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I would like to ask something about convex function and convex model.

For example, the function $f(x,y)=\frac{x^2}{y}$ is convex when $x\geqslant0$ and $y>0$.

For a convex model (minimization), the constraints must satisfy

$$g(x)\leqslant0$$

where $g(x)$ is convex function.

So if I have constraints:

$$\frac{x^2}{y}-z\leqslant0$$

$$x\geqslant0$$

$$y>0$$

$x, y, z$ are variables, and the objective function and the other constraints are linear. Is it correct that my model is a convex model?

Thank you.

Dylan

Dylan Lan
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1 Answers1

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The optimization problem in standard form:

$min_x \quad f_0(x)$

$\quad \quad f_i(x) \le 0, \quad i=1,\cdots, m$

$\quad \quad h_i(x) = 0 \quad i=1,\cdots,p$

is called a convex optimization problem if:

  • The objective function $f_0$ is convex.

  • The functions defining the inequality constraints, $f_i, i=1,\cdots,m$ are convex.

  • The functions defining the equality constraints, $h_i, i=1,\ldots,m$ are affine.

It seems in your problem all conditions are satisfied so ---> It is convex. (See Boyd Book) for more detailed explanation.

TNM
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