# How to linearize the following constraints

Given the following two expressions:

1. $$x - \frac{1}{T}\sum_{i} y_{i}$$
2. $$x - \frac{1}{Q}\sum_{i} \beta_{i} y_{i}$$

where $$x \in \mathbb{Z}_{+}$$, $$y \in \mathbb{R}_{+}$$, and $$T$$, $$Q$$ and $$\beta_{i}$$ are given constants.

I need to find the following:
a) The min between these two expressions $$F = \min (x - \frac{1}{T}\sum_{i} y_{i}, x - \frac{1}{Q}\sum_{i} \beta_{i} y_{i})$$
b) if $$F > 0$$ I want to enforce on variable $$w \in \mathbb{Z}_{+}$$ the floor of the min, $$i.e.$$ $$w \leq \lfloor{F}\rfloor$$
c) if $$F < 0$$ I want to enforce on another variable $$v \in \mathbb{Z}_{+}$$ the ceil of the absolute value of $$F$$ (since the minimum could be negative) $$i.e. v \geq \lceil {|F|}\rceil$$.

What I have done so far is the following:

1. To find the min function, I did the same as in here How to linearize min function as a constraint?. I define two aux variables equals each expression.
$$r^{1} = x - \frac{1}{T}\sum_{i} y_{i}$$
$$r^{2} = x - \frac{1}{Q}\sum_{i} \beta_{i} y_{i}$$

a binary variable $$z$$ and the following constraints are introduced to find the min:
$$r^{2} - r^{1} \leq Mz$$
$$r^{1} - r^{2} \leq M(1-z)$$
$$F \leq r^{2}$$
$$F \leq r^{1}$$
$$F \geq r^{1} - M(1 - z)$$
$$F \geq r^{2} - M(z)$$

1. To find the absolute value $$abs = |F|$$. As proposed by here https://groups.google.com/g/gurobi/c/YUWqUfk3PSc I introduce two continuous variables $$p$$ and $$n$$ and a binary variable $$\delta$$ used as an indicator and the following constraints:
$$F = p -n$$
$$p \leq M \delta$$
$$n \leq M (1 - \delta)$$
$$abs = p + n$$

Now my question is I have difficulties to formulate the third constraints. I introduced a binary variable $$\delta^{'}$$ that equals 1 when $$F > 0$$ and 0 otherwise. Then this can be formulate as following:
$$F \leq M\delta^{'}$$
$$F \geq M(1 - \delta^{'})$$

Now I want to say:
if $$\delta^{'} = 1$$ then $$w \leq \lfloor{F}\rfloor$$ ($$v$$ is already enforced to be 0 so we don't have to worry about it in this case)
if $$\delta^{'} = 0$$ then $$v \geq \lceil{abs}\rceil$$ (similarly $$w$$ is already enforced to be 0)

Given the variable $$\delta \in \{0,1\}$$ that decides if the minimum function $$F$$ is positive ($$\delta = 1$$) or negative ($$\delta = 0$$) The formulation is rather straightforward.
$$w \geq F - \epsilon - M(1 - \delta)$$
$$w \leq F + M(1 - \delta)$$
$$v \geq |F| - M\delta$$
$$v \leq F + \epsilon + M\delta$$
where $$\epsilon$$ is a small tolerance, and $$M$$ a big valid constant.