We have the optimization problem:

Minimize $f(x,y) = (x+1)^2 + y^2$ subject to $g(x,y) = -x^3 + y^2 \leq 0$.

We would like to find multipliers $\lambda_0, \lambda_1$ satisfying the Fritz-John optimality conditions, and show that KKT fails here.

In regards to Fritz-John, we have the following system if I am not mistaken

$\begin{cases}2\lambda_0 (x+1)-3\lambda_1x^2 = 0 \\ 2\lambda_0y+2\lambda_1y = 0\\ -x^3+y^2 \leq 0 \\ \lambda_1(-x^3+y^2) = 0\end{cases}$

Option 1: Suppose that the feasability constraint $-x^3 + y^2 \leq 0$ is not binding, meaning $-x^3 + y^2 < 0$. Then $\lambda_1 = 0$ and our system reduces to $\begin{cases}2\lambda_0 (x+1) = 0\\2\lambda_0y = 0\end{cases}$

If $\lambda_0 = 0$ then all our lambdas are zero which violates the FJ conditions, so $\lambda_0 \neq 0$, and so $x = -1 $ and $y=0$. Sadly $(x,y) = (-1, 0)$ is not a feasable point since it violates $-x^3 + y^2 < 0$

So all in all, option 1 is not relevant.

Option 2: the feasability constraint is binding, so $-x^3 + y^2 = 0$ or $x = y^{\frac{2}{3}}$. Now our system reduces to

$\begin{cases}2\lambda_0 (y^{\frac{2}{3}}+ 1) - 3\lambda_1 y^{\frac{4}{3}} = 0 \\ 2\lambda_0 y + 2\lambda_1y = 0\end{cases}$

I'm unsure what to do now. There's no way to find all three variables. I was specifically asked for the lambdas. If we assume that $y = 0$ then there's no way to find the lambdas, they could be anything.

If we assume that $y \neq 0$ then we have $\lambda_1 = -\lambda_0$ and our first equation becomes $2\lambda_0(y^{\frac{2}{3}} + 1) +3\lambda_0 y^{\frac{4}{3}} = \lambda_0 (2(y^{\frac{2}{3}} + 1) + 3y^{\frac{4}{3}}) = 0$.

Now again we must have $\lambda_0 = 0$ which means all lambdas are zero but FJ doesn't allow for that.



It is straightforward to show that a solution exists and the constraint must be active

The Fritz John conditions give $\lambda_0 Df((x,y)) + \lambda_1 Dg((x,y)) = 0$, with $(\lambda_0,\lambda_1) \neq 0$.

We get $2(\lambda_0+\lambda_1)y = 0$, so either $y=0$ or $\lambda_0+\lambda_1 = 0$.

If $\lambda_0 + \lambda_1 = 0$ then we obtain $2(x+1)+3x^2 = 0$ which has no (real) solution hence this is impossible.

Consequently $y=0$ from which we obtain $x=0$ (since $y^2 = x^3$) which gives a cost of $1$. Since $2 \lambda_0 = 0$ we see that $\lambda_0 = 0$ and so $\lambda_1>0$.

Hence any multiplier of the form $(0,t)$, with $t>0$ will do.

As an aside, the multipliers (which are non negative) are often normalised to lie on the simplex, so in this case, the solution would be $\lambda=(0,1)$.

Note that $Dg((0,0)) = 0$, so KKT does not apply (that is, it does not fail!).

  • $\begingroup$ "Consequently $y=0$ from which we obtain $x=0$" is wrong. We know $-x^3 + y^2 \leq 0$, not that they are equal to zero. $\endgroup$ – Oria Gruber Dec 16 '19 at 20:25
  • $\begingroup$ If the constraint is active (see first sentence) then we must have equality. How is that reasoning wrong? $\endgroup$ – copper.hat Dec 16 '19 at 20:30
  • $\begingroup$ By active you mean strict? As in it's an equality constraint rather than inequality $\endgroup$ – Oria Gruber Dec 16 '19 at 20:54
  • $\begingroup$ A constraint $g(x) \le 0$ is active at $x^*$ iff $g(x^*) = 0$. $\endgroup$ – copper.hat Dec 16 '19 at 21:17
  • $\begingroup$ Note that I should have just dismissed the $\lambda_0+\lambda_1 = 0$ possibility immediately as we have $\lambda_k \ge 0$ and $(\lambda_0,\lambda_1) \neq 0$. $\endgroup$ – copper.hat Dec 17 '19 at 4:26

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