## 4 Computational Efficiency of the Algorithms
\begin{table}
\begin{tabular}{l l l} \hline \hline
**Method** & **Computational Cost (\(C_{method}\))** & **Computational Efficiency** \\ \hline \(2^{nd}NM\) & \(\frac{1}{3}n^{3}+2n^{2}+\frac{2}{3}n\) & \(2^{\frac{1}{5^{th}}}\)\({}^{4^{th}NR}\) \\ \(4^{th}NR\) & \(\frac{2}{3}n^{3}+4n^{2}+\frac{4}{3}n\) & \(4^{\frac{1}{5^{th}}}\)\({}^{4^{th}NR}\) \\ \(5^{th}ACT\) & \(\frac{2}{3}n^{3}+5n^{2}+\frac{7}{3}n\) & \(5^{\frac{1}{5^{th}}}\)\({}^{4^{th}MBJ}\) \\ \(5^{th}SD\) & \(\frac{1}{3}n^{3}+8n^{2}+\frac{8}{3}n\) & \(5^{\frac{1}{5^{th}}}\)\({}^{4^{th}SD}\) \\ \(5^{th}PJ\) & \(\frac{2}{3}n^{3}+8n^{2}+\frac{7}{3}n\) & \(5^{\frac{1}{5^{th}}}\)\({}^{4^{th}SA}\) \\ \(8^{th}SA\) & \(\frac{1}{3}n^{3}+10n^{2}+\frac{23}{3}n\) & \(8^{\frac{1}{5^{th}}}\)\({}^{4^{th}MBJ}\) \\ \(8^{th}MBJ\) & \(\frac{1}{3}n^{3}+9n^{2}+\frac{17}{3}n\) & \(8^{\frac{1}{5^{th}}}\)\({}^{4^{th}SD}\) \\ \(8^{th}SD\) & \(\frac{1}{3}n^{3}+15n^{2}+\frac{1}{3}n\) & \(8^{\frac{1}{5^{th}}}\)\({}^{4^{th}PJ}\) \\ \(8^{th}PJ\) & \(\frac{2}{3}n^{3}+10n^{2}+\frac{19}{3}n\) & \(8^{\frac{1}{5^{th}}}\)\({}^{4^{th}PJ}\) \\ \hline \hline \end{tabular}
\end{table} 
Table 1: Table of Computational Cost and Computational Efficiency. It is clear that when \(R_{method1,method2}>1\), the iterative method \(1\) is more efficient than method \(2\). \(5^{th}pJ\) versus \(2^{nd}NM\): The ratio (23) is given by  
\[R_{5^{th}pJ;2^{nd}NM}=\frac{(\frac{1}{3}n^{3}+2n^{2}+\frac{2}{3}n)\ log(5)}{( \frac{2}{3}n^{3}+8n^{2}+\frac{2}{3}n)\ log(2)}.\]  
Based on the computation, we have \(R_{5^{th}pJ;2^{nd}NM}>1\) for \(n\geq 32\). Thus, we conclude that \(CE_{5^{th}pJ}>CE_{2^{nd}NM}\) for \(n\geq 32\). \(5^{th}pJ\) versus \(4^{th}NR\): In this case, the ratio (23) is given by  
\[R_{5^{th}pJ;4^{th}NR}=\frac{(\frac{2}{3}n^{3}+4n^{2}+\frac{4}{3}n)\ log(5)}{( \frac{2}{3}n^{3}+8n^{2}+\frac{2}{3}n)\ log(4)}.\]  
It has been verified that \(R_{5^{th}pJ;4^{th}NR}>1\) for \(n\geq 32\). Hence, we have \(CE_{5^{th}pJ}>CE_{4^{th}NR}\) for \(n\geq 32\). \(5^{th}pJ\) versus \(8^{th}SA\): Here the ratio (23) is given by  
\[R_{5^{th}pJ;8^{th}SA}=\frac{(\frac{1}{3}n^{3}+10n^{2}+\frac{23}{3}n)\ log(5)}{( \frac{2}{3}n^{3}+8n^{2}+\frac{2}{3}n)\ log(8)}.\]  
It can be checked that \(R_{5^{th}pJ;8^{th}SA}>1\) for \(n=2\) which implies that \(CE_{5^{th}pJ}>CE_{8^{th}SA}\) for \(n=2\). \(5^{th}pJ\) versus \(8^{th}SD\): Here the ratio (23) is given by  
\[R_{5^{th}pJ;8^{th}SD}=\frac{(\frac{1}{3}n^{3}+15n^{2}+\frac{17}{3}n)\ log(5)}{( \frac{2}{3}n^{3}+8n^{2}+\frac{2}{3}n)\ log(8)}.\]  
It has been verified that \(R_{5^{th}pJ;8^{th}SD}>1\) for \(2\leq n\leq 15\). Hence, we have \(CE_{5^{th}pJ}>CE_{8^{th}SD}\) for \(2\leq n\leq 15\). \(8^{th}pJ\) versus \(2^{nd}NM\): The ratio (23) is given by  
\[R_{8^{th}pJ;2^{nd}NM}=\frac{(\frac{1}{3}n^{3}+2n^{2}+\frac{2}{3}n)\ log(8)}{( \frac{2}{3}n^{3}+10n^{2}+\frac{13}{3}n)\ log(2)}.\]  
It can be checked that \(R_{8^{th}pJ;2^{nd}NM}>1\) for \(n\geq 13\). Thus, we conclude that \(CE_{8^{th}pJ}>CE_{2^{nd}NM}\) for \(n\geq 13\). \(8^{th}pJ\) versus \(4^{th}NR\): The ratio (23) is given by  
\[R_{8^{th}pJ;4^{th}NR}=\frac{(\frac{2}{3}n^{3}+4n^{2}+\frac{4}{3}n)\ log(8)}{( \frac{2}{3}n^{3}+10n^{2}+\frac{13}{3}n)\ log(4)}.\]  
It has been verified that \(R_{8^{th}pJ;4^{th}NR}>1\) for \(n\geq 13\). Hence, we conclude that \(CE_{8^{th}pJ}>CE_{4^{th}NR}\) for \(n\geq 13\). \(8^{th}pJ\) versus \(5^{th}ACT\): The ratio (23) is given by  
\[R_{8^{th}pJ;5^{th}ACT}=\frac{(\frac{2}{3}n^{3}+5n^{2}+\frac{7}{3}n)\ log(8)}{( \frac{2}{3}n^{3}+10n^{2}+\frac{13}{3}n)\ log(5)}.\]  
Based on the computation, we have \(R_{8^{th}pJ;5^{th}ACT}>1\) for \(n\geq 19\). Thus, we conclude that \(CE_{8^{th}pJ}>CE_{5^{th}ACT}\) for \(n\geq 19\). \(8^{th}pJ\) versus \(5^{th}pJ\): The ratio (23) is given by  
\[R_{8^{th}pJ;5^{th}pJ}=\frac{(\frac{2}{3}n^{3}+8n^{2}+\frac{7}{3}n)\ log(8)}{( \frac{2}{3}n^{3}+10n^{2}+\frac{13}{3}n)\ log(5)}.\]  
It can be checked that \(R_{8^{th}pJ;5^{th}pJ}>1\) for \(n\geq 2\), which implies that \(E_{8^{th}pJ}>E_{5^{th}pJ}\) for \(n\geq 2\).