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Is bases the same as basis ? (Simplex Algorithm)

  1. Sep 24, 2011 #1
    1. The problem statement, all variables and given/known data

    [PLAIN]http://img193.imageshack.us/img193/3662/unledmcg.png [Broken]

    3. The attempt at a solution

    I rewrote the whole thing in dictionary

    [tex]x_3 = 15 - 8x_1 - 4x_2[/tex]
    [tex]x_4 = 7 - 2x_1 - 6x_2[/tex]
    [tex]z = 0 + 22x_1 - 12x_2[/tex]

    [tex]x_i \geq 0 [/tex]
    [tex]1\leq i \leq 4[/tex]

    a) So my basis/bases is x = (x_3,x_4)

    b) Do I have to solve the above using the Simplex Algorithm first?

    c) I have to set x_1 or x_2 = 0 and the other one not 0.

    So doing so I get (I chose x_1 and let x_2 = t)

    [tex]x_3 = 15 - 4x_2[/tex]
    [tex]x_4 = 7 - 6x_2[/tex]
    [tex]z = -12x_2[/tex]

    So I get a line of

    x = (0,t|15-t,7-6t)^t

    So my answer is a line with [tex]t \in (-\infty, \frac{7}{6})[/tex]

    I have infinity in my interval...something does not feel right
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Sep 24, 2011 #2


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    Staff Emeritus
    Science Advisor
    Gold Member

    Bases is the plural of basis. One basis, two bases. I have to go and do something else now, so I don't have time to help you with the actual problem, but I'm sure someone else will.
  4. Sep 24, 2011 #3
    How could there be more than one basis for any LOP?
  5. Sep 28, 2011 #4
    I need to make a correction

    When they mean bases, they just mean all combination of the basic variables.

    I happen to have 4 variables and I could only only choose 2 slacks, so that's 6 combination namely

    [tex]\beta = \left \{ 1,2 \right \}, \left \{ 1,3 \right \}, \left \{ 1,4 \right \}, \left \{ 2,3 \right \}, \left \{ 2,4 \right \}, \left \{ 3,4 \right \}[/tex]
  6. Oct 2, 2011 #5
    I just have one question, it's given that at least one basic solution HAS to be an optimal right? There can't be two optimal?
  7. Oct 2, 2011 #6


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    Homework Helper

    There may be multiple optimal solutions, but yes at least one (if the Weierstrass theorem is satisfied).
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