## 3 System Model
### Line-of-Sight Probability  
It is assumed that the sensor moves on a preassigned path and would transmit data to \(R\) at random time intervals. Thus, the probability of an LoS existing for the transmission would depend on the location of the sensor at a given time \(t_{1}\) and the height of the boxes stored in the path of the signal.  
For the calculation of LoS probability, the following assumptions related to the physical structure of the warehouse are made:  
* The stacks are huge and thick and can be assumed as walls;
* The stacks are placed in equal distances;
* The roof is of fixed height \(h\).  
The warehouse is considered to be of size \(C_{i}\) with dimensions \(W_{i}(m)\times L_{i}(m)\times H_{i}(m)\), and it is assumed that the space between each stack is a rectangular cavity and the system can be divided into \(N\) equal rectangular cavities, with area \(S_{i}\). At a given time, if \(S\) and \(R\) are located in the same rectangular cavity, the link is an LoS link. The transmission path from \(S\) through relay node \((RN)\) to the destination node \((DN)\) is shown in Figure 2.  
For a link of distance \(d_{SR}\), the horizontal distance between \(S\) and \(R\) is denoted as \(d^{\prime}\), where \(d^{\prime}=\sqrt{d_{SR}^{2}-h^{2}}\), and the angle of \(d^{\prime}\) is expressed as \(\theta\). \(x\) and \(y\) are the distances from the mid-point of \(d^{\prime}\) to the sides of lengths \(L(m)\) and \(W(m)\), respectively. The values of \(x\), \(y\), and \(\theta\) are independent, and if \(U\) denotes the uniform distribution, \(x\sim U(0,\frac{L_{i}}{2})\), \(y\sim U(0,\frac{W_{i}}{2})\), and \(\theta\sim U\big{(}0,\frac{\pi}{2}\big{)}\), the joint probability density function (PDF) \(f(x,y,\theta)\) is given as follows:  
\[f(x,y,\theta)=\begin{cases}\frac{8}{L_{i}M_{i}\pi},&(x,y,\theta)\in A\\ 0,&\text{else}\end{cases}\]  
where \(A=\{(0\leq x<\frac{L_{i}}{2})\cap(0\leq y<\frac{W_{i}}{2})\cap(0\leq\theta< \frac{\pi}{2})\}\).  
For an LoS link to exist in a given rectangular box, the following condition has to be met:  
\[\bigg{(}x>\frac{d^{\prime}cos\theta}{2}\bigg{)}\cap\bigg{(}y>\frac{d^{\prime} sin\theta}{2}\bigg{)}.\]  
As presented in [27], the LoS probability changes with the horizontal length of the link \(d^{\prime}\), and for simplicity, we assume that in the presented system, this length is always less than \(L_{i}\). Then, the LoS probability for a single rectangular space is given as follows:  
\[P_{LoS}(d)=\begin{cases}\frac{d^{2}-2d(L_{i}+M_{i})+L_{i}M_{i}\pi}{L_{i}M_{i} \pi},&\text{if }d<M_{i}\\ \frac{-L_{i}^{2}+2 dM_{i}(\sqrt{1-(L_{i}/d)^{2}}-1)}{L_{i}M_{i}\pi}+\frac{2L_{i }M_{i}\arcsin(L_{i}/d)}{L_{i}M_{i}\pi},&\text{if }M_{i}\leq d<L_{i}\\ 0,&\text{otherwise}\end{cases} \tag{1}\]  
In the complete warehouse with \(N\) rectangular spaces, the LoS probability can be found using Equation (2),  
\[P_{LoS}(d^{\prime})=\sum_{i=1}^{N}\frac{S_{i}}{\sum_{i=1}^{N}S_{i}}P_{i,LoS}( d^{\prime}). \tag{2}\]  
Figure 1: Structure of the indoor warehouse environment.  
Thus, based on \(P_{LoS}\), the Rician factor can be expressed as follows:  
\[K=\frac{P_{i,LoS}(d^{\prime})}{1-P_{i,LoS}(d^{\prime})}. \tag{3}\]