1. URBAN TRANSPORTATION NERWORKS
FORMULATING THE ASSIGNMENT PROBLEM

2. OBJECTIVE Some notations The basic transformation
Equivalency conditions
Uniqueness conditions
The system – optimization formulation
User equilibrium and System Optimum

3. 1.NOTATIONS Present some network notations 𝑁: node set 𝐴: arc set
𝑅: set of origin nodes
𝑆: set of destination nodes
𝐾 𝑟𝑠 : set of path connecting O-D pairs 𝑟−𝑠

4. 𝑥 𝑎 flow on arc 𝑎
𝒙=(…, 𝑥 𝑎 , …)
𝑡 𝑎 travel time on arc 𝑎
𝒕= …, 𝑡 𝑎 ,…
𝑓 𝑘 𝑟𝑠 flow on path 𝑘 connecting O-D pair 𝑟−𝑠
𝒇 𝑟𝑠 = …,𝑓 𝑘 𝑟𝑠 ,…. ;𝒇=(…, 𝒇 𝑟𝑠 , …)

5. 𝑐 𝑘 𝑟𝑠 travel time on path 𝑘 connecting O-D pair 𝑟−𝑠
𝒄 𝑟𝑠 = …, 𝑐 𝑘 𝑟𝑠 ,…. ;𝒄=(…, 𝒄 𝑟𝑠 , …)
𝑞 𝑟𝑠 trip rate between O-D pair 𝑟−𝑠
𝒒=(…, 𝑞 𝑟𝑠 , …)
𝛿 𝑎,𝑘 𝑟𝑠 indicator variable
𝛿 𝑎,𝑘 𝑟𝑠 = 1: arc a on path 𝑘 between 𝑟−𝑠 0:otherwise
(∆ 𝑟𝑠 ) 𝑎,𝑘 = 𝛿 𝑎,𝑘 𝑟𝑠 ; ∆=( …,(∆ 𝑟𝑠 ) 𝑎,𝑘 ,…)

6. Path-arc incidence relationship
Travel time on path and travel time on link
𝑐 𝑘 𝑟𝑠 = 𝑎 𝑡 𝑎 𝛿 𝑎,𝑘 𝑟𝑠 (1.1a)
Link flow and path flow
𝑥 𝑎 = 𝑟,𝑠,𝑘 𝑓 𝑘 𝑟𝑠 𝛿 𝑎,𝑘 𝑟𝑠 (1.1.b)

7. In the matrix form:
𝒄=𝒕∆ and 𝒙=𝒇 ∆ 𝑻

8. Example

9. Assume: from 1 to 4: the first path (1,3)
the second path (1,4)
from 2 to 4: the first path (2,3)
the second path (2,4)
The path-arc incidence relationships
𝑐 1 14 = 𝑡 1 𝛿 1, 𝑡 2 𝛿 2, 𝑡 3 𝛿 3, 𝑡 4 𝛿 4,1 14 = 𝑡 1 + 𝑡 2
𝑥 3 = 𝑓 𝛿 3, 𝑓 𝛿 3, 𝑓 𝛿 3, 𝑓 𝛿 3,2 24
= 𝑓 𝑓 1 24

10. 2.THE BASIC TRANSFORMATION
Mathematical program: (Beck-mann’s transformation)
Min 𝑧 𝒙 = 𝑎 0 𝑥 𝑎 𝑡 𝑎 (𝜔) 𝑑𝜔 (2.1𝑎)
(The meaning of this objective function?)
Subject to
𝑓 𝑘 𝑟𝑠 =𝑞 𝑟𝑠 ∀𝑟,𝑠 (2.1b)
𝑓 𝑘 𝑟𝑠 ≥ ∀𝑘,𝑟,𝑠 (2.1𝑐)
𝑥 𝑎 = 𝑟,𝑠,𝑘 𝑓 𝑘 𝑟𝑠 𝛿 𝑎,𝑘 𝑟𝑠 ∀𝑎 (2.1c)

11. Some assumptions
𝜕 𝑡 𝑎 ( 𝑥 𝑎 ) 𝜕 𝑥 𝑎 >0 ∀𝑎 (2.2𝑎)
𝜕 𝑡 𝑎 ( 𝑥 𝑎 ) 𝜕 𝑥 𝑏 =0 ∀𝑎≠𝑏 (2.2𝑏)

12. EXAMPLE 𝑡 1 =2+ 𝑥 1 ; 𝑡 2 =1+2 𝑥 2 Trip rate 𝑞=5⇒ 𝑥 1 + 𝑥 2 =5
UE condition: 𝑡 1 = 𝑡 2 ⇒ 𝑥 1 =3 and 𝑥 2 =2 flow units , 𝑡 1 = 𝑡 2 =5(𝑡ime units)

13. Use program (2.1):
min 𝑧 𝒙 = 0 𝑥 𝜔 𝑑𝜔+ 0 𝑥 𝜔 𝑑𝜔
Subject to
𝑥 1 + 𝑥 2 =5
𝑥 1 , 𝑥 2 ≥0
Solution:
𝑥 1 =3 & 𝑥 2 =2; 𝑡 1 = 𝑡 2 =5

14. 3.EQUIVALENCE CONDITIONS

15. 𝑓 𝑘 𝑟𝑠 𝜕𝐿(𝒇,𝒖) 𝜕 𝑓 𝑘 𝑟𝑠 =0; 𝜕𝐿(𝒇,𝒖) 𝜕 𝑓 𝑘 𝑟𝑠 ≥0 ∀ 𝑘,𝑟,𝑠 (3.2𝑎)
SOLVE PROGRAM (2.1)
Consider Lagrange’s function:
𝐿 𝒇,𝒖 =𝑧 𝒙 𝒇 + 𝑟,𝑠 𝑢 𝑟𝑠 ( 𝑞 𝑟𝑠 − 𝑘 𝑓 𝑘 𝑟𝑠 ) 3.1a
𝑓 𝑘 𝑟𝑠 ≥0∀ 𝑘,𝑟,𝑠 (3.1b)
The first-order condition:
𝑓 𝑘 𝑟𝑠 𝜕𝐿(𝒇,𝒖) 𝜕 𝑓 𝑘 𝑟𝑠 =0; 𝜕𝐿(𝒇,𝒖) 𝜕 𝑓 𝑘 𝑟𝑠 ≥0 ∀ 𝑘,𝑟,𝑠 (3.2𝑎)
𝜕𝐿(𝒇,𝒖) 𝜕 𝑢 𝑟𝑠 =0∀ 𝑟,𝑠 (3.2𝑏)

16. The general first conditions (3.2a)
𝑐 𝑘 𝑟𝑠 − 𝑢 𝑟𝑠 =0 ∀ 𝑘, 𝑟, 𝑠 (3.3b)
𝑘 𝑓 𝑘 𝑟𝑠 = 𝑞 𝑟𝑠 ∀ 𝑟, 𝑠 (3.3c)
𝑓 𝑘 𝑟𝑠 ≥0 ∀ 𝑘, 𝑟, 𝑠 (3.3d)

17. RELATED TO U-E PROBLEM From 3.3a & (3.3b):
𝑐 𝑘 𝑟𝑠 = 𝑢 𝑟𝑠 = min 𝑘 𝑐 𝑘 𝑟𝑠 ∀ 𝑘 s.t: 𝑓 𝑘 𝑟𝑠 >0
⇒ state the U-E principle

18. 4. UNIQUENESS CONDITIONS
Prove that: problem have unique solution w.r.t link flow:
Objective function(2.1a) is strictly convex w.r.t 𝒙
Feasible region 2.1b & 2.1c is convex
Hessian matrix is positive define:
𝛻 2 𝑧 𝒙 =diag( 𝑑 𝑡 1 𝑥 1 𝑑 𝑥 1 , 𝑑 𝑡 1 𝑥 2 𝑑 𝑥 2 ,…, 𝑑 𝑡 1 𝑥 𝑛 𝑑 𝑥 𝑛 )

19. EXAMPLE

20. Equilibrium link flow:
𝑥 1 =2, 𝑥 2 =3, 𝑥 3 =3, 𝑥 4 =2, 𝑥 5 =5
Equilibrium path flow (use (1.1b))
𝑓 1 15 =2𝛼, 𝑓 2 15 =2 1−α
𝑓 1 25 =3−2𝛼, f 2 25 =2α
0≤𝛼≤1

21. ⇒ Program (2.1) has not unique solution w.r.t path flow 𝒇.
Reason?

22. 5. THE SYSTEM- OPTIMIZATION FORMULATION
Program:
min 𝑧 (𝒙) = 𝑎 𝑡 𝑎 𝑥 𝑎 ( 𝑡 𝑎 ) 𝑎
(The meaning of objective function?)
Subject to
𝑎 𝑓 𝑘 𝑟𝑠 =𝑞 𝑟𝑠 ∀𝑟,𝑠 (5.1b)
𝑓 𝑘 𝑟𝑠 ≥ ∀𝑘,𝑟,𝑠 (5.1𝑐)
𝑥 𝑎 = 𝑟,𝑠,𝑘 𝑓 𝑘 𝑟𝑠 𝛿 𝑎,𝑘 𝑟𝑠 ∀𝑎 (5.1c)

23. Characteristic:
Solution of S-O problem does not present equilibrium situation.
At S-O situation, divers can decrease their travel time by unilaterally changing routes

24. SOLVE S-O PROGRAM Equivalency conditions:
𝑐 𝑘 𝑟𝑠 − 𝑢 𝑟𝑠 =0 ∀ 𝑘, 𝑟, 𝑠 (5.2b)
𝑘 𝑓 𝑘 𝑟𝑠 = 𝑞 𝑟𝑠 ∀ 𝑟, 𝑠 (5.2c)
𝑓 𝑘 𝑟𝑠 ≥0 ∀ 𝑘, 𝑟, 𝑠 (5.2d)
Where: 𝑡 𝑎 = 𝑡 𝑎 𝑥 𝑎 + 𝑑 𝑡 𝑎 ( 𝑥 𝑎 ) 𝑑 𝑥 𝑎

25. 6. USER EQUILIBRIUM & SYSTEM OPTIMUM

26. SPECIAL CASE
Congestion is ignored: 𝑡 𝑎 𝑥 𝑎 = 𝑐 𝑎 =const ∀𝑎 𝑡 𝑎 𝑥 𝑎 = 𝑐 𝑎 =const ∀𝑎 ⇒ SO≡UE
General case: 𝑚𝑖𝑛 𝑧 𝒙 >𝑚𝑖𝑛𝑧(𝒙)

27. UE solution can be obtain by solve SO problem
𝑡 𝑎 𝑥 𝑎 = 1 𝑥 𝑎 𝑜 𝑥 𝑎 𝑡 𝑎 𝜔 𝑑𝜔
And vice versa.

28. BRAESS’S PARADOX

29. U-E solution: 𝑓 1 = 𝑓 2 =3 (flow units)
𝑡 1 = 𝑡 2 =53, 𝑡 3 = 𝑡 4 =30 (time units)
Total travel time: 498 (flow units)

30. Adding a new path to improve flow

31. UE solution: 𝑥 1 = 𝑥 2 = 2,𝑥 3 = 𝑥 4 =4, 𝑥 5 =2(flow units)
𝑐 1 = 𝑐 2 = 𝑐 3 =92 (time units)
Total travel time: 552 (time units)

32. Compare Explain: Mathematically: Total travel time 552>498
Travel time by each traveller in network: 92>83
Explain:
Rooted in the property of the UE
Individual choice of routes effects on other networks users.
Mathematically:
UE objective function decrease: 399→386

33. THANK YOU