Topics
- Review
- Congestion Games
- Potential Games
Recap from Last Week
A game consists of 3 elements:
- Set of players N
- Sets of strategies {Si}i∈N
- Sets of payoffs {ui}i∈N
Given a game, what do we do?
- Maximize the sum of everyone’s payoff.
- Socially optimal outcome (what we want to happen).
- Maximize individual’s payoff conditional on everyone else’s strategy and find a fixed point.
- Nash equilibrium (what we think will happen).
1 & 2 are often different because there are strategic interactions and individual incentives are unaligned.
Some games have useful structures, which impose useful restrictions on {Si} and {ui}.
- Dynamic games
- Games on networks
Congestion Games
Paths are labeled (edge, traffic, cost/duration):
![](/courses/14-15j-networks-spring-2018/8b8ce3581ee51d9fa952ff7b9d51a045_MIT14_15JS18_rec8d.jpg)
- Directed network (J, E) = ({A, B, C}, {1, 2, 3, 4}).
- Set of paths P = {p1, p2, p3} = {(1), (2,3), (2,4)}.
- Traffic on paths {xp1, xp2, xp3} = {x1, x3, x4}.
- Total (social) cost: x1*L1(x1) + x2*L2(x2) + x3*L3(x3) + x4*L4(x4).
- Minimizing this with respect to x1, … x4 gives socially optimal traffic.
As a game, this can be written:
- N = [0, 1]
- Si = {p1, p2, p3}
- ui(Pi, everyone else’s strategy) = ui(Pi, x1…x4)
However, each individual driver incurs:
- L1(x1) if he takes p1.
- L2(x2) + L3(x3) if he takes p
- L2(x2) + L4(x4) if he takes p3.
So, his best response correspondence is to choose a path that gives minimum individual cost.
- In equilibrium, we must have L1(x1) = L2(x2) + L3(x3) = L2(x2) + L4(x4).
Example:
Paths are labeled (edge/path, traffic, individual cost):
![](/courses/14-15j-networks-spring-2018/9a415541da92a0359c9fe3fb90f215fb_MIT14_15JS18_rec8c.jpg)
- Model constraint: x1 + x2 = 1.
- Total cost: x1*L1(x1) + x2*L2(x2) = x12 + x2 = x12 + (1 — x1) = (x1 — ½)2 + ¾.
- Socially optimal traffic: (x1S, x2S) = (½ , ½).
Each individual choose path with lower cost, so in equilibrium:
- L1(x1E) = L2(x2E), so (x1E, x2E) = (1, 0).
- Equilibrium total cost is x1E*L1(x1E) + x2E*L2(x2E) = 1 ≥ ¾.
How to solve this?
- Impose toll C on p1.
- Then, ui(p1, x1, x2) = x1 + C.
- We want L1(x1S) + C = L2(x2S).
- C = ½.
Potential Games
In physics, particles move along the unique potential field.
![](/courses/14-15j-networks-spring-2018/770315e0e12be93ddf1324991317bfce_MIT14_15JS18_rec8a.jpg)
In society, people move along their own incentives.
![](/courses/14-15j-networks-spring-2018/c4284b0043eabefd7d88fd6dc2a0704f_MIT14_15JS18_rec8b.jpg)
However, there are cases where people’s incentives are so similar that they move as if there is a unique potential field.
Example: Traffic Congestion
Definition: A game is an exact potential game if there is exists Φ (unique potential field) such that:
- ui(Si, S-i) — ui(Si’, S-i) = Φ(Si, S-i) — Φ(Si’, Si).
- That is, i’s incentive matches the potential’s gradient.