# Modelling a binary variable in LPs

I need your help. I'm setting up an LP and I'm trying to find constraints to introduce the binary varibale $$b_{ij}$$. So it should take the value 0 if the sum of all $$a_{ij}$$ values to the period t are $$\le n~ \forall i \in I \text{ and } j \in J$$. $$n$$ being a not yet defined whole number. Otherwise, it is supposed to take the value 1. I think one models the whole thing with Big-M, but how exactly I do not know unfortunately.

• Is $a_{ij}$ a decision variable? If so, what type? Is $n$ a decision variable or an input parameter? How do $i$ and $j$ relate to $t$? Commented Jun 16, 2023 at 17:36
• $a_{ij}$ is also either 0 or 1. It indicates whether seller $i$ sold all of his products in period $j$. If so, it is $=1$ Commented Jun 16, 2023 at 17:38
• So you want to model the following for all $i$ and $t$? $$b_{it} = 0 \iff \sum_{j \le t} a_{ij} \le n$$ Commented Jun 16, 2023 at 17:40
• Yep that's correct Commented Jun 16, 2023 at 17:42

You want to enforce $$b_{it} = 0 \iff \sum_{j \le t} a_{ij} \le n.$$ You can enforce $$b_{it} = 0 \implies \sum_{j \le t} a_{ij} \le n$$ with big-M constraint $$\sum_{j \le t} a_{ij} - n \le M_1 b_{it},$$ where $$M_1 = |\{j\in J: j \le t\}| - n$$.
You can enforce $$\sum_{j \le t} a_{ij} \le n \implies b_{it} = 0,$$ equivalently its contrapositive $$b_{it} = 1 \implies \sum_{j \le t} a_{ij} \ge n + 1,$$ with big-M constraint $$n + 1 - \sum_{j \le t} a_{ij} \le M_2 (1 - b_{it}),$$ where $$M_2 = n + 1$$.
Another way to think of this is that you want $$\sum_{j \le t} a_{ij}$$ to be between $$\color{red}{0}$$ and $$\color{red}{n}$$ if $$b_{it}=0$$ and between $$\color{blue}{n+1}$$ and $$\color{blue}{U_t}=|\{j\in J: j \le t\}|$$ if $$b_{it}=1$$, yielding $$\color{red}{0}(1-b_{it}) + (\color{blue}{n+1})b_{it} \le \sum_{j \le t} a_{ij} \le \color{red}{n}(1-b_{it}) + \color{blue}{U_t} b_{it}$$
$$\sum_{j\le t} a_{i,j} \le n + Mb_{i,j}$$
$$n+\epsilon \le \sum_{j \le t} a_{i,j} + M(1-b_{i,j})$$
with $$\epsilon$$ as a small number, depending upon expected values of $$n$$
• @RobPratt summing over $t$ Commented Jun 16, 2023 at 19:06