Linear Programming Lecture Notes - Penn State Personal Web Server

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If a selfish profit maximizer wishes to buy these items, then we will seek a price per resource that minimizes the total he could pay for all the items, that is: m (9.46) min i=1 wibi The Strong Duality Theorem asserts that the optimal solution to this problem will produce fair shadow prices that force the total amount an individual could purchase the resources of the company for to be equal to the amount the company could make in manufacturing products itself. Example 9.13. Assume that a leather company manufactures two types of belts: regular and deluxe. Each belt requires 1 square yard of leather. A regular belt requires 1 hour of skilled labor to produce, while a deluxe belt requires 2 hours of labor. The leather company receives 40 square yards of leather each week and a total of 60 hours of skilled labor is available. Each regular belt nets $3 in profit, while each deluxe belt nets $5 in profit. The company wishes to maximize profit. We can compute the fair price the company could sell its time or labor (or the amount the company should be willing to pay to obtain more leather or more hours). The problem for the leather manufacturer is to solve the linear programming problem: max 3x1 + 5x2 s.t. x1 + 2x2 ≤ 60 x1 + x2 ≤ 40 x1, x2 ≥ 0 The dual problem for this problem is given as: max 60w1 + 40w2 s.t. w1 + w2 ≥ 3 2w1 + w2 ≥ 5 w1, w2 ≥ 0 If we solve the primal problem, we obtain the final revised simplex tableau as: ⎡ ⎤ z 2 1 160 x1 ⎣ −1 2 20 ⎦ 1 −1 20 x2 Note that both x1 and x2 are in the basis. In this case, we have w1 = 2 and w2 = 1 from Row 0 of the tableau. We can likewise solve the dual-problem by converting it to standard form and then using the simplex algorithm, we would have: max 60w1 + 40w2 s.t. w1 + w2 − v1 = 3 2w1 + w2 − v2 = 5 w1, w2, v1, v2 ≥ 0 150
In this case, it is more difficult to solve the dual problem because there is no conveniently obvious initial basic feasible solution (that is, the identity matrix is not embedded inside the coefficient matrix). The final full simplex tableau for the dual problem would look like: z w1 w2 ⎡ ⎢ ⎣ z w1 w2 v1 v2 RHS 1 0 0 −20 −20 160 0 1 0 1 −1 2 0 0 1 −2 1 1 We notice two things: The reduced costs of v1 and v2 are precisely the negatives of the values of x1 and x2. This was to be expected, these variables are duals of each other. However, in a minimization problem, the reduced costs have opposite sign. The second thing to notice is that w1 = 2 and w2 = 1. These are the same values we determined in the primal simplex tableau. Lastly, let’s see what happens if we increase the amount of leather available by 1 square yard. If w2 (the dual variable that corresponds to the leather constraint) is truly a shadow price, then we should predict our profit will increase by 1 unit. Our new problem will become: max 3x1 + 5x2 s.t. x1 + 2x2 ≤ 60 x1 + x2 ≤ 41 x1, x2 ≥ 0 At optimality, our new revised simplex tableau will be: z x1 ⎡ 2 ⎣ −1 1 2 ⎤ 161 22 ⎦ 1 −1 19 x2 Thus, if our Leather Manufacturer could obtain leather for a price under $1 per yard. He would be a fool not buy it. Because he could make an immediate profit. This is what economists call thinking at the margin. Shadow Prices Under Degeneracy. We have asserted that the value of the dual variable does not necessarily provide a shadow price at a degenerate constraint. To see this, consider the example of the degenerate toy maker problem. Example 9.14. max 7x1 + 6x2 s.t. 3x1 + x2 + s1 = 120 x1 + 2x2 + s2 = 160 x1 + s3 = 35 7 4 x1 + x2 + s4 = 100 x1, x2, s1, s2, s3, s4 ≥ 0 ⎤ ⎥ ⎦ 151
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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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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
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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 160 and 161: Figure 9.2. The simplex algorithm b
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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