- #1
flyingpig
- 2,579
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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
[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.
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