## 3 Problem Formulation and Solutions
### Solution: Power Allocation (PA) and Power Splitting (PS)  
The sum rate can be maximized by exploiting the joint optimization of PA and PS. For addressing the above optimization dilemma, the problem definition (17) can be transformed and can be defined as  
\[\max_{p_{d,k},\ \rho_{d,k}}=\sum_{d=1}^{D}\sum_{k=1}^{|\Xi_{d}|}R_{d,k} \tag{20}\]  
s.t. (17b)-(17e)It is very much clear that (20) is a non-convex optimization problem and, considering the stated constraints, it is still a challenging task. \[ \begin{split}\bar{y}_{d,k}^{ID}=\left(\bar{\mathbf{h}}_{d,k}^{H} \bar{\mathbf{f}}_{d}^{BB}\sqrt{p_{d,k}}\,x_{d,k}+\bar{\mathbf{h}}_{d,k}^{H}\bar {\mathbf{f}}_{d}^{BB}\sum_{j=1}^{k-1}\sqrt{p_{d,j}}\,x_{d,j}+\right.\\ \left.\bar{\mathbf{h}}_{d,k}^{H}\sum_{i\
eq d}\sum_{j=1}^{|\Xi_{i }|}\mathbf{f}_{i}^{BB}\sqrt{p_{i,j}}\,x_{i,j}+n_{d,k}\right)+\frac{u_{d,k}}{ \sqrt{\rho_{d,k}}}\,,\end{split}  \tag{21}\]  
Following [3], the mean square error (MSE) related to the estimation of \(x_{d,k}\) can be defined as  
\[ \begin{split} e_{d,k}=1-2\mathbb{R}\Big{(}c_{d,k}\sqrt{\rho_{d,k} }\bar{\mathbf{h}}_{d,k}^{H}\bar{\mathbf{f}}_{d}^{BB}\Big{)}+\\ \left|c_{d,k}\right|^{2}\left(p_{d,k}\left\|\bar{\mathbf{h}}_{d, k}^{H}\bar{\mathbf{f}}_{d}^{BB}\right\|_{2}^{2}+\chi_{d,k}\right)\,,\end{split}  \tag{22}\]  
where \(c_{d,k}\) represents the equivalent channel estimation coefficient. It is proven that optimal \(c_{d,k}\) leads to the minimization of the \(e_{d,k}\). Based on [3], the optimized \(c_{d,k}\) can be articulated as  
\[c_{d,k}^{o}=\left(\sqrt{p_{d,k}}\bar{\mathbf{h}}_{d,k}^{H}\bar{\mathbf{f}}_{d} ^{BB}\right)^{H}\left(p_{d,k}\left\|\bar{\mathbf{h}}_{d,k}^{H}\bar{\mathbf{f}} _{d}^{BB}\right\|_{2}^{2}+\chi_{d,k}\right)^{-1}, \tag{23}\]  
and the minimum \(e_{d,k}\) can be obtained as  
\[e_{d,k}^{o}=1-\left(p_{d,k}\left\|\bar{\mathbf{h}}_{d,k}^{H}\bar{\mathbf{f}}_{ d}^{BB}\right\|_{2}^{2}\right)\left(p_{d,k}\left\|\bar{\mathbf{h}}_{d,k}^{H} \bar{\mathbf{f}}_{d}^{BB}\right\|_{2}^{2}+\chi_{d,k}\right)^{-1}, \tag{24}\]  
Using the Sherman-Morrison-Woodbury formula for the matrix inversion, we can have  
\[(1+\gamma_{d,k})^{-1}=\min_{c_{d,k}}e_{d,k} \tag{25}\]  
As in [3], the optimization dilemma/problem can be reformulated as  
\[\max_{p_{d,k},\rho_{d,k}}\sum_{d=1}^{D}\sum_{k=1}^{\Xi_{d}}\max_{a_{d,k}>0}\max _{c_{d,k}}\left(-\frac{a_{d,k}\,e_{d,k}}{ln2}+log_{2}\,a_{d,k}+\frac{1}{ln2}\right) \tag{26}\]  
s.t. (17b)-(17e)  
Here, \(a_{d,k}>0\) is introduced as a slack variable. The optimization problem, problem (26), can be iteratively resolved through exploiting the alternating optimization (AO) algorithm after dividing (26) into four sub-problems related to \((c_{d,k})\), \((a_{d,k})\), \((p_{d,k})\), \((\rho_{d,k})\). Particularly, for a given solutions related to \(p_{d,k}^{i-1}\) and \(\rho_{d,k}^{i-1}\) in the \((i-1)\)th iteration, the optimal solution at the \((i)\)th iteration for \(c_{d,k}^{i}\) leads to the solution of the optimization problem,  
\[\min_{c_{d,k}}e_{d,k} \tag{27}\]  
After achieving the optimal solution for \(c_{d,k}\), the next step is to solve the optimization problem as stated below:  
\[\max_{a_{d,k}>0}\left(-\frac{a_{d,k}\,e_{d,k}}{ln2}+log_{2}\,a_{d,k}+\frac{1}{ ln2}\right) \tag{28}\]This will lead to the generation of an optimal solution for \(a_{d,k}\). As in [3], the optimal solution \(a_{d,k}^{i}\) at the \(i\)th iteration can be articulated as  
\[a_{d,k}^{i}=\frac{1}{e_{d,k}^{\rho}} \tag{29}\]  
Following achieving the optimal solution for \(c_{d,k}\) and \(a_{d,k}\), the optimization dilemma/ problem can be reformulated as  
\[\max_{\rho,\lambda,k}\sum_{d=1}^{D}\sum_{k=1}^{\Xi_{d}}\left(-\frac{a_{d,k}\,e _{d,k}}{ln2}+log_{2}\,a_{d,k}+\frac{1}{ln2}\right) \tag{30}\]  
s.t. (17e)  
Now, to obtain to obtain the optimal \(\rho_{d,k}\), a new slack variable \(\tau_{d,k}\geq\frac{1}{\rho_{d,k}}\) is introduced, and the optimization dilemma/problem (30) can be equivalently represented as  
\[\max_{\rho_{d,k}}\sum_{d=1}^{D}\sum_{k=1}^{\Xi_{d}}\tau_{d,k} \tag{31}\]  
s.t.