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Transportation Engineering 2010 Fall Semester

Route Choice

Route Choice

Zone 1

Zone 2

500 trips • Trip Distribution

Zone 1

Zone 2

Car: 300 trips • Mode Choice

Bus: 50 trips

Subway: 150 trips

Zone 1

Zone 2

Among 300 trips of Car

• Route Choice Route 1 : 50 trips

Route 2 : 150 trips

Route 3 : 100 trips

• 100 trips • From here • To Sinseol-Dong Station

• Using Car • Which Route?

Route 1

Route 2

Route 3

Source: http://map.google.com

Route Choice

• All-or-Nothing Assignment • User Equilibrium (UE) Assignment • System Optimal (SO) Assignment

Finding Shortest Path

• Why do we need to find the shortest path? – In the trip assignment (or route choice), the

model assumes that people try to travel the minimum-travel-time paths

– The problem is finding the minimum-travel- time paths connecting each O-D pair for a given set of link travel time.

Dijkstra’s Algorithm

1 3 8 10

2 4 7 9

5 6

1 : Node 1

: Link with Travel Time 10 10

9 8 9 8

4 11 14

6 12 10

13 15

11

13

• Find Shortest Path from Zone 5 to Zone 10

Dijkstra’s Algorithm

1 3 8 10

2 4 7 9

5 6

21(4) 25(3) 40(8)

19(4) 25(4) 35(7)13(5)

11(1)

26(4)

• Result

25(3) : Travel time from node 5 is 25 and the previous node is 3

Shortest Path from Node 5 to Node 10

All-or-Nothing Assignment

• All trips are assigned on the shortest route which is the minimum travel time or cost between zones

• Simple and inexpensive to perform • Does not take account of congestion effect

– Assumes there is no travel time change due to increased traffic

– Flow patterns could be unrealistic – Can be used for a special cases (significantly under-

saturated traffic etc.)

Link Performance Function • Link travel time vs. Flow demand • Travel time increases with flow demand • Grows exponentially if the demand is larger than the capacity • BPR (Bureau of Public Road) Function

1 : Link Travel Time : Free Flow Travel Time : Link Flow Demand : Capacity (Max. flow Rate) : Parameter to be calibrated

(default: 0.15) : Parameter to be calibrated

(default: 4)

0

0.5

1

1.5

2

2.5

3

3.5

4

0 1000 2000 3000 4000 5000 /

Capacity

Link Performance Function

• Can flow exceed “capacity”? • On a link, the capacity is thought of as a maximum

“outflow.” • Demand is inflow. • If inflow > outflow for a period of time, there is

queueing (and delay). • For Example: For a 1 hour period, if 2100 cars arrive

and 2000 depart (=capacity), 100 are still there. • This equation tries to represent that phenomenon in a

simple way.

User Equilibrium

• Wardrop’s First Principal – “The travel times in all routes actually used are

equal and less than those which would be experienced by a single vehicle on any unused route”

– Each user acts to minimize his/her own cost, subject to every other user doing the same

– Travel times are equal on all used routes. – Thus, at equilibrium, no one can reduce his or

her travel cost by choosing another choice set.

UE Example

O/D Trip = 5 The link perfomance functions

t1=2+x1 and t2=1+2x2.

Flow conservation: x1+x2 = 5

UE status: Travel time for path 1 =

Travel time for path 2 t1 = t2 2+x1 = 1+2x2

Solve: 2+5-x2=1+2x2 x2 = 2, x1 = 3

Route 2

Route 1

t1=2+x1

t2=1+2x2

UE Formulation

SO Assignment

• System Optimal – Purpose: Minimize the sum of total travel time

of all people • Wardrop’s Second Principal

– “At equilibrium the average journey time is minimum”

– Ideal Condition • Each user behaves cooperatively in choosing his own

route to ensure the most efficient use of the whole system

– Not a real world case – Can be achieved using road pricing (toll) etc.

SO Assignment

This can be formulated as a mathematical program

Min

Subject to , for all k, and O-D pairs

, for all k, and O-D pairs

Where, for all links

a a a

Z = X t

k k

f = q
f 0*k *

a k ak OD K

X = f δ

Example

O/D Trip = 5 The link performance functions

t1=2+x1 and t2=1+2x2.

Flow conservation: x1+x2 = 5Route 2

Route 1

t1=2+x1

t2=1+2x2

• Result

**X1 X2 T1 T2 Total
Travel Time
**

AoN 0 5 2 11 55

UE 3 2 5 5 25

SO 3.17 1.83 5.17 4.66 24.92

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