Linear Programming Lecture Notes - Penn State Personal Web Server

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for all λ > 0. This can only be true if Ad ≤ 0. Likewise:x + λd ≥ 0 holds for all λ > 0 if and only if d ≥ 0. This completes the proof. Corollary 4.25. If (4.20) P = {x ∈ R n : Ax = b, x ≥ 0} and d is a direction of P , then d must satisfy: (4.21) Ad = 0, d ≥ 0, d = 0. Exercise 45. Prove the corollary above. Example 4.26. Consider the polyhedral set defined by the equations: x1 − x2 ≤ 1 2x1 + x2 ≥ 6 x1 ≥ 0 x2 ≥ 0 This set is clearly unbounded as we showed in class and it has at least one direction. The direction d = [0, 1] T pointing directly up is a direction of this set. This is illustrated in Figure 4.7. In this example, we have: (4.22) A = Figure 4.7. An Unbounded Polyhedral Set: This unbounded polyhedral set has many directions. One direction is [0, 1] T . 1 −1 −2 −1 Note, the second inequality constraint was a greater-than constraint. We reversed it to a less-than inequality constraint −2x1 − x2 ≤ −6 by multiplying by −1. For our chosen direction d = [0, 1] T , we can see that: 1 −1 0 −1 (4.23) Ad = = ≤ 0 −2 −1 1 −1 58
Clearly d ≥ 0 and d = 0. 6. Extreme Points Definition 4.27 (Extreme Point of a Convex Set). Let C be a convex set. A point x0 ∈ C is a extreme point of C if there are no points x1 and x2 (x1 = x0 or x2 = x0) so that x = λx1 + (1 − λ)x2 for some λ ∈ (0, 1). 2 An extreme point is simply a point in a convex set C that cannot be expressed as a strict convex combination of any other pair of points in C. We will see that extreme points must be located in specific locations in convex sets. Definition 4.28 (Boundary of a set). Let C ⊆ R n be (convex) set. A point x0 ∈ C is on the boundary of C if for all ɛ > 0, Bɛ(x0) ∩ C = ∅ and Bɛ(x0) ∩ R n \ C = ∅ Example 4.29. A convex set, its boundary and a boundary point are illustrated in Figure 4.8. BOUNDARY POINT INTERIOR BOUNDARY Figure 4.8. Boundary Point: A boundary point of a (convex) set C is a point in the set so that for every ball of any radius centered at the point contains some points inside C and some points outside C. Lemma 4.30. Suppose C is a convex set. If x is an extreme point of C, then x is on the boundary of C. Proof. Suppose not, then x is not on the boundary and thus there is some ɛ > 0 so that Bɛ(x0) ⊂ C. Since Bɛ(x0) is a hypersphere, we can choose two points x1 and x2 on the boundary of Bɛ(x0) so that the line segment between these points passes through the center of Bɛ(x0). But this center point is x0. Therefore x0 is the mid-point of x1 and x2 and since x1, x2 ∈ C and λx1 + (1 − λ)x2 = x0 with λ = 1/2 it follows that x0 cannot be an extreme point, since it is a strict convex combination of x1 and x2. This completes the proof. Most important in our discussion of linear programming will be the extreme points of polyhedral sets that appear in linear programming problems. The following theorem establishes the relationship between extreme points in a polyhedral set and the intersection of hyperplanes in such a set. 2 Thanks to Bob Pakzad-Hurson who fixed a typo in this definition in Version ≤ 1.4. 59
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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
Page 20 and 21: Figure 1.5. A Level Curve Plot with
Page 22 and 23: and thus our vector is v = (x1 −
Page 24 and 25: Figure 1.7. Gradients of the Bindin
Page 26 and 27: finishing toys and 120 hours per we
Page 28 and 29: 2. Graphically Solving Linear Progr
Page 30 and 31: S r x 0 Br( x0) Figure 2.2. A Bound
Page 32 and 33: S Every point on this line is an al
Page 34: 6. Problems with Unbounded Feasible
Page 37: to “prove” your example works.
Page 40 and 41: Definition 3.6 (Matrix Transpose).
Page 42 and 43: 3. Matrices and Linear Programming
Page 44 and 45: Remark 3.16. We can deal with const
Page 46 and 47: Theorem 3.20. Each elementary row o
Page 48 and 49: Gauss-Jordan Elimination Computing
Page 50 and 51: and right hand side vector b = [7,
Page 52 and 53: Gauss-Jordan elimination to solve t
Page 54 and 55: which clearly has a solution for al
Page 56 and 57: 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 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 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 :=
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This leads to a vector of reduced c
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2.1. Lexicographic Minimum Ratio Te
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Proof. The initial basis Im yieds a
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for i = 1, . . . , n. Then we have:
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Revised Simplex Algorithm (1) Ident
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simplex tableau). We obtain: z1 −
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Proof. We can prove Farkas’ Lemma
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A1· Half-space cy > 0 A2· c Am·
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3. The Karush-Kuhn-Tucker Condition
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Remark 8.9. The expressions: ∗ A
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(x ∗ 1,x ∗ 2) = (16, 72) 3x1 +
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Since w1, w2 ≥ 0, we know that w
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Example 8.13. Consider the followin
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Figure 8.5. This figure illustrates
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(9.4) Proof. Rewrite Problem D as:
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This vector will be related to c in
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3. Strong Duality Lemma 9.7. Proble
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Likewise beginning with the KKT con
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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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