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

2. Metric Facility Location2
facility 3
client 2
F : set of facilities.
D : set of clients.
Facility i has facility cost fi.
cij: distance between i and j in V.
Client j wants to be connected to rj distinct facilities.
SODA Talk, 01/2003

3. 1) Pick a set S of facilities to open.
We want to:
open
facility
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 3
facility 2
client
SODA Talk, 01/2003

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

9. 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.
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.
Sj vj = Sj (cij + wij)
= Sj cij + fi + zi
So, summing over all i,
Sj njvj = (cost) + Si zi.
i : yi = 1
cij
wij j vj
j: xij > 0
SODA Talk, 01/2003

10. 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 smallest vj.
Form cluster M Í Fj with SiÎM yi = r’j. 2 1 5
active client
Cluster M
facility in some Fj j 2
SODA Talk, 01/2003

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

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

13. Theorem: Total cost ≤ 4·OPT. Proof: Cost of phase 1 = Sj njvj – Si zi
Phase 2 facility 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 a set of facilities S, wt(S) = SiÎS yi
Open facilities with yi = 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’j clusters.
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. A Sample Run Theorem: Total cost ≤ (2+2/e)·OPT. open facility facility
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 phases 2 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 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 approximation.
Fault-tolerant k median with rj = r: primal-dual algorithm gives a approx. using Lagrangean relaxation.
A detailed version will be available soon at
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