trip generation and mode choice
DESCRIPTION
Trip Generation and Mode Choice. CEE 320 Steve Muench. Outline. Trip Generation Mode Choice Survey. 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 - PowerPoint PPT PresentationTRANSCRIPT
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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
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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.