Einführung in Verkehr und Logistik - Fakultät für ... · PDF...

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Einführung in Verkehr und Logistik

(Bachelor)

Transport Demand

Univ.-Prof. Dr. Knut Haase

Institut für Verkehrswirtschaft

Wintersemester 2013/2014, Dienstag 10:15-11:45 Uhr, Phil E

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Service Production Process in Public Transport

1. Demand Model

2. Infrastructure (e.g. tracks and stations)

3. Tariff zone planning

4. Line planning (routes and frequencies)

5. Timetabling

6. Vehicle scheduling

7. Crew scheduling

8. Rostering

9. Dispatching (vehicles and staff)

I Simultaneous planning approaches, e.g. simultaneous vehicle and crewscheduling, provide significant cost reduction potentials.1

I Demand effects should be integrated in line planning and timetabling.2

1See [HDD01].2See [KH08].

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Transport Demand

What are discrete choices?

I Course of studyI Transport modeI Port choice by a shipownerI Beer brandI Shopping facilityI ...

Discrete choice appear in every day of life!

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An example

The choice between taking the bus or car might be influenced byI CostI Travel-timeI IncomeI FlexibilityI InterchangesI SecurityI Eco friendlinessI ...

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How to get the data and what to do with it?

Empirical analysis

I Sample; we distinguish in generalI revealed preferences (chosen alternative)I stated preferences (stated alternative)

I A sample is used toI Specification of utility functionsI Demand forecastI Analysis of willingness to payI Evaluation of infrastructural improvementsI Service design

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Numerical exampleTransport mode choice of forwarders in a region:

Would you choose combined cargo if the time for intial leg and final legisI 1 hour?I 2 hours?I 3 hours?

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Contingency table

time for initial and final legCombined cargo 1h (k = 1) 2h (k = 2) 3h (k = 3)

yes (i = 1) 100 100 50 250no (i = 2) 200 300 250 750

300 400 300 1000

p(i = 1) = 250=1000 = 0:25

p(i = 1; k = 2) = 100=1000 = 0:10

p(i = 1) =3P

k=1p(i = 1; k) = 100

1000 + 1001000 + 50

1000 = 0:25

p(k = 3ji = 1) = 50=250 = 0:2

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p(i = 1; k = 2) = p(i = 1)p(k = 2ji = 1) = 0:25 � 0:40 = 0:10

= p(k = 2)p(i = 1jk = 2) = 0:40 � 0:25 = 0:10

Let p(i) > 0 and p(k) > 0:

p(kji) = p(i ; k)=p(i)

p(i jk) = p(i ; k)=p(k)

Assumption: p(i jk) is constant over all periodsp(k) might change over periods!

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A behavioural model

3 Parameter p(i = 1jk = 1) = �1

p(i = 1jk = 2) = �2

p(i = 1jk = 3) = �3

p(i = 2jk) = 1� p(i = 1jk) !not an additional parameter (binary case:y=n)

Estimate a simple model

�1 = 100=300 = 0:333

�2 = 100=400 = 0:250

�3 = 50=300 = 0:167

Hypothesis: The smaller the time for initial leg and final leg the larger thechoice probability of combined cargo

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Standardfehler der Schätzung

I Bernoulli-distributed variable (urn experiment)

I Sum of Bernoulli-distributed variable is binomial distributed.

I Standard error of the mean of a random variable:p

�2=N�2 : Variance N : Sample size

I Variance of a Bernoulli-distributed variable: �(1� �)

� : Probability of occurrence

I Estimator for standard errors: sk =p

�k(1� �k)=Nk

I Approximation for the 95% confidence intervall (c.i.): [�k � 2 � sk ]

I t-statistic : �k=skI t-value > 2 !significant different from zero

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Results for our small example

�k sk 95% c.i. t-valuek = 1 0.333 0.027 [0.279,0.388] 12.247k = 2 0.250 0.022 [0.207,0,293] 11.547k = 3 0.167 0.022 [0.124,0.210] 7.746

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Apply our model

Assume we obtain the following data for a region (potentials forcombined cargo):

time of initial transport weight in tons per yearand final leg total combined cargo (estimates)1 h 1,200,000 400,0002 h 2,400,000 600,0003 h 3,000,000 500,000

6,600,000 1,500,000

Market share: 22.73%

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Construction of additional terminals

time of inital transport weight in tons per yearand final leg total combined cargo (estimates)1 h 1,500,000 500,0002 h 2,700,000 675,0003 h 2,400,000 400,000

6,600,000 1,775,000

!Market share increases up to 29%

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Properties of estimators

I UnbiasedI EfficientI For non-linear models we have only asymptotical properties:

- consistent: N !1 estimator = true value- asymptotically efficient: there is no consistent estimator with asmall standard error

I Discrete choice models are non-linear models!

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Assumptions

I Individuals (n = 1; : : : ;N) choose between only two alternatives (i = 1; 2)

I Individual (or observation) n chooses alternative i with largest utility

I Assume a linear utility function Uni that consists of

I deterministic (representative) utility Vni andI stochastic utility �ni , such that

I Uni = Vni + �ni

I �ni is independent and identically extreme value distributed (IID EV) with

f (�ni ) = e��ni e�e��ni

F (�ni ) = e�e��ni

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Density function and cumulative density function of EV

-2

-1,5 -1

-0,5 0

0,5 1

1,5 2

2,5 3

3,5 4

4,5 5

0

0,25

0,5

0,75

1

f(!ni) = e!!nie!e!ni

F (!ni) = e!e!ni

!

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Central assumption

The difference of two independently extreme value distributed variables�n = �ni 0 � �ni , is logistically distributed:

f (�n) =�e��n

(1+ e���n )2� > 0

F (�n) =1

1+ e���n

� is a scale parameter

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Choice probabilities I

n chooses i = 1, iffUn1 > Un2

Since Uni is random (due to �ni ) we can only compute the propability ofUn1 > Un2

Pn(i = 1) = P(Un1 > Un2)

= P(Vn1 + �n1 > Vn2 + �n2)

= P(�n1 � �n2 > Vn2 � Vn1)

= P(�n2 � �n1 < Vn1 � Vn2)

= P(�n < Vn1 � Vn2)

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Choice probabilities II

Pn(i = 1) = P(�n2 � �n1 < Vn1 � Vn2)

= P(�n < Vn1 � Vn2)

= F (Vn1 � Vn2)

=1

1+ e��(Vn1�Vn2)

=e�Vn1

e�Vn1 + e�Vn2

Binary Logit Model

Pn(i = 1) =e�Vn1

e�Vn1 + e�Vn2

� is not identified and has to be fixed to an arbitrary value (1 e.g.)

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NormalisationWe can not measure the absolute value of utility. A multiplication ofutility of each alternative with a positive constant does not change choiceprobabilities (and thus choice process).!Normalisation of � = 1!Variance of �ni : �2=6

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Reference alternativeAn addition of a constant to the utility of each alternative does notinfluence the choice probabilities!Constrain coefficients of the refernce alternative to a fixed value (e.g.0)

Un1 = �1 + �3TC1 + �4TT1 + �6INCn + �n1 (1)

Un2 = �2 + �3TC2 + �5TT2 + �7INCn + �n2 (2)

�2 = �7 = 0

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Choice probability of alternative 1 P(i = 1)

-5 -4 -3 -2 -1 0 1 2 3 4 50

0,25

0,5

0,75

1

Pn(i = 1) =1

1 + e!Vn1

Vn2 = 0

Pn(i = 1)

Vn1

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Multinomial Logit Model (MNL)

Assumptions

I sames as binary logit

I �ni is iid EV over all i and n

f (�ni ) = e��ni e�e��ni

F (�ni ) = e�e��ni

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MNL choice probabilities3

Pn(i) = P[ 8j 6= i : Uni > Unj ]

= P[ 8j 6= i : Vni + �ni > Vnj + �nj ]

= P[ 8j 6= i : �nj � �ni � Vni � Vnj ]

=

ZI ( 8j 6= i : �nj � �ni � Vni � Vnj )f (�n)d�n

if �ni iid EV

=eVniPj eVnj

Comments:I I (�) Indicator functionI We may solve the integral even if �ni is not iid EV

! Nested-, Cross-Nested, Mixed Logit

3See [Tra03, pp. 38-41]

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Some commentsI We may write MNL choice probabilities as

Pn(i) =e�VniP

j2Cne�Vnj

� Scale parameter

Cn Choice set of individual n

� = 0 ) even distributed choice probabilities

�!1 ) deterministic choices

j Cn j= 2 ) binary logit

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Independence from irrelevant alternatives (IIA)4

I Stems directly from the ”independent” part of the iid EV assumptionon �ni

I Ratio of choice probabilities of two alternatives is independent fromthe presence or absence of another alternative (or their attributes)

I Might be inadequat some times. If so, then make other assumptionabout distribution of �ni

I !Nested-Logit, Mixed Logit etc.

Proof IIAPn(i)Pn(i 0)

=eVniPj eVnj

=eVni 0Pj eVnj

=eVni

eVni 0

= eVni�Vni 0

4See [Tra03, pp. 49-54]

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Red bus - blue bus paradoxon5

Example

Pn(’Car’) =12; Pn(’Red Bus’) =

12

Now add a blue bus to Cn

Pn(’Car’) =13; Pn(’Red Bus’) =

13; Pn(’Blue Bus’) =

13

We would expect

Pn(’Car’) =12; Pn(’Red Bus’) =

14; Pn(’Blue Bus’) =

14

5See [BAL85, pp. 51-55]

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IIA holds strictly only on individual level

Example: choice of shopping facility

Group CBD Outlet park 1with car 20 % 80 %no car 80 % 20 %frequency 50 % 50 %

Add a new alternative Outlet park 2

Group CBD Outlet park 1 Outlet park 2with car 11.11 % 44.44 % 44.44 %no car 66.66 % 16.66 % 16.66 %frequency 38.88 % 30.56 % 30.56 %

) The more socio-economic variables we use the less critical is IIA on”market-share”-level

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Specification and identification

Un1 = �1 + �4TC1 + �4TT1 + �6INCn + �n1

Un2 = �2 + �4TC2 + �5TT2 + �7INCn + �n2

Un3 = �3 + �4TC3 + �6TT3 + �8INCn| {z }Vn3

+�n3

I �1, �2, �3: alternative-specific constants (ASCs)I Only jCnj � 1 ASCs are identifiedI ASCs reflect the expectation value of �ni only if a full set of ASCs

(jCnj � 1) is employed.I Alternative-specific specifications of socio-economic variables

(depend on n only!) - like income - need normalisation, too.

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Specification of utility with categorial variables

(k) Attribute (variable) Value (l), reference category

(1) Age teen (t), adult (a), old (o)(2) Income high (h), low (l)(3) Gender female (f ), male (m)

Zkl : Dummy variable of attribute k with value l

�i0 + �itZ1;t + �iaZ1;a + �ihZ2;h + �if Z3;f

Segment of population (old, low income, male): �0

ASC for segment (adult, high income, male): �0 + �a + �h

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Estimation of �’s by Maximum Likelihood6

How do we get the �’s?

Definitionyni =1, if n has chosen i (0, otherwise)

Re-write logit choice probabilities as

Pn(i) =JY

i=1

(Pn(i))yni

Since �ni are iid over n as well we can write the likelihood function as

L(�) =NY

n=1

JYi=1

(Pn(i))yni

Note, � is the coefficient vector of interest! That is, all �ik to beestimated.

6See [Tra03, pp. 64-67]

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The log-Likelihood makes estimation easier and obtains the same extremepoints as L(�)

L(�) = ln L(�) = ln

NY

n=1

JYi=1

(Pn(i))yni

!=

NXn=1

JXi=1

yni lnPn(i)

Substitute Pn(i) with eVniPjeVnj

!

L(�) =

NXn=1

JXi=1

yni ln

eVniPj eVnj

!

=

NXn=1

JXi=1

yniVni �

NXn=1

JXi=1

yni ln

Xj

eVnj

!

=

NXn=1

JXi=1

yni

Xk

�ik � znik

!�

NXn=1

JXi=1

yni ln

Xj

e�P

k�jk �znjk

�!

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Maximise L(�)

Start � = 0 : L(�) = 0� N � ln J(if Cn is the same for all n)

�� Vector of estimated coefficients that maximises L(�)

Goodness-of-fitPseudo-R2: LR = 1� l(��)=l(0)

Software for estimation

I BIOGEMEI LIMDEP, NLOGITI SPSSI R

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BIerlaire Optimization toolbox for GEv Model Estimation7

MNL is a special case of a Generalised Extreme Value (GEV) Model(nowadays: Multivariate Extreme Value (MEV))

http://biogeme.epfl.ch

public domain, cross plattform

Invoke via console (f.e. DOS prompt: cmd.exe): biogeme mymodelmysample.dat

I We write the model to mymodel.modI .mod file may be manipulated by any texteditor (notepad++ is

advised)I Data for estimation has to be stored in mysample.datI Results can be found in output mymodel.rep

7See [Bie08].

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Example

A choice problemStudents (n) with no car available chose on their commute to university eitherto walk (i = 1) or go by bike (i = 2) or to take the bus (i = 3). The choicedepends on travel-time in minutes dni and gender sn with female studentsindicated by 1. Now we write our deterministic utilities

Vn1 = �0;1 + �1;1d1;n + �2;1snVn2 = �0;2 + �1;2d2;n + �2;2snV3n = �0;3 + �1;3d3;n + �2;3sn

For identification purposes:

�0;1 = �1;1 = �2;1 = 0

That is, Vn1 = 0

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BIOGEME syntax

mymodel.mod

[Choice]Choice

[Beta]// Name Value LowerBound UpperBound status (0=variable, 1=fixed)b01 0 -10000 10000 1b02 0 -10000 10000 0b03 0 -10000 10000 0b11 0 -10000 10000 1b12 0 -10000 10000 0b13 0 -10000 10000 0b21 0 -10000 10000 1b22 0 -10000 10000 0b23 0 -10000 10000 0

[Utilities]// Id Name Avail linear-in-parameter expression1 Foot av1 b01 * one + b11 * d1 + b21 * s2 Bike av2 b02 * one + b12 * d2 + b22 * s3 Bus av3 b03 * one + b13 * d3 + b23 * s

[Expressions]one = 1av1 = 1av2 = 1av3 = 1

[Model]$MNL

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Data

mysample.dat

ID Choice d1 d2 d3 s1 1 6 8 13 02 1 8 9 13 13 1 12 9 15 14 2 21 12 17 05 1 6 7 12 0: : : : : :

95 3 51 21 24 096 1 18 11 14 197 1 6 9 12 098 2 33 15 16 199 1 19 11 13 0

100 1 15 12 16 1

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Results: model statistics

mymodel.rep

Model: Multinomial LogitNumber of estimated parameters: 6

Number of observations: 100Number of individuals: 100

Null log-likelihood: -109.861Cte log-likelihood: -88.991

Init log-likelihood: -109.861Final log-likelihood: -68.002

Likelihood ratio test: 83.718Rho-square: 0.381

Adjusted rho-square: 0.326Final gradient norm: +4.082e-07

Diagnostic: Convergence reached...Iterations: 11

Run time: 00:00

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Results: coefficient estimates

Utility parameters******************Name Value Std err t-test p-val Rob. std err Rob. t-test Rob. p-val---- ----- ------- ------ ----- ------------ ----------- ----------b01 0.00 --fixed--b02 -5.57 1.15 -4.84 0.00 1.01 -5.50 0.00b03 -10.4 2.17 -4.78 0.00 2.02 -5.13 0.00b11 0.00 --fixed--b12 0.324 0.0746 4.35 0.00 0.0665 4.87 0.00b13 0.554 0.123 4.52 0.00 0.115 4.81 0.00b21 0.00 --fixed--b22 -0.128 0.617 -0.21 0.84 * 0.580 -0.22 0.83 *b23 -0.317 0.678 -0.47 0.64 * 0.666 -0.48 0.63 *

Vn1 = 0

Vn2 = �5:57+ 0:324 � d2;n � 0:128 � snV3n = �10:4+ 0:554 � d3;n � 0:317 � sn

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Commute-to-school mode choice modelling for students located inDresden, Saxony. 8

Choice set

i =

8>><>>:

1 foot2 bike3 transit4 car

8See [MTH08].

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Exogenous Variables

I Alternative-specific constant (k = 0)I Distance between school and student’s location in [km] (k = 1)I Car availability (k = 2)

zni2 =

�1 Car always available0 else

I Season=weather condition (k = 3)

zni3 =

�1 Winter=bad weather0 else

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Estimation results (N = 9300)Mode Variable (Coeff.) Estimate SE t-statisticfoot ASC (�10) 10.774 0.234 46.042(i = 1) Distance (�11) -4.376 0.183 -37.008

Car-avail (�12) -5.279 0.1634 -3.609Season (�13) -0.591 0.2385 -22.129

bike ASC (�20) 6.570 0.190 34.596(i = 2) Distance (�21) -0.904 0.033 -27.365

Car-avail (�22) -4.772 0.184 -22.599Season (�23) -2.081 0.147 -14.160

transit ASC (�30) 4.477 0.171 26.210(i = 3) Distance (�31) -0.052 0.017 -2.966

Car-avail (�32) -5.553 0.143 -38.870Season (�33) -0.489 0.135 -3.621

Null log-likelihood: -12 892.50 Final log-likelihood: -4792.14Likelihood ratio test: 16200.80 Rho-square: 0.628301

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Utility Functions

I Car-availability ”always” (zni2 = 1)I Summer term (zni3 = 0)

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I Car-availability ”not always” (zni2 = 0)I Summer term (zni3 = 0)

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How to compute utility values?

I Distance to school 0.5 km (zni1 = 0:5)I Car is always available (zni2 = 1)I Summer term (zni3 = 0)

Vn1 = 10; 774� 4; 376 � 0; 5� 5; 2796 � 1 = 3; 307

Vn2 = 6; 570� 0; 904 � 0; 5� 4; 772 � 1 = 1; 346

Vn3 = 4; 477� 0; 052 � 0; 5� 5; 553 � 1 = �1; 102

Vn4 = 0

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Corresponding choice probabilities

I Distance to school 0.5 km (zni1 = 0:5)I Car is always available (zni2 = 1)I Summer term (zni3 = 0)

Pn1 =e3;307

e3;307 + e1;346 + e�1;102 + e0 = 0; 841 = 84; 1%

Pn2 =e1;346

e3;307 + e1;346 + e�1;102 + e0 = 0; 118 = 11; 8%

Pn3 =e�1;102

e3;307 + e1;346 + e�1;102 + e0 = 0; 010 = 1; 0%

Pn4 =e0

e3;307 + e1;346 + e�1;102 + e0 = 0; 031 = 3; 1%

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Transit choice probabilities

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Literature I

M. Ben-Akiva and S.R. Lerman.Discrete choice analysis, theory and applications to travel demand.MIT Press, Cambridge, MA, 1985.

M. Bierlaire.An introduction to BIOGEME (Version 1.8), 2008.

K. Haase, G. Desaulniers, and J. Desrosiers.Simultaneous vehicle and crew scheduling in urban mass transit systems.Transportation Science, 35(3):286, 2001.

M. Klier and K. Haase.Line optimization in public transport systems.In J. Kalcsics and S. Nickel, editors, Operations Research Proceedings 2007,pages 473–478. Springer, 2008.

S. Müller, S. Tscharaktschiew, and K. Haase.Travel-to-school mode choice modelling and patterns of school choice in urbanareas.Journal of Transport Geography, 16(5):342–357, 2008.

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Literature II

A. Schöbel.

Line planning in public transportation: models and methods.

OR Spectrum, 34:491–510, 2012.

E. Train, Kenneth.

Discrete choice methods with simulation.

Cambridge University Press, 2003.