Exam Question - Convex Optimisation

Question

I had an exam question today and want to check if my answer is right or if there was a mistake...

Consider the following convex optimisation problem:

$$\begin{array}{ll} \text{minimize} & - \ln(1+x_1)- \ln(\frac{3}{2}+x_2)- \ln(4+x_3)\\ \text{subject to} & x_1+x_2+x_3=1\\ & x_1, x_2, x_3\geq0\end{array}$$

I got that the dual has optimal solution $\lambda =\frac{2}{5}$ and $x=(\frac{3}{2},1, -\frac{3}{2})$. However, obviously, the third variable violates its constraint. I'm thinking either I got the answer wrong or the question was wrong! Can someone help me here?

Answer 1

Question is correct but your answer is not! If you put more detail here about your solution I can explain why you got wrong answer. Anyway here it is my guess about your solution:

The problem is a convex optimization over compact constraint, and slater CQ holds therefore, $x^* $ is an optimal solution iff $x^*$ is KKT point. so if you set up KKT sys, you get a simple system of non linear equations and inequalities including the primal inequality i.e., $x=(x_1,x_2,x_3) \ge 0$. So if you have taken into account the later inequality you should have came up right answer which satisfies non-negativity condition! I bet you didn't consider this condition....

Answer 2

I am sorry to tell you this, but you did not get the correct answer (which I guess you already know since you have $x_3<0$ in your solution). The correct solution is

$$ x_1 = \frac{3}{4}\\ x_2 = \frac{1}{4}\\ x_3 = 0 $$

You can find this answer just by reasoning. How? Well as you know, $\ln(x)$ has a negative second derivative. This means that an increase in $x$, say $\Delta x$, will have less impact on $\ln(x)$ for larger values of $x$ than for smaller. You can interpret the first constraint as a resource constraint. Thus, you have a total of $1$ to distribute over the three non-negative variables. If you give all your resources to $x_1$ it will still not "catch up" to $\ln(4)$. The same goes for $x_2$. So there is little point in giving any resources to $x_3$, thus $x_3=0$. Now the best way of distributing the resources between $x_1$ and $x_2$ is to make the two first terms equal (due to the negative second derivative). Thus solving $$ \ln(1+x_1)=\ln(\frac{3}{2}+x_2) \text{ for $x_1+x_2=1$, and $x_1,x_2\geq0$} $$ yielding $$ x_1=\frac{3}{4}\\ x_2= \frac{1}{4} $$ So the result can be derived just by realizing that the second derivative of $\ln{x}$ is negative. Hope this was helpful!