# Fast and Robust Asynchronous Rendezvous Scheme for Cognitive Radio Networks
## 4 Proposed Channel-Hopping Scheme
### FRARS for Asymmetric System  
I extend the FRARS algorithm to a general asymmetric algorithm by allowing each SU to have different available channels. When the geographical locations of the SUs are far, it is very likely that each SU has different available channels because of their relative locations to PUs. Suppose that the number of available channels for each sender and receiver are \(|m_{1}|\) and \(|m_{2}|\), the total number of channels is \(M\), and the number of common available channels is \(G\). The sender and receiver do not know each other's available channels, but at least one common channel is assumed to be available, otherwise rendezvous is impossible. Each SU must take total \(M\) channels for each round of its CH sequence, thus the length of one round is \(2M-1\). In the first round of the receiver CH sequence, the randomly selected one channel among \(m_{2}\) can rendezvous within \(2M-1\) time slots; however, it may not belong to \(m_{1}\). Then the rendezvous cannot be achieved. However, the receiver does not have to consider unavailable channels that are not in \(m_{2}\), and the rendezvous can be guaranteed within \(|m_{2}|\) rounds of CH sequence, therefore the upper bound of MTTR for asymmetric system is \((2M-1)(|m_{2}|+1-G)\). When the rendezvous happens in the first round of receiver's CH sequence, in other words, the receiver selects one of the commonly available channels in the first round of CH sequence, and the TTR value will be the same as the symmetric EETTR value \(ETTR_{sym}=\frac{3}{4}M+\frac{1}{4M}\). For the second-round and third-round rendezvous cases, the TTR values are \((2M-1)+ETTR_{sym}\) and \((2M-1)2+ETTR_{sym}\), respectively. Therefore, the expected TTR value for the asymmetric model can be computed as:  
\[ETTR_{sym}=\sum_{i=0}^{|m_{2}|-G}\left((2M-1)i+ETTR_{sym}\right)\cdot\frac{1 }{|m_{2}|+1-G} \tag{6}\]  
where \(\frac{1}{|m_{2}|+1-G}\) is the probability of rendezvous of each round. This can be further simplified as \(\frac{(2M-1)(|m_{2}|-G)}{2}+ETTR_{sym}\). In the enhanced FRARS system, I also implement a replacement process to increase rendezvous probability by replacing unavailable channels in the CH sequence with randomly selected available channels. In the EJS system, I investigate that the distribution of replacement channel selection is not uniformly distributed, and hence the performance of the asymmetric EJS system can be degraded. For example, suppose that \(M=\{1,2,3,4,5\}\), \(m_{1}=\{2,4\}\), and \(G=\{4\}\). Then, according to the replacement algorithm, the unavailable channels 1, 3, 5 will be replaced by channel 2, thus the rendezvous probability is not improved. In addition, it is very likely that the performance of the EJS system can be degraded under a jamming attack. To avoid this problem, I introduce a random-selection scheme for the replacement process in the enhanced FRARS system, i.e., the unavailable channels are replaced with randomly selected channels among the available channels. Using the same example as above, the unavailable channels 1, 3, 5 are replaced with randomly selected channels among \(m_{1}=\{2,4\}\). Consequently, the probability of rendezvous will be enhanced by the replacement process. While the upper bound of MTTR for asymmetric system will be the same, the \(ETTR_{asym}\) value is decreased by that random replacement scheme. Since the probability of rendezvous of each round, \(\frac{1}{|m_{2}|+1-G}\), does not depend on \(m_{1}\), the random replacement process can only affect \(ETTR_{sym}\) value in the whole \(ETTR_{asym}\) value in (6). To compute the modified \(ETTR_{sym}\), I need to consider two different rendezvous cases: case 1--endezvous on a randomly selected replaced channel; case 2--endezvous on a non-replaced channel. Let \(P_{R}\) denote the probability of case 1 and the ETTR values for each case are denoted as \(ETTR_{case1}\) and \(ETTR_{case2}\), respectively. The probability \(P_{R}\) simply represents the number of random replaced channels over the whole one round time slots \(2M-1\).