Fault-Tolerant Facility Location

1. Fault-Tolerant Facility Location Chaitanya Swamy David Shmoys Cornell University

2. F : set of facilities. D : set of clients. Facility i has facility costfi. cij: distance between i and j in V. Client j wants to be connected to rjdistinct facilities. Metric Facility Location 2 facility 3 client 2 SODA Talk, 01/2003

3. 1) Pick a set S of facilities to open. 2) Assign each client j to rj open facilities. Goal: Minimizetotalfacility costofS + sum of distances(service cost). open facility facility We want to: 2 3 2 client SODA Talk, 01/2003

4. Previous Work • rj=1: • LP rounding: Lin & Vitter; Shmoys, Tardos & Aardal; Chudak & Shmoys; Sviridenko (Sv). • Primal-dual algorithms: Jain & Vazirani; Markakis, Mahdian, Saberi & Vazirani (MMSV01); Jain, Mahdian & Saberi (JMS02). Best approx.-Mahdian, Ye & Zhang: 1.52. • Uniform requirements, rj=r: • Markakis et al. (MMSV01) : 1.861. • Non-uniform requirements, rj: • Jain & Vazirani :O(log rmax). • Guha, Meyerson & Munagala :2.47. SODA Talk, 01/2003

5. Our Results • Non-uniform rj: get a 2.076-approx. – LP rounding. • rj=r: can extend JMS02, Mahdian et al. to get a 1.52-approx. – primal-dual + greedy improvement. • Fault tolerant k-median with rj=r: get a 4-approx. – above primal-dual + Lagrangean relaxation. SODA Talk, 01/2003

6. vj wij j cij i Max. j rjvj - i zi(Dual) s.t. vj ≤ wij + cij i, j j wij ≤ fi + zii vj, wij, zi ≥ 0 i, j LP Formulation Min.i fiyi + j,i cijxij (Primal) s.t. i xij≥ rjj 0 ≤ xij ≤ yi ≤ 1i, j yi : indicates if facility i is open. xij: indicates if client j is connected to facility i. SODA Talk, 01/2003

7. 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 Strong Duality: Primal optimum = Dual optimum. We bound the cost using both primal and dual optimum solutions. SODA Talk, 01/2003

8. 4-approximation : outline ≤ vj ≤ vj 2 view as rjcopies 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·vjÞ total service cost ≤ρ·Sj rjvj. Problem:Have –zisin thedual. But zi > 0 Þyi = 1! So can open these facilities and charge all of this cost to the LP SODA Talk, 01/2003

9. Cost = Si:yi=1 (fi + Sj:xij>0 cij) = Sj njvj – Si zi. Proof: Each i with zi > 0 is opened, all j s.t. wij > 0 are connected to it. j: xij > 0 Sj vj =Sj (cij+wij) =Sj cij+fi + zi So, summing over all i, Sj njvj = (cost) + Si zi. wij vj j cij i : yi = 1 The Algorithm 1. Taking care of –zis. Open all i s.t. yi = 1. For any j, if xij > 0 and yi = 1, connect a copy of j to i. Remove all these i. nj = no. of copies of j connected. SODA Talk, 01/2003

10. j Cluster M The Algorithm (contd.) 2. Clustering: Ensure that each copy j(c) has a open facility nearby. r’j= residual reqmt. of j = rj – nj. j is active if r’j > 0 Fj = { i : xij > 0 and yi < 1} in ­fi order. • Pick j with smallestvj. • Form cluster M ÍFj with SiÎM yi = r’j. 2 1 5 active client facility in some Fj 2 SODA Talk, 01/2003

11. 0 inactive client 3 X X X 0 facility opened from M facility removed from Fj X • Open r’jcheapest facilities in M. • For k s.t. FkÇ M ¹ f, connect r’j copies to opened facilities. Decrease r’k, set Fk=Fk-M. active client 2 1 5 facility in some Fj 2 j Cluster M SODA Talk, 01/2003

12. Clustering Phase • Lemma: Facility cost ≤ Si fiyi. • Proof: We create cluster M ÍFj with SiÎM yi = r’j. Cost of r’j cheapest facilities in M ≤r’j· (avg. cost) = SiÎM fiyi. Lemma: Service cost of copy k(c) ≤3vk. Proof: Cost ≤ vk + 2vj vj ≤ vk since j was chosen as cluster center. k(c) ≤ vk ≤ vj ≤ vj j Cluster M Analysis Solution is feasible : each j is connected to rj distinct facilities. SODA Talk, 01/2003

13. Theorem: Total cost ≤ 4·OPT. Proof: Costof phase 1 = Sj njvj – Si zi Phase 2facility cost≤ Si fiyi service cost≤3·Sj (rj – nj)vj Total cost≤ Si fiyi + 3·(Sj rjvj – Si zi ) ≤4·OPT. SODA Talk, 01/2003

14. Randomized Rounding Idea : Open facility i with probability ρ·yi . First cluster facilities and open ≥ 1 facility in each cluster – will serve as a backup facility. Also open non-cluster facility i with probability ρ·yi . High probability that some facility is open; if none are open use a backup facility. Expected service cost decreases. j(c) Expected facility cost≤ρ·Si fiyi SODA Talk, 01/2003

15. Algorithm Outline Notation: For aset of facilities S, wt(S) = SiÎS yi • Open facilities withyi = 1. • Also open all i s.t. yi ≥ ½. For any j, if xij ≥ ½, connect a copy of j to i. • a) Form clusters: Ensure that • · Clusters are disjoint, • ·½ ≤ wt(cluster)≤ 1, and • · Each j is connected to r’jclusters. • b) Open facilities: First open exactly 1 facility in each cluster M. This is used as a backup facility. Now open every facility i with prob. proportional to yi, so that total Pr[i is opened] =2yi. SODA Talk, 01/2003

16. open facility A Sample Run facility Theorem: Total cost ≤ (2+2/e)·OPT. client SODA Talk, 01/2003

17. How to improve this? • Better analysis: better bound on the max. distance in a cluster – decreases the ‘backup cost’. • Balance costs of phases2 and 3. • Use pipage rounding (Sv) to decrease the prob. that no facility serving a copy is open. SODA Talk, 01/2003

18. Summary of Results • Give a 2.076-approx. algorithm for non-uniform rj. 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 median with rj = r: primal-dual algorithm gives a 4-approx. using Lagrangean relaxation. A detailed version will be available soon at www.orie.cornell.edu/~shmoys. SODA Talk, 01/2003

19. Open Questions • Is the fault-tolerant case harder than rj=1 case? Better hardness results? Approximation-preserving reduction? • 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 median with non-uniform rj. SODA Talk, 01/2003