# Optimization-Based Approaches to Control of Probabilistic Boolean Networks
## 4 Integer Programming Approach
### Solution Method for Problem 2-A
Then, the system (8) can be derived as_  
\[x_{1}(k+1) = \delta_{1,1}(k)\Big{\{}x_{3}(k)(1-u(k))\Big{\}}+\delta_{1,2}(k) \Big{\{}(1-x_{3}(k))(1-u(k))\Big{\}},\] \[x_{2}(k+1) = \delta_{2,1}(k)\Big{\{}x_{1}(k)(1-x_{3}(k))\Big{\}},\] \[x_{3}(k+1) = \delta_{3,1}(k)\Big{\{}x_{1}(k)(1-x_{2}(k))\Big{\}}+\delta_{3,2}(k )\Big{\{}x_{2}(k)+u(k)-x_{2}(k)u(k)\Big{\}}.\]  
_The equality constraint (9) can be derived as_  
\[\sum_{j=1}^{2}\delta_{1,j}(k)=1,\ \ \delta_{2,1}(k)=1,\ \ \sum_{j=1}^{2}\delta_{3,j}(k)=1.\]  
_The state equation in the system (10) can be derived as_  
\[x_{1}(k+1) = z_{1,1}(k)-z_{1,2}(k)+\delta_{1,2}(k)-z_{1,3}(k)-z_{1,4}(k)+z_{1, 5}(k),\] \[x_{2}(k+1) = x_{1}(k)-z_{2,1}(k),\] \[x_{3}(k+1) = z_{3,1}(k)-z_{3,2}(k)+z_{3,3}(k)+z_{3,4}(k)-z_{3,5}(k),\]  
_where_  
\[z_{1,1} = \delta_{1,1}x_{3},\ \ z_{1,2}=\delta_{1,1}x_{3}u,\ \ z_{1,3}=\delta_{1,2}u,\ \ z_{1,A}=\delta_{1,2}x_{3},\ \ z_{1,5}=\delta_{1,2}x_{3}u,\] \[z_{2,1} = x_{1}x_{3},\] \[z_{3,1} = \delta_{3,1}x_{1},\ \ z_{3,2}=\delta_{3,1}x_{1}x_{2},\ \ z_{3,3}=\delta_{3,2}x_{2},\ \ z_{3,A}=\delta_{3,2}u,\ \ z_{3,5}=\delta_{3,2}x_{2}u.\]  
_Using Lemma 2, these relations can be equivalently transformed into a set of linear inequalities. Then, \(p=11\) holds. Thus, the linear inequality in the system (10) can be derived._  
Using the system (10), consider rewriting Problem 2-C as an ILP problem. First, equality constraint (9) is embedded in the inequality constraint in the system (10). From the state equation in the system (10), we can obtain  
\[x_{2}(k+1) = \delta_{2,1}(k)\Big{\{}x_{1}(k)(1-x_{3}(k))\Big{\}},\]0  
Using this expression, we can obtain  
\[x_{2}(k+1) = \delta_{2,1}(k)\Big{\{}x_{1}(k)(1-x_{3}(k))\Big{\}},\]1  
where \(\bar{x}:=[\ x^{\top}(0)\ \ x^{\top}(1)\ \ \cdots\ \ x^{\top}(N)\ ]^{\top}\), \(\bar{v}:=[\ v^{\top}(0)\ \ v^{\top}(1)\ \ \cdots\ \ v^{\top}(N-1)\ ]^{\top}\) and  
\[\bar{A}=\left[ \begin{array}{c}I\\ A\\ A^{2}\\ \vdots\\ A^{N}\end{array} \right],\ \ \bar{B}=\left[ \begin{array}{cccc}0&0&\cdots&0\\ B&0&\cdots&0\\ AB&\ddots&\ddots&\vdots\\ \vdots&\ddots&\ddots&0\\ A^{N-1}B&\cdots&AB&B\end{array} \right].\]  
From the linear inequality in the system (10), we can obtain  
\[\bar{C}\bar{x}+\bar{D}\bar{v}\leq\bar{E}, \tag{12}\]where  
\[\bar{C}= \begin{bmatrix}C&&0\\ &\ddots&&\vdots\\ &&C&0\end{bmatrix} ,\;\;\bar{D}= \begin{bmatrix}D&&\\ &\ddots&\\ &&D\end{bmatrix} ,\;\;\bar{E}= \begin{bmatrix}E\\ \vdots\\ E\end{bmatrix} .\]  
Next, consider the linear inequality \(\sum_{k=0}^{N-1}L\delta(k)\geq\ln\rho\) in Problem C. From \(\delta(k)=Wv(k)\), \(W:=[ \begin{array}{cccc}0&I_{l}&0&I_{l\times m}\end{array} ]\), this inequality can be rewritten as  
\[-L\tilde{W}\bar{v}\leq-\ln\rho,\]  
where \(\bar{L}=[ \begin{array}{cccc}L&\cdots&L\end{array} ]\) and  
\[\tilde{W}= \begin{bmatrix}W&&\\ &\ddots&\\ &&W\end{bmatrix} .\]  
Cost function (6) can be rewritten as  
\[J=\bar{Q}\bar{x}+\bar{R}\bar{v}, \tag{13}\]  
where \(\bar{Q}:=[ \begin{array}{cccc}Q&\cdots&Q&Q_{f}\end{array} ]\) and \(\bar{R}:=[ \begin{array}{cccc}R&\cdots&R\end{array} ]\). By substituting the state Equation (11) into the inequality (12) and the cost function (13), we can obtain the ILP problem. In [23], this result was obtained as the following theorem. **Theorem 3**: _Problem 2-C is equivalent to the following ILP problem._  
**Problem 2-D:**  
\[ \begin{array}{ll}\text{find}&\bar{v}\in\{0,1\}^{(m+l+p)N}\\ \text{min}&(\bar{R}+\bar{Q}\bar{B})\bar{v}+\bar{Q}\bar{A}x_{0}\\ \text{subject to}&\left[\begin{array}{c}\bar{C}\bar{B}+\bar{C}\\ -L\tilde{W}\end{array}\right]\bar{v}\leq\left[\begin{array}{c}\bar{E}-\bar{ C}\bar{A}x_{0}\\ -\ln\rho\end{array}\right].\end{array} \]  
Problem 2-D can be solved by using a suitable free/commercial solver. Finally, we explain a solution method for Problem 2-B. In Problem 2-B, we assume that \(u(k)\) (i.e., certain elements of \(\bar{v}\)) is given. Under this assumption, Problem 2-B can be transformed into an ILP problem by replacing \"min\" with \"max\" in Problem 2-D.