Given min $x_1-4x_2+x_3$
s.t $x_1+2x_2+2x_3=-2$
$||x||^2\leq1$
(i)Given a KKT point of the problem above,must it be an optimal solution?
(ii) Find the optimal solution of the problem using KKT conditions.
for (i) taking the point $(0,-0.89,-0.11)$ we satisfy slater's condition and the objective function is convex and the first contrains is affine and second is convex with inequality. therefore a KKT point is optimal point.
for (ii) I know I just need to find 1 KKT point
$L(x,\lambda,\eta)=x_1-4x_2+x_3+\lambda(x_1+2x_2+2x_3+2)+\eta(x_1^2+x_2^2+x_3^2-1)$
(1) $\frac{\partial L}{\partial x_1}=1+\lambda+2\eta x_1 = 0$
(2) $\frac{\partial L}{\partial x_2}=-4+2\lambda+2\eta x_2 = 0$
(3) $\frac{\partial L}{\partial x_3}=1+2\lambda+2\eta x_3 = 0$
(4) $\lambda(x_1+2x_2+2x_3+2)=0$
(5) $\eta(x_1^2+x_2^2+x_3^2-1)=0$
(6) $(x_1+2x_2+2x_3+2)=0$
(7) $x_1^2+x_2^2+x_3^2-1\leq0$
if $\lambda=\eta=0$ we get from (1) that $1=0$ therefore not feasible
if $\lambda\neq0,\eta=0$ from (1) we get that $\lambda=-1$ and from (2) we get that $\lambda=2$ not feasible again.
if $\eta\neq0,\lambda=0$ we get from (1) and (3) that $x_1=x_3=\frac{-1}{2\eta}$ and $x_2=\frac{2}{\eta}$ plugging into (6) we get that $\eta=\frac{-5}{4}<0$ not possible because $\eta\geq0$ therefore not feasible again.
if $\lambda,\eta\neq0$ from (2)-(3) we get $\frac{-5}{2\eta}+x_2=x_3$ and (2)-2*(1) we get that $\frac{-6}{4\eta}+\frac{1}{2}x_2=x_1$ plugging into (6) we get that $\frac{1}{\eta}=\frac{9}{13}x_2+\frac{4}{13}$ plugging all that we got into (5) we know that $x_1^2+x_2^2+x_3^2=1$ because $\eta\neq0$ and I get that $x_2=0.0503$ or $x_2=-0.764$ the second options is not feasible because $\eta\geq0$ and from the first one I don't get the right answer(checked in matlab)
I wonder what am I doing wrong here, any help please?
Opt vector is $(-0.5194,0.1074,-0.8478)$ opt val $-1.7969$
Best Answer
Something went wrong in this step. I put (5) into Wolfram Alpha and got $x_2 \approx 0.1074$ as one of the solutions.