General quad over lin

Uncategorized
Nov 25, 2020
I

Hi, I have a kind of straightforward question and maybe my mind has been stuck somewhere.

I have a objective function appear in the form \frac{ax^2 + bx + c}{y}, where x,y > 0 are the variables, and a,b,c > 0 are known coefficient, which is convex. I did not come up how to implement the code properly to use it as my objective function.

Many thanks for the help and reminder.

M

It can always be written as \frac{a(x+p)^2+q}{y}.

If q< 0 it is not convex (at least not for all y>0).

If q\geq 0 it is a*quad_over_lin(x+p,y)+q*inv_pos(y), modulo my syntax errors.

I
Replying to #2

Many thanks for the reply.

Exactly as you mentioned, it can be tackled this way, yet the feasibility depends on q.

While in a general sense, \frac{ax^2 + bx + c}{y}, with x,y > 0 are the variables, and a,b,c > 0 are known coefficient, is convex, as it Hessian is positive semidefinite.

So I wonder if there is a way to tackle \frac{ax^2 + bx + c}{y} universally without dependence on q, as my program may repeat many times with different values of coefficients.

Thanks again for the help.

M

If a and q don’t have the same sign, it is not convex. If a > 0 as you wrote, that means q > 0 is needed for convexity.

Please read Why isn't CVX accepting my model? READ THIS FIRST!

M
Replying to #3

To make a concrete example, the function \frac{x^2+4x+1}{y} is not convex on x,y>0.

More generally, if \frac{a(x+p)^2+q}{y} is convex on x,y>0 then it remains convex on the line x=y-p and that is just ay+q/y on the set y>p which is convex if and only if q>0.

I

Hi, Mark and Michal,

Many thanks for your detailed reply. You are definitely right. I was taking granted for the positive definiteness of a matrix with positive diagonals and negative non-diagonals.