Linear Programming Lecture Notes - Penn State Personal Web Server

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2.1. Lexicographic Minimum Ratio Test. Suppose we are considering a linear programming problem and we have chosen an entering variable xj according to a fixed entering variable rule. Assume further, we are given some current basis matrix B and as usual, the right-hand-side vector of the constraints is denoted b, while the coefficient matrix is denoted A. Then the minimum ratio test asserts that we will chose as the leaving variable the basis variable with the minimum ratio in the minimum ratio test. Consider the following set: (7.4) I0 = r : br bi = min : i = 1, . . . , m and aji > 0 ajr aji In the absence of degeneracy, I0 contains a single element: the row index that has the smallest ratio of bi to aji , where naturally: b = B−1b and aj = B−1A·j. In this case, xj is swapped into the basis in exchange for xBr (the rth basic variable). When we have a degenerate basic feasible solution, then I0 is not a singleton set and contains all the rows that have tied in the minimum ratio test. In this case, we can form a new set: (7.5) I1 = r : a1r ajr a1i = min aji : i ∈ I0 Here, we are taking the elements in column 1 of B −1 A·1 to obtain a1. The elements of this (column) vector are then being divided by the elements of the (column) vector aj on a index-by-index basis. If this set is a singleton, then basic variable xBr leaves the basis. If this set is not a singleton, we may form a new set I2 with column a2. In general, we will have the set: (7.6) Ik = r : akr ajr aki = min aji : i ∈ Ik−1 Lemma 7.5. For any degenerate basis matrix B for any linear programming problem, we will ultimately find a k so that Ik is a singleton. Exercise 57. Prove Lemma 7.5. [Hint: Assume that the tableau is arranged so that the identity columns are columns 1 through m. (That is aj = ej for i = 1, . . . , m.) Show that this configuration will easily lead to a singleton Ik for k < m.] In executing the lexicographic minimum ratio test, we can see that we are essentially comparing the tied rows in a lexicographic manner. If a set of rows ties in the minimum ratio test, then we execute a minimum ratio test on the first column of the tied rows. If there is a tie, then we move on executing a minimum ratio test on the second column of the rows that tied in both previous tests. This continues until the tie is broken and a single row emerges as the leaving row. Example 7.6. Let us consider the example from Beale again using the lexicographic minimum ratio test. Consider the tableau shown below. Tableau I: z x1 x2 ⎡ z ⎢ 1 ⎢ 0 ⎢ ⎣ 0 x1 0 1 0 x2 0 0 1 x3 0 0 0 x4 3/4 1/4 1/2 x5 −20 −8 −12 x6 1/2 −1 −1/2 x7 −6 9 3 ⎤ RHS 0 ⎥ 0 ⎥ 0 ⎦ x3 0 0 0 1 0 0 1 0 1 112
Again, we chose to enter variable x4 as it has the most positive reduced cost. Variables x1 and x2 tie in the minimum ratio test. So we consider a new minimum ratio test on the first column of the tableau: 1 0 (7.7) min , 1/4 1/2 From this test, we see that x2 is the leaving variable and we pivot on element 1/2 as indicated in the tableau. Note, we only need to execute the minimum ratio test on variables x1 and x2 since those were the tied variables in the standard minimum ratio test. That is, I0 = {1, 2} and we construct I1 from these indexes alone. In this case I1 = {2}. Pivoting yields the new tableau: Tableau II: ⎡ ⎤ z x1 x2 x3 x4 x5 x6 x7 RHS z ⎢ 1 0 −3/2 0 0 −2 5/4 −21/2 0 ⎥ x1 ⎢ 0 1 −1/2 0 0 −2 −3/4 15/2 0 ⎥ x4 ⎣ 0 0 2 0 1 −24 −1 6 0 ⎦ x3 0 0 0 1 0 0 1 0 1 There is no question this time of the entering or leaving variable, clearly x6 must enter and x3 must leave and we obtain 1 : Tableau III: ⎡ z x1 x4 x6 ⎢ ⎣ z x1 x2 x3 x4 x5 x6 x7 RHS 1 0 −3/2 −5/4 0 −2 0 −21/2 −5/4 0 1 −1/2 3/4 0 −2 0 15/2 3/4 0 0 2 1 1 −24 0 6 1 0 0 0 1 0 0 1 0 1 Since this is a minimization problem and the reduced costs of the non-basic variables are now all negative, we have arrived at an optimal solution. The lexicographic minimum ratio test successfully prevented cycling. 2.2. Convergence of the Simplex Algorithm Under Lexicographic Minimum Ratio Test. Lemma 7.7. Consider the problem: ⎧ ⎪⎨ max c P ⎪⎩ T x s.t. Ax = b x ≥ 0 Suppose the following hold: (1) Im is embedded in the matrix A and is used as the starting basis, (2) a consistent entering variable rule is applied (e.g., largest reduced cost first), and (3) the lexicographic minimum ratio test is applied as the leaving variable rule. Then each row of the sequence of augmented matrices [b|B −1 ] is lexicographically positive. Here B is the basis matrix and b = B −1 b. 1 Thanks to Ethan Wright for finding a small typo in this example, that is now fixed. 113 ⎤ ⎥ ⎦
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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
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 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 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
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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