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### Linearization the product of three variables (two binary & one continuous)

More compactly, you want to enforce $$-x|y| \le p \le x|y|$$ Assume $L \le y \le U$ for some constants $L\le 0$ and $U\ge 0$, and let $M=\max(-L,U)$. Introduce continuous decision variable $w$ (to ...
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### Linearization the product of three variables (two binary & one continuous)

Basically you are trying the following: $p \le xp_{max}$: ensures $p=0$ if $x=0$, else $p$ is free, $p_{max}$ is upper bound of $p$ And $y(2z-1) \le p \le y(1-2z)$ For $y, z$ switching I think ...
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1 vote

### Linearization the product of three variables (two binary & one continuous)

By updating the question, the mentioned conditional inequalities can be linearized by introducing new indicator variables, $z_i$, for each part and coupling those to make linearization. For example ...
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### Linearizing a disjunctive expression into MILP

From the modeling language point of view, the above disjunctive form can be written by CPLEX in the following form: ...
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1 vote

### Maximization problem with preferences on variables

How is this following constraint Either $b_z + y \le b_x+z +b_y\le x + b_z +yb_y$ Or let's break it $b_x+z \le x + b_z$ $b_z + y \le z +b_y$
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Constr(7) is saying from every slot $s$ if any container (that's why you have summation, whenever you want to work on conditions like any/atleast) is picked up, the seaside crane's position $x_{2t}$ ...