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The Simplex Algorithm 虞台文 大同大學資工所 智慧型多媒體研究室
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Content Basic Feasible Solutions The Geometry of Linear Programs Moving From Bfs to Bfs Organization of a Tableau
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Linear Programming Basic Feasible Solutions 大同大學資工所 智慧型多媒體研究室
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The Goal of the Simplex Algorithm Any LP can be converted into the standard form
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Convex Polytope defines a convex set F F may be 1.Bounded 2.Unbounded 3.Empty
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The Basic Idea of Simplex Algorithm basic feasible solution at a corner Finding optimum by moving around the corner of the convex polytope in cost descent sense
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The Basic Idea of Simplex Algorithm basic feasible solution at a corner Finding optimum by moving around the corner of the convex polytope in cost descent sense How to find an initial feasible solution? How to move from corner to corner?
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Assumption 1 Assume that A is of rank m. There is a basis Independent has an inverse.
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Basic Solution The basis solution corresponding to B is: The basis solution may be infeasible.
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Example x1x1 x2x2 x3x3 x4x4 x5x5 x6x6 x7x7
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x1x1 x2x2 x3x3 x4x4 x5x5 x6x6 x7x7 feasible
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Example x1x1 x2x2 x3x3 x4x4 x5x5 x6x6 x7x7 infeasible
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Basic Feasible Solution defines an nonempty convex set F If a basic solution is in F, it is called a basic feasible solution (bfs).
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The Existence of BFS’s F may be 1.Bounded 2.Unbounded 3.Empty If F is nonempty, at least one bfs. defines an nonempty convex set F
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The Existence of BFS’s F may be 1.Bounded 2.Unbounded 3.Empty If F is nonempty, at least one bfs. How to find it?
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Assumptions 1. A is of rank m. 2. F is nonempty. 3. c ’ x is bounded below for x F. 1. A is of rank m. 2. F is nonempty. 3. c ’ x is bounded below for x F. defines an nonempty convex set F
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Assumptions 1. A is of rank m. 2. F is nonempty. 3. c ’ x is bounded below for x F. 1. A is of rank m. 2. F is nonempty. 3. c ’ x is bounded below for x F. There exist bfs’s. defines an nonempty convex set F Ensure there is a bounded solution. Are all bfs’s the vertices of the convex polytope defined by F ?
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The Simplex Alogrithm The Geometry of Linear Programs 大同大學資工所 智慧型多媒體研究室
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Linear Subspaces of R d S R d is a subspace of R d if it is closed under vector addition and scalar multiplication. S defined below is a subspace of R d. a set of homogenous linear equations
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Dimensions
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Affine subspaces A translated linear subspace. E.g., an affine subspace a set of nonhomogenous linear equations an affine subspace
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Dimensions an affine subspace
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Subsets of R d The following subsets are not subspace or affine subspace. A line segment The first quadrant A halfspace
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Dimensions The following subsets are not subspace or affine subspace. A line segment The first quadrant A halfspace The dimension of any subset of R d is the smallest dimension of any affine subspace which contains it. Dim = 1 Dim = 2
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The Feasible Spaces of LP 1. A is of rank m. 2. F is nonempty. 3. c ’ x is bounded below for x F. 1. A is of rank m. 2. F is nonempty. 3. c ’ x is bounded below for x F. defines an nonempty convex set F
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Hyperplane/Halfspace An affine subspace of R d of dimension d 1 is called a hyperplane. A hyperplane defines two closed halfspaces: Fact: Halfspaces are convex.
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Convex Polytopes The intersection of a finite number of halfspaces, when it is bounded and nonempty is called a convex polytope, or simply a polytope. Fact: Halfspaces are convex.
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x1x1 x2x2 x3x3 Example (0, 0, 3) (1, 0, 3) (2, 0, 2) (2, 0, 0) (2, 2, 0)(0, 2, 0) (0, 1, 3) (0, 0, 0)
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More on Polytopes Geometric views of a polytope – The convex hull of a finite set of points. – The intersection of a finite number of halfspaces, when it is bounded and nonempty, i.e., defined by The algebraic view of a polytope The feasible space defined by LP (in standard form). Any relation?
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LP Polytopes Assume m columns n m columns x1x1 x2x2 xnmxnm x n m+1 xnxn A can always be in this form if rank(A)=m. LP
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LP Polytopes Assume m columns n m columns x1x1 x2x2 xnmxnm x n m+1 xnxn LP
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LP Polytopes 00 LP
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Polytopes LP m inequalities m equalities
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Polytopes LP m inequalities Slack variables
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Polytopes LP Slack variables
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Polytopes and LP Slack variables H I H
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Polytopes and LP The answer
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Polytopes and LP The answer
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Polytopes & F of LP defines a polytope defines a feasible set Are there any relations?
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Polytopes & F of LP defines a polytope defines a feasible set Some points in P are vertices. Some points in F are bfs’s. Are there any relations?
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Theorem 1 a convex polytope a feasible set of the corresponding LP is a vertex of P is a bfs.
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Theorem 1 Pf) “”“” See textbook.
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Theorem 1 Pf) “”“” Fact:is a vertex, then it cannot be the strict combination of points of P, i.e.,
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Theorem 1 Pf) “”“” Fact:is a vertex, then it cannot be the strict combination of points of P, i.e.,
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Theorem 1 Pf) “”“” Fact: Define B = {A j : x j > 0, 1 j n} Let We want to show at most m nonzero x i ’s. We want to show that vectors in B are linearly independent. Suppose not. Then, some d j 0. >0 if is sufficiently small.
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Theorem 1 Pf) “”“” Fact: Define B = {A j : x j > 0, 1 j n} Let We want to show at most m nonzero x i ’s. We want to show that vectors in B are linearly independent. Suppose not. Then, some d j 0. >0 if is sufficiently small. Define is sufficiently small.
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Theorem 1 Pf) “”“” Fact: Define B = {A j : x j > 0, 1 j n} Let We want to show at most m nonzero x i ’s. A j are linearly independent. | B | m. Since rank(A) = m, we can always augment B to include m linearly independent vectors. Using B to form basic columns renders x a bfs.
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Discussion Pf) “”“” Fact: Define B = {A j : x j > 0, 1 j n} Let We want to show at most m nonzero x i ’s. A j are linearly independent. | B | m. Since rank(A) = m, we can always augment B to include m linearly independent vectors. Using B to form basic columns renders x a bfs. If | B | < m, there may be many ways to augment to m linearly independent vectors. Two different B and B ’ may corresponds to the same bfs. If | B | < m, there may be many ways to augment to m linearly independent vectors. Two different B and B ’ may corresponds to the same bfs. Two different, their corresponding B and B ’ must be different.
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Example x1x1 x2x2 x3x3 (0, 0, 3) (1, 0, 3) (2, 0, 2) (2, 0, 0) (2, 2, 0)(0, 2, 0) (0, 1, 3) (0, 0, 0)
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Example Corresponding F of LP
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Example Corresponding F of LP
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Example Corresponding F of LP B and B ’ determine the same bfs, i.e, x = (2, 2, 0, 0, 0, 3, 0) ’. B and B ’ determine the same bfs, i.e, x = (2, 2, 0, 0, 0, 3, 0) ’.
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Degeneration A bfs is called degenerate if it contains more than n m zeros.
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x x Theorem 2 A bfs is called degenerate if it contains more than n m zeros. If two distinct bases corresponding to the same bfs x, then x is degenerate. Pf) Let B and B ’ be two distinct bases which determine the same bfs x. B B’B’ 0 0 00 n m m
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x x Theorem 2 A bfs is called degenerate if it contains more than n m zeros. Pf) Let B and B ’ be two distinct bases which determine the same bfs x. B B’B’ 0 0 00 n m m 00 x has more than n m zeros and, hence, is degenerate. If two distinct bases corresponding to the same bfs x, then x is degenerate.
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Discussion A bfs is called degenerate if it contains more than n m zeros. If two distinct bases corresponding to the same bfs x, then x is degenerate. Changing basis may keep bfs unchanged.
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More on Theorem 1 and 2 Vertices of polytopeBfs’s of LP Change vertexChange bfs ? A bfs is determined by a chosen basis. Change verticesChange basis ?
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Costs ? ? ?
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Costs constant
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Theorem 3 There is an optimal bfs in any instance of LP. Furthermore, if q bfs’s are optimal, their convex combinations are also optimal.
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Theorem 3 There is an optimal bfs in any instance of LP. Furthermore, if q bfs’s are optimal, their convex combinations are also optimal. Let x 0 P be the optimal solution, and let x 1, …, x N be the vertices of P. with Let j be the index corresponding to the vertex with lowest cost. x j is the optimum. Pf) and
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Theorem 3 There is an optimal bfs in any instance of LP. Furthermore, if q bfs’s are optimal, their convex combinations are also optimal. Assume y 1, …, y q be the optimal vertices. Pf) Let withand
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Linear Programming Moving from Bfs to Bfs 大同大學資工所 智慧型多媒體研究室
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Facts The optimal solution can be found from vertices of the corresponding polytope. The bfs’s of LP and the vertices of polytope are close correlated. The algorithm to solve LP: – Move from vertex to vertex; or – Move from bfs to bfs
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The BFS’s Basis columns Nonbasis columns The bfs is determined from the set of basis columns, i.e., B. Move from bfs to bfs:
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Move from Bfs to Bfs Let x 0 be the bfs determined by B. Denote the basic components of x 0 as x i0, i = 1, …, m. For any, we have
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Move from Bfs to Bfs m + 1 columns are involved. Choose one A j from to enter B. Choose one A B(i) to leave B. Who enters? Who leaves?
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Move from Bfs to Bfs m + 1 columns are involved. Choose one A j from to enter B. Choose one A B(i) to leave B. Who enters? Who leaves? must be positive. Make one of them zero, and keep others positive. The one that we want to make zero must have x ij > 0. Why?
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Move from Bfs to Bfs Suppose that A j wants to enter B. Then, we choose Make one of them zero, and keep others positive.
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Move from Bfs to Bfs Suppose that A j wants to enter B. Then, we choose Make one of them zero, and keep others positive. 1.How about if x i0 = 0 for some i ? 2.How about if all x ij 0 ? 1.How about if x i0 = 0 for some i ? 2.How about if all x ij 0 ?
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Example x1x1 x2x2 x3x3 (0, 0, 3) (1, 0, 3) (2, 0, 2) (2, 0, 0) (2, 2, 0)(0, 2, 0) (0, 1, 3) (0, 0, 0)
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Example Corresponding F of LP
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Example Corresponding F of LP 2214
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Example Corresponding F of LP 2214 Choosing =1 makes A 6 leaves B.
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Example Corresponding F of LP 224 Choosing =1 makes A 6 leaves B. 01 11 33 11 00 33 = 0 = 1
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Linear Programming Organization of a Tableau 大同大學資工所 智慧型多媒體研究室
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Example Ansx1x1 x2x2 x3x3 x4x4 x5x5 1 3 4 3 5 2 2 1 5 1 1 1 0 1 0 0 0 1 x1x1 x2x2 x3x3 x4x4 x5x5 1 2 3 3 2 11 2 11 3 1 0 0 0 1 0 0 0 1 R1 R2 R1 R3 R1 Elementary Row operations
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Example Ansx1x1 x2x2 x3x3 x4x4 x5x5 1 2 3 3 2 11 2 11 3 1 0 0 0 1 0 0 0 1 R1 R2 R1 R3 R1 B Suppose that we want to choose A 1 to enter the basis, i.e., choosing A 1 as the pivot column. 1/3 2/2 min
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Example Ansx1x1 x2x2 x3x3 x4x4 x5x5 1 2 3 3 2 11 2 11 3 1 0 0 0 1 0 0 0 1 R1 R2 R1 R3 R1 B 1/3 2/2 min
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Example Ansx1x1 x2x2 x3x3 x4x4 x5x5 1 2 3 3 2 11 2 11 3 1 0 0 0 1 0 0 0 1 1/3 2/2 min Ansx1x1 x2x2 x3x3 x4x4 x5x5 1/3 4/3 10/3 1 0 0 2/3 7/3 11/3 1/3 2/3 1/3 0 1 0 0 0 1
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