Editing Nash equilibrium (section)

=== Network traffic ===
{{See also|Braess's paradox}}
[[File:Nash graph equilibrium.png|thumb|250px|Sample network graph. Values on edges are the travel time experienced by a "car" traveling down that edge. <math>x</math> is the number of cars traveling via that edge.]]
An application of Nash equilibria is in determining the expected flow of traffic in a network. Consider the graph on the right. If we assume that there are <math>x</math> "cars" traveling from {{math|A}} to {{math|D}}, what is the expected distribution of traffic in the network?

This situation can be modeled as a "[[game theory|game]]", where every traveler has a choice of 3 strategies and where each strategy is a route from {{math|A}} to {{math|D}} (one of {{math|ABD}}, {{math|ABCD}}, or {{math|ACD}}). The "payoff" of each strategy is the travel time of each route. In the graph on the right, a car travelling via {{math|ABD}} experiences travel time of <math>1+\frac{x}{100}+2</math>, where <math>x</math> is the number of cars traveling on edge {{math|AB}}. Thus, payoffs for any given strategy depend on the choices of the other players, as is usual. However, the goal, in this case, is to minimize travel time, not maximize it. Equilibrium will occur when the time on all paths is exactly the same. When that happens, no single driver has any incentive to switch routes, since it can only add to their travel time. For the graph on the right, if, for example, 100 cars are travelling from {{math|A}} to {{math|D}}, then equilibrium will occur when 25 drivers travel via {{math|ABD}}, 50 via {{math|ABCD}}, and 25 via {{math|ACD}}. Every driver now has a total travel time of 3.75 (to see this, a total of 75 cars take the {{math|AB}} edge, and likewise, 75 cars take the {{math|CD}} edge).<!-- 25 drivers travel via {{math|ABD}} and 50 via {{math|ABCD}}, so 75 cars travel on edge {{math|AB}}. Similarly, 75 cars travel on edge {{math|CD}}. It takes <math>(1 + \frac{75}{100}) + 2 = 3.75</math> to travel via {{math|ABD}}. The travel time of the other routes is the same. -->

Notice that this distribution is not, actually, socially optimal. If the 100 cars agreed that 50 travel via {{math|ABD}} and the other 50 through {{math|ACD}}, then travel time for any single car would actually be 3.5, which is less than 3.75. This is also the Nash equilibrium if the path between {{math|B}} and {{math|C}} is removed, which means that adding another possible route can decrease the efficiency of the system, a phenomenon known as [[Braess's paradox]].