1 / 27

Mechanism Design without Money

Mechanism Design without Money. Lecture 4. Price of Anarchy simplest example. X. G is given R oute 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. 0. x. A. B. 1. Price of Anarchy: general functions. X. c(x).

violet
Download Presentation

Mechanism Design without Money

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Mechanism Design without Money Lecture 4

  2. Price of Anarchy simplest example X • 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 0 x A B 1

  3. Price of Anarchy: general functions X c(x) • 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 0 A B c(1)

  4. 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)

  5. 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.

  6. 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 =

  7. 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=feand y=f*e ) - =

  8. 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

  9. 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

  10. 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

  11. Proof • Reduction from 2DPP: Given a graph G, two sources s1,s2 and two targets t1,t2 are there two vertex disjoint paths s1t1 and s2t2 • If there are no two disjoint paths you will always have a path s2t1 G s1 t1 1 x s t 1 s2 t2 x

  12. 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

  13. The Graph

  14. Bad Nash flow

  15. After edge removal

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

  17. No pure Nash • P1 routes 1 unit from s to t • P2routes 2 units from s to t

  18. 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 allplayers have the sameamount of flow • We will prove a weaker boundwhen players control different amounts of flow s1,s2 t1,s3 ,t4 0 V U x 0 x x x W t2,t3s4

  19. Atomic flow for affine functions • Edge e has CPU aex + be • Player i sends ri • Let f be Nash, f* be OPT. We have for every player i:

  20. Summing over the players:

  21. Manipulations • You get • Solving x2-3x+10 gives (3 + 51/2)/2

  22. Questions?

  23. Extra Slides

  24. Chicken

  25. Road example • 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 1 hour N minutes A B 1 hour N minutes

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

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

More Related