Question 1 |
In a three-phase signal system design for a four-leg intersection, the critical flow ratios for each phase are 0.18, 0.32, and 0.22. The total loss time in each of the phases is 2s. As per Webster's formula, the optimal cycle length (in s, round off to the nearest integer) is _______________
28 | |
46 | |
50 | |
58 |
Question 1 Explanation:
\begin{aligned} L &=2 \times 3=6 \text { seconds } \\ n &=3 \\ y &=(0.18+0.32+0.22) \\ C_{0} &=\left(\frac{1.5 L+5}{1-y}\right)=\left(\frac{1.5 \times 6+5}{1-(0.18+0.32+0.22)}\right) \\ &=50 \text { seconds } \end{aligned}
Question 2 |
Relationship between traffic speed and density is described using a negatively sloped straight line. If v_{f} is the free flow speed then the speed at which the maximum flow occurs is
0 | |
\frac{v_{f}}{4} | |
\frac{v_{f}}{2} | |
v_{f} |
Question 2 Explanation:
Speed at maximum flow =\frac{V_{f}}{2}
Question 3 |
Spot speeds of vehicles observed at a point on a highway are 40, 55, 60, 65 and 80 km/h. The space-mean speed (in km/h, round off to two decimal places) of the observed vehicles is ______________
56.99 | |
25.36 | |
47.85 | |
96.36 |
Question 3 Explanation:
\begin{aligned} V_{1}=40, V_{2}=55, V_{3}=60, & V_{4}=65, V_{5}=80 \\ V_{S}=&\left(\frac{n}{\frac{1}{V_{1}}+\frac{1}{V_{2}}+\ldots+\frac{1}{V_{n}}}\right)=\frac{1}{40}+\frac{1}{55}+\frac{1}{60}+\frac{1}{65}+\frac{1}{80} \\ =& 56.99 \mathrm{kmph} \end{aligned}
Question 4 |
Vehicular arrival at an isolated intersection follows the Poisson distribution. The mean vehicular arrival rate is 2 vehicle per minute. The probability (round off to two decimal places) that at least 2 vehicles will arrive in any given 1-minute interval is _____
0.59 | |
0.47 | |
0.24 | |
0.89 |
Question 4 Explanation:
P(n,t)=\frac{(\lambda t)^n e^{-\lambda t}}{n!}
\lambda =2 \;v/m
t = 1 \text{ minute}
probability of at least 2 vehicles arriving in 1 ? minute is given by,
1-(P(0,1)+P(1,1))
\begin{aligned} &= 1-\left \{ \frac{(2 \times 1)^0 e^{-2(1)}}{0!}+\frac{(2 \times 1)^1 e^{-2(1)}}{1!} \right \}\\ &=1-\left \{ \frac{1(e^{-2})}{1}+ \frac{2(e^{-2})}{1} \right \} \\ &= 1-(e^{-2}+2e^{-2})\\ &=1-3e^{-2}=0.59 \end{aligned}
\lambda =2 \;v/m
t = 1 \text{ minute}
probability of at least 2 vehicles arriving in 1 ? minute is given by,
1-(P(0,1)+P(1,1))
\begin{aligned} &= 1-\left \{ \frac{(2 \times 1)^0 e^{-2(1)}}{0!}+\frac{(2 \times 1)^1 e^{-2(1)}}{1!} \right \}\\ &=1-\left \{ \frac{1(e^{-2})}{1}+ \frac{2(e^{-2})}{1} \right \} \\ &= 1-(e^{-2}+2e^{-2})\\ &=1-3e^{-2}=0.59 \end{aligned}
Question 5 |
A signalized intersection operates in two phases. The lost time is 3 seconds per phase. The maximum ratios of approach flow to saturation flow for the two phases are 0.37 and 0.40. The optimum cycle length using the Webster's method (in seconds, round off to one decimal place) is ______________
44.8 | |
92.6 | |
25.8 | |
60.9 |
Question 5 Explanation:
\begin{aligned} \mathrm{n} &=2, \mathrm{~L}=3 \times 2=6 \mathrm{sec} \\ \mathrm{y}_{1} &=0.37, \mathrm{y}_{2}=0.40 \\ \mathrm{Y} &=\mathrm{y}_{1}+\mathrm{y}_{2} \\ &=0.37+0.40 \\ &=0.77 \\ \mathrm{C}_{0} &=\left(\frac{1.5 \mathrm{~L}+5}{1-\mathrm{Y}}\right)=\frac{1.5 \times 6+5}{1-0.77} \\ &=60.87 \mathrm{sec} \simeq 60.9 \mathrm{sec} \end{aligned}
Question 6 |
The flow-density relationship of traffic on a headway is shown in the figure.

The correct representation of speed-density relationship of the traffic on this highway is


The correct representation of speed-density relationship of the traffic on this highway is

A | |
B | |
C | |
D |
Question 7 |
24-h traffic count at a road section was observed to be 1000 vehicles on a Tuesday
in the month of July. If daily adjustment factor for Tuesday is 1.121 and monthly
adjustment factor for July is 0.913, the Annual Average Daily Traffic (in veh/day, round
off to the nearest integer) is _______.
1121 | |
1023 | |
996 | |
1212 |
Question 7 Explanation:
\begin{aligned} T_{24} &=1000 veh \; (Tuesday) \\ DAF&=1.121 \\ AADT&=? \\ MAF&= 0.913\\ T_{week}&=T_{24} \times DAF \\ &=1000 \times 1.121=1121 \\ AADT&=MAF \times ADT \\ &= 0.913 \times 1121\\ &= 1023.473\\ &=1023 VPD \end{aligned}
Question 8 |
The traffic starts discharging from an approach at an intersection with the signal turning
green. The constant headway considered from the fourth or fifth headway position is
referred to as
discharge headway | |
effective headway | |
intersection headway | |
saturation headway |
Question 9 |
Traffic volume count has been collected on a 2 lane road section which needs upgradation
due to severe traffic flow condition. Maximum service flow rate per lane is observed as
1280 veh/h at level of service 'C'. The Peak Hour Factor is reported as 0.78125. Historical
traffic volume count provides Annual Average Daily Traffic as 122270 veh/day. Directional
split of the traffic flow is observed to be 60 : 40. Assuming that traffic stream consists
of 'All Cars' and all drivers are 'Regular Commuters', the number of extra lane(s) (round
off to the next higher integer) to be provide, is ________.
2 | |
8 | |
6 | |
4 |
Question 9 Explanation:
Number of the needed given LOS =\frac{DDHV}{PHF \times MSF \times F_{HV} \times f_p}
where,
DDHV = Directional distribution hour volume
PHF = Peak hour factor
MSF = Maximum service flow rate of Los
F_{HV} = Heavy vehicle familiarity adjustment factor
f_{p} = Road under familiarity adjustment factor
DDHV = 122270 x 0.6 = 7362 vah/day (0.6 - max. directional distributed)
F_{HV}=f_p=1 as per given condition
N=\frac{7362}{0.78125 \times 1280 \times 1 \times 1} =7.362 = 8 \text{ lanes}
Number of extra lanes = 8-2 = 6 \text{ lanes}
where,
DDHV = Directional distribution hour volume
PHF = Peak hour factor
MSF = Maximum service flow rate of Los
F_{HV} = Heavy vehicle familiarity adjustment factor
f_{p} = Road under familiarity adjustment factor
DDHV = 122270 x 0.6 = 7362 vah/day (0.6 - max. directional distributed)
F_{HV}=f_p=1 as per given condition
N=\frac{7362}{0.78125 \times 1280 \times 1 \times 1} =7.362 = 8 \text{ lanes}
Number of extra lanes = 8-2 = 6 \text{ lanes}
Question 10 |
The relationship between traffic flow rate (q) and density (D) is shown in the figure.

The shock wave condition is depicted by

The shock wave condition is depicted by
flow with respect to point 1 (q_1 = q_{max}) | |
flow changing from point 2 to point 6 (q_2 \gt q_6) | |
flow changing from point 3 to point 7 (q_3 \lt q_7) | |
flow with respect to point 4 and point 5(q_4 = q_5) |
Question 10 Explanation:

There are 10 questions to complete.