All-or-nothing assignment model

In this model it is assumed that (i) the travel time on links do not vary with link flows, i.e. $\tau_a(x_a) = \tau_a$ and (ii) all trip makers (users) have precise knowledge of the travel time on the links. Based on these assumptions about travel times and the postulate that a trip maker will choose that path (or route) which minimizes his / her travel time this assignment model assigns all the trips between a particular origin and destination pair to that route (or path) which offers the minimum travel time.

The exact nature of the assignment model is presented through the following algorithm.

Step 1:

For every $i-j$ pair (i.e., origin - destination pair) with $t_{ij} > 0$ determine the minimum travel time path (or route) using $\tau_a$ as the link travel times. The minimum path determination can be done using any of the various existing algorithms like Flyod's algorithm or Djkastra's algorithm. Detailed description of these algorithms can be found in Teodorovic [#!teo1!#] or any other book on theory of networks. Also initialize all $x_a^0= 0$.

Step 2:

Set iteration counter $k=1$. Select a particular $i-j$ pair.

Step 3:

Assign the entire $t_{ij}$ to the minimum path between the $i-j$ pair. If link $a$ is a part of the minimum path set, $x_a^k=x_a^{k-1} + t_{ij}$ else set $x_a^k=x_a^{k-1} + 0$.

Step 4:

If $k=N$ (where $N$ is the total number of $i-j$ pairs with $t_{ij} > 0$) then report $x_a^k$ as $x_a$. Else, select another $i-j$ pair; set $k=k+1$ and go back to Step 3.

Example

For the network shown in Figure  and the trip distribution matrix given in Table  determine the link flows using the all-or-nothing assignment technique. Note that the numbers on the links of the network denote the travel times and the numbers in the circles denote the zone numbers.

図: Network for example problem on all-or-nothing assignment technique.

表: Trip distribution matrix (O-D matrix) for the example problem on all-or-nothing assignment model.

Origin	Destination zone
zone	1	2	3	4	5
1	0	0	200	100	150
2	0	0	300	300	50
3	200	300	0	100	100
4	100	300	100	0	0
5	150	50	100	0	0

Solution

Note there are 25 possible zone pairs out of which 9 have $t_{ij}= 0$. Hence N= 16.

Step 1 calculations:
The minimum path for the 16 zone pairs (obtained using Djkastra'a algorithm) are as follows:

$i-j$ pair 		 		 Min. path 		 		 $i-j$ pair 				 Min. path 


1-3 		 : 		 1 $\rightarrow$ 3 		 		 1-5 		 : 		 1 $\rightarrow$3 $\rightarrow$ 5 


3-1 		 : 		 3 $\rightarrow$ 1 		 		 5-1 		 : 		 5$\rightarrow$ 3 $\rightarrow$ 1 


1-4 		 : 		 1 $\rightarrow$ 3$\rightarrow$ 4 		 		 2-3 		 : 		 2 $\rightarrow$ 3 


4-1 		 : 		 4$\rightarrow$ 3 $\rightarrow$ 1 		 		 3-2 		 : 		 3 $\rightarrow$ 2 


2-4 		: 		 2 $\rightarrow$ 3 $\rightarrow$ 4 		 		  2-5 		 : 		 2 $\rightarrow$ 3$\rightarrow$ 5 


4-2 		 : 		 4 $\rightarrow$ 3 $\rightarrow$ 2 		 		 5-2 		: 		 5 $\rightarrow$ 3 $\rightarrow$ 2 


3-4 		 : 		 3 $\rightarrow$ 4 		 		3-5 		 : 		 3 $\rightarrow$ 5 


4-3 		 : 		 4 $\rightarrow$ 3 		 		 5-3 		:		 5 $\rightarrow$ 3 

Step 2 calculations:
$k=1$. Consider the zone pair $1-3$.

Step 3 calculations
$x_{1 \rightarrow 3}^1 = 0 + 200 = 200$; the rest of the $x_a^k$ remain zero.

Step 4 calculations
Since $k \neq 16$, set $k=2$ and select zone pair $1-5$ as the next pair and go back to Step 3.

Step 3 calculations
$x_{1 \rightarrow 3}^2 = 200 + 150 = 350$; $x_{3 \rightarrow 5}^2 = 0 + 150 = 150$; the rest of the $x_a^k$ remain zero.

Step 4 calculations
Since $k \neq 16$, set $k=3$ and select zone pair $3-1$ as the next pair and go back to Step 3.

In this manner Steps 3 and 4 are repeated till all the zone pairs are chosen (i.e., $k=16$). Finally, the following assignment is obtained.

$x_{1 \rightarrow 3}$ =	450		$x_{3 \rightarrow 5}$ =	300
$x_{2 \rightarrow 3}$ =	650		$x_{4 \rightarrow 2}$ =	0
$x_{2 \rightarrow 4}$ =	0		$x_{4 \rightarrow 3}$ =	500
$x_{3 \rightarrow 1}$ =	450		$x_{5 \rightarrow 3}$ =	300
$x_{3 \rightarrow 2}$ =	650		$x_{5 \rightarrow 4}$ =	0
$x_{3 \rightarrow 4}$ =	550		$x_{4 \rightarrow 5}$ =	0