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I want to linearize the following function

$f(x)=\frac{\theta}{2(1-\rho)}$

where, $\theta$ and $\rho$ are continuous variables. When I checked the convexity, I found that it is not convex since one of the eigenvalues is positive and the other is negative. In my function $f(x)$,

$\theta=\sum_{j\in J}\frac{\lambda_j}{\mu_j^2}$

and

$\rho=\sum_{j\in J}\frac{\lambda_j}{\mu_j}$.

Here, all $\lambda_j$ and $\mu_j$ are known parameters, however, the set $J$ is variable, which makes $\theta$ and $\rho$ variable. Assume $N=3$, then the set $J$ can have 7 or ($2^N-1$) combinations, e.g., $\{1\}, \{2\}, \{3\}, \{1,2\}, \{1,3\}, \{2,3\}, \{1,2,3\}$ excluding ${\emptyset}$. The problem is solvable with small $N$, however, in my case I have $N=5000$ or more. So, I have $2^{5000}-1$ different sets of $J$ to calculate. I will use $f(x)$ in an optimization problem to minimize a cost function.

Is there any way to linearize or find an approximation for $f(x)$?

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1 Answer 1

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You mention $f(x)$ but I don't see any $x$? I assume this is just sloppy notation.

Let's start at the bottom. I believe $\theta$ and $\rho$ can be expressed as:

$$\begin{align} &\theta = \sum_j \alpha_j x_j\\ &\rho = \sum_j \beta_j x_j\\ &\sum_j x_j \ge 1\\ &x_j \in \{0,1\} \end{align} $$

where $\alpha_j, \beta_j$ are constants and $x_j$ are binary variables.

The equation $$ z = \frac{\theta}{2(1-\rho)} $$ can now be rewritten as: $$ 2(1-\rho)z = \theta $$ or $$ 2z - 2 \sum_j \beta_j (z x_j) = \theta $$ The terms $z x_j$ can be easily linearized (multiplication of a binary and a continuous variable).

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