GELE Adjustment Computations (Least Squares) — Least Squares — Observation EquationsStudy Notes
Complete study notes for Least Squares — Observation Equations, written for GELE aspirants. Unlike generic notes, these focus on what Professional Regulation Commission (PRC) — Board of Geodetic Engineering actually tests in the GELE Adjustment Computations (Least Squares) section: high-yield concepts, common question types, and the worked examples that match recent exam patterns.
Exam context
The Geodetic Engineer Licensure Examination is conducted by Professional Regulation Commission (PRC) — Board of Geodetic Engineering and is scheduled for September 2026. The Adjustment Computations (Least Squares) subtest is marked as "Core" in the official pattern, and Least Squares — Observation Equations appears in position 2nd of 5 in the GELE Adjustment Computations (Least Squares) review rotation. Passing mark: 70% weighted average, no sub-test below 50%. Recent GELE 2026 papers have drawn roughly a meaningful share of questions from this subject.
Least Squares — Observation Equations - Study Notes
The method of least squares is the fundamental rigorous approach used in geodetic engineering to adjust redundant observations and determine the most probable values of unknown parameters. When you have more observations than unknowns (a condition called redundancy), you cannot satisfy all observations exactly—there will always be residuals or corrections needed. The observation-equation (parametric) form of least squares models each observation as a function of unknown parameters, then solves for those parameters in a way that minimizes the weighted sum of squared residuals. This chapter develops the mathematical foundation of observation equations, the matrix formulation, and practical solutions including the important single-parameter case that reduces to a weighted mean. Understanding this method is essential for all geodetic adjustments: leveling networks, traverse computations, coordinate adjustments under WGS84/PRS92, and regulatory compliance under RA 4374 (Surveying Profession) and RA 8560 (Professional Regulation Commission).
Summary
The least-squares method via observation equations is the standard and most rigorous approach to adjust redundant observations in geodetic surveys. By formulating each observation as a function of unknown parameters and linearizing nonlinearities, you construct a design matrix A and a weight matrix P reflecting observational precision. Solving the normal equations (AᵀPA)·Δx = AᵀPl yields parameter corrections; adding these to approximate values gives adjusted parameters x̂. The key insight is redundancy (r = n − u > 0): only with more observations than unknowns can you assess data quality via the reference variance σ̂₀² and compute reliable uncertainties for the final parameters. When there is a single unknown, the solution reduces to the weighted mean—a powerful sanity check. Professional practice under PRC regulations (RA 4374, RA 8560) and Philippine standards (WGS84, PRS92, PPCS) requires you to document the adjustment, report parameter uncertainties, and validate the reference variance. Mastery of observation-equation least squares is essential for the PRC Geodetic Engineer Licensure Examination and for defensible surveying practice. The method is universal: from simple leveling networks to complex GPS processing to traverse adjustments, the same matrix equations apply. By working through board-style problems and understanding why the method works (minimizing squared residuals), you'll be prepared for conceptual and computational questions on the exam.
Sections
The least-squares method rests on a simple but powerful idea: given more observations than unknowns, find the unknown parameters that make the observed data 'fit' as well as possible. The measure of fit is the weighted sum of squared residuals, denoted Σwᵢvᵢ², where vᵢ is the residual (the correction needed to each observation) and wᵢ is its weight (reciprocal of variance). Why squared residuals? Because: 1. Squaring penalizes large errors more heavily than small ones, forcing the solution to avoid outliers. 2. The method produces unbiased estimates and, under normal error distribution, is statistically optimal (maximum likelihood). 3. It is mathematically tractable—the solution is linear in the observations. Why weights? In geodetic work, observations have different precisions. A distance measured with a high-precision EDM should carry more weight than one measured with a tape. Weights are inversely proportional to variance: wᵢ = 1/σᵢ². A weight matrix P (usually diagonal) incorporates all weights formally. The least-squares solution gives the most probable values under the assumption that random errors follow a normal (Gaussian) distribution with mean zero. This assumption is standard in geodetic practice and is validated by the reference variance check after adjustment.
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1. The Least-Squares Principle and Philosophy
Examples
- Example 1.1: Three independent measurements of a distance are 100.234 m, 100.241 m, and 100.238 m. Without additional information, the simple mean is (100.234 + 100.241 + 100.238)/3 = 100.238 m. If the first measurement came from a less precise instrument, a weighted mean would give it less influence. Least squares formalizes this weighting.
- Example 1.2: In a leveling network, all routes between two points have different horizontal distances and thus different standard deviations. Least squares combines all routes, weighting each by 1/distance, to get the best estimate of the height difference.
Key Points
- Minimize Σwᵢvᵢ² to find the most probable unknowns.
- Weights reflect observation precision: wᵢ = 1/σᵢ².
- Residuals are corrections: vᵢ = (computed value − observed value).
- Assumes random, normally distributed errors.
- Provides a statistical basis for error propagation and quality assessment.
In the observation-equation (parametric) model, each observation is related to unknowns through a functional equation. For the ith observation: lᵢ + vᵢ = fᵢ(x₁, x₂, ..., xᵤ) where lᵢ is the observed value, vᵢ is the residual (unknown correction), and xⱼ are the unknown parameters we wish to estimate. For a nonlinear function, we linearize about approximate values x₀ⱼ: fᵢ(x) ≈ fᵢ(x₀) + (∂fᵢ/∂x₁)·Δx₁ + (∂fᵢ/∂x₂)·Δx₂ + ... + (∂fᵢ/∂xᵤ)·Δxᵤ Denoting the partial derivatives as aᵢⱼ = ∂fᵢ/∂xⱼ and collecting into matrices: **v = A·Δx − l** where: - **v** is the n × 1 vector of residuals - **A** is the n × u design matrix (or Jacobian), with entry aᵢⱼ - **Δx** is the u × 1 vector of parameter corrections (what we solve for) - **l** is the n × 1 vector of "observed minus computed" (given the approximate values x₀) - n is the number of observations, u is the number of unknowns The weight matrix **P** is an n × n diagonal matrix: **P** = diag(w₁, w₂, ..., wₙ) where wᵢ = 1/σᵢ². For observations with equal precision, all weights are equal (or can be set to 1). The normal equations are derived by minimizing **v**ᵀ**P****v**: (**A**ᵀ**P****A**) · **Δx** = **A**ᵀ**P****l** Solving for the corrections: **Δx** = (**A**ᵀ**P****A**)⁻¹ **A**ᵀ**P****l** The adjusted parameter values are: **x̂** = **x₀** + **Δx**
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2. Observation Equations: The Matrix Formulation
Examples
- Example 2.1 (Simple 1-Unknown Case): A point's elevation H is observed three times: l₁ = 152.34 m, l₂ = 152.30 m, l₃ = 152.36 m. The functional equation for each is simply: lᵢ + vᵢ = H. Start with an approximate value H₀ = 152.30 m. Then l = [0.04, 0.00, 0.06]ᵀ (observed minus computed). The design matrix is **A** = [1, 1, 1]ᵀ (one column, since there's one unknown). With equal weights w = 1, the normal equation becomes: (AᵀA)·ΔH = Aᵀl, or 3·ΔH = 0.10, giving ΔH = 0.0333 m. Thus Ĥ = 152.30 + 0.0333 = 152.333 m.
- Example 2.2 (Two-Unknown Case): Suppose two unknowns (e.g., a horizontal shift ΔX and a vertical shift ΔZ) affect three observed distances. The design matrix A would be 3 × 2, with each row containing the partial derivatives of the respective distance function with respect to ΔX and ΔZ. The normal matrix AᵀPA becomes 2 × 2, solvable for two corrections.
Key Points
- Observation equation: lᵢ + vᵢ = fᵢ(x₁, ..., xᵤ).
- Linearize nonlinear functions using partial derivatives.
- Design matrix A contains partial derivatives (aᵢⱼ = ∂fᵢ/∂xⱼ).
- Weight matrix P is diagonal: P = diag(1/σ₁², 1/σ₂², ..., 1/σₙ²).
- Normal equations: (AᵀPA)·Δx = AᵀPl.
- Solution: Δx = (AᵀPA)⁻¹·AᵀPl.
- Final estimates: x̂ = x₀ + Δx.
A key concept in least-squares adjustment is redundancy (or degrees of freedom): **r = n − u** where n is the number of observations and u is the number of unknowns. • If r = 0 (n = u): The system has a unique solution, but no redundancy. There is no way to check for errors or assess the quality of the solution. This is called a critically determined system. • If r > 0 (n > u): The system is overdetermined. Least-squares adjustment is necessary and meaningful. You can assess internal consistency and detect blunders. • If r < 0 (n < u): The system is underdetermined. There are infinite solutions, and adjustment is impossible. In geodetic practice, redundancy is essential. It allows: 1. Detection of blunders or outliers in observations. 2. Statistical testing (reference variance) to check whether the assumed weights and errors are realistic. 3. Rigorous uncertainty estimates (variance-covariance matrices) for adjusted parameters. Example: A leveling network connecting four points requires three independent measurements (three heights relative to a datum) to determine all four elevations—thus u = 3. If you have 8 measurements, then r = 8 − 3 = 5 redundant observations. These five extras allow you to 'vote' on the best estimate and flag inconsistent data.
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3. Redundancy and Degrees of Freedom
Examples
- Example 3.1 (Leveling Network): Four points need elevations determined. A datum defines one (or one height difference is external). Thus u = 3 unknowns. If you have 6 independent leveling routes, r = 6 − 3 = 3 redundancies. You can use least squares and check for systematic errors.
- Example 3.2 (Traverse Adjustment): A closed traverse has 15 measurements (distances and angles). The unknowns are typically 2 scale factors, 1 rotation, and 2 translation shifts (5 unknowns), giving r = 15 − 5 = 10 redundancies. This high redundancy allows robust detection of blunders and reliable quality assessment.
Key Points
- Redundancy r = n − u; must have r > 0 for a meaningful adjustment.
- r = 0: Unique solution, no error check possible.
- r > 0: Overdetermined system; least squares applies and quality control is possible.
- r < 0: Impossible; system has insufficient constraints.
- Redundancy enables blunder detection and reference variance testing.
After solving for adjustments **Δx**, you compute the adjusted residuals: **v̂** = **A**·**Δx** − **l** The reference variance (or unit variance) is: **σ̂₀² = (v̂ᵀ·P·v̂) / r** where r = n − u is the redundancy. Interpretation: - If σ̂₀² ≈ 1 (or close to a baseline value if weights have been scaled), the observations fit the model well. The assumed standard deviations and weights are realistic. - If σ̂₀² >> 1, residuals are larger than expected. This may indicate blunders, model deficiencies, or weights that are too optimistic (σᵢ assumed too small). - If σ̂₀² << 1, residuals are smaller than expected. The weights may be too conservative (σᵢ assumed too large). A rough guideline: σ̂₀² should lie in the range 0.5 to 2 for a well-conditioned adjustment. Outside this range, investigate your observations and model. Uncertainties of adjusted parameters can then be estimated: **Cov(x̂)** = σ̂₀² · (**AᵀPA**)⁻¹ The standard deviation of the ith adjusted parameter is σ(x̂ᵢ) = √[Cov(x̂)]ᵢᵢ. This propagates the observational uncertainties through to the final estimates—a requirement under PRC standards and Philippine surveying regulations (RA 4374).
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4. The Reference Variance and Quality Assessment
Examples
- Example 4.1: After adjusting a network with r = 5 redundancies, you compute v̂ᵀ·P·v̂ = 5.2. Then σ̂₀² = 5.2 / 5 = 1.04 ≈ 1, indicating an excellent fit. The assumed observation precisions are validated.
- Example 4.2: A traverse adjustment yields σ̂₀² = 3.5, well above the acceptable range. This alerts you to possible gross errors, such as a misread angle or a distance contaminated by systematic error (e.g., temperature correction forgotten). Recheck the raw observations before accepting the adjustment.
- Example 4.3: Given Cov(x̂) from the normal matrix inverse, if the first diagonal entry is 0.000004 m², the standard deviation of the first parameter is σ(x̂₁) = √0.000004 = 0.002 m = 2 mm. This quantifies the positional accuracy achieved by the adjustment.
Key Points
- Reference variance: σ̂₀² = (v̂ᵀ·P·v̂) / (n − u).
- σ̂₀² ≈ 1 indicates good fit; >> 1 or << 1 signals problems.
- Variance-covariance matrix: Cov(x̂) = σ̂₀² · (AᵀPA)⁻¹.
- Use Cov(x̂) to estimate standard deviations of adjusted unknowns.
- Reference variance is a diagnostic tool for model validation.
One of the most important and frequently tested scenarios in geodetic adjustment is the single-unknown case. When you have one unknown (such as an elevation, a distance, or a single parameter) observed multiple times with different precisions, the least-squares solution reduces elegantly to the **weighted mean**. Setup: - One unknown: x (e.g., elevation H, distance d) - n observations: l₁, l₂, ..., lₙ - Weights: w₁, w₂, ..., wₙ (proportional to precision, or 1/σᵢ²) - Functional form: lᵢ + vᵢ = x (each observation estimates x with error vᵢ) The observation-equation matrix form is: - **A** = [1, 1, ..., 1]ᵀ (n × 1 column of ones) - **l** = [l₁, l₂, ..., lₙ]ᵀ - **P** = diag(w₁, w₂, ..., wₙ) The normal equation becomes: (**AᵀPA**) · x̂ = **AᵀP****l** (Σwᵢ) · x̂ = Σwᵢlᵢ Solving: **x̂ = Σwᵢlᵢ / Σwᵢ** This is the **weighted mean**. Special cases: - If all wᵢ = 1 (equal weights), x̂ = (Σlᵢ) / n, the simple arithmetic mean. - If weights are inversely proportional to distance (common in leveling), longer routes have less weight. - If weights are inversely proportional to variance, more precise observations are favored. For the variance of the weighted mean: σ(x̂)² = 1 / Σwᵢ The uncertainty decreases as you add more (or more precise) observations—a fundamental principle of error propagation. This reduction to the weighted mean is more than a computational trick; it validates the least-squares method. Examiners often ask, "What does least squares give you when there's only one unknown?" The answer: the weighted mean—simple, intuitive, and correct.
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5. The Single-Unknown Case: Weighted Mean
Examples
- Example 5.1 (Leveling from Three Routes): A benchmark's elevation is determined by three independent leveling routes: Route 1 (2 km): 152.342 m; Route 2 (1 km): 152.298 m; Route 3 (4 km): 152.364 m. Assuming weight is inversely proportional to distance (wᵢ ∝ 1/dᵢ), normalize weights to sum to 1: w₁ = 1/2 / (1/2 + 1/1 + 1/4) = 0.5 / 1.75 ≈ 0.286; w₂ = 1 / 1.75 ≈ 0.571; w₃ = 0.25 / 1.75 ≈ 0.143. Weighted mean: Ĥ = 0.286(152.342) + 0.571(152.298) + 0.143(152.364) = 43.57 + 86.96 + 21.79 = 152.32 m. (Exact: Ĥ ≈ 152.323 m.) Variance: σ(Ĥ)² = 1 / 1.75 ≈ 0.571, so σ(Ĥ) ≈ 0.756 mm.
- Example 5.2 (Distance from Four Tapes): A distance is measured four times with tapes of different ages and calibrations. Measurements: 100.234, 100.241, 100.238, 100.236 m with estimated standard deviations 2, 1, 1.5, 2 mm, respectively. Weights: w₁ = 1/4 = 0.25; w₂ = 1/1 = 1; w₃ = 1/2.25 ≈ 0.444; w₄ = 1/4 = 0.25; Σw = 1.944. Weighted mean: d̂ = [0.25(100.234) + 1(100.241) + 0.444(100.238) + 0.25(100.236)] / 1.944 ≈ 100.239 m. The most precise tape (w₂) dominates the result.
Key Points
- One unknown: x̂ = Σwᵢlᵢ / Σwᵢ (weighted mean).
- Equal weights: x̂ = (Σlᵢ) / n (arithmetic mean).
- Variance of weighted mean: σ(x̂)² = 1 / Σwᵢ.
- Weights wᵢ = 1/σᵢ² (inverse variance weighting).
- The single-unknown case is the most common practical application and board-exam scenario.
A systematic approach ensures accuracy and compliance with professional standards (PRC-governed under RA 4374, RA 8560): **Step 1: Define Unknowns and Observations** - Clearly identify all unknowns (u), e.g., coordinates, elevations, scale factors. - List all observations (n) and their measured values. - Confirm redundancy r = n − u > 0. **Step 2: Choose Approximate Values** - For each unknown, provide a reasonable initial estimate x₀ⱼ (from a quick calculation, previous survey, or external reference). - Linearization about these values is essential for nonlinear models. **Step 3: Compute the Design Matrix A** - For each observation i and unknown j, compute aᵢⱼ = ∂fᵢ/∂xⱼ evaluated at x₀. - Record the n × u matrix A with all partials. **Step 4: Compute Observed Minus Computed** - Using approximate values x₀, compute the 'computed' value of each observation: fᵢ(x₀). - Compute the residual vector: l = observation − computed = lᵢ − fᵢ(x₀). - Arrange into the n × 1 vector **l**. **Step 5: Assign Weights and Form P** - Determine or estimate the standard deviation σᵢ of each observation. - Compute weights: wᵢ = 1/σᵢ² (or normalize as needed). - Form the weight matrix: **P** = diag(w₁, w₂, ..., wₙ). **Step 6: Form and Solve Normal Equations** - Compute **N** = **AᵀPA** (u × u symmetric matrix). - Compute **b** = **AᵀPl** (u × 1 vector). - Solve **N·Δx** = **b** using Gaussian elimination, LU decomposition, or matrix inversion: **Δx** = **N⁻¹·b**. **Step 7: Compute Adjusted Parameters** - **x̂** = **x₀** + **Δx**. **Step 8: Compute Residuals and Reference Variance** - **v̂** = **A·Δx** − **l**. - σ̂₀² = (**v̂ᵀ·P·v̂**) / (n − u). - Check if σ̂₀² is close to 1. If not, investigate. **Step 9: Compute Variance-Covariance Matrix** - **Cov(x̂)** = σ̂₀² · **N⁻¹**. - Extract standard deviations: σ(x̂ᵢ) = √[Cov(x̂)]ᵢᵢ. **Step 10: Document and Report** - Report adjusted parameters x̂ with their standard deviations. - Report the reference variance σ̂₀². - Summarize the redundancy, residuals, and any notes on data quality. - Compliance: ensure documentation meets RA 4374 (Professional Regulation) and RA 8560 standards for accreditation and legal defensibility.
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6. Practical Steps for an Observation-Equation Adjustment
Examples
- Example 6.1 (Simple Benchmark Adjustment): A new benchmark's elevation is observed via three leveling routes. (This is a single-unknown case.) Step 1: u = 1 (elevation H), n = 3 (three routes), r = 2. Step 2: x₀ = 152.30 m (rough estimate). Step 3: A = [1, 1, 1]ᵀ (constant partial). Step 4: l = [0.04, 0.00, 0.06]ᵀ m (three observed values minus computed). Step 5: Weights w = [0.5, 1.0, 0.25] (routes of different lengths). Step 6: N = AᵀPA = 1.75; b = AᵀPl = 0.07; Δx = 0.07 / 1.75 ≈ 0.04 m. Step 7: Ĥ = 152.30 + 0.04 = 152.34 m. Step 8: Residuals are computed; σ̂₀² is checked. Final uncertainty σ(Ĥ) = √(1/1.75) ≈ 0.76 mm.
- Example 6.2 (Traverse Adjustment): A 5-point traverse has 15 measured angles and distances; 5 unknowns (2 coordinates + 3 nuisance parameters). r = 10. Follow steps 1–10: set approximate coordinates, compute partials (trigonometric functions), assign weights based on instrument precision, solve the 5 × 5 normal system, assess σ̂₀², and report final coordinates with mm-level uncertainties.
Key Points
- Ten-step systematic procedure ensures repeatable, defensible adjustments.
- Always confirm r > 0 before attempting adjustment.
- Choose reasonable x₀ to ensure convergence in nonlinear problems.
- Weights must reflect observational precision; inadequate weighting is a common error.
- The reference variance check validates the model and weights.
- Report uncertainties (standard deviations) alongside adjusted values.
- Professional documentation is required by PRC regulations.
**Pitfall 1: Incorrect Residual Sign** Definition matters. If you define v = computed − observed, you'll get opposite signs to v = observed − computed. Be consistent and state your convention clearly. The least-squares principle minimizes Σwᵢvᵢ² regardless of sign, but the matrix equation changes: - If v = observed − computed (as in these notes): **v** = **A·Δx** − **l**. - If v = computed − observed: **v** = **l** − **A·Δx** (and normal equations reverse sign). Both are correct if used consistently. On exams, show your definition. **Pitfall 2: Forgetting to Linearize** For nonlinear functional relationships (e.g., intersection of two circles, distance-angle relations), you must linearize using Taylor series. Using the original nonlinear equation without linearization will give wrong answers. Always expand ∂f/∂x around approximate values. **Pitfall 3: Weight Matrix Confusion** Don't confuse weights and variances: - Weight wᵢ = 1/σᵢ² (inverse of variance). - Variance σᵢ² (square of standard deviation). - Standard deviation σᵢ (uncertainty of observation i). If told an observation has σᵢ = 2 mm, then wᵢ = 1/4 mm⁻², not 2 or 1/2. Incorrect weighting leads to biased solutions. **Pitfall 4: Insufficient Redundancy** If r ≤ 0, the system is underdetermined or just-determined. You cannot use least squares; you cannot check for blunders. Always verify r > 0. **Pitfall 5: Ignoring the Reference Variance** A reference variance σ̂₀² = 10 is a red flag. Don't ignore it; investigate. Common causes: gross errors in observations, incorrect model (missing variables), or unrealistic weight assumptions. Professional practice (and the PRC) expect you to diagnose and resolve these issues. **Pitfall 6: Single-Unknown Confusion** When there is one unknown, least squares simplifies to the weighted mean. Don't over-formalize: if asked "adjust a single measurement with three observations," the answer is just x̂ = Σwᵢlᵢ / Σwᵢ. A full matrix solution is correct but unnecessary and time-wasting on an exam. **Pitfall 7: Rounding and Numerical Precision** Keep sufficient decimal places (at least 0.1 mm for coordinates, 1 mm for elevations) during intermediate steps. Premature rounding propagates errors. For normal matrix inversion, use double precision and be aware of ill-conditioning if AᵀPA is nearly singular. **Pitfall 8: Coordinate Systems and Datums** When adjusting coordinates under WGS84 or PRS92, ensure all observations and unknowns are in the same reference frame. Mixing datums or confusing PPCS/UTM zones is a common error on exams and in practice. Reference RA 4374 and PD 1529 for datum requirements. **Exam Tip 1: Board-Style Problems** Exam questions often give simple numerical cases (e.g., three observations of one unknown with specified weights). Draw up the design matrix, compute normal equations by hand, and solve. Show all steps. Examiners look for method, not just the final answer. **Exam Tip 2: Conceptual Understanding** Questions like "Explain why redundancy is necessary" or "What does the reference variance tell you?" test understanding. Be ready to explain not just 'how' but 'why.' A good answer explains the statistical reasoning and practical implications. **Exam Tip 3: Mixed Cases** Don't assume all problems are single-unknown. Some may involve two unknowns (e.g., X and Y coordinate shifts) or require you to set up (but not fully solve) a larger system. Know how to formulate A and interpret the normal equations even if exact numerical solution is beyond the scope. **Exam Tip 4: Practical Context** Problems may mention leveling networks, traverses, or intersection surveys. Recognize the problem type, identify the unknowns and observations, and apply the standard 10-step procedure. Context helps, but the mathematical approach is universal.
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7. Common Pitfalls and Exam Tips
Examples
- Example 7.1 (Sign Convention): Two people solve the same adjustment using opposite residual definitions. Person A gets Ĥ = 152.34 m, Person B also gets Ĥ = 152.34 m. The least-squares principle ensures the same final answer. But intermediate calculations (the design matrix setup, normal equations) differ. On an exam, this is acceptable as long as the convention is stated and used consistently.
- Example 7.2 (Nonlinear Model): Suppose you measure the angle α subtended by a baseline of known length L at an unknown distance d. The functional relation is d = L / (2 tan(α/2)). To linearize, compute ∂d/∂α and evaluate at an approximate d₀. Forgetting to do this and plugging the nonlinear formula directly into the normal equations will yield an incorrect answer.
- Example 7.3 (Weight Errors): An observation with σ = 5 mm should have weight w = 1/25 = 0.04. A student mistakenly uses w = 1/5 = 0.2, 'inflating' the observation's influence by a factor of 5. The resulting solution is biased toward that observation, violating least-squares optimality.
Key Points
- Define residual sign clearly and use it consistently throughout.
- Always linearize nonlinear functional relationships.
- Weight = 1/variance, not inverse of standard deviation alone.
- Confirm r > 0; if r ≤ 0, least squares is invalid.
- Single unknown → use weighted mean (fast and foolproof).
- Reference variance is diagnostic; investigate σ̂₀² >> 1 or << 1.
- Keep numerical precision high during computation.
- On exams, show all work; clarity and method matter as much as the answer.
- Understand PRC standards (RA 4374, RA 8560) regarding documentation and datum.
For rapid recall during exams, here is a condensed summary of essential formulas: **Observation-Equation Model:** - Observation i: lᵢ + vᵢ = fᵢ(x₁, ..., xᵤ) - Linearized form: **v** = **A·Δx** − **l** **Matrices:** - **A** (n × u): Design matrix; entry aᵢⱼ = ∂fᵢ/∂xⱼ - **l** (n × 1): Observed minus computed at x₀ - **P** (n × n): Weight matrix, P = diag(w₁, ..., wₙ), wᵢ = 1/σᵢ² - **v** (n × 1): Residuals (computed after solving) **Normal Equations:** - **N·Δx** = **b**, where **N** = **AᵀPA**, **b** = **AᵀPl** **Solution:** - **Δx** = **N⁻¹·b** = (**AᵀPA**)⁻¹·**AᵀPl** - **x̂** = **x₀** + **Δx** (adjusted parameters) **Residuals and Quality:** - **v̂** = **A·Δx** − **l** - σ̂₀² = (**v̂ᵀ·P·v̂**) / (n − u) (reference variance) - **Cov(x̂)** = σ̂₀² · **N⁻¹** (variance-covariance matrix) - σ(x̂ⱼ) = √[Cov(x̂)]ⱼⱼ (standard deviation of jth parameter) **Special Case: Single Unknown** - x̂ = Σwᵢlᵢ / Σwᵢ (weighted mean) - σ(x̂)² = 1 / Σwᵢ - σ(x̂) = 1 / √(Σwᵢ) **Redundancy:** - r = n − u (degrees of freedom) - Require r > 0 for least-squares adjustment **Key Relationships:** - Higher precision (smaller σᵢ) → higher weight wᵢ → larger influence on solution. - More observations → larger **A**, more equations, better overdetermination, smaller uncertainty on x̂. - Better weights (more realistic σᵢ) → σ̂₀² closer to 1. This summary is a reference checklist for problem-solving and self-grading.
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8. Matrix Formulation Summary and Quick Reference
Examples
- Example 8.1 (Quick Check): You've solved an adjustment and computed σ̂₀² = 0.95. Is this good? Yes—it's very close to 1, indicating the model and weights fit the data well. If you had gotten σ̂₀² = 5, you'd need to recheck observations and assumptions.
- Example 8.2 (Weighted Mean Verification): Three height readings: 100.0, 100.2, 100.1 m with weights 1, 2, 1. Quick: ĥ = (1×100.0 + 2×100.2 + 1×100.1) / (1+2+1) = 300.5 / 4 = 100.125 m. Using the full matrix approach with A = [1,1,1]ᵀ, P = diag(1,2,1), l = [0, 0.2, 0.1], you'd get the same answer—confirming understanding.
- Example 8.3 (Uncertainty Propagation): After adjusting a network, you find σ(x̂₁) = 0.05 m (50 mm) and σ(x̂₂) = 0.02 m (20 mm). These are the positional uncertainties achievable by the survey. They're reported in your technical report and define the map's accuracy under RA 4374.
Key Points
- v = A·Δx − l (linearized observation equation).
- N = AᵀPA, b = AᵀPl; solve N·Δx = b.
- x̂ = x₀ + Δx (final adjusted parameters).
- σ̂₀² = (v̂ᵀ·P·v̂) / (n − u) diagnoses fit.
- Cov(x̂) = σ̂₀² · N⁻¹ gives parameter uncertainties.
- Single unknown simplifies to weighted mean.
- Redundancy r = n − u must be > 0.
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