cee 320 fall 2008 trip generation and mode choice cee 320 anne goodchild
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Trip Generationand Mode Choice
CEE 320Anne Goodchild
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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 (households)– 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)• Time consuming and expensive
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Trip Generation Models
• Linear (simple)
– Number of trips is a function of user characteristics
• Poisson (a bit better)
nnxxxT ...22110
nni xxx ...ln 22110
!TeTP
T
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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
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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
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0.2
0.3
0.4
0.5
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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
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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
• Buying a golf ball– Price– Driving distance– Life expectancy
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Typical ranges
Consumer’s ideal• Average Driving Distance: 225-275 yards • Average Ball Life: 18-54 holes• Price: $1.25-$1.75
Producer’s ideal• Average Driving Distance: 225-275 yards • Average Ball Life: 18-54 holes• Price: $1.25-$1.75
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Would you rather have:
• 250 yards• 54 holes• $1.75
• 225 yards• 36 holes• $1.25
If we ask enough of these questions, and respondent has some underlying rational value system, we can identify the
values for each attribute that would cause them to answer the way that they did.
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Utilities
• What is the respondent willing to pay for one additional yard of driving distance?
• What is the respondent willing to pay for one additional hole of ball life?
• Assume no value for the bottom of the range
• Calculate their utility for any ball offering, then the probability they will choose each offering.
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Concerns
• All attributes must be independent• Must capture everything that is important
(or use constant)• Can present a set of choices and predict
population’s response– Offer complete set of choices– Red bus/blue bus
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My Results
i
U
U
i i
i
e
eP
8201.13265.02636.01075.1 eeees
U sk
1815.08201.1
1075.1
e
e
eP
i
U
U
bus i
i
4221.08201.1
2636.0
e
e
eP
i
U
U
car i
i
3964.08201.1
3265.0
e
e
eP
i
U
U
walk i
i
Ubus = – 1.1075
Ucar = – 0.2636
Uwalk = – 0.3265
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