GELE Adjustment Computations (Least Squares) — Least Squares — Observation EquationsMisconception Buster
Mistake patterns in Least Squares — Observation Equations — the trap questions GELE sets and the wrong assumptions reviewers make. This page walks through each misconception, why it is wrong, and how Professional Regulation Commission (PRC) — Board of Geodetic Engineering turns it into a tempting but incorrect answer choice.
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 - Misconception Buster
Adjustment Computations is one of the most technically demanding subjects in the PRC Geodetic Engineer Licensure Examination. Students who misunderstand the least-squares observation-equation method often lose critical marks on problems that appear straightforward. This guide targets the exact wrong beliefs that cause exam failures — from sign-convention errors in residuals to misidentifying when a problem reduces to a simple weighted mean. Each misconception is paired with a trap question modeled after actual board-exam item styles. Master these corrections and you protect yourself against the most common sources of lost marks in this subject.
Summary
The following are the most critical takeaways for avoiding lost marks in Adjustment Computations — Observation Equations on the PRC Geodetic Engineer Licensure Examination: (1) Always use v = computed − observed (not observed minus computed) in the parametric formulation. (2) Redundancy r = n − u must be positive (r ≥ 1) for a valid least-squares adjustment; n = u gives an exact, un-adjustable solution and σ̂₀² is undefined. (3) In differential leveling, weights are inversely proportional to route length: w ∝ 1/K — longer routes get lower weight. (4) The design matrix A contains partial derivatives (the Jacobian), not raw observations; the discrepancy vector l = ℓ(observed) − f(x₀), not the raw observations themselves. (5) A large reference variance σ̂₀² signals blunders or wrong weights — not a computational error in the normal equations. (6) The least-squares solution reduces to the weighted mean only for a single unknown with direct observations; multiple unknowns always require solving the full normal equations. (7) The observation-equation and condition-equation methods are mathematically equivalent — identical data and weights must produce identical results. (8) The weight matrix P is diagonal only for independent observations; correlated observations (e.g., GPS baselines from the same session) require the full inverse covariance matrix. Internalizing these eight principles will protect you against the most common sources of wrong answers in this subject area.
Misconceptions
The residual v is defined as (observed − computed), so a larger observed value always gives a positive residual.
Tags
- sign_convention
- common_error
- formula_confusion
Topic
Residual Definition and Sign Convention
Severity
critical
Exam Impact
A sign reversal on v propagates into v^T P v, giving a numerically correct magnitude but possibly wrong sign checks, and the residual vector reported in a solution will be the negative of the expected answer — directly costing marks on show-solution items.
The Reality
In the standard observation-equation (parametric) formulation used in Philippine geodetic engineering practice, the residual is defined as v = (computed − observed), i.e., v = Ax̂ − l. This sign convention is required so that the normal equations A^T P A x̂ = A^T P l are derived correctly. Reversing the sign flips the residual vector and can give a wrong reference variance. Always check the convention used in your derivation before substituting.
Trap Question
Question
An elevation of a benchmark is observed as 152.34 m. The least-squares adjusted value is 152.30 m. Using the standard observation-equation convention, what is the residual v for this observation?
Explanation
In the parametric least-squares model, v = computed − observed = 152.30 − 152.34 = −0.04 m. The adjusted (computed) value is smaller than what was observed, so the residual is negative. Students who memorize 'observed minus true' from basic surveying get the wrong sign.
Wrong Answer
v = 152.34 − 152.30 = +0.04 m
Correct Answer
v = 152.30 − 152.34 = −0.04 m
Misconception Id
M1
Correct Vs Incorrect
Correct Approach
v_i = (A x̂)_i − ℓ_i (computed minus observed). The normal equations are derived from minimizing v^T P v under this definition. Always write the observation equation as ℓ + v = f(x̂), so v = f(x̂) − ℓ.
Incorrect Approach
v_i = ℓ_i − (A x̂)_i → student reports v = +0.04 m when it should be −0.04 m.
Why Students Believe It
Students carry over the intuitive idea from basic surveying error analysis where 'error = measured minus true.' They apply this casually to the least-squares context without checking how their textbook or derivation defines the sign convention.
If there are exactly as many observations as unknowns (n = u), you can still perform a least-squares adjustment.
Tags
- redundancy
- conceptual_gap
- degrees_of_freedom
Topic
Redundancy and Degrees of Freedom
Severity
critical
Exam Impact
Board questions often ask for the degrees of freedom or whether a network can be adjusted. Answering 'yes, adjustment is possible' when n = u is a direct wrong answer. Also, computing σ̂₀² with n − u = 0 in the denominator is mathematically undefined.
The Reality
Least-squares adjustment requires redundant observations: n > u (number of observations greater than number of unknowns). When n = u, the system is exactly determined — there is a unique solution but no redundancy, so the matrix A^T P A may be square and invertible, but there is zero degree of freedom. This means you cannot compute a meaningful reference variance σ̂₀² = v^T P v / (n − u) because the denominator is zero. All residuals are forced to zero. There is nothing to 'adjust' — the solution is exact, not a least-squares estimate.
Trap Question
Question
A leveling network has 6 height-difference observations and 6 unknown benchmark elevations. A geodetic engineer sets up the normal equations and solves for x̂. He then computes σ̂₀² = v^T P v / (6 − 6). What is wrong with this procedure?
Explanation
The reference variance σ̂₀² = v^T P v / (n − u) is only meaningful when n − u ≥ 1. With n = u, the system is exactly determined, residuals are all zero by definition, and the formula yields 0/0 — undefined. The network needs at least one additional observation to have redundancy.
Wrong Answer
Nothing is wrong; the procedure is valid and σ̂₀² = 0 means a perfect fit.
Correct Answer
The denominator (n − u = 0) makes σ̂₀² undefined. With zero redundancy, all residuals are forced to zero — this is not a least-squares adjustment but an exact solution. No quality check on the observations is possible.
Misconception Id
M2
Correct Vs Incorrect
Correct Approach
Redundancy r = n − u = 8 − 8 = 0. No adjustment is possible. The solution is unique (exact), all residuals = 0, and σ̂₀² is undefined. For a valid least-squares adjustment, you need n > u, i.e., r ≥ 1.
Incorrect Approach
Network has 8 observations and 8 unknowns → 'We can do least squares, all formulas apply.' Student proceeds to compute v^T P v / 0 and gets a math error or undefined answer.
Why Students Believe It
Students see the normal-equation formula x̂ = (A^T P A)^{-1} A^T P l and assume it always works as long as A is square and invertible. They do not connect the concept of 'redundancy' to the feasibility of adjustment.
The weight of an observation is directly proportional to the distance (or route length) in leveling — longer routes get higher weight.
Tags
- weighting
- common_error
- leveling
- formula_confusion
Topic
Weighting of Observations — Leveling
Severity
critical
Exam Impact
Assigning weights proportional (instead of inversely proportional) to distance completely reverses the weighting scheme, giving the least reliable observation the most influence on the adjusted value. This is a guaranteed wrong answer in weighted-mean leveling problems on the board exam.
The Reality
In differential leveling, the standard deviation of a route's misclosure grows with route length K (in km) because errors accumulate. Therefore σ² ∝ K, which means weight w = 1/σ² ∝ 1/K. Longer routes are LESS reliable and receive LOWER weight. This is the PPCS/PRS92 conventional practice for geodetic leveling networks in the Philippines: w_i = 1/K_i (or w_i = c/K_i for any common constant c).
Trap Question
Question
Three leveling routes give the elevation of BM-Pasig as 18.450 m (K = 1 km), 18.460 m (K = 2 km), and 18.440 m (K = 4 km). What is the weighted-mean elevation?
Explanation
Weight in leveling is inversely proportional to route length (w ∝ 1/K) because uncertainty accumulates with distance. The 1 km route is most reliable and dominates the weighted mean. Using w = K instead of w = 1/K gives the wrong answer of 18.447 m.
Wrong Answer
Using w = K: w = 1, 2, 4; Σw = 7; Ĥ = [1(18.450) + 2(18.460) + 4(18.440)] / 7 = 128.930/7 ≈ 18.447 m
Correct Answer
Using w = 1/K: w = 1, 0.5, 0.25; Σw = 1.75; Ĥ = [1(18.450) + 0.5(18.460) + 0.25(18.440)] / 1.75 = [18.450 + 9.230 + 4.610] / 1.75 = 32.290 / 1.75 ≈ 18.451 m
Misconception Id
M3
Correct Vs Incorrect
Correct Approach
Route A: 2 km → w_A = 1/2 = 0.50; Route B: 4 km → w_B = 1/4 = 0.25. Route B is longer, so its measurement is less precise, hence lower weight.
Incorrect Approach
Route A: 2 km → w_A = 2; Route B: 4 km → w_B = 4. Student gives Route B more weight because it is longer.
Why Students Believe It
Students reason that a 'longer' survey represents more effort and therefore more trustworthy data, confusing the physical length with reliability. The word 'proportional' in the weighting rule is memorized without understanding its direction.
The design matrix A contains the actual observed values, not the partial derivatives of the observation equations.
Tags
- design_matrix
- conceptual_gap
- matrix_setup
Topic
Design Matrix Construction
Severity
critical
Exam Impact
Incorrectly populating A with raw observations instead of partial derivatives produces a completely wrong normal-equation system. The resulting x̂ is meaningless. This is a fundamental setup error that makes the entire solution wrong.
The Reality
The design matrix A (also called the coefficient matrix or Jacobian) contains the partial derivatives of each observation equation with respect to each unknown, evaluated at the approximate values x₀. The vector l contains the discrepancy (observed minus computed at x₀), not the raw observations themselves. For a linear observation equation f(x) = a₁x₁ + a₂x₂, the row of A is [a₁, a₂] and l = ℓ_observed − f(x₀). A never holds raw observation values.
Trap Question
Question
An observation equation is: Elevation of Point P = 100.00 + Δh, where Δh is observed as 5.23 m and the approximate elevation of P is 105.00 m. What is the correct entry in the design matrix A and the discrepancy vector l for this observation?
Explanation
A contains the partial derivative (here = 1, since the equation is linear in the unknown). l is the observed value minus the computed value at the approximation: l = 5.23 − 5.00 = 0.23 m. The raw observed value 5.23 goes into the computation of l, not directly into A.
Wrong Answer
A = [5.23], l = [100.00]
Correct Answer
A = [1] (partial derivative of the observation equation with respect to the unknown correction δx), l = 5.23 − (105.00 − 100.00) = 5.23 − 5.00 = 0.23 m
Misconception Id
M4
Correct Vs Incorrect
Correct Approach
Observation equation: H_B − H_A = Δh₁₂. If H_B is the unknown (x₁) and H_A is known, then ∂f/∂x₁ = 1. So the A row = [1] and l = Δh₁₂(observed) − (H_B₀ − H_A) using the approximate value H_B₀.
Incorrect Approach
Observation: H_B = H_A + Δh₁₂; observed Δh₁₂ = 3.45 m. Student writes A row = [3.45] and l = [H_A approximate]. Wrong — A holds partials, not observed values.
Why Students Believe It
When students first see the matrix equation v = Ax̂ − l, they confuse 'A' with the coefficient matrix of the raw observations. In simple linear systems from algebra, the coefficient matrix directly holds numerical coefficients from the equations — students transfer this idea incorrectly.
Least squares gives the 'exact true value' of the unknowns — the adjusted values are free of error.
Tags
- conceptual_gap
- statistical_interpretation
- precision
Topic
Interpretation of Adjusted Values
Severity
major
Exam Impact
Questions on precision of adjusted values, standard errors of adjusted coordinates, and confidence intervals require understanding that x̂ still has uncertainty. Treating adjusted values as exact leads to wrong answers on error-propagation and quality-assessment questions.
The Reality
Least squares gives the Best Linear Unbiased Estimates (BLUE) under the Gauss-Markov theorem — they are the most probable values given the observations and the assumed error model. But they still carry uncertainty. The covariance matrix of the adjusted parameters is Σ_x̂ = σ̂₀² (A^T P A)^{-1}, and each adjusted value has a standard deviation that can be computed from the diagonal of Σ_x̂. The adjusted values are the best we can do, not error-free truths.
Trap Question
Question
After a least-squares adjustment of a geodetic leveling network (PRS92), the adjusted elevation of BM-Manila is reported as 5.234 m. A student concludes that this elevation is now exact and no uncertainty needs to be reported. Is this correct?
Explanation
Least squares minimizes v^T P v and produces BLUE estimates under Gauss-Markov conditions, but these estimates are not exact. The precision is quantified by the covariance matrix of the adjusted parameters. Reporting adjusted values without their standard errors is incomplete and does not meet geodetic survey reporting standards.
Wrong Answer
Yes, least squares gives the most probable value, so the adjusted elevation is the true value and needs no uncertainty statement.
Correct Answer
No. The adjusted elevation 5.234 m is the most probable estimate, but it still has an associated standard deviation computed from σ̂₀² (A^T P A)^{-1}. A proper survey report must include the standard error of the adjusted elevation.
Misconception Id
M5
Correct Vs Incorrect
Correct Approach
After adjustment, the elevation of BM-7 is 234.567 m ± σ_Ĥ, where σ_Ĥ = σ̂₀ × √[(A^T P A)^{-1}_{ii}]. The adjusted value is the most probable estimate but still has quantifiable uncertainty.
Incorrect Approach
After adjustment, the elevation of BM-7 is 234.567 m — this is now exact. No further uncertainty exists.
Why Students Believe It
The phrase 'most probable values' sounds absolute and definitive. Students interpret 'adjustment' as 'correction to the truth,' not realizing that the method minimizes a statistical criterion and still produces estimates with uncertainty.
A large reference variance σ̂₀² always means the adjustment failed or the data is bad.
Tags
- statistical_interpretation
- reference_variance
- conceptual_gap
Topic
Reference Variance Interpretation
Severity
major
Exam Impact
Exam questions ask students to interpret σ̂₀² and diagnose network problems. Saying 'the adjustment failed' when the correct answer is 'the weights are too optimistic' or 'a blunder is suspected' loses marks on diagnostic and interpretation questions.
The Reality
The reference variance σ̂₀² = v^T P v / (n − u) is a statistical test of the assumed weighting scheme against the actual residuals. A value significantly greater than the expected value (e.g., σ̂₀² >> 1 when weights are set so σ₀ = 1) flags one or more of: (a) blunders (gross errors) in observations, (b) weights that are too optimistic (underestimated σ_i), (c) a misspecified observation model. It does NOT mean the adjustment algorithm failed — it means the data or the weight model needs re-examination. Conversely, σ̂₀² << 1 suggests weights are too pessimistic.
Trap Question
Question
After adjusting a GPS baseline network, σ̂₀² = 6.2. A geodetic engineer concludes that the normal equations were solved incorrectly. Is this conclusion correct? What does σ̂₀² = 6.2 actually indicate?
Explanation
The reference variance is a goodness-of-fit statistic for the weighting model. A value >> 1 triggers investigation of blunders (using methods like data snooping) or revision of the weight matrix. The normal equations may have been solved perfectly correctly and still yield σ̂₀² ≠ 1.
Wrong Answer
Yes, the computation is wrong because σ̂₀² should equal 1 for a correct least-squares adjustment.
Correct Answer
No. σ̂₀² = 6.2 indicates that the actual residuals are larger than what the assumed weights predict. This points to possible blunders in the observations, over-optimistic precision assumptions, or model errors — not a computational mistake in the normal equations.
Misconception Id
M6
Correct Vs Incorrect
Correct Approach
σ̂₀² = 4.8 >> 1 → 'The reference variance is significantly larger than the a priori value of 1. This suggests either a blunder in one or more observations, or that the observation weights (precisions) were over-estimated. Investigate outlier residuals using data-snooping tests.'
Incorrect Approach
σ̂₀² = 4.8 → 'The least-squares algorithm produced an incorrect result; redo the computation.'
Why Students Believe It
Students see σ̂₀² as a pass/fail indicator. Since a value 'near 1' is described as good, any large value is interpreted as a failure of the adjustment process itself.
When all observations have equal weight, the least-squares solution is always the simple arithmetic mean, regardless of the number of unknowns.
Tags
- weighted_mean
- conceptual_gap
- multiple_unknowns
Topic
Weighted Mean vs. Full Least-Squares Solution
Severity
major
Exam Impact
Students who apply the arithmetic mean to a two-unknown problem (e.g., adjusting both easting and northing corrections simultaneously) get completely wrong answers. This is a common trap in problems involving multiple unknowns with equal-weight observations.
The Reality
The least-squares solution reduces to the simple arithmetic mean ONLY when (1) there is a single unknown AND (2) all observations directly and equally measure that unknown AND (3) all weights are equal. As soon as there are two or more unknowns, or observations are combinations of unknowns (e.g., angle observations in a traverse), the normal equations must be set up and solved — even if all weights are equal. Equal weights simply mean P = I (the identity matrix), but the system A^T A x̂ = A^T l still needs to be solved.
Trap Question
Question
Four observations of two unknown elevations H_A and H_B are made with equal weights. Can you find the least-squares solution by simply taking the arithmetic mean of the observations that mention H_A and separately those that mention H_B?
Explanation
Equal weights (P = I) simplify the weight matrix but do not eliminate the need to solve the normal equations when multiple unknowns are involved. The system A^T A x̂ = A^T l must be solved. Only the special case of n independent observations of one unknown with equal weights reduces to the arithmetic mean.
Wrong Answer
Yes. Equal weights mean P = I, so the arithmetic mean of each group gives the least-squares estimate of each unknown.
Correct Answer
No. With two unknowns and mixed observations (each observation may involve both H_A and H_B), you must set up and solve the full normal equations A^T A x̂ = A^T l. The arithmetic mean reduction only applies when each observation exclusively measures a single unknown.
Misconception Id
M7
Correct Vs Incorrect
Correct Approach
Set up three observation equations: α + k = ℓ₁, α + k = ℓ₂, α + k = ℓ₃ (if k affects all). With P = I, normal equations A^T A x̂ = A^T l yield the correct system. For a true single-unknown case (k = 0), the arithmetic mean is correct, but only then.
Incorrect Approach
Three angle observations of angle α (equal weights): 45°10′20″, 45°10′24″, 45°10′18″. Two unknowns: α and a systematic correction k. Student takes arithmetic mean of all three observations and assigns result to α.
Why Students Believe It
Students learn the special case 'one unknown, equal weights → arithmetic mean' and overgeneralize it to all least-squares problems with equal weights. They forget that the arithmetic mean only applies when each observation directly measures the same single quantity.
The weight matrix P must always be diagonal — off-diagonal terms are always zero.
Tags
- weight_matrix
- correlated_observations
- conceptual_gap
Topic
Weight Matrix — General Form
Severity
minor
Exam Impact
Board exam problems almost universally assume independent observations, so diagonal P is correct in test contexts. However, theory questions may ask about the general form of P, and answering 'always diagonal' is wrong.
The Reality
P is diagonal only when observations are statistically independent (zero correlation). In practice, correlated observations (e.g., GPS baselines derived from the same session, or repeated measurements with systematic effects) have non-zero off-diagonal terms in the covariance matrix Σ_ℓ, giving a full (non-diagonal) P = σ₀² Σ_ℓ^{-1}. Using a diagonal P when correlations exist introduces bias into the adjustment. For board exam purposes, most problems assume independence (diagonal P), but knowing when this breaks down is essential for professional practice.
Trap Question
Question
In which of the following cases is it most appropriate to use a non-diagonal weight matrix P in a least-squares adjustment? (A) Independent leveling routes of different lengths; (B) GPS baselines derived from the same reference station in the same observing session; (C) Independently measured horizontal angles with different standard deviations; (D) Unweighted distance observations from different instruments.
Explanation
A non-diagonal P (equivalently, non-diagonal Σ_ℓ) is required when observations are correlated. Different weights on independent observations still give a diagonal P. GPS baselines from the same session are a classic source of correlated observations in geodetic network adjustment.
Wrong Answer
(A) — longer routes have different weights so P must be non-diagonal.
Correct Answer
(B) — GPS baselines from the same session share common satellite geometry and atmospheric conditions, introducing correlations. The full covariance matrix Σ_ℓ is non-diagonal, so P = σ₀² Σ_ℓ^{-1} is also non-diagonal.
Misconception Id
M8
Correct Vs Incorrect
Correct Approach
P is diagonal when observations are independent. In the general case, P = σ₀² Σ_ℓ^{-1}, where Σ_ℓ is the full covariance matrix of the observations. For independent observations with equal precision, P = σ₀²/σᵢ² × I.
Incorrect Approach
'The weight matrix P is always diagonal because each observation has one weight.' — Student cannot answer theory questions about correlated observations.
Why Students Believe It
Introductory textbooks present P as a diagonal matrix because independent observations are assumed. Students memorize this and never consider the case of correlated observations, which is common in real geodetic network adjustments.
The number of unknowns u in a least-squares adjustment always equals the number of unknowns explicitly listed in the problem.
Tags
- datum_deficiency
- rank_deficiency
- network_design
- conceptual_gap
Topic
Datum Deficiency and Network Configuration
Severity
major
Exam Impact
Problems asking for degrees of freedom in a free network are traps. Simply computing n − u gives wrong results if the datum deficiency is not accounted for. This also affects whether A^T P A is invertible.
The Reality
The design matrix A must have full column rank for (A^T P A) to be invertible. If the network is datum-deficient (no fixed control points), the rank of A is less than u, and the system has infinitely many solutions. In Philippine geodetic practice under RA 8560 and NAMRIA standards, at least one fixed point (or minimum constraint) must be introduced to define the datum. The effective number of independent unknowns is u minus the datum deficiency d. Redundancy = n − (u − d) = n − u + d for a minimum-constrained network.
Trap Question
Question
A 2D horizontal control network has 12 direction observations and 5 unknown station positions (10 coordinate unknowns). A student computes redundancy = 12 − 10 = 2. What critical factor has the student overlooked?
Explanation
In a 2D geodetic network, the datum defines the origin (2 parameters), orientation (1 parameter), and in some cases scale (1 parameter). These datum parameters are not observable from direction/distance measurements alone and must be fixed externally. Ignoring datum deficiency leads to a singular normal-equation matrix.
Wrong Answer
Nothing. Redundancy = n − u = 12 − 10 = 2 is correct.
Correct Answer
The student ignored the datum deficiency. A 2D network requires a minimum of 4 constraints (fix 2 coordinates of one station + 1 azimuth + 1 scale, or equivalently fix both coordinates of two stations) to define the datum. Without fixing these, A^T P A is rank-deficient. With minimum constraints applied, the effective redundancy is correctly computed.
Misconception Id
M9
Correct Vs Incorrect
Correct Approach
A 2D free network has datum deficiency d = 4 (2 translations + 1 rotation + 1 scale in a similarity transformation, or 3 for a rigid body). Effective redundancy = 10 − 8 + d. If minimum constraints are applied (fixing one point and one azimuth), d = 0 and redundancy = 10 − 8 = 2 for that constrained network.
Incorrect Approach
A 2D horizontal network has 10 angle observations and 4 unknown points (8 coordinates). Redundancy = 10 − 8 = 2. Student does not check if the network is datum-defined.
Why Students Believe It
Students count the unknowns as stated in the problem narrative without accounting for datum defects. A free network (no fixed points) has a datum deficiency that reduces the effective number of solvable unknowns — or equivalently, A^T P A becomes rank-deficient and cannot be inverted.
Minimizing the sum of residuals (Σvᵢ = 0) is the same as least squares (minimizing Σwᵢvᵢ²).
Tags
- criterion
- common_error
- weighted_mean
- formula_confusion
Topic
Least-Squares Criterion vs. Summation of Residuals
Severity
major
Exam Impact
Students who use Σv = 0 as their primary adjustment criterion instead of minimizing Σwv² will produce wrong results in all weighted problems. The correct approach is always to minimize the weighted sum of squares.
The Reality
Σvᵢ = 0 is a property specific to the unweighted arithmetic mean — it means the residuals sum to zero. Least squares minimizes Σwᵢvᵢ², which is a completely different criterion. In the weighted case, Σwᵢvᵢ may not equal zero (depending on the constraint). Even in unweighted cases, Σv = 0 is merely a consequence of the least-squares solution for one unknown, not the defining criterion. The correct criterion is the minimization of the quadratic form v^T P v.
Trap Question
Question
Three weighted observations of a distance d are: 100.05 m (w=1), 100.10 m (w=2), 100.08 m (w=1). A student adjusts by finding d̂ such that v₁ + v₂ + v₃ = 0. Is this approach correct for least squares?
Explanation
Least squares minimizes Σwᵢvᵢ², not Σvᵢ. The weighted mean correctly applies this criterion. Setting Σv = 0 without weights gives the unweighted arithmetic mean = 100.077 m, which ignores the differential precision of the observations.
Wrong Answer
Yes, Σv = 0 is the least-squares condition.
Correct Answer
No. The least-squares condition is to minimize Σwᵢvᵢ² = 1(d̂−100.05)² + 2(d̂−100.10)² + 1(d̂−100.08)². The correct solution is the weighted mean: d̂ = [1(100.05)+2(100.10)+1(100.08)] / (1+2+1) = 400.33/4 = 100.083 m. The Σv = 0 criterion gives a different (wrong) answer when weights differ.
Misconception Id
M10
Correct Vs Incorrect
Correct Approach
Minimize Φ = v^T P v = Σwᵢvᵢ² with respect to the unknowns. Setting ∂Φ/∂x̂ = 0 gives the normal equations A^T P A x̂ = A^T P l. The condition Σv = 0 may or may not hold depending on the problem structure.
Incorrect Approach
To adjust, set Σvᵢ = 0 and solve for the unknown. This works for the arithmetic mean but fails for weighted problems or multi-unknown systems.
Why Students Believe It
Students remember that 'errors cancel out' in random-error theory, and that Σv = 0 is a property of the arithmetic mean. They confuse this necessary condition of the mean with the least-squares criterion, thinking both methods are equivalent.
The observation-equation method and the condition-equation method always give different adjusted values.
Tags
- method_comparison
- conceptual_gap
- equivalence
Topic
Observation-Equation vs. Condition-Equation Methods
Severity
minor
Exam Impact
Theory questions asking whether the two methods give the same result are common. Answering 'different results' loses marks. Also, if a student's computation gives different results from both methods on the same data, it signals a computational error somewhere.
The Reality
Both the observation-equation (parametric) method and the condition-equation (conditional) method are equivalent formulations of the same least-squares principle. When applied correctly to the same network with the same weights, they must yield identical adjusted observations, identical residuals, and identical reference variance. The choice between them is purely one of computational convenience — condition equations can be more efficient for highly redundant networks with few unknowns, while observation equations are more systematic and easier to program.
Trap Question
Question
A student adjusts a leveling loop using the observation-equation method and obtains a reference variance σ̂₀² = 1.8. Another student adjusts the same loop using the condition-equation method and obtains σ̂₀² = 2.3. Which student made a computational error?
Explanation
The observation-equation and condition-equation methods are dual formulations of the same least-squares problem. They are mathematically equivalent and must produce identical adjusted values, residuals, and reference variance. Any numerical discrepancy indicates a mistake in one or both computations.
Wrong Answer
Both results can be correct because the two methods use different mathematical approaches and may give different reference variances.
Correct Answer
At least one student made a computational error. Both methods, applied correctly to the same data and weights, must yield the same residuals and the same reference variance. The discrepancy means at least one solution is wrong.
Misconception Id
M11
Correct Vs Incorrect
Correct Approach
If both methods are applied correctly to the same data with the same weights, they must yield Ĥ = 152.32 m (or the same value). A discrepancy between the two methods signals a computational error in one or both solutions.
Incorrect Approach
'I used the observation-equation method and got Ĥ = 152.32 m. My classmate used the condition-equation method and got 152.30 m. Both can be correct because they use different formulations.'
Why Students Believe It
The two methods (parametric/observation-equation vs. conditional/condition-equation) look very different mathematically, so students assume they yield different numerical results.
In the observation equation v = Ax̂ − l, the vector l contains the raw observed values.
Tags
- discrepancy_vector
- linearization
- common_error
- formula_confusion
Topic
Discrepancy Vector l — Correct Computation
Severity
critical
Exam Impact
Substituting raw observations into l instead of discrepancies gives completely wrong normal equations in nonlinear problems. For the common linear case in leveling, the distinction is less critical, but for angle-based or GPS problems (which involve nonlinear observation functions), this error invalidates the entire solution.
The Reality
In the linearized observation-equation model, l is the discrepancy vector, not the raw observations. Specifically, l_i = ℓ_i(observed) − f_i(x₀), where f_i(x₀) is the observation function evaluated at the initial approximate parameter values x₀. The unknown vector x̂ then represents corrections to x₀, not the parameters themselves. After solving, the adjusted parameters are x̂_final = x₀ + x̂. If the observation equations are exactly linear (no nonlinearity), then x₀ = 0 is a valid choice, and l reduces to the raw observations — but this is a special case, not the general rule.
Trap Question
Question
In a trilateration network, the approximate distance between two stations is computed as 5,000.00 m. The observed distance is 5,000.85 m. What is the correct entry in the discrepancy vector l for this observation?
Explanation
In the linearized parametric model, l = ℓ(observed) − f(x₀). The vector x̂ solved from the normal equations gives corrections to x₀. After solving, the final adjusted parameter = x₀ + x̂. Using the raw observation (5,000.85 m) as l instead of the discrepancy (0.85 m) is incorrect in the linearized formulation.
Wrong Answer
l = 5,000.85 m (the raw observed value)
Correct Answer
l = 5,000.85 − 5,000.00 = 0.85 m (observed minus computed at approximate values)
Misconception Id
M12
Correct Vs Incorrect
Correct Approach
Using approximate coordinates, compute d₁₂(approximate) = 1523.12 m. Then l = d₁₂(observed) − d₁₂(approximate) = 1523.45 − 1523.12 = 0.33 m. The vector x̂ gives the corrections to the approximate coordinates, not the coordinates themselves.
Incorrect Approach
Observation: measured distance d₁₂ = 1523.45 m. Student writes l = [1523.45] directly, treating the raw observation as the discrepancy vector.
Why Students Believe It
Students see 'l' described as 'the observation vector' in some textbooks and directly substitute the raw field measurements into l without computing the discrepancy from approximate values.
Quick Self Check
In the standard parametric formulation, v = computed − observed (i.e., v = Ax̂ − l). This sign convention is needed to derive the normal equations correctly. The residual is positive when the computed value exceeds the observed value.
Statement
The residual in the standard observation-equation method is defined as v = observed − computed.
When n = u, the redundancy is zero (n − u = 0). The system is exactly determined, all residuals are forced to zero, and σ̂₀² = 0/0 is undefined. A least-squares adjustment (with a meaningful reference variance) requires n > u.
Statement
A leveling network with n = 10 observations and u = 10 unknown benchmark elevations can be rigorously adjusted using least squares, and the reference variance can be computed.
Weight w ∝ 1/K for leveling, because the variance of a route's misclosure grows with distance (σ² ∝ K). Longer routes are less precise and must have lower weights.
Statement
In differential leveling, observations from longer routes (larger K in km) should receive smaller weights in a weighted-mean adjustment.
A is the Jacobian matrix of partial derivatives: A_{ij} = ∂f_i/∂x_j evaluated at x₀. It never holds raw observation values. This is the fundamental structure of the linearized least-squares model.
Statement
The design matrix A in the observation-equation method contains the partial derivatives of the observation equations with respect to the unknowns, evaluated at the approximate parameter values.
A large σ̂₀² indicates that residuals are larger than the weighting scheme predicts — pointing to blunders, over-optimistic weights, or model errors. The normal equations may have been solved perfectly correctly. Diagnosing a large σ̂₀² requires data snooping and weight model review, not recomputing the normal equations.
Statement
If σ̂₀² is significantly greater than 1 after a least-squares adjustment, it proves that the normal equations were solved incorrectly.
Both methods are mathematically equivalent formulations of the same least-squares problem. They differ only in computational approach. Identical data and weights must produce identical adjusted values, residuals, and σ̂₀².
Statement
The observation-equation method and the condition-equation method, when applied correctly to the same data with the same weights, must yield identical adjusted values and the same reference variance.
l_i = ℓ_i(observed) − f_i(x₀). This discrepancy vector, combined with the design matrix A of partial derivatives, gives the linearized system. The solved x̂ gives corrections to x₀, not the parameters directly.
Statement
In the linearized observation-equation model, the vector l is computed as (observed value) minus (the observation function evaluated at approximate parameter values).
The arithmetic mean reduction only applies when there is a single unknown and each observation directly and exclusively measures that unknown. With equal weights and multiple unknowns (or mixed observation equations), you must still solve the full normal equations A^T A x̂ = A^T l.
Statement
If all observations have equal weight, the least-squares solution always reduces to the simple arithmetic mean of all observations.
Previous chapter
Theory of Errors, Weights and Most Probable Value
Next chapter
Condition Equations and Figure Adjustment
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