CIV2282 Transport and Traffic Engineering Week 12, 2019
Prac Class 12 – Managed Motorways and Non-Motorised Transport Page 1 of 4
| Group | Participating Group Members |
MONASH UNIVERSITY
DEPARTMENT OF CIVIL ENGINEERING
CIV2282: Transport and Traffic Engineering
Semester 2/2019, Week 12 beginning Monday 21st October
Practical Class 12: Managed Motorway and Non-motorised Transport
(With Answers)
The exercises in this practical class draw on the material presented in the lecture slides
“Week 12 – Managed Motorways”.
You will probably need a spreadsheet such as Excel to complete these exercises,
remember to bring a computer with spreadsheet software into the practical class.
Exercise 1 – Ramp metering
A freeway on-ramp uses ramp metering to regulate the flow of traffic entering the freeway.
The on-ramp has two lanes, one for single passenger vehicles (SPVs) which must wait at the
ramp meter, and an HOV (high occupancy vehicle) bypass lane where vehicles may enter the
freeway mainline without stopping at the meter. Vehicles arrive at the ramp at a rate of 300
vehicles per hour during the whole day, except for during one peak hour where the rate is 1000
vehicles per hour. Across the entire day, 20% of all vehicles are HOVs each carrying 2 persons.
The freeway mainline has a capacity of 6000 vehicles per hour and carries a demand of 5400
vehicles per hour over the whole day. The ramp queue can hold as many vehicles as needed.
a) Determine the optimal metering rate β for the SPV lane, i.e. the proportion that the onramp flow must be reduced to.
Outside the peak hour there is no need for ramp metering since the mainline capacity (6000
veh/h) is greater than the total demand (5700 veh/h), which is the sum of the mainline
demand (5400 veh/h) and the ramp demand (300 veh/h).
During the peak hour, the total demand (5400 + 1000 = 6400 veh/h) exceeds the mainline
capacity (6000 veh/h), so we need ramp metering to prevent congestion in the mainline.
To calculate the optimal metering rate β, we equate the mainline capacity to the demand
restricted by ramp metering:
6000 veh/h = 5400 veh/h + (0.2 + β × 0.8) × 1000 veh/h.
Solving the above equation for β, we obtain β = 0.5.
The optimum metering rate during the peak hour is 0.5.
CIV2282 Transport and Traffic Engineering Week 12, 2019
Prac Class 12 – Managed Motorways and Non-Motorised Transport Page 2 of 4
b) Assuming a D/D/1 queueing system, what will be the total delay (in vehicle hours
travelled) for the SPV users?
We can calculate the delay graphically by drawing the queue accumulation polygon of the
SPV lane, as shown in Figure 1. At the beginning of the peak hour when we start to employ
ramp metering with β = 0.5 as found in a), a part of the demand for the SPV lane will be
accumulating in a queue. At the end of the peak hour, ramp metering will not be active (as
there is no need) and thus the queue will start dissolving and will disappear completely
around 2.3 hours after the start of the peak hour.
Figure 1 – Queue Accumulation Polygon, SPVs metered
The figure can be constructed as follows:
During the peak hour, the SPV demand (0.8 × 1000 = 800 veh/h) will be entering the SPV
lane (blue line, slope = 800 veh/h), whereas only half of it (β × 800 = 400 veh/h) is allowed
to exit into mainline (orange line, slope = 400 veh/h). At the end of the peak hour, there are
400 vehicles in the queue at the on-ramp behind the metering signal.
After the peak hour ends, the SPV demand is 0.8 × 300 = 240 veh/h (yellow line, slope =
240 veh/h), whereas the capacity available to the SPV demand is 6000 – 5400 – 0.2 × 300
= 540 veh/h (purple line, slope = 540 veh/h).
The queue at the on-ramp behind the metering signal reduces to zero after a time of 400 /
(540 – 240) = 11/3 hours (1 hour, 20 minutes) after the end of the peak hour.
The total delay is found by the area between the arrival (blue/yellow) and departure
(orange/purple) curves = (½ × 1 × 400) + (½ × 11/3 × 400) = 466.67
The total delay experienced by the SPV users is 467 vehicle-hours
0
200
400
600
800
1000
1200
1400
0 0.5 1 1.5 2 2.5
Cumulative Number of Vehicles
time (h)
CIV2282 Transport and Traffic Engineering Week 12, 2019
Prac Class 12 – Managed Motorways and Non-Motorised Transport Page 3 of 4
c) What will be the time saving (in passenger hours travelled) compared to a scenario where
all vehicles are treated the same (i.e. ramp metering is applied to both HOVs and SPVs)?
As before, we need to draw the queue accumulation polygon of the ramp for the case where
ramp metering is applied to both the HOV and SPV lanes (i.e., the vehicles are treated the
same), as shown in Figure 2.
Figure 2 – Queue Accumulation Polygon, SPVs and HOVs metered
In this case, the optimal metering rate β1 can be calculated from the following equation:
6000 veh/h = 5400 veh/h + β1 × 1000 veh/h.
Solving the above equation for β1, we obtain β1 = 0.6.
The figure can be constructed as follows: During the peak hour, the ramp demand (1000
veh/h) will be entering the ramp (blue line, slope = 1000 veh/h), whereas only 60% of it
(β1 × 1000 = 600 veh/h) is allowed to exit into mainline (orange line, slope = 600 veh/h).
At the end of the peak hour, there are 400 vehicles in the queue at the on-ramp behind the
metering signal.
After peak hour ends, the ramp demand is 300 veh/h (yellow line, slope = 300 veh/h),
whereas the capacity available to the ramp demand is 6000 – 5400 = 600 veh/h (purple line,
slope = 600 veh/h).
The queue at the on-ramp behind the metering signal reduces to zero after a time of 400 /
(600 – 300) = 11/3 hours (1 hour, 20 minutes) after the end of the peak hour.
The total delay is found by the area between the arrival (blue/yellow) and departure
(orange/purple) curves = (½ × 1 × 400) + (½ × 11/3 × 400) = 466.67 vehicle-hours.
As SPVs carry 1 passenger each and HOVs carry 2, the total delay (in person hours
travelled) can be calculated as follows:
0
200
400
600
800
1000
1200
1400
0 0.5 1 1.5 2 2.5
Cumulative Number of Vehicles
time (h)
CIV2282 Transport and Traffic Engineering Week 12, 2019
Prac Class 12 – Managed Motorways and Non-Motorised Transport Page 4 of 4
DAllVehs = DSPV + DSPV
= 0.8 × 1 pax/veh × 466.67 veh.h + 0.2 × 2 pax/veh × 466.67 veh.h
= 560 pax.h,
This can be compared to the total delay if only SPVs were metered (from part b):
DSPVonly = 1 pax/veh × 466.67 veh.h = 466.67 pax.h
From where we obtain the time saving as follows:
DAllVehs – DSPVonly = 560 – 466.67 = 93.33 pax.h.
The travel time saving is 93 person-hours.
d) Determine the maximum queue length that will occur during the day.
The maximum queue length will occur at the end of the peak hour, and can be calculated
as follows (for b)):
1h × (800 veh/h – 400 veh/h) = 400 veh.
Note that the same answer is obtained for the maximum queue length if all vehicles had to
pass through the metering signal (part c).
The maximum queue length is 400 vehicles.