Introduction to Transportation Engineering Instructor Dr. Norman Garrick Hamed Ahangari 6th March...

33
Introduction to Transportation Engineering Instructor Dr. Norman Garrick Hamed Ahangari 6th March 2014

Transcript of Introduction to Transportation Engineering Instructor Dr. Norman Garrick Hamed Ahangari 6th March...

Page 1: Introduction to Transportation Engineering Instructor Dr. Norman Garrick Hamed Ahangari 6th March 2014.

Introduction to Transportation Engineering

Instructor Dr. Norman Garrick

Hamed Ahangari6th March 2014

Page 2: Introduction to Transportation Engineering Instructor Dr. Norman Garrick Hamed Ahangari 6th March 2014.

Trip Distribution

Page 3: Introduction to Transportation Engineering Instructor Dr. Norman Garrick Hamed Ahangari 6th March 2014.

Gravity Model

Page 4: Introduction to Transportation Engineering Instructor Dr. Norman Garrick Hamed Ahangari 6th March 2014.

Example 1

• P=800• A= 10

• P=1500• A= 10

• P=200• A= 20

• P=1200• A= 2

Zone 1 Zone 2

Zone 4Zone 3

Calibration Factor, c = 1.5 Socio-economic factor, kij = 1.0

Page 5: Introduction to Transportation Engineering Instructor Dr. Norman Garrick Hamed Ahangari 6th March 2014.

Target-Year Inter-zonal Impedances, {Wij}-Given

TAZ 1 2 3 4

1 5 10 20 10

2 10 5 25 100

3 20 25 5 50

4 10 100 50 5

Page 6: Introduction to Transportation Engineering Instructor Dr. Norman Garrick Hamed Ahangari 6th March 2014.

Step 1- Calculate Friction Factors, {Fij}

TAZ 1 2 3 4

1 0.0894 0.0316 0.0112 0.0316

2 0.0316 0.0894 0.0080 0.0010

3 0.0112 0.0080 0.0894 0.0028

4 0.0316 0.0010 0.0028 0.0894

0894.0)5(

115.111 F

WF

cij

ij

Page 7: Introduction to Transportation Engineering Instructor Dr. Norman Garrick Hamed Ahangari 6th March 2014.

Step 2- Find Denominator of Gravity Model Equation {AjFijKij}

TAZ 1 2 3 4 TotalΣ(AjFijKij)

1 0.17889 0.63246 0.1118 0.31623 1.23937

2 0.06325 1.78885 0.08 0.01 1.9421

3 0.02236 0.16 0.89443 0.02828 1.10507

4 0.06325 0.02 0.02828 0.89443 1.00596

Page 8: Introduction to Transportation Engineering Instructor Dr. Norman Garrick Hamed Ahangari 6th March 2014.

Step 3- Find Probability that Trip i will be attracted to Zone j, {pij}

Σ(AjFijKij)pij =

AjFijKij

TAZ 1 2 3 4 Total

1 0.144 0.510 0.090 0.255 1

2 0.033 0.921 0.041 0.005 1

3 0.020 0.145 0.809 0.026 1

4 0.063 0.020 0.028 0.889 1

Page 9: Introduction to Transportation Engineering Instructor Dr. Norman Garrick Hamed Ahangari 6th March 2014.

Step 4- Find Trip Interchanges, {Tij}

TAZ 1 2 3 4 Total

1 173 612 108 306 1200

2 7 184 8 1 200

3 30 217 1214 38 1500

4 50 16 22 711 800

Tij = Pipij

Page 10: Introduction to Transportation Engineering Instructor Dr. Norman Garrick Hamed Ahangari 6th March 2014.

Zone 1 Zone 2

Zone 3 Zone 4

173

6127

108 30 30650

1 16

38

22

Page 11: Introduction to Transportation Engineering Instructor Dr. Norman Garrick Hamed Ahangari 6th March 2014.

Example 2 UConn.

UConn

Buckland Hill

Westfarm Mall

Eastbrook Mall

Page 12: Introduction to Transportation Engineering Instructor Dr. Norman Garrick Hamed Ahangari 6th March 2014.

Uconn.P1=2000

EastbrookA1: 10

t1: 10 min

Buckland HillA2: 60

t2: 20 min

WestfarmA3: 80

t3: 40 min

Calibration Factorc = 2.0

Socioeconomic Adj. FactorK = 1.0

Page 13: Introduction to Transportation Engineering Instructor Dr. Norman Garrick Hamed Ahangari 6th March 2014.

Calculations-1

J Aj t1j F1j=tij^(-2) Aj*F1j

1 10 10 1/100=0.01 0.10

2 60 20 1/400=0.0025 0.15

3 80 40 1/1600=0.000625 0.05

Total 0.30

Page 14: Introduction to Transportation Engineering Instructor Dr. Norman Garrick Hamed Ahangari 6th March 2014.

Calculations-2

J Aj*F1j P1j= Aj*F1j/(∑Aj*F1j) T1j= P1*p1j

1 0.10 0.1/0.30=0.333 2000*0.33=667

2 0.15 0.15/0.3=0.5 2000*0.5=1000

3 0.05 0.05/0.3=0.167 2000*0.167=333

Total 0.30 1.000 2000

Page 15: Introduction to Transportation Engineering Instructor Dr. Norman Garrick Hamed Ahangari 6th March 2014.

Trip Distribution Results

UConn

Buckland Hill

Westfarm Mall

Eastbrook Mall

1000 trips

333 trips667 trips

Page 16: Introduction to Transportation Engineering Instructor Dr. Norman Garrick Hamed Ahangari 6th March 2014.

Mode Choice

Page 17: Introduction to Transportation Engineering Instructor Dr. Norman Garrick Hamed Ahangari 6th March 2014.

Example 1

Calculate the mode shares for auto and light rail transit (LRT) using the utility function given.

• uk = ak – 0.045 Ta – 0.03 Tw – 0.01 Tr– 0.005 C

where: Ta - access time, Tw - waiting time,

Tr - riding time, C - out of pocket cost

Page 18: Introduction to Transportation Engineering Instructor Dr. Norman Garrick Hamed Ahangari 6th March 2014.

Modes Specifications

Mode a(k) Ta Access time

TwWaiting

time

TrRiding time

COut of

pocket costLRT -0.04 15 15 50 40

Auto -0.025 10 0 20 100

Page 19: Introduction to Transportation Engineering Instructor Dr. Norman Garrick Hamed Ahangari 6th March 2014.

Key formulas

𝑈𝑚=𝑎𝑚−𝑎1∗𝑋 1−𝑎2∗𝑋 2

Page 20: Introduction to Transportation Engineering Instructor Dr. Norman Garrick Hamed Ahangari 6th March 2014.

Utility Calculations

Mode Calculations Utility

LRT =-0.04-0.045*15-0.03*15 - .01*50- 0.005*40 -1.865

Auto =-0.025-0.045*10-0.03*0 - .01*20- 0.005*100 -1.190

uk = ak – 0.045 Ta – 0.03 Tw – 0.01 Tr– 0.005 C

Page 21: Introduction to Transportation Engineering Instructor Dr. Norman Garrick Hamed Ahangari 6th March 2014.

Mode choice Estimation

Mode Uk e^(Uk) Probability of selecting

LRT -1.865 0.1548 0.337

Auto -1.190 0.3042 0.663

Total 0.4841 1.000

Page 22: Introduction to Transportation Engineering Instructor Dr. Norman Garrick Hamed Ahangari 6th March 2014.

Policy AnalysisWhich policy may promote transit?

1-Applying congestion pricing for Auto mode( Increase the cost of auto to 120)

2- Improving transit by decreasing waiting time to 5 minute

Page 23: Introduction to Transportation Engineering Instructor Dr. Norman Garrick Hamed Ahangari 6th March 2014.

Scenario Analysis

Mode Uk e^(Uk) Prob. of selecting

LRT -1.865 0.1548 0.360

Auto -1.290 0.2752 0.640

Total 0.4301 1.000

Mode Uk e^(Uk) Prob. of selecting

LRT -1.565 0.2091 0.407

Auto -1.190 0.3042 0.593

Total 0.5233 1.000

Congestion Pricing Increase Train Frequencies

Page 24: Introduction to Transportation Engineering Instructor Dr. Norman Garrick Hamed Ahangari 6th March 2014.

Comparing Results

Base ca

se

Congestion Pric

ing

Increase

frequencie

s0

10

20

30

40

50

60

70

Mod

e Sh

are

%

LRT

Auto

LRT

Auto

LRT

Auto

Page 25: Introduction to Transportation Engineering Instructor Dr. Norman Garrick Hamed Ahangari 6th March 2014.

Example 2Mode choice: Storrs to NYC

• Available modes:• Auto (Private car )

• Bus

• Train (From New haven)uk = ak – 0.0075 X1 – 0.012 X2

X1: Travel Cost

X2: travel Time

Page 26: Introduction to Transportation Engineering Instructor Dr. Norman Garrick Hamed Ahangari 6th March 2014.

Mode Characteristics - AutoCost Analysis

Gasoline Cost(300 miles round trip)

35 $

Tool Cost 15 $

Parking Cost 40 $

Total Cost 80 $

Time AnalysisRound trip 150*2=300 min

Parking 20Total time 320 min

Page 27: Introduction to Transportation Engineering Instructor Dr. Norman Garrick Hamed Ahangari 6th March 2014.

Mode Characteristics - Bus

Cost Analysis

Ticket CostRound trip

25*2= 50 $

Time AnalysisWaiting time 15*2= 30 min

Round trip Travel time 210*2=420Total time 440 min

Page 28: Introduction to Transportation Engineering Instructor Dr. Norman Garrick Hamed Ahangari 6th March 2014.

Mode Characteristics - TrainCost Analysis

Gasoline Price 12 $

Parking 8 $

Train Ticket 27 $

Total Cost 47 $

Time AnalysisDriving from Storrs to New Haven 55*2=110 min

Parking 20 minTransfer time 2*15= 30 min

Train from New Haven to NYC 100*2=200 minTotal time 360 min

Page 29: Introduction to Transportation Engineering Instructor Dr. Norman Garrick Hamed Ahangari 6th March 2014.

Modes CharacteristicsMode Total Cost Total Time a(k)

Auto 80 320 -0.3

Bus 50 460 -0.5

Train 47 340 -0.2

Page 30: Introduction to Transportation Engineering Instructor Dr. Norman Garrick Hamed Ahangari 6th March 2014.

Utility CalculationsMode Calculations Utility

Auto =-0.3-0.0075*80-0.012*320 -4.740

Bus =-0.5-0.0075*50-0.012*450 -6.275

Train =-0.2-0.0075*47-0.012*340 -4.873

Page 31: Introduction to Transportation Engineering Instructor Dr. Norman Garrick Hamed Ahangari 6th March 2014.

Mode choice Estimation

Mode Uk e^(Uk) Probability of selecting

Auto -4.740 0.008738 0.478

Bus -6.375 0.001883 0.103

Train -4.773 0.007654 0.418

Total 0.018275 1.000

Page 32: Introduction to Transportation Engineering Instructor Dr. Norman Garrick Hamed Ahangari 6th March 2014.

Scenario 2- New Utility Function(Sensitive to Out of Vehicle time)

uk = ak – 0.0075 X1 – 0.012 X2 – 0.036 X3

X1: Travel Cost,X2: in vehicle travel Time

X3:out of vehicle travel Time

Page 33: Introduction to Transportation Engineering Instructor Dr. Norman Garrick Hamed Ahangari 6th March 2014.

Modal Choice in Scenario 2

Mode Uk e^(Uk) Probability of selecting

Auto -5.22 0.005407 0.627

Bus -6.995 0.000916 0.106

Train -6.0725 0.002305 0.267

Total 0.008629 1.000