How to solve system of inequalities?

Question

Consider the following inequalities: $$ z \geq f(y) \;\; \mathsf{or} \;\; z \leq f(y)$$ and $$ g(y) \geq x \;\; \mathsf{or} \;\; g(y) \leq x $$ Is there a systematic way to solve for $z$ as a function of $x$? i.e., find some relationship: $$ z \;\; ?? \;\; h(x)$$ The issue I am finding is that I can isolate $y$, but depending on the direction of the inequality, I might not be able to substitute the expression for $y$ in $f(y)$.

Here is a very simple example: $$z \geq y$$ $$y \geq x$$ In this case, we can easily write $z \geq x$, but now consider the example: $$z \leq y$$ $$y \geq x$$ Now, I don't think there any way to write a relationship between $z$ and $x$ without $y$. If not,how do we solve this system of inequalities?

Answer 1

All you can say in the second case is $y \ge \max(x, z) $. Nothing can be said relating $x$ and $z$.

Answer 2

If all the inequalities are linear, use a Linear Programming solver, with wither no objective function, or "0" as the objective function. The result will either be a solution, or a conclusion that no solution exists (i.e., the problem is infeasible).

If one or more of the inequalities are nonlinear, use a Nonlinear Programming (optimization) solver. If you use a local Nonlinear Programming (optimization) solver which finds a solution, you are done. But if it fails to find a solution, there may still be one. If you use a (branch and bound) global optimization solver, and it successfully completes, you will either get a solution or a conclusion that there is no solution; however, it is possible that it will not successfully complete.

Answer 3

Caution: when $z\ge y\text{ and }y\ge x$, it is wrong to say that $z\ge x$ is the solution. For example, take $y=0$. Then $x=1,z=2$ fulfills $z\ge x$ but $2\ge0\text{ and }0\ge 1$ does not hold.

Now, let us consider the general case of $z\ge g(y)\land f(y)\ge x$. The locus of the points $(g(y),f(y))$ describes a curve (not necessarily continuous). In general, the set $x\le a\land z\ge b$ is the top-right quadrant defined by the lines $x=a,y=b$. If you move the apex $(a,b)$ of this quadrant, you get the locus of the admissible points $(x,z)$.

In the figure below, we have considered an arbitrary curve. The boundary of the allowed area is made of arcs of the curve and line segments/half lines that tangent the curve. So the description of this boundary must be piecewise.