Linear Programming Lecture Notes - Penn State Personal Web Server

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Since w1, w2 ≥ 0, we know that w is unrestricted in sign. Thus we have: ⎧ ⎪⎨ wA − v Dual Feasibility ⎪⎩ ∗ = c w ∗ unrestricted v ∗ (8.51) ≥ 0 This completes the proof. Exercise 63. Use the trick of converting a minimization problem to a maximization problem to identify the KKT conditions for the following problem: (8.52) min cx s.t. Ax ≥ b x ≥ 0 [Hint: Remember, Ax ≥ b is the same as writing −Ax ≤ −b. Now use the KKT conditions for the maximization problem to find the KKT conditions for this problem.] 4. Relating the KKT Conditions to the Tableau Consider a linear programming problem in Standard Form: ⎧ ⎪⎨ max cx (8.53) P s.t. Ax = b ⎪⎩ x ≥ 0 with A ∈ R m×n , b ∈ R m and (row vector) c ∈ R n . The KKT conditions for a problem of this type assert that wA − v = c vx = 0 at an optimal point x for some vector w unrestricted in sign and v ≥ 0. (Note, for the sake of notational ease, we have dropped the ∗ notation.) Suppose at optimality we have a basis matrix B corresponding to a set of basic variables xB and we simultaneously have non-basic variables xN. We may likewise divide v into vB and vN. Then we have: (8.54) wA − v = c =⇒ w B N − vB vN = cB cN (8.55) vx = 0 =⇒ vB vN xB = 0 xN We can rewrite Expression 8.54 as: (8.56) wB − vB wN − vN = cB cN 132
This simplifies to: wB − vB = cB wN − vN = cN Let w = cBB −1 . Then we see that: (8.57) wB − vB = cB =⇒ cBB −1 B − vB = cB =⇒ cB − vB = cB =⇒ vB = 0 Since we know that xB ≥ 0, we know that vB should be equal to zero to ensure complementary slackness. Thus, this is consistent with the KKT conditions. We further see that: (8.58) wN − vN = cN =⇒ cBB −1 N − vN = cN =⇒ vN = cBB −1 N − cN Thus, the vN are just the reduced costs of the non-basic variables. (vB are the reduced costs of the basic variables.) Furthermore, dual feasibility requires that v ≥ 0. Thus we see that at optimality we require: (8.59) cBB −1 N − cN ≥ 0 This is precisely the condition for optimality in the simplex tableau. We now can see the following facts are true about the Simplex Method: (1) At each iteration of the Simplex Method, primal feasibility is satisfied. This is ensured by the minimum ratio test and the fact that we start at a feasible point. (2) At each iteration of the Simplex Method, complementary slackness is satisfied. After all, the vector v is just the reduced cost vector (Row 0) of the Simplex tableau. If a variable is basic xj (and hence non-zero), then the its reduced cost vj = 0. Otherwise, vj may be non-zero. (3) At each iteration of the Simplex Algorithm, we may violate dual feasibility because we may not have v ≥ 0. It is only at optimality that we achieve dual feasibility and satisfy the KKT conditions. We can now prove the following theorem: Theorem 8.12. Assuming an appropriate cycling prevention rule is used, the simplex algorithm converges in a finite number of iterations to an optimal solution to the linear programming problem. Proof. Convergence is guaranteed by the proof of Theorem 7.10 in which we show that when the lexicographic minimum ratio test is used, then the simplex algorithm will always converge. Our work above shows that at optimality, the KKT conditions are satisfied because the termination criteria for the simplex algorithm are precisely the same as the criteria in the Karush-Kuhn-Tucker conditions. This completes the proof. 133
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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
Page 94 and 95: Based on this information, we can c
Page 96 and 97: 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 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:
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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