Suppose I have a model for creating a nurse's duty roster. The model has the indices $i$ for the person, $t$ for the day and $s$ for the shift. I have the binary variable $x_{its}$ which indicates whether a person is working or not. Now I want to "capture" the monotony of working hours in a new variable. For example, if a nurse works the same shift for 14 days, with or without break days, then this should be stored in a new binary variable. How could I do this?
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$\begingroup$ Is a shift a sequence of hours, or days? $\endgroup$– KuifjeCommented May 11 at 11:21
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$\begingroup$ A sequence of hours, a day has three shifts $\endgroup$– manofthousandnamesCommented May 11 at 11:24
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$\begingroup$ Do you mean exactly 14 days or at least 14 days? $\endgroup$– RobPrattCommented May 11 at 13:39
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$\begingroup$ @RobPratt for exactly 14 days, but I want to define this monotony variable $y_{id}$ for every $i$ and day $t$, so f.e on day 15, the shifts were the same for person $i$, which yield $y_{i15}=1$, but now on day 15 the shift changes, which means $y_{i16}=0$, as the 14 days prior to day 16 were not always the same. So I want to show each day for each person whether the last 14 days have been monotonous. $\endgroup$– manofthousandnamesCommented May 12 at 7:23
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1 Answer
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Define a new binary variable $y_{its}$ that takes value $1$ if nurse $i$ works the same shift $s$ for $n$ consecutive days (exactly), starting on day $t$.
You want to enforce: $$ \neg x_{i,t-1,s} \wedge \left(\bigwedge_{j=t}^{t+n-1} x_{ijs}\right) \wedge \neg x_{i,t+n,s} \implies y_{its}\quad \forall i,s,t $$ Or equivalently via CNF: \begin{align} &\neg \left(\neg x_{i,t-1,s} \wedge \left(\bigwedge_{j=t}^{t+n-1} x_{ijs}\right) \wedge \neg x_{i,t+n,s}\right) \vee y_{its}\quad &\forall i,s,t \\ & x_{i,t-1,s} \vee \left(\bigvee_{j=t}^{t+n-1} \neg x_{ijs}\right) \vee x_{i,t+n,s} \vee y_{its}\quad &\forall i,s,t \\ &x_{i,t-1,s} + \sum_{j=t}^{t+n-1} (1- x_{ijs}) + x_{i,t+n,s} + y_{its} \ge 1\quad &\forall i,s,t \\ & \sum_{j=t}^{t+n-1} x_{ijs} \le x_{i,t-1,s} + x_{i,t+n,s} + y_{its} + (n-1) \quad &\forall i,s,t \end{align}
If you need to enforce the converse: $$ y_{its} \implies \neg x_{i,t-1,s} \wedge \left(\bigwedge_{j=t}^{t+n-1} x_{ijs}\right) \wedge \neg x_{i,t+n,s} \quad \forall i,s,t $$ The same method yields: \begin{align} y_{its} + x_{i,t-1,s} &\le 1 &\forall i,s,t\\ y_{its} + x_{i,t+n,s} &\le 1 &\forall i,s,t\\ y_{its} &\le x_{ijs} &\forall i,s,t, j\in\{t,\cdots,t+n-1\}\\ \end{align}
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$\begingroup$ Thank you very much. Can I also definie $y_{is}$ as $y_{it}$, indicating that the monotony occurs in day $t$? And for which indices is the upper constraint defined? I guess it is not $\forall t\in T$? $\endgroup$ Commented May 11 at 11:41
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$\begingroup$ you could use $y_{is}^t$ to indicate that the monotony occurs for nurse $i$ with shift $s$ on days $t,...,t+n-1$. $\endgroup$– KuifjeCommented May 11 at 11:44
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$\begingroup$ Indeed, the upper constraint is defined for $t=1,\cdots,|T|-n+1$. You need to be careful at the borders. For example, for $t=1$, replace $x_{i,0,s}$ by $0$. And same for $t=|T|-n+1$: replace $x_{i,|T|+1,s}$ by $0$. $\endgroup$– KuifjeCommented May 11 at 12:23
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$\begingroup$ Thanks. Can I also definie $z_{it}=\sum_{s\in S} y_{is}^t$, would this not add up? And can I also define the constraints for the last $n$ days and not to following one? $\endgroup$ Commented May 11 at 13:10
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$\begingroup$ In the first of the three inequality constraints, should $t$ be $t-1$? Also, since the inequalities test a streak of exactly $n$ days with the same shift starting on day $t$, shouldn't $y$ have a subscript $t$? $\endgroup$– prubin ♦Commented May 11 at 15:09