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Consider an objective function in the form of minimization of maximization that is the sum of $N$ similar functions $f\left(x\right)\ge 0$, $\ \forall x$. The summation of all variables is constant (e.g., 1), like a limited resource:

${\mathrm{min}} \sum^N_{j=1}{f\left(x_j\right)}$

or

${\mathrm{max}} \sum^N_{j=1}{f\left(x_j\right)}$

s.t.

$\sum^N_{j=1}{x_j}=1$

$x_j\ge 0,\ \forall j$

The function $f(x)$ is highly non-linear and non-convex. I examined the model computationally for a given function, and in the optimal solution, I can see that

  • For the minimization case, there is the most unbalance: all terms in the objective function except one become zero;

  • For the maximization case, there is the highest balance: all terms in the objective function get a value close to each.

I cannot analytically prove these observations. Are there any theorems or properties about $f(x)$ to show this fact?


This is the main problem with at least two terms in the objective function. Each term represents a facility. So, they can be more than two.

$min Z(x_1,y_1,x_2,y_2)=\frac{{\beta }_c{x_1}^2+{\beta }_mx_1y_1+{\beta }_n{y_1}^2}{(h_1-r_cx_1-r_ny_1)}+\frac{{\beta }_c{x_2}^2+{\beta }_mx_2y_2+{\beta }_n{y_2}^2}{(h_2-r_cx_2-r_ny_2)}$

s.t.

$x_1+x_2=1$

$y_1+y_2=1$

$0\le x_1,y_1,x_2,y_2\le 1$

All parameters are non-negative. The denominators are also positive.

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    $\begingroup$ Did you visualize it using may be just 3 variable? $\endgroup$ Commented Dec 11, 2022 at 15:36
  • $\begingroup$ What exactly does "similar" mean? If you actually showed us the function(s), maybe (maybe) someone could say something useful about the problem. Do you mean the function(s) are identical, so that the objective is additive in the components of x? $\endgroup$ Commented Dec 11, 2022 at 15:47
  • $\begingroup$ If $f$ is linear, then all solutions are equal. If $f$ can be any "highly non-linear and non-convex" function then I'm afraid you won't get a general answer. Maybe if $f$ can be written piecewise, split into a few pieces with each piece being either convex or concave, then you can arrive at a conclusion. $\endgroup$
    – Stef
    Commented Dec 11, 2022 at 20:25

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