sec7_1 - 7 Finding an initial feasible tableau To begin the...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
7 Finding an initial feasible tableau To begin the simplex method for solving a linear program in standard equal- ity form, we need to find an initial feasible tableau. Sometimes, this is easy. For example, the first linear program we studied was the problem maximize x 1 + x 2 subject to x 1 2 x 1 + 2 x 2 4 x 1 , x 2 0 . We transformed this problem to standard equality form by introducing slack variables: maximize x 1 + x 2 = z subject to x 1 + x 3 = 2 x 1 + 2 x 2 + x 4 = 4 x 1 , x 2 , x 3 , x 4 0 . Our initial system of equations is then z - x 1 - x 2 = 0 x 1 + x 3 = 2 x 1 + 2 x 2 + x 4 = 4 , and, as we observed, this system is already a feasible tableau, corresponding to the basis [3 , 4]. More generally, given any linear program in standard inequality form, maximize c T x subject to Ax b x 0 , if we transform to standard equality form by introducing slack variables, then the original system of constraints constitute a tableau for the trans- formed problem, where the basic variables are exactly the slack variables. Furthermore, this tableau is feasible, provided b 0. In general, however, it may not so easy to find an initial feasible tableau, or even to decide whether a system of constraints has a feasible solution. We next introduce a technique to resolve this difficulty. We illustrate with an example. Consider the linear program (7.1) maximize 2 x 1 + x 2 subject to x 1 + x 2 = 3 - x 1 + x 2 - x 3 = 1 x 1 , x 2 , x 3 0 . 40
Image of page 1

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

View Full Document Right Arrow Icon
For the moment, we ignore the objective function, and simply try to decide whether or not the problem has a feasible solution, and, if so, to find one. With this in mind, we introduce two new artificial variables that measure the error in each of the equality constraints: ( * ) x 4 = 3 - ( x 1 + x 2 ) x 5 = 1 - ( - x 1 + x 2 - x 3 ) . We will try to force these errors to zero by solving the linear program (7.2) max - x 4 - x 5 : ( * ) holds and x 1 , . . . , x 5 0 . Central to the success of this approach is the following property relationship between the linear programs (7.1) and (7.2): (7.1) is feasible (7.2) has an optimal solution with value zero. To see this, notice that, given any feasible solution for (7.1), setting x 4 = x 5 = 0 gives a feasible solution for (7.2) with value zero, and clearly no feasi- ble solution for (7.2) can have greater objective value than zero. Conversely, if a feasible solution for (7.2) has objective value zero, then both x 4 and x 5
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    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.

    Student Picture

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

  • Left Quote Icon

    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.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    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.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern