CIVL2111

Transportation Engineering

Land use and socio-

Trip Distribution
Transportation system
economic projections specification

Trip generation

Trip distribution

◼ How do trips generated Modal choice

from each TAZ allocate Trip assignment

to other? Direct user impacts

What is the input to

this step?
 1
3
2 5

Trip Distribution
4 8
7
6

Zone 1 ? Zone 1
?
Input ? ? Input
Zone 2 Zone 2
TAZ Productions TAZ Attractions
?
1 12 ? 1 9
2 19 2 12
3 35 3 4
4 4 4 38
5 5 ? 5 45
6 10 ? 6 6
7 13
8 22
? 7 4
Zone 8 Zone 8 8 2

TAZ 1 2 3 4 5 6 7 8
1
2
? ? ? ? ?
Output 3
Zone 1
4
5
Trip Matrix
6
7 2
8
 A general form of a trip matrix
interzonal trip volume or

Trip production
constraint

Trip attraction T
j
ij = Pi
constraint

T
i
ij = Aj

3
 The gravity model – simplest
form Higher trip production/attraction, higher
interzonal volume?
F, Tij
M1 , Pi M2, Aj
Higher impedance, higher interzonal
r, Wij volume?

Pi A j M1 M 2
Tij = k F=k
Wijc r2
Tij = interzonal volume F = gravitational force of attraction
k, c = parameters k = proportionality constant
Pi = trip production of zone i M1 = mass of body 1
Aj = trip attraction of zone j M2 = mass of body 2
Wij = interzonal impedance r = distance between two bodies
Gravity model Newton’s law of gravitation

◼ Impedance: travel time, distance, travel cost, etc.

4
 Rationale of gravity model
Attraction
2
zone W12
2
Production
zone
1 W13
1
Attraction 3
3 zone
T12
T12 A2
= =?
T13 A3 T13
A2 W12 − c
T12 = P1 T12 = P1
A2 + A3 W12 − c + W13− c 5
 Rationale of gravity model
3 W31
3
Attraction
Production 1
1 zone
zone
W21

2 2

T21 W21− c
T21
=? =
T31 W31− c
T31
P2 W21− c
T21 = A1 T21 = A1
P2 + P3 W21− c + W31− c 6
 Production constrained
gravity model
attraction  attractiveness
Zone 1 2
Pi A ' j
Tij = k ' Commercial floor 10000 20000
Wijc area (arces)
Tij = interzonal volume Relative 1 2
Wij = interzonal impedance attractiveness
k’, c = parameters
Pi = trip production of zone i
Will the interzonal
A’j = attractiveness or relative attractiveness of zone j
volume be changed
if 10 and 20 are
used?
Use page 5 to
explain the
shopping centre
example. 7
 Production constrained
gravity model
Pi Aj
Tij = k ' −1
Wijc  Ax 
k ' =  c 
Pi =  Tix  x Wix  Tix
x Zone X
Zone i
 
 Aj / Wij 
c
the proportion of zone i trips
Tij = Pi 
  ( A 
x / W ix )
c 

that will be allocated to
zone j
x
1
friction factor Fij =
Wijc  
 
Aj Fij 
Tij = Pi
  A F 
 x ix 
 x  8
 Production constrained gravity
model
◼ A set of socioeconomic adjustment factors Kij are
introduced to incorporate effects that are not captured by
the limited number of explanatory variables:
 
 
Aj Fij K ij 
Tij = Pi = Pp
  Ax Fix K ix  i ij
 x 
pij = the proportion of production zone i trips
associated with zone j.
Sum of the proportion equals 1?
The trip attraction constraint is satisfied?

See question 7 of problem set 1 for example 9

 Example
◼ Given c = 2.0, Kij =1
Skim table Wij 
Zone Productions Relative
attractiveness i\j 1 2 3 4
1 1500 0 1 5 10 15 20
2 0 3 2 10 5 10 15
3 2600 2 3 15 10 5 10
4 0 5 4 20 15 10 5

Apply the production-constrained gravity model to estimate the

attraction of each zone
Zone 1 is purely residential?
Zone 2 is purely non-residential?
10
Zone 3?
 i\j 1 2 3 4
1 5 10 15 20
2 10 5 10 15
Example 3 15 10 5 10
4 20 15 10 5
Solution: For i = 1, Pi = 1500
1 Aj F1 j K1 j
Aj F1 j = 2 K1 j 
Aj F1 j K1 j p1 j = T1 j = P1 p1 j
j W1 j  A F
x
x 1x K1x

1 0 1/52 = 0.0400 1.0 0.0 0.0/0.0514 0

=0
2 3 1/102 =0.0100 1.0 0.0300 0.03/0.0514 875
=0.584
3 2 1/152 =0.0044 1.0 0.0089 0.0089/0.0514 260
=0.173
4 5 1/202 =0.0025 1.0 0.0125 0.0125/0.0514 365
=0.243
Total 0.0514 1.00 1500
11
 i\j 1 2 3 4
1 5 10 15 20
2 10 5 10 15
Example 3 15 10 5 10
4 20 15 10 5
For i = 3, Pi = 2600

Aj
1
F3 j = 2 K3 j Aj F3 j K3 j p3 j = Aj F3 j K3 j T3 j = P3 p3 j
j W3 j  A F
x
x 3x K3 x

1 0 1/152 = 0.0044 1.0 0.0 0.0/0.16 0

=0
2 3 1/102 = 0.0100 1.0 0.03 0.03/0.16 488
=0.188
3 2 1/52 = 0.0400 1.0 0.08 0.08/0.16 1300
=0.500
4 5 1/102 =0.0025 1.0 0.05 0.05/0.16 812
=0.312
Total 0.16 1.00 2600
12
 Example
Aj =  Tyj
y
A2 = 875 + 488 = 1363
A3 = 260 + 1300 = 1560
A4 = 365 + 812 = 1177
i\j 1 2 3 4 Sum
(production)
1 0 875 260 365 1500
2 0 0 0 0 0
3 0 488 1300 812 2600
4 0 0 0 0 0
Sum (Attraction) 0 1363 1560 1177 4100
13
Ratio of attraction is not equal to the ratio of relative attractiveness!
 Attraction constrained gravity
model
PiAj
◼ Based on Aj =  Tyj and Tij = k 
y Wijc
 
 PF   
i ij K ij
Tij = Aj   = Aj pij Tyj

 y
PyFyj K yj
 Zone y
Zone j
pij = the proportion of attraction zone j trips
associated with zone i
Pi = population of zone i (  production)
K’ij = socioeconomic adjustment factors

The trip attraction constraint is satisfied?

The trip production constraint is satisfied?
14
 Calibration of the gravity
model i\j
Compare with regression
model, page 23 of ch 2
3 4 5
Trip distribution 1 ? ? ?
estimation: 2 ? ? ?
future/target year trip
distribution
c = 2, Kij= 1 Gravity model i\j 3 4 5
1 350 100 50
Future/target year Pi & Aj 2 200 600 200

Future year skim table

Calibration: c = ? and Kij = ?

Base year trip distribution
Calibration
i\j 3 4 5
1 300 150 50
procedure via trip c = 2, Kij= 1
2 180 600 220
length comparison
Base year Pi & Aj
Base year skim table
15
 Calibration of the gravity
model - Example Residential
(a) Base-year generation Non-residential
3 (10)
i Pi A'j 2
1 500 0 Non- (5)
2 1000 0 residential
(5) 4
3 0 2 (15)
4 0 3
(10) Non-
residential
5 0 5 1 (15) 5
(b) Base-year distribution
i\j 3 4 5 Residential Travel time in
1 300 150 50 min
2 180 600 220
find the value of c to reproduce the observed base-year data
16
 Calibration of the gravity
model - Example
f
Solution: 0.6

0.5

Step 0: determine observed freq distribution 0.4

0.3
Table b: base year trip distribution 0.2

that ln F = −c3lnW , one can 4calibrate the model

Notei\j 5 0.1

either1through c or300 150

F with W fixed. 50 0

2 180 600 220

5 10 15

Based on Table (b) and the distance between zones, W

the distribution of the total trips is given below: 3 (10)
2

W QT ijIJ
f, frequency = (5)
column 2/sum (5) 4
5 300+600=900 600/1500=0.60
900
(10)
(15)

10 150+180=330 330/1500=0.22
15 50+220=270 270/1500=0.18 1 (15) 5

sum = 1500 1.00 17

 Step 3: compare the observed and
calculated frequency distributions
f 0.7
0.6

0.5

Step 1: assume a value of c

0.4

0.3

Assuming that c=2.0, the following are obtained: 0.2

Step 2: determine calculated frequency f 0.1

0
iI\J
\j 3 4 5 5 10 15
W
1 303 114 83
2 123 741 136 Step 4: adjust F
2 / 52
where 303 = 500  2 and so on  Observed f 
2 / 5 + 3 / 102 + 5 / 152 F* = F  
 calculated f 
W  TQIJ f  0.6  1  0.6 
F5 = F5   = 2   = 0.034
ij *
5 1044 1044/1500=0.70  0.7  5  0.7 
10 237 237/1500=0.16
F10* =  2 
1 0.22 
15 219 219/1500=0.14   = 0.01375
sum = 1500 1.00  10  0.16 
F15* =  2 
where 1044=303+741 and so on 1 0.18 
  = 0.00571
 15  0.14  18
 Calibration of the gravity
model - Example
Step 2: determine calculated frequency f Step 3: compare the observed and
calculated frequency distributions
Using the adjusted F’s, the following results are
obtained: f
0.7

0.6

iI\J
\j 3 4 5 0.5

1 251 145 104 0.4

2 176 654 170 0.3

2  0.0343
= 
0.2
where 251 500 ,
2  0.0343 + 3  0.01375 + 5  0.00571 0.1

etc. 0
5 10 15

W QTijIJ f W
the trip length distributions
5 905 905/1500=0.60 are similar between observed
10 321 321/1500=0.21 and calculated values, so stop
15 274 274/1500=0.19 adjusting F
sum = 1500 1.00
19
where 905=251+654 and so on
 Calibration of the gravity
model - Example
Step 5: determine c
W 5 10 15
F 0.0343 0.01375 0.00571
X=Ln W 1.6094 2.3026 2.7081
Y=Ln F -3.3814 -4.2867 -5.1655
0
0 1 2 3
-1
Y
-2
Ln F

Predicted Y
-3

-4 Linear (Predicted
ln F = −1.9245ln W Y)
-5 c = 1.9245

-6
Ln W 20
 Singly constrained gravity
model: other forms
How to incorporate
Pi A j socioeconomic factor?
Tij = k c = kPi Aj Fij
W ij

Tij = kPi Aj Fij Tij = kPi Aj Fij

Production
1
constrained k=
gravity model
T = Pi  Aj Fij k=
1
j
ij j
 Aj Fij
j

Tij = kPi Aj Fij Tij = kPi Aj Fij

Attraction 1
k=
constrained
gravity model
T = Aj  Pi Fij k=
1
i
ij i
 Pi Fij
i 21
 Doubly constrained gravity
model
▪ The balancing factors are interdependent.
Tij = Ci Pi D j Aj Fij ▪ Solving them by an iterative process
(Furness procedure).
1
Ci = ▪ For the first iteration, set all Dj=1,
j
D j Aj Fij solve Ci,
▪ then use the current Ci values to re-
estimate all Dj,
1
Dj = ▪ For the second iteration, use the Dj
i
Ci Pi Fij value obtained in the last iteration to
re-estimate all Ci,
▪ then use the current Ci values to re-
estimate all Dj,
Ci, Dj = balancing factors
▪ repeat until convergence is 2achieved
(e.g. i (Ci − Ci ) + j ( D j − D j )  0.00001 )
k k −1 2 k k −1

See question 8 of problem or the maximum number of iterations

is reached K = iteration number 22
set 1 for example.
 Friction factor models
Fij
◼ Inverse Power: Fij = Wij , c  0 −c

◼ Exponential: ij
F = e
− bW
,b  0
ij

◼ Gamma: ij F = aW − c − bW
ij e , a , b, c  0
ij

What happen when c, b, or a <=0?

Wij
Parameters for gamma friction factor model
Trip Purpose a b c
HBW 28507 0.123 0.020
HBO 139173 0.094 1.285
NHB 219113 0.010 1.332
ref. NCHRP 365 / TransCAD UTPS Manual pg. 80

23
 Travel impedance: Generalized
travel cost - example
5 min waiting time

Bus fare: HK$6.9

5 min walk
45 min on bus
10 min walk
Value of in-vehicle travel time
= HK$ 50 per hour
Values of walking and waiting
times = HK$ 100 per hour.

W=
10/60*100+5/60*100+6.9+45
24
/60*50+5/60*100=HK$77.73
 Limitations of gravity model
◼ Simplistic nature of impedance - not flow
dependent Travel
time
◼ Lack of a behavioral basis

◼ Too much reliance on K-Factors

How to remove these limitations?

Traffic
flow

25