maximizing the negative norm is it convex problem?

Question

I have an objective function with two decision variables x1 and x2

\begin{equation} \begin{array}{cl} \underset{\alpha_{1},\alpha_{2}}{ \max} a_1x_1+ a_2x_2 -c\sqrt{h_{1}(x) ^2 + h_{2}(x)^2} \\ \text { s.t. } x_{1}+x_{2}=1 \end{array} \end{equation}

I tried to introduce new variables $b_1=h_1(x) ,b_2=h_2(x) $ so my objective function will be

\begin{array}{cl} \underset{\alpha_{1},\alpha_{2}}{ \max} a_1x_1+ a_2x_2 -cu \\ \text { s.t. } x_{1}+x_{2}=1 ,\\ h_1(x)=b_1, \\ h_2(x)=b_2 ,\\ \sqrt{b_1^2 + b_2^2 } \leq u \end{array}

my questions:

1. does the reformulation with the new constraints correct ?

2. because I am maximizing the negative norm does that mean it is a convex function ?

3. how can I solve my problem in MATLAB

Thanks in advance

Answer 1

Assuming $c$ is positive, it is a simple SOCP-representable convex problem. I would not be surprised if you can solve it analytically but otherwise solve it using an SOCP solver (or just use any nonlinear solver, it is easy enough)

Using YALMIP in MATLAB, you would just write

sdpvar x1 x2
h1 = ...
h2 = ...
Constraints= [x1+x2==1];
Objective = (a1*x1+a2*x2 - c*norm([h1;h2]);
optimize(Constraints,-Objective);

Answer 2

If I understand the problem correctly the optimization is over $(x_{1},x_{2})$ and not over $(a_{1},\,a_{2})$ because in the last case max=$+\infty$.

Just use the linear constraint and write $x_{2}=1-x$. Then we get

max $a_{1}x+a_{2}(1-x)-c\sqrt{H_{1}^{2}(x)+H_{2}^{2}(x)}$

where $H_{1}(x)=h_{1}(x,1-x)$ and $H_{2}(x)=h_{2}(x,1-x)$

which can be solved in the standard way by considering derivative of the function and finding stationary points, which is:

$a_{1}-a_{2}-c\dfrac{1}{2\sqrt{H_{1}^{2}(x)+H_{2}^{2}}(x)}(2H_{1}(x)H_{1}'(x)+2H_{2}(x)H_{2}'(x))=0$

Solving the last equality you get some stationary points and by considering second derivative you will decide which is the maximum! But it all depends on functions $\, h_{1},\,h_{2}$.