Linear Programming Lecture Notes - Penn State Personal Web Server

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Likewise beginning with the KKT conditions for the dual problem (Expression 9.15 - 9.17) we can write: ∗ ∗ w A − v ≥ c Primal Feasibility w ∗ ≥ 0 ⎧ ⎪⎨ Ax Dual Feasibility ⎪⎩ ∗ + s ∗ = b x ∗ ≥ 0 s ∗ ≥ 0 ∗ ∗ (v ) x = 0 Complementary Slackness w ∗ s ∗ = 0 Here we substitute v ∗ for w ∗ A − c in the complementary slackness terms. Thus we can see that the KKT conditions for the primal and dual problems are equivalent. This completes the proof. Remark 9.8. Notice that the KKT conditions for the primal and dual problems are equivalent, but the dual feasibility conditions for the primal problem are identical to the primal feasibility conditions for the dual problem and vice-versa. Thus, two linear programming problems are dual to each other if they share KKT conditions with the primal and dual feasibility conditions swapped. Exercise 67. Compute the dual problem for a canonical form minimization problem: ⎧ ⎪⎨ min cx P s.t. Ax ≥ b ⎪⎩ x ≥ 0 Find the KKT conditions for the dual problem you just identified. Use the result from Exercise 63 to show that KKT conditions for Problem P are identical to the KKT conditions for the dual problem you just found. Lemma 9.9 (Strong Duality). There is a bounded optimal solution x ∗ for Problem P if and only if there is a bounded optimal solution w ∗ for Problem D. Furthermore, cx ∗ = w ∗ b. Proof. Suppose that there is a solution x ∗ for Problem P . Let s ∗ = b − Ax ∗ . Clearly s ∗ ≥ 0. By Theorem 8.7 there exists dual variables w ∗ and v ∗ satisfying dual feasibility and complementary slackness. Dual feasibility in the KKT conditions implies that: (9.26) v ∗ = w ∗ A − c We also know that w ∗ , v ∗ ≥ 0. Complementary Slackness (from Theorem 8.7) states that v ∗ x ∗ = 0. But v ∗ is defined above and we see that: (9.27) v ∗ x ∗ = 0 =⇒ (w ∗ A − c) x = 0 Likewise, since we have s ∗ = b − Ax ∗ . Complementary slackness assures us that w ∗ (b − Ax ∗ ) = 0. Thus we see that: (9.28) w ∗ (b − Ax ∗ ) = 0 =⇒ w ∗ s ∗ = 0 144
Thus it follows from Lemma 9.7 that w ∗ is an optimal solution to Problem D since it satisfies the KKT conditions. The fact that Problem P has an optimal solution when Problem D has an optimal solution can be proved in a similar manner starting with the KKT conditions given in Lemma 9.7 and applying the same reasoning as above. Finally, at optimality, we know from Lemma 9.7 that: (9.29) (w ∗ A − c) x ∗ = 0 =⇒ w ∗ Ax ∗ = cx ∗ We also know from Theorem 8.7 that: (9.30) w ∗ (b − Ax ∗ ) = 0 =⇒ w ∗ b = w ∗ Ax ∗ Therefore we have: w ∗ b = cx ∗ . This completes the proof. Corollary 9.10. If Problem P is infeasible, then Problem D is either unbounded or infeasible. If Problem D is infeasible, then either Problem P is unbounded or infeasible. Proof. This result follows by contrapositive from Lemma 9.9. To see this, suppose that Problem P is infeasible. Then Problem P has no bounded optimal solution. Therefore, Problem D has no bounded optimal solution (by Lemma 9.9). If Problem D has no bounded optimal solution, then either Problem D is unbounded or it is infeasible. A symmetric argument on Problem D completes the proof of the Lemma. Exercise 68. Consider the problem max x1 + x2 s.t. x1 − x2 ≥ 1 − x1 + x2 ≥ 1 x1, x2 ≥ 0 (1) Show that this problem is infeasible. (2) Compute its dual. (3) Show that the dual is infeasible, thus illustrating Corollary 9.10. The following theorem summarizes all of the results we have obtained in the last two sections. Theorem 9.11 (Strong Duality Theorem). Consider Problem P and Problem D. Then exactly one of the following statements is true: (1) Both Problem P and Problem D possess optimal solutions x ∗ and w ∗ respectively and cx ∗ = w ∗ b. (2) Problem P is unbounded and Problem D is infeasible. (3) Problem D is unbounded and Problem P is infeasible. (4) Both problems are infeasible. 4. Geometry of the Dual Problem The geometry of the dual problem is, in essence, exactly the same the geometry of the primal problem insofar as they are both linear programming problems and thus their feasible 145
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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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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 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
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Page 160 and 161: Figure 9.2. The simplex algorithm b
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Page 164 and 165: Recall the optimal full tableau for
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