Chapter 1. Preliminaries


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1 Introduction This dissertation is a reading of chapter 4 in part I of the book : Integer and Combinatorial Optimization by George L. Nemhauser & Laurence A. Wolsey. The chapter elaborates links between Polyhedral Theory, Linear Programming and Integer Programming. Chapter 1 contains preliminaries, with some results on elementary linear algebra. The relation between affine independence and linear independence is proved. At last, we state Farkas lemma. Chapter 2 is Polyhedral Theory. First result we prove is: For a polyhedron P, dim (P) rank (A =, b = ) = n where P = {x / Ax b} and (A =, b = ) be the rows of (A, b) (constraints) for which strict equality holds for all x P. For a valid inequality x =, a face of P is a set of all points satsfying the inequality. Next, we define a Facet: A face with dimension one less than dimension of P. We show that for a facet there exists some inequality x for k representing the facet. Other way, we also prove that one of the inequalities representing the facet is necessary in the description of P. Further it is proved that the facets are sufficient for the description of P i.e., every inequality x for r that represents a face of P of dimension less than dim (P) 1 is irrelavant to the description of P. A theorem which characterizes facets and determines when a valid inequality is a facet is also proved. Next, a representation of polyhedra in terms of lowestdimensional faces (extreme points) is given. It is also proved that a polyhedron has a finite number of extreme points and extreme rays. We show that an optimal solution of a linear programming problem exists at an extreme point, if the problem is unbounded then it has an extreme ray. One of the fundamental results on the representation of polyhedra called Minkowski s theorem is proved. Its converse is Weyl s theorem. The basic results for the dual pair of linear programs in terms of extreme points and extreme rays are also established. 1
2 Chapter 3 is on Polarity. A polar of P is a set of all valid inequalities ( ) of P. First, we give characterization of using extreme points and extreme rays of P. Further, we show that ( ) is an extreme ray of if and only if ( ) defines a facet of P. Next, we show that defines a facet of if and only if is an extreme point of P, and 0 defines a facet of if and only if is an extreme ray of P. The 1polar is defined as. The complete symmetry between P and is : P = { for t T, where are the extreme points of } = { for k K, where are the extreme points of P} Moreover, each of inequalities in descriptions define facets. Chapter 4 gives polyhedral ties between linear and integer programs. We write S = and show that it is a rational polyhedron. Infact, we shall show that conv(s) can be generated from a finite number of points in S and a finite number of integer valued rays. Hence we can solve the integer program where S = (IP) by solving linear program max { (CIP) It is proved that if CIP has a bounded optimal value then it has an optimal solution (at an extreme point of ) that is an optimal solution to IP. Conversely if is an optimal solution to IP then is an optimal solution to CIP. Next, IP is unbounded if and only if CIP is unbounded. Thus IP is either infeasible or unbounded or has an optimal solution. 2
3 Chapter 1 Preliminaries Proposition 1.1: The following statements are equivalent: a. {x / Ax = b}. b. Rank (A) = rank (A, b). Definition 1.2: A set of points x k are affinely independent if the unique solution of = 0, = 0 is = 0, for i = 1,, k. Note 1.3: Observe that linear independence implies affine independence, but the converse is not true. Proposition 1.4: The following statements are equivalent: a. x k are affinely independent. b.   are linearly independent. c. (,  1),, (,  1) are linearly independent. Proof: [ (a) (b) ] Consider = 0 i.e., = 0 Clearly sum of coefficients is zero. Hence by (a), = 0, i = 2,, n. This means  are linearly independent. [ (b) (c) ] Consider (,  1) = 0. = 0 and = 0 = 0 Now, use (b) to conclude (c). [ (c) (a) ] Consider = 0, = 0 (,  1) = 0. = 0 i [use (c)] Note 1.5: By above proposition there can not be more than n + 2 affinely independent vectors, otherwise there would be at least n + 1 linearly independent 3
4 vectors in. Hence the maximum number of affinely independent points in is n + 1. Proposition 1.6: If {x Ax = b}, the maximum number of affinely independent solutions of Ax = b is n + 1 rank(a). Proof: We know that maximum number of linearly independent solutions of Ax = b is the dimension of ker(a) which is, by rank nullity theorem, n rank(a). Hence by above note the maximum number of affinely independent solutions is n rank(a) + 1. Example 1.7: Suppose (A, b) = Then rank(a) = rank(a, b) = 1. By proposition 1.6, the maximum number of affinely independent solutions of Ax = b is 31 = 2. Two such solutions are = (5 2) and = (1 1). Definition 1.8: (Orthogonal Subspace) For a subspace H the set = {x / xy = 0 for y H} is called the orthogonal subspace of H. Note 1.9: is a subspace of. Example 1.10: H = {x / = 2 } is a subspace of. It is a line passing through origin. Clearly here A = (1,  2), since  2 = 0. As = 2, H = {x / x = (2, 1) u, u }. Therefore B = (2, 1). Now, = {x / xy = 0, y H} = {x / (, ) = 0, u } = {x / 2 + = 0} Definition 1.11: The projection of p on H is the vector q H such that p  q. Further, we define (S) = {q / q is the projection of p on H for some q S} Definition 1.12: Given a set S, a point x is a convex combination of points of S if there exists a finite set of points in S and a with 4
5 = 1 and x =. The convex hull of S, denoted by conv(s), is the set of all points that are convex combinations of points in S. Definition 1.13: (Extreme Point) is an extreme point of P if there do not exist such that Definition 1.14: Let. If P = then is called a ray of P. Note 1.15: If then for each Therefore a point r is a ray of P if and only if for any point the set { Definition 1.16: A ray r of P is an extreme ray if there do not exist rays for any such that Theorem 1.17 (Farkas Lemma): Let A be a real m n matrix and let c be a real nonzero n vector. Then either the primal system Ax 0 and has a solution for x or the dual system = c and y has a solution for, y but never both. 5
6 Chapter 2 Polyhedral Theory Polyhedra and its dimension Definition 2.1: (Polyhedron) A polyhedron P is the set of points that satisfy a finite number of linear inequalities. i.e.; P = {x / Ax b}, where (A, b) is an m (n + 1) matrix. If entries in (A, b) are rational then P is called a rational polyhedron. Note 2.2: We shall consider only rational polyhedra. Definition 2.3: (Bounded Polyhedron) A polyhedron P is bounded if there exist an w such that P {x /  w w for j = 1 n}. Definition 2.4: (Polytope) A bounded polyhedron is called a polytope. Remark 2.5: A polyhedron is a convex set. Definition 2.6: (Cone) C is a cone if x C implies λx C for all λ. Remark 2.7: The polyhedron {x / Ax 0} is a cone. Definition 2.8: (Dimension of a Polytope) If the maximum number of affinely independent points in polyhedron P is k + 1 then we define dim(p) = k. Definition 2.9: A polyhedron is full dimensional if dim(p) = n. Note 2.10: We will show that if P is not full dimensional, then at least one of the inequalities is satisfied at equality by all points of P. Notations : Let M = {1, 2 m}, = {i M / = for all x P} and = {i M / for some x P} = M  Let ( ), ( ) be the corresponding rows of (A, b). We write the equality and inequality sets of the representation (A, b) of P, as P = {x /, }. Note 2.11: If i then obviously ( ) can not be written as a linear combination of the rows of ( ). 6
7 Definition 2.12: (Inner Point) x P is called an inner point of P if for all i. Definition 2.13: (Interior Point) x P is called an interior point of P if for all i M. Proposition 2.14: Every non empty polyhedron P has an inner point. Proof: If = {i M / for some x P} =, then we say every point of P is inner otherwise, for each i there exist a point P with. Then = (P is convex) Since for all i, is an inner point. Note : Henceforth, we will always assume that P. Following proposition relate the dimension of P to the rank of its equality matrix ( ). Proposition 2.15: If P, then dim(p) + rank( ) = n. Proof: Suppose rank ( ) = rank ( ) = n  k where 0 k n. Hence by proposition 1.6 there are k + 1 affinely independent solutions of x = 0. Let be any such solutions and let be an inner point of P. Now, for sufficiently small, + for i = 1,, k+1 are affinity independent points in P. Thus dim(p) k and dim(p) + rank (A =, ffgb = ) n. Now, suppose that dim(p) = k and that x 1,, x k+1 are affinely independent points of P. Since for j = 1,, k+1, by proposition 1.6 we have rank (A =, b = ) (n + 1)  (k + 1) = n  k dim(p) + rank ( ) n. Corollary 2.16: A polyhedron P is full dimensional if and only if it has an interior point. Proof: dim(p) = n rank ( ) = 0 = 7
8 = M P has an interior point (definition 2.13 ). Example 2.17: Suppose is given by 1 2 Observe that the points, lie in P and are affinely independent, since are linearly independent. Hence Further, all points of P satisfy the equality by proposition Therefore. Hence 8
9 Thus Describing Polyhedra by Facets In this section, we shall find out which of the inequalities x are necessary in the description of P and which can be dropped. Further, we shall show that a full dimensional polyhedron P has a unique minimal representation by a finite set of linear inequalities. Definition 2.18: (Valid inequality) The inequality x [or ( )] is called a valid inequality for P if it is satisfied by all points in P. Note 2.19: Clearly, ( ) is a valid inequality if and only if P lies in the half space {x / x } if and only if max { x / x P}. Definition 2.20: Let ( ) be a valid inequality for P, and let F = {x P / x = }. Then F is called a face of P. We say ( ) represents F. A face F is said to be proper if F and F P. Note 2.21: It is obvious that the face F represented by ( ) is non empty if and only if max { x / x P} =. Definition 2.22: When a face of a valid inequality ( ) is nonempty, we say ( ) supports P. 9
10 Remark 2.23: As a first step in discarding superfluous inequalities, note that we can discard inequalities x that are not supports of P. Hence from now we suppose that all the inequalities x for i M support P and therefore represent nonempty faces. Proposition 2.24: If P = {x / Ax b} with equality set M, and F is a nonempty face of P, then F is a polyhedron and F = {x / x = for i, x for i }, where M = and = M . The number of distinct faces of P is finite. Proof: Suppose F is the set of optimal solutions to the linear program = max { x / Ax b}. Let u * be an optimal solution to the dual linear program min {ub / ua =, u }, and let I * = {i M / }. Note that by complementary slackness ( x ) = 0 for all i. Now consider the polyhedron F * = { x / = for i I *, for i M I * }. We claim that F = F *. If x F *, then x = u * Ax = = = x F i.e., F * F Now, if x P F *, then for some k I *, so. Then we have x = < = [since ] x F This gives F F *. Thus, F = F * and F is a polyhedron. Since F P, the equality set of F must have the required property:. 10
11 Next, since M is finite, the possible equality subsets [corresponding to the rows of ( )] are finite in number, so the number of distinct faces is finite. Note 2.25: By the proposition 2.15, dim(p) + rank (, ) = n. As, if F is a proper face then dim(f) < dim(p). Definition 2.26: (Facet) A face F of P is a facet of P if dim(f) = dim(p)  1. Proposition 2.27: If F is a facet of P, there exists some inequality for k representating F. Proof: We have dim(f) = dim(p)  1, and by proposition 2.15 we have dim(p) + rank(, ) = n and dim(f) + rank( ) = n This gives rank ( ) = rank(, ) + 1 Hence some constraint from must be representing F. Example 2.28 (See example 2.17): Suppose is given by 1 2 As max{ / x P} = 1, ( ) = (11 1, 1) is a valid inequality for P. Further, = {x P / = 1} = {(0, 0, 1)} is a face of P. Observe that is not generated by any of the inequalities in the description of P. Now consider the face generated by the valid inequality i.e., = {x P / = 2} 11
12 The two points (1, 0, 0) and (0, 0, 1) lie in and are affinely independent. Next, the point (0, 1, 0) P does not lie on. Therefore P. Since dim P = 2 and dim F 1, dim = 1. Thus is a facet of P. Proposition 2.27 implies that one of the initial inequalities must represent. In fact, both + 1 and  0 represent the facet. Finally consider ( ) = (0 0 1, 2). Now max { / x P} = 1 <, so 2 is a valid inequality but not a support of P. Hence 2 can be discarded from the description of P. Proposition 2.29: For each facet F of P, one of the inequalities representing F is necessary in the description of P. Proof: Let be the polyhedron obtained from P by removing all the inequalities representing F. We shall show that P. Let be an inner point of the facet F and let be an inequality representing F. By note 2.11 can not be written as a linear combination of the rows of. i.e., is linearly independent of the rows of. From the farkas lemma, there exists y such that y = 0 and y. Because is an inner point of F, for all inequalities i that do not represent F. But now + y P for sufficiently small > 0 ( y = 0). Now, we shall show that the facets are sufficient for the description of P. Proposition 2.30: Every inequality for r that represents face of P of dimension less than dim(p)  1 is irrelevant to the description of P. Proof: Suppose represents a face F of P with dim(f) = dim(p)  k, k > 1 and the inequality is not irrelevant. Then there exists such that, i  {r} and. Let be an inner point of P. Consider a convex combination z = + (1  ) F, (0, 1). There is a z such that, and i  {r} and 12
13 Hence the equality set of F is (, ) and (, ), which is of rank n dim(p) + 1. Therefore the dimension of F is dim(p) 1, which is a contradiction. Example 2.31: Consider a polyhedron in example (2.28) and consider the face = {x / + = 1, x P}. The equality set of is ( ) = The rank of the matrix is 2. Hence, by proposition 2.15 dim(p) = n  rank( ) = 31 = 2 and hence dim( ) = dim (P) 1 = 2 1 = 1. Thus is a facet represented either by + 1 or  0. In fact, since = 2 ( +  1)  7, is also represented by , which is the representation we gave earlier. Now, consider = { x P / = 0}. The equality set of is ( ) = ( ) whose rank is 1. Therefore again dim is 1 and is a facet. Now, consider a support of P and let = {x P / = 2 } The equality set is ( ) = which has rank 3. Hence dim = 0. In fact = = {(0, 0, 1)}, and hence is redundant. Therefore a minimal description of P is given by + + = = 10 OR
14 x x Remark 2.32: The example raises the question as to when two inequalities (e.g. + 1,  0) are equivalent. This happens if {x / x =, x } = {x / x =, ( + u ) x + u } for all > 0 and u. Definition 2.33: (Equivalent inequalities) We say the inequalities ( ) and ( ) are equivalent, or identical inequalities with respect to P if (, ) = ( ) + u (, ) for some > 0 and u. From the definition of valid inequality to remark (2.32) gives following result. Theorem 2.34 : a. A full dimensional polyhedral P has a unique (to within scalar multiplication) minimal representation by a finite set of linear inequalities. In particular, for each facet of P there is an inequality x (unique to within scalar multiplication) representing and P = { x / x for i = 1,, t} b. If dim(p) = n k with k > 0 then P = { x / x = for i = 1,, k, x, i = k + 1,, k +t}. For i = 1,, k, (a i, ) are a maximal set of linearly independent rows of ( ) and for i = k + 1,, k + t, (a i, ) is any inequality from the equivalence class of inequalities representing the facet. Following theorem characterizes facets and is useful in establishing when a valid inequality is a facet. Theorem 2.35 Let ( ) be the equality set of P and let F = {x P / πx = } be a proper face of P. The following two statements are equivalent: 1. F is a facet of P. 2. If x = for all x F then (, ) = ( π + u, + u ) for some and some u. Proof : [ (2) (1) ] Let 14
15 L = {(, ) = ( π + u, + u ) for some and some u.} and let = {(, ) / x = for all x F}. Since πx + u x = + u for all x F L By hypothesis, L, Hence L =. Suppose that dim(p) = n  k so that by proposition 2.15 we know dim(p) = n rank ( ) this implies rank ( ) = k. Since F is a proper face, (π, ) is not a linear combination of the rows of ( ). Hence (π, ) and k linearly independent rows of ( ) are k + 1 linearly independent Points in L which is closed under scalar multiplication and vector addition. Thus L is a (k + 1) dimensional subspace. Now let,, be a maximal set of affinely independent points in F and let D = be an r ( n + 1 ) matrix. Clearly r n k. Since =  = 0 (, ) = 0. By proposition 1.6, the maximum number of affinely independent solutions of (, ) = 0 is (n + 1) + 1 rank(d) = n + 2 r. Thus is an ( n + 1 r ) dimensional subspace. Since L =, (n + 1) r = k + 1 n k = r. Hence F is a facet of P. [ (1) (2) ] As above, L. Here we need to show L =. Suppose dim(p) = n k, since F is a facet of P, F contains n  k affinely independent points. 15
16 Thus, as in the proof of above (2) (1), dim( ) = k + 1. Since dim(l) = k + 1 and L, L =. Describing polyhedra by extreme points and extreme rays: Now, we give description of polyhedra by extreme points (lowest dimensional faces) & extreme rays (Minkowski s theorem). Proposition 2.36: If P = { and, then P has a face of dimension k and has no proper face of lower dimension. Proof: For any face of P, We have as in proposition 2.15, Now, let F be a face of P of minimum dimension. If there is nothing to prove. So, suppose Let be an inner point of F. Since there exists some other point y of F. Consider the line joining and y, that is some Suppose that the line intersects for some Let lies in and Then because is an inner point. Thus = is a face of P of smaller dimension F, which is a contradiction. Therefore the line does not intersect for any But this means that Since Thus Since proposition 2.15 implies that 16
17 Example 2.37: Let We have By above proposition, gives Thus a face of minimum dimension is the one  dimensional face Note 2.38: If we consider a polyhedra lying within the nonnegative orthant such that By proposition 2.36, these polyhedra have zero dimensional faces. For this reason and for simplicity, henceforth we assume that Note also that if is a polytope, then Proposition 2.39: x is an extreme point of P if and only if x is a zero dimensional face of P. 17
18 Proof: Suppose x is a zero dimensional face of P. by proposition 2.15,. Let be a submatrix of with and nonsingular, so If, then since for Hence so x is an extreme point. If is not a zero dimensional face of P, then again by proposition 2.15 we have rank But now there exist satisfying and for sufficiently small and Now so x is not an extreme point. Proposition 2.40: If, r is an extreme ray of P if and only if is a one dimensional face of. Proof: Let. Let be a one dimensional face of. By proposition 2.15, By rank nullity theorem, has dimension 1. Hence all solutions of are of the form If we obtain a contradiction as in the proposition (2.39). If and ( i.e., nullity ), there exist such that The rays show that r is not an extreme ray. Corollary 2.41: A polyhedron has a finite number of extreme points and extreme rays. Example 2.42 (See example 2.17): The inequalities describing P include,  where matrix is which has The face has the equality set. 18
19 = 0 Note that the negative of the first equation is omitted. Since is of rank 3. Note that is a zero dimensional face, or extreme point. satisfies and for all other constraints. Since is of rank 2, is an extreme ray. A similar argument shows that ray. is another extreme 19
20 Remark 2.43: A polyhedron can be represented in terms of its extreme points and extreme rays. Below, we shall prove this fundamental result known as Minkowski s theorem. Theorem 2.44: If, and is finite, then there is an optimal solution that is an extreme point. Proof: The set of optimal solution is a nonempty face By proposition 2.36, F contains an dimensional face. By proposition 2.39, F contains an extreme point. Theorem 2.45: For every extreme point of P, there exist such that is the unique optimal solution of Proof: Let be the equality set of Let Since the are rational vectors, there exist such that c. Since is a zero dimensional face of P, for all there exists an such that Hence for, cx = = = c Theorem 2.46: If and max is unbounded then P has an extreme ray with Proof: Since max is unbounded, by linear programming duality, the set. By Farkas lemma, there exists an such that and Now consider the linear program max. By theorem 2.44, this linear program has an optional extreme point solution. An optimal extreme point is a point such that the equality set is of rank and Now by proposition 2.40, is an extreme ray of P. Following is the one of the fundamental result on the representation of polyhedra. 20
21 Theorem 2.47 (Minkowski s theorem): If then P =, for k K, j} where is the set of extreme points of P and is the set of extreme rays of P. Proof: Let Q = 0, for k K, j} Since and P is convex, where Also since are rays. Hence Now suppose that so there exist In other words there do not exist satisfying = y  = 1 Then by Farkas lemma, there exists such that and Now consider the linear program max. If it has a finite optimal value, by theorem 2.44 the optimal value is obtained at an extreme point. However, and for all extreme points This gives a contradiction. On the other hand, if the linear program has an unbounded optimum, by theorem 2.46, then exists an extreme ray with  Again this is a contradiction. Hence 21
22 Example 2.48 (See example 2.17): Note that P has one extreme point and two extreme rays Thus P can be described as P = Note 2.49: We recall the basic results for the dual pair of linear programs with (primal) and (dual) Notation: Let and be the sets of extremes points of P and Q, respectively, and let and be the sets of extreme rays of and. respectively. Following theorem is partially a repeat of theorems Theorem 2.50: 1. The following are equivalent: a. The primal problem is feasible, that is b. 2. The following are equivalent when the primal problem is feasible: a. z is unbounded from above: b. There exists an extreme ray of P with c. The dual problem is feasible, that is 3. If the primal problem is feasible and z is bounded, then Proof: 1.By Farkas lemma, if and only if for all with By Minkowski s theorem, = Hence for all if and only if 2. Again by Minkowski s theorem, as the primal is feasible. P =, 22
23 for k K, j}. By proposition 2.46, z is bounded if and only if The equivalence of (b) and (c) is the equivalence in (1) to the dual problem. 3. This follows from strong duality & Minkowski s theorem applied to P and Q. Now we consider the projection of a polyhedron. Note first that projection of a point onto the subspace is the point Therefore it is natural to consider a projection of a polyhedron onto as a projection from the space to the xspace, denoted by Combining Minkowski s theorem and linear programing duality leads to a characterization of certain projections of polyhedra. It also leads to an important converse to Minkowski s theorem, which says that every set obtained as a convex combination of a finite set of vectors in plus a non negative combination of some other finite set of vectors in is a polyhedron. Theorem 2.51: Let Then where are the extreme rays of. Proof: If then Now applying statement of theorem 2.50, to gives Corollary: The projection of a polyhedron is a polyhedron. Remark 2.52: Given two polyhedra following corollary answers the question when Corollary 2.53: If and where D is then if and only if (i) For is a valid inequality for P. (ii) For each there exist a such that Proof: (i) is equivalent to and 23
24 (ii) is equivalent to Following is the converse of Minkowski s theorem: Theorem 2.54 If A is a rational matrix matrix, B is a rational matrix and Q = Then Q is a rational polyhedron. Proof: Observe that when. 24
25 Chapter 3 Polarity Here we consider a polyhedron whose feasible points are the valid inequalities of P. We will characterize the facets of P in terms of the extreme rays of and establish a duality between P and Definition 3.1: is called the polar of the polyhedron Note 3.2: Observe that if and only if is a valid inequality for P. For simplicity, assume that Proposition 3.3: Given a nonempty polyhedron with is a polyhedron cone described by where are the extreme points and extreme rays of P. Proof: Let } Suppose. By Minkowski s theorem, for any any and all Therefore, If 0 then can be increased so that the inequality becomes invalid. Therefore Hence Thus i.e., Conversely, if then by theorem 2.47, (Minkowski s theorem) for some satisfying Hence x = = Therefore So 25
26 Example 3.4 For the given polyhedra, the extreme point is extreme rays are Hence polyhedral description is The main result on polarity is the following. Theorem 3.5: If then is an extreme ray of if and only if defines facet of P. Proof: By Proposition 2.15 is an extreme ray of if and only if its extreme set is of Using the description of from proposition 3.3, this means there exists such that and and is of rank n. [Note that since would imply This implies that the vectors are linearly independent. Hence by proposition 1.4,,,,,, are affinely independent. Therefore defines a facet of P. 26
27 Conversely if defines a facet of P, there exists n affinely independent points of P with for But now considering the polyhedral cone the equality set of includes (, 1), (, 1) and hence is of rank at least n. If the equality set is of rank then by proposition 2.15 dim facet will be 0, so Hence the rank of the equality set of is n. So is one dimensional face of By proposition 2.40, is an extreme ray of A dual of above theorem is Theorem 3.6: Let. Then 1. defines a facet of if and only if is an extreme point of P. 2. defines a facet of if and only if is an extreme ray of P. Proof: By proposition 3.3, every facet of is either of the required form or To show that each of these inequalities defines a facet. Recall that is an extreme point then its equality set is of rank n.. Hence there exist such that are linearly independent, and Now these n vectors plus are affinely independent, and hence defines a facet. A similar argument shows that defines a facet. Remark 3.7: Suppose P is a fulldimensional polytope. By translation we can take the origin 0 to be an interior point. So if is an equality describing P then Hence we can write 27
28 where Then 1polar of P is Similarly, every valid inequality can also be written. By theorem 2.46 P = for k K} and by proposition 3.3 Proposition 3.8: If is a full dimensional polytope, then is a full dimensional polytope and P is the 1polar of. Proof: Since 0 is an interior point of. Now by corollary 2.16 is full dimensional. Suppose has a ray. So that This implies that is a valid inequality for P, which is contradiction. Hence is bounded. Now consider. If then for all and hence Suppose Then there exists no soluation to y = 1 =  So there exists a such that Since 0 we can again normalize so that Then which is a contradiction. Hence Note 3.9: Following theorem shows that there is complete symmetry between P and. Theorem 3.10: Let P be full dimensional and bounded, and let 0 be an interior point of P. Then a. P = { for t T, where are the extreme points of }, and b. = { for k K, where are the extreme points of P}. 28
29 Moreover, each of the inequalities in descriptions a & b define facets. Proof: (b) is proved in remark 3.7 and (a) is proved in above proposition 3.8. Corrollary 3.11: If P is described in theorem 3.8 Then if and only if max{ and if and only if max{ Proof: if and only if 1 for all if and only if max{.the second statement is the dual. Note 3.12: Above corollary gives equivalence between separation & optimization. Example 3.13: (P) ( )
30 Chapter 4 Polyhdral Ties Between Linear and Integer Programs In this chapter, we will show that that an integer program can, in theory, be reduced to a linear program. Given is an integer matrix, and. We shall show that conv(s) is a rational polyhedron. Further whenever P is bounded, S is either empty or a finite set of points. So by theorem 2.54, conv(s) is a rational polyhedron. When S contains an infinite number of points, we will show that conv(s) can be generated from a finite number of points in S and a finite number of integral valued rays. The idea of the proof is shown in following figure. Geometrically, we see that conv(s) is the polyhedron generated by convex combinations of the points {(1,2), (2,1) (4,0)} plus non negative linear combinations of the rays, which are the extreme rays of P. The important step in the proof is to show that the set of integer points in a polyhedron can be finitely generated. We will give a finite set ( in figure above the integral points in the shaded region of P) and then show that S can be generated by taking a point in Q plus a nonnegative integer linear combination of the extreme rays of P. 30
31 Theorem 4.1: Let P = and S =. (A, b) is an integer matrix. Then (1) There exist a finite set of points of S and a finite set of rays of P such that S = (2) If P is a cone ( b = 0), there exist a finite set of rays of P such that S = Proof : (1) let be the finite set of extreme points of P and let be the finite set of extreme rays of P. Since P is a rational polyhedron, all of these extremal vectors have rational coordinates. We have, by Minkowski s theorem P = Without loss of generality, we can assume that are integer vectors (denominations of Q = ). Let It is a finite set, since and are finite in number. Let. Clearly. Observe that if and only if.we can write = { + } +{ } 0 for k K and j J The first term is a point of Q, so there exist Hence the result. such that ( ) (2) If P is a cone, then implies. Therefore it suffices to take, as in part (1), 31
32 =. Now, we prove that conv(s) is a rational polyhedron. Theorem 4.2: Let P = { where (A, b) be an integer matrix and let. Then conv(s) is a rational polyhedron. Proof: Observe that any point can be written in the form ( ). Hence any convex combination of points { can be written as x = = = + = where for and = Hence Conv(S) = { for and. Where for and. Therefore by theorem 2.54(weyl s theorem) is a rational polyhedron. Remark 4.3: It is clear that the above proof can be extended to mixed integer sets with rational data. Therefore all of the following results apply to mixed integer sets and mixed integer programs. Note 4.4: The above proof also shows that if then the extreme rays of { and conv( coincides. Remark 4.5: The theorem also suggests that we can solve the integer program where S = (IP) by solving linear program max { (CIP) Following results shows this. Theorem 4.6: Suppose S =, and any. Then 32
33 (3) The objective value of IP is unbounded from above if and only if the objective value of CIP is unbounded from above. (4) If CIP has a bounded optimal value, then it has an optimal solution (namely, an extreme point of conv(s) ) that is an optimal solution to IP. (5) If is an optimal solution to IP then is an optimal solution to CIP. Proof: Let and be the optimal values of IP and CIP, respectively. If one of the objective is unbounded from above then or As conv(s) a. The inequality implies that if then =. On the otherhand if =, then there is an integral extreme point and a ray such that and for all But then for all which implies that b. Since conv(s) is a polyhedron. If CIP has an optimal solution, then it has an extreme point optimal solution say. Thus so. Therefore. c. This follows from points (a) and (b) as. Clearly, above theorem implies following corollary. Corollary 4.7: IP is either infeasible or unbounded or has an optimal solution. Remark 4.8: Therefore 4.6 states that we can solve the integer program IP by solving the linear program CIP. This is easy if we know polyhedron representation of conv(s) in terms of linear inequalities. But generally it is not known. Hence we formulate integer program using some polyhedron P such that S =. Definition 4.9: we say that is a valid inequality for a set S if for all. Proposition 4.10: If is valid for S, it is also valid for. Proof: Consider an. Then where and and Hence = 33
34 = Below we show that to know the dimensionality of a facet of, it suffices to consider points of S. Theorem 4.11: If defines a face of dimension k 1 of then there are k affinely independent points such that for Proof: By definition, there are k affinely independent points such that for If for, there is nothing to prove. Suppose. Then = where and for all and Now and for imply that for. Since are affinely independent, there exists such that are affinely independent. We repeat this process if otherwise check whether Consider the problem max (LP) We know how to relate IP to CIP. Now we relate IP to LP. Let z(d) = msax and so that and z(b) = max { cx / x with. Proposition 4.12: a. 34
35 b. if and only if c. if and only if d. If then or z(b) is finite. e. If, then or z(b) = Proof: Clearly If then by duality and hence If then from theorem 2.50 there exists an extreme ray of P with c Since can be taken to be integer This proves statements a, b and c. If then it follows, by duality that or is finit. Hence statements d follows. Similarly if then it follows by duality that or If then from statement c it follows that Hence statement e follows. Corollary 4.13: a. If then. b. If is finite then or z(b) is finite. c. If then or z(b) =. 35
36 REFERENCE Integer and Combinational Optimization; Nemhauser G. L., Wolsey L. A., A Wiley Interscience publication, John wiley & sons Inc.,
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