Fall 2017 - MTAT .08.043

Transportation Theory and Applications

Lecture VI: Traffic assignment

A. Hadachi

Topics❖ Overview

❖ Link cost function

❖ Traffic assignment models

❖ All or nothing (AON)

❖ User equilibrium assignment

❖ System optimum assignment

❖ Other methods

General Overview

Definition:Traffic assignment is the process of allocating given set of trip interchanges to the specified transportation system

The major objective behind traffic assignment are:

1. Estimate the volume of traffic on the road segments of the network and obtain aggregate network measures.

2. Estimate inter zone travel cost3. Analyse the travel pattern of each O-D pairs4. Identify congested road segments and collect useful traffic data for

designing future junctions.

Link Cost Function ❖ Two link problem with constant travel time function

Flow (x)

travel time

Link Cost Function

Definition:

The relation between the link flow and link impedance is called the link cost function

It is given by the following formula

t = t0[1 + α(x

k)β ]

t is the travel timex is the flowto is the free flow travel timek is practical capacity model parameters α & β

Traffic assignment Method 1

All-or-nothing assignment (AON)

All-or-nothing

Step 1: Step 2: Step 3:Find the

shortest path between

zones

Assign all trips to links

in the shortest path

Continue till trips between all zones pairs

have been assigned

All-or-nothingExample:

1 2t1= 10 min

t2= 15minx2

x1

Let’s consider a case where travel time is not a function flow but it a constant as shown in the figure

#Trips: 50

All-or-nothingExample:

1 2t1= 10 min

t2= 15minx2

x1

Let’s consider a case where travel time is not a function flow but it a constant as shown in the figure

#Trips: 50

All-or-nothingProblem:

1 2 3

4 5 6

7 8 9

A B

C D

33 4

4 3

4

57 9

12

12

8

10

7

5 5

A B C DA - 100 55 66B 22 - 44 77C - - - -D - - - -

Execute an AOM assignment.

Trip Exchange

All-or-nothing

SimpleInexpensive

Results easy to comprehend

Assumes all traffic will travel on shortest path

Generate an realistic flow patterns

+ -

Traffic assignment Method 1

User equilibrium assignment (UE)

User equilibrium assignment

Definition:

Statement:

The user equilibrium assignment is based on Wardrop’s first principle

Wardrop’s first principle

no driver can unilaterally reduce his/her travel costs by shifting to another route. t1 = t2

User equilibrium assignmentUE assignment conditions can be expressed given an OD pair as follows:

∀k : fk(ck − u) = 0

∀k : ck − u ≥ 0

is the flow on path kis the travel cost on path kis the minimum cost

fk

ck

u

(a)(b)

Equation states:

if ck − u = 0eq: (a)

ck − u ≥ 0if eq: (a)fk ≥ 0 This means all paths will have same

travel time

fk = 0

User equilibrium assignmentAssumption in UE:

1.User has perfect knowledge of the path cost2.Travel time on a given link is a function of the flow on that link only3.Travel time function are positive and increasing.

Hence, the UE assignment comes to an optimisation problem that can be formulated as follows:

Z =!

a

" xa

0ta(xa)dx

!

k

frsk = qrs : !r, s

Minimize:

Subject to:

xa =!

r

!

s

!

k

δrsa,kfrsk : !a

frsk ≥ 0 : ∀k, r, s

xa ≥ 0 : a ∈ A

is the pathequilibrium flow in link atravel time on link aflow on path k connecting OD pairs r-strip rate between r and sis constraint function defined as follows:

k

xa

ta

frsk

qrs

δrsa,k

δr,sa,k =

!1 :0 : otherwise

if link a belongs to path k

User equilibrium assignmentExample:

1 2x2

x1

t1 = 6 + 4x1

t2 = 4 + x22

t1 & t2: average travel time on route 1 and 2 (min)x1 & x2: traffic flow on routes 1 and 2 (Veh/hour)

Assuming user equilibrium conditions and hourly flow rate of 4500 Veh/hr, determine the following:

Travel time on each routeTraffic volumes on each route

Total system travel time

User equilibrium assignmentExample:

The basic flow conservation identity: q = x1 + x2 = 4.5

q: total traffic flow between OD pairs in 1000s of Veh/hr

Step 1: Check route usability

if all traffic is assigned to route 1

t1(4.5) = 25min

t2(0) = 4min

if all traffic is assigned to route 2

t1(0) = 6min

t2(4.5) = 24.25min

User equilibrium assignmentExample:

Wardrop’s equilibrium rule: t1 = t2

6 + 4x1 = 4 + x22

From flow conservation, we have: x1 + x2 = 4.5

x1 = 4.5− x2

6 + 4(4.5− x2) = 4 + x22

x2 = 2.89

x1 = 4, 5− x2 = 1, 6

Computing average travel times:

t1 = 6 + 4(1.6) = 12.4min

t2 = 4 + (2.89)2 = 12.4min

User equilibrium assignmentExample:

Total System Travel time:

Z(x) = x1t1(x1) + x2t2(x2)

= 2899veh/hr(12.4min) + 1601veh/hr(12.4min)

= 930veh/hr

Traffic assignment Method 1

System optimum assignment

System optimum assignment Statement:

System optimum assignment is based on Wardrop’s second principle.

Definition: Wardrop’s second principle

drivers cooperate with one another in order to minimise total system travel time.

System optimum assignment

!

k

frsk = qrs : !r, s

Minimize:

Subject to:

xa =!

r

!

s

!

k

δrsa,kfrsk : !a

frsk ≥ 0 : ∀k, r, s

xa ≥ 0 : a ∈ A

is the pathequilibrium flow in link atravel time on link aflow on path k connecting OD pairs r-strip rate between r and sis constraint function defined as follows:

k

xa

ta

frsk

qrs

δrsa,k

δr,sa,k =

!1 :0 : otherwise

if link a belongs to path k

Z =!

a

xata(xa)

System optimum assignment Example:

Using the previous example t1 = 6 + 4x1 t2 = 4 + x22

Z(x) = x1t1(x1) + x2t2(x2)

Z(x) = x1(6 + 4x1) + x2(4 + x22)

Z(x) = 6x1 + 4x21 + 4x2 + x3

2

From flow conservation, we have: x1 + x2 = 4.5

x1 = 4.5− x2

Z(x) = 6(4.5� x2) + 4(4.5� x2)2 + 4x2 + x3

2

Z(x) = x32 + 4x2

2 � 38x2 + 108

System optimum assignment Example:

To minimise we take derivatives equal to zero:

Z(x)

dx2= 3x2

2 + 8x2 � 38 = 0 x2 = 2.46

x1 = 4.5� x2 = 2.03

d(Z(x))

Remark Results are not equal to user equilibrium travel time

For System optimum travel times:

t1 = 6 + 4(2.03) = 14.13min

t2 = 4 + (2.46)2 = 10.08min

System optimum assignment Example:

Total travel time:

Z(x) = x1t1(x1) + x2t2(x2)

= 2033veh/hr(14, 13min) + 2467veh/hr(10.08min)

= 893.2veh/hr

Before in UE we got 930veh/hr, hence we saved 36.8veh/hr

Summary ❖ Completed 4 Steps in Travel demand Model

Trip generation

Trip distribution

Modal split

Assignment

Land use data

zone to zone,travel time

cost, etc

Highway and transit network

Traffic volumes

Highway and transit network

1

2

3

4