1. Facility Location with Nonuniform Hard Capacities Tom Wexler Martin Pál Éva Tardos Cornell University ( A 9-Approximation)

2. 2 Metric Facility Location Given: Client and facility nodes. Facility cost f i for facility i. Distances c ij for all nodes. Need to: 1) Pick set S of facilities to open. 2) Assign each client to facility in S. c f (S): Cost to open S. (facility cost) c s (S): Sum of client-facility distances. (service cost) Goal: Minimize c f (S) + c s (S)

3. 3 Metric Facility Location 5 4 2 3 client facility 5 4 2 3 client facility opened facility Problem Instance Feasible Solution

4. 4 Capacities Every facility i has a capacity u i. Hard: can’t open multiple copies, can’t violate capacities. Results for capacitated FL Chudak & Shmoys ‘99, Jain & Vazirani ‘99: … LP techniques give c-approx. for …soft capacities. Soft: (e.g.) can open k copies of facility i and serve k·u i clients. Problem: Known LPs have large integrality gap for hard capacities.

5. 5 Further Results Using Local Search Techniques Korupolu et al ‘98, Chudak & Williamson ‘99: …Local search gives c-approx. for …uniform, hard capacities. Arya et al ‘01: …Local search gives c-approx. for …nonuniform, soft capacities. Our Result: Pal, Tardos, Wexler ‘01: …Local search gives 9-approx. for …nonuniform, hard capacities.

6. 6 Local Search 1) Start with a feasible solution S. 2) Improve solution with “local operations”. 3) Stop when there are no remaining operations that lower the cost. Need To: Define Operations. Prove a Local Optimum is good.

7. 7 add Operation 1) add(s) – Open facility s. Given S, where do we send clients? Just an assignment problem. Can solve optimally w/ min cost flow. Thm: [Korupolu et al] If no add(s) op. improves solution S, c s (S) ≤ c s (S OPT ) + c f (S OPT ).

8. 8 open & close Operations 2) open(s,T) – Open facility s, send clients from T to s, and close T. All reassigned clients are sent to s. Need: cap(s) ≥ demand served by T. 3) close(s,T) – Close facility s, send clients from s to T, and open T. Need: cap(T) ≥ demand served by s. Too many such operations, but we can find some that are good enough.

9. 9 open & close Operations t1t1 t2t2 t3t3 t1t1 t2t2 t3t3 s s open(s,T)close(s,T)

10. 10 Onto Analysis We’ve defined local operations. Algorithm outputs solution S. We want to compare S to S OPT. We’ll prove… Theorem: S local opt. c(S) ≤ 9c(S OPT ). add op. bounds the serive cost c s (S). Need to bound facility cost c f (S).

11. 11 The Plan In a local opt., feasible operations give bounds. Consider close(s, T). c f (s) ≤ c f (T) + Δ benefit of closing s cost of opening T change in cost of shipping to T We’ll pick some operations to bound c f (S) in terms of c f (S OPT ).

12. 12 The Plan (cont’) c f ( T ) ≤ c f ( T ’ ) + Δ (closed) (opened) Pick operations so that each s in S is closed exactly once. each s in S OPT is opened ≤ k times. reassignment costs are small. Where do operations come from?

13. 13 The Exchange Graph S S OPT c ij Flow in = demand served by S Flow out ≤ capacity of facilities in S OPT y = min cost flow from S to S OPT. Fact: cost(y) ≤ c s (S) + c s (S OPT ) flow = clients

14. 14 The Flow We can turn a min cost flow into a tree by augmenting cycles. S S OPT How do we get operations from this? flow of clients

15. 15 Naïve Approach S S OPT flow of clients Just try close(s,T), using the neighbors of s for T. s Problem: Expensive facilities in S OPT may be opened many times.

16. 16 Other Approaches S S OPT

17. 17 Result Thm: Our local search algorithm gives a 9-approx. for facility location with hard, nonuniform capacities. Note 1: Actually a (9 + є)-approx. Note 2: Scaling lowers it to 8.53… Note 3: At best, algorithm is a 4-approx.