Linear Programming Lecture Notes - Penn State Personal Web Server

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Gauss-Jordan Elimination Computing an Inverse (1) Let A ∈ Rn×n . Let X = [A|In]. (2) Let i := 1 (3) If Xii = 0, then use row-swapping on X to replace row i with a row j (j > i) so that Xii = 0. If this is not possible, then A is not invertible. (4) Replace Xi· by (1/Xii)Xi·. Element (i, i) of X should now be 1. (5) For each j = i, replace Xj· by −Xji Xii Xi· + Xj·. (6) Set i := i + 1. (7) If i > n, then A has been replaced by In and In has been replaced by A−1 in X. If i ≤ n, then goto Line 3. Algorithm 5. Gauss-Jordan Elimination for Matrix Inversion Step 5 (i = 1): We multiply row 1 of X by −3 and add the result to row 2 of X to obtain: 1 X := 0 2 −2 1 −3 0 1 Step 6: i := 1 + 1 = 2 and i = n so we return to Step 3. Steps 3 (i = 2): The new element A22 = −2 = 0. Therefore, no swapping is required. Step 4 (i = 2): We replace row 2 of X with row 2 of X multiplied by −1/2. 1 2 1 0 X := 0 1 3 2 − 1 2 Step 5 (i = 2): We multiply row 2 of X by −2 and add the result to row 1 of X to obtain: 1 X := 0 −2 1 0 1 3 2 1 − 2 Step 6 (i = 2): i := 2 + 1 = 3. We now have i > n and the algorithm terminates. Thus using Algorithm 5 we have computed: A −1 −2 1 = 3 2 1 − 2 This value should be the result you obtained in Exercise 27. Exercise 28. Does the matrix: ⎡ ⎤ 1 2 3 A = ⎣4 5 6⎦ 7 8 9 have an inverse? [Hint: Use Gauss-Jordan elimination to find the answer.] Exercise 29 (Bonus). Implement Gauss-Jordan elimination in the programming language of your choice. Illustrate your implementation by using it to solve the previous exercise. [Hint: Implement sub-routines for each matrix operation. You don’t have to write them as matrix multiplication, though in Matlab, it might speed up your execution. Then use these subroutines to implement each step of the Gauss-Jordan elimination.] 36
6. Solution of <strong>Linear</strong> Equations Let A ∈ R n×n , b ∈ R n . Consider the problem: (3.26) Ax = b There are three possible scenarios: (1) There is a unique solution x = A −1 b. (2) There are no solutions; i.e., there is no vector x so that Ax = b. (3) There are infinitely many solutions; i.e., there is an infinite set X ⊆ R n such that for all x ∈ X , Ax = b. We can use Gauss-Jordan elimination to find a solution x to Equation 3.26 when it occurs. Instead of operating on the augmented X = [A|In] in Algorithm 5, we use the augmented matrix X := [A|b]. If the algorithm terminates with A replaced by In, then the solution vector resides where b began. That is, X is transformed to X := [In|A −1 b]. If the algorithm does not terminate with X := [In|A −1 b], then suppose the algorithm terminates with X := [A ′ |b ′ ]. There is at least one row in A ′ with all zeros. That is A ′ has the form: (3.27) A ′ ⎡ 1 0 . . . ⎤ 0 ⎢ 0 ⎢ ⎢. = ⎢ ⎢0 ⎢ ⎣. 1 0 . . . . .. . . . . .. 0 ⎥ . ⎥ 0 ⎥ . ⎦ 0 0 . . . 0 In this case, there are two possibilities: (1) For every zero row in A ′ , the corresponding element in b ′ is 0. In this case, there are an infinite number of alternative solutions to the problem expressed by Equation 3.26. (2) There is at least one zero row in A ′ whose corresponding element in b ′ is not zero. In this case, there are no solutions to Equation 3.26. We illustrate these two possibilities in the following example: Example 3.26. Consider the system of equations: x1 + 2x2 + 3x3 = 7 4x1 + 5x2 + 6x3 = 8 7x1 + 8x2 + 9x3 = 9 This yields matrix: ⎡ 1 A = ⎣4 2 5 ⎤ 3 6⎦ 7 8 9 37
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 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 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
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
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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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