The Braess’ Paradox

Assume a network with a single O/D pair (1,4) . There are 2 paths available to travelers: p1=(a,c) and p2=(b,d).

For a travel demand of 6, the equilibrium path flows are xp1* = xp2* = 3 and

The equilibrium path travel cost is Cp1= Cp2= 83.

3

2

1

4

a

c

b

d

ca(fa)=10 fa cb(fb) = 10fb+50

cc(fc) = fc+50 cd(fd) = 10fd

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