Linear Programming Lecture Notes - Penn State Personal Web Server

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x1 x5 x2 x λx2 +(1− λ)x3 x4 x3 x = µx5 +(1− µ)(λx2 +(1− λ)x3) x2 x1 λx2 +(1− λ)x3 d1 x3 x x = λx2 +(1− λ)x3 + θd1 Figure 4.11. The Cartheodory Characterization Theorem: Extreme points and extreme directions are used to express points in a bounded and unbounded set. 68
CHAPTER 5 The Simplex Method 1. <strong>Linear</strong> <strong>Programming</strong> and Extreme Points In this section we formalize the intuition we’ve obtained in all our work in two dimensional linear programming problems. Namely, we noted that if an optimal solution existed, then it occurred at an extreme point. For the remainder of this chapter, assume that A ∈ R m×n with full row rank and b ∈ R m let (5.1) X = {x ∈ R n : Ax ≤ b, x ≥ 0} be a polyhedral set over which we will maximize the objective function z(x1, . . . , xn) = cT x, where c, x ∈ Rn . That is, we will focus on the linear programming problem: ⎧ ⎪⎨ max c (5.2) P ⎪⎩ T x s.t. Ax ≤ b x ≥ 0 Theorem 5.1. If Problem P has an optimal solution, then Problem P has an optimal extreme point solution. Proof. Applying the Cartheodory Characterization theorem, we know that any point x ∈ X can be written as: k l (5.3) x = λixi + i=1 i=1 µidi where x1, . . . xk are the extreme points of X and d1, . . . , dl are the extreme directions of X and we know that k λi = 1 (5.4) i=1 λi, µi ≥ 0 ∀i We can rewrite problem P using this characterization as: (5.5) max s.t. k i=1 λic T xi + k λi = 1 i=1 λi, µi ≥ 0 ∀i l i=1 µic T di 69
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Linear Programming: Penn State Math
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8. Caratheodory Characterization Th
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List of Figures 1.1 Goat pen with u
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4.6 Convex Direction: Clearly every
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Preface Stop! Stop right now! This
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x Goat Pen Figure 1.1. Goat pen wit
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We have formulated the general maxi
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Figure 1.3. Contour Plot of z = x 2
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Figure 1.5. A Level Curve Plot with
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and thus our vector is v = (x1 −
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Figure 1.7. Gradients of the Bindin
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finishing toys and 120 hours per we
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2. Graphically Solving Linear Progr
Page 30 and 31: S r x 0 Br( x0) Figure 2.2. A Bound
Page 32 and 33: S Every point on this line is an al
Page 34: 6. Problems with Unbounded Feasible
Page 37: to “prove” your example works.
Page 40 and 41: Definition 3.6 (Matrix Transpose).
Page 42 and 43: 3. Matrices and Linear Programming
Page 44 and 45: Remark 3.16. We can deal with const
Page 46 and 47: Theorem 3.20. Each elementary row o
Page 48 and 49: Gauss-Jordan Elimination Computing
Page 50 and 51: and right hand side vector b = [7,
Page 52 and 53: Gauss-Jordan elimination to solve t
Page 54 and 55: which clearly has a solution for al
Page 56 and 57: which is a contradiction. Case 2: S
Page 58 and 59: Definition 3.46 (Basic Variables).
Page 60 and 61: This leads to the following linear
Page 62 and 63: This is not in a form Matlab likes,
Page 64 and 65: Example 4.5. Figure 4.1 illustrates
Page 66 and 67: Figure 4.3. A hyperplane in 3 dimen
Page 68 and 69: 4. Rays and Directions Recall the d
Page 70 and 71: for all λ > 0. This can only be tr
Page 72 and 73: Theorem 4.31. Let P ⊆ R n be a po
Page 74 and 75: Figure 4.9. A Polyhedral Set: This
Page 76 and 77: Figure 4.10. Visualization of the s
Page 78 and 79: It is clear that P is bounded. In f
Page 82 and 83: If there is some i such that c T di
Page 84 and 85: Consider now the fact xj = 0 for al
Page 86 and 87: Example 5.9. Consider the Toy Maker
Page 88 and 89: using this information, we can comp
Page 90 and 91: Figure 5.1. The Simplex Algorithm:
Page 92 and 93: Using this information, we can see
Page 94 and 95: Based on this information, we can c
Page 96 and 97: Using the rule you developed in Exe
Page 98 and 99: Exercise 52. Consider the diet prob
Page 100 and 101: Suppose we choose s4 as the leaving
Page 103 and 104: CHAPTER 6 Simplex Initialization In
Page 105 and 106: A basic feasible solution for our a
Page 107 and 108: variables. This is because we start
Page 109 and 110: To remove the artificial variables
Page 111 and 112: Remark 6.6. In the case of a minimi
Page 113 and 114: Example 6.13. Suppose we solve the
Page 115 and 116: That is, s1 = −12 and s2 = −20
Page 117 and 118: The complete problem describing McL
Page 119: * data section */ data; set DAY :=
Page 122 and 123: This leads to a vector of reduced c
Page 124 and 125: 2.1. Lexicographic Minimum Ratio Te
Page 126 and 127: Proof. The initial basis Im yieds a
Page 128 and 129: for i = 1, . . . , n. Then we have:
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Revised Simplex Algorithm (1) Ident
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simplex tableau). We obtain: z1 −
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Proof. We can prove Farkas’ Lemma
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A1· Half-space cy > 0 A2· c Am·
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3. The Karush-Kuhn-Tucker Condition
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Remark 8.9. The expressions: ∗ A
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(x ∗ 1,x ∗ 2) = (16, 72) 3x1 +
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Since w1, w2 ≥ 0, we know that w
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Example 8.13. Consider the followin
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Figure 8.5. This figure illustrates
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(9.4) Proof. Rewrite Problem D as:
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This vector will be related to c in
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3. Strong Duality Lemma 9.7. Proble
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Likewise beginning with the KKT con
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egions are polyhedral sets 1 . Cert
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Figure 9.2. The simplex algorithm b
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If a selfish profit maximizer wishe
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Recall the optimal full tableau for
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For the sake of space, we will prov
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We can choose either s1 or s2 as a
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