Midterm_Math164Winter2006_SOL

Midterm_Math164Winter2006_SOL - MATH 164/2 Optimization...

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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 ~ 0 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 0 (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 = ~ 0 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 = ~ 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 > 0 (inactive). Therefore the point x = (2 , 4 , 2 , 1) T satisfies all eight constraints of the problem, so it is a feasible point, x S . (b2) Notice that d = (1 , 2 , 1 , 1) T 6 = ~ 0 and d ~ 0. Therefore, using the proof for (a), it is sufficient to show that Ad ~ 0, where A is the matrix corresponding to the
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