Linear Programming Lecture Notes - Penn State Personal Web Server

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Remark 8.9. The expressions: ∗ Ax ≤ b Primal Feasibility x ∗ (8.37) ≥ 0 ⎧ ⎪⎨ w Dual Feasibility ⎪⎩ ∗ A − v ∗ = c w ∗ ≥ 0 v ∗ (8.38) ≥ 0 ∗ ∗ w (Ax − b) = 0 (8.39) Complementary Slackness v ∗ x ∗ = 0 are called the Karush-Kuhn-Tucker (KKT) conditions of optimality. Note there is one element in w for each row of the constraints Ax ≤ b and one element in the vector v for each constraint of the form x ≥ 0. The vectors w and v are sometimes called dual variables and sometimes called Lagrange Multipliers. We can think of dual feasibility as expressing the following interesting fact: at optimality, the gradient of the objective function c can be expressed as a positive combination of the gradients of the binding constraints written as less-than-or-equal-to inequalities. That is, the gradient of the constraint Ai·x ≤ bi is Ai· and the gradient of the constraint −xj ≤ 0 is −ej. More specifically, the vector c is in the cone generated by the binding constraints at optimality. Example 8.10. Consider the Toy Maker Problem (Equation 2.8) with Dual Variables (Lagrange Multipliers) listed next to their corresponding constraints: ⎧ max z(x1, x2) = 7x1 + 6x2 Dual Variable s.t. 3x1 + x2 ≤ 120 (w1) ⎪⎨ x1 + 2x2 ≤ 160 (w1) ⎪⎩ x1 ≤ 35 (w3) x1 ≥ 0 (v1) x2 ≥ 0 (v2) In this problem we have: ⎡ 3 A = ⎣1 ⎤ 1 2⎦ ⎡ b = ⎣ 1 0 120 ⎤ 160⎦ c = 35 7 6 Then the KKT conditions can be written as: ⎧ ⎡ ⎤ ⎡ 3 1 x1 ⎪⎨ ⎣1 2⎦ ≤ ⎣ x2 Primal Feasibility 1 0 ⎪⎩ 120 ⎤ 160⎦ 35 x1 0 ≥ 0 x2 128
⎧ ⎪⎨ Dual Feasibility ⎪⎩ w1 w2 w3 w1 w2 w3 v1 v2 ≥ 0 0 ⎡ 3 ⎣1 ⎤ 1 2⎦ − 1 0 v1 v2 = 7 6 ≥ 0 0 0 ⎧ ⎛⎡ ⎤ ⎪⎨ 3 1 w1 w2 w3 ⎝⎣1 2⎦ Complementary Slackness 1 0 ⎪⎩ v1 v2 x1 x2 = 0 x1 x2 ⎡ ≤ ⎣ 120 ⎤ ⎡ 160⎦ − ⎣ 35 120 ⎤⎞ 160⎦⎠ 35 Recall that at optimality, we had x1 = 16 and x2 = 72. The binding constraints in this case where 3x1 + x2 ≤ 120 x1 + 2x2 ≤ 160 To see this note that if 3(16) + 72 = 120 and 16 + 2(72) = 160. Then we should be able to express c = [7 6] (the vector of coefficients of the objective function) as a positive combination of the gradients of the binding constraints: ∇(7x1 + 6x2) = 7 6 ∇(3x1 + x2) = 3 1 ∇(x1 + 2x2) = 1 2 That is, we wish to solve the linear equation: (8.40) w1 w2 3 1 1 = 2 7 6 Note, this is how Equation 8.32 looks when we apply it to this problem. The result is a system of equations: 3w1 + w2 = 7 w1 + 2w2 = 6 A solution to this system is w1 = 8 5 and w2 = 11 5 . This fact is illustrated in Figure 8.4. Figure 8.4 shows the gradient cone formed by the binding constraints at the optimal point for the toy maker problem. Since x1, x2 > 0, we must have v1 = v2 = 0. Moreover, since x1 < 35, we know that x1 ≤ 35 is not a binding constraint and thus its dual variable w3 is also zero. This leads to the conclusion: ∗ x1 = x ∗ 2 16 72 and the KKT conditions are satisfied. ∗ w1 w∗ 2 w∗ ∗ 3 = 8/5 11/5 0 v1 v∗ 2 = 0 0 129
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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
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
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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 142 and 143: (x ∗ 1,x ∗ 2) = (16, 72) 3x1 +
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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