## 3 Custom Interior-Point Method for Solving FLPRC
### Complexity Analysis  
In this section, we analyze the calculation complexity. The complexity is measured by the number of floating operations (flops), and all of the calculation in this section is based on [PERSON] (2007). Let \(n_{v}=\text{dim}(v)=4N+N(N-1)\) denote the dimension of \(v\). The size of the coefficient matrix of the original Equation (36) is \((n_{v}+n_{\lambda})\)Using the standard LU decomposition, the computational complexity of the Equation (36) is \((2/3)(n_{v}+n_{\lambda})^{3}\). Expanding this, the amount of calculation is expressed as follows:  
\[\begin{array}{rcl}(2/3)(n_{v}+n_{\lambda})^{3}&=&\frac{343}{8}N^{6}+(\frac{2793 }{8}+\frac{147}{4}M)N^{5}+(\frac{7581}{8}+\frac{399}{2}M+\frac{21}{2}M^{2})N^ {4}\\ &&+(\frac{6859}{3}+\frac{1083}{4}M+\frac{57}{2}M^{2}+M^{3})N^{3}\end{array} \tag{48}\]  
On the other hand, the calculation effort after 1 st elimination is performed is \(2n_{v}^{2}n_{\lambda}+(2/3)n_{v}^{3}\). In this equation, the complexity is proportional to \(n_{\lambda}\). Since \(n_{\lambda}>n_{v}\), the amount of calculation is greatly reduced. The amount of calculation after 1 st elimination is performed is as follows:  
\[\begin{array}{rcl}2n_{v}^{2}n_{\lambda}+(2/3)n_{v}^{3}&=&\frac{17}{3}N^{6}+(4 7+2M)N^{5}+(129+12M)N^{4}+(117+18M)N^{3}\end{array} \tag{49}\]  
The calculation effort in the second term can be further reduced by performing 2 nd elimination. Let \(n_{r}=\dim(r)=4N\) be the dimensions of \(r\), and let \(n_{d}=\dim(d)=N(N-1)\) denote the dimensions of \(d\). The calculation effort after 1 st elimination is performed is \((2/3)n_{v}^{3}=(2/3)(n_{r}+n_{d})^{3}\). The calculation effort after 2 nd elimination is performed is as shown in \(2n_{r}^{2}n_{d}+(2/3)n_{r}^{3}\). In this equation, the complexity is proportional to \(n_{d}\). Since \(n_{d}>n_{r}\) in the range of \(N>4\), the amount of calculation is greatly reduced. The amount of calculation after 2 nd elimination that is performed is as follows:  
\[\begin{array}{rcl}2n_{v}^{2}n_{\lambda}+2n_{r}^{2}n_{d}+(2/3)n_{r}^{3}&=& \frac{4}{5}N^{6}+(\frac{79}{5}+2M)N^{5}+(\frac{526}{5}+12M)N^{4}+(\frac{329}{ 3}+18M)N^{3}\end{array} \tag{50}\]  
The calculation effort in the second term can be further reduced by performing 3 rd elimination. Let \(n_{xy}=\dim(x)+\dim(y)=2N\) denote the dimensions of \(x\) and \(y\), and \(n_{vh}=\dim(w)+\dim(h)=2N\) denote the dimensions of \(w\) and \(h\). The calculation effort after 2 nd elimination is performed is as shown in \((2/3)n_{r}^{3}=(2/3)(n_{xy}+n_{vh})^{3}\). The calculation effort after 3 rd elimination is performed is \(n_{vh}+2n_{xy}^{3}n_{vh}+(2/3)n_{xy}^{3}\). Here, the first term is the amount of calculation when sparse factorization is used by utilizing the fact that the matrix \(E_{22}\) of Equation (47) is a symmetric and sparse matrix. The third term is a block diagonal matrix. Therefore, it can be decomposed into equations related to \(x\) and \(y\) and the calculation effort is \((2/3)N^{3}\times 2=(4/3)N^{3}\). The amount of calculation after 3 rd elimination is performed is as follows:  
\[\begin{array}{rcl}2n_{v}^{2}n_{\lambda}+2n_{r}^{2}n_{d}+n_{vh}+2n_{xy}^{2}n_ {vh}+(4/3)N^{3}\\ =&\frac{4}{5}N^{6}+(\frac{79}{5}+2M)N^{5}+(\frac{526}{5}+12M)N^{4}+(\frac{52}{ 3}+18M)N^{3}+2N\end{array} \tag{51}\]  
These complexity is summarized in Table 1. From the above analysis, the problem can be reduced to solving smaller equations by decomposing the matrix rather than solving the original matrix.  
\begin{table}
\begin{tabular}{l c c} \hline \hline
**Equation** & \multicolumn{2}{c}{**Complexity (\# of Flops)**} \\ \hline Original Newton Equation (36) & \((2/3)(n_{v}+n_{\lambda})^{3}\) & \(\frac{343}{8}N^{6}+(\frac{2793}{5}+\frac{147}{2}M)N^{5}\) \\  & & \(+(\frac{7581}{8}+\frac{79}{27}M+\frac{1}{2}M^{2})N^{4}\) \\  & & \(+(\frac{689}{3}+\frac{1683}{4}M+\frac{57}{2}M^{2}+M^{3})N^{