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}<br /> -6 &amp; 4&amp; 10\\ <br /> 0&amp; 7&amp; 15<br /> \end{pmatrix} This is 2 by 3

And my y^t is \begin{pmatrix}<br /> 0\\ <br /> 0<br /> \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 &lt; 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}<br /> -6 &amp; 4&amp; 10\\ <br /> 0&amp; 7&amp; 15<br /> \end{pmatrix} This is 2 by 3

And my y^t is \begin{pmatrix}<br /> 0\\ <br /> 0<br /> \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[ <br /> \begin{array}{cc|c} <br /> 1 &amp; 0 &amp; 0 \\ <br /> 4 &amp; 7 &amp; 5 \\ <br /> 10 &amp; 15 &amp; 14 \\ <br /> \end{array} <br /> \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