Linear Programming Lecture Notes - Penn State Personal Web Server

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S r x 0 Br( x0) Figure 2.2. A Bounded Set: The set S (in blue) is bounded because it can be entirely contained inside a ball of a finite radius r and centered at some point x0. In this example, the set S is in R 2 . This figure also illustrates the fact that a ball in R 2 is just a disk and its boundary. Algorithm for Solving a Linear Programming Problem Graphically Bounded Feasible Region, Unique Solution (1) Plot the feasible region defined by the constraints. (2) Plot the level sets of the objective function. (3) For a maximization problem, identify the level set corresponding the greatest (least, for minimization) objective function value that intersects the feasible region. This point will be at a corner. (4) The point on the corner intersecting the greatest (least) level set is a solution to the linear programming problem. Algorithm 1. Algorithm for Solving a Two Variable Linear Programming Problem Graphically–Bounded Feasible Region, Unique Solution Case Exercise 13. Use the graphical method for solving linear programming problems to solve the linear programming problem you defined in Exercise 10. 4. Problems with Alternative Optimal Solutions We’ll study a specific linear programming problem with an infinite number of solutions by modifying the objective function in Example 2.3. 18
Example 2.7. Suppose the toy maker in Example 2.3 finds that it can sell planes for a profit of $18 each instead of $7 each. The new linear programming problem becomes: ⎧ max z(x1, x2) = 18x1 + 6x2 s.t. 3x1 + x2 ≤ 120 ⎪⎨ x1 + 2x2 ≤ 160 (2.10) x1 ≤ 35 ⎪⎩ x1 ≥ 0 x2 ≥ 0 Applying our graphical method for finding optimal solutions to linear programming problems yields the plot shown in Figure 2.3. The level curves for the function z(x1, x2) = 18x1 + 6x2 are parallel to one face of the polygon boundary of the feasible region. Hence, as we move further up and to the right in the direction of the gradient (corresponding to larger and larger values of z(x1, x2)) we see that there is not one point on the boundary of the feasible region that intersects that level set with greatest value, but instead a side of the polygon boundary described by the line 3x1 + x2 = 120 where x1 ∈ [16, 35]. Let: S = {(x1, x2)|3x1 + x2 ≤ 120, x1 + 2x2 ≤ 160, x1 ≤ 35, x1, x2 ≥ 0} that is, S is the feasible region of the problem. Then for any value of x ∗ 1 ∈ [16, 35] and any value x ∗ 2 so that 3x ∗ 1 + x ∗ 2 = 120, we will have z(x ∗ 1, x ∗ 2) ≥ z(x1, x2) for all (x1, x2) ∈ S. Since there are infinitely many values that x1 and x2 may take on, we see this problem has an infinite number of alternative optimal solutions. Based on the example in this section, we can modify our algorithm for finding the solution to a linear programming problem graphically to deal with situations with an infinite set of alternative optimal solutions (see Algorithm 2): Algorithm for Solving a Linear Programming Problem Graphically Bounded Feasible Region (1) Plot the feasible region defined by the constraints. (2) Plot the level sets of the objective function. (3) For a maximization problem, identify the level set corresponding the greatest (least, for minimization) objective function value that intersects the feasible region. This point will be at a corner. (4) The point on the corner intersecting the greatest (least) level set is a solution to the linear programming problem. (5) If the level set corresponding to the greatest (least) objective function value is parallel to a side of the polygon boundary next to the corner identified, then there are infinitely many alternative optimal solutions and any point on this side may be chosen as an optimal solution. Algorithm 2. Algorithm for Solving a Two Variable Linear Programming Problem Graphically–Bounded Feasible Region Case Exercise 14. Modify the linear programming problem from Exercise 10 to obtain a linear programming problem with an infinite number of alternative optimal solutions. Solve 19
Page 1: Linear Programming: Penn State Math
Page 4 and 5: 8. Caratheodory Characterization Th
Page 7 and 8: List of Figures 1.1 Goat pen with u
Page 9 and 10: 4.6 Convex Direction: Clearly every
Page 11: Preface Stop! Stop right now! This
Page 14 and 15: x Goat Pen Figure 1.1. Goat pen wit
Page 16 and 17: We have formulated the general maxi
Page 18 and 19: 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 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 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
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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
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Remark 6.6. In the case of a minimi
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Example 6.13. Suppose we solve the
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That is, s1 = −12 and s2 = −20
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The complete problem describing McL
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* 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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