GELE Adjustment Computations (Least Squares) — Least Squares — Observation EquationsExam Answer Templates

Marks

1

Topic

Least-Squares Principle and Residuals

Difficulty

easy

Template Id

T1

Examiner Tip

Examiners want the sign convention stated explicitly. The phrase 'computed minus observed' is the standard; using it verbatim earns the mark instantly.

Model Answer

A residual (v) is the correction applied to an observation such that v = (adjusted/computed value) − (observed value). Residuals are the differences between the adjusted and raw measurements that result from the least-squares minimization process.

Question Type

very_short_answer

Answer Structure

• Line 1: State the symbol and define residual as computed minus observed [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing v = observed minus computed (wrong sign convention loses the mark)
• Defining residual vaguely as 'error' without the computed-minus-observed relationship
• Omitting the symbol v entirely

Key Phrases To Include

• residual
• v = computed minus observed
• correction to observation
• adjusted value

Marks

1

Topic

Least-Squares Principle

Difficulty

easy

Template Id

T2

Examiner Tip

The formula Σwᵢvᵢ² must appear. A verbal statement alone without the mathematical expression is insufficient for full credit in a computation-focused subject.

Model Answer

The least-squares criterion states that the best estimates of the unknowns are those that minimize the weighted sum of squared residuals: minimize Σ(wᵢvᵢ²), where wᵢ is the weight of the i-th observation and vᵢ is its residual.

Question Type

very_short_answer

Answer Structure

• Line 1: State the criterion (minimize weighted sum of squared residuals) and write Σwᵢvᵢ² [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing Σvᵢ² without weights (unweighted criterion — loses context)
• Using absolute values |vᵢ| instead of squared residuals vᵢ²
• Stating 'minimize errors' without the mathematical expression

Key Phrases To Include

• minimize
• weighted sum of squared residuals
• Σwᵢvᵢ²
• weight
• residual

Marks

1

Topic

Redundancy and Degrees of Freedom

Difficulty

easy

Template Id

T3

Examiner Tip

Always define n (number of observations) and u (number of unknowns) explicitly before writing r = n − u. This earns method marks even if a careless arithmetic error is made.

Model Answer

Redundancy r = n − u = 15 − 9 = 6. The network has 6 degrees of freedom, meaning 6 redundant observations are available for the least-squares adjustment.

Question Type

numerical

Answer Structure

• Line 1: Write the formula r = n − u [½ mark]
• Line 2: Substitute and compute: 15 − 9 = 6 [½ mark]

Scoring Breakdown

Common Mark Deductions

• Subtracting in reverse (u − n) giving a negative answer
• Omitting the formula and just stating the number
• Not labeling n and u before substituting

Key Phrases To Include

• redundancy
• r = n − u
• degrees of freedom
• redundant observations

Marks

2

Topic

Weight Matrix in Least Squares

Difficulty

medium

Template Id

T4

Examiner Tip

The statement 'diagonal P means observations are uncorrelated/independent' is a key phrase examiners look for in 2-mark questions. Pair it with pᵢ = 1/σᵢ² for full marks.

Model Answer

The weight matrix P reflects the relative reliability (precision) of each observation. For uncorrelated (independent) observations, P is a diagonal matrix with entries pᵢ = 1/σᵢ², where σᵢ is the standard deviation of the i-th observation. A diagonal P implies that observations are statistically independent — no cross-correlations exist between them. Observations with smaller standard deviations receive larger weights and thus have greater influence on the adjusted result.

Question Type

short_answer

Answer Structure

• Line 1: Define P as a measure of relative reliability/precision of observations [1 mark]
• Line 2: State that diagonal P means independent observations with pᵢ = 1/σᵢ²; explain influence on result [1 mark]

Scoring Breakdown

Common Mark Deductions

• Saying P = σᵢ² (inverse is missing — common sign error)
• Confusing P with the covariance matrix (P is the inverse of the covariance matrix)
• Not explaining what a diagonal structure implies about correlations

Key Phrases To Include

• weight matrix P
• pᵢ = 1/σᵢ²
• diagonal matrix
• independent/uncorrelated observations
• relative reliability
• greater influence

Marks

2

Topic

Single Unknown — Weighted Mean Application

Difficulty

easy

Template Id

T5

Examiner Tip

For levelling, always state explicitly 'weights are inversely proportional to route length, w = 1/K' before computing. This single phrase signals to the examiner that you understand the physical basis of weighting.

Model Answer

Given: w ∝ 1/K (weights inversely proportional to route length). Step 1 — Compute weights: w₁ = 1/3 = 0.3333 w₂ = 1/2 = 0.5000 w₃ = 1/6 = 0.1667 Σw = 0.3333 + 0.5000 + 0.1667 = 1.0000 Step 2 — Compute weighted mean (least-squares solution for one unknown): Ĥ = Σ(wᵢHᵢ) / Σwᵢ Ĥ = [0.3333(85.412) + 0.5000(85.406) + 0.1667(85.418)] / 1.0000 Ĥ = [28.4706 + 42.7030 + 14.2373] / 1.0000 Ĥ = 85.4109 / 1.0000 Ĥ ≈ 85.411 m The least-squares adjusted elevation of BM-PHIL01 is 85.411 m.

Question Type

numerical

Answer Structure

• Step 1: Compute individual weights wᵢ = 1/Kᵢ and their sum [1 mark]
• Step 2: Apply weighted mean formula Ĥ = Σ(wᵢHᵢ)/Σwᵢ and evaluate [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using w = K (proportional instead of inversely proportional to length) — loses both marks
• Omitting the unit 'metres' in the final answer
• Using simple (unweighted) arithmetic mean instead of weighted mean

Key Phrases To Include

• w = 1/K
• weighted mean
• Ĥ = Σ(wᵢHᵢ)/Σwᵢ
• least-squares solution reduces to weighted mean

Marks

2

Topic

Observation Equation Matrix Form

Difficulty

medium

Template Id

T6

Examiner Tip

The notation v = Ax̂ − l is standard in most Philippine geodesy textbooks. Always write dimensions in parentheses (n×1, n×u, u×1, n×1) — this demonstrates matrix literacy and earns the second mark.

Model Answer

The general observation equation in matrix (parametric) form is: v = Ax̂ − l where: v = (n × 1) vector of residuals A = (n × u) design matrix (coefficient matrix of partial derivatives) x̂ = (u × 1) vector of unknown parameter corrections l = (n × 1) vector of observed-minus-computed values (also called the misclosure vector) n = number of observations u = number of unknowns This is the linearized form obtained by expanding each observation equation as a first-order Taylor series about approximate values of the unknowns.

Question Type

short_answer

Answer Structure

• Line 1: Write the matrix equation v = Ax̂ − l [1 mark]
• Line 2: Define all symbols (v, A, x̂, l) with correct matrix dimensions [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing v = l − Ax̂ (sign reversal — wrong convention as defined in the parametric form)
• Calling A the 'observation matrix' instead of 'design matrix' without defining it
• Omitting matrix dimensions

Key Phrases To Include

• v = Ax̂ − l
• design matrix
• residual vector
• misclosure vector
• n × u
• linearized

Marks

3

Topic

Normal Equations and Least-Squares Solution

Difficulty

medium

Template Id

T7

Examiner Tip

Even in a 3-mark derivation, show three distinct steps: (1) state the objective, (2) differentiate, (3) write the solution. One mark per step is the standard marking scheme for derivation questions.

Model Answer

Starting from the observation equations: v = Ax̂ − l Step 1 — State the least-squares criterion: Minimize: Φ = vᵀPv = (Ax̂ − l)ᵀP(Ax̂ − l) Step 2 — Expand and differentiate with respect to x̂, set equal to zero: ∂Φ/∂x̂ = 2AᵀP(Ax̂ − l) = 0 Step 3 — Rearrange to obtain the Normal Equations: AᵀPAx̂ = AᵀPl Let N = AᵀPA (the normal matrix, u × u) and t = AᵀPl (the right-hand side vector, u × 1). Step 4 — Solve for x̂: x̂ = (AᵀPA)⁻¹AᵀPl = N⁻¹t This is the least-squares solution that minimizes the weighted sum of squared residuals vᵀPv.

Question Type

short_answer

Answer Structure

• Step 1: State objective function Φ = vᵀPv to be minimized [1 mark]
• Step 2: Differentiate and set ∂Φ/∂x̂ = 0 [1 mark]
• Steps 3–4: Rearrange to normal equations AᵀPAx̂ = AᵀPl and state solution x̂ = (AᵀPA)⁻¹AᵀPl [1 mark]

Scoring Breakdown

Common Mark Deductions

• Skipping the differentiation step and jumping directly to normal equations (loses 1 mark for missing derivation)
• Writing (APA)⁻¹ without the transpose superscripts on A
• Forgetting that N = AᵀPA must be square (u × u) and invertible

Key Phrases To Include

• minimize vᵀPv
• normal equations
• AᵀPAx̂ = AᵀPl
• x̂ = (AᵀPA)⁻¹AᵀPl
• normal matrix N = AᵀPA

Marks

3

Topic

Reference Variance and Quality Control

Difficulty

medium

Template Id

T8

Examiner Tip

The interpretation statement earns the third mark. Many students compute σ̂₀² correctly but lose 1 mark by not writing what the value means. Always include: 'A value near 1 indicates the weighting is realistic.'

Model Answer

The reference variance (also called the a posteriori variance of unit weight) is a statistical measure that tests whether the residuals are consistent with the assumed observation weights. Formula: σ̂₀² = vᵀPv / (n − u) where: vᵀPv = weighted sum of squared residuals n = number of observations u = number of unknowns (n − u) = degrees of freedom (redundancy) Purpose: (1) It provides an estimate of the variance of unit-weight observations. (2) A value of σ̂₀² ≈ 1 (when observations are weighted by 1/σᵢ²) indicates that the weighting model is realistic and the adjustment is internally consistent. (3) A large σ̂₀² (>> 1) signals the presence of gross errors (blunders), underestimated measurement uncertainties, or incorrect model assumptions. A value << 1 suggests over-pessimistic (too large) standard deviations were assigned.

Question Type

short_answer

Answer Structure

• Statement 1: Define reference variance as a posteriori variance of unit weight [1 mark]
• Formula: σ̂₀² = vᵀPv/(n − u) with all symbols defined [1 mark]
• Purpose: Explain interpretation (σ̂₀² ≈ 1 means realistic weighting; large value flags blunders) [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing σ̂₀² = vᵀPv/n (dividing by n instead of the redundancy n − u)
• Computing the value but not interpreting what it means for the adjustment quality
• Confusing the reference variance with the standard deviation (forgetting to take the square root if σ̂₀ is requested)

Key Phrases To Include

• a posteriori variance of unit weight
• σ̂₀² = vᵀPv/(n − u)
• degrees of freedom
• realistic weighting
• gross errors/blunders
• internal consistency

Marks

2

Topic

Special Case: Weighted Mean and Arithmetic Mean

Difficulty

medium

Template Id

T9

Examiner Tip

The two conditions are a package — examiners want both. Write them as a numbered list so the examiner can award marks for each condition independently.

Model Answer

The least-squares solution reduces to the simple (unweighted) arithmetic mean when: (1) There is a single unknown x observed directly n times, AND (2) All observations have equal weight (i.e., equal precision, so wᵢ = w = constant for all i). In this case, the design matrix A is a column of ones, P = wI (scalar multiple of identity), and the normal equation AᵀPAx̂ = AᵀPl simplifies to: wn·x̂ = w·Σℓᵢ → x̂ = Σℓᵢ/n which is the arithmetic mean of all observations.

Question Type

short_answer

Answer Structure

• Condition 1: Single unknown observed multiple times [½ mark]
• Condition 2: All weights equal (equal precision) [½ mark]
• Show/state that the formula reduces to x̂ = Σℓᵢ/n [1 mark]

Scoring Breakdown

Common Mark Deductions

• Stating only one condition (single unknown) without mentioning equal weights
• Not writing the formula x̂ = Σℓᵢ/n
• Confusing 'arithmetic mean' with 'weighted mean'

Key Phrases To Include

• single unknown
• equal weights
• equal precision
• arithmetic mean
• x̂ = Σℓᵢ/n
• design matrix A = column of ones

Marks

5

Topic

Complete Weighted Adjustment — Single Unknown

Difficulty

hard

Template Id

T10

Examiner Tip

For 5-mark numerical problems, the examiner expects exactly 5 identifiable steps. Write 'Step 1', 'Step 2', etc. explicitly. Even with a calculation error, you can earn 4 out of 5 if all other steps are correct (method marks).

Model Answer

Given: n = 4 observations, u = 1 unknown (distance D), redundancy r = 4 − 1 = 3. Step 1 — Compute weights (wᵢ = 1/σᵢ² using σ in mm for consistency): w₁ = 1/(3²) = 1/9 ≈ 0.1111 mm⁻² w₂ = 1/(6²) = 1/36 ≈ 0.0278 mm⁻² w₃ = 1/(3²) = 1/9 ≈ 0.1111 mm⁻² w₄ = 1/(6²) = 1/36 ≈ 0.0278 mm⁻² Σw = 0.1111 + 0.0278 + 0.1111 + 0.0278 = 0.2778 mm⁻² Step 2 — Compute weighted mean (least-squares solution for one unknown): D̂ = Σ(wᵢ · ℓᵢ) / Σwᵢ Using working epoch values (subtract 1254.800 for convenience, in mm): ℓ₁ = +31 mm; ℓ₂ = +26 mm; ℓ₃ = +35 mm; ℓ₄ = +29 mm Σ(wᵢ · ℓᵢ) = 0.1111(31) + 0.0278(26) + 0.1111(35) + 0.0278(29) = 3.4444 + 0.7222 + 3.8889 + 0.8056 = 8.8611 mm D̂ (offset) = 8.8611 / 0.2778 = 31.90 mm D̂ = 1254.800 + 0.03190 = 1254.832 m Step 3 — Compute residuals vᵢ = D̂ − ℓᵢ (in mm): v₁ = 31.90 − 31 = +0.90 mm v₂ = 31.90 − 26 = +5.90 mm v₃ = 31.90 − 35 = −3.10 mm v₄ = 31.90 − 29 = +2.90 mm Step 4 — Compute weighted sum of squared residuals vᵀPv: vᵀPv = 0.1111(0.90²) + 0.0278(5.90²) + 0.1111(3.10²) + 0.0278(2.90²) = 0.1111(0.81) + 0.0278(34.81) + 0.1111(9.61) + 0.0278(8.41) = 0.0900 + 0.9677 + 1.0678 + 0.2338 = 2.3593 mm² Step 5 — Reference variance and standard deviation: σ̂₀² = vᵀPv / (n − u) = 2.3593 / 3 = 0.7864 mm² σ̂₀ = √0.7864 = 0.887 mm ≈ 0.9 mm Result: Adjusted distance D̂ = 1254.832 m; Reference standard deviation σ̂₀ ≈ 0.9 mm. Since σ̂₀ < 1 mm (close to 1 when considering 3 mm and 6 mm SDs), the weighting model is approximately consistent.

Question Type

numerical

Answer Structure

• Step 1: Compute all weights wᵢ = 1/σᵢ² and Σw [1 mark]
• Step 2: Apply weighted mean formula D̂ = Σ(wᵢℓᵢ)/Σwᵢ correctly [1 mark]
• Step 3: Compute all four residuals vᵢ = D̂ − observed [1 mark]
• Step 4: Compute vᵀPv (weighted sum of squared residuals) [1 mark]
• Step 5: Compute σ̂₀² = vᵀPv/(n−u) and σ̂₀; state interpretation [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using wᵢ = σᵢ² instead of 1/σᵢ² (inverse relationship missed)
• Forgetting to divide by (n − u) = 3 for the variance (using n instead)
• Computing residuals as observed minus adjusted (sign error — acceptable if consistent but must be stated)
• Omitting units (mm or m) throughout
• Not interpreting σ̂₀ at the end

Key Phrases To Include

• wᵢ = 1/σᵢ²
• weighted mean
• redundancy r = n − u = 3
• vᵀPv
• σ̂₀² = vᵀPv/(n−u)
• reference standard deviation

Marks

3

Topic

Setting Up Design Matrix A

Difficulty

medium

Template Id

T11

Examiner Tip

Always label each row of A with the observation it represents. Write 'Row 1 → Obs 1 (x+y=15.3)' beside the matrix. This makes it easy for the examiner to verify correctness and awards you full marks even if one entry is wrong.

Model Answer

Let unknowns: x̂ = [x, y]ᵀ (u = 2 unknowns); n = 3 observations; redundancy r = 3 − 2 = 1. Observation equations (each observation = function of unknowns + residual): Obs 1: x + y + v₁ = 15.3 → v₁ = (1)x + (1)y − 15.3 Obs 2: x + v₂ = 8.7 → v₂ = (1)x + (0)y − 8.7 Obs 3: y + v₃ = 6.4 → v₃ = (0)x + (1)y − 6.4 Matrix form v = Ax̂ − l: A = | 1 1 | | 1 0 | | 0 1 | x̂ = | x | | y | l = | 15.3 | | 8.7 | | 6.4 | v = | v₁ | | v₂ | | v₃ | Check: r = n − u = 3 − 2 = 1 redundancy — adjustment is possible.

Question Type

numerical

Answer Structure

• Statement: Identify n = 3, u = 2, r = 1 and list the two unknowns [½ mark]
• Write all three observation equations in expanded form (with coefficients shown) [1 mark]
• Assemble A (3×2), x̂ (2×1), l (3×1) in matrix form with correct entries [1½ marks]

Scoring Breakdown

Common Mark Deductions

• Writing l = [observed − computed] with wrong sign (l should be the observed values when using v = Ax̂ − l form)
• Incorrect coefficients in A (e.g., putting 1 in a row where a variable does not appear)
• Omitting the matrix dimensions or not labeling which row corresponds to which observation

Key Phrases To Include

• design matrix A
• observation vector l
• v = Ax̂ − l
• coefficients of partial derivatives
• redundancy r = 1
• 3 × 2 matrix

Marks

2

Topic

Reference Variance and Quality Control

Difficulty

medium

Template Id

T12

Examiner Tip

The contrast must be explicit: 'a priori = assumed before; a posteriori = computed after from residuals.' This parallel structure earns 2 marks in under 4 lines.

Model Answer

A priori standard deviation (σ₀): This is the assumed (pre-adjustment) standard deviation of unit-weight observations, assigned before the adjustment is performed. It represents the expected precision based on instrument specifications, survey procedures, or prior experience. A posteriori standard deviation (σ̂₀): This is computed after the adjustment from the residuals using the formula σ̂₀ = √[vᵀPv/(n − u)]. It reflects the actual fit of the observations to the adjusted model and serves as a check on the assumed precisions. Key distinction: A priori is assumed before; a posteriori is computed after using the residuals. If σ̂₀ ≈ σ₀, the weighting model is validated.

Question Type

short_answer

Answer Structure

• Define a priori σ₀ — assumed before adjustment, based on instrument specs [1 mark]
• Define a posteriori σ̂₀ — computed after adjustment from residuals; formula and purpose stated [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using 'before' and 'after' without specifying what each is based on
• Omitting the formula for σ̂₀
• Not mentioning that comparing the two validates the weighting model

Key Phrases To Include

• a priori
• a posteriori
• assumed before adjustment
• computed after adjustment
• residuals
• σ̂₀ = √[vᵀPv/(n−u)]
• validate weighting model

Marks

3

Topic

Single Unknown — Weighted Mean Application

Difficulty

medium

Template Id

T13

Examiner Tip

The Σwᵢvᵢ = 0 check is a powerful self-verification tool. Always include it — it earns a half mark and demonstrates professional practice.

Model Answer

Given: n = 3, u = 1 (one unknown angle θ), r = 3 − 1 = 2. Step 1 — Convert to seconds from a datum (use 47°23'00" as datum, working in seconds): ℓ₁ = 18"; ℓ₂ = 22"; ℓ₃ = 15"; weights: w₁ = 2, w₂ = 3, w₃ = 1; Σw = 6. Step 2 — Compute weighted mean: θ̂ = Σ(wᵢℓᵢ) / Σwᵢ = [2(18) + 3(22) + 1(15)] / 6 = [36 + 66 + 15] / 6 = 117 / 6 = 19.5" Adjusted angle: θ̂ = 47°23'19.5" Step 3 — Compute residuals vᵢ = θ̂ − ℓᵢ: v₁ = 19.5 − 18 = +1.5" v₂ = 19.5 − 22 = −2.5" v₃ = 19.5 − 15 = +4.5" Check (weighted residuals should sum to zero): Σwᵢvᵢ = 2(+1.5) + 3(−2.5) + 1(+4.5) = 3.0 − 7.5 + 4.5 = 0.0 ✓

Question Type

numerical

Answer Structure

• Step 1: Extract seconds, list weights, state Σw = 6 [½ mark]
• Step 2: Compute weighted mean θ̂ = 117/6 = 19.5" → 47°23'19.5" [1 mark]
• Step 3: Compute all three residuals with correct signs [1 mark]
• Check: Verify Σwᵢvᵢ = 0 [½ mark]

Scoring Breakdown

Common Mark Deductions

• Working directly in DMS without converting to seconds first (arithmetic errors very likely)
• Not performing the Σwᵢvᵢ = 0 check
• Using simple mean (117/6 = 19.5 is the same here by coincidence, but the method must be weighted)

Key Phrases To Include

• weighted mean
• Σwᵢvᵢ = 0 (check)
• residual vᵢ = adjusted minus observed
• degrees-minutes-seconds
• datum seconds

Marks

3

Topic

Redundancy and Degrees of Freedom

Difficulty

medium

Template Id

T14

Examiner Tip

The distinction between 'unique solution (n = u)' and 'least-squares solution (n > u)' is a fundamental concept. Frame your answer around the word 'redundancy' — it is the key term examiners look for.

Model Answer

When n = u (observations equal unknowns), the normal matrix AᵀPA is square and invertible, and the system AᵀPAx̂ = AᵀPl has exactly one solution. However, this solution perfectly satisfies all observation equations — the residuals are zero (v = 0) — meaning there is no redundancy to detect or absorb measurement errors. No statistical quality control (reference variance) is possible. When n > u, the system is overdetermined: there are more equations than unknowns, so no exact solution exists. Least squares resolves this by finding x̂ that minimizes vᵀPv, distributing the misclosures as residuals. This is essential for error detection and reliability assessment. For u = 5 unknowns: minimum n = u = 5 gives a unique solution with no redundancy. For a least-squares adjustment with at least 1 degree of freedom: n_min = u + 1 = 5 + 1 = 6 observations. In practice, a minimum redundancy of 3–5 is recommended for meaningful quality control.

Question Type

short_answer

Answer Structure

• Explain n = u case: unique solution, v = 0, no redundancy, no quality control possible [1 mark]
• Explain n > u case: overdetermined, least squares minimizes vᵀPv, enables error detection [1 mark]
• State minimum n = 6 for u = 5 (with at least r = 1); note practical recommendation [1 mark]

Scoring Breakdown

Common Mark Deductions

• Saying n = u 'cannot be solved' (it can be solved; it just has no redundancy — this is the key distinction)
• Not connecting overdetermination to the ability to detect errors
• Stating minimum n = 5 instead of n = 6 (confusing unique solution with adjustable solution)

Key Phrases To Include

• n = u: unique solution
• v = 0: no redundancy
• n > u: overdetermined
• minimize vᵀPv
• degrees of freedom r = n − u ≥ 1
• quality control

Marks

5

Topic

Complete Weighted Adjustment — Single Unknown

Difficulty

hard

Template Id

T15

Examiner Tip

For 5-mark problems with multiple parts, plan your time: (a) = 1 min, (b) = 1.5 min, (c) = 2 min, (d) = 0.5 min. Part (d) is qualitative but earns a full mark — never skip it. Write 'Route ___ has the largest residual and may warrant re-observation.'

Model Answer

Given: n = 4, u = 1, r = 4 − 1 = 3. Weights w = 1/K. (a) Compute weights and weighted mean: w₁ = 1/4 = 0.2500; w₂ = 1/2 = 0.5000; w₃ = 1/8 = 0.1250; w₄ = 1/1 = 1.0000 Σw = 0.2500 + 0.5000 + 0.1250 + 1.0000 = 1.8750 Work in mm from datum 234.550 m: ℓ₁ = 11 mm; ℓ₂ = 5 mm; ℓ₃ = 13 mm; ℓ₄ = 8 mm Σ(wᵢℓᵢ) = 0.2500(11) + 0.5000(5) + 0.1250(13) + 1.0000(8) = 2.750 + 2.500 + 1.625 + 8.000 = 14.875 mm Ĥ (offset) = 14.875 / 1.8750 = 7.933 mm Ĥ = 234.550 + 0.007933 = 234.5579 m ≈ 234.558 m (b) Compute residuals vᵢ = Ĥ − ℓᵢ (in mm): v₁ = 7.933 − 11.000 = −3.067 mm v₂ = 7.933 − 5.000 = +2.933 mm v₃ = 7.933 − 13.000 = −5.067 mm v₄ = 7.933 − 8.000 = −0.067 mm Check: Σwᵢvᵢ = 0.25(−3.067) + 0.50(+2.933) + 0.125(−5.067) + 1.00(−0.067) = −0.767 + 1.467 − 0.633 − 0.067 = 0.000 ✓ (c) Reference standard deviation: vᵀPv = 0.2500(3.067²) + 0.5000(2.933²) + 0.1250(5.067²) + 1.0000(0.067²) = 0.2500(9.406) + 0.5000(8.603) + 0.1250(25.674) + 1.0000(0.004) = 2.352 + 4.302 + 3.209 + 0.004 = 9.867 mm² σ̂₀² = vᵀPv / (n − u) = 9.867 / 3 = 3.289 mm² σ̂₀ = √3.289 = 1.814 mm ≈ 1.8 mm (reference std dev per km) (d) Quality Assessment: The reference standard deviation of σ̂₀ ≈ 1.8 mm per √km indicates that the observed elevation differences are somewhat inconsistent with purely random errors. However, no single route dominates the residuals excessively, and Σwᵢvᵢ = 0 confirms mathematical consistency. The adjustment is accepted, but a re-observation of Route 3 (largest residual of 5.1 mm) is recommended.

Question Type

numerical

Answer Structure

• Part (a): Weights w = 1/K; weighted mean formula applied → Ĥ ≈ 234.558 m [1 mark]
• Part (b): All four residuals computed with correct signs; Σwᵢvᵢ = 0 check shown [1 mark]
• Part (c): vᵀPv computed step by step; σ̂₀² = vᵀPv/3; σ̂₀ stated [2 marks]
• Part (d): Interpretation of σ̂₀; identify suspect observation; accept/recommend re-obs [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using w = K (directly proportional) instead of w = 1/K
• Not showing the Σwᵢvᵢ = 0 check
• Dividing vᵀPv by n instead of (n − u) for the variance
• Skipping the quality assessment part (d) — worth 1 full mark
• Not converting correctly between metres and millimetres

Key Phrases To Include

• w = 1/K
• weighted mean
• Σwᵢvᵢ = 0
• vᵀPv
• σ̂₀ = √[vᵀPv/(n−u)]
• quality assessment
• suspect observation