# An Improved Dynamic Joint Resource Allocation Algorithm Based on SFR
## 2 System Model and Problem Formulation  
### System Model  
We consider an OFDM-based LTE downlink network where the number of multi-cells is \(C\). Figure 1 illustrates the SFR deployment. Each cell has the same number of users and uniform user distribution is assumed in the whole coverage area. Each cell is divided into two regions, the colored areas are the edge region and the colorless area is the center region. We define a user with the worst channel gain as a virtual user. As seen from the Figure 1, virtual center user and virtual edge user are the users located in their corresponding border area. In our system model, \(P_{\text{max}}\) denotes the maximum transmission power of the base station (BS). The cell-interior radius and the cell radius are \(r\) and \(R\), respectively. The received noise power is \(N_{0}\). \(B\) is the total system bandwidth, which is divided into \(N\) blocks, and each RB occupies a bandwidth of \(B_{sub}\). In SFR, RBs are grouped to cell-edge blocks with a bandwidth of \(B^{major}\) and cell-center block with a bandwidth of \(B^{\min{or}}\), where \(B^{major}\bigcup B^{\min{or}}=B\). Other symbols are defined in Table 1.  
Figure 1: System networking. In cellular networks, the signal power received gets weaker with the increasing of the propagation distance. For the convenience of discussion, we only consider the signals tha come from three BSs in terms of 802.16e standard, including a serving BS and two interfering BSs. For a user in cell \(i\), the signal-to-interference-noise ratio on RB \(j\) is given as  
\[SINR_{i}^{j}=\frac{p_{i}^{j}s_{i,j}^{j}}{\sum\limits_{m\in C_{adj}^{i}}p_{m}^{ j}g_{i,m}^{j}+N_{o}} \tag{1}\]  
where \(C_{adj}^{i}\) is the set of interfering cells which are adjacent to cell \(i\). If \(j\) is the edge-block, \(j\) is equal to \(o\). If \(j\) is the center-block, \(j\) is equal to \(in\). In the SFR system, the edge-user interference on RB \(j\) comes from the signals on RB \(j\) used by center-users in adjacent cells. Thus, we submit Equation (1) into the Shannon capacity formula [17]. The data rate of the edge-user situated in cell \(i\) is expressed as [16]  
\[ \begin{split} R_{i}^{o}&=\sum\limits_{j\in B_{i}^{ major}}B_{sub}\text{log}_{2}(1+SINR_{i}^{j})\\ &=N_{i}^{maj\sigma}B_{sub}\text{log}_{2}(1+\frac{p_{i}^{o}s_{i, j}^{o}}{\sum\limits_{m\in C_{adj}^{i}}p_{m}^{im}s_{i,m}^{o}+N_{o}})\end{split}  \tag{2}\]  
The center-user interference on RB \(j\) has two sources: signals reusing the same resource by adjacent cell edge-user or adjacent cell center-user. For simplifying notation, the adjacent cell interference power is unified into the transmit power of cell-edge RBs, _i.e._, \(p_{m}^{o}\). Therefore, the date rate of center-user in cell \(i\) can be approximately given as:  
\[ \begin{split} R_{i}^{in}&=\sum\limits_{j\in B_{i}^{ major}}B_{sub}\text{log}_{2}(1+SINR_{i}^{j})\\ &\approx N_{i}^{\text{min}\sigma}B_{sub}\text{log}_{2}(1+\frac{p _{i}^{in}s_{i,j}^{in}}{\sum\limits_{m\in C_{adj}^{i}}p_{m}^{o}s_{i,m}^{in}+N_{ o}})\end{split}  \tag{3}\]  
We already know if the data rate requirement of the virtual edge user in each cell is met; the data rate of all users in the same cell can therefore be satisfied. Thus, we take the minimum throughput demands for the virtual users as the minimum throughput requirements for the individual users situated in that cell zone. Thus, the minimum data rate for the edge and center zone in the target cell \(i\) is  
\[R_{i}^{o\text{min}}=\sum\limits_{k\in K_{i}^{o}}\gamma_{i,k} \tag{4}\]