## 3 Scenario and Problem Formulation
### Problem Formulation  
In the scenario shown in Figure 4, we have \(i\) users with \(i=\{1,2,3,\ldots,\mathbb{X}\}\) and we can find the received signal with Equation (1) [6].  
\[\mathbb{R}_{u}=\mathbb{H}_{2}\times\Theta\times\mathbb{H}_{1\mathbb{X}}+\eta_ {u} \tag{1}\]  
where \(\mathbb{H}_{1}\) denotes the channel matrix between BS and the RIS and \(\mathbb{H}_{1}\in\mathbb{C}^{\mathbb{K}_{x}\mathbb{N}}\), \(\mathbb{H}_{2}\in\mathbb{C}^{1\times\mathbb{K}}\) denotes channel matrix between RIS and the user \(u\). \(\Theta\) represents the actual phase shift,  
Figure 4: RIS-aided network scenario.  
Figure 3: ABC Ltd. Scenario.  
and \(\eta_{u}\) denotes the noise. \(\eta_{u}\) is generated by transmission. \(\chi\) represents the transmitted signal from the BS and can be evaluated by Equation (2) [6].  
\[\chi=\sum_{u=1}^{\mathbb{X}}\sqrt{p_{u}\times g_{u}\times s_{u}} \tag{2}\]  
where \(g_{u}\in\mathbb{C}^{*1}\) denotes precoding vector. \(s_{u}\) denotes the unit-power from the discrete constellation set and \(p_{u}\) denotes the unit-power from total power (\(P_{Tx}\)). The SINR (\(\gamma\)) can be calculated by Equation (3) [6].  
\[\gamma_{i}\triangleq\frac{p_{i}\times|h_{2,i}\times\Theta\times\mathbb{H}_{1 }\times g_{i}|^{2}}{\sum_{j=1,j\
eq i}^{\mathbb{K}}p_{i}\times|h_{2,i}\times \Theta\times\mathbb{H}_{1}g_{j}|^{2}+\sigma^{2}} \tag{3}\]  
Using Equation (3), we can calculate the throughput of the network with help of Shannon capacity formula as in Equation (4).  
\[Throughput\ (\Omega)=\sum_{i=1}^{\mathbb{X}}\log_{2}(1+\gamma_{i}) \tag{4}\]  
where \(\gamma_{i}\) is the SINR of \(i\)th users. The instantaneous power \(P_{i}\) can be calculated by (5).  
\[\mathbb{P}_{i}=\mathbb{P}_{UE}+\mathbb{P}_{BS}+\mathbb{P}_{RIS} \tag{5}\]  
Here \(\mathbb{P}_{UE}\) denotes static power of \(U_{E}\), \(\mathbb{P}_{BS}\) represents the static power of BS and \(\mathbb{P}_{RIS}\) shows total hardware static power absorbed by RIS. Equation (6) gives the power of RIS with \(\mathbb{K}\) reflecting element.  
\[P_{RIS}=\mathbb{P}(\theta)*\mathbb{K} \tag{6}\]  
where \(\mathbb{P}(\theta)\) is the dissipated power of phase-shifter. The total power \((P_{T})\) of the network can be calculated by Equation (7) [6].  
\[\mathbb{P}_{T}=\sum_{i=1}^{\mathbb{X}}\mathbb{P}_{UE}+\sum_{i=1}^{\mathbb{N}} \mathbb{P}_{BS}+\sum_{i=1}^{\mathbb{K}}\mathbb{P}_{RIS} \tag{7}\]  
The \(\mathbb{E}\mathbb{E}\) of the network can is the ratio of data rate and \(P_{T}\). Mathematically it can written by Equation (8a). This objective function is subject to constraints from (8b) to (8g).  
\[\mathbb{E}\mathbb{E}_{max}=\frac{\Omega}{P_{T}}\] (8a) subject to: \[P_{min}\leq P_{c}\] (8b) \[SINR_{min}\leq SINR_{c}\] (8c) \[0\leq P_{c}\leq p^{max}\] (8d) \[R^{c}\geq R^{min}=\sum_{c\in C}X_{c  
No power is allocated to non-connected \(U_{E}\). \(P_{c}\) denotes the maximum transmission power of \(c\)th link while \(p_{c}^{u}\) represents the \(c\)th channel power having \(u\)\(U_{E}\) (SSf).
6. The minimum power of \(c\)th \(U_{E}\) must be greater than or equal to 0 (SSg).  
The upcoming section briefly explains the proposed algorithm to optimize the problem in Equation (8a).