## 3 System Model Analysis
### Service Time Model
#### 3.3.3 Total Service Time  
In this paper, we computed the total service time, \(\mathcal{T}_{sd}(t)\), which is shown in Equation (15):  
\[\mathcal{T}_{sd}(t)=\mathcal{T}_{n}(t)+max\Big{\{}V_{n(T_{s})}\mathcal{T}_{i}( t),\mathcal{T}_{i,j}(t)+V_{b(T_{s})}\mathcal{T}_{j}(t)\Big{\}}, \tag{15}\]  
where \(\mathcal{T}_{n}(t)\) is the offloading time from the SNs to BS, and i.\(\mathcal{T}_{n}(t)\) is the forwarding time from BS i to BS j. To divide the tasks into multiple sub-components, the following three scenarios are considered:  
1. In the first case, BS i computes the total computational task, and BS j does not take part in computation. In such a case, \(V_{b(T_{s})}=0\). Hence, \(\mathcal{T}_{j}(t)=0\). Therefore, the maximum function, i.e., \([V_{b(T_{s})}\mathcal{T}_{i}(t)]\), is used.
2. In the second case, BS j computes the complete computational task, and BS i does not take part in the computation. In such a case, \(V_{n(T_{s})}=0\), and, consequently, \(V_{n(T_{s})}\mathcal{T}_{i}(t)=0\). Thus, the maximum function, i.e., \([\mathcal{T}_{i,j}(t)+V_{b(T_{s})}\mathcal{T}_{j}(t)]\), is considered.
3. In the third scenario, both BSs, i.e., node \(b_{i},i\in\mathcal{B}\) and node \(b_{j},j\in\mathcal{B}\) take part in the computation by dividing the task between them. In this case, \([V_{n(T_{s})}\mathcal{T}_{i}(t),\mathcal{T}_{i,j}(t)]\
eq 0\), \([V_{b(T_{s})}\mathcal{T}_{j}(t)]\
eq 0\). Therefore, the greater values between \([V_{b(T_{s})}\mathcal{T}_{i}(t),\mathcal{T}_{i,j}(t)]\) and \([V_{b(T_{s})}\mathcal{T}_{j}(t)]\) are selected as the maximum function with which the BSs compute the task in parallel.  
When the signal between the primary BS node and its coalition BS node is weak and the noise is strong, the division and offloading between two coalition BSs may fail, so we define a threshold value, which is denoted by \(\Gamma\), for the signal-to-noise ratio (SNR). The primary BS node offloads and forwards the sub-component of the computational task to the coalition BS node only when its SNR satisfies the following premise given by Equation (16) or Equation (17):  
\[\text{S}\geq V_{n(T_{s})}(t)\Gamma \tag{16}\]  
and  
\[\text{S}\geq V_{b(T_{s})}(t)\Gamma \tag{17}\]  
where S is the SNR of a BS node.  
During a service session, if one BS moves out of the transmission range of the SN node which has offloaded the computation task to the BS node, the service session for the SN node fails. The success or failure of a service session for an SN node within a \(\mathrm{T_{s}}\) is represented by a variable. Thus, \(\mathcal{F}_{s}\) is shown in Equation (18).  
\[\mathcal{F}_{s}=s.t.\left\{\begin{array}{cc}1.&\quad\quad\quad\text{if }S\ \geq V_{b(T_{s})}\ \Gamma\\ 0.&\quad\quad\quad\quad\text{Otherwise}\end{array}\right. \tag{18}\]  
The distance between the service providing node BS and the request node SNs within a \(T_{s}\) is \(r_{i,j}(t+\mathcal{T}_{sd})\), which is considered to be the distance when the task is completed and the results are returned to the SN.  
To increase the service reliability for task division and results migrating, a shortest-distance search method is proposed to find the best coalition BS. In the next section, we formulate the problem of cooperative computational offloading using coalition game theory. In addition, we adopt an artificial neural network (ANN) to divide the task intelligently and obtain efficient computational offloading.