## 3 The Modified BA Protocol (MBA)
Thus, \(D_{BT}(n_{r})\) indicates the number of consumed slots for one BT cycle under C.E probability of \(\alpha\) and unrecognized tags (\(n_{r}\)) located in reader zone. When C.E happens at \(\alpha\) probability of one, the number of consumed slots in each BT cycle is one. Due to this, BT does not expand the corresponding node. On the contrary, when C.E does nothappen at a slot, the unrecognized \(n_{r}\) tags will collide, and these tags are randomly divided into two subsets. Consequently, \(D_{BT}(n_{r})\) can be formulated as follows:  
\[D_{BT}(n_{r})=\ \alpha\ +\ (1\ -\ \alpha)2^{-n_{r}}\sum_{i=0}^{n_{r}}\left( \begin{array}{c}n_{r}\\ i\end{array} \right)\left[1+D_{BT}(i)+D_{BT}(n_{r}-i)\right]\ \frac{1+2^{1-n_{r}}\ (1 -\ \alpha)\ \sum_{i=0}^{n_{r}-1}\left( \begin{array}{c}n_{r}\\ i\end{array} \right)D_{BT}(i)}{1-2^{1-n_{r}}\ (1-\ \alpha)} \tag{3}\]  
in which \(D_{BT}(0)\) and \(D_{BT}(1)\) = 1. Similarly, the number of readable slots, \(D_{BT}^{1}(n_{r})\), can be found by the recursive equation:  
\[D_{BT}^{1}(n_{r})=\alpha+(1-\alpha)\ 2^{-n_{r}}\sum_{i=0}^{n_{r}}\left( \begin{array}{c}n_{r}\\ i\end{array} \right)\ \left[D_{BT}^{1}(i)+D_{BT}^{1}(n_{r}-i)\right] \tag{4}\]  
in which \(D_{BT}(0)\) = 0 and \(D_{BT}(1)\) = 1. Furthermore, to calculate the identification delay of GBT, the number of unrecognized tags participated in the beginning of \(j\)-th BT cycle, \(u_{j}\), must be known. Let \(r_{j}\) indicate the number of recognized tags in the \(j\)-th cycle. Thus, \(u_{j}\) and \(r_{j}\) are obtained as follows:  
\[u_{j}=n_{r}-\sum_{i=1}^{j-1}r_{i} \tag{5}\]  
\[r_{j}=D_{BT}^{1}(u_{j}) \tag{6}\]  
Alternatively, from Equations (5) and (6), \(u_{1}\), \(r_{1}\), \(u_{2}\), \(r_{2}\), \(u_{3}\), \(r_{3}\), \(\ldots\), \(u_{m}\), \(r_{m}\) are obtained, in which \(u_{1}=n_{r}\) and \(r_{m}=u_{m}\). Lastly, the identification delay of GBT is the sum of the slots of all BT cycles plus one as follows:  
\[D_{GBT}(n_{r})=\ \sum_{j=1}^{m}[D_{BT}(u_{j})]+1 \tag{7}\]  
The term (1) in \(D_{GBT}(n_{r})\) represents the last cycle in GBT, which consists of one idle slot to ensure that all the tags in reader zone have been identified. According to [15], the number of slots required to identified \(r_{j}\) tags in j-th cycle, \(D_{BT}(u_{j})\), is equal to  
\[1+2C_{BT}(r_{j}) \tag{8}\]  
in which \(C_{BT}(rj)\) denotes the number of collision slots required to identify \(r_{j}\) tags in \(j\)-th cycle. Thus, substituting (8) in (7) yields  
\[D_{GBT}(n_{r})=\ \sum_{j=1}^{m}[1+2C_{BT}(r_{j})]+1 \tag{9}\]  
Now, let \(r\) denote the number of cycles required to recognize \(n_{r}\) tags (\(r=m+1\)) by GBT, so  
\[D_{GBT}(n_{r})=\ \sum_{j=1}^{r-1}[1+2C_{BT}(r_{j})]+1=\ r-1+2\sum_{j=1}^{r-1}[C_{ BT}(r_{j})]+\ 1=2C_{BT}(n_{r})\ +\ r \tag{10}\]  
in which \(C_{BT}(n_{r})\) denotes the total collision slots required to identify \(n_{r}\) tags through (\(r-1\)) cycles of GBT, and \(C_{BT}(r_{j})\), which is calculated according to [15] as follows:  
\[C_{BT}(r_{j})=\sum_{k=0}^{\infty}C_{BT}(r_{j},k) \tag{11}\]  
in which \(C_{BT}(r_{j},k)\) is the number of collision slots at \(k\)-th tree level of BT \(j\)-th cycle. Moreover, at \(\alpha\) = 0 (C.E probability of zero), the number of GBT cycles (\(r\)) is two only. Therefore,  
\[D_{GBT}(n_{r})=2C_{BT}(n_{r})\ +\ 2 \tag{12}\]as \(\alpha\) increases the collision slots of each BT cycle are reduced, and thus the identification delay is reduced. In fact, MBA protocol consists of two phases. In phase I, the staying tags are identified without colliding with each other through number of successful (for staying tags) and idle (for leaving) cycles (denoted by slots in BT and GBT).