Linear Programming Lecture Notes - Penn State Personal Web Server

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Figure 8.5. This figure illustrates the optimal point of the problem given in Example 8.13. Note that at optimality, the objective function gradient is in the dual cone of the binding constraint. That is, it is a positive combination of the gradients of the left-hand-sides of the binding constraints at optimality. The gradient of the objective function is shown in green. gradient of the objective function (c) is a positive combination of the gradients of the binding constraints at optimality. [Hint: Use the vector w you should have identified.] 136
CHAPTER 9 Duality In the last chapter, we explored the Karush-Kuhn-Tucker (KKT) conditions and identified constraints called the dual feasibility constraints. In this section, we show that to each linear programming problem (the primal problem) we may associate another linear programming problem (the dual linear programming problem). These two problems are closely related to each other and an analysis of the dual problem can provide deep insight into the primal problem. 1. The Dual Problem Consider the linear programming problem ⎧ ⎪⎨ max c (9.1) P ⎪⎩ T x s.t. Ax ≤ b x ≥ 0 Then the dual problem for Problem P is: ⎧ ⎪⎨ min wb (9.2) D s.t. wA ≥ c ⎪⎩ w ≥ 0 Remark 9.1. Let v be a vector of surplus variables. Then we can transform Problem D into standard form as: ⎧ min wb ⎪⎨ s.t. wA − v = c (9.3) DS w ≥ 0 ⎪⎩ v ≥ 0 Thus we already see an intimate relationship between duality and the KKT conditions. The feasible region of the dual problem (in standard form) is precisely the the dual feasibility constraints of the KKT conditions for the primal problem. In this formulation, we see that we have assigned a dual variable wi (i = 1, . . . , m) to each constraint in the system of equations Ax ≤ b of the primal problem. Likewise dual variables v can be thought of as corresponding to the constraints in x ≥ 0. Lemma 9.2. The dual of the dual problem is the primal problem. 137
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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:
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Using this information, we can see
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Based on this information, we can c
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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 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 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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