Past Year Questions: Traffic Engineering - 2 | Transportation Engineering - Civil Engineering (CE) PDF Download

Question 1. An observer counts 240 veh/h at a specific highway location. Assume that the vehicle arrival at the location is Poisson distributed, the probability of having one vehicle arriving over a 30-second time interval is _______ .     [2014 : 2 Marks, Set-II]
Solution:

Ans: 2 e^-2 ≈ 0.271
Sol:
Arrival rate λ = 240 veh/hr.
Time interval t = 30 s = 30/3600 hr = 1/120 hr.
Mean arrivals in t: λt = 240 × (1/120) = 2.
For a Poisson process, P(1 arrival) = (λt) e^-λt = 2 e^-2 ≈ 0.271.
Question 2. The average spacing between vehicles in a traffic stream is 50 m, then the density (in veh/km) of the stream is __________ .      [2014 : 1 Mark, Set-II]
Solution: 

Ans: 20 veh/km
Sol:
Spacing s = 50 m = 0.05 km per vehicle.
Density k = 1/s = 1/0.05 = 20 veh/km.
Question 3.  An isolated three-phase traffic signal is designed by Webster's method. The critical flow ratios for three phases are 0.20,0.30, and 0.25 respectively and lost time per phase is 4 seconds. The optimum cycle length (in seconds) is ___________ .      [2014 : 2 Marks, Set-I]
Solution: Sum of the flow,
y = y₁ + y₂ + y₃ 
= 0.2 + 0.3 + 0.25 = 0.75
Total lost time in a cycle L = 4 × 3 = 12 s.
Optimum cycle length,
Ans: 92 s
Sol:
Webster's formula: C₀ = (1.5L + 5)/(1 - y).
Substitute L = 12 s and y = 0.75:
C₀ = (1.5 × 12 + 5)/(1 - 0.75) = (18 + 5)/0.25 = 23/0.25 = 92 s.

Question 4.  The speed-density (u-k) relationship on a single lane road with unidirectional flow is u = 70-0.7 k, where u is in km/hr and k is in veh/km. The capacity of the road (in veh/hr) i s ___________ .     [2014 : 2 Marks, Set-I]
Solution: 
Method-I

Method-II

Ans: 1750 veh/hr
Sol:
Flow q = u × k = (70 - 0.7k) k = 70k - 0.7k².
Differentiate w.r.t. k and set to zero for maximum: dq/dk = 70 - 1.4k = 0 ⇒ k = 70/1.4 = 50 veh/km.
Corresponding speed u = 70 - 0.7 × 50 = 70 - 35 = 35 km/hr.
Capacity q_max = u × k = 35 × 50 = 1750 veh/hr.
Question 5.  The minimum value of 15 minute peak hour factor on a section of a road is     [2014 : 1 Mark, Set-I]
(a) 0.10 
(b) 0.20 
(c) 0.25 
(d) 0.33 
Answer: (c)
Solution:
Ans: (c)
Explanation:
The 15-minute peak hour factor (PHF₁₅) is defined as V/(4 × V₁₅), where V is the hourly volume and V₁₅ is the maximum 15-minute volume within that hour.
The maximum possible value of V₁₅ is the full hour concentrated in one 15-minute interval, giving PHF₁₅ = 1. The minimum practically meaningful value occurs when all hourly flow is evenly concentrated into one 15-minute quarter of the hour, giving PHF₁₅ = 0.25. Normal observed range is about 0.7-0.98 for typical traffic; the theoretical minimum is 0.25.


Question 6. For two major-roads with divided carriage way crossing at right angle, a full clover leaf interchange with four indirect ramps is provided. Following statements are made on turning movement of vehicles to all direction from both road, identity the correct statement.   [2013 : 1 Mark]
(a) Merging from left is not possible, but diverging to left not possible.
(b) Merging from left and diverging to left is possible.
(c) Merging from left is possible but diverging is not possible.
(d) Neither merging from left nor diverging to left is possible.
Answer: (b)
Solution:
Ans: (b)
Explanation:
A full cloverleaf provides loop ramps (clover lefts) that allow vehicles to make left-turn movements via indirect paths, so merging from the left is achievable using the loop. Diverging to the left is possible using the indirect ramp geometry. Hence both merging from left and diverging to left are possible.

Question 7. It was observed that 150 vehicle crossed a particular location of highway in 30 minutes. Assume that vehicle arrival follow a negative exponential distribution. The number of time headways greater than 5 sec. in above observation is    [2013 : 1 Mark]
Solution:

Where, P(x) is the probability of x events (vehicle arrivals) in some time interval (t).
X is the mean arrival rate in that interval.

Now, the probability that zero vehicle arrive in an interval t, denoted as P(0), will be same as the probability that the headway (into arrival time) greater than or equal to t.

Ans: 99 vehicles (approximately)
Sol:
Total observed vehicles N = 150 in T = 30 min = 1800 s.
Arrival rate λ = 150/1800 = 0.083333... veh/s.
Probability that a headway exceeds 5 s: P(T > 5) = e^-λ·5 = e^-0.083333×5 = e^-0.416667 ≈ 0.6592.
Expected number of headways > 5 s = 150 × 0.6592 ≈ 98.9 ≈ 99.
Question 8. A two-lane urban road with one-way traffic has a maximum capacity of 1800 vehicles/hour. Under the jam condition, the average length occupied by the vehicles is 5.0 m. The speed versus density relationship is linear. For a traffic volume of 1000 vehicles/hour, the density (in vehicles/km) is    [2012 : 2 Marks]
(a) 52
(b) 58
(c) 67
(d) 75
Answer: (c)
Solution:
Method-II
Capacity per lane


∴ k = 33.32 Veh./km (for single lane)
So, for two lane urban road,
k = 33.32 × 2 = 66.64 ∼ 67
Ans: (c) 67
Explanation:
Given total capacity for two lanes = 1800 veh/hr ⇒ capacity per lane q_max,lane = 900 veh/hr.
Jam density per lane k_j = 1000 m/km ÷ 5 m/veh = 200 veh/km.
For a linear (Greenshields) relationship, q_max,lane = u_f × k_j/4 ⇒ 900 = u_f × 200/4 ⇒ u_f = 18 km/hr.
For a flow of 1000 veh/hr total ⇒ per lane q = 500 veh/hr. Solve q = k u_f (1 - k/k_j) with u_f=18 and k_j=200:
500 = 18 k (1 - k/200) ⇒ 0.09 k² - 18 k + 500 = 0.
Solving gives k ≈ 33.33 veh/km per lane (choose the stable lower root). For two lanes, total density ≈ 33.33 × 2 = 66.67 ≈ 67 veh/km.


Question 9. Two major roads with two lanes each are crossing in an urban areas to form an un-controlied intersection. The number of conflict points when both roads are one way is "X' and when both roads are two-way is " Y'. The ratio of  X to Y is     [2012 : 1 Mark]
(a) 0.25
(b) 0.33
(c) 0.50
(d) 0.75
Answer: (a)
Solution: When both roads are one-way then total no. of conflict points, x = 6
When both roads are two-way then total no . of conflict points, y = 24


Ans: (a)
Explanation:
For the given intersection geometry, the number of basic vehicular conflict points is 6 when both approaches operate one-way, and 24 when both approaches are two-way. Thus X/Y = 6/24 = 0.25.
Question 10.  The cumulative arrival and departure curve of one cycle of an approach lane of a signalized intersection is shown in the adjoining figure. The cycle time is 50 s and the effective red time is 30s and the effective green time is 20 s. What is the average delay?   [2011 : 2 Marks]

(a) 15 s
(b) 25 s
(c) 35 s
(d) 45 s
Answer: (a)
Solution:
Method-I
Average delay

Method-II

Vehicle 1 arrives at 0 second and departs at 30 seconds.
Vehicle 40 arrivers at 50 seconds and departs at the same time.
Delay for vehicle 1 = 30 second
Delay for vehicle 40 = 0 second

Ans: (a) 15 s
Explanation:
The delays of queued vehicles decrease linearly from the maximum delay (30 s) for the first queued vehicle to zero for the last vehicle cleared in the cycle. The average delay is the area of this right triangle divided by the number of vehicles:
Average delay = (1/2 × base × height)/number = (1/2 × 40 × 30)/40 = 15 s.
Question 11. If the jam density is given as k- and the free flow speed is given as u_f, the maximum flow for a linear traffic speed-density model is given by which of the following options?    [2011 : 2 Marks]
(a)

(b)

(c)

(d)

Answer: (a)
Ans: (a)
Explanation:
For a linear speed-density (Greenshields) model, u = u_f(1 - k/k_j). Flow q = k u = u_f k (1 - k/k_j). This quadratic attains its maximum at k = k_j/2, giving q_max = u_f k_j/4. Hence option (a) is correct.
Question 12. The probability that k number of vehicles arrive (i.e. cross a predefined line) in time t is given as (λt) e^-λt/k! where λ is the average vehicle arrival rate. What is the probability that the time headway is greater than or equal to time t₁?   [2011 : 1 Mark]
(a)

(b)

(c)

(d)

Answer: (d)
Solution: Probability of arriving 'n' number of vehicles in time f,

If time headway is greater than or equal to time f, then number of vehicles arriving is zero,

Ans: (d)
Explanation:
Time headway ≥ t₁ means zero arrivals in interval t₁. For a Poisson process P(0 arrivals) = e^{-λ t₁}, which corresponds to option (d).
Question 13. As per IRC: 67-2001, a traffic sign indicating the Speed Limit on a road should be of   [2010 : 1 Mark]
(a) circular shape with white background and red border
(b) triangular shape with white background and red border
(c) triangular shape with red background and white border
(d) circular shape with red background and white border
Answer: (a)
Solution: Speed limit sign is a regulatory or mandatory sign and should be of circular shape with white background and red border.