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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 + d S , for any x S and any 0. Let x S and 0 arbitrary. Then Ax b , and using the assumptions on d , we have: A ( x + d ) = Ax + A ( d ) = Ax + Ad b + ~ 0 = b . Therefore A ( x + d ) b (1) Also, because x ~ 0, 0 and d ~ 0, we have x + d ~ 0 + ~ 0 = ~ (2) From (1) and (2), we deduce that x + d S for any x S and any 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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 Spring '09
 WEISBART
 Math

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