Example of stochastic assignment

Table 140 shows the key input data used in the example network. If the following parameters are chosen for the search, then in a single external iteration, all 3 conceivable routes will be found:

  • Number of search iterations = 5
  • σ= 8 • R0.5
  • Compared to the "objective" impedances (resulting from impedance definitions and VDFs), the impedances of the network objects are changed for alternative shortest path searches. They are drawn randomly from a normal distribution which has the objective impedance R as mean value and whose standard deviation σ is given as a function of R.

LinkNo

Type

v0 [km/h]

Length [m]

Capacity [car units]

R0* [min]

R0* [s]

1

20

100

5000

1200

3:00 AM

180

2

20

100

5000

1200

3:00 AM

180

3

20

100

5000

1200

3:00 AM

180

5

20

100

5000

1200

3:00 AM

180

6

20

100

5000

1200

3:00 AM

180

7

20

100

5000

1200

3:00 AM

180

8

30

80

16000

800

12:00 PM

720

9

30

80

5000

800

3:45 AM

225

10

40

60

10000

500

10:00 AM

600

11

40

60

5000

500

5:00 AM

300

Table 140: Impedance in unloaded network, input parameters for stochastic assignment

Route

Links

Length [m]

R0* [min]

R0* [s]

1

1+8+9

26000

12:18:45 AM

1125

2

1+2+3+5+6+7

30000

12:18:00 AM

1080

3

10+11+5+6+7

30000

12:24:00 AM

1440

Input parameters

  • BPR function with a = 1, b = 2, c = 1
  • Δbottom = 0,5, Δtop = 0.5 ⇒ Δ = 0.5
  • Assignment with Logit, β= 0.001

After completing the search, the correction factor for the independence of each route is determined according to Cascetta. It is based on the similarity of the individual route pairs with reference to time t0 or to the length. Table 141 shows the commonality factors C. These are used to calculate the correction factor CFr of route r.

  • Route 1

  • Route 2

  • Route 3

Table 141: Calculation of commonality factor C for all route pairs

Route pair

t0ij

t0i

t0j

Cij

1.1

1125

1125

1125

1.00

1.2

180

1125

1080

0.16

1.3

0

1125

1440

0.00

2.1

180

1080

1125

0.16

2.2

1080

1080

1080

1.00

2.3

540

1080

1440

0.43

3.1

0

1440

1125

0.00

3.2

540

1440

1080

0.43

3.3

1440

1440

1440

1.00

The share by route is calculated from the correction factor according to Cascetta and from the impedance Rmin0 in the unloaded network.

For Route 1, the portion is calculated using the Logit model as follows:

In the same way, the shares for routes 2 and 3 shown in Table 142 are calculated. The product of share P and demand F is the volume of each route qr1 in the first iteration step. For Route 1, the calculation is as follows: 0.425 • 2000 = 849.4 PCU. Based on the route volumes, the link volumes and thus the network impedances can be calculated (Illustration 109). This results in the impedances R1 of the routes. These interim results can be reproduced in Visum if the maximum number of internal iterations are set to M = 1 in the assignment parameters.

Table 142: Volumes in the first internal iteration step m = 1

Route

E

Rmin0

exp(Rmin0)•E

Portion P

qr1

R1

1

0.8596

1125

0.279079049

0.425

849.4

2470

2

0.6264

1080

0.212737561

0.324

647.5

1961

3

0.6978

1440

0.165335421

0.252

503.2

2848

Total

 

 

0.657152032

1.000

2000.0

 

Illustration 109: Volumes and link run times after the first internal iteration step m=1

For the route choice in the second iteration step, an estimated impedance Rmin1 is calculated. Since Δ = 0.5, this impedance results from the formation of the mean value of Rmin0 and R1. On the basis of Rmin1, as in the first iteration step, the assignment is then made for the 3 routes. For each route, the interim result is qr2. To smooth the volumes between two iteration steps, the MSA method (Method of Successive Averages) is used.

For m = 2, this results in the following for the volume of Route 1:

This route volume then leads to the link volumes and impedances of the second iteration step (Table 143). The iterations are repeated until the termination criteria are met.

Table 143: Volumes in the second internal iteration step m = 2

Route

E

Rmin1

exp(R)•E

Portion P

qr2

qr2

R2

1

0.8596

1797.6

0.142432

0.3944

788.8

819.1

2405.2

2

0.6264

1520.7

0.136919

0.3791

758.3

702.9

2016.0

3

0.6978

2144.0

0.081775

0.2264

452.9

478.0

2785.6

Total

 

 

0.361126

 

 

2000