Mode Choice Lecture 10 Norman W. Garrick

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Mode Choice Lecture 10 Norman W. Garrick

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Mode Choice Lecture 10 Norman W. Garrick. Mode Choice. The introduction of congestion charging in London in 2003 is one example of a situation where mode choice modeling is needed. The fundamental question is this case was “ Will charging work to reduce congestion? ” - PowerPoint PPT Presentation

Transcript of Mode Choice Lecture 10 Norman W. Garrick

Page 1: Mode Choice Lecture 10 Norman W. Garrick

Mode ChoiceLecture 10

Norman W. Garrick

Page 2: Mode Choice Lecture 10 Norman W. Garrick

Norman W. Garrick

Mode Choice

The introduction of congestion charging in London in 2003 is one example of a situation where mode choice modeling is needed.

The fundamental question is this case was “Will charging work to reduce congestion?”

The expectation was that people would switch modes

Mode choice modes are used to predict how many people will switch

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Norman W. Garrick

Mode Choice: London Congestion Charging

0.75 mile

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Norman W. Garrick

Mode Choice: London Congestion Charging

These are some of London's congestion charging camera's. They monitor all vehicles entering through the London congestion charge boundary zone and automatically issue a bill to owner of the vehicle. There are thousands of these all around the perimeter of the "C" zone.

Source: http://www.flickr.com/photos/astrolondon/224826598/

http://www.tfl.gov.uk/tfl/roadusers/congestioncharge/whereandwhen/

Transport for London Link

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Norman W. Garrick

Mode Choice

The introduction of the new AVE service between Madrid and Barcelona is another example of a situation were model choice modeling would be needed.

How many people would switch from plane or driving to train?

http://www.renfe.es/video.html

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Norman W. Garrick

Factors affecting Mode Choice

Earlier in the semester we talked about some of the factors affecting mode choice. These factors include

1. Type and purpose of trip 2. Car ownership status 3. Cost (mostly out of pocket cost) 4. Door-to-door travel time 5. Convenience/service/comfort 6. Prestige 7. Availability 8. Accessibility of mode 9. Land use characteristics of start and end point

Obviously, not all of these factors can be effectively incorporated into a quantitative model of mode choice. We need to be cognizant of those factors that are important in influencing choice but are not fully accounted for in mode choice models.

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Norman W. Garrick

Mode Split or Modal Choice Models

Mode Choice Models are used to try to predict travelers mode choice

Contemporary models are based on using UTILITY or DISUTILITY functions

These functions are meant to express the level of satisfaction (for utility functions) or dissatisfaction (for disutility functions) with a given mode

Once the utility function is calculated for each mode, the probability that a given mode will be chosen can then be calculated

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Norman W. Garrick

Utility Function

A utility function takes the following form

uk = ak + a1 X1 + a2 X2 + ….. ar Xr + ε0

uk – utility function for mode k

ak – modal constant

Xr – variables measuring modal attributes such as cost or time of travel

ar – coefficient associated with each attribute

Where

ε0 – error term

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Norman W. Garrick

Multinomial Logit Model

If utility function, uk, is assumed to be a Weibull Probability Distribution then the Multinomial Logit Model is used to calculate the probability that a traveler will chose a given mode

Multinomial Logit Model

p(k) = euk / Σ eu

k

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Norman W. Garrick

Example

The mode available between Zone I and J are i) Automobile (A), ii) Bus (B)

Find the market share for each mode given the attribute table (on next page) for the modes.

The utility function is

uk = ak – 0.025 X1 – 0.032 X2 – 0.015 X3– 0.002 X4

wherex1 – access plus egress time (min)x2 – waiting time (min)x3 – line haul time (min)x4 – out of pocket cost (cents)

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Norman W. Garrick

Example … continue

The attribute table for each mode is given below

And aa = -0.10ab = +0.00

Therefore

u(A) = -0.625u(B) = -1.530

Probability of selecting PC, p(A) = e(-0.625) / [e(-0.625)+e(-1.530)] = 0.71

x1 x2 x3 x4

Auto 5 0 20 100

Bus 10 15 40 50