{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

HW3SolutionKey

# HW3SolutionKey - DMOR OEM 2007 Homework 3 Solution Key...

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

DMOR OEM 2007 Homework 3 Solution Key Solution for problem 1: (questions, comments, suggestions please to Zeki Caner Taskin, [email protected] ) a) Since there are 4 variables and 3 equality constraints, there is a single nonbasic variable and 3 basic variables. First let x 1 be the nonbasic variable. Setting x 1 = 0 essentially eliminates it from the system, and we are left with: x 2 - x 3 = 1 2 x 2 = 3 2 x 2 + x 3 + x 4 = 3 This is a linear system of 3 variables and 3 equalities with a unique solution. Therefore our first basic solution is ( x 1 , x 2 , x 3 , x 4 ) = (0 , 1 . 5 , 0 . 5 , - 0 . 5). Since x 4 < 0, this is not a basic feasible solution. Similarly we set x 2 = 0 and solve for the basic variables to get ( x 1 , x 2 , x 3 , x 4 ) = (3 , 0 , 2 , 1). Since all variables are nonnegative, this corresponds to a basic feasible solution and hence an extreme point. Setting x 3 = 0 leads to ( x 1 , x 2 , x 3 , x 4 ) = ( - 1 , 2 , 0 , - 1), which is infea- sible, and finally setting x 4 = 0 leads to ( x 1 , x 2 , x 3 , x 4 ) = (1 , 1 , 1 , 0), which is another extreme point. b) The underlying idea is the same as part a) except the way we treat x 4 . x 4 is still a nonbasic variable, but here we treat it as a constant and move to the right hand side of the equations (instead of setting it equal to 0 and removing from the system altogether). x 1 + x 2 - x 3 = 1 x 1 + 2 x 2 = 3 2 x 2 + x 3 = 3 - x 4 Adding the first and the last equalities yields ( x 1 + 3 x 2 = 4 - x 4 ). Subtracting the second equation from this, we get ( x 2 = 1 - x 4 ). Plugging

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