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I have three variables $x_{1},x_{2},x_{3}$ and a function $f : D \rightarrow \mathbb{R}$ where $D$ is defined as such :

$$D = (x_{1},x_{2},x_{3})$$ such that $$x_{1}+x_{2}+x_{3}=1$$ and $$x_{1}>x_{2}>x_{3}$$ Therefore, D is the intersection of a plane and a box, and hence is a plane.

The objective is to find $(x_{1},x_{2},x_{3}) \in D$ such as $f(x_{1},x_{2},x_{3}) = c \in \mathbb{R}$. The function $f$ is injective and so for a fixed $c \in \mathbb{R}$, there can be multiple solutions $(x_{1},x_{2},x_{3})$. Among all the solutions, I want to find the one that has the highest $x_{1}$, and then once the highest $x_{1}$ is found, the highest $x_{2}$. How do I specify this constraint ?

Note that the function $f$ doesn't have a closed-form solution and I am using minimize from scipy to find the $(x_{1},x_{2},x_{3})$.

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If I understood your problem correctly, you want to find $(x_1,x_2,x_3)$ such that:

  • $(x_1,x_2,x_3) \in D$
  • $f(x_1,x_2,x_3) = c$, where $c$ is a given parameter
  • $x_1>x_2 >x_3$

Also, you want to maximize $x_1$ and $x_2$, but $x_1$ has "priority" over $x_2$. So you can use the following objective function:

$$ \max \; \omega_1 x_1 + \omega_2 x_2 $$

where $\omega_1$ is weight larger than $\omega_2$, e.g., $\omega_1 = 10 \omega_2$. Choosing adequate values for $\omega_1$ and $\omega_2$ depends on $D$.

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  • $\begingroup$ In this case simply add the terms to your existing objective function. $\endgroup$
    – Kuifje
    Jun 25, 2020 at 10:20
  • $\begingroup$ EDIT : Sorry I made a mistake in the last comment. The objective function is rather $$\min f(x_{1},x_{2},x_{3})-c$$ so the transformed objective function would be $$\min f(x_{1},x_{2},x_{3})-c-\lambda (w_{1}x_{1}+w_{2}x_{2})$$ with $\lambda >0$. In this case, the heuristic choice of $\lambda$ makes the problem randomish, no ? $\endgroup$
    – FredNgu
    Jun 25, 2020 at 10:33
  • $\begingroup$ Why don't you add $f(x_1,x_2,x_3)=c$ as a constraint ? Or if you cannot have the exact equality, $f(x_1,x_2,x_3) \le c +\varepsilon_1$, and $f(x_1,x_2,x_3) \ge c -\varepsilon_2$ $\endgroup$
    – Kuifje
    Jun 25, 2020 at 10:33
  • $\begingroup$ I couldn't have the exact equality so my first intuition was to minimize the difference. But yes actually adding a small $\epsilon$ and adding it as a inequality constraint is a great idea and should work, thank you ! I'll try this out. $\endgroup$
    – FredNgu
    Jun 25, 2020 at 10:37
  • $\begingroup$ When I first heard of 'Operations Research Beta' I immediately thought to look you up, Kuifje! $\endgroup$
    – BCLC
    Jun 26, 2020 at 7:48

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