Trying to define a variable as the index in another (vector) variable (MATLAB,YALMIP)

回答済み

Pira Perry

2023年04月17日 16:00

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Hello everyone.

I want to define a decision variable that can indicate the first entry in another variable that is equal to  specified value. For example,

a = binvar(4,1); % 4x1 binary variable
b = intvar(1);

I want to add the constraint that let b change with a and indicate the index of first entry in a that is equal to 1, eg. a=(1,1,1,0) => b = 1; a=(0, 1, 1, 0) =>  b=2; a=(0,0,1,0) => b=3; etc.... The relation between a and b is just like function  b=find(a(:) == 1) 

Does anyone know how express this constraint in MATLAB? Or any similar constraint? Very much thanks!

関連

• MATLAB
• YALMIP

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  Riley Clement

  • Gurobi Staff

  2023年04月19日 06:51 編集済み

  Hi Pira,

  Say you have binary variables \(a_1, \ldots, a_n\) and integer variable \(b\), then introduce binary variables \(x_1, \ldots, x_n\) and the following constraints.

  \begin{eqnarray}
  \sum_{i=1}^nx_i & \leq & 1\\
  x_i & \leq & a_i, \quad \forall i \in \{1,\ldots,n\}\\
  \sum_{j=1}^i x_j & \geq & a_i, \quad \forall i \in \{1,\ldots,n\}\\
  b & = & \sum_{i=1}^nix_i
  \end{eqnarray}

  The \(x\) variable corresponding to the leading 1, will be 1, all else will be 0.

  An example:

  \begin{eqnarray}
  && a = (0,0,1,0,1,1)\\
  && x = (0,0,1,0,0,0)\\
  && b = 1\cdot0 + 2\cdot0 + 1\cdot3 + 0\cdot4 + .... = 3\\
  \end{eqnarray}

  If \(a\) is all zero then \(b\) will be zero.

  - Riley

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  Pira Perry

  2023年04月19日 03:12

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  Hi Riley,

  Thanks for your immediate reply.
  Seems like there's no existed function to express this constraint, but it's nice to use the index i to weight and sum. I will try it.

  Thanks again and best regards.
  Pira

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  Riley Clement

  • Gurobi Staff

  2023年04月19日 03:29

  Hi Pira,
  
  No problem, another formulation to try also:
  
  Introduce binary variables \(x_1, \ldots, x_n\) and the following constraints.

  \begin{eqnarray}
  a_i & \leq & x_i, \quad \forall i \in \{1,\ldots,n\}\\
  x_i & \leq & x_{i+1}, \quad \forall i \in \{1,\ldots,n-1\}\\
  \sum_{j=1}^ia_j & \geq & x_i \quad \forall i \in \{1,\ldots,n\}\\
  b & = & n+1 - \sum_{i=1}^nx_i
  \end{eqnarray}

  An example:

  \begin{eqnarray}
  && a = (0,0,1,0,1,1)\\
  && x = (0,0,1,1,1,1)\\
  && b = 7 - \sum_ix_i = 3\\
  \end{eqnarray}

  If \(a\) is all zero then \(b\) will be \(n+1\).
  
  I'm not sure which approach is best without running tests, but feel free to share if you find out.

  - Riley

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