Linear Programming Lecture Notes - Penn State Personal Web Server

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Exercise 52. Consider the diet problem we covered in Example 3.49. I wish to design a diet consisting of Raman noodles and ice cream. I’m interested in spending as little money as possible but I want to ensure that I eat at least 1200 calories per day and that I get at least 20 grams of protein per day. Assume that each serving of Raman costs $1 and contains 100 calories and 2 grams of protein. Assume that each serving of ice cream costs $1.50 and contains 200 calories and 3 grams of protein. (1) Develop a linear programming problem that will help me minimize the cost of my food intake. (2) Remembering to transform the linear programming problem you found above into standard form, use the simplex algorithm to show that this problem has an infinite set of alternative optimal solutions. (3) At an optimal extreme point, find an expression for the set of infinite alternative optimal exteme points like the one shown in Equation 5.43. (4) Plot the feasible region and the level curves of the objective function. Highlight the face of the polyhedral set on which the alternative optimal solutions can be found. 7. Degeneracy and Convergence In this section we give an example of degeneracy and its impact on the simplex algorithm. Example 5.15. Consider the modified form of the toy maker problem originally stated in Example 4.37: (5.44) ⎧ ⎪⎨ ⎪⎩ max 7x1 + 6x2 s.t. 3x1 + x2 ≤ 120 x1 + 2x2 ≤ 160 x1 ≤ 35 7 4 x1 + x2 ≤ 100 x1, x2 ≥ 0 The polyhedral set and level curves of the objective function are shown Figure 5.4. We can convert the problem to standard form by introducing slack variables: (5.45) ⎧ ⎪⎨ ⎪⎩ max 7x1 + 6x2 s.t. 3x1 + x2 + s1 = 120 x1 + 2x2 + s2 = 160 x1 + s3 = 35 7 4 x1 + x2 + s4 = 100 x1, x2, s1, s2, s3, s4 ≥ 0 86
Figure 5.4. An optimization problem with a degenerate extreme point: The optimal solution to this problem is still (16, 72), but this extreme point is degenerate, which will impact the behavior of the simplex algorithm. Suppose we start at the extreme point where x1 = 35 and x2 = 15 and s2 = 95 and s4 = 23.75. In this case, the matrices are: ⎡ 3 ⎢ B = ⎢ 1 ⎣ 1 1 2 0 0 1 0 ⎤ 0 0 ⎥ 0⎦ 7/4 1 0 1 ⎡ ⎤ 1 0 ⎢ N = ⎢0 0 ⎥ ⎣0 1⎦ 0 0 B −1 ⎡ ⎤ 35 ⎢ b = ⎢15 ⎥ ⎣95⎦ B−1 ⎡ ⎤ 0 1 ⎢ N = ⎢ 1 −3 ⎥ ⎣−2 5 ⎦ 95 4 −1 5 4 cBB −1 b = 335 cBB −1 N − cN = 6 −11 The tableau representation is: (5.46) z x1 x2 s2 ⎡ z ⎢ 1 ⎢ 0 ⎢ 0 ⎢ ⎣ 0 x1 0 1 0 0 x2 0 0 1 0 s1 6 0 1 −2 s2 0 0 0 1 s3 −11 1 −3 5 s4 0 0 0 0 ⎤ RHS 335 ⎥ 35 ⎥ 15 ⎥ 95 ⎦ s4 0 0 0 −1 0 5/4 1 95 4 MRT (s3) From this, we see that the variable s3 should enter (because its reduce cost is negative). In this case, there is a tie for the leaving variables: we see that 95/5 = 19 = (95/4)/(5/4), therefore, either s2 or s4 could be chosen as the leaving variable. This is because we will move to a degenerate extreme point when s3 enters the basis. 87 35 − 19 19
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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
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 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 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 144 and 145: Since w1, w2 ≥ 0, we know that w
Page 146 and 147: 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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