# Dynamic RACH Partition for Massive Access of Differentiated M2M Services
## 3 System Model
### Random Access Procedures
On the basis of the illustration above, the number of devices successfully completing the access attempts \(S_{N}\) is defined as:

\[S_{N}=A_{s}f_{s}\times e^{-\frac{A_{s}f_{s}}{M_{n}}}+A_{n}f_{n}\times e^{- \frac{A_{n}f_{n}}{M_{n}}} \tag{6}\]

Then, substitute Equations (3) and (4) into Equation (6), and we can easily get:

\[S_{N}=A_{s}f_{s}\times e^{-\frac{A_{s}f_{s}(1+\beta)}{c\beta}}+A_{n}f_{n}\times e ^{-\frac{A_{n}f_{n}(1+\beta)}{c\beta}} \tag{7}\]Obviously, our ultimate purpose is to maximize \(S_{N}\). However, from the curve of \(S_{N}\)_versus_\(\beta\), we can observe that when \(\beta\) is relatively smaller, \(S_{N}\) probably reaches its maximum, which implies that \(S_{N}\) could reach its maximum in the case of a suitable value of \(\beta\). In other words, we can maximize the \(S_{N}\) by limiting the number of preambles allocated for delay-sensitive devices, which relates to the number of sensitive devices successfully accessing the network. Consequently, by intentionally setting the bounds of the sensitive devices successfully accessing the network, our scheme is formulated as the optimization problems depicted in Equation (8). \[\left\{ \begin{array}{ll}\max\limits_{\beta}&\left\{A_{s}f_{s}\cdot e^{-\frac{ A_{s}f_{s}}{c}(1+\frac{1}{\beta})}+A_{nf}f_{n}\cdot e^{-\frac{A_{nf}f_{n}(1+ \beta)}{c}}\right\}&\text{if }A_{s}f_{s}\in(0,1]\\ \max\limits_{A_{s}f_{s}e^{-\frac{A_{s}f_{s}}{c}(1+\frac{1}{\beta})}\geq 1}& \left\{A_{s}f_{s}\cdot e^{-\frac{A_{s}f_{s}}{c}(1+\frac{1}{\beta})}+A_{nf}f_{n} \cdot e^{-\frac{A_{nf}f_{n}(1+\beta)}{c}}\right\}&\text{if }A_{s}f_{s}\in(1,3)\\ \max\limits_{A_{s}f_{s}e^{-\frac{A_{s}f_{s}}{c}(1+\frac{1}{\beta})}\geq 2}& \left\{A_{s}f_{s}\cdot e^{-\frac{A_{s}f_{s}}{c}(1+\frac{1}{\beta})}+A_{nf}f_{n} \cdot e^{-\frac{A_{nf}f_{n}(1+\beta)}{c}}\right\}&\text{if }A_{s}f_{s}\in[3,+\infty)\end{array} \right. \tag{8}\]

The objective function in Equation (8) maximizes the total number of devices successfully completing the access procedure under three restrictions of \(\beta\). Then, we analyze the formulated optimization problems and correspondingly find their optimal solutions (denoted as \(\beta^{*}\)), respectively. **Case 1:**\(A_{s}f_{s}\in(0,1]\)

When \(A_{s}f_{s}\in(0,1]\), our formulated problem becomes an unconstrained optimization problem. First, we get the first-order derivative of the objective function depicted in Equation (9):

\[\frac{A_{s}^{2}f_{s}^{2}}{c\beta^{2}}\times e^{-\frac{A_{s}f_{s}(1+\beta)}{c ^{2}}}-\frac{A_{nf}^{2}f_{n}^{2}}{c}\times e^{-\frac{A_{nf}f_{n}(1+\beta)}{c}}\equiv 0 \tag{9}\]

and let Equation (9) be equal to zero. By simplifying Equation (9), we have:

\[\frac{A_{s}f_{s}}{c}\left(1+\frac{1}{\beta}\right)-\frac{A_{nf}f_{n}}{c}(1+ \beta)+2\log\beta=\log\left(\frac{A_{s}^{2}f_{s}^{2}}{A_{nf}^{2}f_{n}^{2}}\right) \tag{10}\]

Since there is no closed solution for Equation (10) through using Mathematica, we decide to find the approximate solution by means of numerical analysis. We adopt the **[PERSON] iteration algorithm** depicted in Algorithm 2 to find the approximate solution \(\beta^{*}\). ```
1:Initialization: endow \(A_{s},A_{n},f_{s},f_{n}\) with any feasible values and then set the initial value of iteration \(\beta_{0}\);
2:Iteration: taking \(\beta_{0}\) as the iteration initial value and \(10^{-5}\) as the given precision, the [PERSON] iterative algorithm is used to search for the first approximate solution \(\beta_{1}\);
3:Recursion: take \(\beta_{1}\) as the iteration initial value and \(10^{-5}\) as the given precision to get the second approximate solution \(\beta_{2}\), and repeat the steps above to get the solution set \(\{\beta_{1},\beta_{2},...\beta_{m}\}\);
4:Confirmation: substitute \(\{\beta_{1},\beta_{2},...\beta_{m}\}\) into the objective function in turn and select the \(\beta^{*}\) for maximizing the objective function;
5:End
```

**Algorithm 2** : [PERSON] iterative algorithm. **Case 2:**\(A_{s}f_{s}\in(1,3)\)

Different from **Case 1**, the problem under this case becomes an optimization problem with an inequality constraint.