## 3 System Model Analysis
### Service Time Model
#### 3.3.1 Communication Time  
We consider the communication time as the offloading time of the task from SN node \(\text{n},\text{n}\in\text{N}\) to BS node \(\text{b},\text{b}\in\text{B}\); the forwarding time of the sub-task from node \(\text{i}\in\text{B}\) to node \(\text{j}\in\text{B}\); and the return time of the final result from \(\text{b}\in\text{B}\) to \(\text{n}\in\text{N}\). The pathloss, d, in the communication for underwater acoustic communication between the SN and BS can be calculated by using [PERSON]'s model [39], as shown in Equation (1):  
\[d=K\times 10 log(r)+\frac{r}{1000}\times 10\;loga(f) \tag{1}\]  
where \(K\) is the spreading factor, and the value of \(K\) is taken as 1.5 for practical spreading; \(r\) is the distance (m) between the SN and BS (or from BS to BS); \(f\) is the frequency of the signal in kHz; and \(a(f)\) is the absorption coefficient and is calculated by using [PERSON]'s formula mentioned in [40]. The absorption coefficient depends on the frequency, pressure, temperature, salinity, PH, and acoustic propagation velocity, as given in [41]. The signal-to-noise ratio (SNR) is calculated at the receiver. It is assumed that losses occur due to the multi-path effect, and the Doppler effect can be neglected. Hence, the SNR \(S\) is calculated as shown in Equation (2) [42].  
\[\text{S}=\frac{\text{T}_{\text{ASL}}}{\text{d}\times\text{\mu}\times\text{W}}, \tag{2}\]  
where d is the path-loss factor, as mentioned in (1); W is the receiver bandwidth; and \(\mu\) is the linear sum of the four ambient noise components, i.e., turbulence, shipping, waves, and thermal noise, which are calculated as shown in Equation (3) [39].  
\[\begin{array}{l}10\text{log}\;\mu_{\text{Tur}}=17-30\text{log}(\text{f})\\ 10\text{log}\;\mu_{\text{Sp}}=40-20(\text{s}-0.5+26\text{log}(\text{f})-60 \text{log}(\text{f}+0.003)\;\;\;\;\;\;\;\;\;\;\;\;0<s<1\\ 10\text{log}\;\mu_{\text{Wav}}=50+7.5\text{l}^{\frac{1}{2}}+20\text{log}(\text{ f})-40\text{log}(\text{f}+0.4)\\ 10\text{log}\;\mu_{\text{Ther}}=-15+20\text{log}(\text{f}).\end{array} \tag{3}\]The \(T_{ASL}\) is used to calculate the transmitter's acoustic signal level and is given by Equation (4):  
\[T_{ASL}=170.8+10 logP+10 log\eta+I, \tag{4}\]  
where P is the transmitted power, and \(\eta\) is the efficiency factor. The efficiency factor includes the losses associated with electro-acoustic conversion, and \(I\) is the projector directivity index [43]. Hence, the data rate from SNs to BS is given by Equation (5):  
\[\mathcal{R}_{n,b}=W^{n}log_{2}(1+S),\forall n\in\mathcal{N},b\in\mathcal{B}, \tag{5}\]  
where \(W^{n}\) denotes the bandwidth reserved for SNs. In addition, the data rate from BS i to BS j for the division of the task is calculated as shown in Equation (6):  
\[\mathcal{R}_{i,j}=W^{i,j}log_{2}(1+S),\forall i,j\in\mathcal{B}, \tag{6}\]  
Here, \(W^{i,j}\) is the bandwidth reserved for transmitting the task offloaded within \(T_{s}\). The data rate, \(\mathcal{R}_{b,n}\), from the BS to the SNs is given by Equation (7):  
\[\mathcal{R}_{b,n}=W^{b,n}log_{2}(1+S),\forall n\in\mathcal{N},\in\mathcal{B}, \tag{7}\]  
Hence, the communication time for offloading the computational task within \(T_{s}\) to \(i\in\mathcal{B}\) is calculated as shown in Equation (8):  
\[\mathcal{T}_{\text{n}}(\text{t})=\frac{V_{n(T_{s})}\times A_{t}}{\mathcal{R}_ {n,b}}. \tag{8}\]  
The communication time for forwarding the computational task from i to j at \(T_{s}\) is calculated as shown in Equation (9):  
\[\mathcal{T}_{\text{ij}}(\text{t})=\frac{V_{b(T_{s})}\times A_{t}}{\mathcal{R}_ {i,j}}. \tag{9}\]  
The return time of the final computation result from the BS to the SNs is calculated as shown in Equation (10):  
\[\mathcal{T}_{\text{b,n}}(\text{t})=\frac{\Phi\times A_{t}}{\mathcal{R}_{b,n}}. \tag{10}\]