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A binary array $t = [t_1, t_2, t_3, t_4, t_5]$ with each element a binary integer variable taking values 0 or 1. You can think this vector as slots with 1 representing the slot being taken and 0 otherwise.

Constraints: Now 2 appointments need to be scheduled with the first one taking 1 slot and the second one taking 2 slots. The second appointment must be scheduled at or after the second slot ($t_2$).

Objective: Maximize the number of consecutive zeros in array $t$.(Intend to leave a long range empty slots for future planning)

Solutions: One of the optimal solutions is to put first appointment into $t_1$ and the second appointment into $t_2$ and $t_3$, $t = [1,1,1,0,0]$, which has a consecutive zero number 2. A feasible but non-optimal solution is to put first appointment into $t_1$ and the second appointment into $t_3$ and $t_4$, $t=[1,0,1,1,0]$, which has a consecutive zero number 1.

Optimal Constraints: How to formulate the question in a linear/integer/mixed-integer way that can be solved by an optimization solver? Constraints can be definitely formulated in a linear integer way but I am having a hard time for the objective.

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  • $\begingroup$ Do you specifically want to maximize the number of consecutive zeros at the end of the sequence (meaning a long string of zeros in the middle of the sequence would not fulfill your goal)? $\endgroup$ – prubin Feb 20 at 21:10
  • $\begingroup$ Either way works. Any of [1,1,1,0,0], [1,0,0,1,1], [0,0,1,1,1] is the optimal solution. $\endgroup$ – MIMIGA Feb 21 at 17:12
  • $\begingroup$ Any other objective that can achieve a similar goal is also helpful. $\endgroup$ – MIMIGA Feb 21 at 18:39
  • $\begingroup$ Hi MIMIGA, Is the length of array t a constant in your problem? How about number of appointment and their duration? $\endgroup$ – Oguz Toragay Feb 21 at 22:01
  • $\begingroup$ @OguzToragay Length of array, number of appointments, and their duration are all fixed. They are not constraints. I assume if the example I gave can be formulated, I can extend to a more general formulation. $\endgroup$ – MIMIGA Feb 22 at 3:53
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Let $N$ be the dimension of your binary vector $x$. Introduce new variables $w_n \in [0,n]$ for $n=1,\dots,N$. Each $w_n$ will capture the number of consecutive zeros culminating at position $n$. So, for instance, if $x=[1,0,0,1,1]$, then $w=[0,1,2,0,0]$. Note that $w$ does not need to be declared integer; the constraints will force it to be integer-valued.

Next, add the following constraints for each $n$ (where $w_0 = 0$): \begin{align*} w_{n} & \le N(1-x_{n})\\ w_{n}-w_{n-1} & \le1\\ w_{n}-w_{n-1} & \ge1-Nx_{n}. \end{align*} If $x_n=1$, the first constraint forces $w_n=0$ and the second and third constraints have no effect. If $x_n=0$, the first constraint has no effect, while the second and third constraints combine to force $w_n=w_{n-1}+1$.

As I understand you, you now want to maximize $$\max_{n=1}^N w_n.$$ To do this, you can introduce binary variables $z_1,\dots,z_N$ along with the constraint $$\sum_{n=1}^N z_n = 1,$$ plus a continuous variable $y$ to represent the objective value. You will maximize $y$ subject the constraints $$y\le w_{n}+N(1-z_{n})\quad \forall n.$$ This last constraint is nonbinding when $z_n=0$. For exactly one $n$, $z_n$ will be 1 and $y$ will be less than or equal to $w_n$. The solver will choose the $n$ corresponding to the largest $w_n$ and set $y=w_n$.

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  • $\begingroup$ it's an interesting answer but, would you please, what does this "$\sum\limits_{n=1}^N z_n = 1$" mean? $\endgroup$ – A.Omidi Feb 23 at 5:30
  • $\begingroup$ @A.Omidi it's for linearization of $y = \max w_n$. You can check the answer here $\endgroup$ – EhsanK Feb 23 at 20:05
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    $\begingroup$ Paul, I think the first paragraph should be modified since $w_n$ (as also obvious in your example) is an integer and $w_n \notin [0,1]$ $\endgroup$ – EhsanK Feb 23 at 20:06
  • $\begingroup$ @EhsanK, that's right. many thanks for your hint. $\endgroup$ – A.Omidi Feb 24 at 5:13
  • $\begingroup$ @EhsanK: Thanks, you're right that $w_n$ is an integer, not binary (except for $n=1$). I've fixed it. $\endgroup$ – prubin Feb 24 at 15:53
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I would say that this objective function would do a decent job:

$$\min \sum_{i = 0}^{n} i^2 t_i$$

where $n$ is the total number of available slots.

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  • $\begingroup$ But this only drives to allocate to early slots, right? $\endgroup$ – MIMIGA Mar 11 at 20:24
  • $\begingroup$ Yes, this would try to group them all in the early slots. If instead minimize we maximize, then it will group them on the late slots. Indeed, this is just a simple objective function that does not consider all possible solutions of your problem. $\endgroup$ – jalopez910 Mar 26 at 18:00

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