## 5 Decentralized Algorithm
### Power Allocation Algorithm  
In this subsection, we modify the power allocation algorithm proposed in Subsection 4.2 to be decentralized, so that it can be processed at each TP independently and simultaneously. The individual power allocation problem of each TP is given by:  
\[ \begin{array}{ll}\max&\sum\limits_{k=1}^{K}\sum\limits_{n=1}^{N_{\text{BB}}} \beta_{m,k}^{n}\frac{R_{k}^{n}}{\bar{R}_{k}}\Bigg{/}\sum\limits_{k=1}^{K}\sum \limits_{n=1}^{N_{\text{BB}}}\beta_{m,k}^{n}p_{m}^{n}\\ \text{s.t.~{}~{}}\text{C1:}&0\leq p_{m}^{n}\leq S,~{}~{}\forall n\\ \text{C3:}&\sum\limits_{n=1}^{N_{\text{BB}}}\sum\limits_{k=1}^{K}\beta_{m,k}^{ n}R_{k}^{n}\leq C_{m}\\ \text{C4:}&R_{k}^{n}\geq R_{\text{thres}},\forall n,k\in\mathcal{U}_{m}\\ \end{array}  \tag{22}\]  
As in Subsection 4.2, we constitute and solve the dual function of Equation (22), instead of solving it directly. The dual function in terms of constraints C3 and C4 is given as:  
\[ \begin{array}{ll}\max\limits_{p_{m}^{n}}&\sum\limits_{n=1}^{N_{\text{BB}}} \Bigg{(}\sum\limits_{k=1}^{K}\beta_{m,k}^{n}\frac{R_{k}^{n}}{\bar{R}_{k}} \Bigg{/}\sum\limits_{k=1}^{K}\beta_{m,k}^{n}p_{m}^{n}\Bigg{)}-\lambda_{m} \left(\sum\limits_{n=1}^{N_{\text{BB}}}\sum\limits_{k=1}^{K}\beta_{m,k}^{n}R_{ k}^{n}-C_{m}\right)\\ &+\sum\limits_{n=1}^{N_{\text{BB}}}\sum\limits_{k=1}^{K}\mu_{n,k}\beta_{m,k}^{ n}\left(R_{k}^{n}-R_{\text{thres}}\right)\\ \text{s.t.~{}~{}}&0\leq p_{m}^{n}\leq S,~{}~{}\forall n\\ \end{array}  \tag{23}\]  
Substituting Equation (15) into Equation (23), we can decompose the power allocation problem into \(N_{\text{RB}}\) independent subproblems. Let \(f_{m,n}\) denote the objective of the subproblem on RB \(n\) of TP \(m\), which is given as:  
\[f_{m,n}=\sum\limits_{k=1}^{K}\beta_{m,k}^{n}\frac{R_{k}^{n}}{\bar{R}_{k}} \Bigg{/}\sum\limits_{k=1}^{K}\beta_{m,k}^{n}p_{m}^{n}-\sum\limits_{k=1}^{K} \lambda_{m}\beta_{m,k}^{n}\left(R_{k}^{n}-\frac{C_{m}}{\beta_{m}}\right)+\sum \limits_{k=1}^{K}\mu_{n,k}\beta_{m,k}^{n}\left(R_{k}^{n}-R_{\text{thres}}\right) \tag{24}\]  
\(\sum\limits_{k=1}^{K}\beta_{m,k}^{n}=0\) means that TP \(m\) does not schedule any transmissions on RB \(n\). In this case, power allocation is not required, _i.e._, \(p_{m}^{n}=0\). In the case where \(\sum\limits_{k=1}^{K}\beta_{m,k}^{n}=1\), transmit power \(p_{m}^{n}\) can be obtained by an iterative method proposed in the rest of this subsection. Let \(k^{*}\) denote the UE of TP \(m\) scheduled on RB \(n\) (_i.e._, \(\sum_{k=1}^{K}\beta_{m,k}^{n}=\beta_{m,k^{*}}^{n}=1\)). Taking the first order derivative of \(f_{m,n}\) with respect to \(p_{m}^{n}\), we obtain:  
\[\frac{\partial f_{m,n}}{\partial p_{m}^{n}}=\left(\frac{1}{\bar{R}_{k^{*}}p_{ m}^{n}}-\lambda_{m}+\mu_{n,k^{*}}\right)\frac{\partial R_{k^{*}}^{n}}{\partial p_{m}^ {n}}-\frac{R_{k^{*}}^{n}}{\bar{R}_{k^{*}}\left(p_{m}^{n}\right)^{2}} \tag{25}\]  
where:  
\[\frac{\partial R_{k^{*}}^{n}}{p_{m}^{n}}=\frac{b\sum\limits_{m\in\mathcal{M}} \beta_{m,k^{*}}^{n}\left\|\mathbf{H}_{m,k^{*}}^{n}\mathbf{w}_{m}^{n}\right\|^ {2}}{\left\|\mathbf{H}_{m,k^{*}}^{n}\mathbf{w}_{m}^{n}\right\|^{2}p_{m}^{n}+ \sum\limits_{m^{\prime}\in\mathcal{M}\backslash\{m\}}\left\|\mathbf{H}_{m^{ \prime},k^{*}}^{n}\mathbf{w}_{m^{\prime}}^{n^{\prime}}\right\|^{2}p_{m^{\prime} }^{n}+\sigma^{2}} \tag{26}\]Let \(\partial f_{n}/\partial p_{m}^{n}=0\). An expression of \(p_{m}^{n}\) in the \(t\)-th iteration can be obtained by solving the equation, which is given as:  
\[p_{m}^{n(t)}=\sqrt{\frac{R_{k^{*}}^{n}}{R_{k^{*}}}}\times\sqrt{ \frac{\left\|\mathbf{H}_{m,k^{*}}^{n}\mathbf{w}_{m}^{n}\right\|^{2}p_{m}^{n(t) }+\sum\limits_{ \begin{subarray}{c}m^{\prime}\in\mathcal{M}\\ m^{\prime}\
eq m\end{subarray} }\left\|\mathbf{H}_{m^{\prime},k^{*}}^{n} \mathbf{w}_{m^{\prime}}^{n}\right\|^{2}p_{m^{\prime}}^{n(t)}+\sigma^{2}}{b \left\|\mathbf{H}_{m,k^{*}}^{n}\mathbf{w}_{m}^{n}\right\|^{2}\left(\frac{1}{R _{k^{*}}p_{m}^{n(t)}}-(\lambda_{m}-\mu_{n,k^{*}})\right)}} \tag{27}\]  
where multipliers \(\lambda_{m}\) and \(\mu_{n,k^{*}}\) should be updated according to Equation (21).