# What do these symbol mean?

1. Sep 22, 2011

### flyingpig

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

Last edited by a moderator: Apr 26, 2017
2. Sep 23, 2011

### 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.

3. Sep 23, 2011

### flyingpig

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

4. Sep 23, 2011

### awkward

A common usage is

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

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

5. Sep 23, 2011

### flyingpig

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. Sep 24, 2011

### awkward

Flyingpig,

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

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

in all cases. You just showed that.

7. Sep 24, 2011

### flyingpig

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

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

8. Sep 24, 2011

### Pyrrhus

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.

9. Sep 24, 2011

### flyingpig

10. Sep 24, 2011

### Pyrrhus

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.

11. Sep 24, 2011

### flyingpig

But this inequality says otherwise and I showed it that y can be negative

12. Sep 24, 2011

### Pyrrhus

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 $(t -a)^{+} = Max(t-a,0)$.

13. Sep 25, 2011

### awkward

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.

14. Sep 27, 2011

### flyingpig

But what a Standard Form LOP, we want x > 0

15. Sep 27, 2011

### flyingpig

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)

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

16. Sep 28, 2011

### Pyrrhus

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
17. Sep 28, 2011

### flyingpig

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 $$y^+ < y^-$$, this is not the goal of maxing LOP (most of the time).

18. Sep 28, 2011

### Pyrrhus

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