## 3 System Model
### Computation Model  
(1) Execution at C  
All the tasks which executed at C share the computation resources of it.  
By defining the service rate of C as \(\eta\), if \(r_{j}\) is selected to be executed at C, the time consumed by completing \(r_{j}\) is  
\[t_{i,j}^{\text{C}}=\frac{p_{i,j}e_{i}c_{i,j}}{\eta-\sum_{i\in\mathcal{M}}\sum_{j \in\mathcal{R}}\big{(}(1-\alpha_{i,j}-\sum_{k\in\mathcal{K}}\beta_{i,j,k})p_{i, j}\lambda_{i}c_{j}\big{)}}, \tag{1}\]  
where the denominator is the stable processing speed [35] (amount of computation processed per second) of C. \(\sum_{i\in\mathcal{M}}\sum_{j\in\mathcal{R}}\big{(}(1-\alpha_{i,j}-\sum_{k \in\mathcal{K}}\beta_{i,j,k})p_{i,j}e_{i}c_{j}\big{)}\) is the total amount of computation of C's tasks executed per second which was determined to be offloaded to C. It can be observed that, with the increase of the tasks executed at C, the processing speed of C is decreasing. Note that \(\eta-\sum_{i\in\mathcal{M}}\sum_{j\in\mathcal{R}}\big{(}(1-\alpha_{i,j}-\sum_{ k\in\mathcal{K}}\beta_{i,j,k})p_{i,j}e_{i}c_{j}\big{)}>0\) is the hard constraint [35] of Formula (3), which means the tasks' arriving rate cannot exceed C's service rate.  
(2) Execution at MEC-BS  
All the tasks which offloaded to a MEC-BS from C share the computation resources of the MEC-BS.  
By defining the service rate of MEC-BS as \(\phi\), if \(r_{j}\) is selected to be executed at MEC-BS, the time consumed by completing \(r_{j}\) is  
\[t_{i,j}^{\text{BS}}=\frac{p_{i,j}e_{i}c_{i,j}}{\phi-\sum_{i\in\mathcal{M}}\sum_ {j\in\mathcal{R}}(\alpha_{i,j}p_{i,j}e_{i}c_{j})}, \tag{2}\]  
where the denominator is the stable processing speed [35] (amount of computation processed per second) of MEC-BS. \(\sum_{i\in\mathcal{M}}\sum_{j\in\mathcal{R}}(\alpha_{i,j}p_{i,j}e_{i}c_{j})\) is the total amount of computation of MEC-BS's tasks executed per second which was determined to be offloaded to MEC-BS. It can be observed that, with the increase of the tasks executed at the MEC-BS, the processing speed of MEC-BS is decreasing. Note that \(\phi-\sum_{i\in\mathcal{M}}\sum_{j\in\mathcal{R}}(\alpha_{i,j}p_{i,j}e_{i}c_{ j})>0\) is the hard constraint [35] of Formula (3), which means the tasks' arriving rate cannot exceed MEC-BS's service rate.  
(3) Execution at MEC-NBS  
All the tasks which offloaded to a MEC-NBS from C share the computation resources of the MEC-NBS.  
By defining the service rate of \(n_{k}\) as \(\phi_{k}\), if \(r_{j}\) is selected to be executed at \(n_{k}\), the time consumed by completing \(r_{j}\) is  
\[t_{i,j,k}^{\text{NBS}}=\frac{p_{i,j}e_{i}c_{i,j}}{\phi_{k}-\sum_{i\in\mathcal{M }}\sum_{j\in\mathcal{R}}(\beta_{i,j,k}p_{i,j}e_{i}c_{j})}, \tag{3}\]  
where the denominator is the stable processing speed [35] (amount of computation processed per second) of \(n_{k}\). \(\sum_{i\in\mathcal{M}}\sum_{j\in\mathcal{R}}(\beta_{i,j,k}p_{i,j}e_{i}c_{j})\) is the total amount of computation of \(n_{k}\)'s tasks executed per second which was determined to be offloaded to MEC-NBS. It can be observed that, with the increase of the tasks executed at the MEC-NBS, the processing speed of \(n_{k}\) is decreasing. Note that \(\phi_{k}-\sum_{i\in\mathcal{M}}\sum_{j\in\mathcal{R}}\big{(}\beta_{i,j,k}p_{i,j}e_{i}c_{j}\big{)}>0\) is the hard constraint [35] of Formula (3), which means the tasks' arriving rate cannot exceed \(n_{k}\)'s service rate.