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CS880: Approximations Algorithms Scribe: Matt Elder Lecturer: Shuchi Chawla Topic: LP Rounding and Randomized Rounding Date: 2/22/
In out last lecture, we discussed the LP Rounding technique for producing approximation algo- rithms. The idea behind LP Rounding is to write the problem as an integer linear program, relax its integrality restraints to efficiently solve the general linear program, and then move the LP so- lution to a nearby integral point in the feasible solution space. The difficulty of this process lies in the rounding step, which demands that a bound on its suboptimality.
We discussed how to apply this method to vertex cover, set cover, and network flow. Here, we give somewhat more complicated rounding methods for facility location, and introduce the technique of randomized rounding in application to set cover and min-congestion rounding.
11.1 Facility Location
Again, the facility location problem gives a collection of facilities and a collection of customers, and asks which facilities we should open to minimize the total cost. We accept a facility cost of fi if we decide to open facility i, and we accept a routing cost of c(i, j) if we decide to route customer j to facility i. Furthermore, we know that the routing costs form a metric.
First, we design a linear program to answer a “relaxed” version of this problem. We let the variable xi denote the extent to which facility i is open, and let yij denote the extent to which customer j is assigned to facility i. The following linear program then expresses the problem:
minimize
i
fixi +
i,j
c(i, j)yij ,
where 0 ≤ yij ≤ xi ≤ 1 ∀i, j
For convenience, let Cf (x) denote the total factory cost induced by x, i.e.,
i fixi. Similarly, let Cr(y) denote the total routing cost induced by y,
i,j c(i, j)yij^. If this LP were modified to require that these variables each equal 0 or 1, this system would be precisely the ILP we need to solve. But, again, solving general ILPs is NP-hard problem, so we solve this related real-valued LP instead.
Let x∗, y∗^ be the optimal solution to this linear program. Since every feasible solution to the original ILP lies in the feasible region of this LP, the cost C(x∗, y∗) is less than the optimal solution to the ILP. Since x∗^ and y∗^ are almost certainly non-integral, we need a way to round this solution to a feasible, integral solution without increasing the cost function much.
To do so, we first employ the filtering technique of Lin and Vitter [1] to produce ˜x, y˜. This filtering will later allow us to put upper bounds on the routing cost that we accept.
- For each customer j, compute the average cost ˜cj =
i c(i, j)y ∗ ij.
- For each customer j, let the Sj denote the set {i | c(i, j) ≤ 2˜cj }.
- For all i and j: if i 6 ∈ Sj , then set ˜yij = 0; else, set ˜yij = y∗ ij /
i∈Sj y
∗ ij.
- For each facility i, let ˜xi = min(2x∗ i , 1).
Lemma 11.1.1 For all i and j, y˜ij ≤ 2 y ij∗.
Proof: If we fix j and treat y∗ ij as a probability distribution, then we can show this by Markov’s inequality. However, the proof of Markov’s Inequality is simple enough to show precisely how it applies here:
c˜j =
i
c(i, j)y∗ ij ≥
i /∈Sj
c(i, j)y∗ ij ≥
i /∈Sj
2˜cj y ij∗ ≥ 2˜cj
i /∈Sj
y ij∗.
So, 1/ 2 ≥
i /∈Sj y ∗ ij. For any fixed^ j,^ y ∗ ij is a probability distribution, so^
i∈Sj y ∗ ij ≥^1 /2. Therefore,
y˜ij = y∗ ij /
i∈Sj y ∗ ij
≤ 2 y ij∗.
Lemma 11.1.2 x,˜ y˜ is feasible, and C(˜x, ˜y) ≤ 2 C(x∗, y∗).
Proof: For any fixed j, the elements ˜yij form a probability distribution. For every i and j, y˜ij ≤ 2 y∗ ij and thus ˜xi ≥
i y˜ij^. It is clear that 0^ ≤^ xi, yij^ ≤^ 1 for all^ i^ and^ j, so ˜x^ and ˜y^ are feasible solutions to the LP.
Now, given ˜x and ˜y, we perform the following algorithm:
- Pick the unassigned j that minimizes ˜cj.
- Open factory i, where i = argmini∈Sj (fi).
- Assign customer j to factory i.
- For all j′^ such that Sj ∩ Sj′^6 = ∅, assign customer j′^ to factory i.
- Repeat steps 1-4 until all customers have been assigned to a factory.
Let L be the set of facilities that we open in this way. We now show that the solution that this algorithm picks has reasonably limited cost.
Lemma 11.1.3 Cf (L) ≤ Cf (x∗) and Cr(L) ≤ 6 Cr(y∗).
Proof: For any two customers j 1 and j 2 that were picked in Step 1, Sj 1 ∩ Sj 2 = ∅.
Consider the facility cost incurred by one execution of Steps 1 through 4. Let j be the customer chosen in Step 1, and let i be the facility chosen in Step 2. Since ˜x is part of a feasible solution, 1 ≤
k∈Sj x˜k.^ So,^ fi^ ≤^ fi
k∈Sj ˜xk; and since^ fi^ is chosen to be minimal,^ fi^ ≤^
k∈Sj fk^ ˜xk. Facility i is the only member of Sj that the algorithm can open.
Let J be the set of all customers selected in Step 1. Considering the above across the algorithm’s whole execution yields
Cf (L) ≤
j∈J
k∈Sj
fk x˜k =
i
fi x˜i = Cf (˜x) ≤ Cf (x∗).
Proof: Let ZS be an indicator variable, which is 1 iff we pick set S in this run of Step 2. We compute:
E[cost of Step 2] = E
[
S
ZS c(S)
]
S
E[ZS ] = c(x∗).
We now need to estimate the number of times that Step 2 is executed. To do so, we estimate the probability that any one element is covered in a particular execution of Step 2. Fix some element a. We know that
S 3 a x ∗ S ≥^ 1. This gives us the following reasoning:
Pr[a is picked] = 1 −
S 3 a
(1 − x∗ S ) ≥ 1 −
S 3 a
exp (−x∗ S ) = 1 − exp
S 3 a
x∗ S
≥ 1 − e−^1.
So, the probability that e is unpicked after k steps is no more than e−k, because each execution of Step 2 is independent. So, the probability that any particular element is unpicked after, say, 2 ln n steps is no more than (1/n^2 ). By the union bound, the probability that there exists an unpicked element after 2 ln n steps is at most n(1/n^2 ) = 1/n.
Thus, with high probability, the number of executions of Step 2 is O(log n). So the expected total cost of the algorithm is c(x∗)O(log n), and this algorithms is a O(log n)-approximation in expectation. Standard methods can convert this to an arbitrarily high-probability result.
References
[1] JH Lin and JS Vitter. Approximation Algorithms for Geometric Median Problems. In Infor- mation Processing Letters, 1992.
