Did I mess up in this inequality?

[flyingpig]

Homework Statement

[PLAIN]http://img703.imageshack.us/img703/7445/unledhpu.png

The Attempt at a Solution

I am having problems with (c), (e) but I will show yu what I did for the others first. I also I forewarn thee that we haven't learned the Simplex Algorithm yet (we might learn it at the end of the week)

(a)I did the whole inequality thing and I got

min

$$w = 1203y_1 + 1551y_2$$

$$0 \geq -6y_1$$
$$5 \geq 4y_1 + 7y_2$$
$$14 \geq 10y_1 + 15y_2$$

$$y_1, y_2 \leq 0$$

(b), I found that y = (0,0)^t worked

(this will be useful for (e))

$$0 \geq 0 = -6(0)$$
$$5 \geq 0 = 4(0) + 7(0)$$
$$14 \geq 0 = 10(0) + 15(0)$$

(c)

Here is the problem, with the first inequality

$$-6t + 4(100) + 10(50) = -6t + 900 \leq 1203$$

$$-6t \leq 303$$

$$t \geq -50.5$$

This doesn't say that all t are positive, I mean okay, all positive values do work because neither the other constraint nor the objective function depend on x_1

Does it make sense to say that "Since $$t \geq -50.5$$ which therefore includes $$t \geq 0$$, hence x = (t, 100,50)^t works for all $$t \geq 0$$"

The other constraints (if you are wondering) all work. So my question is, do I have to worry about the values $$-50.5 \leq t < 0$$?

I excluded 0 because it works.

(d) Nothing fancy here, just z = 5(100) + 14(50) = 1200

(e) If P is feasible, there exists a feasible $$x = (x_1, x_2, x_3)^t$$.

Since I confirmed that my $$y = (0,0)^t$$ (or at least I think yu should believe me) is D-feasible, by the Weak Duality Thrm

I have the inequality

$$c^t \leq y^t A \leq y^t b$$

However there is a problem.

My A matrix is $$\begin{pmatrix} -6 & 4& 10\\ 0& 7& 15 \end{pmatrix}$$ This is 2 by 3

And my y^t is $$\begin{pmatrix} 0\\ 0 \end{pmatrix}$$

How can they multiply together?

(f) need to answer (e) before

(g) I think it follows from (f)? Intuition.

 

[Ray Vickson]

Homework Statement

[PLAIN]http://img703.imageshack.us/img703/7445/unledhpu.png

The Attempt at a Solution

I am having problems with (c), (e) but I will show yu what I did for the others first. I also I forewarn thee that we haven't learned the Simplex Algorithm yet (we might learn it at the end of the week)

(a)I did the whole inequality thing and I got

min

$$w = 1203y_1 + 1551y_2$$

$$0 \geq -6y_1$$
$$5 \geq 4y_1 + 7y_2$$
$$14 \geq 10y_1 + 15y_2$$

$$y_1, y_2 \leq 0$$

(b), I found that y = (0,0)^t worked

(this will be useful for (e))

$$0 \geq 0 = -6(0)$$
$$5 \geq 0 = 4(0) + 7(0)$$
$$14 \geq 0 = 10(0) + 15(0)$$

(c)

Here is the problem, with the first inequality

$$-6t + 4(100) + 10(50) = -6t + 900 \leq 1203$$

$$-6t \leq 303$$

$$t \geq -50.5$$

This doesn't say that all t are positive, I mean okay, all positive values do work because neither the other constraint nor the objective function depend on x_1

Does it make sense to say that "Since $$t \geq -50.5$$ which therefore includes $$t \geq 0$$, hence x = (t, 100,50)^t works for all $$t \geq 0$$"

The other constraints (if you are wondering) all work. So my question is, do I have to worry about the values $$-50.5 \leq t < 0$$?

I excluded 0 because it works.

(d) Nothing fancy here, just z = 5(100) + 14(50) = 1200

(e) If P is feasible, there exists a feasible $$x = (x_1, x_2, x_3)^t$$.

Since I confirmed that my $$y = (0,0)^t$$ (or at least I think yu should believe me) is D-feasible, by the Weak Duality Thrm

I have the inequality

$$c^t \leq y^t A \leq y^t b$$

However there is a problem.

My A matrix is $$\begin{pmatrix} -6 & 4& 10\\ 0& 7& 15 \end{pmatrix}$$ This is 2 by 3

And my y^t is $$\begin{pmatrix} 0\\ 0 \end{pmatrix}$$

How can they multiply together?

(f) need to answer (e) before

(g) I think it follows from (f)? Intuition.

My dual problem is a lot different from yours. Certainly, (y1,y2) = (0,0) does not work.

RGV

 

[flyingpig]

Nope... i messed up my leq and geq again...

I swear it was the other way around last night when I typed this up in preview...I swear...

(a)

$$w = 1203y_1 + 1551y_2$$

$$0 \leq -6y_1$$

$$5 \leq 4y_1 + 7y_2$$

$$14 \leq 10y_1 + 15y_2$$

$$y_1, y_2 \geq 0$$

(b)

A feasible solution would be $$y = (0,1)^t$$

$$w = 1203(0) + 1551(1) = 1551$$

$$0 \leq 0 = -6(0)$$

$$5 \leq 7 = 4(0) + 7(1)$$

$$14 \leq 15 = 10(0) + 15(1)$$

$$0,1 \geq 0$$

 

[Ray Vickson]

This is OK now.

RGV

 

[flyingpig]

Not okay yet

(c) Still unsure

(e) Can't do anything about this

 

[flyingpig]

Omg I got it.

Admittedly I am still stuck on (c), but the others are clear

(e)

I just need to test the x given in (d)

So z= 5(100) + 14(50) = 1200

So as $$t \to \pm \infty$$, $$z \to 1200$$ for all t

(f) No, P is bounded, the limit in (e) shows

(e) No, since P is bounded, so must D.

 

[flyingpig]

I need some comments...anything is okay.

Ray...I love you? Please come back.

 

[flyingpig]

I take back everything I said...

[PLAIN]http://img851.imageshack.us/img851/5065/unledurp.png

The left one is drawn by me and the right one is drawn by computer

From

$$-6y_1 \geq 0$$

It follows that

$$y_1 \leq 0$$

I have also found that the two lines intersect only once and it is for x > 0, http://www.wolframalpha.com/input/?i=RowReduce{{4%2C7%2C5}%2C{10%2C15%2C14}}

So D is unbounded.

How do I put this into Math? Like it didn't ask me to draw a graph and make this analysis.

How do I do this algebraically?

 

[flyingpig]

Also, from Linear Algebra and intuition I have the matrix for the dual

$$\left[ \begin{array}{cc|c} 1 & 0 & 0 \\ 4 & 7 & 5 \\ 10 & 15 & 14 \\ \end{array} \right]$$

I have 3 rows and 2 columns, I must have a free variable.

 

[flyingpig]

I tried e) again and it had NOTHING to do with the dual, at all...e) I need to find a number K such that

$$|x_1| \leq K$$, $$|x_2| \leq K$$, $$|x_3| \leq K$$

Here is the problem, I am not sure how to find this K and what to do with it after I found it. Like how to show that P is unbounded with the K