Design of Approximation Algorithms

7.7 Lagrangean relaxation and the k-median problem 189Our algorithm then splits into two cases, a simple case when α 2 ≥ 1 2and a more complicatedcase when α 2 < 1 2 . If α 2 ≥ 1 2 , we return S 2 as our solution. Note that |S 2 | < k, and thus is afeasible solution. Using α 2 ≥ 1 2and Lemma 7.15, we obtainc(S 2 ) ≤ 2α 2 c(S 2 ) ≤ 2 (α 1 c(S 1 ) + α 2 c(S 2 )) ≤ 2(3 + ϵ) OPT k ,as desired.Before we give our algorithm for the remaining case, we let c(j, S) = min i∈S c ij , so that∑j∈D c(j, S) = c(S). Now for each facility i ∈ S 2, we open the closest facility h ∈ S 1 ; that is,the facility h ∈ S 1 that minimizes c ih . If this doesn’t open |S 2 | facilities of S 1 because somefacilities in S 2 are close to the same facility in S 1 , we open some arbitrary facilities in S 1 so thatexactly |S 2 | are opened. We then choose a random subset of k − |S 2 | of the |S 1 | − |S 2 | facilitiesof S 1 remaining, and open these. Let S be the resulting set of facilities opened.We now show the following lemma.Lemma 7.16: If α 2 < 1 2 , then opening the facilities as above has cost E[c(S)] ≤ 2(3+ϵ) OPT k.Proof. To prove the lemma, we consider the expected cost of assigning a given client j ∈ D toa facility opened by the randomized algorithm. Let us suppose that the facility 1 ∈ S 1 is theopen facility in S 1 closest to j ; that is, c 1j = c(j, S 1 ); similarly, let 2 ∈ S 2 be the open facilityin S 2 closest to j. Note that with probability k−|S 2||S 1 |−|S 2 |= α 1 , the facility 1 ∈ S 1 is opened inthe randomized step of the algorithm if it has not already been opened by the first step of thealgorithm. Thus with probability at least α 1 , the cost of assigning j to the closest open facilityin S is at most c 1j = c(j, S 1 ). With probability at most 1−α 1 = α 2 , the facility 1 is not opened.In this case, we can at worst assign j to a facility opened in the first step of the algorithm; inparticular, we assign j to the facility in S 1 closest to 2 ∈ S 2 . Let i ∈ S 1 be the closest facilityto 2 ∈ S 2 ; see Figure 7.9. Thenc ij ≤ c i2 + c 2jby triangle inequality. We know that c i2 ≤ c 12 since i is the closest facility in S 1 to 2, so thatc ij ≤ c 12 + c 2j .Finally, by triangle inequality c 12 ≤ c 1j + c 2j , so that we havec ij ≤ c 1j + c 2j + c 2j = c(j, S 1 ) + 2c(j, S 2 ).Thus the expected cost of assigning j to the closest facility in S isE[c(j, S)] ≤ α 1 c(j, S 1 ) + α 2 (c(j, S 1 ) + 2c(j, S 2 )) = c(j, S 1 ) + 2α 2 c(j, S 2 ).By the assumption α 2 < 1 2 , so that α 1 = 1 − α 2 > 1 2, we obtainE[c(j, S)] ≤ 2(α 1 c(j, S 1 ) + α 2 c(j, S 2 )).Then summing over all j ∈ D and using Lemma 7.15, we obtainE[c(S)] ≤ 2(α 1 c(S 1 ) + α 2 c(S 2 )) ≤ 2(3 + ϵ) OPT k .Electronic web edition. Copyright 2011 by David P. Williamson and David B. Shmoys.To be published by Cambridge University Press