Linear Programming Lecture Notes - Penn State Personal Web Server

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Using this information, we can see that either x1 or x2 can enter. We can compute the minimum ratio test (MRT) next to the RHS column. If we chose x2 as the entering variable, then the MRT tells us s2 will leave. We put a box around the element on which we will pivot: (5.29) z s1 s2 s3 ⎡ ⎢ ⎣ z x1 x2 s1 s2 s3 RHS 1 −7 −6 0 0 0 0 0 3 1 1 0 0 120 0 1 2 0 1 0 160 0 1 0 0 0 1 35 ⎤ ⎥ ⎦ MRT (x2) If we pivot on this element, then we transform the column corresponding to x2 into the identity column: ⎡ ⎤ 0 ⎢ (5.30) ⎢0 ⎥ ⎣1⎦ 0 This process will correctly compute the new reduced costs and B−1 matrix as well as the new cost information. The new tableau becomes: ⎡ ⎤ z x1 x2 s1 s2 s3 RHS z ⎢ 1 −4 0 0 3 0 480 ⎥ (5.31) s1 ⎢ 0 2.5 0 1 −0.5 0 40 ⎥ x2 ⎣ 0 0.5 1 0 0.5 0 80 ⎦ 0 1 0 0 0 1 35 s3 We can see that x1 is a valid entering variable, as it has a negative reduced cost (−4). We can again place the minimum ratio test values on the right-hand-side of the matrix to obtain: (5.32) z s1 x2 s3 ⎡ ⎢ ⎣ z x1 x2 s1 s2 s3 RHS 1 −4 0 0 3 0 480 0 2.5 0 1 −0.5 0 40 0 0.5 1 0 0.5 0 80 0 1 0 0 0 1 35 ⎤ ⎥ ⎦ 120 80 − MRT (x1) 16 160 35 We now pivot on the element we have boxed to obtain the new tableau1 : ⎡ ⎤ z x1 x2 s1 s2 s3 RHS z ⎢ 1 0 0 1.6 2.2 0 544 ⎥ (5.33) x1 ⎢ 0 1 0 0.4 −0.2 0 16 ⎥ x2 ⎣ 0 0 1 −0.2 0.6 0 72 ⎦ 0 0 0 −0.4 0.2 1 19 s3 All the reduced costs of the non-basic variables (s1 and s2) are positive and so this is the optimal solution to the linear programming problem. We can also see that this solution agrees with our previous computations on the Toy Maker Problem. 1 Thanks to Ethan Wright for catching a typo here. 80
5. Identifying Unboundedness We have already identified a theorem for detecting unboundedness. Recall Theorem 5.7: In a maximization problem, if aji < 0 for all i = 1, . . . , m, and zj − cj < 0, then the linear programming problem is unbounded. This condition occurs when a variable xj should enter the basis because ∂z/∂xj > 0 and there is no blocking basis variable. That is, we can arbitrarily increase the value of xj without causing any variable to become negative. We give an example: Example 5.11. Consider the <strong>Linear</strong> programming problem from Example 2.9: ⎧ max z(x1, x2) = 2x1 − x2 ⎪⎨ s.t. x1 − x2 ≤ 1 ⎪⎩ 2x1 + x2 ≥ 6 x1, x2 ≥ 0 We can convert this problem into standard form by adding a slack variable s1 and a surplus variable s2: ⎧ ⎪⎨ ⎪⎩ max z(x1, x2) = 2x1 − x2 s.t. x1 − x2 + s1 = 1 2x1 + x2 − s2 = 6 x1, x2, s1, s2 ≥ 0 This yields the matrices: ⎡ ⎤ ⎡ ⎤ 2 x1 ⎢ c = ⎢−1⎥ ⎢x2⎥ ⎥ ⎣ 0 ⎦ x = ⎢ ⎥ ⎣s1⎦ 0 s2 A = 1 −1 1 0 2 1 0 −1 We have both slack and surplus variables, so the case when x1 = x2 = 0 is not a valid initial solution. We can chose a valid solution based on our knowledge of the problem. Assume that s1 = s2 = 0 and so we have: B = 1 −1 2 1 In this case we have: This yields: x1 xB = x2 B −1 b = 7/3 4/3 N = s1 xN = s2 1 0 0 −1 B −1 N = cB = We also have the cost information: cBB −1 b = 10 3 cBB −1 N = 4 3 2 −1 cN = 1/3 −1/3 −2/3 −1/3 0 0 b = 1 6 1 − 3 cBB −1 N − cN = 4 3 81 1 − 3
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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).
Page 42 and 43: 3. Matrices and Linear Programming
Page 44 and 45: Remark 3.16. We can deal with const
Page 46 and 47: Theorem 3.20. Each elementary row o
Page 48 and 49: Gauss-Jordan Elimination Computing
Page 50 and 51: and right hand side vector b = [7,
Page 52 and 53: Gauss-Jordan elimination to solve t
Page 54 and 55: which clearly has a solution for al
Page 56 and 57: which is a contradiction. Case 2: S
Page 58 and 59: Definition 3.46 (Basic Variables).
Page 60 and 61: This leads to the following linear
Page 62 and 63: This is not in a form Matlab likes,
Page 64 and 65: Example 4.5. Figure 4.1 illustrates
Page 66 and 67: Figure 4.3. A hyperplane in 3 dimen
Page 68 and 69: 4. Rays and Directions Recall the d
Page 70 and 71: 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 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
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(x ∗ 1,x ∗ 2) = (16, 72) 3x1 +
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Since w1, w2 ≥ 0, we know that w
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Example 8.13. Consider the followin
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Figure 8.5. This figure illustrates
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(9.4) Proof. Rewrite Problem D as:
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This vector will be related to c in
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3. Strong Duality Lemma 9.7. Proble
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Likewise beginning with the KKT con
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egions are polyhedral sets 1 . Cert
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Figure 9.2. The simplex algorithm b
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If a selfish profit maximizer wishe
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Recall the optimal full tableau for
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For the sake of space, we will prov
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We can choose either s1 or s2 as a
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