Error Propagation, Variance-Covariance and Error Ellipses - Exam Answer Templates
Proper answer writing is the single most decisive factor between a passing and failing score in the PRC Geodetic Engineer Licensure Examination. In Adjustment Computations, questions on Error Propagation, Variance-Covariance, and Error Ellipses test both conceptual understanding and computational skill. Examiners award marks for specific phrases, correct formula application, and clearly structured solutions — not just for arriving at a numerical answer. A student who shows complete, step-by-step work, labels every formula, and interprets results in engineering context consistently earns full marks, while one who writes only a final answer risks losing partial-credit marks even when the number is correct. These templates show you exactly how to structure your responses for every mark level so you can maximize your score on exam day.
Answer templates
State the Law of Propagation of Variances for a function of independent measured quantities.
Examiner tip
The examiner checks for the squared partial derivative (∂f/∂xᵢ)² — this is the most commonly dropped element. Write the formula exactly as given and include the summation sign.
Model answer
For a function y = f(x₁, x₂, …, xₙ) of statistically independent measured quantities x₁, x₂, …, xₙ, the variance of y is:
σ_y² = (∂f/∂x₁)²σ₁² + (∂f/∂x₂)²σ₂² + … + (∂f/∂xₙ)²σₙ²
where ∂f/∂xᵢ are the partial derivatives of f with respect to each measured quantity.
Answer structure
- State the formula σ_y² = Σ(∂f/∂xᵢ)²σᵢ² with correct notation [1 mark]
Scoring breakdown
Criteria
Correctly written propagation formula with partial derivative notation and summation over all measured quantities
Common mark deductions
- Writing σ_y = Σ(∂f/∂xᵢ)σᵢ instead of squaring the partial derivatives and the standard deviations
- Omitting the independence condition, which justifies the sum form
- Confusing variance (σ²) with standard deviation (σ) in the formula
Key phrases to include
- Law of Propagation of Variances
- partial derivative
- σ_y² = Σ(∂f/∂xᵢ)²σᵢ²
- independent measured quantities
Two independent horizontal angles, each measured with σ = 5", are added to form a total angle. Find the standard deviation of the total angle.
Examiner tip
For a 1-mark numerical, write the formula and the final answer — no elaborate working needed. The mark is for both the correct method (quadrature) and correct number.
Model answer
Given: σ₁ = σ₂ = 5"
For the sum y = x₁ + x₂ (independent quantities):
σ_y = √(σ₁² + σ₂²) = √(5² + 5²) = √(25 + 25) = √50
σ_y = 7.07"
Answer structure
- State the formula for the sum [0.5 mark]
- Substitute values and compute correct numerical answer with unit ["] [0.5 mark]
Scoring breakdown
Criteria
Correct formula σ_y = √(σ₁² + σ₂²) applied and numerically correct answer of 7.07" with unit
Common mark deductions
- Adding standard deviations linearly: 5 + 5 = 10" (wrong — this assumes perfect correlation)
- Omitting the arc-second (") unit on the final answer
- Not writing the formula before substituting
Key phrases to include
- add in quadrature
- √(σ₁² + σ₂²)
- 7.07"
- independent
Differentiate between the variance of a sum and the variance of the mean for n observations of equal precision. Briefly explain when each applies.
Examiner tip
The examiner specifically looks for both formulas AND the context where each applies — a formula alone without context earns only half the mark at this level.
Model answer
Variance of the sum (y = x₁ + x₂ + … + xₙ):
σ_sum = σ√n
This applies when the total value is the accumulation of n independent measurements, e.g., the total length of a traverse leg taped in n equal bays.
Variance of the mean (ȳ = (x₁ + x₂ + … + xₙ)/n):
σ_mean = σ/√n
This applies when a single quantity is measured n times and averaged to improve precision, e.g., repeated distance measurements with an EDM.
Answer structure
- State the formula σ_sum = σ√n with a correct example [1 mark]
- State the formula σ_mean = σ/√n with a correct example, clearly distinguishing from the sum [1 mark]
Scoring breakdown
Criteria
Correct formula σ_sum = σ√n and a relevant engineering example showing it increases with n
Criteria
Correct formula σ_mean = σ/√n and a relevant engineering example showing it decreases with n, plus an explicit statement that these are different situations
Common mark deductions
- Confusing the two formulas — this is the most critical conceptual error in this chapter
- Writing one formula and claiming it applies to both situations
- No engineering example to contextualise each formula
Key phrases to include
- σ√n (sum of n equal observations)
- σ/√n (mean of n observations)
- accumulation
- repeated measurement
- independent
A rectangular lot is surveyed. The length L = 80.00 m is measured with σ_L = 0.03 m and the width W = 40.00 m with σ_W = 0.02 m. Compute the standard deviation of the computed area A = L × W.
Examiner tip
Always compute and label ∂A/∂L and ∂A/∂W explicitly before substituting — examiners award a mark for the correct partial derivatives alone, independent of the arithmetic.
Model answer
Given: L = 80.00 m, σ_L = 0.03 m; W = 40.00 m, σ_W = 0.02 m
Function: A = L × W
Partial derivatives:
∂A/∂L = W = 40.00 m
∂A/∂W = L = 80.00 m
Law of Propagation of Variances:
σ_A² = (∂A/∂L)²σ_L² + (∂A/∂W)²σ_W²
σ_A² = (40.00)²(0.03)² + (80.00)²(0.02)²
σ_A² = 1600(0.0009) + 6400(0.0004)
σ_A² = 1.44 + 2.56 = 4.00 m⁴
σ_A = √4.00 = 2.00 m²
Answer structure
- Write the function A = L × W and compute both partial derivatives [0.5 mark]
- Write the propagation formula with both terms [0.5 mark]
- Substitute numerical values correctly [0.5 mark]
- State correct final answer 2.00 m² with proper unit [0.5 mark]
Scoring breakdown
Criteria
Correct partial derivatives ∂A/∂L = W = 40 m and ∂A/∂W = L = 80 m with propagation formula written
Criteria
Correct numerical computation and final answer σ_A = 2.00 m² with squared unit (m²)
Common mark deductions
- Using σ_L and σ_W as the partial derivatives instead of W and L
- Writing the unit as m instead of m² for area standard deviation
- Computing σ_A² = (∂A/∂L)σ_L + (∂A/∂W)σ_W (forgetting to square)
Key phrases to include
- ∂A/∂L = W
- ∂A/∂W = L
- Law of Propagation of Variances
- σ_A² = (W)²σ_L² + (L)²σ_W²
- 2.00 m²
Define the variance-covariance matrix Σ_x̂ of adjusted parameters in a least-squares adjustment. State its formula and identify each component.
Examiner tip
Examiners specifically check for σ₀² as the multiplier — its absence loses a half-mark. Also identify each matrix symbol to show you understand the least-squares framework.
Model answer
The variance-covariance matrix of adjusted parameters, denoted Σ_x̂, is a square symmetric matrix that fully describes the precision of and correlation among the adjusted unknown parameters from a least-squares adjustment.
Formula:
Σ_x̂ = σ₀² (AᵀPA)⁻¹
where:
- σ₀² = a posteriori reference variance (estimated unit-weight variance)
- A = design matrix (coefficient matrix of observation equations)
- P = weight matrix of the observations
- (AᵀPA)⁻¹ = inverse of the normal equation matrix
Diagonal elements = variances of individual adjusted parameters (σ_x², σ_y², …)
Off-diagonal elements = covariances between pairs of adjusted parameters (σ_xy, …)
Answer structure
- Define Σ_x̂ in words as a symmetric matrix describing precision and correlation [0.5 mark]
- Write the formula Σ_x̂ = σ₀²(AᵀPA)⁻¹ [0.5 mark]
- Identify each symbol in the formula [0.5 mark]
- State what diagonal and off-diagonal elements represent [0.5 mark]
Scoring breakdown
Criteria
Correct formula Σ_x̂ = σ₀²(AᵀPA)⁻¹ with all symbols identified
Criteria
Correct interpretation of diagonal (variances) vs. off-diagonal (covariances) elements
Common mark deductions
- Writing Σ_x̂ = (AᵀPA)⁻¹ without the σ₀² scalar multiplier
- Not identifying what the diagonal and off-diagonal elements represent
- Confusing A (design matrix) with the normal equation matrix AᵀPA
Key phrases to include
- variance-covariance matrix
- Σ_x̂ = σ₀²(AᵀPA)⁻¹
- a posteriori reference variance
- design matrix
- weight matrix
- diagonal = variances
- off-diagonal = covariances
A traverse leg is measured in 6 equal segments, each with σ = 0.010 m. Calculate the standard deviation of the total traversed distance.
Examiner tip
State explicitly that you are computing the error of a SUM (not a mean) before writing the formula — this explanation is what separates a 2/2 from a 1/2 answer.
Model answer
Given: n = 6 segments, σ per segment = 0.010 m
The total distance D = d₁ + d₂ + … + d₆ is a sum of n independent equal observations.
Applying the propagation formula for a sum of n equal independent observations:
σ_D = σ√n = 0.010 × √6 = 0.010 × 2.4495
σ_D = 0.0245 m ≈ 24.5 mm
Answer structure
- Identify that total distance is a SUM of n independent segments (not a mean) [0.5 mark]
- Write the formula σ_D = σ√n [0.5 mark]
- Substitute: σ√n = 0.010√6 [0.5 mark]
- State correct answer 0.0245 m with unit [0.5 mark]
Scoring breakdown
Criteria
Correct formula σ√n (not σ/√n) with justification that total is a sum of independent segments
Criteria
Correct numerical computation: 0.010 × √6 = 0.0245 m with proper unit
Common mark deductions
- Using σ/√n = 0.010/√6 = 0.00408 m — this is the error of the mean, not the total
- Not explaining why σ√n applies (it is a sum, not an average)
- Incorrect arithmetic for √6
Key phrases to include
- sum of n independent equal observations
- σ_D = σ√n
- 0.0245 m
- error accumulates
Explain, in engineering terms, what a standard error ellipse is. Include: (a) what it represents geometrically, (b) its associated probability level, and (c) what an elongated ellipse indicates about the survey geometry.
Examiner tip
For a 3-mark question divided into three parts, allocate one clear paragraph per part — examiners mark by part, so a well-structured answer with three distinct blocks maximises marks even if one part is weak.
Model answer
(a) Geometric meaning:
The standard error ellipse is an ellipse centred on a computed 2-D position (e.g., a UTM coordinate pair in PRS92/PPCS) that represents the locus of equal uncertainty of that position. Its semi-major axis a and semi-minor axis b are derived from the eigenvalues of the 2×2 position sub-matrix of the variance-covariance matrix Σ_x̂. The orientation angle θ of the major axis (measured from the North or Easting axis) is determined by the eigenvectors of that sub-matrix.
(b) Probability level:
The standard error ellipse corresponds to approximately 39% probability — meaning there is a 39% chance that the true position lies within the ellipse. For practical engineering and cadastral survey reporting (e.g., under PD 1529 or RA 8560 standards), the ellipse is scaled by a factor k = 2.45 to achieve the 95% confidence ellipse.
(c) Elongated ellipse:
An elongated error ellipse (a >> b) indicates weak survey geometry in the direction of the major axis. This means the network configuration — the arrangement of control points and observation directions — provides poor constraint in that direction. In a PPCS-based horizontal control network, this typically arises from unfavorable azimuth distribution of baselines or a lack of redundant observations in a particular direction.
Answer structure
- (a) Define the standard error ellipse: centre, semi-axes from eigenvalues of covariance sub-matrix, orientation from eigenvectors [1 mark]
- (b) State 39% probability for standard ellipse; mention 95% confidence ellipse with k = 2.45 scale factor [1 mark]
- (c) Interpret elongated ellipse as weak geometry/poor constraint in the direction of the major axis [1 mark]
Scoring breakdown
Criteria
Correct geometric description: ellipse centred on computed position, semi-axes from eigenvalues of covariance sub-matrix, orientation from eigenvectors
Criteria
Correct probability level of ~39% for standard ellipse AND mention of 95% confidence scaling (k ≈ 2.45)
Criteria
Correct engineering interpretation: elongated ellipse = weak geometry / poor constraint in the major-axis direction
Common mark deductions
- Stating the standard ellipse represents 68% or 95% probability — both are incorrect; it is ~39%
- Confusing error ellipse with error circle (circle only applies when σ_x = σ_y and σ_xy = 0)
- No reference to eigenvalues/eigenvectors — describing the ellipse without its mathematical origin loses the first mark
- Not linking an elongated ellipse to network geometry in an engineering context
Key phrases to include
- eigenvalues of covariance sub-matrix
- semi-major axis a, semi-minor axis b
- orientation angle θ
- 39% probability
- 95% confidence ellipse
- scale factor k = 2.45
- weak geometry
- poor constraint
Describe the matrix form of error propagation. Write the formula, define all terms, and explain its advantage over the scalar form.
Examiner tip
The most common error is reversing J and Jᵀ. Remember: J is on the left, Jᵀ is on the right — the formula is analogous to the change-of-basis rule in linear algebra.
Model answer
Matrix Form of Error Propagation:
Formula: Σ_y = J Σ_x Jᵀ
where:
- Σ_y = variance-covariance matrix of the output quantities y (m×m matrix for m outputs)
- J = Jacobian matrix of partial derivatives, with Jᵢⱼ = ∂yᵢ/∂xⱼ (m×n matrix for m outputs and n inputs)
- Σ_x = variance-covariance matrix of the input quantities x (n×n matrix)
- Jᵀ = transpose of the Jacobian
Derivation logic:
If y = Jx (linearised relation), then by covariance propagation rules, Σ_y = J Σ_x Jᵀ.
Advantage over scalar form:
- Handles correlated inputs automatically — the scalar form assumes independent inputs (σ_xy = 0), but the matrix form propagates off-diagonal covariance terms through the product J Σ_x Jᵀ.
- Simultaneously propagates multiple output quantities, yielding a complete Σ_y matrix that includes both the output variances (diagonal) and output covariances (off-diagonal).
- More compact and computationally efficient for large systems, as used in least-squares network adjustments.
Answer structure
- State the formula Σ_y = J Σ_x Jᵀ with correct dimensions/notation [1 mark]
- Define all terms: J (Jacobian with ∂yᵢ/∂xⱼ elements), Σ_x, Σ_y, Jᵀ [1 mark]
- State at least two clear advantages over the scalar form, particularly the handling of correlations [1 mark]
Scoring breakdown
Criteria
Correct matrix propagation formula Σ_y = J Σ_x Jᵀ
Criteria
Complete and correct definition of J as the Jacobian of partial derivatives, and identification of Σ_x and Σ_y dimensions
Criteria
Two meaningful advantages stated: (1) handles correlated inputs and (2) propagates multiple outputs simultaneously
Common mark deductions
- Writing Σ_y = Jᵀ Σ_x J (wrong order of transpose — must be J on left, Jᵀ on right)
- Not defining J as the Jacobian of partial derivatives
- Stating only one advantage without elaborating on the correlation-handling capability
Key phrases to include
- Σ_y = J Σ_x Jᵀ
- Jacobian matrix
- Jᵢⱼ = ∂yᵢ/∂xⱼ
- correlated inputs
- off-diagonal covariance
- multiple output quantities
A computed area involves length L and width W, where σ_L = 0.04 m, σ_W = 0.03 m, and the covariance σ_LW = +0.001 m². If L = 60 m and W = 30 m, find σ_A using the full matrix propagation formula.
Examiner tip
The presence of a non-zero covariance σ_LW in the problem statement is a signal to use the full matrix form — examiners include it specifically to test whether you can handle correlated inputs.
Model answer
Given: L = 60 m, W = 30 m; σ_L = 0.04 m, σ_W = 0.03 m, σ_LW = +0.001 m²
Function: A = f(L, W) = L × W
Step 1 — Jacobian (row vector):
J = [∂A/∂L ∂A/∂W] = [W L] = [30 60]
Step 2 — Input covariance matrix:
Σ_x = | σ_L² σ_LW | = | 0.0016 0.001 |
| σ_LW σ_W² | | 0.001 0.0009|Step 3 — Apply Σ_A = J Σ_x Jᵀ:
First compute J Σ_x:
[30 60] × |0.0016 0.001 | = [30(0.0016) + 60(0.001) , 30(0.001) + 60(0.0009)]
|0.001 0.0009| [0.048 + 0.060 , 0.030 + 0.054 ]
= [0.108 , 0.084]
Then Σ_A = (J Σ_x) × Jᵀ = [0.108 0.084] × |30| = 0.108(30) + 0.084(60)
|60|
= 3.240 + 5.040 = 8.280 m⁴
Step 4 — Standard deviation:
σ_A = √8.280 = 2.877 m²
Answer structure
- Compute Jacobian J = [W L] = [30 60] [0.5 mark]
- Write the 2×2 input covariance matrix Σ_x with σ_LW in off-diagonal [0.5 mark]
- Apply Σ_A = J Σ_x Jᵀ with correct matrix multiplication showing intermediate steps [1 mark]
- State σ_A = √8.280 = 2.877 m² with correct unit [1 mark]
Scoring breakdown
Criteria
Correct Jacobian J = [30, 60] and covariance matrix Σ_x with σ_LW = 0.001 in off-diagonal positions
Criteria
Correct matrix multiplication J Σ_x Jᵀ with intermediate results shown
Criteria
Correct final answer σ_A = 2.877 m² (accepting 2.88 m²) with proper unit
Common mark deductions
- Ignoring σ_LW = 0.001 and computing as if inputs are independent (this changes the answer to √(σ_L²W² + σ_W²L²))
- Placing σ_LW on the diagonal instead of the off-diagonal positions
- Using J as a column vector instead of a row vector, leading to wrong matrix dimensions
Key phrases to include
- J = [∂A/∂L ∂A/∂W] = [W L]
- Σ_A = J Σ_x Jᵀ
- off-diagonal covariance σ_LW
- 2.877 m²
In a least-squares horizontal control network adjustment, the 2×2 covariance sub-matrix for a newly determined PRS92 station is:
Σ = |0.0025 0.0010|
|0.0010 0.0049|
(in m²)
(a) Determine the semi-axes of the standard error ellipse. (b) Compute the orientation angle of the major axis.
Examiner tip
Show the quadrant determination of θ explicitly — write the signs of the numerator and denominator of tan(2θ) and state which quadrant 2θ falls in. Examiners deduct for a correct magnitude but wrong quadrant.
Model answer
Given:
Σ = |σ_x² σ_xy| = |0.0025 0.0010|
|σ_xy σ_y²| |0.0010 0.0049|
(a) Semi-axes of the standard error ellipse:
The semi-axes are derived from the eigenvalues λ of the covariance matrix:
Characteristic equation: λ² - (σ_x² + σ_y²)λ + (σ_x²σ_y² - σ_xy²) = 0
Trace: σ_x² + σ_y² = 0.0025 + 0.0049 = 0.0074
Determinant: σ_x²σ_y² - σ_xy² = (0.0025)(0.0049) - (0.0010)² = 0.00001225 - 0.0000100 = 0.00000225
λ² - 0.0074λ + 0.00000225 = 0
Using the quadratic formula:
λ = [0.0074 ± √(0.0074² - 4 × 0.00000225)] / 2
λ = [0.0074 ± √(0.00005476 - 0.00000900)] / 2
λ = [0.0074 ± √0.00004576] / 2
λ = [0.0074 ± 0.006765] / 2
λ₁ = (0.0074 + 0.006765) / 2 = 0.014165 / 2 = 0.007082 m²
λ₂ = (0.0074 - 0.006765) / 2 = 0.000635 / 2 = 0.000318 m²
Semi-major axis: a = √λ₁ = √0.007082 = 0.0841 m ≈ 84.1 mm
Semi-minor axis: b = √λ₂ = √0.000318 = 0.0178 m ≈ 17.8 mm
(b) Orientation angle of the major axis:
Using the formula:
tan(2θ) = 2σ_xy / (σ_x² - σ_y²) = 2(0.0010) / (0.0025 - 0.0049) = 0.0020 / (-0.0024) = -0.8333
2θ = arctan(-0.8333) = -39.81° → adjusted to the correct quadrant
Since σ_xy > 0 and (σ_x² - σ_y²) < 0, the major axis lies in the second/fourth quadrant.
2θ = 180° - 39.81° = 140.19°
θ = 70.1° (orientation of major axis from the x-axis / Easting direction)
Conclusion:
Standard error ellipse: a = 84.1 mm, b = 17.8 mm, θ = 70.1° from Easting
Answer structure
- Identify σ_x², σ_y², σ_xy from the given matrix [0.5 mark]
- Write the characteristic equation for eigenvalues [0.5 mark]
- Compute trace and determinant correctly [0.5 mark]
- Solve quadratic and obtain λ₁ = 0.007082 m² and λ₂ = 0.000318 m² [1 mark]
- Compute semi-axes: a = √λ₁ ≈ 84.1 mm, b = √λ₂ ≈ 17.8 mm [1 mark]
- Apply tan(2θ) formula and compute orientation angle θ ≈ 70.1° with quadrant check [1.5 marks]
Scoring breakdown
Criteria
Correct setup of characteristic equation using trace and determinant of the 2×2 covariance matrix
Criteria
Correct eigenvalues λ₁ = 0.007082 m² and λ₂ = 0.000318 m² obtained by solving the quadratic
Criteria
Correct semi-axes a = √λ₁ ≈ 0.0841 m and b = √λ₂ ≈ 0.0178 m with units in metres or mm
Criteria
Correct formula tan(2θ) = 2σ_xy/(σ_x² - σ_y²) applied with correct substitution
Criteria
Correct orientation angle θ ≈ 70.1° with quadrant check and clear final summary statement
Common mark deductions
- Using a = λ₁ and b = λ₂ (eigenvalues directly as axes) instead of taking the square roots
- Omitting the quadrant check for θ — arctan gives values in (-90°, 90°) and may need adjustment
- Swapping σ_x² - σ_y² and σ_y² - σ_x² in the tan(2θ) formula
- Not stating the units (metres or millimetres) of the semi-axes in the final answer
- Computing the 95% confidence ellipse axes without being asked (standard ellipse uses eigenvalues directly)
Key phrases to include
- eigenvalues of the covariance matrix
- characteristic equation λ² - (trace)λ + (determinant) = 0
- semi-major axis a = √λ₁
- semi-minor axis b = √λ₂
- tan(2θ) = 2σ_xy/(σ_x² - σ_y²)
- orientation angle θ
- quadrant check
What is the probability coverage of the standard error ellipse? How is it scaled to obtain a 95% confidence error ellipse?
Examiner tip
This is a factual 1-mark question — memorise both values (39% and k = 2.45) as a pair. The examiner expects both in a single sentence.
Model answer
The standard error ellipse covers approximately 39% probability. To obtain the 95% confidence ellipse, the semi-axes a and b are multiplied by the scale factor k = 2.45, giving: a₉₅ = 2.45a and b₉₅ = 2.45b.
Answer structure
- State 39% for the standard ellipse AND state k = 2.45 for the 95% ellipse [1 mark]
Scoring breakdown
Criteria
Correct probability of ~39% for the standard ellipse AND scale factor k = 2.45 for the 95% confidence ellipse
Common mark deductions
- Stating 68% (the 1-sigma probability for a 1-D normal distribution, not the 2-D ellipse)
- Stating 95% as the probability of the standard error ellipse without the scaling step
- Providing only the probability without the scale factor, or only the scale factor without the probability
Key phrases to include
- 39% probability
- standard error ellipse
- k = 2.45
- 95% confidence ellipse
In general terms, what does a near-circular error ellipse indicate about a survey network compared to an elongated error ellipse?
Examiner tip
Always frame your answer in terms of geometry and observation distribution — these are the engineering-context terms that earn the mark.
Model answer
A near-circular error ellipse (a ≈ b) indicates well-conditioned geometry — the network provides approximately equal precision in all directions. An elongated ellipse (a >> b) indicates weak geometry and poor constraint in the direction of the major axis, meaning observations are insufficient or poorly distributed in that direction.
Answer structure
- Contrast near-circular (well-conditioned, equal precision all directions) vs. elongated (weak geometry, poor constraint in major-axis direction) [1 mark]
Scoring breakdown
Criteria
Correct contrast: near-circular = good/equal geometry; elongated = weak geometry in the major-axis direction
Common mark deductions
- Stating a circular ellipse is 'perfect' without linking to geometry or observation distribution
- Only describing one case (either circular or elongated) without comparing both
Key phrases to include
- well-conditioned geometry
- equal precision in all directions
- weak geometry
- poor constraint
- major axis direction
A total distance D is derived from three independently taped segments: D₁ = 200 m (σ₁ = 0.02 m), D₂ = 150 m (σ₂ = 0.03 m), D₃ = 250 m (σ₃ = 0.04 m). Compute: (a) the total distance D, and (b) the standard deviation of D.
Examiner tip
When σᵢ values are unequal, always use the full √(Σσᵢ²) formula — the σ√n shortcut only applies when all partial standard deviations are identical.
Model answer
(a) Total distance:
D = D₁ + D₂ + D₃ = 200 + 150 + 250 = 600.00 m
(b) Standard deviation of D:
Since D₁, D₂, D₃ are independent:
σ_D = √(σ₁² + σ₂² + σ₃²)
σ_D = √(0.02² + 0.03² + 0.04²)
σ_D = √(0.0004 + 0.0009 + 0.0016)
σ_D = √0.0029
σ_D = 0.0539 m ≈ 53.9 mm
Answer structure
- (a) Compute D = 200 + 150 + 250 = 600.00 m [0.5 mark]
- (b) Write formula σ_D = √(σ₁² + σ₂² + σ₃²) [0.5 mark]
- Substitute and compute correctly: σ_D = 0.0539 m [1 mark]
Scoring breakdown
Criteria
Correct total D = 600.00 m
Criteria
Correct formula for propagation: sum in quadrature for three independent segments
Criteria
Correct numerical result σ_D = 0.0539 m (or 53.9 mm) with proper unit
Common mark deductions
- Adding standard deviations linearly: 0.02 + 0.03 + 0.04 = 0.09 m
- Using σ√n formula (valid only when all σᵢ are equal, which they are not here)
- Forgetting to take the square root of the sum of squares
Key phrases to include
- independent segments
- √(σ₁² + σ₂² + σ₃²)
- add in quadrature
- 0.0539 m
Derive the standard deviation of the area of a triangle computed as A = (1/2)base × height, where base b = 40.00 m (σ_b = 0.05 m) and height h = 25.00 m (σ_h = 0.04 m). Assume independence.
Examiner tip
Always differentiate the full function, including any scalar multipliers like the 1/2 in the triangle area formula — a common mistake is to drop this coefficient when computing partial derivatives.
Model answer
Given: A = (1/2)bh; b = 40.00 m (σ_b = 0.05 m), h = 25.00 m (σ_h = 0.04 m)
Step 1 — Partial derivatives:
∂A/∂b = (1/2)h = (1/2)(25.00) = 12.50 m
∂A/∂h = (1/2)b = (1/2)(40.00) = 20.00 m
Step 2 — Law of Propagation of Variances (independent quantities):
σ_A² = (∂A/∂b)²σ_b² + (∂A/∂h)²σ_h²
σ_A² = (12.50)²(0.05)² + (20.00)²(0.04)²
σ_A² = 156.25 × 0.0025 + 400.00 × 0.0016
σ_A² = 0.390625 + 0.640000
σ_A² = 1.030625 m⁴
Step 3 — Standard deviation:
σ_A = √1.030625 = 1.015 m²
Check: Area = (1/2)(40)(25) = 500.00 m²; σ_A/A = 1.015/500 = 0.203% (relative precision)
Answer structure
- Compute both partial derivatives: ∂A/∂b = h/2 = 12.50 m and ∂A/∂h = b/2 = 20.00 m [1 mark]
- Write and apply propagation formula: σ_A² = (12.50)²(0.05)² + (20.00)²(0.04)² [1 mark]
- Compute σ_A = 1.015 m² with correct unit [1 mark]
Scoring breakdown
Criteria
Both partial derivatives ∂A/∂b = 12.50 m and ∂A/∂h = 20.00 m correctly computed
Criteria
Propagation formula written with both squared terms substituted correctly
Criteria
Correct final answer σ_A = 1.015 m² with correct unit (m²)
Common mark deductions
- Computing ∂A/∂b = h instead of (1/2)h (forgetting the 1/2 coefficient in the function)
- Reporting the unit as m instead of m²
- Squaring the standard deviations but forgetting to square the partial derivatives
Key phrases to include
- ∂A/∂b = (1/2)h
- ∂A/∂h = (1/2)b
- σ_A² = (∂A/∂b)²σ_b² + (∂A/∂h)²σ_h²
- 1.015 m²
In a least-squares adjustment of a PPCS/UTM horizontal control network under PRS92, the adjusted covariance matrix for two station coordinates (x, y) is:
Σ_x̂ = σ₀² × Q, where σ₀² = 4 mm² and Q = (AᵀPA)⁻¹
The cofactor matrix Q for station P₁ is:
Q = |0.50 0.15|
|0.15 0.80|
(a) Compute the variance-covariance matrix Σ for station P₁.
(b) Determine the standard deviations σ_x and σ_y.
(c) State whether x and y are positively or negatively correlated, and find the correlation coefficient ρ_xy.
Examiner tip
PRC board exam questions on this topic frequently require computation of ρ_xy — memorise ρ_xy = σ_xy/(σ_x σ_y) and note that |ρ| ≤ 1. A ρ_xy near ±1 means the coordinate errors are highly linearly dependent; near 0 means nearly independent.
Model answer
Given: σ₀² = 4 mm² = 4 × 10⁻⁶ m²; Q = |0.50 0.15|
|0.15 0.80|
(a) Variance-covariance matrix:
Σ = σ₀² × Q = (4 × 10⁻⁶) × |0.50 0.15|
|0.15 0.80|
Σ = |(4×10⁻⁶)(0.50) (4×10⁻⁶)(0.15)| = |2.00×10⁻⁶ 0.60×10⁻⁶|
|(4×10⁻⁶)(0.15) (4×10⁻⁶)(0.80)| |0.60×10⁻⁶ 3.20×10⁻⁶|
In mm²: Σ = |2.00 0.60| mm²
|0.60 3.20|
(b) Standard deviations:
σ_x = √(Σ₁₁) = √(2.00 mm²) = 1.414 mm ≈ 1.4 mm
σ_y = √(Σ₂₂) = √(3.20 mm²) = 1.789 mm ≈ 1.8 mm
(c) Correlation coefficient:
σ_xy = Σ₁₂ = 0.60 mm² (positive off-diagonal)
Since σ_xy = +0.60 mm² > 0, x and y are positively correlated.
ρ_xy = σ_xy / (σ_x × σ_y) = 0.60 / (1.414 × 1.789) = 0.60 / 2.529 = 0.237
Conclusion: Station P₁ has σ_x = 1.4 mm, σ_y = 1.8 mm, and the coordinates are weakly positively correlated (ρ_xy = 0.237).
Answer structure
- (a) Multiply σ₀² × Q element-by-element to get Σ with correct values and units [1.5 marks]
- (b) Compute σ_x = √(2.00 mm²) = 1.414 mm and σ_y = √(3.20 mm²) = 1.789 mm [1 mark]
- (c) State positive correlation from sign of σ_xy, and compute ρ_xy = 0.237 using the formula [2.5 marks]
Scoring breakdown
Criteria
Correct Σ = σ₀² × Q multiplication giving 2×2 matrix with correct values in mm² (2.00, 0.60, 0.60, 3.20)
Criteria
Correct standard deviations σ_x = 1.414 mm and σ_y = 1.789 mm from square roots of diagonal
Criteria
Correctly state positive correlation from sign of σ_xy = +0.60 mm²
Criteria
Correct formula ρ_xy = σ_xy/(σ_x × σ_y) and numerical answer ρ_xy = 0.237
Common mark deductions
- Not multiplying by σ₀² — treating Q directly as the covariance matrix
- Taking σ_x = Σ₁₁ (the variance) instead of σ_x = √Σ₁₁ (the standard deviation)
- Confusing the sign of σ_xy for correlation direction — off-diagonal is always the covariance, not the correlation coefficient
- Not stating the engineering interpretation (weak/strong, positive/negative) alongside the numerical ρ value
Key phrases to include
- Σ = σ₀² × Q
- diagonal elements = variances
- off-diagonal = covariance σ_xy
- σ_x = √(Σ₁₁)
- ρ_xy = σ_xy/(σ_x × σ_y)
- positively correlated
- 0.237
Mark wise strategy
State the key definition, formula, or factual answer directly — no elaboration needed. If numerical, write the formula, substitute, and state the answer with unit. Never pad with background information.
Do
- Write the specific formula or definition immediately
- Include the correct SI unit on numerical answers
- State the key technical term (e.g., 'Law of Propagation of Variances', '39% probability')
- For numerical, show: formula → substitution → answer in three compact lines
Don't
- Do not write an introductory sentence like 'Error propagation is a method that...'
- Do not show detailed derivation steps for a 1-mark numerical
- Do not leave out the unit — always include m, mm, m², arc-seconds ('')
Structure your answer to earn 1 mark per distinct element. For numericals: formula → partial derivatives → substitution → answer. For conceptual: definition → distinguish/contrast → example.
Do
- Write the governing formula before substituting any values
- Show each partial derivative explicitly and label it
- Give a brief engineering context or example for conceptual questions
- Write intermediate steps clearly so partial credit is possible even if the final answer has arithmetic errors
Don't
- Do not jump directly from given data to the final answer without showing the formula
- Do not confuse σ√n (sum) with σ/√n (mean) — state which applies and why
- Do not abbreviate unit symbols incorrectly (m² not sq.m, '' not sec)
Allocate 1 mark per logical component. For three-part questions, write one clear paragraph per part. For numericals: given → formula → partial derivatives → propagation computation → answer → engineering interpretation.
Do
- Label sections clearly: '(a)', '(b)', '(c)' or 'Step 1', 'Step 2', 'Step 3'
- For error ellipse/covariance questions, write the 2×2 matrix in full block form with row/column labels
- Include an engineering interpretation of the numerical result (e.g., relative precision, what the σ value means in practice)
- Explicitly state and check the quadrant when computing orientation angles
Don't
- Do not write one long unbroken paragraph — examiners award marks per logical point
- Do not present only the final number for a 3-mark question — intermediate steps are mandatory
- Do not confuse the standard error ellipse (~39%) with the 95% confidence ellipse — state the probability
Treat a 5-mark question as a mini-report: list all given data first, identify the method, execute all steps in order, box the final answer, and write one conclusion sentence. Examiners specifically look for the quadrant check on angles, proper unit management, and an engineering interpretation of the result.
Do
- Write all given data in a 'Given:' block at the top before starting the solution
- Write the governing formula symbolically before substituting numerical values
- Show every intermediate computation (trace, determinant, eigenvalues, eigenvectors/angle) as a numbered step
- Box or underline the final answer and include all required units
- Write a brief conclusion statement connecting the numerical result to the engineering context (e.g., 'The elongated ellipse with a/b ratio of 4.7 indicates weak geometry in the NE direction of the PPCS network')
- For covariance/ellipse problems, write the covariance matrix in standard 2×2 block format
Don't
- Do not skip steps — even if a step seems trivial (e.g., computing the trace), write it explicitly as the examiner marks by step
- Do not use σ_x² directly as the semi-axis of the error ellipse — axes come from the eigenvalues of the covariance matrix, which are different from the diagonal entries when σ_xy ≠ 0
- Do not omit the probability level when describing error ellipses
- Do not mix mm and m within a single computation — choose one unit and be consistent throughout
General answer writing tips
- Always begin a conceptual question with a one-sentence definition that includes the technical term itself — examiners are trained to award the first mark for a precise definition.
- For numerical questions, write the governing formula first (before substituting values), then substitute, then compute — this three-step presentation earns partial credit even if arithmetic is wrong.
- Use proper SI units throughout: metres (m), seconds (arc-seconds, ''), square metres (m²); include units on every intermediate line, not just the final answer.
- When applying the Law of Propagation of Variances, explicitly write each partial derivative and label it — never silently combine terms, as examiners check intermediate steps.
- Add in quadrature for independent errors: always write σ_total = √(σ₁² + σ₂²), never σ₁ + σ₂, and state that the quantities are statistically independent to justify this form.
- For error ellipse questions, always name both semi-axes (a and b), state the orientation angle θ, and specify the probability level (e.g., 39% for the standard ellipse, 95% for the confidence ellipse).
- Distinguish σ/√n (standard deviation of the mean) from σ√n (standard deviation of a sum) — writing the wrong formula is the single most common reason for zero marks on propagation problems.
- When the question involves a covariance matrix, write it in full matrix notation, label rows and columns with the parameter names (e.g., x, y), and identify which entries are variances and which are covariances.
Diagrams
Chapter Concept Map: Error Propagation, Variance-Covariance and Error Ellipses
A hierarchical mind map showing all major topics of the chapter: the scalar and matrix forms of the Law of Propagation of Variances, special cases (sum, mean, scaled), the variance-covariance matrix structure, and error ellipse computation and interpretation. Use this as a visual study guide before attempting any template.
Decision Flowchart: Which Error Propagation Formula to Apply
A step-by-step decision flowchart guiding students to select the correct error propagation formula based on the type of function, independence of inputs, and whether a sum or mean is being computed. This directly addresses the most common source of errors on board exam problems.
Step-by-Step: Computing the Standard Error Ellipse from a 2x2 Covariance Matrix
A comprehensive flowchart showing every step required to compute the standard error ellipse from a 2×2 covariance sub-matrix: trace and determinant, characteristic equation, eigenvalues, semi-axes, orientation angle with quadrant check, and optional scaling to the 95% confidence ellipse. Follow this exact sequence in board exam answers for full marks.
Probability Coverage of Common Error Ellipse Confidence Levels
A pie chart illustrating that the standard error ellipse covers only ~39% of the probability distribution — far less than commonly assumed. Students frequently confuse this with the 1-sigma (68%) or 95% levels. The chart visually reinforces why the scale factor k = 2.45 is needed to reach the 95% confidence ellipse used in practical geodetic reporting.