Price of stability
= Price of stability
For a cost-minimization <game> with positive social optimum and a nonempty set $\mathcal E$ of the specified <Nash equilibria>, the <price of stability> is
$$
\operatorname{PoS}=\frac{\min_{s\in\mathcal E}C(s)}{\min_sC(s)}.
$$
It measures the cost of requiring stability even when the best equilibrium can be selected. The equilibrium class, such as pure or mixed, and social-cost convention must be stated.
In a symmetric two-player game with actions $A,B$, let each player's cost be one at $(A,A)$, two at $(B,B)$ and three when the actions differ. Both coordinated choices are <Nash equilibria>; the best equilibrium has total cost two, matching the optimum, so the <price of stability> is one. The worst equilibrium instead costs four, giving <price of anarchy> two. The distinction matters when a designer can recommend an outcome but cannot force anyone to remain at an unstable one.<The Price of Stability for Network Design with Fair Cost Allocation by Elliot Anshelevich and others>[Introduction, best equilibrium and stability interpretation]{ref}