Solving a system of inequations and equations

Hello,

I need to solve a system of inequations and equation, as follows:

The variables need to be solved include: x, y1, y2

• exp(x) <= V(y1), and If exp(x) < V(y1), then y1=M1;
• exp(x) <= V(y2), and If exp(x) < V(y2), then y2=M2;

And: (1-b)*x + c*(p1*y1+ p2*y2)= d

All parameters a1, a2, b, c, d; p1, p2, M1, M2, function V(.) are known. So, basically, I have 3 equations to solve for 3 variables, which should be able to solve.

But I am having difficulties how to coding the two above inequations which have some conditions go along with them.

Is there anyone having an idea?

Really appreciate your help!

Thank you

[John D'Errico]

Huuh?

"The variables need to be solved include: x, y1, y2

• exp(x) <= a1, and If exp(h) < a1, then y1=M1;"

so we are told of x,y1,y2 as unknowns.

I understand this: exp(x) <= a1

But, suddenly we see something called h appear out of thin air. where did h come from? And what does h have to do with anything?

Next, you have ONE equality.

(1-b)*x + c*(p1*y1+ p2*y2)= d

So a single linear equality in the three unknowns.

As well, the only other thing you have told us is what appear to be 2 strange inequalities, that involve the magic number h. h appears in no other place in anything you have written.

you also state "All parameters a1, a2, b, c, d; p1, p2, beta are known". So then what are M1 & M2 & h? Chopped liver?

Regardless, even if you do have one equality, and TWO inequalities, you cannot solve it. There will generally be an infinite number of solutions.

You need to clearly explain what you have. And even then, you almost certainly need to accept that there is no solution to the problem you have not explained, even if you do explain it.

[heidi pham]

Thank you, since I tried to simplify the problem I am facing, then some notations were mistakenly typed.

I just edited the question.

[John D'Errico]

Ok, so the h disappeared now. Far better. Regardless, you need to face the fact that you have asystem of ONE linear equality in three unknowns.

That expression is equivalent to the equation of a plane in R^3.

Two inequalities are just that. They constrain the set of allowable solutions in the plane. So the result will be a planar region, with apparently some strange boundaries. Regardless, if a solution exists at all, then there will be infinitely many solutions.

Oh, and you still have not been terribly clear. What is the function V? Let me look at those inequalities. We see this:

- exp(x) <= V(y1), and If exp(x) < V(y1), then y1=M1;

What does that mean? Pick some value of y1. That seems to yield an upper limit on x, as a function of y1. But then we are told that IF x is less than that upper bound, then y1 is fixed.

x <= log(V(y1))

So x is not bounded as a function of y1. In fact, we can then write the bound on x as

x <= log(M1)

Likewise, we can use the second "constraint" to determine that

x <= log(M2)

That tells me that x is simultaneously bounded by the smaller of M1, M2.

x <= log(min(M1,M2))

What would happen if the above inequality was an equality instead? We know that x cannot be greater than ONE of M1 and M2, but less than the other, since both inequalities must hold. So lets assume that we have

x == log(min(M1,M2))

AND wlog, that M1 <= M2.

Then we have x fixed, at log(min(M1,M2)). Then y1 can apparently vary. What do we know about y2? Well, if M1 < M2, then y2 is fixed at M2.

Sorry, but clear as mud.