KKT condition just binding and inactive

Question

I have read the textbook saying that if both KKT and Lagrangian multiplier $\lambda$ are $0$, then the constraint is just binding, whereas if KKT multiplier is equal to 0, and Lagrangian multiplier is non-zero value, it is inactive constraint.

As far as I know, KKT and Lagrangian function looks the same. How do we have different multiplier value from KKT and Lagrangian?

Answer 1

Take, for example, only inequality constraints $$ \min f(x)\qquad\text{subject to}\qquad g(x)\le 0,\ x\in X. $$ KKT conditions are conic gradient conditions on $\nabla f$ and $\nabla g_k$ for active constraints $g_k$. It can be formalized as

1. $\nabla L=0$,
2. Complementary slackness $\lambda_k g_k=0$ meaning that inactive constraints do not contribute any nonzero terms in 1,
3. Feasibility constraints $\lambda\ge 0$, $x\in X$, $g(x)\le 0$.

"Just binding" constraint means that it is not important for 1 and 2 (multiplied by zero $\lambda_k$), but only for 3 (feasibility). Often it is because the corresponding constraint is inactive at the point ($g_k<0$), however, it is possible geometrically that an active constraint has $\lambda_k=0$, just in 2 it happens to be $0\cdot 0=0$, but it is very special coincidence. Summarizing,

• $g_k<0$ (inactive) $\Rightarrow$ $\lambda_k=0$ (just bindning)
• $\lambda_k=0$ does not imply $g_k$ is inactive in general.