# Modeling and Analyzing Preemption-Based Service Prioritization in 5G Networks Slicing Framework
## 3 Mathematical Model
A depiction of the system transition diagram, for an arbitrary state \(\mathbf{n}\), i.e., \(\mathbf{n}\in\Omega\), is given in Figure 3.  
Introduce the stationary probabilities of the Markov process \(\mathbf{X}(t)\)  
\[P(\mathbf{n})=\lim_{t\to\infty}P\{\mathbf{X}(t)=\mathbf{n}\},\quad\mathbf{n} \in\Omega,\]  
and describe them using the system of equilibrium equations  
\[\Omega=\Big{\{}\mathbf{n}\in\mathbb{N}^{S}:(\mathbf{n}-\mathbf{N}^{\text{max} })\cdot\mathbf{j}\leq 0\wedge\mathbf{n}\cdot\mathbf{b}\leq C\Big{\}}, \tag{9}\]0 \[\Omega=\Big{\{}\mathbf{n}\in\mathbb{N}^{S}:(\mathbf{n}-\mathbf{N}^{\text{max} })\cdot\mathbf{j}\leq 0\wedge\mathbf{n}\cdot\mathbf{b}\leq C\Big{\}}, \tag{9}\]1 \[\Omega=\Big{\{}\mathbf{n}\in\mathbb{N}^{S}:(\mathbf{n}-\mathbf{N}^{\text{max} })\cdot\mathbf{j}\leq 0\wedge\mathbf{n}\cdot\mathbf{b}\leq C\Big{\}}, \tag{9}\]2 (14a) where \[\Omega=\Big{\{}\mathbf{n}\in\mathbb{N}^{S}:(\mathbf{n}-\mathbf{N}^{\text{max} })\cdot\mathbf{j}\leq 0\wedge\mathbf{n}\cdot\mathbf{b}\leq C\Big{\}}, \tag{9}\]3 represents the stationary probability that the system is in the state \[\Omega=\Big{\{}\mathbf{n}\in\mathbb{N}^{S}:(\mathbf{n}-\mathbf{N}^{\text{max} })\cdot\mathbf{j}\leq 0\wedge\mathbf{n}\cdot\mathbf{b}\leq C\Big{\}}, \tag{9}\]4 and \[\Omega=\Big{\{}\mathbf{n}\in\mathbb{N}^{S}:(\mathbf{n}-\mathbf{N}^{\text{max} })\cdot\mathbf{j}\leq 0\wedge\mathbf{n}\cdot\mathbf{b}\leq C\Big{\}}, \tag{9}\]5 represents the Indicator function ([[https://mathworld.wolfram.com/CharacteristicFunction.html](https://mathworld.wolfram.com/CharacteristicFunction.html)]([https://mathworld.wolfram.com/CharacteristicFunction.html](https://mathworld.wolfram.com/CharacteristicFunction.html)), accessed 1 September 2022): \[L_{A}\{a\}= \begin{cases}1,&\quad a\in\mathcal{A},\\ 0,&\quad\text{otherwise}.\end{cases}  \tag{14b}\]  
Figure 3: Transition diagram fragment for an arbitrary system state \(\mathbf{n}\), i.e., \(\mathbf{n}\in\Omega\). Note that the Markov process describing the system behavior is not reversible. Therefore, compute the system stationary probability distribution \(\mathbf{P}=\left[P(\mathbf{n})\right]_{\mathbf{n}\in\Omega}\), i.e., a size \(|\Omega|\times 1\) matrix, using an iterative method [38; 39; 40] to solve the system of equilibrium equations, rewritten as  
\[\mathbf{A}^{\top}\mathbf{P}=\mathbf{0},\quad\mathbf{P}\cdot\mathbf{j}=1, \tag{15}\]  
where \(\mathbf{A}\) represents the infinitesimal generator of Markov process, i.e., a size \(|\Omega|\times|\Omega|\) matrix, whose entries \(\mathsf{A}(\mathbf{n}\in\Omega,\mathbf{\hat{n}}\in\Omega)\) are computed as follows:  
\[\mathsf{A}(\mathbf{n},\mathbf{\hat{n}})=\left\{ \begin{array}{ll}\mathbf{ \lambda}\cdot\mathbf{e}_{s},\,\text{if}\,\mathbf{\hat{n}}=\mathbf{n}+ \mathbf{e}_{s},&\text{s.t.}\,\,\mathbf{n}\in\Omega_{\text{s}}^{\text{add}}, \\ \text{elseif}\,\mathbf{\hat{n}}=\mathbf{n}+\mathbf{e}_{s}-\mathbf{u}^{(s, \mathbf{n})},&\text{s.t.}\,\,\mathbf{n}\in\Omega_{\text{s}}^{\text{vpad}}, \\ (\mathbf{n}\odot\mu)\cdot\mathbf{e}_{s},\,\text{if}\,\mathbf{\hat{n}}=\mathbf{ n}-\mathbf{e}_{s},&\text{s.t.}\,\,\mathbf{\hat{n}}\in\Omega_{\text{s}}^{\text{add}}, \\ 0,\,\text{otherwise},&\text{i.e.,}\,\,\mathbf{\hat{n}}\in\Omega\setminus\{ \mathbf{n}\},\\ &\text{i.e.,}\,\,s=1,\ldots,S,\end{array} \right.\] (16a) when \[\mathbf{n}=\mathbf{\hat{n}}\], \[\mathsf{A}(\mathbf{n},\mathbf{n})=-\sum_{\mathbf{\hat{n}}\in\Omega\setminus\{ \mathbf{n}\}}\mathsf{A}(\mathbf{n},\mathbf{\hat{n}}). \tag{16b}\]  
Compute the stationary probability distribution \(P(\mathbf{n})\) from (15), then calculate the system Key Performance Indicators (KPIs) using expressions in analytic form.