I have a problem with inequality constraint. And I'm going to merge the constraints.I know that we can add surplus and slack variables in order to achieve equality constraints. And then use below method.
I know Merge constraints for equal constraints ,And I'll explain it further. my trouble is i have more than two constraint .
my problem
\begin{equation}
\begin{split}
\min &\; \sum_{i=1}^N \sum_{k=1}^K \sum_{l=1}^L b_{kl} Y_{ik}Y_{il}+\sum_{i=1}^N\sum_{j=1}^M c_{i,j} X_{i,j}\\
\text{s.t.} & \; \sum_{i=1}^N X_{ij} =1 ,\ \ j=1.2,...,M \\
& \; a_i Z_i \le \sum_{j=1}^M X_{ij} \le b_i Z_i ,\ \ i=1.2,...,N\\
& \;p_i Z_i \le \sum_{k=1}^K Y_{i,k} \le q_i Z_i ,\ \ i=1.2,...,N\\
&\; X_{ij} \le Y_{ik} ,\ \ i=1.2,...,N , \ \ j=1.2,...,M , j\in t_k \\
&\; Y_{ik} \le \sum _{j \in J} X_{i,j} ,\ \ i=1.2,...,N , \ \ K=1.2,...,k \\
&\; X_{ij} \in \{0,1\} \ \ j=1.2,...,M , i=1,...,N\\
&\; Z_{i} \in \{0,1\} \ \ i=1,...,N\\
&\; Y_{ik} \in \{0,1\} \ \ K=1.2,...,k , i=1,...,N\\
\end{split}
\label{OP9}
\end{equation}
where i,j,k is set, i=1,..,32 Represents the number of classes, j=1,..,512 Represents the number of student, k=1,..,40 Represents the type of student. $t_k$ is subset of $j$. Which indicates student jth belongs to which type. for example , if we consider type k=1, then all student with type ′1′ are t1={2,10,78,69,114,500}.
$c_{ij} ,b_{k,l} ,a_i,b_i,p_i,q_i $ are parameter.
i want to use below method , To merge these constraints :
$a_i Z_i \le \sum_{j=1}^M X_{ij} \le b_i Z_i ,\ \ i=1.2,...,N$
or this
$p_i Z_i \le \sum_{k=1}^K Y_{i,k} \le q_i Z_i ,\ \ i=1.2,...,N$ .
How can I use below method for more than two constraints?
****method for merge****
i know merging constraint for equality as below : suppose we have two integer constrain in bounded variable
\begin{equation} \begin{split} \max &\; p_j x_j\\ \text{s.t.} & \; \sum_{j=1}^n w_{1j} x_j =c_1\\ & \;\sum_{j=1}^n w_{2j} x_j =c_2\\ & \; 0\le x_j\le u_j , &\; x_j , \text integer , j=1.2,...,n \end{split} \label{OP} \end{equation}
let
\begin{equation} \begin{split} \\ g(x)=c_1-\sum_{j=1}^n w_{1j} x_j\\ & \; \ \\ h(x) = c_2-\sum_{j=1}^n w_{2j} x_j \\ & \; \end{split} \label{OP1} \end{equation} by using bounds on $x_j$ we derive the following bound on $g(x)$ :
\begin{equation} \begin{split} \\ c_1-\sum_{j=1}^n w_{1j}^+ x_j \le g(x) \le c_1-\sum_{j=1}^n w_{1j}^- x_j\\ & \; \end{split} \label{OP3} \end{equation}
where $w_{1j}^+ = \max \{0,w_{ij}\}$ and $w_{1j}^- = \min \{0,w_{ij}\}$
if we choose a positive integer $\lambda$ satiating :
$\lambda > max \{ c_1-\sum_{j=1}^n w_{1j}^- u_{j}, c_1-\sum_{j=1}^n w_{1j}^+ u_{j} \}$
then we have $|g(x)| < \lambda$
now multiply the constrain $\sum_{j=1}^n w_{2j} x_j =c_2$ by $\lambda$ and it to the first constrain $\sum_{j=1}^n w_{1j} x_j =c_1$ boating the following :
\begin{equation} \begin{split} \max &\; p_j x_j\\ \text{s.t.} & \; \sum_{j=1}^n (w_{1j} + \lambda w_{2j} )x_j =c_1 +\lambda c_2\\ \\ & \; 0\le x_j\le u_j , &\; x_j , \text integer , j=1.2,...,n \end{split} \label{OP5} \end{equation}
It's proven that if the $\lambda$ is chosen like the above, the two problem are same.
Take $p_i Z_i \leq \sum_{k=1}^K Y_{i,k} \leq q_i Z_i$, which is equivalent to: $$\sum_{k=1}^K Y_{i,k} - s_{i} - p_i Z_i = 0, \quad \sum_{k=1}^K Y_{i,k} + t_{i} - q_i Z_i = 0$$ The slack variables $s_i$ and $t_i$ are integer and range between 0 and $K$. Let me focus on the first constraint. The absolute value of the left hand side is bounded by $2K$, so you can use any $\lambda_2>2K$ in your method to merge the constraints for $i=1$ and $i=2$: $$ \begin{align} \left( \sum_{k=1}^K Y_{1,k} - s_{1} - p_1 Z_1 \right) + \lambda_2 \left( \sum_{k=1}^K Y_{2,k} - s_{2} - p_2 Z_2 \right) = 0. \end{align}$$ The left hand side is bounded in absolute values by $(1+\lambda_2)2K$, so you can use your method to combine the constraint above with the constraint for $i=3$ using any $\lambda_3 > (1+\lambda_2)2K$: $$ \begin{align} & \left( \sum_{k=1}^K Y_{1,k} - s_{1} - p_1 Z_1 \right) + \lambda_2 \left( \sum_{k=1}^K Y_{2,k} - s_{2} - p_2 Z_2 \right) + \lambda_3 \left( \sum_{k=1}^K Y_{3,k} - s_{3} - p_3 Z_3 \right)= 0. \end{align}$$ Now take $\lambda_4 > (1+\lambda_2+\lambda_3)2K$ to continue the pattern.