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What do these symbol mean?

  1. Sep 22, 2011 #1
    http://web.mit.edu/15.053/www/AMP-Chapter-04.pdf

    Go to page 8/45

    Where it tackles on the case of y being an unrestricted variable. They have the following

    [tex]y_i = y_i^+ - y_i^-[/tex]

    WHat do the plus and minus thing mean? It says they are both positive in page 9/45.

    Thank you
     
    Last edited by a moderator: Apr 26, 2017
  2. jcsd
  3. Sep 23, 2011 #2

    Mark44

    Staff: Mentor

    I don't think it means anything other than as a way to distinguish two sets with m variables, all using essentially the same name.

    IOW, y1+, y2+, ..., ym+ are the dual variables associated with the first m constraints, while y1-, y2-, ..., ym- are the dual variables associated with the second m constraints.
     
  4. Sep 23, 2011 #3
    My prof said something about representing their absolute values, but it doesn't make sense to have subtraction
     
  5. Sep 23, 2011 #4
    A common usage is

    [tex]x^+ = \begin{cases}
    x &\text{if } x \ge 0\\
    0 &\text{otherwise}
    \end{cases}[/tex]

    [tex]x^- = \begin{cases}
    -x &\text{if } x \le 0\\
    0 &\text{otherwise}
    \end{cases}[/tex]
     
  6. Sep 23, 2011 #5
    So if x => 0, then x^+ is x and x^-1 = 0

    So x^+ - 0 = x^+

    If x <=0, then x^- = -x and x^+ = 0

    0 - (-x) = x, but x<=0, so that doesn't work
     
  7. Sep 24, 2011 #6
    Flyingpig,

    I don't understand what you mean by "doesn't work".

    [tex]x = x^+ - x^-[/tex]

    in all cases. You just showed that.
     
  8. Sep 24, 2011 #7
    No but x can't be negative. I just showed in

    0 - (-x) = x, but x <=0,
     
  9. Sep 24, 2011 #8

    Pyrrhus

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    It depends on the context. For the LP, especially if you are using Simplex or another method that requires the decision variables to be positive, it is a way of allowing a variable that can also be negative.

    [tex] \sum_{i}^{m} b_{i} y_{i} = \sum_{i}^{m} b_{i} y^{+}_{i} - \sum_{i}^{m} b_{i} y^{-}_{i} [/tex]

    So now [itex] y_{i} [/itex] is unrestricted.
     
  10. Sep 24, 2011 #9
    Why can't you do addition?
     
  11. Sep 24, 2011 #10

    Pyrrhus

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    You have [itex] y_{i} [/itex] by the algorithm is required that [itex] y_{i} \geq 0 \forall i [/itex], but for your problem [itex] y_{i} [/itex] is unrestricted, so you can write that as the difference of two other variables that are nonpositive.
     
  12. Sep 24, 2011 #11
    But this inequality says otherwise and I showed it that y can be negative

     
  13. Sep 24, 2011 #12

    Pyrrhus

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    This is not right, you are simply replacing a decision variable with 2 other decision variables where the relation is [itex] y = y^{+} - y^{-} [/itex].

    There is no relation that when [itex] y > 0 \rightarrow y^{+} = y [/itex] That is wrong. Y is unrestricted and will take a positive sign only when [itex] y^{+} > y^{-} [/itex] and negative only when [itex] y^{+} < y^{-} [/itex]. Also, [itex] y^{+} \geq 0 [/itex] and [itex] y^{-} \geq 0 [/itex] by the assumptions of the problem.

    Awkward wrote another convention for the + that is unrelated to your LP problem. I am surprised you did not notice this?. Another convention for + is [itex] (t -a)^{+} = Max(t-a,0) [/itex].
     
  14. Sep 25, 2011 #13
    Yes, you showed [tex]x^+ - x^-[/tex] is equal to x when x is positive or zero, and it's equal to x when x is negative. So it's equal to x in all cases.

    That's the point.
     
  15. Sep 27, 2011 #14
    But what a Standard Form LOP, we want x > 0
     
  16. Sep 27, 2011 #15
    Oh wait...

    If x happens to even contradict one of these conditons

    [tex]x = x^+[/tex]
    [tex]x = x^-[/tex]

    Because the other one becomes 0! But that still doesn't explain the preference for subtraction over addition.

    Here take this one for instance

    Max

    [tex]z = 5x_1 + 4x_2 + 3x_3[/tex]

    s.t
    [tex]3x_1 + 3x_2 + x_3 \leq 5[/tex]
    [tex]4x_1 + 6x_2 + 3x_3 \geq 2[/tex]
    [tex]x_1 + 2x_3 = 4[/tex]

    [tex]x_1, x_2, \geq 0[/tex]

    So for that annoying x_3 constraints I get (by putting this in Standard Form)

    [tex]3x_1 + 3x_2 + (x_3 ^+ + x_3^-) \leq 5[/tex]
    [tex]-4x_1 - 6x_2 - 3(x_3 ^+ + x_3^-) \leq -2[/tex]
    [tex]x_1 + 2(x_3 ^+ + x_3^-) \leq -4[/tex]
    [tex]-x_1 - 2(x_3^+ + x_3^-) \geq 4[/tex]
    [tex]x_3 = x_3 ^+ - x_3 ^-[/tex]
     
  17. Sep 28, 2011 #16

    Pyrrhus

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    flyingpig, do you even read what I've said?

    Go back to your book and read about the inclusion of unrestricted decision variables in a linear program, and also re-read Simplex then come back for questions again. You've not grasped anything I've said.

    What part of this you don't understand? BOTH ARE NONNEGATIVE by the assumption in the LP!.

     
    Last edited: Sep 28, 2011
  18. Sep 28, 2011 #17
    Our book doesn't talk about special examples and yes I have read everyone else is saying, but it may not have gotten through my head. I am sorry, I am slow.

    Exactly, y can be negative. In the case of [tex]y^+ < y^-[/tex], this is not the goal of maxing LOP (most of the time).
     
  19. Sep 28, 2011 #18

    Pyrrhus

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    LOP? you mean the objective function? The goal of Max the objective function is to obtain the highest value within the constraint set.
     
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