I am an engineer who is currently working with some network optimization problem. In my work, I encounter a strange optimization problem that seems to be convex but it has some cubic and quartic constraint. The problem look like this
$\begin{array}{*{20}{c}} {\mathop {\min }\limits_{x,y,z,t,y} }&{x + y + z + t + y}\\ {}&{{a_1}{x^3} + {a_2}{y^3} + {a_3}{z^3} - t \le 0}\\ {}&{{b_1}{x^4} + {b_2}{y^4} + {b_3}{z^4} - y \le 0}\\ {}&{x + y + z + t + y \ge 1}\\ {}&{...{\rm{and}}\,{\rm{some}}\,{\rm{linear}}\,{\rm{constraints}}} \end{array}$
Here all of $a_1,a_2,a_3,a_4$ and $b_1,b_2,b_3,b_4$ are just some positive number. The decision variable $x,y,z,t,y$ are non negative.
What should I do to deal with this type of problem ?.
Edit:
In some situations, the objective function is not linear but may be a sum of quadratic and cubic terms ${x^2} + {y^2} + {z^2} + {t^3} + {y^3}$
Thank you very much !
For a model to be convex, both the objective function and the feasible region must be convex. The objective function $x + 2y + z + t$ is convex because it is linear, so we can get that out of the way. For a constraint to be convex, its eigenvalues for the hessian must be positive. In other words, there are two definitions we need to use:
A matrix is called positive semi-definite if it is symmetric and all its eigenvalues are non-negative.
If the Hessian matrix is positive semi-definite at all points in a set, then the function is convex in that set.
Now let us look at each constraint individually,
The constraint $x + 2y + z + t \ge 1$ is linear, so that is done. For $a_1x^3+a_2y^3+a_3z^3−t\le0$, the Hessian is:
\begin{bmatrix} 6a_1x & 0 & 0 & 0\\ 0 & 6a_2y & 0 & 0\\ 0 & 0 & 6a_3z & 0\\ 0 & 0 & 0 & 0\\ \end{bmatrix}
In which the eigenvalues (the diagonals) are non-negative since $x,y,z\ge0$ and $a_1,a_2,a_3>0$, therefore this is convex. Likewise, for constraint $b_1x^4+b_2y^4+b_3z^4−y$, the Hessian is:
\begin{bmatrix} 12b_1x^2 & 0 & 0 \\ 0 & 12b_2y^2 & 0 \\ 0 & 0 & 12b_3z^2 \end{bmatrix}
In which the eigenvalues (the diagonals) are non-negative since $x,y,z\ge0$ and $b_1,b_2,b_3>0$, therefore this is convex.
Therefore, this model is convex with the original problem for the particular region space. For the additional objective function, the Hessian of that is,
\begin{bmatrix} 2 & 0 & 0 & 0\\ 0 & 6y+2 & 0 & 0\\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 6t \end{bmatrix}
The same logic applies, and it is therefore convex.