5-Modal Split & Traffic Assignment-( Transportation and Traffic Engineering Dr. Sheriff El-Badawy )

1
‫والتكنولوجيا‬ ‫للهندسة‬ ‫العالي‬ ‫مصر‬ ‫معهد‬–‫المدنية‬ ‫الهندسة‬ ‫قسم‬-‫المنصورة‬
Dr. Sherif El-Badawy
3rd Year Civil
Transportation and Traffic Engineering
Modal Split and Traffic
Assignment
Objectives & Goals
Future planning & land
use parameters
Future transport
facilities
Transportation Generation
Transportation distribution
Model split
Traffic assignment
Is the level of
service satisfactory
Are the objectives
achived
YES
YES
No
ChangeFuturetransportfacilities
No
ChangeLanduse
&policies
Present planning &
socio – economic data
Present transport
facilities
Present travel pattern
End
Survey phase
Comprehensive
Urban Transport
Planning Process

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3
Modal Split
Definition:
• Is the process of separating person-trips by the
mode of travel.
• Usually expressed as a fraction, ratio, or
percentage of the total number of trips.
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Characteristics of the trip
Trip purpose, trip length.
Household Characteristics
Income, Car ownership, family size and composition.
Zonal Characteristics
Residential density, concentration of workers,
distance from CBD.
Network Characteristics
Accessibility ratio, travel time ratio, travel cost ratio.
Factors Affecting Modal Split

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5
Mode choice depends on:
• Travel time (in Vehicle + waiting)
• Travel cost
• Socio-economic status of user (income,
family size,….)
• Level of travel service (comfort, Safety,
speed, ……..)
The variable defining the model is a function
called the Utility Function
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Utility Function (Multi-Nomial Logit function)
P%(m) = the probability of choosing mode (m) of transportation
e = exponential base
um = the utility function of different modes
if we assume an urban area with bus, car and taxi modes
% for bus mode will be:
% for car mode will be:
% for Taxi mode will be:

4
7
Utility Function
The general utility function is usually of the form:
Where:
Km = mode specific constant
tm = in vehicle time (minutes-one way)
xm = out of vehicle time (minutes-one way)
d = distance (miles - one way)
cm = cost (piasters - one way)
Y= annual income (pounds)
m is A for Auto, B for Bus, T for Taxi, t for train
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Considering a user choosing between two modes an auto
and a public Bus considering the following situation:
If d was 7 miles. And it
is also given that the
annual income of the user
is 3000 pounds.
BusAuto
tB= 14 mintA= 11 min
xB = 8 minxA = 5 min
cB = 50 piasterscA = 122 piasters
Example
And his value parameters as collected from data collection are:
K1 = -0.03 k2 = -0.34 k3 = -50
The values of the mode specific constants are assumed to be
kA = -0.13 kB = zero

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9
Solution
uA = -0.13 – 0.03 (11) - 0.34 (5/7) -50 (122/3000)
= -0.13 – 0.33 - 0.243 – 2.03 = -2.74
uB = 0 – 0.03 (14) - 0.34 (8/7) -50 (50/3000)
= 0 – 0.42 – 0.39 – 0.83 = -1.64
P%A = 25% and in the same way we get P%B = 75%
BusAuto
tB= 14 mintA= 11 min
xB = 8 minxA = 5 min
cB = 50cA = 122
kB =-0kA =-0.13
k1 = -0.03 k2 = -0.34
k3 = -50
Review
(Modal Split)
(Traffic Assignment )

6
Which route
will be chosen
by the user?
12
Traffic Assignment
Definition: is the stage in the transportation
planning process where in the trip interchanges
are allocated to different parts of the network
forming the transportation system.
In this stage:
i. The route to be travelled is determined.
ii. The inter-zonal flows are assigned to the
selected routs.

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13
General principals
• All assignment techniques are based on route
selection.
• The choice of the route is made on the basis of
a number of criteria as Journey time, length,
cost, comfort, convenience, safety.
 (Travel Resistance)
• Journey time is often considered the sole
criterion since length and cost can be
considered as a function of time in most cases.
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Transport Link
As a first step, the highway network is described by a
system of links and nodes.
A link is a section of a highway network between two
intersections.
A node is either a centroid of a zone or the intersection of
two or more links.

8
Link-Node Map for Highway System
16
Traffic Assignment
Methods
All or Nothing
Capacity
Restraint

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17
All or Nothing Method
Most commonly used method.
Example: Minimum Path Tree
All traffic will choose the route where the travel
resistance is least.
Travel Resistance:
• Time
• Distance
• Cost
• Generalized Cost
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22
2523 24 26
15
11
18
19
20 21
16
12
17
1
13
14
(4)
(4)
(2)
(2)
(2) (2)
(2) (2) (2)
(2)
(2)
(1)
(1)
(1)
(5)
(3)
(3)
(3)
(3)(3)
(5)
(5)
Example
(3)

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19
Starting from centroid 1 we go to each connecting link and
choose the least travel time
T1-20 = 3 T1-17 = 3
the time is the same , if we begin with the node with lower
number node 17 is noted:
T1-17-19 = 5 T1-17-16 = 5 T1-17-16 = 6
The next closest node to centroid 1 is 20
T1-20-19 = 4 T1-20-25 = 6 T1-20-21 = 7
There are two routes to reach 19 from centroid 1, i.e. 1-17-19
and 1-20-19. the rout 1-20-19 is shorter in time, therefore is
chosen
20
22
2523 24 26
15
11
18
19
20 21
16
12
17
1
13
14
(4)
(4)
(2)
(2)
(2) (2)
(2) (2) (2)
(2)
(2)
(1)
(1)
(1)
(5)
(3)
(3)
(3)
(3)(3)
(5)
(5)
The process is repeated until all nodes have been
covered by the shortest path. The minimum path
tree for this highway network is given in figure
(3)

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All or Nothing Assignment:
• All trips are assigned on the shortest route which is the
minimum travel time, cost, distance, …..between zones.
• Simple and inexpensive to perform.
• Does not take into account the effect of traffic congestion:
• Assumes there is no travel time change due to increased traffic.
• Flow patterns could be unrealistic.
• Can be used for special cases (under saturated traffic, ..etc)
1 2 3 4
5 6 7 8
9 10 11 12
14 15 1613
631
232
334
444
132 1
132 1
242 1
Find the shortest path from node 1 to all other nodes (from Garber and Hoel)
Yellow numbers represent link travel times in minutes3
Here’
s
how
…

12
1 2 3 4
5 6 7 8
9 10 11 12
14 15 1613
631
232
334
444
132 1
132 1
242 1
STEP 1
1
2
1 2 3 4
5 6 7 8
9 10 11 12
14 15 1613
631
232
334
444
132 1
132 1
242 1
STEP 2
1
2
4
5

13
1 2 3 4
5 6 7 8
9 10 11 12
14 15 1613
631
232
334
444
132 1
132 1
242 1
STEP 3
1
2
4
5
4
4
1 2 3 4
5 6 7 8
9 10 11 12
14 15 1613
631
232
334
444
132 1
132 1
242 1
STEP 4
1
2
4
4
Eliminate
5 >= 4
4
5

14
1 2 3 4
5 6 7 8
9 10 11 12
14 15 1613
631
232
334
444
132 1
132 1
242 1
STEP 5
1
2
4
4
4 10
6
1 2 3 4
5 6 7 8
9 10 11 12
14 15 1613
631
232
334
444
132 1
132 1
242 1
STEP 6
1
2
4
4
4 1
0
6
7
Eliminate
7 >= 6
7

15
6
1 2 3 4
5 6 7 8
9 10 11 12
14 15 1613
631
232
334
444
132 1
132 1
242 1
STEP 7
1
2
4
4
4 1
0
6
Eliminate
8 >= 7
8
7
7
8
6
1 2 3 4
5 6 7 8
9 10 11 12
14 15 1613
631
232
334
444
132 1
132 1
242 1
STEP 8
1
2
4
4
4 1
0
7
6

16
1
0
7
8
6
1 2 3 4
5 6 7 8
9 10 11 12
14 15 1613
631
232
334
444
132 1
132 1
242 1
STEP 9
1
2
4
4
4 1
0
7
6
1
0
7
8
6
1 2 3 4
5 6 7 8
9 10 11 12
14 15 1613
631
232
334
444
132 1
132 1
242 1
STEP 10
1
2
4
4
4 1
0
7
6
1
0Eliminate
10 >= 7
1
0
Eliminate
10 >= 10

17
7
8
6
1 2 3 4
5 6 7 8
9 10 11 12
14 15 1613
631
232
334
444
132 1
132 1
242 1
STEP 11
1
2
4
4
4 10
7
6
10
1
0
8
7
8
6
1 2 3 4
5 6 7 8
9 10 11 12
14 15 1613
631
232
334
444
132 1
132 1
242 1
STEP 12
1
2
4
4
4 10
7
6
1
0
8
9
91
0
Eliminate 10 > 9
Eliminate
10 >= 9

18
7
8
6
1 2 3 4
5 6 7 8
9 10 11 12
14 15 1613
631
232
334
444
132 1
132 1
242 1
STEP 13
1
2
4
4
4
7
6
10
8
9
9
12 12
Eliminate
12 >= 10
7
8
6
1 2 3 4
5 6 7 8
9 10 11 12
14 15 1613
631
232
334
444
132 1
132 1
242 1
STEP 14
1
2
4
4
4
7
6
10
8
9
9
12 10
Eliminate
12 >= 10

19
7
8
6
1 2 3 4
5 6 7 8
9 10 11 12
14 15 1613
FINAL
1
2
4
4
4
7
6
10
8
9
9
10
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Capacity Restraint Method
• Capacity restraint assignment is a process in
which the travel resistance of a link is
increased according to a relation between the
practical capacity of the link and the volumes
assigned to the link.
• Because of the iterative nature of the
calculations involved, the capacity restrained
method is carried out entirely by computer.

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39
Capacity Restraint Techniques
Smock Method Bureau of Public Roads Method
Where:
To original travel time
TA adjusted travel time
e exponential base
V assigned volume
C link capacity
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DHV = k * ADT
Where
DHV = Design Hourly Volume
k = Constant called Peak hour factor
ADT = Average Daily traffic
We get the required number of lanes according to the
Design Hourly Volume from the equation
No of lanes = DHV ÷ Lane capacity

5-Modal Split & Traffic Assignment-( Transportation and Traffic Engineering Dr. Sheriff El-Badawy )

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5-Modal Split & Traffic Assignment-( Transportation and Traffic Engineering Dr. Sheriff El-Badawy )