CE 2710
Introduction to Transportation Engineering
Homework 5 Solution
1.
Given the following data use the gravity model to calculate all the interchange volumes (in other words, get the trip distribution matrix).
Zone
1
2
3
Production, P i
Attractiveness, A j
2000 2
1000
2000
5
1
1
2
I
3
Impedance Factor, W ij
1
5
20
J
2
20
5
10 10
3
10
10
5
Calibration Factor, c = 1.5 Socio-economic factor, k ij
= 1.0
Solution
Step 1- Calculate Friction Factors
Based on the
F ij

1
W ij c equation we can calculate friction factor for example:
F
11

1
( 5 )
1 .
5

0 .
0894
Then the following matrix is prepared:
F ij (friction factor)
TAZ
I
1
2
1
J
2 3
0.0894 0.0112 0.0316
0.0112 0.0894 0.0316
3 0.0316 0.0316 0.0894
Step 2- Find Denominator of Gravity Model Equation {AjFijKij}
Find denominator of Gravity Model equations for each pair of origin and destination by this relation A j
F ij
K ij

For example for i=1 and j=2 we have :
A j
F ij
K ij

A
2
F
12
K
12

( 5 )( 0 .
0112 )( 1 .
0 )

0 .
0559
Based on the calculations the following matrix is completed:
A j
F ij
K ij
TAZ
I
1
2
J
1 2 3 Σ
0.1789 0.0559 0.0316 0.2664
0.0224 0.4472 0.0316 0.5012
3 0.0632 0.1581 0.0894 0.3108
Step 3- Find Probability that Trip i will be attracted to Zone j, {pij}
In this step, by using the following formula and based on the calculated values in step 2 we can find the probability of travelled that will be attracted between each pair of TAZ.
For example the probability of travel between TAZ 1 and 2 is calculates as below: p ij


A
A j
F ij j
F ij
K ij
K ij

A
2
F
12
K
12
( A
1
F
11
K
11

A
2
F
12
K
12

A
3
F
13
K
13
) p
11

0 .
0559

0 .
2098
0 .
2664
Consequently we can complete the following matrix:
p ij
TAZ
I
1
2
1
J
2 3
0.6715 0.2098 0.1187
0.0446 0.8923 0.0631
3 0.2035 0.5087 0.2878
Step 4- Find Trip Interchanges, {Qij}
Based on the total amount of trip production in each TAZ (given number) and the calculated probabilities in previous step we can find “Trip Interchanges”, Q ij
Q ij
Q
12

P i p ij

P
1 p
12

( 2000 )( 0 .
2098 )

420

The matrix of trip interchanges is as below:
Q ij
TAZ 1 2
J
I
1
2
3
Σ
1343
45
407
1795
420
892
1017
2329
3
237
63
576
876
Σ
2000
1000
2000
5000

2.
Use the gravity model to estimate the trip distribution matrix for this planning year.
Given:
Zone Production, P i
Attractiveness, A j
1
2
2000
5000
10
8
I
Impedance Factor, W ij
I
1
1
2
J
2
10
2 9 1
Calibration Factor, c = 1.5
K ij
J
1
2
1
1
1.2
2
1.2
1
Solution:
Step 1- Calculate Friction Factors
F ij

1
W ij c

F
11

1
( 2 )
1 .
5

0 .
3536
F ij
J
TAZ 1 2
1 0.3536 0.0316
I
2 0.0370 1.0000
Step 2- Find Denominator of Gravity Model Equation {AjFijKij}
A j
F ij
K ij

A
1
F
11
K
11

( 10 )( 0 .
3536 )( 1 .
0 )

3 .
5355
A j
F ij
K ij

A
2
F
12
K
12

( 8 )( 0 .
0316 )( 1 .
2 )

0 .
3036

A j
F ij
K ij
I
TAZ
1
2
1
3.5355
0.4444
J
2
0.3036
8.0000
Σ
3.8391
8.4444
Step 3- Find Probability of travel between TAZ i and TAZ j, {pij} p ij


A j
A j
F ij
F ij
K ij
K ij

( A
1
F
11
A
1
K
11
F
11
K

A
11
2
F
12
K
12
) p
11

3 .
5355
3 .
8391

0 .
9209
I
The probability matrix
P ij
J
TAZ
1
2
1
0.9209
0.0526
2
0.0791
0.9474
Step 4- Find Trip Interchanges, {Qij}
Q ij

P i p ij
Q
11

P
1 p
11

( 2000 )( 0 .
9209 )

1842
Q ij
I
TAZ
1
2
Σ
1
1842
263
2105
J
2
158
4737
4895
Σ
2000
5000
7000

4.
Calculate the market shares for the following modes using the utility function given. u k
= a k
– 0.003 X
1
– 0.04 X
2 a k
, modal
constant
-0.20
X
1
, travel cost
in cents
120
X
2
, travel time in
minutes
30 Auto
BRT
Regular Bus
-0.40
-0.60
60
30
45
55
Solution:
Step 1- Calculation utility values for each mode: u auto u auto

0 .
04

X
2

1 .
76 u
BRT u
BRT

 a
BRT

0 .
4


0
0
.
.
003
18
X
1

1 .
8


0 .
04 X

2 .
38
2 u u
Bus
Bus


 a auto

0 .
2


0 .
003
0 .
36

X
1
1

.
2 a
Bus

0 .
6

0 .
003

0 .
09
X

1
2

.
2
0 .
04
 
X
2 .
2
89





0
0
.
.
2
4


( 0 .
003
( 0 .
003
* 120 )
* 60

0 .
6

( 0 .
003 * 30 )
)


( 0 .
04
( 0 .
04
* 30
* 45 )

( 0 .
04 * 55 )
)
Step 2- Finding market share of each mode: prob ( Auto prob ( Auto )
)
 e

1 .
76 

53 .
72 % e

1 .
76 e

2 .
38  e

2 .
89

( 0 .
172

0 .
172
0 .
289

0 .
0556 )

0 .
172
0 .
32017 prob ( BRT )
 e

1 .
76  e

2 .
38 e

2 .
38  e

2 .
89 prob ( BRT )

28 .
92 % prob ( Bus )
 e

1 .
76  e

2 .
89 e

2 .
38  e

2 .
89 prob ( Bus )

17 .
36 %

( 0 .
172
0 .
09255

0 .
289

0 .
0556 )

0 .
09255
0 .
32017

( 0 .
172
0 .
0556

0 .
289

0 .
0556 )

0 .
0556
0 .
32017

5.
Calculate the market shares for auto and light rail using the utility function given. u k
= a k
– 0.05 T a
– 0.04 T w
– 0.02 T r
– 0.01 C a k
, modal
constant
-0.05
T a
– access time
10
T w
– waiting time
10
T r
– riding time
45
C – out of pocket cost
50 Light Rail
Auto
Solution:
-0.05 5 5 30 100
Step 1- Calculation utility values for each mode: u u rail rail

 a rail


0 .
05
0 .

05 T a
( 0 .
05

*
0 .
04 T w
10 )

( 0

0
.
04
.
*
02 T r
10 )


0 .
01 C
( 0 .
02 * 45 ) u rail
 
0 .
05

0 .
5

0 .
4

0 .
9

0 .
5
 
2 .
35 u u auto auto

 a auto

0 .

05

0 .
05 T a
( 0 .
05 *

5
0 .
04 T w
)

( 0 .

04
0 .
02 T r
* 5 )


0
( 0 .
02
.
01 C
* 30 )


( 0 .
01 * 50
( 0 .
01 * 100 )
) u auto
 
0 .
05

0 .
25

0 .
2

0 .
6

1
 
2 .
1
Step 2- Finding market share of each mode: prob ( Rail )
 e e

2 .
35

2 .
35  e

2 .
1

( 0 .
0 .
09537
09537

0 .
1224 ) prob ( Rail )

43 .
78 %

0 .
09537
0 .
2178 prob ( Auto prob ( Auto )
)
 e

2 e

2 .
1
.
35  e

2 .
1

56 .
22 %

0 .
1224
( 0 .
09537

0 .
1224 )

0 .
1224
0 .
2178