Braess's Paradox - How Far From Optimal Is Traffic at Equilibrium

How Far From Optimal Is Traffic At Equilibrium

At worst, traffic in equilibrium is twice as bad as socially optimal

Proof

= starting point for car j

= target for car j

Strategies for car j are possible paths from to

Each edge e has a travel function for some

Energy on edge e with x drivers:

Total time spent by all drivers on that edge:

((where there are x terms))

E(e) is less than or equal to T(e) and

\begin{align}
L_e(1) + \cdots + L_e(x) &= a_e(1+2+\cdots+x) + b_e x \\
& = a_e \tfrac{x (x+1)}{2} + b_e x \\
& = x ( a_e \tfrac{x+1}{2} + b_e ) \\
& \geq \tfrac{1}{2} x ( a_e x + b_e ) \\
& = \tfrac{1}{2} T(e) \\
\end{align}

Resulting Inequality

If Z is a traffic pattern:

If we start from a socially optimal traffic pattern Z and end in an equilibrium pattern Z':

\begin{align}
SocialCost(Z') & \leq 2 E(Z') \\
& \leq 2 E(Z) \\
& \leq 2 SocialCost(Z) \\
\end{align}

Thus we can see that worst is twice as bad as optimal.

Read more about this topic:  Braess's Paradox

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