## 4 Illustrative Examples
\begin{array}{cccccccc}1&1/3&*&1&1/4&2&*\\ 3&1&1/2&*&*&3&3\\ *&2&1&4&5&6&5\\ 1&*&1/4&1&1/4&1&2\\ 4&*&1/5&4&1&*&1\\ 1/2&1/3&1/6&1&*&1&*\\ *&1/3&1/5&1/2&1&*&1\end{array} \right)\]  
Figure 1: The contour plot of the objective function and the first 18 iterations of the algorithm for minimization (11): (**a**) contour plot on [1/9,9]; (**b**) function value vs. iteration. _where \(*\) represents the missing comparisons. First, replacing the missing comparisons by six variables \(x_{i}\,(i=1,2,\ldots,6)\), we can rewrite the revised matrix with variables as \(\mathbf{A}(\mathbf{x})\):_  
\[\mathbf{A}(\mathbf{x})= \begin{pmatrix}1&1/3&x_{1}&1&1/4&2&x_{5}\\ 3&1&1/2&x_{2}&x_{3}&3&3\\ 1/x_{1}&2&1&4&5&6&5\\ 1&1/x_{2}&1/4&1&1/4&1&2\\ 4&1/x_{3}&1/5&4&1&x_{4}&1\\ 1/2&1/3&1/6&1&1/x_{4}&1&x_{6}\\ 1/x_{5}&1/3&1/5&1/2&1&1/x_{6}&1\end{pmatrix} \]  
_where \(\mathbf{x}=(x_{1},x_{2},x_{3},x_{4},x_{5},x_{6})\)._  
Now, in order to obtain the completion by respecting the restriction \([1/9,9]\), let us first formulate a constrained eigenvalue minimization as  
\[ \begin{array}{ll}\min&\lambda_{max}(\mathbf{A}(\mathbf{x}))\\ \mbox{s.t.}&1/9\leq x_{i}\leq 9\;\;for\;\;i=1,2,3,4,5,6.\end{array}  \tag{13}\]  
Then, solving minimization (13) with the initial value \(\mathbf{x}_{0}=(1,1,1,1,1)\), using the proposed algorithm, the optimal solution is:  
\[\mathbf{x}^{*}=(0.1618,2.7207,1.3465,2.5804,0.8960,0.7731),\]  
\[\lambda_{max}(\mathbf{A}(\mathbf{x}^{*}))=7.4067.\]  
\(CR=0.0504\). Hence, the reconstructed matrix \(\mathbf{A}\) is acceptable, referring to [PERSON]'s 0.1 cut-off rule. In addition, the first 44 iterations and values of the variables for each iteration are provided in Table 3 and in Figure 2, respectively. The first graph in Figure 2 shows the evolution of the variables with respect to the number of iterations. Moreover, the convergence of the variables is not monotone. This is mainly due to the _contraction_ step. For example, when we look at the value of \(x_{2}\) at iterations 11, 18, and 27, it fluctuates. Furthermore, the value of \(\lambda_{max}\) in Table 3 at iterations 2 and 6 drops significantly due to the _expansion_ step. In this example, we used the values for termination: \(\mathtt{TolX}=10^{-4}\), \(\mathtt{TolFun}=10^{-4}\), \(\mathtt{MaxIter}=44\) and \(\mathtt{MaxFunEvals}=77\). There is no significant difference in the \(\lambda_{max}\) values after the 44 th iteration. Using the _method of cyclic coordinates_ in [15], the optimal solution to problem (13) is \(\mathbf{x}^{*}=(0.1618,2.7206,1.3464,2.5804,0.8960,0.7731)\). Similarly, using the _least square method (LSM)_ in [20], the solution is \(\mathbf{x}^{*}=(0.1651,2.7098,1.3498,2.5638,0.9103,0.7851)\). Both approaches provide the same \(CR\) value in this problem, i.e., \(CR=0.0504\). \begin{table}
\begin{tabular}{c c c c c c} \hline \hline
**Iter** & **Min \(\lambda_{max}\)** & **Procedure** & **Iter** & **Min \(\lambda_{max}\)** & **Procedure** \\ \hline
0 & 8.00047 & - & 23 & 7.41505 & contract inside \\
1 & 7.90912 & initial simplex & 24 & 7.4145 & reflect \\
2 & 7.54525 & expand & 25 & 7.4145 & contract inside \\
3 & 7.54525 & reflect & 26 & 7.4145 & reflect \\
4 & 7.54525 & reflect & 27 & 7.4129 & contract outside \\
5 & 7.54525 & reflect & 28 & 7.41214 & contract outside \\
6 & 7.45024 & expand & 29 & 7.4112 & contract inside \\
7 & 7.45024 & reflect & 30 & 7.40973 & reflect \\ \hline \hline \end{tabular}
\end{table} 
Table 3: Iteration number, incumbent optimal value of \(\lambda_{max}\), and simplex procedure for minimization (13).