Outline. Combinatorial Optimization 2. Finite Systems of Linear Inequalities. Finite Systems of Linear Inequalities. Theorem (Weyl s theorem :)
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1 Outline Combinatorial Optimization 2 Rumen Andonov Irisa/Symbiose and University of Rennes 1 9 novembre 2009 Finite Systems of Linear Inequalities, variants of Farkas Lemma Duality theory in Linear Programming Application 1 : The shortest path problem (SPP) and its dual The primal simplex algorithm briefly The dual simplex algorithm briefly Application 2 : Max-Flow Min-Cut (MFMC) Total Unimodularity Rumen Andonov (Irisa) Combinatorial Optimization 2 9 novembre / 33 Rumen Andonov (Irisa) Combinatorial Optimization 2 9 novembre / 33 Finite Systems of Linear Inequalities Finite Systems of Linear Inequalities We assume : A R m n,b R m 1,c R 1 n,x R n 1,y R 1 m A polytope is conv(x) for some finite set X R n. A polytope can also be described as the solution set of a finite system of linear inequalities, i.e. : (Weyl s theorem :) If P is a polytope, then P = {X R n : for some m Z + and a ij,b i R. n j=1 a ij x j b i,for i = 1,...,m} We assume : A R m n,b R m 1,c R 1 n,x R n 1,y R 1 m A polytope is conv(x) for some finite set X R n. A polytope can also be described as the solution set of a finite system of linear inequalities, i.e. : (Weyl s theorem :) If P is a polytope, then P = {X R n : for some m Z + and a ij,b i R. n j=1 a ij x j b i,for i = 1,...,m} (of the Alternative for Linear Inequalities :) {x R n : Ax b} /0 iff {y R m + : ya = 0, yb < 0} = /0 Rumen Andonov (Irisa) Combinatorial Optimization 2 9 novembre / 33 Rumen Andonov (Irisa) Combinatorial Optimization 2 9 novembre / 33
2 Finite Systems of Linear Inequalities (Farkas Lemma ) An equivalent result is the following : given A and b, (Farkas Lemma :) Finite Systems of Linear Inequalities (Farkas Lemma ) An equivalent result is the following : given A and b, (Farkas Lemma :) b) Either {x R n + : Ax = b} /0 or (excl.) { y R m : ya 0, yb < 0} /0 b) Either {x R n + : Ax = b} /0 or (excl.) { y R m : ya 0, yb < 0} /0 Either b cone{a 1,...,A n } (where A i are the columns of A), or a vector y that makes an acute angle with A i and an obtuse angle with b, but not both. A b A b A 1 A y Rumen Andonov (Irisa) Combinatorial Optimization 2 9 novembre / 33 Rumen Andonov (Irisa) Combinatorial Optimization 2 9 novembre / 33 Variants of Farkas Lemma Variants of Farkas Lemma : b) Either {x R+ n : Ax = b} /0 or (excl.) { y R m : ya 0, yb < 0} /0 a) Either {x R+ n : Ax b} /0 or (excl.) { y R+ m : ya 0, yb < 0} /0 c) Either {x R n : Ax b} /0 or (excl.) { y R+ m : ya = 0, yb < 0} /0 d) If P = {r R+ n : Ar = 0}, either P\{0} /0, or {y R m : ya > 0} /0 Duality theory in Linear Programming Let A R m n,b R m 1,c R 1 n, a i is the i-th row and A j is the j-th col. of A. To any primal (P) is associated its dual (D) by associating a dual variable to each primal constraint and following the rules : Primal Dual max min variables x constraints constraints variables y objective coefficients c constraint right hand sides c constraint right hand sides b objective coefficients b a i x b i y i 0 a i x b i y i 0 a i x = b i y i unconstrained x j 0 ya j c j x j 0 ya j c j x j unconstrained ya j = c j Rumen Andonov (Irisa) Combinatorial Optimization 2 9 novembre / 33 Rumen Andonov (Irisa) Combinatorial Optimization 2 9 novembre / 33
3 Duality theory : Canonical Forms Duality theory : Primal/Dual relationship (Illustration) Let be given a primal (P) in a canonical form (i.e.) A primal problem and its dual Its dual (D) is as follows. (P) z = max{cx Ax b,x R n,x 0} (1) (D) w = min{yb ya c,y R m,y 0} (2) maxz = 6x 1 + x 2 + x 3 4x 1 + 3x 2 2x 3 = 1 6x 1 2x 2 + 9x 3 9 2x 1 + 3x 2 + 8x 3 5 x 1 0,x 2 0,x 3 unrestricted minw = y 1 + 9y 2 + 5y 3 4y 1 + 6y 2 + 2y 3 6 3y 1 2y 2 + 3y 3 1 2y 1 + 9y 2 + 8y 3 = 1 y 1 unrestricted,y 2 0,y 3 0 In general, the following relationships hold : primal/dual constraint dual/primal constraint (3) consistent with canonical form variable 0 (4) reversed from canonical form variable 0 (5) eqaulity constraint variable unrestricted (6) Lemma : It is easy to verify that the dual of the dual is the primal. Rumen Andonov (Irisa) Combinatorial Optimization 2 9 novembre / 33 Rumen Andonov (Irisa) Combinatorial Optimization 2 9 novembre / 33 Duality theory in Linear Programming (cont.) Consider the following couple primal-dual (it just just a matter of convenience. (D) is given in the so called standard form to which any LP form can be transformed). (P) z = max{cx Ax b,x R n } (7) (D) w = min{yb ya = c,y 0,y R m } (8) We suppose rank(a) = m n (all redundant equations were removed). Weak Duality : If x is feasible to P and y is feasible to D, then cx yb. Hence 1 if cx = yb, then x and y are optimal 2 if either is unbounded, then the other is infeasible. Proof. Trivial (cx = yax yb). Rumen Andonov (Irisa) Combinatorial Optimization 2 9 novembre / 33 Duality theory in Linear Programming (cont.) Strong Duality : If P and D have feasible solutions, then they have optimal solutions x,y with cx = yb. If either is infeasible, then the other is either infeasible or unbounded. Proof. Can be proved using one of the variant of Farkas lemma. Here we will discuss its geometrical insight only. Remark : The of the Alternative for Linear Inequalities can be proven from the Strong Duality. Hint : Consider the program (D0) min{yb : ya = 0,y R m +} (9) First argue that either the optimal objective value for D0 is zero or D0 is unbounded. Rumen Andonov (Irisa) Combinatorial Optimization 2 9 novembre / 33
4 Duality theory in Linear Programming (cont.) Geometrically : Suppose a finite maximum δ (i.e.) δ = cx = max{cx Ax b}. Visualizing Duality c cx*= δ a1 a2 x* a1x*= β P a2x*=β2 Set A = (a 1,a 2 ) t, b = (β 1,β 2 ) and let a 1 x = β 1,a 2 x = β 2. Then λ 1,λ 2 0 s.t. c = λ 1 a 1 + λ 2 a 2. Hence δ = λ 1 β 1 + λ 2 β 2. This implies : Question (How to find the shortest path from S to T?) max{cx Ax b} = δ = λ 1 β 1 + λ 2 β 2 min{yb ya = c,y 0} (10) (since the λ i give a feasible solution for the minimum). On the other hand, from the weak duality we have max{cx Ax b} min{yb ya = c,y 0}. Hence max{cx Ax b} = min{yb ya = c,y 0}. Rumen Andonov (Irisa) Combinatorial Optimization 2 9 novembre / 33 Rumen Andonov (Irisa) Combinatorial Optimization 2 9 novembre / 33 Visualizing Duality Visualizing Duality Question (How to find the shortest path from S to T?) Question (How to find the shortest path from S to T?) Answer (Just pull S away from T until the gadget is taut.) Answer (Just pull S away from T until the gadget is taut.) Why does it work? Rumen Andonov (Irisa) Combinatorial Optimization 2 9 novembre / 33 Rumen Andonov (Irisa) Combinatorial Optimization 2 9 novembre / 33
5 Visualizing Duality Visualizing Duality Question (How to find the shortest path from S to T?) Question (How to find the shortest path from S to T?) Answer (Just pull S away from T until the gadget is taut.) Answer (Just pull S away from T until the gadget is taut.) Why does it work? Because by pulling S away from T we solve the dual of the shortest-path problem! Why does it work? Because by pulling S away from T we solve the dual of the shortest-path problem! Exercice (Prove the above formally.) Rumen Andonov (Irisa) Combinatorial Optimization 2 9 novembre / 33 Rumen Andonov (Irisa) Combinatorial Optimization 2 9 novembre / 33 Application 1 : The shortest path problem and its dual The shortest path problem (SPP) and its dual Finding the shortest path from S to T is equivalent to pulling S away from T as much as possible. We will prove it by writing the primal and the dual problem of the above modeling. Rumen Andonov (Irisa) Combinatorial Optimization 2 9 novembre / 33 Rumen Andonov (Irisa) Combinatorial Optimization 2 9 novembre / 33
6 The shortest path problem (SPP) SPP : Primal-Dual LP formulation Given a directed graph G = (V,U) and weights w j 0 associated with each arc u j, find a path from s to t with the minimum total weight. s u1 u2 [1] [2] u3 [2] a b u4 [3] u5 [1] t A = s t a b A weighted directed graph and the corresponding node-arc incidence matrix A. The weights are in squared brackets. A path from s to t can be thought of as a flow φ of unit 1 leaving s and entering t. The primal (P) of (SPP) and its dual (D) where a variable π i is assigned to each node i, are given below. Aφ = (P) min wφ (11) +1 row s 1 row t 0 otherwise φ 0 (12) (D) max π s π t (13) π i π j w ij (ij) U (14) π i unrestricted i V (15) Rumen Andonov (Irisa) Combinatorial Optimization 2 9 novembre / 33 Rumen Andonov (Irisa) Combinatorial Optimization 2 9 novembre / 33 Complementary-Slackness in Duality theory Complementary-Slackness in Duality theory : II Let us denote by a i the i-the raw of the matrix A and by A j its j-the column. Points x R n and y R m are complementary, with respect to P and D, if y i (b i a i x) = 0, for i = 1,...m (16) Weak Complementary-Slackness If feasible solutions x and y are complementary, then they are optimal solutions. Proof. 0 = m i=1 y i (b i a i x) = m i=1 y i b i n m j=1 i=1 y i a ij x j = yb cx If x and y are complementary and feasible, then yb = cx and by the Weak Duality they are optimal. Let us denote by a i the i-the raw of the matrix A and by A j its j-the column. Points x R n and y R m are complementary, with respect to P and D, if y i (b i a i x) = 0, for i = 1,...m (17) Strong Complementary-Slackness If feasible solutions x and y are optimal solutions to P and D, respectively, then they are complementary. Note : Similar considerations hold for the other LP duality equation, i.e. x j (c j ya j ) = 0, for j = 1,...n (18) Rumen Andonov (Irisa) Combinatorial Optimization 2 9 novembre / 33 Rumen Andonov (Irisa) Combinatorial Optimization 2 9 novembre / 33
7 Complementary-Slackness in Duality theory : III Variant : A pair (x, y) feasible in P and D respectively, is optimal iff. y i (b i a i x) = 0, for i = 1,...m x j (c j ya j ) = 0, for j = 1,...n (19) Geometrically : Suppose δ = max{cx Ax b}. As we saw then λ 1,λ 2 0 s.t. c = λ 1 a 1 + λ 2 a 2, where a 1 x = β 1,a 2 x = β 2. Note that if ax < β for some inequality in Ax b, then the corresponding component in y is zero. cx*= δ a a1x*= β P c x* a2 a2x*=β2 The primal simplex algorithm briefly Given the following linear program in standard form : Definition (P) z = max{cx Ax = b,x 0,x R n } (20) A basis B is a m m non-singular submatrix of A. Then A = (B,N). Let x = (x B,x N ) and Bx B + Nx N = b. x B + B 1 Nx N = B 1 b = b. x B = B 1 b and x N = 0 is called a basic solution of P. If B 1 b 0, then (x B,x N ) is called a basic primal feasible solution and B is called a primal feasible basis. A point x is an extreme point of the set {x Ax = b,x 0} iff it is a basic feasible solution. ( Fundamental theorem of linear programming) If (P) has a finite optimum then it has an optimal basic feasible solution. Rumen Andonov (Irisa) Combinatorial Optimization 2 9 novembre / 33 Rumen Andonov (Irisa) Combinatorial Optimization 2 9 novembre / 33 The dual simplex algorithm The following is the dual of problem (P) (D) w = min{yb ya c,y R m } (21) Let c = (c B,c N ) and z = c B x B + c N x N = c B b+(c N c B B 1 N)x N. Set y = c B B 1 R m. y is complementary to (x B,x N ) since ya c = c B B 1 (B,N) (c B,c N ) = (0,c B B 1 N c N ) and x N = 0. Max-Flow Min-Cut (MFMC) Definition If c B B 1 N c N, then B is called a dual feasible basis. If B is primal and dual feasible then x = (x B,x N ) = (B 1 b,0) is an optimal solution to P and y = c B B 1 is an optimal solution to D. Rumen Andonov (Irisa) Combinatorial Optimization 2 9 novembre / 33 Rumen Andonov (Irisa) Combinatorial Optimization 2 9 novembre / 33
8 Application 2 : Max-Flow Min-Cut (MFMC) Application 2 : Max-Flow Min-Cut (MFMC) Let N = (s,t,v,e,c) be a flow network with n = V nodes, m = E arcs, and arc-capacity c(i,j). Let the flow in arc i,j be denoted by φ i,j. Flow conservation holds for any node except s,t where e Γ + (s) φ e = e Γ (t) φ e = φ 0 (flow value) (22) Find a flow φ = [φ 1,φ 2,...φ m ] s.t. 0 φ e c e and maximizing φ 0. Let N = (s,t,v,e,c) be a flow network with n = V nodes, m = E arcs, and arc-capacity c(i,j). Let the flow in arc i,j be denoted by φ i,j. Flow conservation holds for any node except s,t where e Γ + (s) φ e = e Γ (t) φ e = φ 0 (flow value) (22) Find a flow φ = [φ 1,φ 2,...φ m ] s.t. 0 φ e c e and maximizing φ 0. Modeling MFMC as a LP problem The node-arc incidence matrix A = [a i,j ] i = 1,2,..., V +1 if arc e j leaves node i and j = 1,2,..., E is defined by a i,j = 1 if arc e j enters node i 0 otherwise Rumen Andonov (Irisa) Combinatorial Optimization 2 9 novembre / 33 Rumen Andonov (Irisa) Combinatorial Optimization 2 9 novembre / 33 MFMC : LP formulation MFMC : Primal-Dual LP formulation Let (as usual) a i denote the ith row of A. The flow conservation at a node i (except s,t) is expressed by : The LP formulation of the MFMC is : let d R n be defined by 1 i = s d i = +1 i = t 0 otherwise a i φ = 0 (23) maxφ 0 (24) dφ 0 + Aφ = 0 (25) φ c (26) φ 0 (27) φ 0 0 (28) Assign variables λ i,i V to (30) and variables γ i,j,(i,j) E to (31). Then we obtain : (P) maxφ 0 (29) dφ 0 + Aφ = 0 (30) φ c (31) φ 0 (32) φ 0 0 (33) (D) min γ i,j c i,j (34) (i,j) E λ i λ j + γ i,j 0 (i,j) E (35) λ s + λ t 1 (36) γ i,j 0 (i,j) E (37) Rumen Andonov (Irisa) Combinatorial Optimization 2 9 novembre / 33 Rumen Andonov (Irisa) Combinatorial Optimization 2 9 novembre / 33
9 MFMC : Max-Flow Min-Cut Relationships Max-Flow Min-Cut Definition An s t cut is a partition (A,Ā) of V, s.t. s A and t Ā. The capacity of an s t cut is C(A,Ā) = c e. e Γ + (A) Every s t cut determines a feasible solution with cost C(A,Ā) to (D) as follows : { { 0 i A, 1 (i,j) s.t. i A,j Ā λ γ i,j = i = 1 i Ā 0 otherwise The value φ 0 is no greater than the capacity C(A,Ā) of any s t cut. Furthermore, the value of the maximum flow equals the capacity of the minimum cut, and a flow φ and cut C(A,Ā) are jointly optimal iff φ i,j = 0 for (i,j) E such that i Ā and j A (38) φ i,j = c i,j for (i,j) E such that i A and j Ā (39) Proof. Trivial, use the complementary slackness. Proof : Just by checking. Note that (35) is strict inequality iff i Ā and j A. Note also that λ s + λ t = 1. Rumen Andonov (Irisa) Combinatorial Optimization 2 9 novembre / 33 Rumen Andonov (Irisa) Combinatorial Optimization 2 9 novembre / 33 Total Unimodularity When solving an LP problem always gives integer solution? Total Unimodularity Definition An integer matrix A is called totally unimodular (TUM) if the determinant of each square submatrix of A equals 0,+1, or -1. P(A) = {x Ax b,x 0,} (40) If A is TUM, then all the vertices of P(A) are integer for any integer vector b. Proof. First prove the case {x Ax = b,x 0,}. Any basic solution x = B 1 b = adj(b)b det(b). Then show that (A I) is TUM. Rumen Andonov (Irisa) Combinatorial Optimization 2 9 novembre / 33 Rumen Andonov (Irisa) Combinatorial Optimization 2 9 novembre / 33
10 Total Unimodularity (suite I) An integer matrix A with a ij = 0,+1, 1 is TUM if no more than two nonzero entries appear in any column, and if the rows of A can be partitioned into two sets I 1 and I 2 s.t. : 1 If a column has two entries of the same sign, their rows are in different sets ; 2 If a column has two entries of the different signs, their rows are in the same sets ; Proof. By induction of the size of submatrices. Let C be a submatrix of size k. For k = 1 it is true. Let k > 1. It is obvious when C has a column of all zeros or with one nonzero. Let C has two nonzero entries in every column. Then the conditions of the theorem imply : i I 1 a ij = i I 2 a ij for every column j. i.e. a linear combination of rows is zero, hence det(c) = 0. Rumen Andonov (Irisa) Combinatorial Optimization 2 9 novembre / 33 Total Unimodularity (suite II) Corollary Any LP in standard or canonical form whose constraint matrix A is either 1 The node-arc incidence matrix of a directed graph, or 2 The node-edge incidence matrix of an undirected bipartite graph, has only integer optimal vertices. This includes the LP formulation of shortest path, max-flow, the transport problem, and weighted bipartite matching. Proof. The matrices in Case 1 satisfy the condition of the theorem with I 2 = /0 ; those in Case 2 with I 1 = V and I 2 = U, where the bipartite graph is G = (V,U,E). Rumen Andonov (Irisa) Combinatorial Optimization 2 9 novembre / 33 Examen June 2009 Exercice 1 : Optimizing an anti-fire system The port company COSMAR wishes to improve its anti-fire security by buying new anti-fire systems SLIC. The company identifies 7 zones of the port to be specially protected. These zones are denoted by X on the below figure. WEST NORTH A B C D E F 1 X 2 X 3 X 4 X X X 5 6 X FIG.: The 7 zones to be protected by the company COSMAR G Optimizing an anti-fire system COSMAR intends to allocate the SLIC systems on the North and West sides of the port. A SLIC system protects the zones situated on the same row or column. For example, a SLIC installed on D covers zones (1,D) et (6,D). It is required that these anti-fire systems cannot be placed on consecutive columns, that at least one SLIC is allocated on row 1 or on column A, and that at least 60% of the SLIC are placed on the North side. Questions : 1 Formulate the corresponding linear program that allows to minimize the number of the anti-fire systems needed to cover the 7 special zones. 2 Since the cost of one SLIC is 8070 euros, COSMAR wishes to buy at most 3 of them. For the non-protected zones COSMAR plans to take out insurance against fire. This insurance costs 2000 euros for each zone on lines 1, 2, 3 and 3000 euros for the other lines. Modify the previous linear model in order to minimize the total cost. 3 Are you aware of software packages permitting to solve such problems? Rumen Andonov (Irisa) Combinatorial Optimization 2 9 novembre / 33 Rumen Andonov (Irisa) Combinatorial Optimization 2 9 novembre / 33
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