Transcript Trip Distribution

Trip Distribution
Transportation Engineering (CIVTREN)
notes of AM Fillone, DLSU-Manila
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Trip Distribution Models
1. GrowthFactor/Fratar Method
• A simple method to distribute trips in a study
area.
• Assumptions of the model
a. the distribution of future trips from a given
origin zone is proportional to the present trip
distribution
b. this future distribution is modified by the
growth factor of the zone to which these trips are
attached
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• The Fratar formula can be written as
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Example: Fratar Method
An origin zone i with 20 base-year trips going to zones a, b, and c numbering
4, 6, and 10, respectively, has growth rates of 2, 3, 4, and 5 for i, a, b, and c,
respectively. Determine the future trips from i to a, b, and c in the future year.
i
Given:
4
a
10
6
b
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20
c
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Solution:
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The Gravity Model
• The most widely used trip distribution
model
• The model states that the number of trips
between two zones is directly proportional
to the number of trip attractions generated
by the zone of destination and inversely
proportional to a function of time of travel
between the two zones.
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• The
gravity model is expressed as
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Single Constrained vs. Doubly Constrained model
• Singly Constrained model – when information is
available about the expected growth trips
originating in each zone only or the other way, trips
attracted to each zone only
• Doubly Constrained model – when information is
available on the future number of trips originating
and terminating in each zone.
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• For
a doubly constrained gravity model, the adjusted
attraction factors are computed according to the formula
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Gravity Model Example
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Solution:
Iteration 1 :
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Calibrating a Gravity Model
• Calibrating of a gravity model is accomplished by
developing friction factors and developing
socioeconomic adjustment factors
• Friction factors reflect the effect travel time of
impedance has on trip making
• A trial-and-error adjustment process is generally
adopted
• One other way is to use the factors from a past
study in a similar urban area
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Three items are used as input to the gravity model for
calibration:
1. Production-attraction trip table for each
purpose
2. Travel times for all zone pairs, including
intrazonal times
3. Initial friction factors for each increment of
travel time
The calibration process involves adjusting the friction factor
parameter until the planner is satisfied that the model
adequately reproduces the trip distribution as represented by
the input trip table – from the survey data such as the triptime frequency distribution and the average trip time.
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The Calibration Process
1. Use the gravity model to distribute trips based on initial
inputs.
2. Total trip attractions at all zones j, as calculated by the
model, are compared to those obtained from the input
“observed” trip table.
3. If this comparison shows significant differences, the
attraction Aj is adjusted for each zone, where a difference
is observed.
4. The model is rerun until the calculated and observed
attractions are reasonably balanced.
5. The model’s trip table and the input travel time table can
be used for two comparisons: the trip-time frequency
distribution and the average trip time. If there are
significant differences, the process begins again.
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Figure 11-7 shows the results
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fourof AM
iterations
comparing
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travel-time frequency.
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Figure 11-9 Smoothed Adjusted Factors, Calibration 2
• An example of smoothed values of F factors in Figure 11-9.
• In general, values of F decreases as travel time increases, and may take the
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form F varies as t-1, t-2, or e-t. Transportation
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• A more general term used for representing travel time (or a
measure of separation between zones) is impedance, and the
relationship between a set of impedance (W) and friction
factors (F) can be written as:
Example:
A gravity model was calibrated with the following results:
Impedance (travel time, mins), W
4
6
8
11
15
Friction factors, F
.035
.029
.025
.021
.019
Using the f as the dependent variable, calculate parameter
A and c of the equation.
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Solution:
The equation can be written as
ln F = ln A – c ln W
ln W
1.39
1.79
2.08
2.40
2.71
ln F
-3.35
-3.54
-3.69
-3.86
-3.96
These figures yield the following values of A = .07 and c = .48.
Hence, F = 0.07/W0.48
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SUMMARY OUTPUT
Since, ln A = -2.73218,
A = e^(-2.73218)
A = .065
and
c = - (-.461)
c = 0.461
Regression Statistics
Multiple R
0.995749
R Square
0.991516
Adjusted R
Square
0.988688
Standard
Error
0.02602
Observations
5
ANOVA
df
Regression
Residual
Total
Intercept
lnw
SS
MS
1 0.237369 0.237369
3 0.002031 0.000677
4
0.2394
Standard
Coefficients
Error
t Stat
-2.73218 0.05194 -52.6023
-0.461 0.024621 -18.7241
F
350.5919
Significance F
0.000333
P-value
Lower 95% Upper 95% Lower 95.0% Upper 95.0%
1.51E-05
-2.89748
-2.56689
-2.89748
-2.56689
0.000333
-0.53935
-0.38265
-0.53935
-0.38265
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