URBAN TRANSPORTATION NERWORKS FORMULATING THE ASSIGNMENT PROBLEM

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

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

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 , …)

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 ,…)

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)

In the matrix form: 𝒄=𝒕∆ and 𝒙=𝒇 ∆ 𝑻

Example

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 1,1 14 + 𝑡 2 𝛿 2,1 14 + 𝑡 3 𝛿 3,1 14 + 𝑡 4 𝛿 4,1 14 = 𝑡 1 + 𝑡 2 𝑥 3 = 𝑓 1 14 𝛿 3,1 14 + 𝑓 2 14 𝛿 3,2 14 + 𝑓 1 24 𝛿 3,1 24 + 𝑓 2 24 𝛿 3,2 24 = 𝑓 1 14 + 𝑓 1 24

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 ≥0 ∀𝑘,𝑟,𝑠 (2.1𝑐) 𝑥 𝑎 = 𝑟,𝑠,𝑘 𝑓 𝑘 𝑟𝑠 𝛿 𝑎,𝑘 𝑟𝑠 ∀𝑎 (2.1c)

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

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)

Use program (2.1): min 𝑧 𝒙 = 0 𝑥 1 2+𝜔 𝑑𝜔+ 0 𝑥 1 1+2𝜔 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

3.EQUIVALENCE CONDITIONS

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)

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)

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

4. UNIQUENESS CONDITIONS Prove that: problem 2.1 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 )

EXAMPLE

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

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

5. THE SYSTEM- OPTIMIZATION FORMULATION Program: min 𝑧 (𝒙) = 𝑎 𝑡 𝑎 𝑥 𝑎 ( 𝑡 𝑎 ) 5.1𝑎 (The meaning of objective function?) Subject to 𝑎 𝑓 𝑘 𝑟𝑠 =𝑞 𝑟𝑠 ∀𝑟,𝑠 (5.1b) 𝑓 𝑘 𝑟𝑠 ≥0 ∀𝑘,𝑟,𝑠 (5.1𝑐) 𝑥 𝑎 = 𝑟,𝑠,𝑘 𝑓 𝑘 𝑟𝑠 𝛿 𝑎,𝑘 𝑟𝑠 ∀𝑎 (5.1c)

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

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

6. USER EQUILIBRIUM & SYSTEM OPTIMUM

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)

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.

BRAESS’S PARADOX

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)

Adding a new path to improve flow

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)

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

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