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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

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Mode choice
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

Norman W. Garrick


Mode choice london congestion charging
Mode Choice: London Congestion Charging

0.75 mile

Norman W. Garrick


Mode choice london congestion charging1
Mode Choice: London Congestion Charging

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

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.

Transport for London Link

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

Norman W. Garrick


Mode choice1
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

Norman W. Garrick


Factors affecting mode choice
Factors affecting Mode Choice

In lecture 2 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.

Norman W. Garrick


Mode split or modal choice models
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

Norman W. Garrick


Utility function
Utility Function

A utility function takes the following form

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

Where

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

ε0 – error term

Norman W. Garrick


Multinomial logit model
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 eux

Norman W. Garrick

ε0 – error term


Example
Example

  • The mode available between Zone I and J are i) Passenger car (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

  • where

    • x1 – access plus egress time (min)

    • x2 – waiting time (min)

    • x3 – line haul time (min)

    • x4 – out of pocket cost (cents)

Norman W. Garrick

ε0 – error term


Example continue
Example … continue

The attribute table for each mode is given below

And

aa = -0.10

ab = +0.00

Therefore

u(A) = -0.625

u(B) = -1.530

Probability of selecting PC, p(A) =

e(-0.625) / [e(-0.625)+e(-1.530)] = 0.71

Norman W. Garrick


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