# Maximizing the Capacity of Edge Networks with Multicasting
## 4 Generalized Representation of Network Capacity  
The classical work concludes that, when the number of nodes remains constant, the system capacity is inversely proportional to the transmission distance, that is, the smaller the transmission distance is, the larger the capacity is [31].  
Let \(L_{i}(C_{i})\) represent the transmission distance of the \(i\)th content with \(C_{i}\) copies on average, and \(p_{i}\) denote its requested probability (The content popularity can follow the Zipf distribution as most related works did [26; 32; 33; 34], or a predicating model can be used [35]). We obtain  
\[L_{m}=min\left(\sum_{i=1}^{m}L_{i}(C_{i})p_{i}\right). \tag{1}\]  
Figure 1: System model.  
s.t._  
\[\left\{\begin{array}{l}C_{i}\leq n\\ \sum\limits_{i=1}^{m}C_{i}\leq n\text{s}\end{array}\right. \tag{2}\]  
The first constraint imposes the limit that the maximum number of copies of each content item in the network cannot be bigger than the total number of SBSs, that is, to ensure that each content item has at most one copy in each SBS. The second constraint limits the total number of content copies, and these cannot exceed the total cache size. Therefore, in addition to content popularity, capacity is mainly dependent on the transmission distance \(L_{i}(C_{i})\), which includes two parts. The first is the cooperative transmission distance \(\overline{L_{i}(C_{i})}\) and the other is the length of the last hop, that is, the distance between the SBS and the user.  
Note that each SBS has a cache size \(s\) and the total size of the content is \(m\); the response probability that the SBS caches the requested content is therefore \(\frac{s}{m}\), and the probability of SBS collaboration is \(\frac{m-s}{m}\)[36]. Then the transmission distance \(L_{m}\) of the content can be presented as:  
\[\begin{array}{lcl}L_{m}&=&\sum\limits_{i=1}^{m}L_{i}(C_{i})p_{i}\\ &=&\frac{m-s}{m}\sum\limits_{i=1}^{m}\overline{L(C_{i})}p_{i}+0(s/m)p_{i}.\\ &=&\frac{m-s}{m}\sum\limits_{i=1}^{m}\overline{L_{i}(C_{i})}p_{i}\end{array} \tag{3}\]  
where we ignore the length of the last hop because it is within the communication range \(R\).  
We replace \(L\) with \(L_{m}\) in the formula \(C=\frac{W}{L\sqrt{n}}\) and obtain  
\[C=\frac{Wm}{(m-s)\sum\limits_{i=1}^{m}\overline{L_{i}(C_{i})}p_{i}\sqrt{n}}. \tag{4}\]  
Before we analyze the average distance in the multicasting transmission, the wireless interference is firstly discussed. Let \(P_{t}\) represent the transmission power of SBSs and \(V(t)\) denote the set of SBSs transmitting signals at time \(t\). For content to be transmitted from \(i\in V(t)\) to \(j\in V(t)\), the following path loss model is used  
\[\frac{\frac{P_{t}}{D_{i,j}^{\text{tr}}}}{N_{0}+\sum_{\varphi\in V(t),\varphi \
eq i}\frac{P_{t}}{D_{\varphi,j}^{\text{tr}}}}\geq\sigma,\]  
where \(\sigma\) denotes the minimal signal-to-interference-plus-noise ratio, \(D_{i,j}^{\text{tr}}\) is the distance between nodes \(i\) and \(j\), \(N_{0}\) is white noise and \(\alpha\) denotes the path-loss exponent, which describes how the signal strength decays with distance and is generally greater than two.