KKT conditions strict inequality constraints

Question

Some people asked questions about KKT conditions with strict inequality constraints, such as

Kuhn Tucker conditions with strict inequality constraints?

Questions about constraints and KKT conditions

https://www.researchgate.net/post/Is_KKT_condition_applicable_for_optimality_when_the_constraints_is_strictly_less_than_type

Let us see an example.

$\min\quad \frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{d} + \frac{d^2}{a}$

$\mathrm{s.t.}\quad a>0,\ b>0,\ c>0,\ d>0,\ a^4+b^4+c^4+d^4 = 4.$

The minimum $4$ is attained at $a=b=c=d=1$. The constraints $a,b,c,d>0$ are strict inequalities. How to deal with them in KKT conditions?

Just as some people said (e.g., the 3rd link above), we simply ignore the strict inequality constraints and use KKT conditions. If the minimum is attainable (that is, min not inf), the solution will satisfy the strict inequalities. For this example, it is the Lagrange multiplier method $L = \frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{d} + \frac{d^2}{a} + \lambda (a^4+b^4+c^4+d^4 - 4)$.

I found the lecture note:

https://ocw.mit.edu/courses/sloan-school-of-management/15-084j-nonlinear-programming-spring-2004/lecture-notes/lec6_constr_opt.pdf

In the lecture note (2nd page), the following problem is considered:

$\min\quad f(x)$

$\mathrm{s.t.}\quad h(x) = 0,\ g(x) \le 0, \ x \in X.$

Here $X$ is an open set. Then KKT conditions are used without including $x\in X$.

I want to make sure if this is the case. Thanks.

Answer 1

Strict inequalities cannot be taken into account in KKT conditions.

To solve the problem, compute stationary points of the problem without these constraints, then throw away all of them that violate these conditions.

Answer 2

I think a common practice is to exclude the strict inequalities from constructing dual problems. Check the counterexample on Page 2 of https://people.eecs.berkeley.edu/~elghaoui/Teaching/EE227A/lecture8.pdf :

A counterexample. Convexity alone is not enough to guarantee strong duality. Consider for example the convex problem $$ \min _{x, y>0} e^{-x}: x^{2} / y \leq 0, $$ with variables $x$ and $y$, and domain $\mathcal{D}=\{(x, y) \mid y>0\}$. We have $p^{*}=1$. The Lagrangian is $L(x, y, \lambda)=e^{-x}+\lambda x^{2} / y$, and the dual function is $$ g(\lambda)=\inf _{x, y>0}\left(e^{-x}+\lambda x^{2} / y\right)= \begin{cases}0 & \lambda \geq 0 \\ -\infty & \lambda<0\end{cases} $$ so we can write the dual problem as $$ d^{*}=\max _{\lambda} 0: \lambda \geq 0 $$ with optimal value $d^{\star}=0$. The optimal duality gap is $p^{\star}-d^{\star}=1$. In this problem, Slater's condition is not satisfied, since $x=0$ for any feasible pair $(x, y)$.