How to express limitation 1/(x^4+x^2) <= y, x>0 in CVX?

Uncategorized
May 17, 2021
L

I need to express in CVX the limitation 1/(x^4+x^2) ≤ f(y), x > 0, where function 1/(x^4+x^2), x>0 is convex with respect to x and f(y) is affine. I have tried the pow_pos and inv_pos, but it seemed that they are failed. Is there any possible way? Thank you!

M

The LHS is convex for all x , not just x > 0.

So if no one has a CVX formulation, perhaps this is another candidate for @Erling 's challenge.

https://twitter.com/J_P_Vielma/status/1014513695605608448

L
Replying to #2

Ok, I see that. Thank you for your reply, and I’ll try some other ways to transform or avoid it.

H

Plenty of ways to break this down to conic atoms. First of, notice that

t \geq 1/(x^4+x^2) \Longleftrightarrow t(x^4+x^2) \geq 1.

Then the breakdown depends on how you see the nonlinear factor. I would personally go for the last one in this list.

When seen as x^4+x^2:

t(r+s) \geq 1\quad\quad (rotated quadratic cone)
r w^3 \geq x^4, w = 1\quad (power cone)
s \geq x^2\quad\quad\quad\quad\quad (quadratic cone)

When seen as x^2(x^2+1):
tr(r+1) \geq 1\quad\quad (geometric mean cone)
r \geq x^2\quad\quad\quad\quad\quad (quadratic cone)

When seen as (x^2)^2+x^2:

t(r+s) \geq 1\quad\quad (rotated quadratic cone)
r \geq s^2\quad\quad\quad\quad\quad (quadratic cone)
s \geq x^2\quad\quad\quad\quad\quad (quadratic cone)

EDIT: Way too careless… Didn’t even check that the introduced variables r and s can’t just be taken to infinity by the solver, rendering all the conic representations above useless. Yup, this might be a challenge…

H

So the inequality is not far from Example 7.3 of https://docs.mosek.com/modeling-cookbook/practical.html#composite-functions.

To reformulate
t \geq 1 / (x^4 + x^2),

we may try to replace the 4th-order denominator by a 4th-order substitute, namely
t \geq 1 / r^4,\quad r^4 \leq x^4 + x^2,

and equalize powers:
t \geq 1 / r^4,\quad r^4 \leq x^4 + s^4,\quad s^2 \leq x.

These constraints can now be represented, in order, using a power cone, a 4-order norm cone, and a rotated quadratic cone.

EDIT: This attempt is also flawed as @Mark_L_Stone points out.

M

I don’t see how to handle
r^4 <= x^4 + s^4

That would be
r <= norm([x;s],4)
which is going in the wrong direction to be convex. CVX handles p-norms “automatically”, but of course they must be used in a convex manner.

Am i misunderstanding what to do with the 4-norm, or is your formulation erroneous?

H

Yeah you are right, my attempts failed and I am out of ideas. I’ll take note of it as usual. Out of curiosity, may I ask what the application is @lk10420?

M

Perhaps 1/(x^4+x^2) is the simplest (in some sense) example satisfying @Erling’s challenge criterion for which a suitable conic formulation has not yet been identified? That should put it at the top of the leaderboard.

Unlike Express 1/(a^2+x^2) in cvx 1/(x^4+x^2) is convex over its entire domain, so does not get demerit points for not being convex over its (natural) domain. Its only “non-niceness” is being singular in the middle of its domain, at x = 0.

L

I am sorry that I didn’t check my emails recently, by the way, thank you for your reply, I have learned a lot! Thank you!
The cause is that I want to solve a non-convex constraint with term (√(1/v^4+1/a^2)-1/a^2)^1/2 in my problem which is complex and hard to be solved by using the KKT conditions even when the problem is changed into convex, then I introduce a slack variable λ>0 and it needs to satisfy
λ>= (√(1/v^4+1/a^2)-1/a^2)^1/2 to keep equivalent. I transform it into
λ^4+λ^2/a^2 >= 1/v^4 , to make it convex I further change the constraint to be convex and have that 1/(λ^4+λ^2/a^2)<=v^4.
where a>0 is a constant and 0<v<b is a variable

H

As promised I took note and returned once the puzzle was solved. This post has gotten an impressive 945 views and deserves it. The solution is found here: