# The Impact of Impulsive Traffic on Cellular Internet of Things Network Performance Indicators
## 5 Mathematical Model
Each incoming group can contain a different number of requests that is not equal to zero; the probability that the group has s requests is equal to \(f_{s}\), where the count of files in the group \(s=1\), \(2\),..., \(n\). For convenience, let's take \(n=v\), so s will vary from \(1\) to \(\upsilon\). Easily, we could know that \(\sum\limits_{s=1}^{v}f_{s}=1\). The average amount of requests in the group is denoted by b and can be calculated by the following equation:  
\[b=\sum\limits_{s=1}^{v}s.f_{s}, \tag{1}\]  
When a group of requests is received, one of three scenarios is possible:  
1. There is sufficient free resource on the chosen resource (slice) to service the whole group. Then, this group will be served at the same time. 2. The slice lacks sufficient free resources to satisfy the whole group. In this situation, a partial group of the whole group will be served, while the rest will be lost and not renewed. Figure 5: The analytical model of an LTE cell simultaneously serving IoT users as well as cellular users, considering the impulsive nature of IoT requests [author’s own processing]. 3. The resources are completely busy. Then no request will be served, and the whole group will be lost without being revived. Let the variance in the total amount of requests for a given incoming group be denoted by \(\sigma^{2}\). As the number of requests in the group moves away from its middle-value b, the value of the variance \(\sigma^{2}\) will increase, and this leads to an increase in the impulsiveness of the incoming requests. \[\sigma^{2}=\sum_{i=1}^{v}f_{i}(i-b)^{2} \tag{2}\]  
Indicate the model under consideration's state space with \(S=\{(i),\ i=0,1,\ldots,v\}\). A random process \(r(t)=i(t)\) explains their changes over time, where \(i(t)\) denotes the number of channels utilized to handle requests at a given point in time \(t\). The process may be categorized as a Markov process since the model's random variables all have an exponential distribution and are independent of other variables. Denote by \(p(i)\) and \(P(i)\) the normalized and unnormalized probability that the model has \(i\) channel in the busy state, respectively. To find these probabilities, a set of equilibrium equations must be created and solved. The usual approach of teletraffic theory is used to tackle this problem. We obtain the following equation system:  
\[ \begin{array}{c}P(0)\lambda=P(1)\mu\ ;\ i=0\\ P(i)(\lambda+i\mu)=(P(0)f_{i}+P(1)f_{i-1}+\ldots+P(i-1)f_{1})\lambda+P(i+1)(i+ 1)\mu\ ;\ \ i=1,\ldots,v-1\\ P(v)v\mu=(P(0)f_{v}+P(1)(f_{v-1}+f_{v})+\ldots+P(v-1)(f_{1}+f_{2}+\ldots+f_{v}) )\lambda;\ \ i=v\end{array}  \tag{3}\]  
A recurring relation relating \(P(j)\) can be obtained via algebraic manipulations of the system of equilibrium equations. It appears as follows:  
\[j\mu P(j)=\lambda\Bigg{(}P(0)\overset{v}{\underset{i=j}{\sum}}f_{i}+P(1)\overset {v}{\underset{i=j-1}{\sum}}f_{i}+\ldots+P(j-1)\overset{v}{\underset{i=1}{ \sum}}f_{i}\Bigg{)} \tag{4}\]  
The quality of service of the model is controlled by these indicators:  
* m--the average number of occupied channels. * \(\pi_{c}\)--the ratio of lost requests. * \(\pi_{t}\)--the portion of time when all channels are busy. The following are the stages in the algorithm for determining \(p(i)\):  
1. Set the value of the unnormalized probability in the case of empty channels (no busy channels) equal to 1, and we can write it as \(P(0)=1\). 2. Find the unnormalized probabilities \(P(j)\) for \(j=1,\ldots,v\) using the recursion in Equation (4). 3. Calculate the normalization constant \(N\): \[N=\sum_{j=0}^{v}P(j)\] (5)
4. Find the normalized probabilities as follows: \[p(j)=\frac{P(j)}{N},\ \ j=0,1,\ldots,v\] (6)
5. Calculate the quality of service indicators as follows: \[\pi_{c}=\frac{1}{b}\sum_{i=1}^{v}p(i)\overset{i-1}{\underset{k=0}{\sum}}f_{v -k}(i-k)\] (7) \[\pi_{t}=p(v)\] (8)\[m=\sum_{i=0}^{v}p(i)i \tag{9}\]