmetic errors occur. (5) CHECK: always verify Σwᵢvᵢ = 0 after computing the weighted mean. (6) COMPUTE σ̂₀² = vᵀPv/r if asked for quality/precision. (7) MANAGE TIME: single-unknown weighted mean problems should take 3–5 minutes; multi-unknown matrix problems may take 8–12 minutes — allocate accordingly. (8) MEMORIZE THESE EXACTLY: x̂ = Σwᵢℓᵢ/Σwᵢ; σ̂₀² = vᵀPv/(n−u); r = n−u; for leveling w = 1/K. These four items cover 80% of board exam points in this topic.
Quick Review Questions
A point's elevation is determined from three leveling routes: H₁ = 45.231 m (K = 1 km), H₂ = 45.245 m (K = 2 km), H₃ = 45.238 m (K = 4 km). Compute the least-squares (most probable) elevation.
Weights: w₁ = 1/1 = 1.00, w₂ = 1/2 = 0.50, w₃ = 1/4 = 0.25; Σw = 1.75. Σwℓ = 1.00(45.231) + 0.50(45.245) + 0.25(45.238) = 45.231 + 22.623 + 11.310 = 79.164. Ĥ = 79.164 / 1.75 = 45.2366 m ≈ 45.237 m. (Carry 4 decimal places: Ĥ = 45.2366 m)
A geodetic network has 15 observations and 9 unknowns. What is the redundancy, and is a least-squares adjustment possible?
r = n − u = 15 − 9 = 6 > 0. Since r > 0, the system is over-determined and a least-squares adjustment is both possible and meaningful. The reference variance will have 6 degrees of freedom.
After a leveling adjustment with n = 4 observations and u = 1 unknown, the weighted sum of squared residuals vᵀPv = 0.0018 m². Compute the reference variance σ̂₀² and the reference standard deviation σ̂₀.
r = n − u = 4 − 1 = 3. σ̂₀² = vᵀPv / r = 0.0018 / 3 = 0.0006 m². σ̂₀ = √0.0006 = 0.02449 m ≈ 24.5 mm. This is the a posteriori standard deviation of unit weight.
For the observation equation model v = Ax̂ − l with two unknowns and four observations, what are the dimensions of A, x̂, l, and v?
The design matrix A always has dimensions n×u = 4×2. The unknowns vector x̂ and misclosure vector l are column vectors of dimensions u×1 = 2×1 and n×1 = 4×1, respectively. The residual vector v has the same dimension as l: n×1 = 4×1.
In the normal equation (AᵀPA)x̂ = AᵀPl, if n = 4 and u = 2, what are the dimensions of the normal matrix N = AᵀPA and the right-hand side AᵀPl?
Aᵀ is u×n = 2×4; P is n×n = 4×4; A is n×u = 4×2. So N = AᵀPA is (2×4)(4×4)(4×2) = 2×2. Similarly, AᵀPl is (2×4)(4×4)(4×1) = 2×1. There are always u normal equations — one per unknown.
After adjustment, you compute Σwᵢvᵢ = 0.003 m instead of 0. What does this indicate?
The condition Σwᵢvᵢ = 0 is a mathematical consequence of the normal equation for the single-unknown case. If it does not equal zero (within rounding tolerance), there is a computational error — typically in computing x̂, computing vᵢ = x̂ − ℓᵢ, or in the weight values. Recheck all steps.
Three direct measurements of an angle have standard deviations σ₁ = 3″, σ₂ = 6″, σ₃ = 3″. Using σ₀ = 3″ as the reference, compute the weights.
wᵢ = σ₀²/σᵢ². w₁ = (3″)²/(3″)² = 9/9 = 1.00. w₂ = (3″)²/(6″)² = 9/36 = 0.25. w₃ = (3″)²/(3″)² = 9/9 = 1.00. The second measurement is four times less precise, so it gets one-quarter the weight.
What is the least-squares solution when only one unknown is measured n times with equal weights?
With equal weights (P = I), the normal equation becomes (AᵀA)x̂ = Aᵀl. For A = column of 1s: AᵀA = n, Aᵀl = Σℓᵢ, so x̂ = Σℓᵢ/n. Least squares with equal weights always reduces to the arithmetic mean — this is why the mean is the 'most probable value' for equal-precision observations.
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