Linear programming (apply different multipliers for negative values )

Question

The optimisation problem [PROBLEM 1]: $$\max \sum_{t=T} x_t $$ subject to: $$\ (1) x_t = (1+\alpha) x_{t-1}\ \forall t ..T$$ $$\ t=\{ 0..T\} , x_0= given$$ $$\ (e.g) x_1 = (1+\alpha) x_0\ $$ $$\ (e.g) x_2 = (1+\alpha) x_1\ $$ $$\ (e.g) x_3 = (1+\alpha) x_2\ $$

$$\ $$

$$\ (2) x_t \ge 0 $$

Now, I would like to allow negative value of the variable "x" and apply different factor. I made the following changes [PROBLEM 2]:

$$\ x_t = x_t^+ - x_t^- $$

$$\ x_t = (1+\alpha) x_{t-1}^+ - (1+\beta) x_{t-1}^- $$

$$\ x_t^+ , x_t^- \ge 0 $$

$$\ x_t \ge -10 $$

The solver gives infeasible(unbounded) solution! Is there an alternative way to solve such kind of problem, particularly, applying different factor (multiplier) for negative values. The model is more complex than what I wrote. But I wanted to just highlight on the problem.

Answer 1

if $x_0$ is free, your problem is unbounded since. $\ x_t = (1+\alpha) x_{t-1}$ implies $\ x_t = (1+\alpha)^t x_0 $ so you can let $ x_0 \to + \infty $ .

if $x_0$ is given and fix then your feasible solution is singleton which is

$$\{ \bigl( x_0, ~ (1+\alpha) x_0, ~ (1+\alpha)^2 x_{0} ,..., ~ (1+\alpha)^T x_{0} \bigr) \}$$

So your optimal solution is trivial . why need to solve problem !

Answer 2

You will likely need to introduce a binary variable $y_t$ for each value of $t$, signaling whether $x_t$ is positive or negative. The definition of $y_t$ would be handled by something like $$-My_t \le x_t \le M(1-y_t)$$ for some adequately large $M>0$. In this version, $y_t=0\implies x_t\ge 0$ and $y_t=1\implies x_t\le 0$. I can't write the $x_t$ : $x_{t+1}$ linkage without knowing the actual model, but it would involve multiple inequalities containing $y_t$.