# Throughput Analysis of IEEE 802.11 WLANs with Inter-Network Interference
## 4 Throughput Analysis of String and Grid Topology Considering Inter-Network Interference
### Analysis of Grid Network Topology
#### 4.2.3 Four Carrier-Sensing Networks  
As shown in Figure 4, Network 5 is in accord with this case. There are Network \(\{2,4,6,8\}\) within carrier-sensing range of Network 5. From (14), the probability that the state of Network 5 is idle can be expressed as (23). \(v(a)\cap v(b)=\{5\}\) means only Network 5 is within the carrier-sensing range of Network \(\{a,b\}\). We can see the pairs of Network \(\{2,8\}\), or Network \(\{4,6\}\) meet this condition. Moreover, when \(\{a,b\}\) is \(\{2,8\}\) then Network c is only Network 4 or Network 6, and vice versa. \[U_{5,a} = 1-\sum\limits_{ \begin{subarray}{c}1-\sum\limits_{k\in v(a) \cap v(b)}\left[x_{k}\prod\limits_{k\in v(k)}\left(1-\gamma_{nk}\right)\right] +\prod\limits_{k\in v(a)\cap v(b)}\left[x_{k}\prod\limits_{k\in v(k)}\left(1- \gamma_{nk}\right)\right](1-Y_{m})^{-1}\\ \prod\limits_{k\in v(1-\gamma_{k},l)}\left(1-\gamma_{k}\right)\end{subarray} }\] \[- \frac{X_{k}}{1-Y_{2}Y_{6}}+\frac{\prod\limits_{x}}{\left(1-\prod \limits_{Y_{c}}\right)\left(1-Y_{2}\right)}+\sum\limits_{ \begin{subarray}{c }X_{k}X_{c}\\ (1-Y_{2}Y_{b})\left(1-Y_{c}\right)\end{subarray} }-\frac{\prod\limits_{l\
eq a} }{\left(1-Y_{2}Y_{b}\right)\prod\left(1-Y_{c}\right)};\] \[v(a)\cap v(b)=\{5\}\,,c\in v(5),a\
eq b,b\
eq c,a\
eq c,m=k,m\
eq 5.\]  
For the easy presentation, we explain the value of the second component in (23) in the case of \(a=2\). Network \(\{5,7,9\}\) are within Network 8's carrier-sensing range. Moreover, Network 2 and Network 8 are in the carrier-sensing state when Network 5 does not transmit. That can be expressed \((1-Y_{2}Y_{8})\). The probability of Network 8 transmitting in this condition is \(\frac{X_{8}}{1-Y_{2}Y_{8}}\). The probability of Network c simultaneous transmitting can be expressed as \(\frac{X_{4}X_{6}}{(1-Y_{4}Y_{6})(1-Y_{2})}\) in the forth component in (23), because c equals 4 or 6. We consider the relationship of Network 2 and Network \(\{4,6\}\). From that, we can know Network \(\{1,3,5\}\) are in the carrier-sensing range of Network \(\{2,4,6\}\). In this condition, \((1-Y_{4}Y_{6})\) means Network 5 does not transmit. \((1-Y_{4}Y_{6})(1-Y_{2})\) means Network \(\{1,3,5\}\) do not transmit. Other components are calculated similarly as the ones in the previous sections. We can express \(U_{5,4}U_{5,6}\), and \(U_{5,8}\) using the similar method. From (2), the carrier-sensing airtime of Network \(i\) could be obtained accurately in (24). \[Y_{5} = \sum\limits_{i\in v(5)}X_{i}(1-\gamma_{5,i}) \tag{24}\] \[- \sum\limits_{j\
eq v(5)}\frac{\prod\limits_{l\in v(j)}X_{l}(1- \gamma_{5,l})}{1-\sum\limits_{k\in v(l)}\left[X_{k}\prod\limits_{h\in v(k)} \left(1-\gamma_{h,k}\right)\right]+\prod\limits_{k\in v(l)}\left[X_{k}\prod \limits_{h\in v(k)}\left(1-\gamma_{h,k}\right)\right](1-Y_{j})^{-1}}\] \[- \sum\limits_{k\in v(1-\gamma_{5,a})X_{b}(1-\gamma_{5,b})}{1-Y_{a} Y_{b}}+\sum\limits_{k\in v(1-\gamma_{5,a})X_{b}(1-\gamma_{5,b})X_{c}(1-\gamma_{5,c})}\] \[- \frac{\prod\limits_{i\in v(5)}X_{i}(1-\gamma_{5,i})}{(1-Y_{a}Y_{b} )\prod(1-Y_{c})};v(a)\cap v(b)=\{5\}\,,c\in v(5),a\
eq b,b\
eq c,a\
eq c.\]  
Specifically, when Network a and Network b transmit simultaneously, only Network 5 is within their carrier-sensing ranges, and we know Network 5 does not have transmission. In (24), \((1-Y_{a}Y_{b})\) means Network 5 does not transmit in this condition. Therefore, the simultaneous transmission time of Network \(a\) and Network \(b\) is \(\frac{X_{a}(1-\gamma_{5,a})X_{b}(1-\gamma_{5,b})}{1-Y_{a}Y_{b}}\). In the case of three networks, for example, Network \(\{a,b,c\}\) transmit simultaneously. When \(a=2\); \(b=4\); \(c=6\), Network \(\{1,3,5\}\) are within the carrier-sensing range of Network \(\{a,b,c\}\). \((1-Y_{2})\) means Network 1 and 3 do not transmit. \((1-Y_{4}Y_{8})\) presents the condition of Network 5 without transmission. \((1-Y_{4}Y_{8})\) presents the condition of Network 5 without transmission. Therefore, the simultaneous transmission time of Network \(\{2,4,6\}\) is \(\frac{X_{2}(1-7_{52})X_{4}(1-7_{54})X_{6}(1-7_{56})}{(1-Y_{4}Y_{6})(1-7_{2})}\).