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Unformatted text preview: MATH 164/2 Optimization, Winter 2006, Midterm Exam Solutions Instructor: Luminita Vese Teaching Assistant: Ricardo Salazar [1] (10 points) (a) Consider the following feasible set S = n x ∈ R n : Ax ≥ b, x ≥ ~ o . Show that if the vector d satisfies d 6 = ~ 0, d ≥ ~ 0 and Ad ≥ ~ 0, then d is a direction of unboundedness for the set S . (b) Consider the following linear programming problem: Minimize z = x 1 5 x 2 + x 3 3 x 4 subject to 3 x 1 x 2 + 2 x 4 ≥ 2 2 x 1 + 3 x 4 ≥  1 x 2 x 3 ≥ 2 5 x 1 3 x 3 ≥ 2 x 1 , x 2 , x 3 , x 4 ≥ (b1) Show that x = (2 , 4 , 2 , 1) T is a feasible point to the problem and label each of the constraints as active or inactive. (b2) Show that the vector d = (1 , 2 , 1 , 1) T is a direction of unboundedness (note that this problem, as given, is not in standard form!) Solution: (a) We know by definition that d is a direction of unboundedness for S if d 6 = ~ and if x + γd ∈ S , for any x ∈ S and any γ ≥ 0. Let x ∈ S and γ ≥ 0 arbitrary. Then Ax ≥ b , and using the assumptions on d , we have: A ( x + γd ) = Ax + A ( γd ) = Ax + γAd ≥ b + γ ~ 0 = b . Therefore A ( x + γd ) ≥ b (1) Also, because x ≥ ~ 0, γ ≥ 0 and d ≥ ~ 0, we have x + γd ≥ ~ 0 + γ ~ 0 = ~ (2) From (1) and (2), we deduce that x + γd ∈ S for any x ∈ S and any γ ≥ 0, therefore d 6 = 0 is a direction of unboundedness for the set S . (b1) We verify that x = (2 , 4 , 2 , 1) T satisfies all eight constraints given. 3 · 2 4 + 2 · 1 = 4 > 2 (1st inactive) 2 · 2 + 3 · 1 = 1 = 1 (2nd active) 4 2 = 2 = 2 (3rd active) 5 · 2 3 · 2 = 4 > 2 (4th inactive) x 1 = 2 > 0 (inactive), x 2 = 4 > 0 (inactive), x 3 = 2 > 0 (inactive), x 4 = 1 > (inactive)....
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This note was uploaded on 08/24/2009 for the course MATH 262447221 taught by Professor Weisbart during the Spring '09 term at UCLA.
 Spring '09
 WEISBART
 Math

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