# How to solve this nonlinear optimisation problem?

I have the following optimization problem where $$K>0$$. \begin{align*} \min_{y_1,\ y_2\ge 0} 2k(\exp(-y_1)+\exp(-\min(y_1,y_2)))+2y_1+y_2. \end{align*}

I divided into two cases:

Case 1: $$y_1\le y_2$$: \begin{align*} \min_{\substack{y_1,\ y_2\ge 0}} 4k\exp(-y_1)+2y_1+y_2. \end{align*} Assuming Lagrange multipliers $$\lambda_1\ge 0$$, $$\lambda_2\ge 0$$, corr. to the constraints $$y_1,y_2\ge 0$$, the KKT conditions: \begin{align*} -4k\exp(-y_1)+2-\lambda_1 &= 0\\ 1-\lambda_2=0 \end{align*} From the second equation, $$\lambda_2=1$$, by complementary slackness, $$y_2^*=0$$, and since $$y_1\le y_2$$, $$y_1^*=0$$, and so $$-4k+2=\lambda_1$$. Since $$\lambda_1\ge 0$$, this solution is possible for $$k\le0.5$$, else there is no solution for this case.

Case 2: $$y_1\ge y_2$$: \begin{align*} \min_{\substack{y_1,\ y_2\ge 0}} 2k\exp(-y_1)+2k\exp(-y_2)+2y_1+y_2. \end{align*}

The KKT conditions: \begin{align*} -2k\exp(-y_1)+2-\lambda_1 &= 0\\ -2k\exp(-y_2)+1-\lambda_2 &= 0 \end{align*}

For $$\lambda_1=0$$ and $$\lambda_2=0$$, I get $$y_1^*=\log(k)$$ and $$y_2^*=\log(2k)$$ and this violates the constraint. If $$y_1^*=\log(k)$$ and $$y_2^*=0$$, $$\lambda_2=-2k+1$$, and this holds only for $$k\le 1/2$$.

I get two solutions $$(0,0)$$ and $$(\log(k),0)$$ for $$k\le 0.5$$ and no solutions for $$k\ge 0.5$$. To find the best among these two solutions, I compare the objective values: $$4k$$ for $$(0,0)$$ and $$2(1+k+\log(k))$$ for $$(\log(k),0)$$. For $$k\le 0.5$$, the latter is smaller always, and the unique solution is $$(\log(k),0)$$.

Is this working correct?

• $(0,0)$ is also a solution to the second case (as it should be, since $y_1=y_2$ falls in both categories). Oct 3, 2018 at 18:22
• you may miss points on the boundary, you should add the constraint $y_1\geq y_2$ with a Lagrange multiplier Oct 3, 2018 at 18:50
• @LinAlg: Thanks, but arguing with three Lagrange multipliers was difficult. Do you find any error in the above working (apart from (0,0))?
– Esha
Oct 3, 2018 at 19:05
• The math checks out, but the method is incorrect since you omit the boundary. The analysis should not be much harder; just distinguish between $\lambda_3=0$ and $\lambda_3>0$. Oct 3, 2018 at 19:17