Linear Programming Lecture Notes - Penn State Personal Web Server

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Two-Phase Simplex Algorithm (1) Given a problem of the form of the general maximization (or minimization) problem from Equation 2.1, convert it to standard form: ⎧ ⎪⎨ max c P ⎪⎩ T x s.t. Ax = b x ≥ 0 with b ≥ 0. (2) Introduce auxiliary variables xa and solve the Phase I problem: ⎧ ⎪⎨ P1 ⎪⎩ min e T xa s.t. Ax + Imxa = b x, xa ≥ 0 (3) If x∗ a = 0, then an initial feasible solution has been identified. This solution can be converted into a basic feasible solution as we discuss below. Otherwise, there is no solution to Problem P . (4) Use the Basic Feasible solution identified in Step 3 to start the Simplex Algorithm (compute the reduced costs given the c vector). (5) Solve the Phase II problem: ⎧ ⎪⎨ max c P ⎪⎩ T x s.t. Ax = b x ≥ 0 Algorithm 7. Two-Phase Simplex Algorithm When we solve the Phase I problem, if x ∗ a = 0 at optimality, then there is no solution. If x ∗ a = 0, then there are two possibilities: (1) The basis consists only of variables in the vector x; i.e., no auxiliary variable is in the basis. (2) There is some auxiliary variable xai = 0 and this variable is in the basis; i.e., the solution is degenerate and the degeneracy is expressed in an auxiliary variable. 2.1. Case I: xa = 0 and is out of the basis. If xa = 0 and there are not elements of the vector xa in the basis, then we have identified a basic feasible solution x = [xB xN] T . Simply allow the non-zero basic elements (in x) to be xB and the remainder of the elements (not in xa) are in xN. We can then begin Phase II using this basic feasible solution. 2.2. Case II: xa = 0 and is not out of the basis. If xa = 0 and there is at least one artificial variable still in the basis, then we have identified a degenerate solution to the Phase I problem. Theoretically we could proceed directly to Phase II, assigning 0 coefficients to the artificial variables as long as we ensure that no artificial variable ever becomes positive again. [BJS04] notes that this can be accomplished by selective pivoting, however it is often more efficient and simpler to remove the artificial variables completely from the basis before proceeding to Phase II. 96
To remove the artificial variables from the basis, let us assume that we can arrange Rows 1 − m of the Phase I simplex tableau as follows: (6.15) xB xBa xN xNa RHS xB Ik 0 R1 R3 b xBa 0 Im−k R2 R4 0 Column swapping ensures we can do this, if we so desire. Our objective is to replace elements in xBa (the basic artificial variables) with elements from xN non-basic, non-artificial variables. Thus, we will attempt to pivot on elements in the matrix R2. Clearly since the Phase I coefficients of the variables in xN are zero, pivoting in these elements will not negatively impact the Phase I objective value. Thus, if the element in position (1, 1) is non-zero, then we can enter the variable xN1 into the basis and remove the variable xBa 1 . This will produce a new tableau with structure similar to the one given in Equation 6.15 except there will be k + 1 non-artificial basic variables and m − k − 1 artificial basic variables. Clearly if the element in position (1, 1) in matrix R2 is zero, then we must move to a different element for pivoting. In executing the procedure discussed above, one of two things will occur: (1) The matrix R2 will be transformed into Im−k or (2) A point will be reached where there are no longer any variables in xN that can be entered into the basis because all the elements of R2 are zero. In the first case, we have removed all the artificial variables from the basis in Phase I and we can proceed to Phase II with the current basic feasible solution. In the second case, we will have shown that: Ik R1 (6.16) A ∼ 0 0 This shows that the m − k rows of A are not linearly independent of the first k rows and thus the matrix A did not have full row rank. When this occurs, we can discard the last m − k rows of A and simply proceed with the solution given in xB = b, xN = 0. This is a basic feasible solution to the new matrix A in which we have removed the redundant rows. Example 6.5. Once execution of the Phase I simplex algorithm is complete, the reduced costs of the current basic feasible solution must be computed. These can be computed during Phase I by adding an additional “z” row to the tableau. In this case, the initial tableau has the form: (6.17) zII z xa1 xa2 ⎡ ⎢ ⎣ z x1 x2 s1 s2 xa1 xa2 RHS 1 −1 −2 0 0 0 0 0 1 3 5 −1 −1 0 0 32 0 1 2 −1 0 1 0 12 0 2 3 0 −1 0 1 20 The first row (zII) is computed for the objective function: (6.18) x1 + 2x2 + 0s1 + 0s2 + 0xa1 + 0xa2, which is precisely the Phase II problem, except we never allow the artificial variables xa1 and xa2 to carry into the Phase II problem. If we carry out the same steps we did in Example 97 ⎤ ⎥ ⎦
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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
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 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: variables. This is because we start
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 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:
Page 152 and 153: This vector will be related to c in
Page 154 and 155: 3. Strong Duality Lemma 9.7. Proble
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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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