Linear Programming Lecture Notes - Penn State Personal Web Server

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Example 5.9. Consider the Toy Maker Problem (from Example 2.3). The linear programming problem given in Equation 2.8 is: ⎧ ⎪⎨ ⎪⎩ max z(x1, x2) = 7x1 + 6x2 s.t. 3x1 + x2 ≤ 120 x1 + 2x2 ≤ 160 x1 ≤ 35 x1 ≥ 0 x2 ≥ 0 We can convert this problem to standard form by introducing the slack variables s1, s2 and s3: ⎧ ⎪⎨ ⎪⎩ max z(x1, x2) = 7x1 + 6x2 s.t. 3x1 + x2 + s1 = 120 which yields the matrices ⎡ ⎤ ⎡ 7 ⎢ ⎢6⎥ ⎢ ⎥ ⎢ c = ⎢ ⎢0 ⎥ x = ⎢ ⎣0⎦ ⎣ 0 x1 + 2x2 + s2 = 160 x1 + s3 = 35 x1, x2, s1, s2, s3 ≥ 0 x1 x2 s1 s2 s3 ⎤ ⎥ ⎦ We can begin with the matrices: ⎡ 1 B = ⎣0 0 1 ⎤ ⎡ 0 3 0⎦ N = ⎣1 ⎤ 1 2⎦ 0 0 1 1 0 In this case we have: ⎡ ⎤ and Therefore: xB = ⎣ s1 s2 s3 ⎦ xN = x1 x2 ⎡ 3 A = ⎣1 1 2 1 0 0 1 ⎤ ⎡ 0 0⎦ b = ⎣ 1 0 0 0 1 120 ⎤ 160⎦ 35 cB = B −1 ⎡ b = ⎣ 120 ⎤ 160⎦ B 35 −1 ⎡ 3 N = ⎣1 ⎤ 1 2⎦ 1 0 ⎡ ⎣ 0 ⎤ 0⎦ cN = 0 7 6 c T BB −1 b = 0 c T BB −1 N = 0 0 c T BB −1 N − cN = −7 −6 74
Using this information, we can compute: c T BB −1 A·1 − c1 = −7 c T BB −1 A·2 − c2 = −6 and therefore: ∂z ∂x1 = 7 and ∂z ∂x2 = 6 Based on this information, we could chose either x1 or x2 to enter the basis and the value of the objective function would increase. If we chose x1 to enter the basis, then we must determine which variable will leave the basis. To do this, we must investigate the elements of B−1A·1 and the current basic feasible solution B−1b. Since each element of B−1A·1 is positive, we must perform the minimum ratio test on each element of B−1A·1. We know that B−1A·1 is just the first column of B−1N which is: B −1 A·1 = ⎡ ⎣ 3 ⎤ 1⎦ 1 Performing the minimum ratio test, we see have: 120 160 35 min , , 3 1 1 In this case, we see that index 3 (35/1) is the minimum ratio. Therefore, variable x1 will enter the basis and variable s3 will leave the basis. The new basic and non-basic variables will be: ⎡ xB = ⎣ s1 s2 x1 ⎤ ⎦ xN = s3 x2 cB = and the matrices become: ⎡ ⎤ ⎡ 1 0 3 0 B = ⎣0 1 1⎦ N = ⎣0 ⎤ 1 2⎦ 0 0 1 1 0 ⎡ ⎣ 0 ⎤ 0⎦ cN = 7 Note we have simply swapped the column corresponding to x1 with the column corresponding to s3 in the basis matrix B and the non-basic matrix N. We will do this repeatedly in the example and we recommend the reader keep track of which variables are being exchanged and why certain columns in B are being swapped with those in N. 0 6 Using the new B and N matrices, the derived matrices are then: B −1 ⎡ b = ⎣ 15 ⎤ 125⎦ B 35 −1 ⎡ −3 N = ⎣−1 ⎤ 1 2⎦ 1 0 The cost information becomes: c T BB −1 b = 245 c T BB −1 N = 7 0 c T BB −1 N − cN = 7 −6 75
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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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Page 40 and 41: 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 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
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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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A1· Half-space cy > 0 A2· c Am·
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3. The Karush-Kuhn-Tucker Condition
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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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