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Did I mess up in this inequality?

  1. Sep 20, 2011 #1
    1. The problem statement, all variables and given/known data

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



    3. 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

    [tex]w = 1203y_1 + 1551y_2[/tex]

    [tex]0 \geq -6y_1[/tex]
    [tex]5 \geq 4y_1 + 7y_2[/tex]
    [tex]14 \geq 10y_1 + 15y_2[/tex]

    [tex]y_1, y_2 \leq 0[/tex]

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

    (this will be useful for (e))

    [tex]0 \geq 0 = -6(0)[/tex]
    [tex]5 \geq 0 = 4(0) + 7(0)[/tex]
    [tex]14 \geq 0 = 10(0) + 15(0)[/tex]

    (c)

    Here is the problem, with the first inequality

    [tex]-6t + 4(100) + 10(50) = -6t + 900 \leq 1203[/tex]

    [tex]-6t \leq 303[/tex]

    [tex]t \geq -50.5[/tex]

    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 [tex]t \geq -50.5[/tex] which therefore includes [tex]t \geq 0[/tex], hence x = (t, 100,50)^t works for all [tex]t \geq 0[/tex]"

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

    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 [tex]x = (x_1, x_2, x_3)^t[/tex].

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

    I have the inequality

    [tex]c^t \leq y^t A \leq y^t b[/tex]

    However there is a problem.

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

    And my yt is [tex]\begin{pmatrix}
    0\\
    0
    \end{pmatrix}[/tex]

    How can they multiply together?

    (f) need to answer (e) before

    (g) I think it follows from (f)? Intuition.
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Sep 21, 2011 #2

    Ray Vickson

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    Science Advisor
    Homework Helper

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

    RGV
     
    Last edited by a moderator: May 5, 2017
  4. Sep 21, 2011 #3
    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)

    [tex]w = 1203y_1 + 1551y_2[/tex]

    [tex] 0 \leq -6y_1[/tex]

    [tex]5 \leq 4y_1 + 7y_2[/tex]

    [tex]14 \leq 10y_1 + 15y_2[/tex]

    [tex]y_1, y_2 \geq 0[/tex]

    (b)

    A feasible solution would be [tex]y = (0,1)^t[/tex]

    [tex]w = 1203(0) + 1551(1) = 1551[/tex]

    [tex] 0 \leq 0 = -6(0) [/tex]

    [tex]5 \leq 7 = 4(0) + 7(1)[/tex]

    [tex]14 \leq 15 = 10(0) + 15(1)[/tex]

    [tex]0,1 \geq 0[/tex]
     
  5. Sep 21, 2011 #4

    Ray Vickson

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    Science Advisor
    Homework Helper

    This is OK now.

    RGV
     
  6. Sep 21, 2011 #5
    Not okay yet

    (c) Still unsure

    (e) Can't do anything about this
     
  7. Sep 22, 2011 #6
    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 [tex]t \to \pm \infty[/tex], [tex]z \to 1200[/tex] for all t

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

    (e) No, since P is bounded, so must D.
     
  8. Sep 24, 2011 #7
    I need some comments...anything is okay.

    Ray...I love you? Please come back.
     
    Last edited: Sep 24, 2011
  9. Sep 27, 2011 #8
    I take back everything I said...

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

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

    From

    [tex]-6y_1 \geq 0[/tex]

    It follows that

    [tex]y_1 \leq 0[/tex]

    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?
     
    Last edited by a moderator: May 5, 2017
  10. Sep 27, 2011 #9
    Also, from Linear Algebra and intuition I have the matrix for the dual

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

    I have 3 rows and 2 columns, I must have a free variable.
     
  11. Sep 28, 2011 #10
    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

    [tex]|x_1| \leq K[/tex], [tex]|x_2| \leq K[/tex], [tex]|x_3| \leq K[/tex]

    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
     
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