Example of the Equilibrium_Lohse procedure

The Equilibrium_Lohse procedure is demonstrated below with a calculation example. Table 133 shows the parameter settings of Equilibrium_Lohse and the impedance for links and routes in the unloaded network. Table 134, Table 135, and Table 136 show three iterations of the calculation process.

LinkNo

Type

Length [m]

v0 [km/h]

Capacity [car units]

R0* [min]

1

20

5,000

100

1,200

03:00

2

20

5,000

100

1,200

03:00

3

20

5,000

100

1,200

03:00

5

20

5,000

100

1,200

03:00

6

20

5,000

100

1,200

03:00

7

20

5,000

100

1,200

03:00

8

30

16,000

80

800

12:00

9

30

5,000

80

800

03:45

10

40

10,000

60

500

10:00

11

40

5,000

60

500

05:00

Table 133: Impedance in unloaded network, input parameters of Equilibrium_Lohse procedure

Route

Links

Length [m]

R0* [min]

1

1+8+9

26,000

0:18:45

2

1+2+3+5+6+7

30,000

0:18:00

3

10+11+5+6+7

30,000

0:24:00

Input parameters:

  • BPR function with a = 1, b = 2, c = 1
  • bottom = 0.15
  • top = 0.5
  • V1 = 2.5
  • V2 = 4
  • V3 = 0.002

LinkNo

Volume 1 [car units]

R1 [min]

TT1

f(TT1)

Delta 1

R1* [min]

1

2,000

11:20

2.78

0.0452

0.4796

07:00

2

2,000

11:20

2.78

0.0452

0.4796

07:00

3

2,000

11:20

2.78

0.0452

0.4796

07:00

5

2,000

11:20

2.78

0.0452

0.4796

07:00

6

2,000

11:20

2.78

0.0452

0.4796

07:00

7

2,000

11:20

2.78

0.0452

0.4796

07:00

8

0

12:00

0.00

0.0450

0.5000

12:00

9

0

03:45

0.00

0.0450

0.5000

03:45

10

0

10:00

0.00

0.0450

0.5000

10:00

11

0

05:00

0.00

0.0450

0.5000

05:00

Table 134: Example of Equilibrium_Lohse: 1. Iteration step

Route

Volume 1

R1

R1*

1

0

0:27:05

0:22:45

2

2,000

1:08:00

0:41:59

3

0

0:49:00

0:35:59

LinkNo

Volume 2 [car units]

R2 [min]

TT2

f(TT2)

Delta 2

R2* [min]

1

2,000

11:20

0.62

0.0450

0.4925

09:08

2

1,000

05:05

0.27

0.0450

0.4962

06:03

3

1,000

05:05

0.27

0.0450

0.4962

06:03

5

1,000

05:05

0.27

0.0450

0.4962

06:03

6

1,000

05:05

0.27

0.0450

0.4962

06:03

7

1,000

05:05

0.27

0.0450

0.4962

06:03

8

1,000

30:45

1.56

0.0451

0.4855

21:06

9

1,000

09:37

1.56

0.0451

0.4855

06:36

10

0

10:00

0.00

0.0450

0.5000

10:00

11

0

05:00

0.00

0.0450

0.5000

05:00

Table 135: Example of Equilibrium_Lohse: 2. Iteration step

Route

Volume 2

R2

R2*

1

1,000

0:51:42

0:36:50

2

1,000

0:36:45

0:39:22

3

0

0:30:15

0:33:08

LinkNo

Volume 3 [car units]

R3 [min]

TT3

f(TT3)

Delta 3

R3* [min]

1

1,333

06:42

0.27

0.0450

0.4963

07:56

2

667

03:56

0.35

0.0450

0.4953

05:00

3

667

03:56

0.35

0.0450

0.4953

05:00

5

1,333

06:42

0.11

0.0450

0.4984

06:22

6

1,333

06:42

0.11

0.0450

0.4984

06:22

7

1,333

06:42

0.11

0.0450

0.4984

06:22

8

667

20:20

0.04

0.0450

0.4994

20:43

9

667

06:21

0.04

0.0450

0.4994

06:28

10

667

27:47

1.78

0.0451

0.4842

18:37

11

667

13:53

1.78

0.0451

0.4842

09:18

Table 136: Example of Equilibrium_Lohse: 3. Iteration step

Route

Volume 3

R3

R3*

1

667

0:33:23

0:35:07

2

667

0:34:40

0:37:03

3

667

1:01:47

0:47:02

Table 133, Table 134, Table 135, and Table 136 illustrate the first three iteration steps of the Equilibrium_Lohse procedure for the example network.

Iteration step 1, n = 1
  • Volume 1

The volume of the first iteration step results from an "all or nothing" assignment onto the lowest impedance route in the unloaded network. For impedance R0*, this is route 2 loaded with 2,000 car trips.

  • Current impedance R1

The current impedance R1 of every link results from the BPR capacity function (a =1, b = 2, c= 1). For link 1, for example, the following can be calculated:

R1 (link 1) = 3 min x (1+(2 000/1 200)²) = 11 min 20s

  • Estimated impedance R1*

The estimated impedance R1* of every link consists of the current impedance R1 and the estimated impedance R0* of the last iteration step. It results from the learning factor Δ. To determine R1* for link 1, the following calculations are necessary:

  • R0* = 3 min = 180 s
  • R1 = 11 min 20s = 680 s
  • TT1 = |R1 - R0*| /R0* = |680 s - 180 s| / 180 s = 2.78

 

  • R1* = R0* + Δ1 • (R1 - R0*) = 180 s + 0.4796 • (680 s - 180 s) = 420 s
Iteration step 2, n = 2
  • Volume 2

The lowest impedance route for R1* is route 1. Now two routes exist, route 1 and 2. Each route is loaded with 1/n, i.e. ½ the demand, so that each route is used by 1,000 cars.

  • Current impedance R2

The current impedance R2 of every link increases on newly loaded links 8 and 9, and it decreases on links 2, 3, 5, 6 and 7.

  • Estimated impedance R2*

The estimated impedance R2* of every link consists of the current impedance R2 and the estimated impedance R1* of the last iteration step.

Iteration step 3, n = 3
  • Volume 3

The lowest impedance route for R2* is route 3. 2,000 car trips are now equally distributed across routes 1, 2 and 3.

  • Current impedance R3

The current impedance R3 again results from the current volume 3 via the VD function.

  • Estimated impedance R3*

The estimated impedance R3* of every link consists of the current impedance R3 and the estimated impedance R2* of the last iteration step.

Iteration step 4, n = 4

The concluding route search based on R3* determines route 1 as the shortest route. Thus, the following route volumes result:

  • Volume route 1 = 2/4 • 2,000 = 1,000 trips
  • Volume route 2 = 1/4 • 2,000 = 500 trips
  • Volume route 3 = 1/4 • 2,000 = 500 trips