## Appendix A
Proof of Theorem 5.: From Equations (18) and (26), it follows  
\[\left(\widetilde{h}_{ij}^{(t+1)}\right)^{\sigma(s)} = \prod_{k=1}^{n}\left(\left(h_{ik}^{(t+1)}\right)^{\sigma(s)} \times\left(h_{jk}^{(t+1)}\right)^{\sigma(s)}\right)^{\frac{1}{n}}\] \[= \prod_{k=1}^{n}\left(\left(h_{ik}^{(t)}\right)^{\sigma(s)} \times\left(\widetilde{h}_{ik}^{(t)}\right)^{\sigma(s)}\right)^{(1-\alpha)} \times\left(\left(h_{jk}^{(t)}\right)^{\sigma(s)}\right)^{\alpha}\times\left( \left(\widetilde{h}_{jk}^{(t)}\right)^{\sigma(s)}\right)^{(1-\alpha)}\right)^ {\frac{1}{n}}\] \[= \left(\widetilde{h}_{ij}^{(t)}\right)^{\sigma(s)}\]  
Thus,  
\[\overline{h}_{ij,1}^{\sigma(s)}\times\overline{h}_{ji,2}^{\sigma(s)}+\overline{h} _{ji,1}^{\sigma(s)}\times\overline{h}_{ij,2}^{\sigma(s)}\geq 2\sqrt{ \overline{h}_{ij,1}^{\sigma(s)}\times\overline{h}_{ji,2}^{\sigma(s)}\times \overline{h}_{ji,1}^{\sigma(s)}\times\overline{h}_{ij,2}^{\sigma(s)}}=2,\]0 \[\overline{h}_{ij,1}^{\sigma(s)}\times\overline{h}_{ji,2}^{\sigma(s)}+\overline{h} _{ji,1}^{\sigma(s)}\times\overline{h}_{ij,2}^{\sigma(s)}\geq 2\sqrt{ \overline{h}_{ij,1}^{\sigma(s)}\times\overline{h}_{ji,2}^{\sigma(s)}\times \overline{h}_{ji,1}^{\sigma(s)}\times\overline{h}_{ij,2}^{\sigma(s)}}=2,\]1So,  
\[\overline{h}_{ij,1}^{\sigma(s)}\times\overline{h}_{ji,2}^{\sigma(s)}+\overline{h} _{ji,1}^{\sigma(s)}\times\overline{h}_{ij,2}^{\sigma(s)}\geq 2\sqrt{ \overline{h}_{ij,1}^{\sigma(s)}\times\overline{h}_{ji,2}^{\sigma(s)}\times \overline{h}_{ji,1}^{\sigma(s)}\times\overline{h}_{ij,2}^{\sigma(s)}}=2,\]2 \[\overline{h}_{ij,1}^{\sigma(s)}\times\overline{h}_{ji,2}^{\sigma(s)}+\overline{h} _{ji,1}^{\sigma(s)}\times\overline{h}_{ij,2}^{\sigma(s)}\geq 2\sqrt{ \overline{h}_{ij,1}^{\sigma(s)}\times\overline{h}_{ji,2}^{\sigma(s)}\times \overline{h}_{ji,1}^{\sigma(s)}\times\overline{h}_{ij,2}^{\sigma(s)}}=2,\]3 \[\overline{h}_{ij,1}^{\sigma(s)}\times\overline{h}_{ji,2}^{\sigma(s)}+\overline{h} _{ji,1}^{\sigma(s)}\times\overline{h}_{ij,2}^{\sigma(s)}\geq 2\sqrt{ \overline{h}_{ij,1}^{\sigma(s)}\times\overline{h}_{ji,2}^{\sigma(s)}\times \overline{h}_{ji,1}^{\sigma(s)}\times\overline{h}_{ij,2}^{\sigma(s)}}=2,\]4  
here by Lemma 1 equality holds only when \(\left(h_{ij}^{(t)}\right)^{\sigma(s)}\times\left(\widetilde{h}_{ji}^{(t)} \right)^{\sigma(s)}=1\), but that lead to \(H^{(t)}=\widetilde{H}^{(t)}\), which is not true, hence  
\[\overline{h}_{ij,1}^{\sigma(s)}\times\overline{h}_{ji,2}^{\sigma(s)}+\overline{h} _{ji,1}^{\sigma(s)}\times\overline{h}_{ij,2}^{\sigma(s)}\geq 2\sqrt{ \overline{h}_{ij,1}^{\sigma(s)}\times\overline{h}_{ji,2}^{\sigma(s)}\times \overline{h}_{ji,1}^{\sigma(s)}\times\overline{h}_{ij,2}^{\sigma(s)}}=2,\]5  
and  
\[\sum_{i=1}^{n-1}\sum_{j=1+1}^{n}\sum_{s=1}^{d}\left( \begin{array}{c}\left(h _{ij}^{(t+1)}\right)^{\sigma(s)}\times\left(\widetilde{h}_{ji}^{(t+1)}\right) ^{\sigma(s)}+\\ \left(h_{ij}^{(t+1)}\right)^{\sigma(s)}\times\left(\widetilde{h}_{ji}^{(t+1)} \right)^{\sigma(s)}\end{array} \right)<\sum_{i=1}^{n-1}\sum_{j=1+1}^{n}\sum_{ s=1}^{d}\left( \begin{array}{c}\left(h_{ij}^{(t)}\right)^{\sigma(s)}\times \left(\widetilde{h}_{ji}^{(t)}\right)^{\sigma(s)}+\\ \left(h_{ij}^{(t)}\right)^{\sigma(s)}\times\left(\widetilde{h}_{ji}^{(t)} \right)^{\sigma(s)}\end{array} \right). \tag{10}\]  
By Equation (26)  
\[\left(p_{ij}^{(t)}\right)^{\sigma(s)}<\left(p_{ij}^{(t+1)}\right)^{\sigma(s)}< \left(\widetilde{p}_{ij}^{(t)}\right)^{\sigma(s)},\]  
with each iteration \(\left(p_{ij}^{(t)}\right)^{\sigma(s)}\) will come closer to \(\left(\widetilde{p}_{ij}^{(t)}\right)^{\sigma(s)}\) depending on the value of \(\alpha\). Table 2 shows the smaller the \(\alpha\) is faster the value of \(\left(p_{ij}^{(t)}\right)^{\sigma(s)}\) coming closer to \(\left(\widetilde{p}_{ij}^{(t)}\right)^{\sigma(s)}\).