### Lecture 5 Bus Rapid Transit, ridership estimation

```Lecture 5
Bus Rapid Transit, ridership
estimation procedures
Lecture Outline
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The Vehicle Operation Cycle
Capacity of the Bus Stop
Scheduling
The Vehicle Operation Cycle
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It is useful to refer to the operation of a vehicle (van, bus, or train) through the
course of a day of transit service; this is commonly referred to as the vehicle
“cycle” because it tends to repeat itself from one day to the next. This cycle is
illustrated in the figure below.
Vehicle Cycle –Explained (1)
• The cycle begins when a vehicle is started from a depot (a garage, yard, or
other location). The depot serves as a common location where vehicles are
stored or maintained.
• The vehicle is positioned from the depot to a location where it can begin
service. This is commonly at the terminus of a route. Such a movement, from
the depot to this location, is called a “pull-out”.
• The vehicle travels from its starting location to another route terminus,
stopping at stations or stops along the route to allow passengers to board and
to alight. We will call such movement of the vehicle a “trip” (or, vehicle trip).
• While moving along the route, the vehicle incurs both running time and dwell
time. Running time is the time spent traveling between stops or stations, and
dwell time is the time spent stopped at locations to allow passengers to board
and alight.
• While moving along the route, the vehicle is engaged in “revenue service”, so
that time and miles spent along the route while providing passenger service
are called “revenue hours” and “revenue miles”.
Vehicle Cycle –Explained (2)
• When it reaches a route terminus, the vehicle is re-positioned for further
service. If it returns along the same route or another route starting from
the same terminus, there may be a short time for recovery before
reentering revenue service, called a “layover”.
• A vehicle may also be moved between termini to start service on a
different route, resulting in what is called a “deadhead” trip, not in
revenue service, between the two termini.
• The vehicle continues in revenue service on the fixed routes, repeating the
process of stopping at stations or stops to allow passengers to board and
alight.
• When the vehicle has reached its final terminus for its set of trips, it
returns to the depot. Such movement from a route terminus to the depot
is called a “pull-in”.
• This vehicle cycle is common to fixed-route service, particularly for bus
and rail transit systems. In the case of demand-responsive service, there
are no formal “termini” for a route; rather, the “termini” represent specific
locations where persons are picked up or dropped off. At any point, if the
Service
• As an example, a route with a 10-minute headway has a
frequency of 6 vehicles per hour.
Bus Seating Capacities
Dwell Time
• The dwell time is the time required to serve passengers at the
busiest door, plus the time required to open and close the
doors. A value of 2 to 5 seconds for door opening and closing
is reasonable for normal operations.
Dwell Time Estimation
CV of Dwell time
• The coefficient of variation of dwell times (the
standard deviation of dwell times divided by
the mean dwell time) typically ranges from
40% to 80%, with 60% recommended as an
appropriate value in the absence of field data
Clearance Time
• Clearance time includes two components, (1)
the time for a bus to start up and travel its
own length while exiting a bus stop, and for
off-line stops, (2) the re-entry delay associated
with waiting for a sufficient gap in traffic to
allow a bus to pull back into the travel lane.
• Start-up and exiting time may be assumed to
be 10 seconds
Failure Rate
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The probability that a queue of buses
will not form behind a bus stop, or
failure rate, can be derived from basic
statistics. The value Za represents the
area under one “tail” of the normal
curve beyond the acceptable levels of
probability of a queue forming at a bus
stop.
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CBD stops. Za values of 1.440 down to
1.040 should be used. They result in
probabilities of 7.5 to 15 percent,
respectively, that queues will develop.
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Outlying stops. A Za value of 1.960
should be provided wherever possible,
especially when buses must pull into
stops from the travel lane. This results
in queues beyond bus stops only 2.5
percent of the time.
Capacity of Bus Stop
Example Problem
• A bus route on a city street is scheduled for 2 min
headways. On the average, 7 passenger per bus
board at a particular stop and 10 passengers alight.
All boarding passengers use the front door and all
alighting passengers use the back door. Assume
clearance time is 15 s, and that maximum allowable
probability of bus queuing is 1 %. The stop is a nearside stop at an intersection with a g/C ratio of 0.50.
How many berths are required. Co-efficient of
Variation of dwell time can be taken as 0.6.
Solution
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Number of Buses:
N= (60 min/h) / 2 (min/bus) = 30 buses/hr
Clearance time= tc= 15 secs
Dwell times:
– Boarding 7 passengers *3.0 s/passenger =21
– Alighting 10 Passengers * 2.5 s/passenger=25
• Headways are influenced by the demand for service,
with the shortest headways being maintained during
the busiest periods.
are just filled at the max. Load point on the route.
• Policy headway: Arbitrarily determined maximum
headways that are intended to represent the
minimum acceptable standard for frequency of
service.
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h= (M x L )/(R x P)
M=bus seating capacity
L=maximum acceptable avg. Load factor (ratio of
passengers to seats)
• P=total patronage for the route in passenger per
hour
• R=ratio of maximum load to passengers boarding
(Usually taken as 0.6 - 0.7)
• A third possibility is to provide headways that
minimise the sum of operating cost and the
value of time spent by passengers waiting for
service.
• The fleet size to operate the route is given by
N= ᶿ/h ( N= number of vehicles in the fleet,
ᶿ=cycle time)
• The operating cost Co= ᵧ N= ᵧ ᶿ/h, ᵧ= operating
cost per vehicle-hour of operation
• The avg. Number of passenger boarding per
hour (P).
• If we assumed that, on the avg. Passengers
wait half a headway for service, the total cost
of the passengers waiting time is
Cw=µ P h / 2, µ=value of passengers
waiting time in Rs./hr
C=Co +Cw= total cost
• Taking derivative of c with respect to h, setting
this equal to zero, and solving for h result in
• h= sqrt of ((2 ᵧ ᶿ)/ (µ P))
Problem
• An urban bus route has a patronage 500 passengers per hour
and a cycle time of 2.5 hr. It is operated with buses having a
seat capacity of 50 passengers. The operator believes that
passengers value waiting time at \$10/hr . The ratio of max.
Load to the total number of passengers boarding is 0.60, and
the operator’s max. Load factor standard is 1.20. Determine
operating cost and passenger time cost, and the actual
Constructing Timetables
• Requirement
– Travel time
– Cycle time
– Number of vehicles
Cycle time: It is consist of the sum of the travel times on the individual
route segments, minimum layovers required to dampen variation in
the running time and provide work breaks for drivers, and excess
layover needed to make the cycle an integral multiple of the headway.
Cy T = T + tL
T= Σ ti + tl
tl =max( tv , tw )
N= nhint (T/h)
tL=Nh – T
problem
• Construct a bus route schedule based on the running
times given below. Headways are 30 min and
minimum layovers are 7 min at each end or 10% of
running time, which ever is greater. The first bus
from A to D should leave at 8:00 am. The schedule
should cover the time block from 8:00 am to 12:00
noon.
• Segments travel times are
– A-B: 20 min, B-C: 15 min, C-D: 40 min, D-C: 38 min, C-B: 18
min, B-A: 22 min
Solution
• Sum of all travel times:
– 20+15+40+38+18+22 = 153 min
tw = 2 x 7 =14 min
tv = 0.10 X 153 = 15.3 min
tl = 15.3 min
T=153 +15.3 = 168.3 min
N= nhint (168.3/30) =6 vehicles
Theta = 6 X 30 = 180 min
Excess lay over =tL= 180-168.3= 11.7 min
Total layover = 11.7 min + 15.3 min = 27 min
• We are using 14 min layover at A and 13min at
D
Bus A
1
2
3
4
5
6
1
2
3
08:00
08:30
09:00
09:30
10:00
10:30
11:00
11:30
12:00
B
08:20
08:50
09:20
09:50
10:20
10:50
11:20
11:50
12:20
C
08:35
09:05
09:35
10:05
10:35
11:05
11:45
12:05
12:35
D Bus D
09:15
09:45
10:15
10:45
11:15
11:45
12:15
12:45
13:15
4
5
6
1
2
3
4
5
6
07:58
08:28
08:58
09:28
C
08:36
09:06
09:36
10:06
B
08:54
09:24
09:54
10:24
A
09:16
09:46
10:16
10:46
• Headways have an enormous impact
on ridership levels above a certain critical waiting time.
• Following Boyle, the effect of changes in headway are
directly proportional to changes in ridership by a
simple conversion factor of 1.5.
• That is, if a headway is reduced from 12 to 10 minutes,
the average rider wait time will decrease by 1 minute,
the overall trip time by the same one minute, so the
ridership increase will be on the order of 1 x 1.5 + 1 or