1 / 8

Exercise 10.10

Exercise 10.10. Prove that the k-server WFA algorithm is k-competitive when the metric space is a real line segment. Solution: All initial points and requests are on the line segment pq. To prove k-competitive, we need to prove that Total extended cost <= (k+1)OPT + const; OPT = w f (C f ).

shauna
Download Presentation

Exercise 10.10

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. Exercise 10.10 Prove that the k-server WFA algorithm is k-competitive when the metric space is a real line segment. Solution: All initial points and requests are on the line segment pq. To prove k-competitive, we need to prove that Total extended cost <= (k+1)OPT + const; OPT = wf(Cf)

  2. How does Double Coverage (DC) handle it? If the request falls to the left of the leftmost server or to the right of the rightmost server, only that server moves since it is the closest server. If the request is in between two adjacent servers, move both these servers toward the request at equal speed until one server reaches it. We can operate DC as a lazy algorithm by essentially keeping track of “virtual” locations of servers. DC is k-competitive.

  3. How does WFA handle it? If wi(X) is the optimal cost of serving r1..ri starting at configuration A0 and ending in configuration X, then to service ri+1, it moves to Ai+1 that contains ri+1 and minimizes wi+1(Ai+1) + d(Ai,Ai+1). If r does not belong to X, then the server at r must’ve been moved to some point in c after the request was serviced. Hence, wi+1(Ai+1) = wi+1(X – c + r) + d(c,r) = wi(X-c+r) + d(c,r) So, we would like to minimize the cost of moving to the next configuration which serves the request while making it as close to the optimal configuration as possible. This is no different when the metric space is a line segment.

  4. Minimizers: Configuration X is a minimizer of r with respect to wi if it minimizes wi+1(X) – ΣxЄX d(r,x) and maximizes the extended cost: maxx{wi+1(X) – wi(X)} We first show that minimizers are extreme configurations i.e. for line segment pq, k+1 configurations which contain m copies of p and k-m copies of q; 0<=m<=k Let r Є pq, wi is any work function. Assume there is a point xЄX on pr. If we move x to p, wi(X) can increase by at most d(p,x) while the distance of x from r will increase by d(p,x). Therefore, X-x+p is also a minimizer of r. Similarly, we can move all points of X to either p or q.

  5. Potential Φ = ΣXw(X) Assume configuration A be the configuration where maximum of wf(X) is achieved. Configuration A would be the configuration where servers reside at either p or q (minimizer). When request r is serviced, ∆Φ >= w’(A) – w(A), where A maximizes the difference w’(X) – w(X). Hence,∆Φ >= EXT(r) Summing over all the request points, Φf – Φ0 >= EXT

  6. Also, Φf <= (k+1) maxXwf(X), (due to k+1 configurations) Hence, wf(A) <= wf(Cf) + D(A,Cf), where D(A,Cf) is maximum distance = d(p,q) So, Φf <= (k+1)wf(Cf) + (k+1) d(p,q), 2nd term is a constant. EXT <= Φf – Φ0 <= (k+1)wf(Cf) + constant, Φ0 is also a constant. Since wf(Cf) is the optimal cost, we have proved that the WFA algorithm is k-competitive if the metric space is a real line segment.

  7. Questions?

  8. References On-line computation & Network Algorithms, Yair Bartal On the competitive ratio of the work function algorithm for the k-server problem, Yair Bartal & Elias Koutsoupias

More Related