What do these symbol mean?

  • Thread starter flyingpig
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  • #1
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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
 
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Answers and Replies

  • #2
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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.
 
  • #3
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My prof said something about representing their absolute values, but it doesn't make sense to have subtraction
 
  • #4
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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]
 
  • #5
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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
 
  • #6
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Flyingpig,

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

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

in all cases. You just showed that.
 
  • #7
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No but x can't be negative. I just showed in

0 - (-x) = x, but x <=0,
 
  • #8
Pyrrhus
Homework Helper
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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.
 
  • #9
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Why can't you do addition?
 
  • #10
Pyrrhus
Homework Helper
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Why can't you do addition?
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.
 
  • #11
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[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]
But this inequality says otherwise and I showed it that y can be negative

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
 
  • #12
Pyrrhus
Homework Helper
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But this inequality says otherwise and I showed it that y can be negative
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].
 
  • #13
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No but x can't be negative. I just showed in

0 - (-x) = x, but x <=0,
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.
 
  • #14
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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.
But what a Standard Form LOP, we want x > 0
 
  • #15
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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]
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]
 
  • #16
Pyrrhus
Homework Helper
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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!.

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.
 
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  • #17
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1
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.
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.

What part of this you don't understand? BOTH ARE NONNEGATIVE by the assumption in the LP!.
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.
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).
 
  • #18
Pyrrhus
Homework Helper
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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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