Linear Programming Lecture Notes - Penn State Personal Web Server

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Figure 4.9. A Polyhedral Set: This polyhedral set is defined by five half-spaces and has a single degenerate extreme point located at the intersection of the binding constraints 3x1 + x2 ≤ 120, x1 + 2x2 ≤ 160 and 28 16x1 + x2 0 so that d = λ1d1 + λ2d2. We have already seen by Theorem 4.24 that is P is a polyhedral set in the positive orthant of R n with form: P = {x ∈ R n : Ax ≤ b, x ≥ 0} then a direction d of P is characterized by the set of inequalities and equations Ad ≤ 0, d ≥ 0, d = 0. Clearly two directions d1 and d2 with d1 = λd2 for some λ ≥ 0 may both satisfy this system. To isolate a unique set of directions, we can normalize and construct the set: (4.28) D = {d ∈ R n : Ad ≤ 0, d ≥ 0, e T d = 1} 62
here we are interested only in directions satisfying e T d = 1. This is a normalizing constraint that will chose only vectors whose components sum to 1. Theorem 4.39. A direction d ∈ D is an extreme direction of P if and only if d is an extreme point of D when D is taken as a polyhedral set. Proof. (⇒)Suppose that d is an extreme point of D (as a polyhedral set) and not an extreme direction of P . Then there exist two directions d1 and d2 of P and two constants λ1 and λ2 with λ1, λ2 ≥ 0 so that d = λ1d1 + λ2d2. Without loss of generality, we may assume that d1 and d2 are vectors satisying e T di = 1 (i = 1, 2). If not, then we can scale them so their components sum to 1 and adjust λ1 and λ2 accordingly. But this implies that: 1 = e T d = λ1e T d1 + λ2e T d2 = λ1 + λ2 Further, the fact that d1 and d2 are directions of P implies they must be in D. Thus we have found a convex combination of element of D that equals d, contradicting our assumption that d was an extreme point. (⇐)Conversely, suppose that d is an extreme direction of P whose components sum to 1 and not an extreme point of D. (Again, we could scale d if needed.) Then d cannot be recovered from a positive combination of directions d1 and d2. But if d is not an extreme point of D, then we know λ1d1 +λ2d2 = d for d1, d2 ∈ D and λ1 +λ2 = 1 and λ1, λ2 ∈ (0, 1). This is clearly contradictory. Every strict convex combination is a positive combination and therefore our assumption that d was an extreme direction was false. Example 4.40. Let’s consider Example 4.26 again. The polyhedral set in this example was defined by the A matrix: 1 −1 A = −2 −1 and the b vector: 1 b = −6 If we assume that P = {x ∈ R n : Ax ≤ b, x ≥ 0}, then the set of extreme directions of P is the same as the set of extreme points of the set D = {d ∈ R n : Ad ≤ 0, d ≥ 0, e T d = 1} Then we have the set of directions d = [d1, d2] T so that: d1 − d2 ≤ 0 −2d1 − d2 ≤ 0 d1 + d2 = 1 d1 ≥ 0 d2 ≥ 0 The feasible region (which is really only the line d1 + d2 = 1) is shown in red in Figure 4.10. The critical part of this figure is the red line. It is the true set D. As a line, it has two extreme points: (0, 1) and (1/2, 1/2). Note that (0, 1) as an extreme point is one of the direction [0, 1] T we illustrated in Example 4.26. 63
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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 −
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
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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
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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 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
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Page 103 and 104: CHAPTER 6 Simplex Initialization In
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Page 111 and 112: Remark 6.6. In the case of a minimi
Page 113 and 114: Example 6.13. Suppose we solve the
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Page 117 and 118: The complete problem describing McL
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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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