1. Price of Stability for Network Design with Fair Cost Allocation
[Anshellevich, Dasgupta, Kleinberg, Tardos, Wexler, Roughgarden : FOCS ’04]
Presented by
Mangesh Gupte

2. Network Design Games Graph G = (V,E) , cost function c : E ̶> R
Users U : user i wants to connect terminals (si , ti) ( buy a si-ti path )
Cost for common edges is shared equally
ci(e) = c(e) / xe
c(e) = cost of edge e
xe = number of users using edge e
We want to find the Price of Stability (Cost of Best Nash / Optimal Cost) of this Game

3. Examples Game with high Price of Anarchy
1+ε
s1, s2,.., sk
t1, t2, .., tk k
Game with high Price of Stability
s1, s2,.., sk 1
1/n
1/2
1+ε t1 t2 tn

4. Price of Stability of the fair connection game is at most H(k)
Construct a potential function
Φ : (S1, S2, …, Sk) ̶> R
Φ(S) – Φ(S’) = ui(S’) – ui(S) : S, S’ differ in strategy for player i
Define : Φ(S) = Σe in E c(e) Hxe
c(S) ≤ Φ(S) ≤ Hk c(S)
SOPT : optimal solution. Start from SOPT and apply best response to get SNASH
c(SNASH) ≤ Φ(SNASH) ≤ Φ(SNASH) ≤ Hk c(SOPT)

5. If the edge cost function is concave, non-decreasing then the PoS is Hk
Define Φ(S) = Σe in E Σx=1..xe ce(x) / x
Again Φ(S) - Φ(S’) = ui(S’) – ui(S) : S, S’ differ in strategy for player i
Σx=1..xe ce(x) / x ≤ Hxe ce(xe)
ce(k) / k ≤ ce(j) / j for j ≤ k
Hence , c(S) = Σe in E ce(x) / xe ≤ Σe in E Σx=1..xe ce(x) / x = Φ(S)
Using same arguments c(S) ≤ Φ(S) ≤ Hxe c(S)

6. More General cost function
Cost of an edge : ce is arbitrary non-decreasing, satisfies conditions :
Φ(S) = Σe in E Σx=1..xe ce(x) / x
c(S) ≤ A Φ(S)
Φ(S) ≤ B c(S)
Then, the Price of Stability is AB
These are called Potential Games.
Example : fe(x) = ce(x) / x + de(x) , where d is a polynomial of degree l.
Then de(1) + de(2) + … + de(t) = 1l + 2l tl = (l+1) tl+1 = (l+1) de(t+1)

7. It is NP-hard to know if there is a Nash Equilibrium of cost at most C
3D Matching : Given sets X, Y, Z and subsets of the form (xi, yj, zk) , can we cover all elements exactly.
Given an instance of 3D matching we create an instance of the connection game
Set C = |X|+|Y|+|Z| and create C users with a distinct source node for each one
Connect as given by diagram, to the unique sink node t

8. t 3 3
Subsets X Y Z

9. It is easy to see that
Every 3D matching is a Nash Equilibrium of the fair connection game
If there is a 3D matching, then a Nash Equilibrium will be a matching
Hence, finding Nash Equilibrium of cost at most C is equivalent to telling if there exists a 3D matching, which is NP-Hard

10. Thank You