Linear Programming Lecture Notes - Penn State Personal Web Server

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Revised Simplex Algorithm (1) Identify an initial basis matrix B and compute B−1 , w, b and z and place these into a revised simplex tableau: w z B−1 b xB (2) For each j ∈ J use w to compute: zj − cj = wA·j − cj. (3) Choose an entering variable xj (for a maximization problem, we choose a variable with negative reduced cost, for a minimization problem we choose a variable with positive reduced cost): (a) If there is no entering variable, STOP, you are at an optimal solution. (b) Otherwise, continue to Step 4. (4) Append the column aj = B −1 A·j to the revised simplex tableau: xB w z B −1 b zj − cj aj (5) Perform the minimum ratio test and determine a leaving variable (using any leaving variable rule you prefer). (a) If aj ≤ 0, STOP, the problem is unbounded. (b) Otherwise, assume that the leaving variable is xBr which appears in row r of the revised simplex tableau. (6) Use row operations and pivot on the leaving variable row of the column: zj − cj aj transforming the revised simplex tableau into: w ′ z ′ 0 x ′ B B ′−1 b ′ er where er is an identity column with a 1 in row r (the row that left). The variable xj is now the r th element of xB. (7) Goto Step 2. Algorithm 8. Revised Simplex Algorithm We can find the optimal number of bugs of each type the software company should fix assuming it wishes to minimize its exposure to risk using a linear programming formulation. Let x1 be the number of non-critical bugs corrected and x2 be the number of critical software bugs corrected. Define: (8.3) (8.4) y1 = 50 − x1 y2 = 5 − x2 Here y1 is the number of non-critical bugs that are not fixed while y2 is the number of critical bugs that are not fixed. The time (in hours) it takes to fix these bugs is: (8.5) 3x1 + 12x2 118
Let: (8.6) y3 = 1 8 (80 − 3x1 − 12x2) Then y3 is a variable that is unrestricted in sign and determines the amount of time (in days) either over or under the two-week period that is required to ship the software. As an unrestricted variable, we can break it into two components: (8.7) y3 = z1 − z2 We will assume that z1, z2 ≥ 0. If y3 > 0, then z1 > 0 and z2 = 0. In this case, the software is completed ahead of the two-week deadline. If y3 < 0, then z1 = 0 and z2 > 0. In this case the software is finished after the two-week deadline. Finally, if y3 = 0, then z1 = z2 = 0 and the software is finished precisely on time. We can form the objective function as: (8.8) z = 100y1 + 1000y2 + 500z2 The linear programming problem is then: (8.9) min z =y1 + 10y2 + 5z2 s.t. x1 + y1 = 50 x2 + y2 = 5 3 8 x1 + 3 2 x2 + z1 − z2 = 10 x1, x2, y1, y2, z1, z2 ≥ 0 Notice we have modified the objective function by dividing by 100. This will make the arithmetic of the simplex algorithm easier. The matrix of coefficients for this problem is: ⎡ ⎤ (8.10) x1 x2 y1 y2 z1 z2 ⎢ 1 ⎣ 0 0 1 1 0 0 1 0 0 0 ⎥ 0 ⎦ 3 8 3 2 0 0 1 −1 Notice there is an identity matrix embedded inside the matrix of coefficients. Thus a good initial basic feasible solution is {y1, y2, z1}. The initial basis matrix is I3 and naturally, B −1 = I3 as a result. We can see that cB = [1 10 0] T . It follows that c T B B−1 = w = [1 10 0]. Our initial revised simplex tableau is thus: z y1 y2 ⎡ 1 ⎢ 1 ⎣ 0 10 0 1 0 0 0 ⎤ 100 50 ⎥ 5 ⎦ 0 0 1 10 (8.11) z1 There are three variables that might enter at this point, x1, x2 and z1. We can compute the reduced costs for each of these variables using the columns of the A matrix, the coefficients of these variables in the objective function and the current w vector (in row 0 of the revised 119
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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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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 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
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 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
Page 148 and 149: Figure 8.5. This figure illustrates
Page 150 and 151: (9.4) Proof. Rewrite Problem D as:
Page 152 and 153: This vector will be related to c in
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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