Linear Programming Lecture Notes - Penn State Personal Web Server

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(9.4) Proof. Rewrite Problem D as: ⎧ ⎪⎨ ⎪⎩ max − b T w T s.t. − A T w T ≤ −c T w T ≥ 0 Let β = −b T , G = −A T , u = w T and κ = −c T . Then this new problem becomes: (9.5) ⎧ ⎪⎨ max βu s.t. Gu ≤ κ ⎪⎩ u ≥ 0 Let x T be the vector of dual variables (transposed) for this problem. We can formulate the dual problem as: (9.6) ⎧ ⎪⎨ min x ⎪⎩ T κ s.t. x T G ≥ β x T ≥ 0 Expanding, this becomes: (9.7) ⎧ ⎪⎨ ⎪⎩ min − x T c T s.t. − x T A T ≥ −b T x T ≥ 0 This can be simplified to: ⎧ ⎪⎨ max c (9.8) P ⎪⎩ T x s.t. Ax ≤ b x ≥ 0 as required. This completes the proof. Lemma 9.2 shows that the notion of dual and primal can be exchanged and that it is simply a matter of perspective which problem is the dual problem and which is the primal problem. Likewise, by transforming problems into canonical form, we can develop dual problems for any linear programming problem. The process of developing these formulations can be exceptionally tedious, as it requires enumeration of all the possible combinations of various linear and variable constraints. The following table summarizes the process of converting an arbitrary primal problem into its dual. This table can be found in Chapter 6 of [BJS04]. 138
CONSTRAINTS VARIABLES MINIMIZATION PROBLEM MAXIMIZATION PROBLEM UNRESTRICTED apple 0 apple 0 apple = = UNRESTRICTED Table 1. Table of Dual Conversions: To create a dual problem, assign a dual variable to each constraint of the form Ax ◦ b, where ◦ represents a binary relation. Then use the table to determine the appropriate sign of the inequality in the dual problem as well as the nature of the dual variables. Example 9.3. Consider the problem of finding the dual problem for the Toy Maker Problem (Example 2.3) in standard form. The primal problem is: max 7x1 + 6x2 s.t. 3x1 + x2 + s1 = 120 (w1) x1 + 2x2 + s2 = 160 (w2) x1 + s3 = 35 (w3) x1, x2, s1, s2, s3 ≥ 0 Here we have placed dual variable names (w1, w2 and w3) next to the constraints to which they correspond. The primal problem variables in this case are all positive, so using Table 1 we know that the constraints of the dual problem will be greater-than-or-equal-to constraints. Likewise, we know that the dual variables will be unrestricted in sign since the primal problem constraints are all equality constraints. The coefficient matrix is: ⎡ ⎤ 3 1 1 0 0 A = ⎣1 2 0 1 0⎦ 1 0 0 0 1 Clearly we have: c = 7 6 0 0 0 ⎡ b = ⎣ 120 ⎤ 160⎦ 35 Since w = [w1 w2 w3], we know that wA will be: wA = 3w1 + w2 + w3 w1 + 2w2 w1 w2 w3 139 0 apple 0 CONSTRAINTS VARIABLES
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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
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Exercise 52. Consider the diet prob
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Page 103 and 104: CHAPTER 6 Simplex Initialization In
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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
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Page 138 and 139: 3. The Karush-Kuhn-Tucker Condition
Page 140 and 141: Remark 8.9. The expressions: ∗ A
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Page 154 and 155: 3. Strong Duality Lemma 9.7. Proble
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Page 164 and 165: Recall the optimal full tableau for
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