Linear Programming Lecture Notes - Penn State Personal Web Server

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Using the rule you developed in Exercise 49, show that the minimization problem has an unbounded feasible solution. Find an extreme direction for this set. [Hint: The minimum ratio test is the same for a minimization problem. Execute the simplex algorithm as we did in Example 5.11 and use Theorem 5.12 to find the extreme direction of the feasible region.] 6. Identifying Alternative Optimal Solutions We saw in Theorem 5.6 that is zj − cj > 0 for all j ∈ J (the indices of the nonbasic variables), then the basic feasible solution generated by the current basis was optimal. Suppose that zj − cj ≥ 0. Then we have a slightly different result: Theorem 5.13. In Problem P for a given set of non-basic variables J , if zj − cj ≥ 0 for all j ∈ J , then the current basic feasible solution is optimal. Further, if zj − cj = 0 for at least one j ∈ J , then there are alternative optimal solutions. Furthermore, let aj be the j th column of B −1 A·j. Then the solutions to P are: (5.39) ⎧ xB = B −1 b − ajxj ⎪⎨ bi xj ∈ 0, min : i = 1, . . . , m, aji aji ⎪⎩ xr = 0, ∀r ∈ J , r = j > 0 Proof. It follows from the proof of Theorem 5.6 that the solution must be optimal as ∂z/∂xj ≤ 0 for all j ∈ J and therefore increasing and xj will not improve the value of the objective function. If there is some j ∈ J so that zj − cj = 0, then ∂z/∂xj = 0 and we may increase the value of xj up to some point specified by the minimum ratio test, while keeping other non-basic variables at zero. In this case, we will neither increase nor decrease the objective function value. Since that objective function value is optimal, it follows that the set of all such values (described in Equation 5.39) are alternative optimal solutions. Example 5.14. Let us consider the toy maker problem again from Example 2.3 and 5.9 with our adjusted objective (5.40) z(x1, x2) = 18x1 + 6x2 Now consider the penultimate basis from Example 5.9 in which we had as basis variables x1, s2 and x2. ⎡ xB = ⎣ x1 ⎤ ⎡ s1 x2⎦ xN = cB = ⎣ s3 18 ⎤ 6 ⎦ 0 cN = 0 0 s2 The matrices become: ⎡ ⎤ ⎡ 3 1 0 1 B = ⎣1 2 1⎦ N = ⎣0 ⎤ 0 0⎦ 1 0 0 0 1 The derived matrices are then: B −1 ⎡ b = ⎣ 35 ⎤ 15⎦ B 95 −1 ⎡ 0 N = ⎣ 1 ⎤ 1 −3⎦ −2 5 84
The cost information becomes: c T BB −1 b = 720 c T BB −1 N = 6 0 c T BB −1 N − cN = 6 0 This yields the tableau: ⎡ z x1 x2 z ⎢ 1 0 0 (5.41) s1 ⎢ 0 1 0 s2 ⎣ 0 0 1 s1 6 0 1 s2 0 0 0 s3 0 1 −3 ⎤ RHS 720 ⎥ 35 ⎥ 15 ⎦ 0 0 0 −2 1 5 95 s3 Unlike example 5.9, the reduced cost for s3 is 0. This means that if we allow s3 to enter the basis, the objective function value will not change. Performing the minimum ratio test however, we see that s2 will still leave the basis: ⎡ ⎤ z x1 x2 s1 s2 s3 RHS z ⎢ 1 0 0 6 0 0 720 ⎥ (5.42) x1 ⎢ 0 1 0 0 0 1 35 ⎥ x2 ⎣ 0 0 1 1 0 −3 15 ⎦ s2 0 0 0 −2 1 5 95 Therefore any solution of the form: s3 ∈ [0, 19] ⎡ (5.43) ⎣ x1 ⎤ ⎡ x2⎦ = ⎣ 35 ⎤ ⎡ 15⎦ − ⎣ 95 1 ⎤ −3⎦ s3 5 s2 MRT (s3) is an optimal solution to the linear programming problem. This precisely describes the edge shown in Figure 5.3. Figure 5.3. Infinite alternative optimal solutions: In the simplex algorithm, when zj − cj ≥ 0 in a maximization problem with at least one j for which zj − cj = 0, indicates an infinite set of alternative optimal solutions. 85 35 − 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
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 92 and 93: Using this information, we can see
Page 94 and 95: Based on this information, we can c
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
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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
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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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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