Linear Programming Lecture Notes - Penn State Personal Web Server

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This leads to a vector of reduced costs given by: (7.3) cB T B −1 N − cN T = 3/4 −20 1/2 −6 These yield an initial tableau with structure: z x1 x2 ⎡ z ⎢ 1 ⎢ 0 ⎣ 0 x1 0 1 0 x2 0 0 1 x3 0 0 0 x4 3/4 1/4 1/2 x5 −20 −8 −12 x6 1/2 −1 −1/2 x7 −6 9 3 ⎤ RHS 0 ⎥ 0 ⎥ 0 ⎦ 0 0 0 1 0 0 1 0 1 x3 If we apply an entering variable rule where we always chose the non-basic variable to enter with the most positive reduced cost (since this is a minimization problem), and we choose the leaving variable to be the first row that is in a tie, then we will obtain the following sequence of tableaux: Tableau I: z x1 x2 x3 ⎡ ⎢ ⎣ Tableau II: ⎡ z x4 x2 x3 ⎢ ⎣ Tableau III: ⎡ z x4 x5 x3 ⎢ ⎣ Tableau IV: ⎡ z x6 x5 x3 ⎢ ⎣ z x1 x2 x3 x4 x5 x6 x7 RHS 1 0 0 0 3/4 −20 1/2 −6 0 0 1 0 0 1/4 −8 −1 9 0 0 0 1 0 1/2 −12 −1/2 3 0 0 0 0 1 0 0 1 0 1 z x1 x2 x3 x4 x5 x6 x7 RHS 1 −3 0 0 0 4 7/2 −33 0 0 4 0 0 1 −32 −4 36 0 0 −2 1 0 0 4 3/2 −15 0 0 0 0 1 0 0 1 0 1 z x1 x2 x3 x4 x5 x6 x7 RHS 1 −1 −1 0 0 0 2 −18 0 0 −12 8 0 1 0 8 −84 0 0 −1/2 1/4 0 0 1 3/8 −15/4 0 0 0 0 1 0 0 1 0 1 z x1 x2 x3 x4 x5 x6 x7 RHS 1 2 −3 0 −1/4 0 0 3 0 0 −3/2 1 0 1/8 0 1 −21/2 0 0 1/16 −1/8 0 −3/64 1 0 3/16 0 0 3/2 −1 1 −1/8 0 0 21/2 1 110 ⎤ ⎥ ⎦ ⎤ ⎥ ⎦ ⎤ ⎥ ⎦ ⎤ ⎥ ⎦
Tableau V: ⎡ z x6 x7 x3 ⎢ ⎣ Tableau VI: ⎡ z x1 x7 x3 ⎢ ⎣ Tableau VII: ⎡ z x1 x2 x3 ⎢ ⎣ z x1 x2 x3 x4 x5 x6 x7 RHS 1 1 −1 0 1/2 −16 0 0 0 0 2 −6 0 −5/2 56 1 0 0 0 1/3 −2/3 0 −1/4 16/3 0 1 0 0 −2 6 1 5/2 −56 0 0 1 z x1 x2 x3 x4 x5 x6 x7 RHS 1 0 2 0 7/4 −44 −1/2 0 0 0 1 −3 0 −5/4 28 1/2 0 0 0 0 1/3 0 1/6 −4 −1/6 1 0 0 0 0 1 0 0 1 0 1 z x1 x2 x3 x4 x5 x6 x7 RHS 1 0 0 0 3/4 −20 1/2 −6 0 0 1 0 0 1/4 −8 −1 9 0 0 0 1 0 1/2 −12 −1/2 3 0 0 0 0 1 0 0 1 0 1 We see that the last tableau (VII) is the same as the first tableau and thus we have constructed an instance where (using the given entering and leaving variable rules), the Simplex Algorithm will cycle forever at this degenerate extreme point. 2. The Lexicographic Minimum Ratio Leaving Variable Rule Given the example of the previous section, we require a method for breaking ties in the case of degeneracy is required that prevents cycling from occurring. There is a large literature on cycling prevention rules, however the most well known is the lexicographic rule for selecting the entering variable. Definition 7.2 (Lexicographic Order). Let x = [x1, . . . , xn] T and y = [y1, . . . , yn] T be vectors in R n . We say that x is lexicographically greater than y if: there exists m < n so that xi = yi for i = 1, . . . , m, and xm+1 > ym+1. Clearly, if there is no such m < n, then xi = yi for i = 1, . . . , n and thus x = y. We write x ≻ y to indicate that x is lexicographically greater than y. Naturally, we can write x y to indicate that x is lexicographically greater than or equal to y. Lexicographic ordering is simply the standard order operation > applied to the individual elements of a vector in R n with a precedence on the index of the vector. Definition 7.3. A vector x ∈ R n is lexicographically positive if x ≻ 0 where 0 is the zero vector in R n . Lemma 7.4. Let x and y be two lexicographically positive vectors in R n . Then x + y is lexicographically positive. Let c > 0 be a constant in R, then cx is a lexicographically positive vector. Exercise 56. Prove Lemma 7.4. 111 ⎤ ⎥ ⎦ ⎤ ⎥ ⎦ ⎤ ⎥ ⎦
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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
Page 72 and 73: Theorem 4.31. Let P ⊆ R n be a po
Page 74 and 75: Figure 4.9. A Polyhedral Set: This
Page 76 and 77: Figure 4.10. Visualization of the s
Page 78 and 79: It is clear that P is bounded. In f
Page 80 and 81: x1 x5 x2 x λx2 +(1− λ)x3 x4 x3
Page 82 and 83: If there is some i such that c T di
Page 84 and 85: 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 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 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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