In [1] following modeling question is posed:
This is not so easy to do. Here is my attempt. The idea behind my approach, is to use another intermediate integer variable \(a_i\) (accumulator) defined by \[a_i = \begin{cases} a_{i-1}+1 & \text{if $x_i=1$ }\\ 0 & \text{otherwise}\end{cases}\]
The calculation of variable \(a_i\) can be stated as a quadratic constraint: \[a_i = x_i(a_{i-1}+1)\]
In terms of indicator constraints (implications as available in many MIP solvers) this would look as follows: \[\begin{align} &x_i=0 \Rightarrow a_i = 0\\ &x_i = 1 \Rightarrow a_i = a_{i-1}+1\end{align}\]
If the MIP solver does not support indicator constraints, we can write this as a bunch of linear inequalities: \[\begin{align} & (a_{i-1}+1) - (1-x_i)M \le a_i \le (a_{i-1}+1) + (1-x_i)M\\ & 0 \le a_i \le x_i M \end{align}\] Here \(M\) is the length of the largest possible string of consecutive 1's, e.g.\(M=\mathit{card}(i)\).
We can also linearize the quadratic constraint, but that leads to extra variables.
This is complex enough, we really have to try this out. The results of the big-\(M\) inequality approach look like:
The MINLP formulation is the most elegant: just two equations.
I doubt the user really needs this. Usually if we know more about the rest of the problem, we can simplify things considerably. For instance, suppose the model really only needs to establish an upper bound on the run-length. If we only need \(z_i \le R\), we can drop a boatload of constraints. One formulation can be: \[ \begin{align} & x_i =1 \Rightarrow z_i = z_{i-1} + 1\\ & z_i \le R\end{align}\] A formulation only involving the original \(x_i\) variables can be: \[\sum_{k=i}^{i+R}x_k \le R \>\> \forall i\]
This is a good example where we need to look at the whole model to give the best possible advice on how to model things. Just looking at a fragment may lead to formulations that are too general and thus way too complicated.
Given binary variables \(x_i \in \{0,1\}\), I want to form an integer variable \(z_i\) that has the length of the preceding string of ones (this is sometimes called the run-length). I.e.:
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| Given x(i) find z(i) |
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| Additional variable a(i) |
Step 1: form \(a_i\)
The calculation of variable \(a_i\) can be stated as a quadratic constraint: \[a_i = x_i(a_{i-1}+1)\]
In terms of indicator constraints (implications as available in many MIP solvers) this would look as follows: \[\begin{align} &x_i=0 \Rightarrow a_i = 0\\ &x_i = 1 \Rightarrow a_i = a_{i-1}+1\end{align}\]
If the MIP solver does not support indicator constraints, we can write this as a bunch of linear inequalities: \[\begin{align} & (a_{i-1}+1) - (1-x_i)M \le a_i \le (a_{i-1}+1) + (1-x_i)M\\ & 0 \le a_i \le x_i M \end{align}\] Here \(M\) is the length of the largest possible string of consecutive 1's, e.g.\(M=\mathit{card}(i)\).
We can also linearize the quadratic constraint, but that leads to extra variables.
Step 2: form \(z_i\)
To go from \(a_i\) to \(z_i\), we can define: \[z_i = \begin{cases} a_{i-1} & \text{if $x_i=0$ and $x_{i-1}=1$ }\\ 0 & \text{otherwise}\end{cases}\] In implication form this is:\[\begin{align} & x_i=0 \text{ and } x_{i-1}=1 \Rightarrow z_i =a_{i-1}\\ & x_i=1 \text{ or } x_{i-1}=0 \Rightarrow z_i=0 \end{align}\] This is not immediately ready for use with indicator constraints (these constraints want a single binary variable at the left of \(\Rightarrow\)). After introducing a new binary variable \(y_i \in \{0,1\}\) and calculating \(y_i = x_{i-1} (1-x_i)\) (this can be linearized easily), we can implement the indicator constraints:\[\begin{align}&y_i \ge x_{i-1}-x_i\\ &y_i \le x_{i-1}\\ & y_i \le 1-x_i\\ & y_i=1 \Rightarrow z_i =a_{i-1}\\ & y_i=0 \Rightarrow z_i=0 \\& y_i \in \{0,1\} \end{align}\]
Written as linear inequalities, we have: \[\begin{align} & a_{i-1} - x_i M - (1-x_{i-1})M \le z_i \le a_{i-1} + x_i M + (1-x_{i-1})M\\ & 0 \le z_i \le 1-x_{i-1} M \\& z_i\le (1-x_i)M \end{align}\]
An MINLP formulation is simpler: \[z_i = a_{i-1}(1-x_i) x_{i-1}\]
Putting things together
To summarize: we can come up with (at least) three different formulations:
| big-M inequalities | Indicator constraints | MINLP formulation |
|---|---|---|
| \[\begin{align}&a_i \le (a_{i-1}+1) + (1-x_i)M\\ & a_i \ge (a_{i-1}+1) - (1-x_i)M\\& a_i \le x_i M\\& z_i \le a_{i-1} + (1-x_{i-1}+x_i)M\\& z_i \ge a_{i-1} - (1-x_{i-1}+x_i)M \\ &z_i \le x_{i-1} M\\ & z_i\le (1-x_i)M\\& a_{i} \in \{0,1,2,\dots\}\\ & z_{i} \in \{0,1,2,\dots\} \end{align}\] | \[\begin{align} & x_i=0 \Rightarrow a_i = 0\\ &x_i = 1 \Rightarrow a_i = a_{i-1}+1\\ & y_i \le 1-x_i\\& y_i \le x_{i-1}\\ &y_i \ge x_{i-1}-x_i \\ & y_i = 0 \Rightarrow z_i = 0\\& y_i=1 \Rightarrow z_i = a_{i-1}\\ &a_{i} \in \{0,1,2,\dots\}\\ &y_{i} \in \{0,1\} \\ & z_{i} \in \{0,1,2,\dots\} \end{align}\] | \[\begin{align} &a_i = x_i(a_{i-1}+1)\\ &z_i = a_{i-1}(1-x_i) x_{i-1}\\& a_{i} \in \{0,1,2,\dots\}\\ & z_{i} \in \{0,1,2,\dots\} \end{align}\] |
This is complex enough, we really have to try this out. The results of the big-\(M\) inequality approach look like:
---- 71 VARIABLE x.L
i3 1.000, i4 1.000, i7 1.000, i8 1.000, i9 1.000
---- 71 VARIABLE a.L
i3 1.000, i4 2.000, i7 1.000, i8 2.000, i9 3.000
---- 71 VARIABLE z.L
i5 2.000, i10 3.000
The MINLP formulation is the most elegant: just two equations.
Maximum run-length
I doubt the user really needs this. Usually if we know more about the rest of the problem, we can simplify things considerably. For instance, suppose the model really only needs to establish an upper bound on the run-length. If we only need \(z_i \le R\), we can drop a boatload of constraints. One formulation can be: \[ \begin{align} & x_i =1 \Rightarrow z_i = z_{i-1} + 1\\ & z_i \le R\end{align}\] A formulation only involving the original \(x_i\) variables can be: \[\sum_{k=i}^{i+R}x_k \le R \>\> \forall i\]
This is a good example where we need to look at the whole model to give the best possible advice on how to model things. Just looking at a fragment may lead to formulations that are too general and thus way too complicated.
References
- Linear/Integer Programming: Counting Consecutive Ones, https://math.stackexchange.com/questions/2732897/linear-integer-programming-count-consecutive-ones