Linear Programming Lecture Notes - Penn State Personal Web Server

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Example 8.13. Consider the following linear programming problem: ⎧ max z(x1, x2) = 3x1 + 5x2 ⎪⎨ s.t. x1 + 2x2 ≤ 60 (w1) x1 + x2 ≤ 40 (w2) ⎪⎩ x1 ≥ 0 (v1) x2 ≥ 0 (v2) Note we have assigned dual variables corresponding to each constraint on the right-hand-side of the constraints. That is, dual variable w1 corresponds to the constraint x1 + 2x2 ≤ 60. We can write this problem in standard form as: ⎧ ⎪⎨ ⎪⎩ max z(x1, x2) = 3x1 + 5x2 s.t. x1 + 2x2 + s1 = 60 (w1) x1 + x2 + s2 = 40 (w2) x1 ≥ 0 (v1) x2 ≥ 0 (v2) s1 ≥ 0 (v3) s2 ≥ 0 (v4) Note we have added two new dual variables v3 and v4 for the non-negativity constraints on slack variables s1 and s2. Our dual variable vectors are: w = [w1 We can construct an initial simplex tableau as: w2] and v = [v1 v2 v3 v4]. z s1 s2 ⎡ ⎢ ⎣ z x1 x2 s1 s2 RHS 1 −3 −5 0 0 0 0 1 2 1 0 60 0 1 1 0 1 40 ⎤ ⎥ ⎦ In this initial configuration, we note that v1 = −3, v2 = −5, v3 = 0 and v4 = 0. This is because s1 and s2 are basic variables. We also notice that complementary slackness is satisfied. That is at the current values of x1, x2, s1 and s2 we have: ⎡ ⎤ v1 v2 v3 v4 x1 ⎢ ⎣ x1 x2 s1 s2 ⎥ ⎦ = 0 Applying the Simplex Algorithm yields the final tableau: z x2 ⎡ z ⎢ 1 ⎣ 0 x1 0 0 x2 0 1 s1 2 1 s2 1 −1 ⎤ RHS 160 ⎥ 20 ⎦ 0 1 0 −1 2 20 The optimal value for v is [0 0 2 1]. Note v ≥ 0 as required. Further, complementary slackness is still maintained. Notice further that the current value of B −1 can be found in 134
the portion of the matrix where the identity matrix stood in the initial tableau. Thus we can compute w as: w = cBB −1 Since cB = [5 3] (since x2 and x1 the basic variables at optimality) we see that: w = 5 3 1 −1 −1 2 = 2 1 That is, w1 = 2 and w2 = 1. Note that w ≥ 0. This is because w is also a dual variable vector for our original problem (not in standard form). The KKT conditions for a maximization problem in canonical form require w ≥ 0 (see Theorem 8.7). Thus, it makes sense that we have w ≥ 0. Note this does not always have to be the case if we do not begin with a problem in canonical form. Last, we can see that the constraints: x1 + 2x2 ≤ 60 x1 + x2 ≤ 40 are both binding at optimality (since s1 and s2 are both zero). This means we should be able to express c = [3 5] T as a positive combination of the gradients of the left-hand-sides of these constraints using w. To see this, note that w1 corresponds to x1 + 2x2 ≤ 60 and w2 to x1 + x2 ≤ 40. We have: 1 ∇(x1 + 2x2) = 2 1 ∇(x1 + x2) = 1 Then: w1 1 + w2 2 1 = (2) 1 1 + (1) 2 1 = 1 Thus, the objective function gradient is in the dual cone of the binding constraint. That is, it is a positive combination of the gradients of the left-hand-sides of the binding constraints at optimality. This is illustrated in Figure 8.5. We can also verify that the KKT conditions hold for the problem in standard form. Naturally, complementary slackness and primal feasibility hold. To see that dual feasibility holds note that v = [0 0 2 1] ≥ 0. Further: 2 1 1 2 1 0 1 1 0 1 3 5 − 0 0 2 1 = 3 5 0 0 Here 3 5 0 0 is the objective function coefficient vector for the problem in Standard Form. Exercise 64. Use a full simplex tableau to find the values of the Lagrange multipliers (dual variables) at optimality for the problem from Exercise 62. Confirm that complementary slackness holds at optimality. Lastly show that dual feasibility holds by showing that the 135
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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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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
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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
Page 142 and 143: (x ∗ 1,x ∗ 2) = (16, 72) 3x1 +
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Page 148 and 149: Figure 8.5. This figure illustrates
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
Page 156 and 157: Likewise beginning with the KKT con
Page 158 and 159: egions are polyhedral sets 1 . Cert
Page 160 and 161: Figure 9.2. The simplex algorithm b
Page 162 and 163: If a selfish profit maximizer wishe
Page 164 and 165: Recall the optimal full tableau for
Page 166 and 167: For the sake of space, we will prov
Page 168 and 169: We can choose either s1 or s2 as a
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