## 5 Numerical Example
### Example 2  
Consider the benchmark model HE3 in [42] as a more large-scale example. Discretizing this model, we can obtain \(A\), \(B\), and \(C\) in (9) as follows:  
\[A= \begin{bmatrix}0.99&0.00&-0.01&-0.00&-0.03&-0.00&-0.98&-0.00\\ -0.01&0.94&0.03&-0.00&-0.02&0.05&0.00&-0.00\\ 0.00&-0.00&0.97&0.00&0.02&0.00&-0.00&0.00\\ 0.00&-0.00&-0.03&0.99&0.00&0.02&-0.00&0.97\\ 0.00&-0.00&-0.07&-0.00&0.92&0.01&-0.00&-0.00\\ -0.00&-0.00&0.01&0.00&-0.02&0.92&0.00&0.00\\ 0.00&-0.00&0.09&0.00&0.00&1.00&0.00\\ 0.00&-0.00&-0.00&-0.00&0.09&0.00&-0.00&0.99\end{bmatrix} ,\]  
\[B= \begin{bmatrix}0.00&0.01&-0.00&-0.00\\ -0.10&0.01&0.00&-0.00\\ -0.00&-0.00&0.00&-0.00\\ -0.00&0.00&0.01&0.01\\ -0.00&0.00&0.02&0.01\\ 0.01&0.00&0.00&-0.03\\ -0.00&-0.00&0.00&-0.00\\ -0.00&0.00&0.00&0.00\end{bmatrix} ,\]  
\[C= \begin{bmatrix}I_{6}&0_{6,2}\end{bmatrix} .\]  
The parameter \(\beta\) in (13) is given by \(\beta=0.15\). We suppose that the structure of a SAN is given by Figure 1. Then, as explained in Section 3.3, \(q\) is given by \(q=3\). We also set the following: \(\mathcal{I}_{1}^{a}=\{1,2\}\), \(\mathcal{I}_{1}^{s}=\{1,2\}\), \(\mathcal{I}_{2}^{a}=\{3\}\), \(\mathcal{I}_{2}^{s}=\{3,4\}\), \(\mathcal{I}_{3}^{a}=\{4\}\), and \(\mathcal{I}_{3}^{s}=\{5,6\}\). First, consider solving Problem 1. Then, we can obtain  
\[A_{c}= \begin{bmatrix}0.56&-0.07&-0.05&-0.03&-0.16&-0.15&-0.87&-0.20\\ 0.05&0.88&0.18&0.00&-0.35&0.13&-1.03&-0.88\\ 0.04&0.01&0.81&0.01&0.05&0.02&-0.31&0.08\\ -0.05&0.14&-0.29&0.62&0.27&0.13&-0.25&1.44\\ -0.00&0.00&-0.09&-0.03&0.73&-0.00&-0.21&-0.49\\ 0.05&-0.02&0.13&0.01&0.04&0.88&0.06&0.06\\ 0.07&0.01&0.16&0.01&0.02&0.02&1.10&0.03\\ -0.02&0.02&-0.06&-0.05&0.15&0.01&-0.06&1.11\end{bmatrix} ,\]  
\[B_{c}= \begin{bmatrix}-0.49&-0.03&0.00&-0.00&-0.00&0.00\\ 0.00&-0.07&-0.00&0.04&0.00&-0.00\\ 0.03&0.00&-0.00&-0.00&0.00&-0.00\\ -0.03&0.12&0.00&-0.40&-0.00&0.00\\ -0.02&0.00&0.00&-0.00&0.00&0.00\\ 0.03&-0.00&-0.00&0.01&0.00&-0.00\\ 0.09&0.01&-0.00&0.00&0.00&-0.00\\ -0.01&0.02&0.00&-0.06&-0.00&0.00\end{bmatrix} ,\]  
\[C_{c}= \begin{bmatrix}0.87&0.06&-2.42&-0.11&-1.60&0.73&-15.10&-4.52\\ 1.83&0.27&-28.12&0.86&2.89&1.15&-50.71&11.44\\ -1.38&0.05&0.60&1.33&7.75&1.05&13.55&20.94\\ 0.90&-0.31&0.82&-0.05&1.82&-0.96&-4.71&2.28\end{bmatrix} .\]  
From this result, we see that \(B_{c}\) is column-sparse but \(C_{c}\) is not row-sparse. Also in this example, it is theoretically guaranteed that the obtained controller can stabilize the closed-loop system. In implementing the controller, the third, fifth, and sixth columns of \(B_{c}\) are replaced with a zero vector. Then, by a numerical simulation from time 0 to 80, we can confirm that (13) with \(\beta=0.15\) holds. In this case, the effect of the truncation error issufficiently small. However, since \(C_{c}\) is not row-sparse, all communication channels must be used. Next, consider solving Problem 2. We suppose that using the channel \(1\) is avoided as much as possible. Based on this purpose, we set \(w_{(1)}=100,000\) and \(w_{(2)}=w_{(3)}=1\). Then, we can obtain  
\[A_{c}= \begin{bmatrix}1.05&0.01&-22.22&-0.24&-37.37&12.90&-2.78&0.34\\ 0.00&0.94&1.43&0.19&0.16&-0.72&-0.59&0.08\\ 0.01&-0.00&0.62&-0.02&-0.07&0.09&-0.17&0.00\\ -0.08&-0.03&5.69&0.69&0.46&-0.26&4.85&-0.40\\ 0.15&-0.05&3.26&-0.07&0.56&0.05&0.90&-1.49\\ 0.67&0.00&-12.57&0.18&0.25&0.80&-18.62&2.53\\ -0.00&-0.00&0.96&0.01&2.10&-0.62&1.13&-0.02\\ -0.01&-0.00&0.38&-0.03&0.19&-0.06&0.04&0.93\\ \end{bmatrix} ,\]  
\[B_{c}= \begin{bmatrix}-0.00&-0.00&-20.96&-0.27&-37.32&12.91\\ -0.00&-0.00&1.78&0.18&0.19&-0.77\\ -0.00&0.00&-0.28&-0.02&-0.08&0.09\\ 0.00&0.00&1.03&-0.23&0.51&-0.30\\ -0.00&-0.00&0.02&-0.00&-0.15&0.04\\ 0.00&0.00&-0.04&-0.04&0.49&-0.08\\ 0.00&0.00&0.77&0.01&2.10&-0.62\\ 0.00&0.00&0.22&-0.02&0.11&-0.06\\ \end{bmatrix} ,\]  
\[C_{c}= \begin{bmatrix}0.00&-0.00&-0.00&0.00&-0.00&-0.00&-0.00&0.00\\ 0.00&0.00&-0.00&0.00&-0.00&-0.00&0.00\\ -16.93&2.17&42.85&-0.28&13.03&0.78&243.70&27.99\\ 15.84&0.30&-316.42&5.19&-4.65&-0.99&-456.53&66.43\\ \end{bmatrix}