Cycle-adjustable Channel Hopping Sequence

One problem of the RRICH scheme is that the time-to-rendezvous (TTR) might be very long when the number of channels N is very large. To solve the long TTR problem for a large number of channels N, we introduce the Cycle-Adjustable Channel Hopping (CACH) scheme in this section. The key idea of CACH is to create another layer of logical channels and have SUs rendezvous on logical channels. By

choosing a modulo operation between logical channels and physical channels, CACH still achieves the maximum degree of overlapping as RRICH and thus it can still be used for solving the PU long-time blocking problem.

To reduce the TTR, we choose a much smaller prime power u for the construction of the first sub-frame in RRICH and have two SUs rendezvous on one of the u logical channels in the first time intervals. As in the construction of RRICH, we find a Galois field GF(u) with the two operations and . Then SU i chooses its parameter set , where is the initial seed and is the hopping seed. However, unlike RRICH, both the initial seed and the hopping seed are chosen in [0,u-1]. In other words, the hopping seed can be the zero element in the GF(u) used here. Each CACH sequence is a periodic sequence with period . For

, we further partition this period of time intervals into N sub-frames, each with time intervals. The last interval in a sub-frame is called the indemnity time interval. Let and be the logical channel and the physical channel used by SU i at the time interval. Suppose that , where q is the quotient of t divided by and r is its remainder. Then and

are determined by the following equation:

(3)

The construction of the sequence is the same as that in (1) except we remove the effect of q. Thus, the sequence is a periodic sequence with period and it repeats itself in every sub-frame. The index q is used in the mapping from a logical channel to a physical channel through the modulo operation in (3). As such, the physical channels used in each sub-frame are different.

Fig. 2.2 gives an example for the construction of CACH sequences for N=5 and

u=3. In this example, the addition in GF(3) is the usual addition with the MOD 3 operation and the multiplication in GF(3) is the usual multiplication with the MOD 3 operation. Since u=3, each sub-frame contains four time intervals with the last time interval in each sub-frame being the indemnity interval. For SU A with parameter set

, we have , , and

. As the last time interval in each sub-frame is the indemnity interval, . Note that the logical channel hopping sequence repeats itself in each sub-frame with the sequence 1, 0, 2, 2. The physical channel hopping sequence then adds 1 with the MOD 5 operation to 1, 0, 2, 2 in each sub-frame and that leads to 1, 0, 2, 2 for the sub-frame, 2, 1, 3, 3 for the 1^st sub-frame, 3, 2, 4, 4 for the sub-frame, 4, 3, 0, 0 for the sub-frame and 0, 4, 1, 1 for the sub-frame. Both the logical channel hopping sequence and the physical channel hopping sequence for SU B with the parameter set are also shown in Fig. 2.2.

Following the same argument as in the proof of Lemma 2.2, we have the following lemma for CACH. Note that this lemma still holds even though we allow the hopping seed to be the zero element.

Lemma 2.5. Consider two SUs with the parameter sets and .

Figure 2.2 CACH sequences for two SUs with 5 physical channels and 3 logical channels (i.e., N =5 and u=3).

(i) If they are assigned with the same hopping seed and the same initial seed, i.e., and , then they will rendezvous at each time interval.

(ii) If they are assigned with the same hopping seed ( ), but with different initial seeds and ( ), they will rendezvous at the indemnity time intervals (the last time interval in each sub-frame), i.e., , .

(iii) If they are assigned with different hopping seeds ( ), they will rendezvous at the time interval in each sub-frame, i.e., , , where

(4)

In Theorem 2.6, we show that CACH also achieves the maximum degree of overlapping as RRICH.

Theorem 2.6. Any two SUs will rendezvous at least once in each sub-frame.

Moreover, the physical channels they rendezvous in the first m sub-frames contain at least m distinct channels, .

Proof. From Lemma 2.5 (i), (ii) and (iii), it is clear that they will rendezvous at least once in each sub-frame. If they are assigned with the same hopping seed ( ), it follows from Lemma 2.5 (i) and (ii) that they will rendezvous in the sub-frame on the logical indemnity channel and thus on the physical channel It

is clear that all the elements in the set are

distinct. On the other hand, if they are assigned with different hopping seeds ( ), we have from Lemma 2.5 (iii) that they will rendezvous at the time interval in each sub-frame, where r is specified in (4). In this case, they will rendezvous on the logical channel in the sub-frame and on the physical channel

Once again, it is clear that all elements in set are distinct. ■

Analogous to proof for Corollary 2.4, one can use the results in Theorem 2.6 to show

that CACH also solves the PU long-time blocking problem.

Corollary 2.7. Suppose that there are m channels that are used by PUs.

Any two SUs will rendezvous within time intervals.

In comparison with RRICH, the worst case TTR for CACH is shorter than that of RRICH if . However, this is at the cost of increasing the system load and thus the possibility of causing the control channel congestion. To see this, recall that the load of a channel at a particular time interval is defined as the probability that an SU hops on that channel at that interval. Note that there are u choices for the initial seeds and u choices for the hopping seeds in CACH. As in RRICH, we simply assume that each SU chooses its initial seed and its hopping seed independently and uniformly over [0,u-1]. Thus, each SU is distributed uniformly to one of the u logical channels in each time interval. Thus, the probability that an SU is distributed in a logical channel (and the corresponding physical channel) is simply , which could be substantially higher than the ideal load when . As a direct consequence of Corollary 2.7, the MCTTR of CACH is which could be substantially smaller than in ACH [26] and ARCH [25]. We also note that it is difficult to create an additional layer of logical channels in the asynchronous setting, where the two SUs do not have the same indices of time intervals.

In particular, if we choose u=2, the worst case TTR for CACH is 3 and its system load is only 1/2, which is lower than 2/3 of M-QCH [18]. For this case, CACH is better than M-QCH as CACH has a lower system load while keeping the same degree of overlapping and the same worst case TTR. Now we compare CACH with L-QCH [18]. If the maximum allowable TTR is , it is shown in Theorem 2 of [18]

that the system load of any QCH system is at least . L-QCH is the QCH system with the system load . Taking , we then derive that the system load of the CACH scheme is only , which is significantly lower than in

L-QCH. In view of these, we conclude that CACH is in general much better than QCH in terms of reducing system load while keeping the same degree of overlapping and the same worst case TTR.

Certainly, choosing the number of logical channels u that optimizes the system performance, e.g., throughput, depends on various system characteristics, e.g., the number of SUs, the number of channels, and the characteristics of PUs. For this, we will perform various computer simulations in the next section. One of our findings from the simulations results is to set the number of logical channels u close to the average number of neighbors for a reasonably good throughput. As the load of CACH is 1/u, the average number of SUs that hop to a rendezvous channel is close to 1 if u is close to the average number of neighbors.

Note that setting u to be close to the average number of neighbors depends on the topology of the network and might be difficult to implement in a dynamic network. If we view the average number of SUs hopping to a rendezvous channel as the utilization of the rendezvous channel, then the key factor for achieving a good throughput is to maintain a reasonably good utilization in a rendezvous channel.

Instead of directly estimating (or tracking) the average number of neighbors, one might estimate the average number of contentions in a rendezvous channel and use that to maintain a reasonably good utilization in a rendezvous channel. Specifically, we can partition the average number of contentions in a rendezvous channel into several levels and associate each level with an appropriate u. By so doing, we can make CACH adapt to the dynamic change of the network. However, how to estimate the number of contentions in a rendezvous channel and establish the mapping between that and u require further study