# Can we simplify (perhaps linearize) this constraint?

We are dealing with a stochastic model and one of the constraints is \begin{align} y_j=\frac{\sum_{i \in I}\sum_{k \in K}\mathbb{E}\left[X_{ik}^2\right]x^k_{ij}}{\sum_{i \in I} \sum_{k \in K} \mathbb{E}\left[X_{ik}\right]x^k_{ij}}. \end{align} Here, decision variables are $$y_j\geq 0$$ and $$x_{ij}^k$$ which is binary and $$X_{ik}$$ is a random variable for which we know its mean and variance.

Is there a way to perhaps linearize this constraint? The only thing that came to mind for me was to use $${Var}[X]=\mathbb{E}[X^2]-\mathbb{E}[X]^2$$, but this was not useful.

I would appreciate some hints so I try to solve it myself.

• Do you also have constraints like $\sum_i x_{ij}^k=1$ or maybe $\sum_k x_{ij}^k=1$? Dec 17, 2021 at 22:35
• We have $\sum_j \sum_k x_{ij}^k=1$ for all $i$. Could you please let me know why this important? Is there another way to simplify the ratio?
– user9659
Dec 18, 2021 at 4:03
• I was thinking that you might be able to apply compact linearization a la Liberti. Where else does $y_j$ appear in the model? Dec 18, 2021 at 4:33
• Thank you. I'm looking it up now. The complete constraint that we modeled is $\mathbb{E}[Y_j]=(y_j z_j)/(1-z_j)$ for all $j$ and $z_j>0$, and we have $y_j$ as mentioned in the question. That is why I was trying to linearize the $RHS$ of $y_j$ by itself.
– user9659
Dec 18, 2021 at 5:59

Assuming the denominator cannot be zero (which would cause the known universe to implode) and that you can provide an upper bound for $$y_j$$, you can multiply both sides of the equation by the denominator. The new right side (the numerator) will be linear. The new left side will be $$\sum_i \sum_k \mu_{ik} x_{ij}^k y_j$$ (where $$\mu$$ is the mean of $$X$$). Now you just need to linearize the product $$x_{ij}^k y_j$$, which is a FAQ. See here for an answer.

• Thank you. Is there a way to simplify the LHS by itself?
– user9659
Dec 18, 2021 at 4:08
• @prubin, Thanks for your useful answer. could you say please, how the new RHS might be linear if $\mathbb{E} X^2_{i,k}$ would be a variable? Is it replaced by its mean that is a parameter? Dec 18, 2021 at 7:56
• @A.Omidi $X_{i,k}$ is a random variable, not a decision (model) variable. As noted by the author, the expectation of the square is the sum of the variance and the square of the mean, both of which I am assuming are parameters to the problem.
– prubin
Dec 18, 2021 at 17:41
• @ZiggyIggy I'm not sure what you mean.
– prubin
Dec 18, 2021 at 17:42
• @prubin, Many thanks for your explanation. Dec 19, 2021 at 4:55