CELE Transportation & Highway Engineering — Traffic Engineering and Highway CapacityExam Answer Templates

Marks

1

Topic

Fundamental Traffic Flow Relationship

Difficulty

easy

Template Id

T1

Examiner Tip

A 1-mark question expects a single, precise statement. The formula alone with correct symbols earns full credit, but naming the variables shows mastery and prevents ambiguity.

Model Answer

The fundamental traffic flow relationship is: q = k × u where q = flow (veh/hr), k = density (veh/km), and u = space-mean speed (km/h).

Question Type

very_short_answer

Answer Structure

• Line 1: Write the formula q = k × u [1 mark]
• Line 2 (bonus clarity): Identify each variable and unit

Scoring Breakdown

Common Mark Deductions

• Writing q = k/u or q = u/k (inverted relationship)
• Using 'speed' without specifying 'space-mean speed'
• Omitting units entirely

Key Phrases To Include

• q = ku
• flow
• density
• space-mean speed
• veh/hr
• veh/km
• km/h

Marks

1

Topic

Fundamental Traffic Variables

Difficulty

easy

Template Id

T2

Examiner Tip

Distinguish 'per unit length' (density) from 'per unit time' (flow) — this distinction is a frequent trick in the exam.

Model Answer

Traffic density (k) is the number of vehicles occupying a unit length of roadway at a given instant. Its SI unit is vehicles per kilometre (veh/km).

Question Type

very_short_answer

Answer Structure

• Line 1: Definition of density — number of vehicles per unit length of road [1 mark]
• Line 2: SI unit — veh/km

Scoring Breakdown

Common Mark Deductions

• Confusing density with flow (flow is per unit time, density is per unit length)
• Stating unit as veh/m or veh/hr instead of veh/km

Key Phrases To Include

• number of vehicles
• unit length of roadway
• given instant
• veh/km

Marks

1

Topic

Spacing and Headway

Difficulty

easy

Template Id

T3

Examiner Tip

Memorise: headway uses 3600 (seconds/hr), spacing uses 1000 (metres/km). Mix them up and you lose the mark.

Model Answer

Given: q = 900 veh/hr Formula: h = 3600 / q h = 3600 / 900 = 4 s The average headway is 4 seconds.

Question Type

numerical

Answer Structure

• Line 1: State the given value
• Line 2: Write the formula h = 3600 / q [1 mark — formula + correct substitution]
• Line 3: Compute the answer with unit

Scoring Breakdown

Common Mark Deductions

• Using h = 1000 / q (confusing headway formula with spacing formula)
• Omitting the unit 'seconds'
• Arithmetic error in division

Key Phrases To Include

• h = 3600 / q
• headway
• seconds
• 3600

Marks

2

Topic

Fundamental Traffic Flow Relationship

Difficulty

easy

Template Id

T4

Examiner Tip

Always show the formula, then substitute. Even if you make an arithmetic slip, you earn the method mark for the correct formula.

Model Answer

Given: k = 50 veh/km u = 40 km/h Required: Traffic flow q Solution: Using the fundamental relation: q = k × u q = 50 × 40 q = 2 000 veh/hr The traffic flow is 2 000 veh/hr.

Question Type

numerical

Answer Structure

• Step 1: List given data with units [0.5 mark]
• Step 2: Identify and write the formula q = ku [0.5 mark]
• Step 3: Substitute and compute [0.5 mark]
• Step 4: State final answer with correct unit [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Forgetting to specify 'space-mean speed' when defining u
• Arithmetic error (e.g., 50 × 40 ≠ 2000)
• Omitting the unit veh/hr in the final answer

Key Phrases To Include

• q = ku
• fundamental relation
• 2 000 veh/hr
• space-mean speed

Marks

2

Topic

Spacing and Headway

Difficulty

easy

Template Id

T5

Examiner Tip

Show both formulas clearly labelled (a) and (b). Examiners mark each part independently — a wrong headway doesn't void the spacing mark.

Model Answer

Given: k = 40 veh/km q = 1 200 veh/hr Required: Average spacing s and average headway h Solution: (a) Average spacing: s = 1 000 / k s = 1 000 / 40 s = 25 m (b) Average headway: h = 3 600 / q h = 3 600 / 1 200 h = 3 s The average spacing is 25 m and the average headway is 3 s.

Question Type

numerical

Answer Structure

• Step 1: Write spacing formula s = 1000/k and compute [1 mark]
• Step 2: Write headway formula h = 3600/q and compute [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using 1000 in the headway formula instead of 3600
• Using 3600 in the spacing formula instead of 1000
• Not labelling which answer corresponds to spacing vs. headway

Key Phrases To Include

• s = 1000/k
• h = 3600/q
• 25 m
• 3 s

Marks

2

Topic

Peak-Hour Factor

Difficulty

easy

Template Id

T6

Examiner Tip

The denominator '4' represents four 15-minute periods in one hour. Explaining this conversion earns full credit and shows conceptual understanding.

Model Answer

The Peak-Hour Factor (PHF) is a dimensionless ratio that measures the temporal distribution of traffic demand within the peak hour. It quantifies how uniformly traffic is spread across the four 15-minute periods of the peak hour. Formula: PHF = V / (4 × V₁₅) where: V = total peak-hour volume (veh) V₁₅ = volume during the highest 15-minute period within the peak hour (veh) PHF ranges from 0.25 (all demand in one 15-min period) to 1.00 (perfectly uniform flow).

Question Type

short_answer

Answer Structure

• Sentence 1: Definition of PHF — temporal distribution of demand within peak hour [1 mark]
• Sentence 2: State formula PHF = V / (4V₁₅) with variable definitions [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing PHF = 4V₁₅ / V (inverted formula)
• Not defining V₁₅ as the highest 15-minute volume
• Omitting the range (0.25 to 1.00)

Key Phrases To Include

• PHF
• peak hour
• 15-minute
• V₁₅
• 4V₁₅
• 0.25
• 1.00
• temporal distribution

Marks

3

Topic

Peak-Hour Factor

Difficulty

medium

Template Id

T7

Examiner Tip

Mention the alternative formula q_design = 4V₁₅ as a cross-check. Examiners value verification steps — it shows engineering discipline.

Model Answer

Given: V = 2 400 veh (peak-hour volume) V₁₅ = 700 veh (peak 15-minute volume) Required: (a) PHF, (b) Design flow rate Solution: (a) Peak-Hour Factor: PHF = V / (4 × V₁₅) PHF = 2 400 / (4 × 700) PHF = 2 400 / 2 800 PHF = 0.857 ≈ 0.86 (b) Design Flow Rate: q_design = V / PHF q_design = 2 400 / 0.857 q_design ≈ 2 800 veh/hr Note: Alternatively, q_design = 4 × V₁₅ = 4 × 700 = 2 800 veh/hr (same result). The PHF is 0.86 and the design flow rate is 2 800 veh/hr.

Question Type

numerical

Answer Structure

• Step 1: List given data [0.5 mark]
• Step 2: Apply PHF = V/(4V₁₅) and compute [1 mark]
• Step 3: Apply design flow = V/PHF and compute [1 mark]
• Step 4: State final answers with units [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Computing PHF correctly but using a wrong formula for design flow (e.g., V × PHF instead of V/PHF)
• Rounding PHF too early and getting a significantly wrong design flow
• Forgetting the unit veh/hr for the design flow rate

Key Phrases To Include

• PHF = V / (4 × V₁₅)
• 2 800 veh/hr
• design flow rate
• q_design = V / PHF

Marks

3

Topic

Fundamental Traffic Variables

Difficulty

medium

Template Id

T8

Examiner Tip

The phrase 'harmonic mean' instantly signals to the examiner that you know the technical definition. Pair it with the phrase 'distance divided by average travel time' for a complete, mark-earning response.

Model Answer

Space-mean Speed (u_s): The harmonic mean of individual vehicle speeds, computed as the total distance travelled divided by the total travel time for all vehicles in the observation space. It represents the average speed experienced over a section of road. Time-mean Speed (u_t): The arithmetic mean of individual vehicle speeds measured at a fixed point during a specific time interval (e.g., radar gun readings). It is always greater than or equal to the space-mean speed. Relationship: u_t ≥ u_s (time-mean speed ≥ space-mean speed) The space-mean speed (u_s) is used in q = ku because: • The flow equation is derived from the conservation of vehicles over a road section and a time interval. • Space-mean speed correctly averages the time each vehicle spends in the section, making it thermodynamically consistent with density k (veh/km). • Using time-mean speed would overestimate the flow for a given density.

Question Type

short_answer

Answer Structure

• Paragraph 1: Define space-mean speed — harmonic mean, distance/time over a section [1 mark]
• Paragraph 2: Define time-mean speed — arithmetic mean at a fixed point [1 mark]
• Paragraph 3: State which is used in q = ku and justify with the conservation argument [1 mark]

Scoring Breakdown

Common Mark Deductions

• Reversing the definitions (calling arithmetic mean 'space-mean')
• Stating time-mean speed is used in q = ku
• No justification for why space-mean is used — just saying 'it is the correct one' is insufficient

Key Phrases To Include

• space-mean speed
• harmonic mean
• time-mean speed
• arithmetic mean
• u_t ≥ u_s
• q = ku
• conservation
• fixed point

Marks

3

Topic

Fundamental Traffic Flow Relationship

Difficulty

medium

Template Id

T9

Examiner Tip

Always convert time to hours before computing flow. A common error is leaving the denominator as '30 minutes' — this yields veh/min, not veh/hr.

Model Answer

Given: Count = 750 veh in 30 min Spacing s = 30 m Required: (a) q, (b) k, (c) u Solution: (a) Flow: q = 750 veh / (30/60 hr) q = 750 / 0.5 q = 1 500 veh/hr (b) Density: s = 1 000 / k → k = 1 000 / s k = 1 000 / 30 k = 33.33 veh/km (c) Space-mean speed: q = k × u → u = q / k u = 1 500 / 33.33 u = 45.0 km/h Summary: q = 1 500 veh/hr, k = 33.33 veh/km, u = 45.0 km/h.

Question Type

numerical

Answer Structure

• Step (a): Convert time and compute q = count / time in hours [1 mark]
• Step (b): Invert spacing formula to get k = 1000/s [1 mark]
• Step (c): Derive u = q/k from fundamental relation [1 mark]

Scoring Breakdown

Common Mark Deductions

• Forgetting to convert 30 minutes to 0.5 hours when computing flow
• Using k = 1000 × s instead of k = 1000/s
• Inconsistent units between parts leading to a wrong speed

Key Phrases To Include

• 1 500 veh/hr
• 33.33 veh/km
• 45 km/h
• k = 1000/s
• u = q/k

Marks

3

Topic

Capacity and Level of Service

Difficulty

medium

Template Id

T10

Examiner Tip

Examiners specifically test whether you know the correct LOS parameter: DENSITY for freeways, DELAY for signalised intersections. State this distinction explicitly for full marks.

Model Answer

Level of Service (LOS) is a qualitative measure that characterises the operating conditions of a traffic stream. For freeways, the Highway Capacity Manual (HCM) defines six levels, LOS A through LOS F, graded primarily by density (pc/km/ln). LOS A — Free-flow conditions. Density is very low (≤ 7 pc/km/ln on basic freeway segments). Drivers can freely select their desired speed with virtually no restriction from other vehicles. Travel is highly comfortable and convenient. LOS F — Forced or breakdown flow. Demand exceeds capacity (typically > 28 pc/km/ln or flow collapse). The traffic stream operates at unstable, stop-and-go conditions. Average speeds drop sharply, delay is excessive, and queues form and propagate upstream. Key parameter for freeways: Density (pc/km/ln). This distinguishes freeways from signalised intersections, where delay (s/veh) is the primary LOS measure.

Question Type

short_answer

Answer Structure

• Sentence 1–2: Define LOS and state it covers A to F, measured by density for freeways [1 mark]
• Paragraph 2: Describe LOS A — free flow, low density, free speed selection [1 mark]
• Paragraph 3: Describe LOS F — breakdown, demand > capacity, stop-and-go [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using delay (s/veh) as the LOS parameter for freeways — this is for signalised intersections
• Reversing LOS A and LOS F descriptions
• Not quantifying LOS A density threshold (≤7 pc/km/ln)

Key Phrases To Include

• Level of Service
• density pc/km/ln
• LOS A
• free-flow
• LOS F
• breakdown
• demand exceeds capacity
• stop-and-go

Marks

2

Topic

Capacity and Level of Service

Difficulty

easy

Template Id

T11

Examiner Tip

The word 'passenger cars' is crucial — capacity is expressed in pc/h/ln because heavy vehicles are converted using Passenger Car Equivalents (PCE). Always use 'pc' not 'veh' for this value.

Model Answer

The ideal capacity of a single freeway lane under ideal conditions is approximately 2 000 passenger cars per hour per lane (pc/h/ln). Three ideal conditions assumed: 1. Lane width of at least 3.6 m. 2. Clearance of at least 1.8 m from obstructions on the right shoulder and 0.6 m on the left (lateral clearance). 3. All passenger cars in the traffic stream (no heavy vehicles — trucks, buses, or recreational vehicles).

Question Type

short_answer

Answer Structure

• Line 1: State capacity ≈ 2 000 pc/h/ln [1 mark]
• Lines 2–4: List three valid ideal conditions [1 mark — at least two required]

Scoring Breakdown

Common Mark Deductions

• Stating 2 000 veh/hr without the qualifier 'pc' (passenger cars) — trucks reduce capacity
• Listing random conditions not related to HCM ideal conditions
• Giving a capacity value far outside the 1 800–2 400 pc/h/ln range without justification

Key Phrases To Include

• 2 000 pc/h/ln
• ideal conditions
• lane width
• lateral clearance
• passenger cars
• no heavy vehicles

Marks

5

Topic

Peak-Hour Factor

Difficulty

hard

Template Id

T12

Examiner Tip

The comment/interpretation portion is worth a full mark in 5-mark questions. Use engineering language: say 'the design flow exceeds the hourly volume because of peaking' — this shows you understand the purpose of PHF in design.

Model Answer

Given: 15-minute period counts: 420, 510, 480, 390 veh Required: (a) V, (b) PHF, (c) design flow rate (a) Peak-Hour Volume: V = 420 + 510 + 480 + 390 V = 1 800 veh/hr (b) Peak 15-Minute Volume: V₁₅ = 510 veh (highest count in the four periods) PHF = V / (4 × V₁₅) PHF = 1 800 / (4 × 510) PHF = 1 800 / 2 040 PHF = 0.882 ≈ 0.88 (c) Design Flow Rate: q_design = V / PHF q_design = 1 800 / 0.882 q_design ≈ 2 040 veh/hr Alternative: q_design = 4 × V₁₅ = 4 × 510 = 2 040 veh/hr ✓ Comment: A PHF of 0.88 indicates that the peak hour is relatively uniform — the busiest 15-minute period (510 veh) represents 88% of the theoretical uniform hourly rate. The design flow rate of 2 040 veh/hr (> 1 800 veh/hr) accounts for the peaking nature of demand; geometric and signal design must be based on this higher value to prevent LOS failure during the peak 15 minutes.

Question Type

numerical

Answer Structure

• Step (a): Sum four 15-min counts to get V = 1 800 veh [1 mark]
• Step (b): Identify V₁₅ = 510 (maximum), apply PHF = V/(4V₁₅) = 0.88 [1.5 marks]
• Step (c): Compute design flow = V/PHF = 2 040 veh/hr [1.5 marks]
• Comment: Interpret PHF value and engineering significance [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using the average 15-min count (450) instead of the maximum (510) for V₁₅
• Skipping the comment or writing a superficial one (e.g., 'the PHF is good') — no marks awarded
• Not converting the design flow to veh/hr units
• Forgetting to identify V₁₅ before computing PHF

Key Phrases To Include

• V₁₅ = 510
• PHF = 0.88
• design flow rate = 2 040 veh/hr
• peaking nature
• q_design = V/PHF
• 4 × V₁₅

Marks

5

Topic

Capacity and Level of Service

Difficulty

hard

Template Id

T13

Examiner Tip

Always write out at least the adjacent LOS thresholds (LOS D upper and LOS E upper boundary) when identifying LOS — this proves you know the standard, not just that you guessed the letter.

Model Answer

Given: q = 1 900 pc/hr/ln k = 25 pc/km/ln V₁₅ = 530 pc (peak 15-minute, per lane) Required: (a) space-mean speed, (b) LOS, (c) PHF (a) Space-Mean Speed: From q = k × u: u = q / k u = 1 900 / 25 u = 76 km/h (b) Level of Service (Density-Based, HCM Basic Freeway Segments): k = 25 pc/km/ln HCM LOS thresholds (basic freeway, free-flow speed ≈ 110–120 km/h): LOS A: k ≤ 7 pc/km/ln LOS B: k ≤ 11 pc/km/ln LOS C: k ≤ 16 pc/km/ln LOS D: k ≤ 22 pc/km/ln LOS E: k ≤ 28 pc/km/ln ← 25 pc/km/ln falls here LOS F: k > 28 pc/km/ln (or flow breakdown) Since 22 < 25 ≤ 28 → LOS E (operating near capacity, unstable flow possible) (c) Peak-Hour Factor: V = q = 1 900 pc/hr/ln (peak-hour volume per lane) PHF = V / (4 × V₁₅) PHF = 1 900 / (4 × 530) PHF = 1 900 / 2 120 PHF = 0.896 ≈ 0.90 Summary: (a) u = 76 km/h (b) LOS E — near-capacity operation (c) PHF = 0.90

Question Type

numerical

Answer Structure

• Step (a): u = q/k = 76 km/h [1 mark]
• Step (b): Show LOS thresholds table or key values; identify k = 25 within LOS E range [2 marks]
• Step (c): PHF = V/(4V₁₅) = 1900/2120 = 0.90 [1 mark]
• Summary with correct labels and units [1 mark]

Scoring Breakdown

Common Mark Deductions

• Identifying LOS D instead of LOS E — must check 22 < 25 ≤ 28 carefully
• Stating LOS without citing the threshold values — loses 1 mark for LOS part
• Using wrong formula for speed (u = k/q instead of u = q/k)
• Confusing per-lane values with total roadway values

Key Phrases To Include

• u = q/k = 76 km/h
• LOS E
• density = 25 pc/km/ln
• near-capacity
• PHF = 0.90
• HCM density thresholds

Marks

5

Topic

Fundamental Traffic Flow Relationship

Difficulty

hard

Template Id

T14

Examiner Tip

For 5-mark long-answer questions, use paragraph headings (Free-Flow Branch, Capacity Point, Congested Branch) — this organises your answer and ensures you don't skip any of the five scoring points.

Model Answer

The Fundamental Diagram of Traffic Flow is a graphical representation of the relationship q = ku, showing how flow (q) varies with density (k) and speed (u) under different traffic states. 1. Free-Flow Branch (k < k_optimal): As density increases from zero, speed remains approximately constant (near free-flow speed u_f) because vehicles are widely spaced and do not interact. Consequently, flow increases nearly linearly: q = k × u_f (approximately) This region corresponds to LOS A through D on freeways. 2. Capacity Point (k = k_optimal, q = q_max): At the optimum or critical density k_opt, the product q = ku is maximised. This is the capacity of the facility (≈ 2 000 pc/hr/ln under ideal conditions). At this point, speed has decreased to its critical value u_opt = q_max / k_opt. 3. Congested Branch (k > k_optimal): Beyond k_opt, increasing density forces drivers to slow down so severely that the loss in speed outweighs the gain from more vehicles — flow decreases despite higher density. In the extreme case (jam density k_j), speed approaches zero and flow = 0. This is the forced-flow or breakdown region (LOS F). Key Observations: • Flow-speed curve: q increases as speed decreases from u_f to u_opt; then q decreases further to zero at standstill. • Speed-density curve: Speed decreases monotonically from u_f (at k = 0) to 0 (at k = k_j). • Same flow value q can exist at two different states — one on each branch — leading to non-unique solutions that are important in traffic incident management. Conclusion: Engineers must design for the free-flow branch (maintain k < k_opt) to ensure stable and predictable traffic operations. Crossing into the congested branch results in LOS E/F and queue formation.

Question Type

long_answer

Answer Structure

• Paragraph 1: Introduce the fundamental diagram and q = ku [1 mark]
• Paragraph 2: Free-flow branch — linear increase in q, near constant speed, LOS A–D [1 mark]
• Paragraph 3: Capacity point — maximum q at optimum density and critical speed [1 mark]
• Paragraph 4: Congested branch — q decreases despite increasing k, LOS F [1 mark]
• Paragraph 5: Key observations — non-uniqueness and design implication [1 mark]

Scoring Breakdown

Common Mark Deductions

• Claiming flow always increases with density — fails to describe the congested branch
• Not mentioning the capacity/optimum density concept
• No mention of LOS linkage or design implication
• Diagram-only answer without explanation — no marks unless description accompanies

Key Phrases To Include

• q = ku
• free-flow branch
• congested branch
• optimum density
• capacity
• jam density
• speed decreases
• flow decreases
• LOS F
• non-unique

Marks

3

Topic

Spacing and Headway

Difficulty

medium

Template Id

T15

Examiner Tip

Verification problems reward students who show the final check explicitly (25 × 36 = 900 ✓). Write the check mark — it signals to the examiner that you intentionally verified the answer.

Model Answer

For Observation 2 (find q₂, k₂, u₂ from s₂ and h₂): k₂ = 1 000 / s₂ = 1 000 / 40 = 25 veh/km q₂ = 3 600 / h₂ = 3 600 / 4 = 900 veh/hr u₂ = q₂ / k₂ = 900 / 25 = 36 km/h Verification of q = ku: k₂ × u₂ = 25 × 36 = 900 = q₂ ✓ For Observation 1 (if k₁ = 30 veh/km and u₁ from context is unknown): q₁ = k₁ × u₁ = 30 u₁ — consistent with q = ku by definition. Conclusion: Observation 2 is internally consistent with the fundamental relation q = ku. The computed values k₂ = 25 veh/km, q₂ = 900 veh/hr, and u₂ = 36 km/h satisfy q = ku exactly.

Question Type

numerical

Answer Structure

• Step 1: Compute k₂ from spacing formula k = 1000/s [1 mark]
• Step 2: Compute q₂ from headway formula q = 3600/h [1 mark]
• Step 3: Compute u₂ = q/k and verify k × u = q [1 mark]

Scoring Breakdown

Common Mark Deductions

• Swapping the spacing and headway formulas
• Not performing the verification multiplication k × u = q
• Failing to state 'consistent' or 'verified' in the conclusion

Key Phrases To Include

• k = 1000/s
• q = 3600/h
• u = q/k
• 25 veh/km
• 900 veh/hr
• 36 km/h
• verified