Mechanism Design without Money Lecture 4 1

Price of Anarchy simplest example G is given Route 1 unit from A to B, through AB,AXB OPT– route ½ on AB and ½ on AXB (check!) NASH – route 1 on AXB Ratio is 4/3 2 BA X 0 1 x

Price of Anarchy: general functions G is given Route 1 unit from A to B, through AB,AXB OPT – choose y to minimize yC (y) + (1 – y)C (1) Nash – route 1 on AXB Ratio is Also called the Pigou Bound 3 BA X 0 c(1) c(x)

Can things be worse? No! For a set of cpu functions C, define In our examples before r was 1 Theorem: For any routing game G, if the cost per unit functions come from C, the Price of Anarchy is at most a(C) 4

Another property of Nash flows Theorem: Let f be a Nash flow. For any other flow f* which routes the same amount, we have Note: The cost per unit of every edge is constant, and we just want to route the flow. 5

Proof Define Note that Therefore, we need to prove that H(f*,f) ≥ H(f,f) This follows by using that if a path has any flow in it in a Nash flow, its cost is minimal 6

Proof that a(C) is the bound on PoA Let C be the set of functions on the edges. Let f* be the optimal solution, and f be the Nash. And in particular setting r=f e and y=f* e 7

How big can a(C) be? Theorem: If C is a set of affine functions, a(C) is at most 4/3 Proof: Do this at home. Hint: compute the derivative. You should get x = r/2 8

Fighting selfishness Braess paradox shows that selfish agents can improve their situation if an edge is removed from the graph Given a graph, which edges should be removed? We are looking at 9

Fighting selfishness is (computationally) hard Problem: Which edges should I cut to improve the worst Nash? It is trivial to get a 4/3 approximation – just cut nothing. The worst Nash for a subgraph is always worse than OPT for the original graph, and the PoA with linear cost functions is 4/3 Thm: It is NP hard to approximate better than 4/3 10

Proof Reduction from 2DPP: Given a graph G, two sources s 1,s 2 and two targets t 1,t 2 are there two vertex disjoint paths s 1  t 1 and s 2  t 2 If there are no two disjoint paths you will always have a path s 2  t 1 11 s t2t2 s1s1 s2s2 t1t1 t G x x 1 1

What happens for non linear cost functions? We said price of anarchy can grow, but what about fighting selfishness? Thm: There exists a graph with n vertices and non linear cost functions, such that removing edges improves the worst Nash by a factor of n/2 12

The Graph 13

Bad Nash flow 14

After edge removal 15

Atomic flows Multiple equilibria (remember the examples) Sometimes there is no pure equilibrium Weaker bounds, different techniques 16

No pure Nash P 1 routes 1 unit from s to t P 2 routes 2 units from s to t 17

Price of anarchy example Not all paths in the equilibrium have the same cost In the example: PoA of 5/2 This is the worst case for affine functions if all players have the same amount of flow We will prove a weaker bound when players control different amounts of flow 18 U V W 0 0 x x x x s 1,s 2 t 2,t 3 s 4 t 1,s 3,t 4

Atomic flow for affine functions 19

Summing over the players: 20

Manipulations You get Solving x 2 -3x+1  0 gives ( /2 )/2 21

Questions? 22

Extra Slides 23

Chicken 24

Road example AB 1 hour N minutes 50 people want to get from A to B There are two roads, each one has two segments. One takes an hour, and the other one takes the number of people on it 25

Nash in road example In the Nash equilibrium, 25 people would take each route, for a travel time of 85 minutes AB 1 hour N minutes 26

Braess’ paradox Now suppose someone adds an extra road which takes no time at all. Travel time goes to 100 minutes AB 1 hour N minutes Free 27