# Constraints that set values to binary variables depending on other binaries

I am trying to write a mathematical problem that involves some conditions based on binary variables. More specifically, I have a set of three binary variables $$d_1$$, $$d_2$$, $$d_3$$ and depending on their values, I want to have some other binary variables $$y_1,\ldots,y_5$$ activated according to the following table:

$$y_1$$ $$y_2$$ $$y_3$$ $$y_4$$ $$y_5$$
$$d_1$$ 1 0 0 1 1
$$d_2$$ 0 1 0 1 1
$$d_3$$ 0 0 1 0 1

For instance, $$y_1$$ is equal to 1 if and only if $$d_1$$ is equal to 1. Same goes for the pairs $$(y_2,d_2)$$ and $$(y_3,d_3)$$. $$y_4$$ becomes equal to 1 if both $$d_1$$ and $$d_2$$ are 1. Finally, $$y_5$$ becomes 1 if all of $$d_1$$, $$d_2$$ and $$d_3$$ are equal to 1.

Moreover, at the same time, only one of $$y_1,\ldots,y_5$$ can be activated, which is to formulate as $$\sum y_i = 1$$.

I am having problems though formulating the other constraints regarding the $$d_i$$, $$y_i$$ variables.

I have tried formulating the following set of constraints for $$y_5$$:

$$y_{\rm intermediate} \geq d_1 + d_2 - 1$$

This will allow $$y_{\rm intermediate}$$ to become 1 if both $$d_1$$ and $$d_2$$ are activated.

Then, I could have: \begin{align}y_5 &\geq y_{\rm intermediate} + d_3 - 1\\y_4&\geq y_{\rm intermediate} + (1 - d_3) - 1\end{align}

Then depending on the value of $$d_3$$ either $$y_5$$ or $$y_4$$ will become equal to 1.

However, I am having trouble formulating the rest of the constraints for $$y_1$$ to $$y_3$$ and I am also not sure if what I already have is good enough.

Does anyone have any pointers or ideas? Any help would be greatly appreciated.

## 1 Answer

If I understand correctly, the following enforces your desired behavior: \begin{align} y_1 &= d_1 \\ y_2 &= d_2 \\ y_3 &= d_3 \\ y_4 &\ge d_1 + d_2 - 1\\ y_5 &\ge d_1 + d_2 + d_3 - 2\\ \end{align} If you also want to enforce $$y_4 \implies (d_1 \land d_2)$$ and $$y_5 \implies (d_1 \land d_2 \land d_3)$$, then include these additional constraints: \begin{align} y_4 &\le d_1 \\ y_4 &\le d_2 \\ y_5 &\le d_1 \\ y_5 &\le d_2 \\ y_5 &\le d_3 \end{align}

On second thought, your comment about only one $$y_i$$ being activated makes me think that, instead of $$y_1 = 1 \iff d_1 = 1$$, you meant $$y_1 = 1 \iff (d_1,d_2,d_3) = (1,0,0)$$. Equivalently, $$y_1 = d_1(1-d_2)(1-d_3)$$, which you can linearize as follows: \begin{align} y_1 &\le d_1 \\ y_1 &\le 1 - d_2 \\ y_1 &\le 1 - d_3 \\ y_1 &\ge d_1 + (1 - d_2) + (1 - d_3) - 2 \end{align} You could similarly set up four linear inequality constraints for each of the other $$y_i$$ and then impose $$y_1+y_2+y_3+y_4+y_5=1$$. But a simpler formulation is just to treat the table entries as the constraint coefficients: \begin{align} y_1 + y_4 + y_5 &= d_1 \\ y_2 + y_4 + y_5 &= d_2 \\ y_3 + y_5 &= d_3 \\ y_1 + y_2 + y_3 + y_4 + y_5 &= 1 \end{align}