Model the constraints of an integer linear program I have to model the costraints of an integer linear program. The objective function is not important here.
Suppose we have $C$ configurations (indexed by $i = 1, \dots, C$) and $T$ time intervals (indexed by $j = 1, \dots, T$). I wish that:
a) only one configuration is active at each time interval;
b) when a configuration has been chosen I must keep it for at least $\bar{t}$ time intervals (with $\bar{t} \leq T$).
So I define the decision variables:
$y(i,j) :=
\begin{cases}
1 &amp; \text{if configuration $i$ is active at time $j$}; \\[2ex]
0 &amp; \text{otherwise}.
\end{cases}$
For request a) I write the constraints:
\begin{equation}
\sum_{i=1}^C y(i,j) = 1 \ \ \ \forall \ j = 1, \dots, T.
\end{equation}
Now, I don't know how to proceed with the request b). Any help or suggestions are appreciated. Thanks in advance.
For part b), you can introduce another binary variable $x_{i,j}$ to indicate the start, together with constraints:
\begin{align}
x_{i,j} &amp;\le y_{i,k} &amp;&amp;\text{for $k\in\{j,\dots,j+\overline{t}-1\}$} \tag1\\
x_{i,j} &amp;\le 1 - y_{i,j-1} \tag2\\
(1 - y_{i,j-1}) + y_{i,j} - 1 &amp;\le x_{i,j} \tag3
\end{align}
Constraint $(1)$ enforces $x_{i,j} = 1 \implies \land_{k=j}^{j+\overline{t}-1} (y_{i,k} = 1)$.
Constraint $(2)$ enforces $x_{i,j} = 1 \implies y_{i,j-1} = 0$.
Constraint $(3)$ enforces $(y_{i,j-1} = 0 \land y_{i,j} = 1) \implies x_{i,j} = 1$.

Alternatively, you can omit $x$ and just impose
$$(1 - y_{i,j-1}) + y_{i,j} - 1 \le y_{i,k} \quad\text{for $k\in\{j+1,\dots,j+\overline{t}-1\}$},$$
equivalently,
$$y_{i,j} - y_{i,j-1} \le y_{i,k} \quad\text{for $k\in\{j+1,\dots,j+\overline{t}-1\}$}, \tag4$$
(where $y_{i,j}$ is treated as $0$ if $j&lt;0$ or $j&gt;T$),
which enforces
$$(y_{i,j-1} = 0 \land y_{i,j} = 1) \implies \bigwedge_{k=j+1}^{j+\overline{t}-1} (y_{i,k} = 1).$$

Yet a third approach is to enforce
$$\left(y_{i,j-1} \land \lnot \bigwedge_{k=1}^{\overline{t}} y_{i,j-k}\right) \implies y_{i,j},$$
equivalently,
$$\left(\lnot y_{i,j-1} \lor \bigwedge_{k=1}^{\overline{t}} y_{i,j-k}\right) \lor y_{i,j},$$
which has conjunctive normal form
$$\bigwedge_{k=1}^{\overline{t}} \left(\lnot y_{i,j-1} \lor y_{i,j-k} \lor y_{i,j}\right),$$
yielding linear constraints
$$1 - y_{i,j-1} + y_{i,j-k} + y_{i,j} \ge 1 \quad \text{for $k \in \{1,\dots,\overline{t}\}$}.$$
Rearranging this, we obtain
$$y_{i,j} \ge y_{i,j-1} - y_{i,j-k} \quad \text{for $k \in \{1,\dots,\overline{t}\}$}. \tag5$$
Now we can recover @toronto hrb's formulation by aggregation:
$$y_{i,j} \ge y_{i,j-1} - \frac{1}{\overline{t}}\sum_{k=1}^{\overline{t}} y_{i,j-k} \tag6$$
So $(6)$ is correct but weaker than $(5)$.
Writing variables as $y(i,j)$ is not conventional. Following what you've written, define
$y_{ij}=
\begin{cases}
1 &amp; \text{if configuration $i$ is active at time $j$}; \\[2ex]
0 &amp; \text{otherwise}.
\end{cases}$
You also need to define binary variables to indicate if a configuration is chosen or not as
$z_i =
\begin{cases}
1 &amp; \text{if configuration $i$ is chosen}; \\[2ex]
0 &amp; \text{otherwise}.
\end{cases}$
Now the constraint of part b) can be written as
$$\sum_{j=1}^{T}y_{ij}\ge \bar{t} \cdot z_i, i=1,\cdots, C$$
If configuration $i$ is chosen, $z_i=1$, at least $\bar t$ time intervals are forced; otherwise, $z_i=0$, which places no constraints on time invervals.
Edit for consecutive time intervals. No need to use $z_i$ any more.
Let's use Configuration 1 with $\bar{t}=3$ as an example. The constraints should be like
$$y_{12}\ge y_{11}-\frac{1}{3}y_{11}$$
$$y_{13}\ge y_{12}-\frac{1}{3}(y_{11}+y_{12})$$
$$y_{14}\ge y_{13}-\frac{1}{3}(y_{11}+y_{12}+y_{13})$$
$$y_{15}\ge y_{14}-\frac{1}{3}(y_{12}+y_{13}+y_{14})$$
$$\cdots$$
$$y_{1,T}\ge y_{1,T-1}-\frac{1}{3}(y_{1,T-3}+y_{1,T-2}+y_{1,T-1})$$
Does this work? If $y_{11}=1$, then the first three constraints force $y_{12}=y_{13}=1$, but $y_{14}$ is flexible.
Putting it in a compact form, it becomes
$$y_{ij}\ge y_{i,j-1}-\frac{1}{\bar{t}}\sum_{k=1}^{\bar{t}}y_{i,j-k}, j=2, \cdots, T; i=1,\cdots, C.$$
$$y_{i,0}=y_{i,-1}=0, i=1, \cdots, C.$$