Linear Programming Lecture Notes - Penn State Personal Web Server

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(x ∗ 1,x ∗ 2) = (16, 72) 3x1 + x2 ≤ 120 x1 +2x2 ≤ 160 x1 ≤ 35 x1 ≥ 0 x2 ≥ 0 Figure 8.4. The Gradient Cone: At optimality, the cost vector c is obtuse with respect to the directions formed by the binding constraints. It is also contained inside the cone of the gradients of the binding constraints, which we will discuss at length later. Exercise 62. Consider the problem: max x1 + x2 s.t. 2x1 + x2 ≤ 4 x1 + 2x2 ≤ 6 x1, x2 ≥ 0 Write the KKT conditions for an optimal point for this problem. (You will have a vector w = [w1 w2] and a vector v = [v1 v2]). Draw the feasible region of the problem. At the optimal point you identified in Exercise 58, identify the binding constraints and draw their gradients. Show that the objective function is in the positive cone of the gradients of the binding constraints at this point. (Specifically find w and v.) The Karush-Kuhn-Tucker Conditions for an Equality Problem. The KKT conditions can be modified to deal with problems in which we have equality constraints (i.e., Ax = b). Corollary 8.11. Consider the linear programming problem: ⎧ ⎪⎨ max cx (8.41) P ⎪⎩ s.t. Ax = b x ≥ 0 130
with A ∈ Rm×n , b ∈ Rm and (row vector) c ∈ Rn . Then x∗ ∈ Rn if and only if there exists (row) vectors w∗ ∈ Rm and v∗ ∈ Rn so that: (8.42) (8.43) (8.44) (8.45) (8.46) ∗ Ax = b Primal Feasibility x ∗ ≥ 0 ⎧ ⎪⎨ w Dual Feasibility ⎪⎩ ∗ A − v ∗ = c w ∗ unrestricted v ∗ ≥ 0 Complementary Slackness v ∗ x ∗ = 0 Proof. Replace the constraints Ax = b with the equivalent constraints: Ax ≤ b −Ax ≤ −b Let w1 be the vector corresponding to the rows in Ax ≤ b and w2 be the vector corresponding to the rows in −Ax ≤ −b. Then the KKT conditions are: ⎧ ⎪⎨ Ax ≤ b Primal Feasibility −Ax ≤ −b ⎪⎩ x ∗ (8.47) ≥ 0 ⎧ (8.48) (8.49) ⎪⎨ Dual Feasibility ⎪⎩ ⎧ ⎪⎨ Complementary Slackness ⎪⎩ w ∗ 1A − w2A − v ∗ = c w ∗ 1 (Ax ∗ − b) = 0 w ∗ 2 (b − Ax ∗ ) = 0 v ∗ x ∗ = 0 w ∗ 1 ≥ 0 w ∗ 2 ≥ 0 v ∗ ≥ 0 The fact that Ax = b at optimality ensures we can re-write primal-feasibility as: Ax = b Primal Feasibility x ∗ ≥ 0 Furthermore, since Ax = b we know that complementary slackness of w ∗ 1 (Ax ∗ − b) = 0 w ∗ 2 (b − Ax ∗ ) = 0 is always satisfied and thus, we only require v ∗ x ∗ = 0 as our complementary slackness condition. Finally, let w = w1 − w2. Then dual feasibility becomes: (8.50) wA − v = w ∗ 1A − w2A − v ∗ = c 131
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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
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S r x 0 Br( x0) Figure 2.2. A Bound
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S Every point on this line is an al
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6. Problems with Unbounded Feasible
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to “prove” your example works.
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Definition 3.6 (Matrix Transpose).
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3. Matrices and Linear Programming
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Remark 3.16. We can deal with const
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Theorem 3.20. Each elementary row o
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Gauss-Jordan Elimination Computing
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and right hand side vector b = [7,
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Gauss-Jordan elimination to solve t
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which clearly has a solution for al
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which is a contradiction. Case 2: S
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Definition 3.46 (Basic Variables).
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This leads to the following linear
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This is not in a form Matlab likes,
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Example 4.5. Figure 4.1 illustrates
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Figure 4.3. A hyperplane in 3 dimen
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4. Rays and Directions Recall the d
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for all λ > 0. This can only be tr
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Theorem 4.31. Let P ⊆ R n be a po
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Figure 4.9. A Polyhedral Set: This
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Figure 4.10. Visualization of the s
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It is clear that P is bounded. In f
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x1 x5 x2 x λx2 +(1− λ)x3 x4 x3
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If there is some i such that c T di
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Consider now the fact xj = 0 for al
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Example 5.9. Consider the Toy Maker
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using this information, we can comp
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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:
Page 130 and 131: Revised Simplex Algorithm (1) Ident
Page 132 and 133: simplex tableau). We obtain: z1 −
Page 134 and 135: Proof. We can prove Farkas’ Lemma
Page 136 and 137: A1· Half-space cy > 0 A2· c Am·
Page 138 and 139: 3. The Karush-Kuhn-Tucker Condition
Page 140 and 141: Remark 8.9. The expressions: ∗ A
Page 144 and 145: Since w1, w2 ≥ 0, we know that w
Page 146 and 147: Example 8.13. Consider the followin
Page 148 and 149: Figure 8.5. This figure illustrates
Page 150 and 151: (9.4) Proof. Rewrite Problem D as:
Page 152 and 153: This vector will be related to c in
Page 154 and 155: 3. Strong Duality Lemma 9.7. Proble
Page 156 and 157: Likewise beginning with the KKT con
Page 158 and 159: egions are polyhedral sets 1 . Cert
Page 160 and 161: Figure 9.2. The simplex algorithm b
Page 162 and 163: If a selfish profit maximizer wishe
Page 164 and 165: Recall the optimal full tableau for
Page 166 and 167: For the sake of space, we will prov
Page 168 and 169: We can choose either s1 or s2 as a
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