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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)

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nni xxx ...ln 22110

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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

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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

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Example Calculations

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PExactly 2:

Less than 2:

More than 2:

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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

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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

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0.8

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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

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ility

of

Exc

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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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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

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mk sk

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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