Solving a system of Inequalities

[timtitan]

Hello, I'm having some trouble with a Queuing Networks question, not the networks but solving a system of inqualities based on the network.

Homework Statement

I have to find the value of α that gives maximum γ, and then use the value. The system is defined by
5\gamma< 1
20\gamma \alpha<1
(60/0.9) \gamma (1-\alpha)<1

Now \alpha is a probability and lies in the region 0<\alpha<1
While \gamma is a rate and is non-zero.

Homework Equations

The Attempt at a Solution

Now I've got so far as to put the system in this form and to solve through to find that in the region
0< \alpha ≤ 10/13, that 0 < \gamma < -3/(200(\alpha-1))
while in the region
10/13 < \alpha <1, that 0 < \gamma < 1/(20 \alpha)

Thus the maximum value of \gamma lies in the region \gamma < 13/200 when \alpha = 10/13.

This much is fine, but I need to use an actual value of \gamma in the next part of the question and I can't think how to get a \gamma = expression. Any help would be gratefully appreciated.

 

[haruspex]

0< \alpha ≤ 10/13, 0 < \gamma < -3/(200(\alpha-1))
while in the region
10/13 < \alpha <1, 0 < \gamma < 1/(20 \alpha)

Thus the maximum value of \gamma lies in the region \gamma < 13/200 when \alpha = 10/13.

γ = 13/200 satisfies all the constraints;
for α < 10/13, γ < 3/(200(1-α)) < 3/(200(1-10/13)) = 13/200, right?
and for α > 10/13 etc.

 

[timtitan]

Yes but isn't that only if \gamma ≤ 13/200 whereas in this case \gamma &lt; 13/200 so surely \gamma must be infinitesimally less than this value?

 

[haruspex]

Ok, now I understand your question properly.
If you take the constraints as strict, there is no maximum value that satisfies them. No matter what value you pick, you can always get a slightly higher one.

 

[Ray Vickson]

Hello, I'm having some trouble with a Queuing Networks question, not the networks but solving a system of inqualities based on the network.

Homework Statement

I have to find the value of α that gives maximum γ, and then use the value. The system is defined by
5\gamma&lt; 1
20\gamma \alpha&lt;1
(60/0.9) \gamma (1-\alpha)&lt;1

Now \alpha is a probability and lies in the region 0&lt;\alpha&lt;1
While \gamma is a rate and is non-zero.

Homework Equations

The Attempt at a Solution

Now I've got so far as to put the system in this form and to solve through to find that in the region
0&lt; \alpha ≤ 10/13, that 0 &lt; \gamma &lt; -3/(200(\alpha-1))
while in the region
10/13 &lt; \alpha &lt;1, that 0 &lt; \gamma &lt; 1/(20 \alpha)

Thus the maximum value of \gamma lies in the region \gamma &lt; 13/200 when \alpha = 10/13.

This much is fine, but I need to use an actual value of \gamma in the next part of the question and I can't think how to get a \gamma = expression. Any help would be gratefully appreciated.

In general, optimization problems subject to strict inequality constraints do not have solutions: you may have an "inf" but not a minimum, or a "sup" but not a maximum. For example, what is the solution of the simple problem
minimize x, subject to x > 0?
Answer: the problem is ill-posed, and does not have a solution!

Often, when students are first introduced to such problems they are sloppy and write ">" when they should write "≥" or they write "<" when they should write "≤".

RGV