Linear Programming Lecture Notes - Penn State Personal Web Server

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Definition 3.46 (Basic Variables). For historical reasons, the variables in the vector xB are called the basic variables and the variables in the vector xN are called the non-basic variables. We can use matrix multiplication to expand the left hand side of this expression as: (3.50) BxB + NxN = b The fact that B is composed of all linearly independent columns implies that applying Gauss- Jordan elimination to it will yield an m × m identity and thus that B is invertible. We can solve for basic variables xB in terms of the non-basic variables: (3.51) xB = B −1 b − B −1 NxN We can find an arbitrary solution to the system of linear equations by choosing values for the variables the non-basic variables and solving for the basic variable values using Equation 3.51. Definition 3.47. (Basic Solution) When we assign xN = 0, the resulting solution for x is called a basic solution and (3.52) xB = B −1 b Example 3.48. Consider the problem: 1 2 3 (3.53) 4 5 6 ⎡ ⎣ x1 ⎤ x2⎦ 7 = 8 x3 Then we can let x3 = 0 and: 1 2 (3.54) B = 4 5 We then solve1 : x1 (3.55) = B −1 x2 7 = 8 −19 3 20 3 Other basic solutions could be formed by creating B out of columns 1 and 3 or columns 2 and 3. and Exercise 38. Find the two other basic solutions in Example 3.48 corresponding to 2 3 B = 5 6 B = 1 3 4 6 In each case, determine what the matrix N is. [Hint: Find the solutions any way you like. Make sure you record exactly which xi (i ∈ {1, 2, 3}) is equal to zero in each case.] 1 Thanks to Doug Mercer, who found a typo below that was fixed. 46
11. Solving <strong>Linear</strong> Programs with Matlab In this section, we’ll show how to solve <strong>Linear</strong> Programs using Matlab. Matlab assumes that all linear programs are input in the following form: ⎧ min z(x) =c T x (3.56) ⎪⎨ ⎪⎩ s.t. Ax ≤ b Hx = r x ≥ l x ≤ u Here c ∈ R n×1 , so there are n variables in the vector x, A ∈ R m×n , b ∈ R m×1 , H ∈ R l×n and r ∈ R l×1 . The vectors l and u are lower and upper bounds respectively on the decision variables in the vector x. The Matlab command for solving linear programs is linprog and it takes the parameters: (1) c, (2) A, (3) b, (4) H, (5) r, (6) l, (7) u If there are no inequality constraints, then we set A = [] and b = [] in Matlab; i.e., A and b are set as the empty matrices. A similar requirement holds on H and r if there are no equality constraints. If some decision variables have lower bounds and others don’t, the term -inf can be used to set a lower bound at −∞ (in l). Similarly, the term inf can be used if the upper bound on a variable (in u) is infinity. The easiest way to understand how to use Matlab is to use it on an example. Example 3.49. Suppose I wish to design a diet consisting of Raman noodles and ice cream. I’m interested in spending as little money as possible but I want to ensure that I eat at least 1200 calories per day and that I get at least 20 grams of protein per day. Assume that each serving of Raman costs $1 and contains 100 calories and 2 grams of protein. Assume that each serving of ice cream costs $1.50 and contains 200 calories and 3 grams of protein. We can construct a linear programming problem out of this scenario. Let x1 be the amount of Raman we consume and let x2 be the amount of ice cream we consume. Our objective function is our cost: (3.57) x1 + 1.5x2 Our constraints describe our protein requirements: (3.58) 2x1 + 3x2 ≥ 20 and our calorie requirements (expressed in terms of 100’s of calories): (3.59) x1 + 2x2 ≥ 12 47
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Linear Programming: Penn State Math
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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 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 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
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
Page 105 and 106: A basic feasible solution for our a
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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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