# Jointly Optimize the Residual Energy of Multiple Mobile Devices in the MEC-WPT System
## 3 System Model
### Task-Offloading Model  
Considering that energy harvesting and task offloading are performed by multiple MDs in a time block, we use the TDMA protocol, which is shown in Figure 2.  
\begin{table}
\begin{tabular}{l c} \hline \hline
**Parameters** & **Notation** \\ \hline The input bit of the task & \(R_{i,j}\) \\ The offload bit & \(R_{i,j}^{off}\) \\ The number of cycles for one bie & \(C_{j}\) \\ The effective capacitance coefficient & \(\kappa_{j}\) \\ The frequency of the CPU & \(f_{i,j}\) \\ The communication bandwidth & B \\ The channel gain from MD to AP & \(h_{i,j}^{m}\) \\ The transmission power from MD to AP & \(p_{i,j}^{m}\) \\ The distance between device and AP & \(d_{i,j}\) \\ The constant circuit power & \(p_{m,j}\) \\ Energy conversion efficiency & \(v_{j}\) \\ The channel gain of downlink channel & \(h_{i,j}^{a}\) \\ The AP transmission power & \(p_{i,j}^{a}\) \\ The energy harvesting time & \(t_{i}^{har}\) \\ \hline \hline \end{tabular}
\end{table}
Table 1: Parameters.  
In Figure 2, \(t0\) represents the energy harvesting time of all MDs in the time block. Multiple MDs parallel the energy harvesting, so the energy harvesting time of each MD in the same time slice is equal. \(t1\) denotes the task-offloading time of all MDs in the time block, and there is no requirement that \(t0\) and \(t1\) must be equal. Here, we suppose that if there is a multi-core CPU perfect well on the edge server, the execution time of the computing task offloaded to the edge server can be ignored. Compared with the size of the offloading task, the size of the computation result is usually much smaller than the offloaded task and can be downloaded directly by the MD, so we also ignore the download time. According to the above assumptions, we can clearly know that the task-offloading model is mainly about offloading communication.  
From the Shannon formula in communication knowledge, we know that the transmission speed of offloading is:  
\[r_{i,j}=B\log_{2}\left(1+\frac{p_{i,j}^{m}h_{i,j}^{m}}{N_{0}d_{i,j}}\right)\quad i \in\mathbf{N},j\in\mathbf{M} \tag{3}\]  
where \(h_{i,j}^{m}\) represents the channel gain from MD \(j\) to AP, which is a constant during the offload duration, \(p_{i,j}^{m}\) represents the transmission power from MD \(j\) to the AP, \(N_{0}\) represents the noise power from device \(j\) to the AP, \(d_{i,j}\) indicates the distance between device \(j\) and AP. \(R_{i,j}^{off}\) is the offloading data size, where \(r_{i,j}=\frac{R_{i,j}^{off}}{l_{i,j}^{eff}}\). For simple expression, we introduce a function \(f(x)=N_{0}d_{i,j}(2^{\frac{x}{B}}-1)\), then the transmission power of offloading from MD \(j\) to the AP is \(p_{i,j}^{m}=\frac{1}{h_{i,j}^{m}}f(\frac{R_{i,j}^{off}}{l_{i,j}^{eff}})\). Note that to avoid the problem of division by zero, we define \(f(\frac{R_{i,j}^{off}}{l_{i,j}^{eff}})=0\) when \(R_{i,j}^{off}\) or \(t_{i,j}^{eff}=0\). In addition, a constant circuit power \(p_{m,j}\) (such as digital-to-analog converter (DAC), filter, etc.) is consumed. We can get the communication energy consumption of the task offload model:  
\[E_{i,j}^{off}=\frac{t_{i,j}^{off}}{h_{i,j}^{m}}f(\frac{R_{i,j}^{off}}{t_{i,j}^ {off}})+p_{m,j}t_{i,j}^{off}\quad i\in\mathbf{N},j\in\mathbf{M} \tag{4}\]  
We consider that the computing power of the MEC server is limited and can help us obtain more realistic results. We suppose \(F\) represents the maximum computing power of the MEC server in each time block. Here we use the maximum number of CPU cycles in a time block to represent:  
\[\sum_{j=1}^{M}R_{i,j}^{off}\mathrm{C}_{j}\leq F\quad i=1,2,\ldots,N \tag{5}\]  
This constraint ensures that the computing delay of the edge server is negligible.  
Figure 2: Time-division multiplexing model for energy harvesting and task offloading.