Linear Programming Lecture Notes - Penn State Personal Web Server

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Recall the optimal full tableau for this problem was: z x1 x2 s2 ⎡ z ⎢ 1 ⎢ 0 ⎢ 0 ⎣ 0 x1 0 1 0 0 x2 0 0 1 0 s1 0 0 0 1 s2 7/5 −2/5 7/10 1/2 s3 0 0 0 0 s4 16/5 4/5 −2/5 −2 ⎤ RHS 544 ⎥ 16 ⎥ 72 ⎥ 0 ⎦ 0 0 0 0 2/5 1 −4/5 19 s3 We can compute the dual variables w for this by using cB and B−1 at optimality. You’ll notice that B−1 can always be found in the columns of the slack variables for this problem because we would have begun the simplex algorithm with an identity matrix in that position. We also know that cB = [7 6 0 0] at optimality. Therefore, we can compute cBB−1 as: cBB −1 = 7 6 0 0 ⎡ 0 ⎢ ⎢1 ⎣0 0 0 1 −2/5 7/10 1/2 0 0 0 ⎤ 4/5 −2/5 ⎥ −2 ⎦ 0 0 2/5 1 −4/5 = 0 7/5 0 16/5 In this case, it would seem that modifying the right-hand-side of constraint 1 would have no affect. This is true, if we were to increase the value by a increment of 1. Suppose however we decreased the value of the right-hand-side by 1. Since we claim that: (9.47) ∂z ∂b1 = w1 there should be no change to the optimal objective function value. However, our new optimal point would occur at x1 = 15.6 and x2 = 72.2 with an objective function value of 542.4, clearly the value of the dual variable for constraint 1 is not a true representation the shadow price of resource 1. This is illustrated in Figure 9.3 where we can see that modifying the right-hand-side of Constraint 1 is transforming the feasible region in a way that substantially changes the optimal solution. This is simply not detected because degeneracy in the primal problem leads to alternative optimal solutions in the dual problem. It should be noted that a true margin price can be computed, however this is outside the scope of the notes. The reader is referred to [BJS04] (Chapter 6) for details. 6. The Dual Simplex Method Occasionally when given a primal problem, it is easier to solve the dual problem and extract the information about the primal problem from the dual. Example 9.15. Consider the problem: min x1 + 2x2 s.t. x1 + 2x2 ≥ 12 2x1 + 3x2 ≥ 20 x1, x2 ≥ 0 152
(a) Original Problem (b) RHS Decreased (c) RHS Increased Figure 9.3. Degeneracy in the primal problem causes alternative optimal solutions in the dual problem and destroys the direct relationship between the resource margin price that the dual variables represent in a non-degenerate problem. To transform this problem to standard form, we would have to introduce surplus variables to obtain: min x1 + 2x2 s.t. x1 + 2x2 − s1 = 12 2x1 + 3x2 − s2 = 20 x1, x2, s1, s2 ≥ 0 In this case there is no immediately obvious initial basic feasible solution and we would have to solve a Phase I problem. Consider the dual of the original maximization problem: max 12w1 + 20w2 s.t. w1 + 2w2 ≤ 1 2w1 + 3w2 ≤ 1 w1, w2 ≥ 0 This is a maximization problem whose standard form is given by: max 12w1 + 20w2 s.t. w1 + 2w2 + v1 = 1 2w1 + 3w2 + v2 = 1 w1, w2, v1, v2 ≥ 0 In this case, a reasonable initial basic feasible solution for the dual problem is to set v1 = v2 = 1 and w1 = w2 = 0 (i.e., w1 and w2 are non-basic variables) and proceed with the simplex algorithm from this point. In cases like the one illustrated in Example 9.15, we can solve the dual problem directly in the simplex tableau of the primal problem instead of forming the dual problem and solving it as a primal problem in its own tableau. The resulting algorithm is called the dual simplex algorithm. 153
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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
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Theorem 4.31. Let P ⊆ R n be a po
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Figure 4.9. A Polyhedral Set: This
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Figure 4.10. Visualization of the s
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It is clear that P is bounded. In f
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x1 x5 x2 x λx2 +(1− λ)x3 x4 x3
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If there is some i such that c T di
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Consider now the fact xj = 0 for al
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Example 5.9. Consider the Toy Maker
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using this information, we can comp
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Figure 5.1. The Simplex Algorithm:
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Using this information, we can see
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Based on this information, we can c
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Using the rule you developed in Exe
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Exercise 52. Consider the diet prob
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Suppose we choose s4 as the leaving
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CHAPTER 6 Simplex Initialization In
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A basic feasible solution for our a
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variables. This is because we start
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To remove the artificial variables
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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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Page 138 and 139: 3. The Karush-Kuhn-Tucker Condition
Page 140 and 141: Remark 8.9. The expressions: ∗ A
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Page 150 and 151: (9.4) Proof. Rewrite Problem D as:
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Page 154 and 155: 3. Strong Duality Lemma 9.7. Proble
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Page 160 and 161: Figure 9.2. The simplex algorithm b
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