# An Energy-Efficient Coverage Algorithm for Macrocell--Small Cell Network Systems
## 3 Problem Description  
### System Model with Novel Configurations  
Consider a two-layer cellular network system, with the focus directed at an MBS containing densely distributed SBSs, where it is assumed that there is an SBS (that is normally switched off) at the approximate centers of each potential coverage hole. The MBS is constantly activated. To facilitate later description, the number of SBSs is denoted by \(K\), and the number of users is denoted by \(M\). Here, downlink transmission under this scenario is discussed, under the assumption that the arrival traffic pattern is always backlogged. Let \(P_{\text{m\_basic}}\) and \(P_{\text{s\_basic}}\) denote the basic power consumption levels of the MBS and an SBS, respectively. Let \(P_{\text{m\_max\_t}}\) denote the transmission power consumption upper limit of the MBS. In order to simplify system operations, SBS transmission power consumption is set to a three-mode adjustable configuration that includes an off state, a low power output state, and a high power output state (i.e., cell-enlarged state), which are closely related to different coverage areas. Let \(P_{\text{s\_t}}^{\text{(low)}}\) and \(P_{\text{s\_t}}^{\text{(high)}}\) denote the low and high transmission power levels of an SBS, respectively. Thus, for an SBS, its low power consumption, denoted as \(P_{\text{s}}^{\text{(low)}}\), and high power consumption, denoted as \(P_{\text{s}}^{\text{(high)}}\), can be shown to be  
\[P_{\text{s}}^{\text{(low)}}=P_{\text{s\_basic}}+P_{\text{s\_t}}^{\text{(low)}} \tag{1}\]  
and  
\[P_{\text{s}}^{\text{(high)}}=P_{\text{s\_basic}}+P_{\text{s\_t}}^{\text{(high)}} \tag{2}\]  
respectively. In addition, assume that if the SBS deployed in the center of a coverage hole is activated, and its power is adjusted to \(P_{\text{s}}^{\text{(low)}}\), its coverage range is equal to the range of the coverage hole, and therefore, enough to fully fill the coverage hole. Assume that the available frequency bandwidth for the MBS is denoted by \(B_{\text{m}}\), that the available frequency bandwidth for an SBS is denoted by \(B_{\text{s}}\), and that the effective frequency reuse mechanism is in use, such that signals among the MBS and SBSs will not interfere with each other. Define \(C_{\text{ave}}\) as the average transmission rate requirement of all users. Define \(P_{\text{rz}}\) as the minimum signal power that can be received by the receiving terminal. According to the well-known Shannon capacity formula [26; 27], the bandwidth required by \(C_{\text{ave}}\) can be obtained as follows:  
\[B_{\text{req}}=\frac{C_{\text{ave}}}{\log_{2}(1+\frac{P_{\text{rz}}}{N_{0}})} \tag{3}\]  
where \(N_{0}\) is the white Gaussian noise power. In this study, the available bandwidths of the MBS and any SBS are divided into equal parts, respectively. Each part represents a channel. Specifically, let \(N_{\text{m}}\) and \(N_{\text{s}}\) denote the number of channels of the MBS and any SBS, respectively. As such, \(N_{\text{m}}\) and \(N_{\text{s}}\) can be expressed as:  
\[N_{\text{m}}=\left\lfloor\frac{B_{\text{m}}}{B_{\text{req}}}\right\rfloor \tag{4}\]  
and  
\[N_{\text{s}}=\left\lfloor\frac{B_{\text{s}}}{B_{\text{req}}}\right\rfloor \tag{5}\]  
respectively. It is assumed that each user uses one channel at a time. For each SBS, we assume that channels have the same transmission power.