1 / 14

Online Multicast with Egalitarian Cost Sharing

Online Multicast with Egalitarian Cost Sharing. Moses Charikar (Princeton) Howard Karloff (AT&T) Claire Mathieu (Brown) Seffi Naor (Technion) Mike Saks (Rutgers) Talk presented by Warren Schudy (Brown ). Introduction.

joshwa
Download Presentation

Online Multicast with Egalitarian Cost Sharing

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. Online Multicast with Egalitarian Cost Sharing Moses Charikar (Princeton) Howard Karloff (AT&T) Claire Mathieu (Brown) Seffi Naor (Technion) Mike Saks (Rutgers) Talk presented by Warren Schudy (Brown)

  2. Introduction • Multicast: n terminal vertices must connect to root vertex r via paths. Edges have costs c(e). Goal is to minimize total cost: OPT Steiner Tree. • Egalitarian cost sharing: an edge e used by the paths of n(e) terminals charges each terminal c(e)/n(e). Terminals are selfish, non-cooperative. Nash equilibrium (N.E.): no terminal wants to change its path if everything else stays the same. Question: how much more costly is the outcome of selfish choices? That is: bound (cost of N.E.)/ OPT?

  3. Impact of selfishness (cost of worst N.E.)/OPT = n [Koutsoupias Papadimitriou‘99]Price of anarchy (cost of best N.E.)/OPT = O(log n/ loglog n) [Anshelevich Dasgupta Kleinberg Tardos Wexler Roughgarden ‘04, Agarwal Charikarr ‘06] Price of stability Question: what about (cost of N.E.)/OPT for N.E. reachable by some process? Best response dynamics: when activated, a terminal always chooses its current cheapest path to root In which order do activations occur?

  4. Two phase model r r Phase 1 Phase 2 t4 t1 t4 t3 t2 t1 t3 t2 • Activation model [Chekuri Chuzhoy Lewin-Eytan Naor Orda ‘06] • Phase 1: Terminals are activated one by one • Phase 2: Re-activated terminals may change their path (arbitrary sequence of re-activations) Ω(log n/ loglog n)≤(cost of resulting N.E.)/OPT≤ O(√n log2 n)[CCLNO] re-fires

  5. New results • For two phase model: Ω(log n) ≤(cost of resulting N.E.)/OPT ≤ O(log3 n) • For General sequence of interleaved activations and re-activations, except that terminal arrivals (first activations) are in random order: (cost of resulting N.E.)/OPT = O(√n polylog(n)) Next 5 slides: proof sketch of O(log3 n) result

  6. Proving O(log3 n) • Potential function • cost ≤  ≤ H(n)*cost • Re-activations decrease  • So, cost after phase 2 ≤  after phase 2 ≤  after phase 1 ≤ O(log n)*cost after phase 1 • Must prove: (cost after phase 1)≤ O(log2 n)*OPT

  7. Analysis of phase 1 • Define “Gap revealing” linear program (cost after phase 1) ≤ Value(LP) • Relax the LP and write dual linear program Value(LP) ≤ Value(Dual) by linear programming duality • Define feasible dual solution… Value(Dual) ≤ Value(solution) • … of value O(log2 n) OPT Value(solution) = O(log2 n) OPT

  8. Gap revealing LP • s(i): cost of i’s path on arrival of ib(i): cost of new edges bought by i •  b(i) =(Cost after phase 1) =  b(i)’s s(i) = • If terminal j arrives after terminal i, then j could go to i and reuse i’s path with discount: s(j)≤ d(j,i)+s(i)-b(i)/2

  9. Relax, take dual Take a tree T over the terminals, such that children arrive after their parent. Relax the linear program by writing the constraint s(j)≤ d(j,i)+s(i)-b(i)/2 for j child of i in T only So, dual has one variable z(j) for each edge of T between j and parent(j) (C(i): children of i in T)

  10. How is T defined? Must have: children arrive after their parent Take Eulerian tour π of min spanning tree of terminals. Note: Cost(π) ≤ 2(OPT Steiner tree) Try to have: parent(j) is in the vicinity of j along π, and so:  d(j,parent(j))=O(log2 n)* Cost(π) Left subtree r Path to root t1 t3 t2 t4 Right subtree

  11. Random Arrivals Result O(√n polylog(n)) proof sketch • Arbitrary interleaving of arrivals and reactivations, but: assume order of arrivals is random • Analyze potential Φ • Reactivations decrease potential • Φ(k): potential right after kth terminal arrives; bound E[Φ(k+1) - Φ(k) given Φ(k)]

  12. Analysis: arrival of j • Path picked by j is difficult to analyze. Instead, • Consider again Eulerian tour π of min spanning tree of terminals. • Pick i randomly from previously arrived terminals in the vicinity of j along π. Connect j to i and follow i’s path.

  13. Conclusion and Open Problems • General theme: Bound cost of solutions reachable by best response dynamics • Obvious open question:analyze arbitrary mix of arrivals and reactivations • Random slow arrivals: • Arrivals in random order + arbitrary interleavings • When new terminal arrives, solution in equilibrium • Other problems ? • Multiple source-sink pairs: linear lower bound

  14. Properties of T r • Children arrive after their parents • Every node has at most 2 children, and nodes with 2 children are at levels integer* log n •  d(j,parent(j))=O(log2 n)*OPT

More Related