Linear Programming Lecture Notes - Penn State Personal Web Server

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A1· Half-space cy > 0 A2· c Am· Positive Cone of Rows of A Figure 8.1. System 2 has a solution if (and only if) the vector c is contained inside the positive cone constructed from the rows of A. non-empty. This set is the cone of vectors perpendicular to the rows of A. This is illustrated in Figure 8.2 A2· Half-space cy > 0 A1· Am· c Non-empty intersection Cone of perpendicular vectors to Rows of A Figure 8.2. System 1 has a solution if (and only if) the vector c is not contained inside the positive cone constructed from the rows of A. Example 8.4. Consider the matrix: 1 0 A = 0 1 and the vector c = 1 2 . Then clearly, we can see that the vector w = 1 2 will satisfy System 2 of Farkas’ Lemma, since w ≥ 0 and wA = c. Contrast this with c ′ = 1 −1 . In this case, we can choose x = 0 1 T . Then Ax = 0 1 T ≥ 0 and c ′ x = −1. Thus x satisfies System 1 of Farkas’ Lemma. These two facts are illustrated in Figure 8.3. Here, we see that c is inside the positive cone formed by the rows of A, while c ′ is not. Exercise 60. Consider the following matrix: 1 0 A = 1 1 124
[0 1] [1 2] System 2 has a solution Positive cone [1 -1] System 1 has a solution Figure 8.3. An example of Farkas’ Lemma: The vector c is inside the positive cone formed by the rows of A, but c ′ is not. and the vector c = 1 2 . For this matrix and this vector, does System 1 have a solution or does System 2 have a solution? [Hint: Draw a picture illustrating the positive cone formed by the rows of A. Draw in c. Is c in the cone or not?] 2.2. Theorems of the Alternative. Farkas’ lemma can be manipulated in many ways to produce several equivalent statements. The collection of all such theorems are called Theorems of the Alternative and are used extensively in optimization theory in proving optimality conditions. We state two that will be useful to us. Corollary 8.5. Let A ∈ R k×n and E ∈ R l×n . Let c ∈ R n be a row vector. Suppose d ∈ R n is a column vector and w ∈ R k is a row vector and v ∈ R l is a row vector. Let: and M = A E u = w v [1 0] Then exactly one of the following systems has a solution: (1) Md ≤ 0 and cd > 0 or (2) uM = c and u ≥ 0 Proof. Let x = −d. Then Md ≤ 0 implies Mx ≥ 0 and cd > 0 implies cx < 0. This converts System 1 to the System 1 of Farkas’ Lemma. System 2 is already in the form found in Farkas’ Lemma. This completes the proof. Exercise 61. Prove the following corollary to Farkas’ Lemma: Corollary 8.6. Let A ∈ R m×n and c ∈ R n be a row vector. Suppose d ∈ R n is a column vector and w ∈ R m is a row vector and v ∈ R n is a row vector. Then exactly one of the following systems of inequalities has a solution: (1) Ad ≤ 0, d ≥ 0 and cd > 0 or (2) wA − v = c and w, v ≥ 0 [Hint: Write System 2 from this corollary as wA − Inv = c and then re-write the system with an augmented vector [w v] with an appropriate augmented matrix. Let M be the augmented matrix you identified. Now write System 1 from Farkas’ Lemma using M and x. Let d = −x and expand System 1 until you obtain System 1 for this problem.] 125
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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
Page 86 and 87: Example 5.9. Consider the Toy Maker
Page 88 and 89: using this information, we can comp
Page 90 and 91: Figure 5.1. The Simplex Algorithm:
Page 92 and 93: 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 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 144 and 145: Since w1, w2 ≥ 0, we know that w
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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