## 3 Resource Allocation Algorithm
### The Proposed Multi-Cell Resource Allocation Algorithm  
If we have the notation \(p(S_{i})=g(x_{i})_{\min}\), the first partial derivative of \(p(S_{i})\) with respect to \(S_{i}\) is expressed as:  
\[\frac{\partial p(S_{i})}{S_{i}}=\frac{I_{i}+S_{i}}{(I_{i}+\beta_{i})\cdot(I_{i }+S_{i}+\beta_{i})e^{\frac{\beta_{i}S_{i}}{(I_{i}+\beta_{i})(I_{i}+\beta_{i})} }}>0 \tag{13}\]  
Therefore, \(p(S_{i})_{\min}=p(S_{i})|_{S_{i}=0}=1\) and \(g(x_{i})_{\min}\geq p(S_{i})_{\min}=1\). Combined with Equation (8), the conclusion \(\frac{\partial f_{0}}{\partial x_{i}}<0\) can be obtained. Therefore, \(\frac{\partial f}{\partial x_{i}}>0\), which means that for a certain terminal, it achieved a higher throughput when it obtains a higher resource ratio. Therefore, the sum of resource ratio factors in a certain cell equaling to 1 is a necessary condition to maximize the throughput performance. The equivalent problem of problem **P2** can be expressed as:  
\[ \begin{split}\textbf{P3}:&\min& f_{0}(x)=\sum\limits_{i=1}^{N}-\alpha_{i}x_{i}\log_{2} \Big{(}1+\frac{S_{i}}{I_{i}+\beta_{i}x_{i}}\Big{)}\\ & s.t.& x_{i}\geq 0,\ i=1,\cdots N\\ &\textbf{1}^{T}\mathbf{x}=1\end{split}  \tag{14}\]Because problem **P3** is a convex optimization problem and it has a feasible solution \(\mathbf{x}=\left\{x_{i}=\frac{1}{N}|i=1,\cdots,N\right\}\), which satisfies the Slater condition, problem **P3** has a strong duality [29]. Because problem **P3** has a strong duality, it satisfies the Karush-Kuhn-Tucher (KKT) condition. By introducing Lagrange multiplier \(\lambda_{i}\) for the inequality constraint \(x_{i}\geq 0\) and \(\
u\) for the equality constraint \(\mathbf{1}^{T}\mathbf{x}=1\), the following KKT condition can be obtained [29]:  
\[\left\{ \begin{array}{ll}x_{i}\geq 0,\ i=1,\cdots,N&(a)\\ \mathbf{1}^{T}\mathbf{x}=1&(b)\\ \lambda_{i}\geq 0,\ i=1,\cdots,N&(c)\\ \lambda_{i}x_{i}=0,\ i=1,\cdots,N&(d)\\ -\alpha_{i}\log_{2}(g(x_{i}))-\lambda_{i}+\
u=0,\ i=1,\cdots,N&(e)\end{array} \right. \tag{15}\]  
In Equation (15e), \(\lambda_{i}\) is a slack variable. By substituting \(\lambda_{i}=\
u-\alpha_{i}\log_{2}(g(x_{i}))\) into Equations (15c) and (15d), we have:  
\[\left\{ \begin{array}{ll}x_{i}\geq 0,\ i=1,\cdots,N&(a)\\ 1^{T}x=1&(b)\\ \
u\geq\alpha_{i}\log_{2}(g(x_{i})),\ i=1,\cdots,N&(c)\\ x_{i}(\
u-\alpha_{i}\log_{2}(g(x_{i})))=0,\ i=1,\cdots,N&(d)\end{array} \right. \tag{16}\]  
According to Equation (16a), \(x_{i}=0\) or \(x_{i}>0\). By observing Equation (16d), there are two cases as follows:  
1. If \(x_{i}=0\), the value of \(\
u-\alpha_{i}\log_{2}(g(0))\) can be an arbitrary real number. Because \(g(0)=1+\frac{S_{i}}{l_{i}}\), the value of \(\
u-\alpha_{i}\log_{2}\Bigl{(}1+\frac{S_{i}}{l_{i}}\Bigr{)}\) can be an arbitrary real number. Combined with Equation (16.c), we have \(\
u\geq\alpha_{i}log_{2}\Bigl{(}1+\frac{S_{i}}{l_{i}}\Bigr{)}\);
2. If \(x_{i}>0\), \(\
u-\alpha_{i}log_{2}(g(x_{i}))=0\). Therefore, \(\
u=\alpha_{i}log_{2}(g(x_{i}))\geq 0\) and the conclusion satisfies Equation (16c). According to the monotonic decreasing property of \(g(x_{i})\), we obtained that \(\
u<\alpha_{i}log_{2}\Bigl{(}1+\frac{S_{i}}{l_{i}}\Bigr{)}\). Because \(\
u=\alpha_{i}log_{2}(g(x_{i}))\), we have \(x_{i}=g^{-1}\Bigl{(}2^{\frac{S_{i}}{l_{i}}}\Bigr{)}\) where \(g^{-1}(\cdot)\) is the inverse function of \(g(\cdot)\). By defining the notation \(\sigma_{i}=\alpha_{i}log_{2}\Bigl{(}1+\frac{S_{i}}{l_{i}}\Bigr{)}\), we have: \[x_{i}=\left\{ \begin{array}{ll}g^{-1}\Bigl{(}2^{\frac{\
u}{l_{i}}}\Bigr{)},&0 \leq\
u<\sigma_{i}\\ 0,&\
u\geq\sigma_{i}\end{array} \right.\] (17) By substituting Equation (17) into Equation (16b), we have: \[q(\
u)=\sum_{i=1}^{N}x_{i}=1\] (18)  
In the left side of Equation (18), \(q(\
u)\) is a piecewise decreasing function with respect to \(\
u\) and the break point is \(\sigma_{i}\). Therefore, Equation (18) has a unique solution with respect to \(\
u\)[29]. If the value of \(\
u\) is obtained, the value of \(x_{i}\) can be determined by Equation (17) and problem **P3** is solved.