How to set a limit for a switch to 0 of a variable for 2 variables combined

Question

I have a follow up question to another question of mine How to set a limit for a switch to 0 of a variable about counting the number of switches to 0 of one decision variable. Now I would like to ask the same question for two combined decision variables. So basically I have the decision variable x(t) and another decision variable y(t). The both quantify the heating output of a heating device for two different thermal storage systems for every timeslot t in [1,...,288]. It should be avoided that the heating device is switched on and off frequently thus I want to set a limit for the switching.

The rule in pseudocode looks like this:

if (x(t-1)>0 AND x(t)=0 AND y(t)=0) OR if (y(t-1)>0 AND y(t)=0 AND x(t)=0) 
then increase count by 1
Constraint: count <= limit

Both variables x(t) and y(t) are continious variables with the boundaries [0,1]. It should also be noted that x(t) and y(t) can't be greater 0 simultaneously (this would mean that the heating device would have heated up 2 storages at the same time which is not possible). For this I just use 2 constraints with binary auxilliary variable h(t)

x(t)<= h(t)
y(t)<= (1- h(t))
with h(t) in {0,1}

Any idea how I can derive constraints for something like this? Next to the pure answer I would also appreciate some general advices as how to approach questions like this (if there is a more or less general way of doing this).

Answer 1

Again, you need to introduce binaries:

• $\delta_t$ takes values $1$ if and only if the device is switched off at time $t$
• $\alpha_t$ is the binary associated with variable $x_t$
• $\beta_t$ is the binary associated with variable $y_t$

The condition can be written in conjunctive normal form as follows:

$$ (\alpha_{t-1}\wedge \lnot \alpha_{t} \wedge \lnot \beta_{t}) \vee (\beta_{t-1}\wedge \lnot \beta_{t} \wedge \lnot \alpha_{t}) \implies \delta_t\\ \lnot \left( (\alpha_{t-1}\wedge \lnot \alpha_{t} \wedge \lnot \beta_{t}) \vee (\beta_{t-1}\wedge \lnot \beta_{t} \wedge \lnot \alpha_{t})\right) \vee \delta_t\\ (\lnot \alpha_{t-1}\vee \alpha_{t} \vee \beta_{t}) \wedge (\lnot \beta_{t-1}\vee \beta_{t} \vee \alpha_{t}) \vee \delta_t\\ (\lnot \alpha_{t-1}\vee \alpha_{t} \vee \beta_{t} \vee \delta_t)\wedge (\lnot \beta_{t-1}\vee \beta_{t} \vee \alpha_{t} \vee \delta_t)\\ (1-\alpha_{t-1}+\alpha_t + \beta_t+ \delta_t \ge 1) \wedge (1-\beta_{t-1}+\beta_t + \alpha_t+ \delta_t \ge 1) $$

And since $x_t$ and $y_t$ cannot simultaneously be positive, the constraints are \begin{align} \alpha_{t-1} &\le \alpha_t + \beta_t+ \delta_t \quad &\forall t \tag{1}\\ \beta_{t-1} &\le \beta_t + \alpha_t+ \delta_t \quad &\forall t \tag{2}\\ \sum_t \delta_t &\le \ell \tag{3} \\ \alpha_t + \beta_t &\le 1 \quad &\forall t \tag{4} \\ x_t &\le M \alpha_t \quad &\forall t \tag{5} \\ y_t &\le M \beta_t \quad &\forall t \tag{6} \\ \end{align}