Linear Programming Lecture Notes - Penn State Personal Web Server

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egions are polyhedral sets 1 . Certain examples have a very nice geometric visualization. Consider the problem: (9.31) ⎧ ⎪⎨ ⎪⎩ max 6x1 + 6x2 s.t. 3x1 + 2x2 ≤ 6 2x1 + 3x2 ≤ 6 x1, x2 ≥ 0 In this problem we have: 3 A = 2 2 3 c = 6 6 b = 6 6 Notice that A is a symmetric matrix and cT = b and the dual problem is: ⎧ min 6w1 + 6w2 ⎪⎨ s.t. 3w1 + 2w2 ≥ 6 (9.32) 2w1 + 3w2 ≤ 6 ⎪⎩ w1, w2 ≥ 0 This results in a geometry in which the dual feasible region is a reflection of the primal feasible region (ignoring non-negativity constraints). This is illustrated in Figure 9.1. Figure 9.1. The dual feasible region in this problem is a mirror image (almost) of the primal feasible region. This occurs when the right-hand-side vector b is equal to the objective function coefficient column vector c T and the matrix A is symmetric. 1 Thanks for Michael Cline for suggesting this section. 146
We can also illustrate the process of solving this problem using the (revised) simplex algorithm. In doing so, we first convert Problem 9.33 to standard form: ⎧ max 6x1 + 6x2 ⎪⎨ s.t. 3x1 + 2x2 + s1 = 6 (9.33) 2x1 + 3x2 + s2 = 6 ⎪⎩ x1, x2 ≥ 0 This yields an initial tableau: ⎡ ⎤ z 0 0 0 (9.34) s1 ⎣ 1 0 6 ⎦ 0 1 6 s2 This is because our initial cB = 0 0 and thus w = 0 0 . This means at the start of the simplex algorithm, w1 = 0 and w2 = 0 and x1 = 0 and x2 = 0, so we begin at the origin, which is in the primal feasible region and not in the dual feasible region. If we iterate and choose x1 as an entering variable, our updated tableau will be: (9.35) z x1 s2 ⎡ ⎣ 2 0 12 1/3 0 2 −2/3 1 2 ⎤ ⎦ Notice at this point, x1 = 2, x2 = 0 and w1 = 2 and w2 = 0. This point is again feasible for the primal problem but still infeasible for the dual. This step is illustrated in Figure 9.2. Entering x2 yields the final tableau: ⎡ ⎤ z 6/5 6/5 72/5 (9.36) x1 ⎣ 3/5 −2/5 6/5 ⎦ −2/5 3/5 6/5 x2 At this final point, x1 = x2 = w1 = w2 = 6, which is feasible to both problems. This 5 step is also illustrated in Figure 9.2. We should note that this problem is atypical in that the primal and dual feasible regions share one common point. In more standard problems, the two feasible regions cannot be drawn in this convenient way, but the simplex process is the same. The simplex algorithm begins at a point in the primal feasible region with a corresponding dual vector that is not in the feasible region of the dual problem. As the simplex algorithm progresses, this dual vector approaches and finally enters the dual feasible region. Exercise 69. Draw the dual feasible region for the following problem. max 3x1 + 5x2 s.t. x1 + 2x2 ≤ 60 x1 + x2 ≤ 40 x1, x2 ≥ 0 Solve the problem using the revised simplex algorithm and trace the path of dual variables (the w vector) in your plot of the dual feasible region. Also trace the path of the primal vector x through the primal feasible region. [Hint: Be sure to draw the area around the dual 147
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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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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 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
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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