## 5 Performance Analysis
### Analytical Model
From the above discussion, we can represent the system state space, S, as:  
\[\text{S}=\{(\text{i},\text{j},\text{k},\text{m},\text{n})|\text{i}\geq 0,\text{ j}\geq 0,\text{ i}+\text{j}+\text{k}\leq\text{C},\text{ }0\leq\text{m}\leq\text{M},\text{ }0\leq\text{n}\leq\text{N}\} \tag{1}\]  
In the proposed system, there are two queues, and based on the queues' status and the total number of free resources, we can classify the system state space (S) of the proposed system into different substates. For those substates, every single substate defines a section of the practical state of the proposed system. Furthermore, we must study the behavior of every single substate and define its balance equation. Hereafter, the detailed balance equation is derived, where the usage of several derived states leads us into the required balance equations. The queuing analysis methods presented in [26; 27] were followed. In order to derive the proper balance equation for each substate, we need to define the number of Boolean indicators to control the transition rates between substates, as shown in Table 4.  
The system state space represents the complete, feasible states of the proposed system. In order to simplify the analysis of the system, the system state space S can be described by using four substates \(\text{S}_{\text{i}}\) (\(i=1,2,3,4\)) such that \(\text{S}=\text{S}_{1}\cup\text{S}_{2}\cup\text{S}_{3}\cup\text{S}_{4}\) and these substates can be defied as follows:  
* \(\text{S}_{1}=\{(\text{i},\text{j},\text{k},\text{m},\text{n})\mid\text{i}=0; \text{j}=0;\text{k}=0;\text{m}=0;\text{n}=0\}\)
* \(\text{S}_{2}=\{(\text{i},\text{j},\text{k},\text{m},\text{n})\mid 0<(\text{i}+\text{j}+ \text{k})<\text{C};\text{m}=0;\text{n}=0\}\)
* \(\text{S}_{3}=\{(\text{i},\text{j},\text{k},\text{m},\text{n})\mid 0<(\text{i}+\text{j}+ \text{k})=\text{C};\text{m}=0;\text{n}=0\}\)
* \(\text{S}_{4}=\{(\text{i},\text{j},\text{k},\text{m},\text{n})\mid C<(\text{i}+ \text{j}+\text{k}+\text{n}+\text{m})\leq(\text{C}+\text{M}+\text{N});\text{ }0\leq\text{m}\leq\text{M};\text{ }0\leq\text{n}\leq\text{N}\}\). \begin{table}
\begin{tabular}{c c} \hline \hline
**Symbol** & **Description** \\ \hline uRLLC & ultra-reliable low-latency communications \\ eMBB & enhanced mobile broadband \\ mMTC & massive machine-type communications \\ C & total number of LTE-A RBs \\ \(\lambda_{\text{i}}\) & the arrival rate of service request i \\ \(\mu_{\text{i}}\) & service rate for i service request \\
2 QmMTC & queue of mMTC \\ M & size of 1 QeMBB \\ N & size of 2 QmMTC \\ PF & forced termination probability \\ BP & eMBB, mMTC blocking probability \\ i & number of ongoing uRLLC services \\ j & number of ongoing eMBB services \\ k & number of ongoing mMTC services \\ m & number of queued eMBBs in 1 QeMBB \\ n & number of queued mMTCs in 2 QmMTC \\ D & eMBB and mMTC average delay time \\ U & eMBB and mMTC resource utilization \\ \hline \hline \end{tabular}
\end{table} 
Table 3: Parameter descriptions. In the following subsections, we briefly describe each single substates and derive their balance equation as follows. Substate 1 (S\({}_{1}\)): These substates represent all the possible states where the whole system is empty, i.e., T=zero. That is, all of the resource units (BBUs) of C-RAN are free with no service. We call this an initial state. Therefore, the state space of S\({}_{1}\) can be written as:  
\[S_{1}=\{(i,j,k,m,n)\text{: }i=0;j=0;k=0;m=0;n=0\}\]  
where all five variables are equal to zero and its balance equation can be defined as:  
\[(\lambda_{1}+\lambda_{2}+\lambda_{3})\pi_{0,0,0,0,0}=\pi_{1,0,0,0,0}+\pi_{0,1,0,0,0}+\pi_{0,0,1,0,0} \tag{2}\]  
Substate 2 (S\({}_{2}\)): These substates represent all the possible states (0 < T < C and m = n = 0) where the total number of requests in the system is greater than zero and less than C without any queued requests. Therefore, these substates show only the number of occupied BBUs at the current moment. Therefore, the total available BBU resources will be greater than zero and less than C when both queues are free or empty.