The objective function is surely convex as its second derivative is (6x*(2x^3 + 1))/(x^3 - 1)^3 which is positive if x>1. However, the cvx failed to optimize the problem. How can I deal with the function? A test code is shoed as follows
cvx_begin
variable x;
minimize inv_pos(x^3-1);
subject to
x>=1;
y<=10;
x+y<=3;
cvx_end
And the error message is
You could
maximize x
instead. That will work.
Thank you for replying. I’m sorry that my example may be not appropriate. What about this objective function? It is still convex. But cvx failed to solve it. How can I optimize it?
cvx_begin
variable x;
variable y;
minimize inv_pos(pow_p(x,3)-1)+inv_pos(pow_p(y,3)-1);
subject to
x>=1;
y>=1;
x+y<=3;
cvx_end
Here is the error message
Thanks a lot again.
You can do
minimize(pow_p(x,-3)+pow_p(y,-3))
As for 1/(x^3-1) for x > 1, it appears that none of the forum readers know how to formulate that in CVX, despite its convexity. Therefore, it is a candidate for @Erling’s challenge. https://twitter.com/J_P_Vielma/status/1014513695605608448
Edit: I stand corrected on the preceding paragraph in light of @Michal_Adamaszek’s post below.
In https://docs.mosek.com/modeling-cookbook/practical.html#composite-functions we do
t\geq \frac{1}{xyz-1}, \ xyz>1
of which the present question is a special case.
