The Facility Location Problem - An algorithm and an ... - Corelab

Facility LocationThe Problem and a SolutionThe Facility Location ProblemMetric Uncapacitated Facility Location:Let G be a bipartite graph with bipartition (F, C), where F is theset of facilities and C is the set of cities. Suppose also that|C| = n c and |F| = n f . Thus, the total number of vertices in thegraph is n = n c + n f and the total number of edges ism = n c · n f . Let f i be the cost of opening facility i, and c ij be thecost of connecting city j to facility i. The connection costssatisfy the triangle inequality. We want to find a subset I ⊆ F offacilities that should be opened and a function φ : C −→ Iassigning cities to open facilities, that minimizes the total cost ofopening facilities and connecting cities to them.Achilleos AntoniosFacility Location

The Problem and a SolutionInteger Program for FLThe Facility Location Problemminimizesubject toy i − x ij ≥ 0x ij , y i ∈ 0, 1∑i∈F ,j∈C c ijxij + ∑ i∈F f iy i∑i∈F x ij ≥ 1∀j ∈ C∀i ∈ F, j ∈ C∀i ∈ F, j ∈ CAchilleos AntoniosFacility Location

LP - relaxationThe Problem and a SolutionThe Facility Location Problemminimizesubject toy i − x ij ≥ 0x ij ≥ 0y i ≥ 0∑i∈F ,j∈C c ijxij + ∑ i∈F f iy i∑i∈F x ij ≥ 1∀j ∈ C∀i ∈ F, j ∈ C∀i ∈ F, j ∈ C∀i ∈ FAchilleos AntoniosFacility Location

And the dual:The Problem and a SolutionThe Facility Location Problemmaximizesubject to a j − b ij ≤ c ij∑j∈Cb ijb ij ≥ 0a i ≥ 0∑j∈C a i∀j ∈ C, i ∈ F∀i ∈ F∀i ∈ F, j ∈ C∀i ∈ FAchilleos AntoniosFacility Location

AlgorithmThe Problem and a SolutionThe Facility Location ProblemAlgorithmWe introduce a notion of time, so that each event can beassociated with the time at which it happened. The algorithmstarts at time 0. Initially, each city is defined to be unconnected,all facilities are closed, and a j is set to 0 for every j. TheAlgorithm has two phases:Achilleos AntoniosFacility Location

AlgorithmThe Problem and a SolutionThe Facility Location ProblemPhase 1For every unconnected city j ∈ C, increase the parameter a juniformly, at unit rate, until one of the following events occurs (iftwo events occur at the same time, we process them inarbitrary order).(a) For some city j, and some open facility i, a j = c ij . In thiscase, (i, j) is declared "‘tight"’. If i is open, the city is connected.If not, from now and on, b ij will be raised uniformly. Every edge(i, j), such that b ij > 0 is called "’special"’.(b) For some closed facility i, we have ∑ j∈C b i,j = f i . Thismeans that the total contribution of the cities is sufficient toopen facility i. In this case, temporarily open this facility, and forevery unconnected city j with tight edges , j is connected and iis the connecting witness of j.Achilleos AntoniosFacility Location

AlgorithmThe Problem and a SolutionThe Facility Location ProblemPhase 2Let F t be the set of temporarily open facilities, T the subgraphof G consisting of all special edges. And let H be the subgraphof T 2 induced on F t . Let I be a maximal independent set on H.All facilities in I are open. If city j has a tight edge (i, j) to I,connect i to j. Else, for a tight edge (i, j), find an i ′ in I, which isa neighbour of i in H and connect j to it.The Algorithm achieves an approximation factor of 3Running time: O(mlogm), where m = n c · n fAchilleos AntoniosFacility Location

Algorithm 2The Problem and a SolutionThe Facility Location ProblemAlgorithm 21. We introduce a notion of time, so that each event can beassociated with the time at which it happened. The algorithmstarts at time 0. Initially, each city is defined to be unconnected,all facilities are closed, and a j is set to 0 for every j.2. While C ≠ ∅, for every city j ∈ C, increase the parameter a jsimultaneously, until one of the following events occurs (if twoevents occur at the same time, we process them in arbitraryorder).Achilleos AntoniosFacility Location

Algorithm 2The Problem and a SolutionThe Facility Location ProblemAlgorithm 2(a) For some unconnected city j, and some open facilityi, a j = c ij . In this case, connect city j to facility i and remove jfrom C.(b) For some closed facility i, we have∑j∈C max(0, a j − c ij ) = f i . This means that the totalcontribution of the cities is sufficient to open facility i. In thiscase, open this facility, and for every unconnected city j witha j ≥ c ij , connect j to i, and remove it from C.Achilleos AntoniosFacility Location

Greedy formThe Problem and a SolutionThe Facility Location ProblemAn other description, as a greedy algorithmIn the beginning all cities are unconnected and all facilities areclosed.While C ≠ ∅ :Among all pairs of facilities and subsets of C, find the most costeffective one, (i, C ′ ), open facility i, if it is not already open, andconnect all cities in C ′ to i.Set f i := 0, C := C C ′ .Achieves same running timeApproximation factorAchilleos AntoniosFacility Location