Midterm_Math164Winter2006_SOL

# Midterm_Math164Winter2006_SOL - MATH 164/2 Optimization...

This preview shows pages 1–2. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern