# 2014 CE 2710 HW 5 Solutions 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