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Lecture 5 Bus Rapid Transit, ridership estimation procedures and headway requirements Dr. Muhammad Adnan Lecture Outline • • • • The Vehicle Operation Cycle Capacity of the Bus Stop Headways and its types Scheduling The Vehicle Operation Cycle • 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 vehicle becomes empty, then it may dead-head to the next pick-up location, or it may return to the depot. Frequency and Headway - Transit 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 • 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. • 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. • 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 • • • • 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 • Headways are influenced by the demand for service, with the shortest headways being maintained during the busiest periods. • Capacity headways: headways such that the vehicles 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. Capacity headway • • • • h= (M x L )/(R x P) h=capacity headway 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) Optimum cost headways • 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 Optimum cost headways contd.. • 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 Optimum cost headways contd.. • Taking derivative of c with respect to h, setting this equal to zero, and solving for h result in optimum headway • 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 the capacity headway, the headway that minimise the sum of operating cost and passenger time cost, and the actual headway to be operated. Constructing Timetables • Requirement – Headways – 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 Ridership and Headway Relationship • 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 about 2.5%. Ridership with Population • Lahore : 10,000,000 • Annual per Capita Transit Pessenger Miles: 500 • Karachi : 15,000,000 • Annual per Capita Transit Pessenger Miles: 800