Our model can be easily adapted to capture the SU activity utilizing multiple channels concurrently (channel bonding). Depending on the policy the SU follows after losing a channel, different scenarios will arise. In the scenario we will illus-trate in the following, after losing the first channel (we assume one PU per channel, but this can easily be generalized) the SU does not initiate the scanning procedure.
Rather, she continues transmitting with a lower number of channels, until she loses all of them. We assume that the maximum number of channels the SU can bond together isM. After getting all of them together, the SU will transmit at the high-est rate. Once the SU starts losing the channels one after another, the data rate will decrease accordingly. After the last available channel is lost, only then will the SU initiate the scanning procedure. In the following, we will consider the interweave mode first.
a)Interweave mode: We assume that the distribution of the PU idle periods is identical exponentially distributed with rateηH. This is in line with our assump-tions in the original single channel model. The arrival process is Poisson with rate λ. The scanning time is exponentially distributed (but now to bondM channels) with rateηL. The scanning time can be also Erlang or hyperexponential, with no major changes in the analysis. The Markov chain to mimic this system has more than two states vertically. Each state is described by the couple{j, i}, wherej is the number of files in the SU buffer, andirepresents the number of channels the SU is currently transmitting on. The 2D Markov chain is shown in Fig. 11. There are a number of possible transitions corresponding to the following events:
New arrival: From any state of any level, the chain moves to the right (horizontally) with rateλ.
File finishes transmission: From any state{j, i},(i > 0), the chain moves to the left (horizontally) with a different rate. This rate depends on the number of chan-nels the user is currently having. If the rate of a channel isµH, then withichannels having identical exponential distribution, the transition rate isiµH. So, going from the first row down, these rates areM µH,(M −1)µH, . . . , µH. Needless to say, when all the channels are lost, the SU initiates scanning and there is no transmis-sion. Hence, no transition backwards in the last row (for states{j,0}).
Losing a channel due to PU return in that channel: When such an event occurs the chain moves vertically from leveli(statej, i) to leveli−1(state{j, i−1}) (tran-sitions only to the neighboring states are possible vertically). The rate at which the chain moves vertically depends on the number of channels she is currently using.
Namely, if the SU is communicating on all theM channels, then the transition rate to levelM−1isM ηH. The reason is that we are looking at the minimum ofM exponentially distributed random variables (the first channel to lose), and the rate of such a random variable is equal to the sum of the rates of the individual random variables, which isM ηH. When transmitting onM−1channels, the chain moves downwards at rate (M −1)ηH. Finally, after being left with only one channel, the chain moves to the last level (row) with rateηH. Obviously, due to the
pol-0, M 1, M 2, M j, M
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Figure 11: The 2D Markov chain for the multichannel interweave model.
icy used that after losing a channel transmission goes on, it is not possible to have transitions vertically upwards. Only when the chain is in one of the states{j,0}a transition upwards is possible. Namely, then the scanning forM channels starts, and the chain moves to the first level after bondingM channels together. We as-sume that this happens according to an exponential distribution with rateηL. This model can also capture, with a slight modification, the Erlang and hyperexponen-tially distributed scanning times too.
Having introduced all the possible transitions with the corresponding transition rates, we can proceed to solve the Markov chain. Its solution will lead to the average file delay in the multichannel interweave mode.
If we look carefully at the chain in Fig. 11, we can infer that this chain is actually a special case of the more general chain we have already solved. That is the chain corresponding to the M different power levels in the underlay mode (Fig. 10).
Our chain is a special case of the chain in Fig. 10 with the following rates:
µM = M µH, µM−1 = (M −1)µH, . . . , µ1 = 0, η1,M =ηL, ηi,i−1 = iηH, and all the otherηi,j = 0.
b)Underlay mode: The Markov chain for the multichannel underlay model is slightly different compared to the interweave model (Fig. 11). There is no scanning time involved here. The rate at which the PU leaves the channel, in line with the single channel case, is exponentially distributed with rateηL. The SU uses always the same set of M channels. Once a channel is lost due to PU return, the rate contribution from that channel isµLand(M −1)µH from the idle channels. The 2D Markov chain is shown in Fig. 12. There are a number of possible transitions corresponding to the following events:
New arrival: From any state of any level, the chain moves to the right (horizontally) with rateλ.
File finishes transmission: From any state{j, i},(i > 0), the chain moves to the left (horizontally) with a different rate. This rate depends on the number of chan-nels the user is currently having idle and busy. If the rate of a channel isµH, then withiidle channels the transition rate isiµH + (M −i)µL. The reason is that if there areiidle channels, the SU will transmit at full power on those, and on the remainingM−ichannels that are busy, she can transmit at the lower rate ofµL. Hence, the total rate in this case isiµH + (M −1)µL.
Losing (getting) a channel due to PU return (departure) in (from) that channel: Since we are dealing with the underlay mode, there is no scanning of other channels in-volved. The SU is stuck with a given set ofM channels. When an event of losing a channel occurs the chain moves vertically from leveli(state{j, i}) to leveli−1 (state{j, i−1}) (transitions only to the neighboring states are possible vertically).
The rate at which the chain moves vertically depends on the number of channels she is currently using. Namely, if the SU is communicating on all theMchannels, the transition to levelM−1isM ηH. The reason is that we are looking at the min-imum ofM exponentially distributed random variables (the first channel to lose), and the rate of such a random variable is equal to the sum of rates of the
individ-0, M 1, M 2, M j, M
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Figure 12: The 2D Markov chain for the multichannel underlay model.
ual random variables, which isM ηH. When transmitting onM −1channels, the chain moves downwards with rate(M −1)ηH. Finally, after being left with only one channel, the chain moves to the last level (row) with rateηH.
In case a channel becomes idle while being at state {j, i}, the chain moves vertically upwards to state{j, i+ 1}. The rate at which the chain moves vertically depends on the number of channels that are occupied by the corresponding PUs.
For instance, if 5 channels are busy, then the rate at which the first one leaves is exponentially distributed with rate5ηL(the minimum of 5 identical exponentially distributed variables). So, if the chain is in state{j, i}, the transition rate to state {j, i+ 1}isiηL. Having introduced all possible transitions with the corresponding transition rates, we can proceed solving this Markov chain. Its solution will lead to the average file delay in the multichannel underlay mode.
This chain is also a special case of the chain of Fig. 10, with the following rates:µM =M µH, µM−1 = (M−1)µH+µL, . . . , M µL, ηi,i−1 =iηH, ηi,i+1 = (M−i)ηL. The solution to this chain is then similar to that of Fig. 10.
We would like to point out again that depending on the scanning policy, dif-ferent chains will arise. Nevertheless, most of them will be special cases of the M-level model.