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Trip Generationand Mode Choice

CEE 320Steve Muench

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Outline

1. Trip Generation

2. Mode Choicea. Survey

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Trip Generation

• Purpose– Predict how many trips will be made– Predict exactly when a trip will be made

• Approach– Aggregate decision-making units – Categorized trip types– Aggregate trip times (e.g., AM, PM, rush hour)– Generate Model

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Motivations for Making Trips

• Lifestyle– Residential choice– Work choice– Recreational choice– Kids, marriage– Money

• Life stage• Technology

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Reporting of Trips - Issues

• Under-reporting trivial trips• Trip chaining• Other reasons (passenger in a car for

example)

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Trip Generation Models

• Linear (simple)

• Poisson (a bit better)

nnxxxT ...22110

nni xxx ...ln 22110

! tripsofnumber

x

xexP

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Poisson Distribution

• Count distribution– Uses discrete values– Different than a continuous distribution

!n

etnP

tn

P(n) = probability of exactly n trips being generated over time t

n = number of trips generated over time t

λ = average number of trips over time, t

t = duration of time over which trips are counted (1 day is typical)

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Poisson Ideas

• Probability of exactly 4 trips being generated– P(n=4)

• Probability of less than 4 trips generated– P(n<4) = P(0) + P(1) + P(2) + P(3)

• Probability of 4 or more trips generated– P(n≥4) = 1 – P(n<4) = 1 – (P(0) + P(1) + P(2) + P(3))

• Amount of time between successive trips

tt

eet

thPP

!0

00

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Poisson Distribution Example

Trip generation from my house is assumed Poisson distributed with an average trip generation per day of 2.8 trips. What is the probability of the following:

1. Exactly 2 trips in a day?2. Less than 2 trips in a day?3. More than 2 trips in a day?

!

trips/day8.2trips/day8.2

n

etnP

tn

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Example Calculations

%84.232384.0

!2

18.22

18.22

e

PExactly 2:

Less than 2:

More than 2:

102 PPnP

21012 PPPnP

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Example Graph

0.00

0.05

0.10

0.15

0.20

0.25

0.30

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Trips in a Day

Pro

bab

ility

of

Occ

ura

nce

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Example Graph

0.00

0.05

0.10

0.15

0.20

0.25

0.30

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Trips in a Day

Pro

bab

ility

of

Occ

ura

nce

Mean = 2.8 trips/day

Mean = 5.6 trips/day

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Example: Time Between Trips

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Time Between Trips (Days)

Pro

bab

ility

of

Exc

edan

ce

Mean = 2.8 trips/day

Mean = 5.6 trips/day

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Example

kidsmarriedinternetagegender

autosbus

bicyclesedanvansportssuv

incomeeducationi

*8

*7

*6

*5*4

*#37*36

*35*34*33*32*31

*2*10ln

Recreational or pleasure trips measured by λi (Poisson model):

Variable Coefficient Value Product

Constant 0 1 0

Education (undergraduate degree or higher) 0.15 1 0.15

Income 0.00002 45,000 0.9

Whether or not individual owns an SUV 0.1 1 0.1

Whether or not individual owns a sports car 0.05 0 0

Whether or not individual owns a van 0.1 1 0.1

Whether or not individual owns a sedan 0.08 0 0

Whether or not individual uses a bicycle to work 0.02 0 0

Whether or not individual uses the bus to work all the time -0.12 0 0

Number of autos owned in the last ten years 0.06 6 0.36

Gender (female) -0.15 0 0

Age -0.025 40 -1

Internet connection at home -0.06 1 -0.06

Married -0.12 1 -0.12

Number of kids 0.03 2 0.06

Sum = 0.49

λi = 1.632 trips/day

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Example

• Probability of exactly “n” trips using the Poisson model:

• Cumulative probability – Probability of one trip or less: P(0) + P(1) = 0.52– Probability of at least two trips: 1 – (P(0) + P(1)) = 0.48

• Confidence level– We are 52% confident that no more than one recreational or

pleasure trip will be made by the average individual in a day

20.0

!0

0632.10 632.1

eP

32.0!1

1632.11 632.1

eP

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Mode Choice

• Purpose– Predict the mode of travel for each trip

• Approach– Categorized modes (SOV, HOV, bus, bike, etc.) – Generate Model

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Dilemma

Explanatory Variables

Qu

alit

ativ

e D

epe

nd

ent V

ari

able

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Dilemma

Home to School Distance (miles)

Wa

lk to

Sch

ool

(ye

s/n

o v

ari

able

)

0

1

0 10

1 =

no,

0 =

yes

= observation

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A Mode Choice Model

• Logit Model

• Final form

mkn

kmnmnmk zV

s

U

U

mk sk

mk

e

eP

Specifiable part Unspecifiable part

n

kmnmnmk zU

s = all available alternativesm = alternative being consideredn = traveler characteristick = traveler

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Discrete Choice Example

Regarding the TV sitcom Gilligan’s Island, whom do you prefer?

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Ginger Model

UGinger = 0.0699728 – 0.82331(carg) + 0.90671(mang) + 0.64341(pierceg) – 1.08095(genxg)

carg = Number of working vehicles in household

mang = Male indicator (1 if male, 0 if female)

pierceg = Pierce Brosnan indicator for question #11 (1 if Brosnan chosen, 0 if not)

genxg = generation X indicator (1 if respondent is part of generation X, 0 if not)

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Mary Anne Model

UMary Anne = 1.83275 – 0.11039(privatem) – 0.0483453(agem) – 0.85400(sinm) – 0.16781(housem) + 0.67812(seanm) + 0.64508(collegem) – 0.71374(llm) + 0.65457(boomm)

privatem = number of years spent in a private school (K – 12)

agem = age in years

sinm = single marital status indicator (1 if single, 0 if not)

housem = number of people in household

seanm = Sean Connery indicator for question #11 (1 if Connery chosen, 0 if not)

collegem = college education indicator (1 if college degree, 0 if not)

llm = long & luxurious hair indicator for question #7 (1 if long, 0 if not)

boomm = baby boom indicator (1 if respondent is a baby boomer, 0 if not)

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No Preference Model

Uno preference = – 9.02430x10-6(incn) – 0.53362(gunsn) + 1.13655(nojames) + 0.66619(cafn) + 0.96145(ohairn)

incn = household income

gunsn = gun ownership indicator (1 if any guns owned, 0 if no guns owned)

nojames = No preference indicator for question #11 (1 if no preference, 0 if preference for a particular Bond)

cafn = Caffeinated drink indicator for question #5 (1 if tea/coffee/soft drink, 0 if any other)

ohairn = Other hair style indicator for question #7 (1 if other style indicated, 0 if any style indicated)

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Results10. Regarding the TV sitcom “Gilligan’s Island” whom do

your prefer?

29

9085

30

88 89

7

112

87

0

20

40

60

80

100

120

Ginger Mary Ann No Preference

# o

f R

es

po

nd

an

ts

Survey

average

Model

Average probabilities of selection for each choice are shown in yellow. These average percentages were converted to a hypothetical number of respondents out of a total of 207.

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My Results

s

U

U

mk sk

mk

e

eP

8201.13265.02636.01075.1 eeees

U sk

1815.08201.1

1075.1

e

e

eP

s

U

U

ginger sk

mk

4221.08201.1

2636.0

e

e

eP

s

U

U

annemary sk

mk

3964.08201.1

3265.0

e

e

eP

s

U

U

preferenceno sk

mk

Uginger = – 1.1075

Umary anne = – 0.2636

Uno preference = – 0.3265

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Primary References

• Mannering, F.L.; Kilareski, W.P. and Washburn, S.S. (2005). Principles of Highway Engineering and Traffic Analysis, Third Edition. Chapter 8

• Transportation Research Board. (2000). Highway Capacity Manual 2000. National Research Council, Washington, D.C.