Linear Programming Lecture Notes - Penn State Personal Web Server

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We can choose either s1 or s2 as a leaving variable. For the sake of argument, suppose we choose s2, the variable that’s most negative as the leaving variable. Then our entering variable is chosen by comparing: z1 − c1 |a21| z2 − c2 |a22| 1 = | − 1| 1 = | − 2| Clearly, 1/2 < 1 and therefore, x2 is our entering variable. ⎡ ⎤ z x1 x2 s1 s2 RHS z ⎢ 1 1/2 0 0 1/2 −1 ⎥ ⎢ ⎥ s1 ⎣ 0 -3/2 0 1 −1/2 −3 ⎦ x2 0 1/2 1 0 −1/2 1 At this point, we see we have maintained dual feasibility, but we still do not have primal feasibility. We can therefore choose a new leaving variable (s1) corresponding to the negative element in the RHS. The minimum ratio test shows that this time x1 will enter and the final simplex tableau will be: z x1 x2 ⎡ ⎢ ⎣ z x1 x2 s1 s2 RHS 1 0 0 1/3 1/3 −2 0 1 0 −2/3 1/3 2 0 0 1 1/3 −2/3 0 It’s clear this is the optimal solution to the problem since we’ve achieved primal and dual feasibility and complementary slackness. It’s also worth noting that this optimal solution is degenerate, since there is a zero in the right hand side. Exercise 70. Prove that the minimum ratio test given in the dual simplex algorithm will maintain dual feasibility from one iteration of the simplex tableau to the next. [Hint: Prove that the reduced costs remain greater than or equal to zero, just as we proved that b remains positive for the standard simplex algorithm.] 156 ⎤ ⎥ ⎦
Bibliography [BJS04] Mokhtar S. Bazaraa, John J. Jarvis, and Hanif D. Sherali, Linear programming and network flows, Wiley-Interscience, 2004. [BTN02] A. Ben-Tal and A. Nemirovski, Robust optimization – methodology and applications, Math. Program. Ser. B 92 (2002), 453–480. [Cul72] C. G. Cullen, Matrices and Linear Transformations, 2 ed., Dover Press, 1972. [Dan60] G. B. Danzig, Inductive proof of the simplex method, Tech. report, RAND Corporation, 1960. [KW94] P. Kall and S. W. Wallace, Stochastic programming, John Wiley and Sons, 1994. [Lan87] S. Lang, Linear Algebra, Springer-Verlag, 1987. [MT03] J. E. Marsden and A. Tromba, Vector calculus, 5 ed., W. H. Freeman, 2003. [PS98] C. H. Papadimitriou and K. Steiglitz, Combinatorial optimization: Algorithms and complexity, Dover Press, 1998. [Ste07] J. Stewart, Calculus, 6 ed., Brooks Cole, 2007. [WN99] L. A. Wolsey and G. L. Nemhauser, Integer and combinatorial optimization, Wiley-Interscience, 1999. 157
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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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Suppose we choose s4 as the leaving
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CHAPTER 6 Simplex Initialization In
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A basic feasible solution for our a
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variables. This is because we start
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To remove the artificial variables
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Remark 6.6. In the case of a minimi
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Example 6.13. Suppose we solve the
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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 140 and 141: Remark 8.9. The expressions: ∗ A
Page 142 and 143: (x ∗ 1,x ∗ 2) = (16, 72) 3x1 +
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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:
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Page 154 and 155: 3. Strong Duality Lemma 9.7. Proble
Page 156 and 157: Likewise beginning with the KKT con
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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
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