## 6 Extension with Side-Lobes/Back-Lobes as Well as the Path Loss Effect
### Physical Channel Model
In Scenario IV, the side-/back-lobes of the interfering node \(X_{k}\) and the interfered node \(X_{j}\) cover each other. Thus, we can calculate the interference range denoted by \(R_{i}(SS)\)  
\[R_{i}(SS)=(\frac{Cg_{s}g_{s}P_{t}}{P_{0}})^{\frac{1}{\alpha}} \tag{14}\]  
where we replace both \(G_{t}\) and \(G_{r}\) in Equation (10) by \(g_{s}\). With regard to \(R_{i}(SS)\), \(R_{i}(MS)\) and \(R_{i}(MM)\), we have the following result, which can be used to compare the different interference ranges under the above scenarios. **Lemma 3**: _When the main beamwidth \(\theta_{m}\) is narrow, we have \(R_{i}(SS)\ll R_{i}(MS)\ll R_{i}(MM)\)._  
**Proof. ** First, we have  
\[\frac{R_{i}(SS)}{R_{i}(MS)}=\frac{(\frac{Cg_{s}g_{s}P_{t}}{P_{0}})^{\frac{1}{ \alpha}}}{(\frac{Cg_{m}g_{s}P_{t}}{P_{0}})^{\frac{1}{\alpha}}}=(\frac{g_{s}}{ g_{m}})^{\frac{1}{\alpha}} \tag{15}\]  
Similarly, we have  
\[\frac{R_{i}(MS)}{R_{i}(MM)}=(\frac{g_{s}}{g_{m}})^{\frac{1}{\alpha}} \tag{16}\]  
As shown in Table 2, when the beamwidth \(\theta_{m}\) is narrow (e.g., \(\theta_{m}\leq 10^{\circ}\)), \(g_{s}\ll g_{m}\). Since the path loss factor \(\alpha\) usually ranges from 2 to 4, it is obvious that \(R_{i}(SS)\ll R_{i}(MS)\ll R_{i}(MM)\). We then follow the similar steps in Theorem 1 and derive the upper bounds on the number of channels with consideration of side-lobes and back-lobes. **Theorem 8**: _If there are \(n\) nodes in a planar area with the density \(D\) and each node is equipped with a directional antenna with the beamwidth \(\theta_{m}\), the main beam gain \(g_{m}\) and the side-/back-lobes gain \(g_{s}\), for any valid link set \(LS\) derived from the \(n\) nodes, the corresponding conflict graph \(G(LS)\) can be colored by using \(2D+\frac{4\pi D}{\theta_{m}}\cdot(\frac{g_{s}}{g_{m}})^{\frac{2}{\alpha}}-1\) colors._  
**Proof. ** Consider link \(l_{ij}\) that consists of two nodes \(X_{i}\) and \(X_{j}\). The distance between \(X_{i}\) and \(X_{j}\) is denoted by \(d\). To ensure that \(X_{i}\) can communicate with \(X_{j}\), we require \(d\leq R_{t}\), where \(R_{t}\) is the transmission range of \(X_{i}\). The area of the interference region (including the interference region of the main beam as well as the interference region of the side-/back-lobes) varies with the different distance \(d\). However, \(d\) cannot be too large, otherwise \(X_{i}\) and \(X_{j}\) cannot communicate with each other. It holds that \(d\leq R_{t}\). To simply our analysis, we consider the following two cases. Case 1 (when \(d=R_{t}\)):  
In this case, the interference region of side-/back-lobes is totally covered by the interference region of the main beam of a directional antenna as shown in Figure 12(a). This is because the interference range of side-/back-lobes denoted by \(R_{i}(SS)\) or \(R_{i}(MS)\) (note it depends on whether the interfering node is pointing its main beam or its side-/back-lobes toward the interfered node) is far less than the interference range of the main beam denoted by \(R_{i}(MM)\) as proved in Lemma 3. Following the proof of Theorem 1, the interference region contains no more than \(2D-2\) nodes excluding nodes \(X_{i}\) and \(X_{j}\). From Lemma 1, the conflict graph can be colored by \(2D-1\) colors. Case 2 (when \(d<R_{t}\)):  
When the distance \(d\) is decreased, the interference region caused by the side-/back-lobes may not be totally covered by the interference region of the main beam. For example, there is an extreme case (when \(d\to 0\)), as shown in Figure 12(b), where the interference region consists of two sectors of main lobes and two circles of side-/back lobes, which cannot be totally covered by the interference region of the main lobes. In this case, the interference region has the maximum coverage area. We then calculate the number of nodes in this interference region. The number of nodes in the two circles is at most \(\frac{D}{\frac{g_{s}}{2}(R_{i}(MM))^{2}}\cdot 2\pi(R_{i}(MS))^{2}=\frac{4\pi D}{ \theta_{m}}\cdot(\frac{g_{s}}{g_{m}})^{\frac{2}{\alpha}}\), which is obtained by Equation (16) of Lemma 3.