Lecture 5 - Traffic Assignment
6 March 2023
Mozhgan Pourmoradnasseri, Ph.D.
References:
OD-Matrix (set of trips)
Set of routes
Travelers choose the available route with the least travel time between their origin and destination, reflecting the idea that travel is rarely a goal in and of itself but instead involves some time, cost, or disutility that travelers would prefer to avoid.
Transportation systems involve interactions among multiple agents. The basic facts are:
“It’s the evening peak period, so the freeway will probably be congested. I’ll take another route.”
Even with a relatively simple model of behavior (choosing the fastest route), we end up with mutual dependencies and circular relationships.
Route Travel Times
Route Choices
In game theory, the Nash equilibrium, named after the mathematician John Nash, is the most common way to define the solution of a non-cooperative game involving two or more players. In a Nash equilibrium, each player is assumed to know the equilibrium strategies of the other players, and no one has anything to gain by changing only one's own strategy (Wikipedia).
At the equilibrium solution, no traveler can reduce their travel time by switching to another route.
For each origin and destination, all used routes between those nodes have equal and minimal travel time.
If 7000 vehicles are choosing these routes, how many choose the top route, and how many choose the bottom?
This method can be generalized in any network with a single OD pair \((r,s)\):
\(d^{rs}\)
\(d^{rs}\)
\(r\)
\(s\)
100 cars travel from A to D. Equilibrium will occur when 50 drivers travel via ABD and 50 via ACD. Every driver has a total travel time of 3.5.
Equilibrium will occur when 25 drivers travel via ABD, 50 via ABCD, and 25 via ACD. Every driver now has a total travel time of 3.75.
At equilibrium, 25 vehicles choose the top route, 5 chose the bottom, and all travel times are 50 minutes.
10 vehicles choose the top route, 20 choose the bottom route, and everybody has a travel time of 50 minutes.
Now, we improve the bottom link so that its cost function is 40 + x/2.
What happens to route choices now?
Nobody has saved any time at all! What happened?
What is improved?
This suggests two possible traffic assignment rules:
minimizes the total system travel time
minimizes travel time of used routes
The user equilibrium solution is unique in link flows.
Equilibrium = Fixed point
Find a value that is unchanged when you go around the loop.
x = 4
10 → 7 → 5.5 → 4.75 → 4.375 → . . . which converges to the correct answer 4.
Picking a different starting value: 1 → 2.5 → 3.25 → 3.625 → . . . also converges to the same answer.
Find a value that is unchanged when you go around the loop.
x = 8
Choosing the same starting value, we have 10 → 12 → 16 → 24 → 40 → . . . which diverges to +∞.
Consider some set X and a function f whose domain is X and whose range
is contained in X. A fixed point of f is a value x ∈ X such that x = f (x).
Equilibrium = Fixed point
\(X\) ← λ\(X^*\) + (1 − λ)\(X\).
6. If “close enough to equilibrium,” stop, otherwise return to step 2.
Simulation can be seen as a sampling experiment on a dynamic real system through a computer model formally representing it.
remember: all models are wrong ...
Validation is an iterative process consisting of:
Traffic data can be classified into two categories:
Source: Pourmoradnasseri, M., Khoshkhah, K. and Hadachi, A. "Real-Time Calibration of Disaggregated Traffic Demand." arXiv preprint arXiv:2210.17315 (2022).
Supilinn, Tartu
Manual counting is sometimes unavoidable :)
The traffic heat map of simulated
trips from 16:00 to 17:00.