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 $$\gamma < 13/200$$ 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
$$5\gamma< 1$$
$$20\gamma \alpha<1$$
$$(60/0.9) \gamma (1-\alpha)<1$$

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.

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