2
$\begingroup$

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 $(x_{1},x_{2},x_{3})$.

$\endgroup$
3
$\begingroup$

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 $D$.

| improve this answer | |
$\endgroup$
  • $\begingroup$ In this case simply add the terms to your existing objective function. $\endgroup$ – Kuifje Jun 25 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 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 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 at 10:37
  • $\begingroup$ When I first heard of 'Operations Research Beta' I immediately thought to look you up, Kuifje! $\endgroup$ – BCLC Jun 26 at 7:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.