Linear Programming Lecture Notes - Penn State Personal Web Server

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Figure 9.2. The simplex algorithm begins at a feasible point in the feasible region of the primal problem. In this case, this is also the same starting point in the dual problem, which is infeasible. The simplex algorithm moves through the feasible region of the primal problem towards a point in the dual feasible region. At the conclusion of the algorithm, the algorithm reaches the unique point that is both primal and dual feasible. feasible region. Your dual vector w will not enter the feasible region until the last simplex pivot.] 5. Economic Interpretation of the Dual Problem Consider again, the value of the objective function in terms of the values of the non-basic variables (Equation 5.11): (9.37) z = cx = cBB −1 b + cN − cBB −1 N xN Suppose we are at a non-degenerate optimal point. We’ve already observed that: ∂z (9.38) = −(zj − cj) = cj − cBB ∂xj −1 A·j We can rewrite all these equations in terms of our newly defined term: (9.39) w = cBB −1 to obtain: (9.40) z = wb + (cN − wN) xN Remember, w is the vector of dual variables corresponding to the constraints in our original problem P . Suppose we fix the values of xN. elements with the property that: Then we can see that the vector w has individual (9.41) ∂z ∂bi = wi 148
That is, the i th element of w represents the amount that the objective function value would change at optimality assuming we could modify the right-hand-side of the constraints. Note, this result holds only in the absence of degeneracy, for reasons we will see in an example. Thus, we can think of wi as the shadow price for resource i (the right-hand-side of the i th constraint). A shadow price is the fair price one would pay for an extra unit of resource i. Example 9.12. Consider a leather company that requires 1 square yard of leather to make a regular belt and a 1 square yard of leather to make a deluxe belt. If the leather company can use up to 40 square yards per week to construct belts, then one constraint it may have is: x1 + x2 ≤ 40 In the absence of degeneracy, the dual variable (say w1) will tell the fair price we would pay for 1 extra yard of leather. Naturally, if this were not a binding constraint, then w1 = 0 indicating that extra leather is worth nothing to us since we already have a surplus of leather. To understand the economics of the situation, suppose that we a manufacturer is to produce products 1, . . . , n and we produce x1, . . . , xn of each. If we can sell each product for a profit of c1, . . . , cn, then we wish to find values for x1, . . . , xn to solve: (9.42) max n j=1 cjxj Simultaneously, suppose that m resources (leather, wood, time etc.) are used to make these n products and that aij units of resource i are used to manufacture product j. Then clearly our constraints will be: (9.43) ai1x1 + · · · + ainxn ≤ bi where bi is the amount of resource i available to the company. Suppose now that the company decides to sell off some of its resources (instead of manufacturing products). Suppose we sell each resource for a price wi (i = 1, . . . , m) we’d like to know what a fair price for these resources would be. Each unit of product j not manufactured would result in a loss of profit of cj. At the same time, we would obtain a gain (from selling the excess resources) of: (9.44) m i=1 aijwi Because we would save aij units of unit i from not manufacturing 1 unit of xj (i = 1, . . . , m). Selling this resource would require us to make more money in the sale of the resource then we could in manufacturing the product, or: (9.45) m i=1 aijwi ≥ cj 149
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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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x1 x5 x2 x λx2 +(1− λ)x3 x4 x3
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If there is some i such that c T di
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Consider now the fact xj = 0 for al
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Example 5.9. Consider the Toy Maker
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using this information, we can comp
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Figure 5.1. The Simplex Algorithm:
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Using this information, we can see
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Based on this information, we can c
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Using the rule you developed in Exe
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Exercise 52. Consider the diet prob
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Suppose we choose s4 as the leaving
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CHAPTER 6 Simplex Initialization In
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A basic feasible solution for our a
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variables. This is because we start
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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
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:
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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 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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