# 10.1 Integer Programming and LP relaxation

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1 CS787: Advanced Algorithms Lecture 10: LP Relaxation and Rounding In this lecture we will design approximation algorithms using linear programming. The key insight behind this approach is that the closely related integer programming problem is NP-hard (a proof is left to the reader). We can therefore reduce any NP-complete optimization problem to an integer program, relax it to a linear program by removing the integrality constraints, solve the linear program, and then round the LP solution to a solution to the original problem. We first describe the integer programming problem in more detail Integer Programming and LP relaxation Definition An integer program is a linear program in which all variables must be integers. As in a linear program, the constraints in an integer program form a polytope. However, the feasible set is given by the set of all integer-valued points within the polytope, and not the entire polytope. Therefore, the feasible region is not a convex set. Moreover, the optimal solution may not be achieved at an extreme point of the polytope; it is found at an extreme point of the convex hull of all feasible integral points. (See Figure ) Figure : The feasible region of an integer linear program. Usually, in designing LP-based approximation algorithms we follow the following approach: 1. Reduce the problem to an integer program. 2. Relax the integrality constraint, that is, allow variables to take on non-integral values. 1

2 3. Solve the resulting linear program to obtain a fractional optimal solution. 4. Round the fractional solution to obtain an integral feasible solution. Note that the optimal solution to the LP is not necessarily integral. However, since the feasible region of the LP is larger than the feasible region of the IP, the optimal value of the former is no worse than the optimal value of the latter. This implies that the optimal value to the LP is a lower bound on OPT, the optimal value for the problem we started out with. While the rounded solution is not necessarily optimal for the original problem, since we start out with the optimal LP solution, we aim to show that the rounded solution is not too far from optimal. These relationships between the different values are illustrated in Figure below. The gap between the optimal LP value and the optimal integral solution is called the integrality gap of the linear program. Figure : The relationship between the optimal LP and ILP values for minimization problems. We now apply the linear programming approach to two problems: vertex cover and facility location Vertex Cover revisited We have already seen a factor of 2 approximation using maximum matchings for the lower bound. Today we will see another factor of 2 approximation based on Linear Programming. This time we consider a weighted version of the problem. Recall that we are given a graph G = (V, E) with weights w : V R +, and our goal is to find the minimum weight subset of vertices such that every edge is incident on some vertex in that subset. We first reduce this problem to an integer program. We have one {0, 1} variable for each vertex denoting whether or not this vertex is picked in the vertex cover. Call this variable x v for vertex v. Then we get the following integer program. The first constraint essentially states that for each 2

3 edge we must pick at least one of its endpoints. Minimize v V w v x v subject to x u + x v 1 x v {0, 1} (u, v) E v V To obtain a linear program, we relax the last constraint to the following: x v [0, 1] v V Next we use a standard LP solver to find the optimal solution to this LP. Let x v denote this optimal fractional solution. By our previous argument, the following holds: Proposition Val(x ) OPT, where OPT is the value of the optimal solution to the vertex cover instance. The example below illustrates that the optimal solution to the LP is not necessarily integral. Figure : An example where the vertex cover LP has an integrality gap of 4/3. The optimal fractional solution sets x v = 1/2 for all vertices, with a total cost of 3/2, while the optimal integral solution has cost 2. It remains to round the fractional solution. For vertex cover, the obvious rounding works: for each x v 1/2, set x v = 1 and include v in the vertex cover. For each x v < 1/2, set x v = 0 and don t include v in the vertex cover. It is easy to see that this is a feasible solution and forms a vertex cover. Consider any edge (u, v) E. Then, by construction, x u + x v 1. Therefore, at least one of x u and x v is at least 1/2, and is picked in the vertex cover. It remains to prove that the solution we obtain has small weight. Lemma v V x vw v 2 v V x vw v Proof: Recall that we set x v to be 1 if and only if x v 1/2, and 0 otherwise. The lemma then follows by noting that x v 2x v for all v. Finally, the weight of our vertex cover is exactly v V x vw v because by definition x v = 1 if and only if v is included in our vertex cover, and 0 otherwise. We therefore have the following theorem. 3

4 Theorem The above algorithm is a 2-approximation to weighted vertex cover Facility location The story behind the facility location problem is that a company wants to locate a number of warehouses such that the cost of opening these warehouses plus the cost of shipping its product to its retail stores is minimized. Formally, 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 f i 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, the routing costs form a metric, that is, they are distance functions satisfying the triangle inequality. First, we reduce this problem to an integer program. We let the variable x i denote whether facility i is open, and let y ij denote whether customer j is assigned to facility i. The following program then expresses the problem. The first constraint says that each customer should be assigned to at least one facility. The second says that is a customer is assigned to a facility, then that facility must be open. minimize i f i x i + i,j c(i, j)y ij subject to y ij 1 i x i y ij x i, y ij {0, 1} j i, j i, j To obtain a linear program, we relax the last constraint to x i, y ij [0, 1]. For convenience, let C f (x) denote the total factory cost induced by x, i.e., i f ix i. Similarly, let C r (y) denote the total routing cost induced by y, i,j c(i, j)y ij. 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 or equal to 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. Note that the y ij variables for a single j form a probability distribution over the facilities. The LP pays the expected routing cost over these facilities. If we could pick the closest facility over all those that j is connected to, our routing cost would be no more than in the LP solution. However, the closest facility is not necessarily the cheapest facility, and this is what makes this rounding process complicated. To get around this problem, we first use a filtering technique that ensures that all facilities that j is connected to have small routing cost, and we pick the cheapest of these in our solution. 1. For each customer j, compute the average cost c j = i c(i, j)y ij. 4

5 2. For each customer j, let the S j denote the set {i : c(i, j) 2 c j }. 3. For all i and j: if i S j, then set ỹ ij = 0; else, set ỹ ij = y ij / i S j y ij. 4. For each facility i, let x i = min(2x i, 1). Lemma For all i and j, ỹ ij 2y ij. Proof: If we fix j and treat yij 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 = c(i, j)yij c(i, j)yij > 2 c j yij 2 c j i i/ S j i/ S j i/ S j y ij. So, 1/2 i/ S j yij. For any fixed j, y ij is a probability distribution, so i S j yij 1/2. Therefore, ( ) ỹ ij = yij / i S j yij 2yij. Lemma x, ỹ is feasible, and C( x, ỹ) 2C(x, y ). Proof: For any fixed j, the elements ỹ ij form a probability distribution. For every i and j, ỹ ij 2y ij and thus x i i ỹij. It is clear that 0 x i, y ij 1 for all i and j, so x and ỹ are feasible solutions to the LP. Now, given x and ỹ, we perform the following algorithm: 1. Pick the unassigned j that minimizes c j. 2. Open factory i, where i = argmin i Sj (f i ). 3. Assign customer j to factory i. 4. For all j such that S j S j, assign customer j to factory i. 5. 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 C f (L) 2C f (x ) and C r (L) 6C r (y ). Proof: For any two customers j 1 and j 2 that were picked in Step 1, S j1 S j2 =. 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 S j x k. So, f i f i k S j x k ; and since f i is chosen to be minimal, f i k S j f k x k. Facility i is the only member of S j 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 C f (L) f k x k = f i x i = C f ( x) 2C f (x ). j J k S j i 5

6 Consider now the routing cost C r, and let C r (j) for a customer j denote the cost of routing j in L. If j was picked in Step 1, then its routing cost is c(i, j) for some facility i S j ; so C r (j) 2 c j. Now, suppose instead that j was not picked in Step 1. By the algorithm, there is some j that picked in Step 1 such that S j S j. Suppose that facility i is in this intersection, and say that facility i is the facility to which customers j and j are routed. Now, at long last, we use the fact that c(i, j) forms a metric: we know that C r (j ) c(i, j ) + c(i, j) + c(i, j). Because i is in both S j and S j, we know by their definition that c(i, j ) 2 c j and that c(i, j) + c(i, j) 4 c j. The customer j was not picked in Step 1, and customer j was, so c j c j, and thus, C r (j ) 6 c j. Now, c j was the routing cost of customer j in the y LP solution. So, C r (L) 6C r (y ). This lemma yields the following as a corollary: Theorem This algorithm is a 6-approximation to Facility Location. Notice that, in the preceding construction, we picked S j to be all i such that the cost c(i, j) 2 c j. The constant 2 is actually not optimal for this algorithm. Suppose we replace it with α, some parameter of the construction. If we redo the above arithmetic, we find that C f (L) (1/(1 α))c f (x ) and that C r (L) (3/α)C r (y ). Thus, if we let α = 3/4 instead of 1/2, this method yields a 4-approximation. If we let α be a variable in the actual computed values of C f (x ) and C r (y ), we would get a somewhat better approximation. Note that the integrality gap of the facility location LP is actually quite a bit smaller than 4. There are several better rounding algorithms known based on the same lower bound that lead to improved approximations. 6

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