PRIMAL-DUAL APPROXIMATION ALGORITHMS FOR METRIC FACILITY LOCATION AND K-MEDIAN PROBLEMS K. Jain V. Vazirani Journal of the ACM, 2001

PRIMAL-DUAL APPROACH We start by constructing a primal problem. On the basis of the primal problem, a dual problem is constructed. We solve the dual problem while maintaining its feasibility in order to find a feasible primal solution.

INTEGER PROGRAMMING PROBLEM min Σ j ε C Σ i ε F c ij x j + Σ i ε F f i y i s.t Σ i ε F x ij ≥ 1 for all j ε C y i - x ij ≥ 0 for all i ε F, j ε C x ij ε {0,1} for all i ε F, j ε C y i ε {0,1} for all i ε F y i denotes whether facility i is open or not x ij denotes whether client j is connected to a facility i or not The first constraint ensures that each city is connected to at least one facility. The second constraint ensures that if a client is assigned to a facility then that facility must be open.

RELAXED LP PRIMAL PROBLEM min Σ j ε C Σ i ε F c ij x j + Σ i ε F f i y i s.t Σ i ε F x ij ≥ 1 for all j ε C y i - x ij ≥ 0 for all i ε F, j ε C 0 ≤ x ij ≤ 1 for all i ε F, j ε C 0 ≤ y i ≤ 1 for all i ε F

DUAL PROBLEM max Σ j ε C α j s.t α j - β ij ≤ c ij for all i ε F, j ε C Σ j ε C β ij ≤ f i for all i ε F α j ≥ 0 for all j ε C β ij ≥ 0 for all i ε F, j ε C α j is the total price paid by a client j. α j = c ij + β ij A part of α j goes towards paying the service cost of facility i, i.e., c ij and β ij is the contribution of j towards the facility opening cost of i.

COMPLIMENTARY SLACKNESS CONDITIONS For Primal for all i ε F, j ε C: x ij > 0 ═> α j - β ij = c ij for all i ε F: y i > 0 ═> Σ j ε C β ij = f i For dual for all j ε C: α j > 0 ═> Σ i ε F x ij = 1 for all i ε F, j ε C: β ij > 0 ═> y i = x ij

TERMS USED IN THE PAPER 1.Tight edge:- An edge (i,j) between a client j and a facility i becomes tight when α j = c ij. 2.Special edge:- An edge (i,j) between a client j and a facility i is special if β ij > 0. 3.Unconnected Client:- A client j is said to be unconnected with a facility i if edge between them is not tight. 4.Connected Client:- A client j is said to be connected with a facility i if edge between them is tight. 5.Connecting Witness:- A facility i is known as connecting witness for all those clients with whom it shares a tight edge.

ALGORITHM Phase I :- Initially all the clients are unconnected and α j = 0 for all the clients. Raise the dual variables α j for each unconnected client j uniformly at a unit rate until an edge between some facility i and client j becomes tight. As soon as an edge (i,j) between client j and a facility i becomes tight, start raising β ij at the same rate as α j A facility is declared temporarily open when its opening cost has been completely paid, i.e., Σ j β ij = f i.

As soon as a facility is declared temporarily open, all the clients that share tight edges with this facility are declared connected and the facility becomes its connecting witness. All such edges for which β ij > 0 are declared special. If a client j gets a tight edge to an already open facility i, j is declared as connected to i, i becomes its connecting witness and its β ij is not raised. Phase I continues until all the clients become connected.

At t = 0 α 1 =0 α 3 = f 1 =1 f 2 =1 f 3 = α 2 =0

At t = 1 α 1 =1 α 3 = f 1 =1 f 2 =1 f 3 = α 2 =1

At t = 2 α 1 =2 α 3 = β 11 =0 β 23 =0 f 1 =1 f 2 =1 f 3 = α 2 =2 Tight edge:

At t = 3 α 1 =3 α 3 = β 11 =1 β 23 =1 f 1 =1 f 2 =1 f 3 = α 2 =3 Tight edge: Special edge:

At t = 4 α 1 =3 α 3 = β 11 =1 β 23 =1 f 1 =1 f 2 =1 f 3 = α 2 =4 Tight edge: Special edge:

At the end of Phase f 1 =1 f 2 =1 4 4 Tight edge: Special edge:

Phase II :- It consist of two parts. Part 1:- To choose facilities that will opened permanently. Part 2:- To connect every client to exactly one facility.

PART 1:- Let F t denote the set of all temporarily open facilities. Consider a graph G with all the special edges of path length at most 2, induced on F t. Construct a maximal independent set I of G. Declare all the facilities in I as permanently open.

Phase 2 - part 1:- Forming independent sets 2 f 1 =1f 2 =1 4 4 Permanent Facility

PART 2:- Case 1:- Assign client j to the facility i є I with which it shares a special edge. Declare client j as directly connected to it. Case 2:- If client j does not have a special edge to any facility i є I then assign j to a facility with which it shares a tight edge such that i was its connecting witness. Declare client j as directly connected to i. Case 3:- If client j has neither a special edge nor a tight edge with any facility in I, then assign j to a facility i* which is a neighbor of i in the graph G such that i* є I. Declare client j as indirectly connected to i*.

Phase 2 - part 2- Assigning clients to facilities Case 1: Directly Connected Case 2: Directly Connected Case 3: Indirectly Connected = Directly connected = Indirectly connected

ANALYSIS Let α j = α j f + α j c, where α j f = β ij and α j c = c ij Σ i ε I f i = Σ j ε C α j f …eqn I Consider an indirectly connected client j to i c ij ≤ 3*α j c …eqn II Adding the results of eqn I (multiplied by 3) and eqn II, we get, Σ i ε I Σ j ε C c ij x ij + 3*Σ i ε I f i y i ≤ 3*Σ j ε C α j