# What do these symbol mean?

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

$$y_i = y_i^+ - y_i^-$$

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

Thank you

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Mark44
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.

My prof said something about representing their absolute values, but it doesn't make sense to have subtraction

A common usage is

$$x^+ = \begin{cases} x &\text{if } x \ge 0\\ 0 &\text{otherwise} \end{cases}$$

$$x^- = \begin{cases} -x &\text{if } x \le 0\\ 0 &\text{otherwise} \end{cases}$$

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

Flyingpig,

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

$$x = x^+ - x^-$$

in all cases. You just showed that.

No but x can't be negative. I just showed in

0 - (-x) = x, but x <=0,

Pyrrhus
Homework Helper
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.

$$\sum_{i}^{m} b_{i} y_{i} = \sum_{i}^{m} b_{i} y^{+}_{i} - \sum_{i}^{m} b_{i} y^{-}_{i}$$

So now $y_{i}$ is unrestricted.

Pyrrhus
Homework Helper
You have $y_{i}$ by the algorithm is required that $y_{i} \geq 0 \forall i$, but for your problem $y_{i}$ is unrestricted, so you can write that as the difference of two other variables that are nonpositive.

$$x^+ = \begin{cases} x &\text{if } x \ge 0\\ 0 &\text{otherwise} \end{cases}$$

$$x^- = \begin{cases} -x &\text{if } x \le 0\\ 0 &\text{otherwise} \end{cases}$$
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

Pyrrhus
Homework Helper
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 $y = y^{+} - y^{-}$.

There is no relation that when $y > 0 \rightarrow y^{+} = y$ That is wrong. Y is unrestricted and will take a positive sign only when $y^{+} > y^{-}$ and negative only when $y^{+} < y^{-}$. Also, $y^{+} \geq 0$ and $y^{-} \geq 0$ 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 $(t -a)^{+} = Max(t-a,0)$.

No but x can't be negative. I just showed in

0 - (-x) = x, but x <=0,
Yes, you showed $$x^+ - x^-$$ 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.

Yes, you showed $$x^+ - x^-$$ 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

A common usage is

$$x^+ = \begin{cases} x &\text{if } x \ge 0\\ 0 &\text{otherwise} \end{cases}$$

$$x^- = \begin{cases} -x &\text{if } x \le 0\\ 0 &\text{otherwise} \end{cases}$$
Oh wait...

If x happens to even contradict one of these conditons

$$x = x^+$$
$$x = x^-$$

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

$$z = 5x_1 + 4x_2 + 3x_3$$

s.t
$$3x_1 + 3x_2 + x_3 \leq 5$$
$$4x_1 + 6x_2 + 3x_3 \geq 2$$
$$x_1 + 2x_3 = 4$$

$$x_1, x_2, \geq 0$$

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

$$3x_1 + 3x_2 + (x_3 ^+ + x_3^-) \leq 5$$
$$-4x_1 - 6x_2 - 3(x_3 ^+ + x_3^-) \leq -2$$
$$x_1 + 2(x_3 ^+ + x_3^-) \leq -4$$
$$-x_1 - 2(x_3^+ + x_3^-) \geq 4$$
$$x_3 = x_3 ^+ - x_3 ^-$$

Pyrrhus
Homework Helper
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 $y^{+} > y^{-}$ and negative only when $y^{+} < y^{-}$. Also, $y^{+} \geq 0$ and $y^{-} \geq 0$ by the assumptions of the problem.

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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.
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 $y^{+} > y^{-}$ and negative only when $y^{+} < y^{-}$. Also, $y^{+} \geq 0$ and $y^{-} \geq 0$ by the assumptions of the problem.
Exactly, y can be negative. In the case of $$y^+ < y^-$$, this is not the goal of maxing LOP (most of the time).

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