Fault Tolerant Facility Location Chaitanya Swamy David Shmoys Cornell University

Metric Facility Location 2 facility 3 client 2 F : set of facilities. D : set of clients. Facility i has facility cost fi. cij : distance between any i and j in D  F. Client j wants to be connected to rj distinct facilities. Theory Seminar, 04/2002

1) Pick a set S of facilities to open. We want to : 1) Pick a set S of facilities to open. 2) Assign each client j to rj open facilities. Goal: Minimize total facility cost of S + sum of distances(service cost). 2 open facility 3 facility 2 client Theory Seminar, 04/2002

Previous Work rj=1 : LP rounding : Shmoys, Tardos & Aardal; Chudak & Shmoys; Sviridenko. Primal-dual algorithms : Jain & Vazirani; Markakis, Mahdian, Saberi & Vazirani(MMSV01); Jain, Mahdian & Saberi(JMS02). Local search : Koropulu, Plaxton & Rajaraman; Guha & Khuller; Charikar & Guha. Best approx. - Mahdian, Ye & Zhang (MYZ02) : 1.52. Theory Seminar, 04/2002

Previous Work (contd.) Non-uniform requirements, rj : Our Results: Uniform requirements rj=r : Markakis et al. (MMSV01) : 1.861. Non-uniform requirements, rj : Jain & Vazirani : O(log rmax). Guha, Meyerson & Munagala : 2.47. Our Results: Non-uniform rj : get a 2.076-approx. rj=r : can extend JMS02, MYZ02 to get a 1.52-approx. Theory Seminar, 04/2002

LP Formulation Primal : Min. i fiyi + j,i cijxij s.t. i xij ≥ rj j xij ≤ yi i, j yi ≤ 1 i xij ≥ 0, yi ≥ 0 i, j Theory Seminar, 04/2002

LP Formulation (contd.) Dual : Max. j rjvj - i zi s.t. vj ≤ wij + cij i, j j wij ≤ fi + zi i vj ≥ 0, wij ≥ 0, zi ≥ 0 i, j Theory Seminar, 04/2002

Complementary Slackness Primal Slackness Conditions : xij > 0  vj = wij + cij yi > 0  j wij = fi + zi Dual Slackness Conditions : vj > 0  j xij = rj wij > 0  xij = yi zi > 0  yi = 1 Theory Seminar, 04/2002

4-approximation : outline ≤ vj ≤ vj 2 view as rj copies j(c) : cth copy j(1), j(2) Basic Idea : vj ‘pays’ for each cij s.t. xij > 0. Bound service cost for each copy of j by ρ·vj Þ total service cost ≤ ρ·Sj rjvj. Problem : Have –zis in the dual. But zi > 0 Þ yi = 1! So can open these facilities and charge all of this cost to the LP Theory Seminar, 04/2002

The Algorithm Phase 1 Clustering : Ensures that each copy j(c) has a nearby open facility. Iterative algorithm. S = { j|rj > 0}, Fj = { i|xij > 0} in ­fi order. 2 1 5 client in S facility in some Fj 2 j Cluster M Start of iteration Pick j with smallest vj. Cluster is M Í Fj with SiÎM yi = rj. Theory Seminar, 04/2002

facility removed from Fj client in S 2 client not in S 3 X facility in some Fj X X facility opened from M j Cluster M facility removed from Fj X Open rj cheapest facilities in M. For k s.t. Fk Ç M ¹ f, connect rj copies to opened facilities. Decrease rk, set Fk=Fk-M. End of iteration Theory Seminar, 04/2002

Analysis : Phase 1 Solution is feasible : each j is connected to rj distinct facilities. Lemma : Facility cost ≤ Si fiyi. Proof : Cost of rj cheapest facilities in M ≤ rj· (avg. cost) = SiÎM fiyi. These facilities don’t get used again. Theory Seminar, 04/2002

Analysis (contd.) Lemma : For any j and c, service cost of copy j(c) ≤ 3vj. Proof : ≤ vj j(c) ≤ vk ≤ vk k Cluster M vk ≤ vj since k was chosen as cluster center. Service cost ≤ vj + 2vk ≤ 3vj. Theory Seminar, 04/2002

The Algorithm (contd.) Phase 2 : Taking care of –zis. X Open all (unopened) i s.t. yi = 1. i with yi = 1 X i with yi < 1 and open j rj = 3 For any j, if xij = yi = 1, disconnect a copy of j and connect it to i. Theory Seminar, 04/2002

Analysis : Phase 2 Let L1 = { i | yi = 1 }, Lemma : Cost of phase 2 = Lj = { i | xij = 1 } Í L1 and lj = |Lj|. Lemma : Cost of phase 2 = S fi + S cij = Sj ljvj – Si zi. Proof : Each i with zi > 0 is opened. For iÎL1, all j s.t. wij > 0 are connected to it. So, S vj = (service cost) + (fi + zi) Þ Sj ljvj = S fi + S cij + Si zi. iÎL1 j,iÎLj j|iÎLj iÎL1 j,iÎLj Theory Seminar, 04/2002

cost for copies connected by phase 1 Finally … Theorem : Total cost ≤ 4 times the optimal cost. Proof : Total cost ≤ Si fiyi + 3·Sj (rj – lj)vj + S fi + S cij iÎL1 j,iÎLj facility cost of phase 1 cost of phase 2 cost for copies connected by phase 1 ≤ Si fiyi + 3·Sj (rj – lj)vj + (Sj ljvj – Si zi ) ≤ Si fiyi + 3·(Sj rjvj – Si zi ) ≤ 4·OPT. Theory Seminar, 04/2002

A Randomized Algorithm Idea : Open i with probability ρ·yi Expected facility cost ≤ ρ·Si fiyi Hope that each copy j(c) has a nearby facility open and service cost decreases. Not quite… : no facility may be open! Cluster facilities, open ≥ 1 facility in each cluster. Theory Seminar, 04/2002

facilities opened in phases 1, 2 Phase 1 : Pruning out –zis. Open all i s.t. yi = 1. For each j, if xij = yi = 1, connect j to i. Let Lj = { i | xij = 1 } and lj = |Lj|. Cost = Sj ljvj – Si zi. facilities opened in phases 1, 2 L’j yi = 1 Lj ½ ≤ yi < 1 yi < ½ 10 Fj Set L1 r’j = residual reqmt. = 6 Phase 2 Open all i s.t. ½ ≤ yi < 1. For each j, let L’j = { i | ½ ≤ xij < 1 }. Connect copies of j to i Î L’j. Lose a factor of 2. Set L2 Theory Seminar, 04/2002

Notation : facwt(S, j) = SiÎS xij Phase 3 Notation : facwt(S, j) = SiÎS xij Form clusters : Each cluster has facwt. ≥ ½. Open facilities : Open ≥ 1 facility in a cluster – used as a backup facility. Open facility i with prob. 2yi. Assign facilities to copies : Each copy j(c) gets a preferred set of facilities – P(j(c)) with facwt. ≥ ½. For c ¹ d, P(j(c)) Ç P(j(d)) = f. Connect clients : Connect j(c) to the nearest i open in P(j(c)), or to a backup facility. Theory Seminar, 04/2002

Initial Fj before any iterations Clustering After phases 1 and 2, Fj = { i | xij < ½ }. Sort these by ­cij and distribute among the r’j copies. Cj(c) = avg. service cost of the cth copy, denote Sc Cj(c) = S cijxij by Cj. iÎFj client j i Î Fj Cj(2) 3 Cj(1) Cj(3) Initial Fj before any iterations Want the following properties : Clusters to be disjoint. Each cluster have facwt. ≥ ½. Each j be connected to r’j clusters. Theory Seminar, 04/2002

Fj after some iterations Iterative algorithm S = { j | r’j > 0 }. aj = ‘active’ copy of j, initially = 1. i removed from Fj 4 X X X i Î Fj, serving j Ĉj(3) (aj) X X 1 aj = 3 facilities serving j Fj after some iterations (aj) Ĉj(aj) = avg. distance to the first k facilities in Fj gathering facwt. ≥ ½; say these facilities ‘serve’ j. Will maintain Ĉj(aj) ≤ Cj(aj) (aj) Theory Seminar, 04/2002

X X X X Start of iteration Choose j in S with minimum vj + Ĉj(aj). Form cluster M = facilities serving j . Note : facilities are not split. (aj) aj = 1 4 client in S facility in some Fj 2 j(3) Cluster M For k s.t. Fk Ç M ¹ f, decrease r’k, advance ak, set Fk = Fk – M. aj = 2 3 facility removed from Fj X X X X 1 aj = 4 Cluster M Theory Seminar, 04/2002

Opening Facilities Non-central facilities k Non-central facilities j open with prob. 2yi independent of other choices Cluster M Central facilities opened in 2 steps : Open exactly 1 facility in M : i opened with prob. qyi. Acts as backup, denoted b(k ), for each k s.t. Fk Ç M ¹ f. Open each i in M indep. with prob. (2-q)yi and independent of step 1. (ak) Theory Seminar, 04/2002

Distributing Facilities Copy c gets a preferred set P(j(c)). Preferred sets are disjoint. Ensure facwt(P(j(c)), j) ≥ ½ for all c. Possible to do so since each xij < ½. P(j(2)) P(j(3)) P(j(1)) facility in Fj j r’j = 3 Let Sj(c) = avg. dist. from j to P(j(c)) = (S cijxij)/facwt(P(j(c)), j) Then Sc Sj(c) ≤ 2×Cj. iÎP(j(c)) Theory Seminar, 04/2002

Analysis Feasibility follows from : Facilities in phases 1, 2 not reused. After clustering j is connected to r’j disjoint clusters Þ backups are distinct. Preferred sets are disjoint. So j connected to rj distinct facilities. Theory Seminar, 04/2002

Facility cost Recall L1 = { i | yi = 1 }. Phase 2 : incur a factor of 2. Phase 3 : each i is opened with probability 2yi. Expected facility cost ≤ 2×S fiyi. for phases 2, 3 iÏL1 Theory Seminar, 04/2002

Service cost I Bounding backup cost, denoted by B : r.v. D : event that no i in P(j(c)) is open. Lemma : E[B|D] ≤ 2vj + Cj(c). Proof : 2 cases : backup = b(j(c)) B 1) j(c) ≤ vj ≤ vk k(d) ≤ Ĉj(d) $ iÎM Ç Fj s.t. cik ≤ Ĉj(d) j(c) 2) ≤ Ĉj(d) in expectation ≤ vj k(d) ≤ vk "iÎM Ç Fj, cik > Ĉj(d) Also vk + Ĉj(d) ≤ vj + Ĉj(c) ≤ vj + Cj(c) Theory Seminar, 04/2002

Service Cost II . Fix j, c. Let X(c) = service cost of j(c). d1 d2 dm P(j(c)) sorted by increasing cij . j(c) i Î P(j(c)) Let di = cij, pi = prob. i is opened = 2yi. B(c) = backup cost. D(c) = event that no iÎP(j(c)) is open. p = Pr[D(c)] = (1-p1)…(1-pm) ≤ e-1 davg = weighted avg. of the dis = (Si pidi)/(Si pi) = Sj(c) Theory Seminar, 04/2002

E[X(c)] = [p1d1 + (1-p1)p2d2 + … + (1-p1)…(1-pm-1)pmdm] Then, E[X(c)] = [p1d1 + (1-p1)p2d2 + … + (1-p1)…(1-pm-1)pmdm] + p×E[B(c)|D(c)] ≤ (1-p)×davg + p×[2vj + Cj(c)] ≤ (1-e-1)×Sj(c) + e-1×[2vj + Cj(c)]. Let X = Sc X(c) = service cost of j. Sc Sj(c) ≤ 2×Cj and Sc Cj(c) ≤ 2×Cj. Summing over all c = 1…r’j, E[X] ≤ (1-e-1)×2Cj + e-1×(2r’jvj + Cj) ≤ 2×Cj + 2e-1×r’jvj. Theory Seminar, 04/2002

Putting it all together Phase 1 : pay the optimal LP cost Phases 2, 3 Facility cost : twice LP facility cost Service cost Lose a factor of 2 for phase 2 Phase 3 cost is 2×(LP service cost)+2e-1×(dual value) Overall cost for ≤ (2+2/e)×(LP cost) phases 2, 3 Total cost ≤ (2+2/e)×OPT. Theory Seminar, 04/2002

How to improve this? Distribute facilities more equitably (in an expected sense) among copies - decreases prob. of ‘bad’ event. Better analysis – maximum distance within a cluster can be bounded by 2Cj(c). Balance phases 2 and 3. Theory Seminar, 04/2002

Summary of Results Give a 2.076-approx. algorithm for non-uniform rjs. Based on LP rounding using complem. slackness. For rj = r, extend the primal-dual algorithm of (JMS02) to get a 1.52-approximation. Fault tolerant k medians with rj = r Primal-dual algorithm (JMS02) gives a 4-approx. using Lagrangean relaxation. LP rounding gives a factor of 8. Theory Seminar, 04/2002

Open Questions Reduce gap between rj = r, non- uniform rj. Combinatorial algorithms for non-uniform rj : primal-dual, local-search. Constant-factor approx. for fault tolerant k medians with non-uniform rjs. Theory Seminar, 04/2002