### Throughput for conventional primary system without relay

For a conventional system based on ARQ mechanism, we consider the scenario where the packet can be transmitted at most by two transmission slots, i.e., only one retransmission. We have the following mutually exclusive events for the transmission of a packet:

Event 1: *C*_{1} = {primary packet is successfully received at PR in the first transmission slot}.

Event 2: *C*_{2} = {primary packet is not successfully received at PR in the first transmission slot, but the retransmission of the same packet from PT is successfully received at PR}.

Event 3: *C*_{3} = {primary packet transmission fails after two transmission slots}.

From (2), the probability of the above events are respectively given by

\begin{array}{ll}Pr\left\{{C}_{1}\right\}& =1-{O}_{0},\phantom{\rule{2em}{0ex}}\end{array}

(13)

\begin{array}{ll}Pr\left\{{C}_{2}\right\}& ={O}_{0}(1-{O}_{0}),\phantom{\rule{2em}{0ex}}\end{array}

(14)

\begin{array}{ll}Pr\left\{{C}_{3}\right\}& ={O}_{0}^{2}.\phantom{\rule{2em}{0ex}}\end{array}

(15)

Thus, the throughput {\eta}_{{\mathrm{P}}_{\text{bits}}} is denoted as

{\eta}_{{\mathrm{P}}_{\text{bits}}}={R}_{\mathrm{P}}{\eta}_{\mathrm{P}},

(16)

We define *η*_{P} as the average number of packets successfully delivered per time slot. That is the ratio between how much of each packet that can be successfully delivered and the number of time slots required.

{\eta}_{\mathrm{P}}=\frac{Pr\left\{{C}_{1}\right\}+Pr\left\{{C}_{2}\right\}}{Pr\left\{{C}_{1}\right\}+2Pr\left\{{C}_{2}\right\}+2Pr\left\{{C}_{3}\right\}}.

(17)

The nominator denotes on average how much of each packet can be successfully delivered, and the denominator denotes the average time slots that are consumed. Therefore, *η*_{P} is the actual throughput normalized by the bit rate. In this paper, we focus on analyzing the normalized throughput for convenience. Thus, the throughput for the conventional primary system is denoted as

{\eta}_{\mathrm{P}}=1-{O}_{0}.

(18)

### Throughput for the primary system in cooperation mode

In cooperation mode, the secondary system is always ready to serve as a relay for the primary system, for which we have the following mutually exclusive events:

Event 1: *E*_{1} = {primary packet is successfully received at PR in the first transmission slot}.

Event 2: *E*_{2} = {primary packet is not successfully received at PR and ST{}_{i},\forall i\in \mathcal{M} in the first transmission slot, PT retransmits the same packet in the subsequent slot and it is successfully received at PR}.

Event 3: *E*_{3} = {primary packet is not successfully received at PR, but successfully received at one or

more ST{}_{i},i\in \mathcal{M}; thus, one ST serves as relay to forward this packet to PR in the subsequent slot and it is successfully received at PR}.

Event 4: *E*_{4} = {primary packet transmission fails after two transmission slots}.

From (2) and (4), the probability for *E*_{1}, *E*_{2}, and *E*_{3} are given by

\begin{array}{ll}Pr\left\{{E}_{1}\right\}& =1-{O}_{0},\phantom{\rule{2em}{0ex}}\end{array}

(19)

\begin{array}{ll}Pr\left\{{E}_{2}\right\}& ={O}_{0}{{O}_{1}}^{M}(1-{O}_{0}),\phantom{\rule{2em}{0ex}}\end{array}

(20)

\begin{array}{ll}Pr\left\{{E}_{3}\right\}& ={O}_{0}{P}^{\text{SR}},\phantom{\rule{2em}{0ex}}\end{array}

(21)

where the superscript SR denotes successful relaying, and *P*^{SR} represents the joint probability that at least one of the ST{}_{i},i\in \mathcal{M} successfully receives the primary packet and successfully relays the packet to PR. In other words, *P*^{SR} indicates the probability that *k* ≥ 1 secondary transmitters can successfully receive the primary packet from PT. Among which, *f* ≥ 1 secondary transmitters can successfully relay the primary packet to PR.

We denote \mathcal{B}=\left\{{\text{ST}}_{i}\right|i\in \mathcal{M},{R}_{1i}>{R}_{\mathrm{p}},{R}_{2i}>{R}_{\mathrm{p}}\}, where \mathcal{B} is the set of secondary transmitters ST{}_{i},i\in \mathcal{M} successfully receiving the primary packet and are able to successfully forward the primary packet to PR in the subsequent slot. Hence, \mathcal{B}\subseteq \mathcal{A}. When the first transmission from PT to PR fails, the secondary system selects ST{}_{p},p\in \mathcal{A} with the best channel gain p=\underset{i\in \mathcal{A}}{arg\; max}\left[{\gamma}_{2i}\right] to relay the primary packet. The joint probability *P*^{SR} can be expressed as

\begin{array}{ll}{P}^{\text{SR}}& =\sum _{k=1}^{M}\sum _{f=1}^{k}Pr\{\left|\mathcal{A}\right|=k,\left|\mathcal{B}\right|=f\}\\ =\sum _{k=1}^{M}\sum _{f=1}^{k}Pr\{\left|\mathcal{B}\right|=f\left|\right.\left|\mathcal{A}\right|=k\}Pr\{\left|\mathcal{A}\right|=k\},\end{array}

(22)

and

\begin{array}{l}Pr\{\left|\mathcal{B}\right|=f\left|\right.\left|\mathcal{A}\right|=k\}=\left(\begin{array}{c}k\\ f\end{array}\right){(1-{O}_{2})}^{f}\underset{2}{\overset{(k-f)}{O}},\phantom{\rule{2em}{0ex}}\end{array}

(23)

\begin{array}{l}Pr\{\left|\mathcal{A}\right|=k\}=\left(\begin{array}{c}M\\ k\end{array}\right){(1-{O}_{1})}^{k}\underset{1}{\overset{(M-k)}{O}}.\phantom{\rule{2em}{0ex}}\end{array}

(24)

Then *P*^{SR} is derived as

\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}{P}^{\text{SR}}=\sum _{k=1}^{M}\sum _{f=1}^{k}\left(\begin{array}{c}M\\ k\end{array}\right)\left(\begin{array}{c}k\\ f\end{array}\right){(1-{O}_{2})}^{f}{O}_{2}^{(k-f)}{(1-{O}_{1})}^{k}{O}_{1}^{(M-k)}.

(25)

Thus, we have

\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}Pr\left\{{E}_{3}\right\}\phantom{\rule{0.3em}{0ex}}=\phantom{\rule{0.3em}{0ex}}{O}_{0}\phantom{\rule{0.3em}{0ex}}\sum _{k=1}^{M}\phantom{\rule{0.3em}{0ex}}\sum _{f=1}^{k}\phantom{\rule{0.3em}{0ex}}\left(\begin{array}{c}M\\ k\end{array}\right)\phantom{\rule{0.3em}{0ex}}\left(\begin{array}{c}k\\ f\end{array}\right){(1\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}{O}_{2})}^{f}{O}_{2}^{(k-f)}\phantom{\rule{0.3em}{0ex}}{(1\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}{O}_{1})}^{k}{O}_{1}^{(M-k)}.

(26)

*E*_{4} represents the outage occurrence, and its probability is given by

Pr\left\{{E}_{4}\right\}={O}_{0}{{O}_{1}}_{0}^{\mathit{\text{MO}}}+{O}_{0}{P}^{\text{FR}},

(27)

where the first term {O}_{0}{{O}_{1}}_{0}^{\mathit{\text{MO}}} denotes the first event that the primary packet fails in the first transmission from PT to PR and all the secondary transmitters ST{}_{i},\forall i\in \mathcal{M} also fail to decode the primary packet, and the retransmission of the packet from PT to PR fails again. The second term *O*_{0}*P*^{FR} denotes the event that the primary packet fails at the first transmission from PT to PR link and the superscript FR denotes failed relaying. So *P*^{FR} indicates the joint probability that *k* ≥ 1 secondary transmitters can successfully receive the primary packet from PT. Among which, zero secondary transmitters can successfully relay the primary packet to PR. However, at least one of the secondary transmitter ST{}_{i},i\in \mathcal{M} successfully decodes the primary packet but fails to relay it to PR, i.e., \mathcal{A}\notin \varnothing and \mathcal{B}\in \varnothing. Thus, *P*^{FR} is also the joint probability that denotes the secondary transmitters successfully receiving the primary packet but failing to relay, for which we have

\begin{array}{ll}{P}^{\text{FR}}& =\sum _{k=1}^{M}Pr\{\left|\mathcal{A}\right|=k,\left|\mathcal{B}\right|=0\}\\ =\sum _{k=1}^{M}Pr\{\left|\mathcal{B}\right|=0\left|\right.\left|\mathcal{A}\right|=k\}Pr\{\left|\mathcal{A}\right|=k\}.\end{array}

(28)

By substituting *f* = 0 into (25), we have from (28)

{P}^{\text{FR}}=\sum _{k=1}^{M}\left(\begin{array}{c}M\\ k\end{array}\right){(1-{O}_{1})}^{k}{O}_{1}^{(M-k)}{{O}_{2}}^{k}.

(29)

Therefore,

\phantom{\rule{-14.0pt}{0ex}}Pr\left\{{E}_{4}\right\}={O}_{0}{{O}_{1}}^{\phantom{\rule{0.3em}{0ex}}M}{O}_{0}+{O}_{0}\sum _{k=1}^{M}\left(\begin{array}{c}M\\ k\end{array}\right){(1-{O}_{1})}^{k}{O}_{1}^{(M-k)}{{O}_{2}}^{k}.

(30)

The throughput for primary system under cooperation mode is thus

\begin{array}{ll}{\eta}_{{\mathrm{C}}_{\mathrm{P}}}& =\frac{Pr\left\{{E}_{1}\right\}+Pr\left\{{E}_{2}\right\}+Pr\left\{{E}_{3}\right\}}{Pr\left\{{E}_{1}\right\}+2Pr\left\{{E}_{2}\right\}+2Pr\left\{{E}_{3}\right\}+2Pr\left\{{E}_{4}\right\}}\\ =(1-{O}_{0})+\frac{{O}_{0}\left[{O}_{0}{P}^{\text{SR}}-{P}^{\text{FR}}(1-{O}_{0})\right]}{1+{O}_{0}}.\end{array}

(31)

#### Lemma 1

An upper bound of {\eta}_{{\mathrm{C}}_{\mathrm{P}}} is given by

\begin{array}{ll}{\eta}_{{\mathrm{C}}_{\mathrm{P}}}^{\ast}& =\underset{M\to \infty}{\text{lim}}{\eta}_{{\mathrm{C}}_{\mathrm{P}}}\\ =(1-{O}_{0})+\frac{{O}_{0}^{2}}{1+{O}_{0}}.\end{array}

(32)

#### Proof

Since *P*^{SR} + *P*^{FR} represents the probability that at least one of the secondary transmitters successfully decodes the primary packet,

{P}^{\text{SR}}+{P}^{\text{FR}}=1-{{O}_{1}}^{M}.

(33)

From the binomial theorem, *P*^{FR} in (29) can be expressed as

\phantom{\rule{-15.0pt}{0ex}}\begin{array}{ll}{P}^{\text{FR}}\phantom{\rule{0.3em}{0ex}}& ={\left[(1-{O}_{1}){O}_{2}+{O}_{1}\right]}^{M}-{O}_{1}^{M}=({O}_{2}-{O}_{1}{O}_{2})\\ \phantom{\rule{1em}{0ex}}\times \left[\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}{\left({O}_{2}\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}{O}_{1}{O}_{2}\phantom{\rule{0.3em}{0ex}}+\phantom{\rule{0.3em}{0ex}}{O}_{1}\right)}^{M-1}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}+\phantom{\rule{0.3em}{0ex}}{\left({O}_{2}\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}{O}_{1}{O}_{2}\phantom{\rule{0.3em}{0ex}}+\phantom{\rule{0.3em}{0ex}}{O}_{1}\right)}^{M-2}{O}_{1}\right.\\ \phantom{\rule{1em}{0ex}}+{\left({O}_{2}\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}{O}_{1}{O}_{2}\phantom{\rule{0.3em}{0ex}}+\phantom{\rule{0.3em}{0ex}}{O}_{1}\right)}^{M-3}\phantom{\rule{0.3em}{0ex}}{O}_{1}^{2}\phantom{\rule{0.3em}{0ex}}+\phantom{\rule{0.3em}{0ex}}\dots \phantom{\rule{0.3em}{0ex}}+\phantom{\rule{0.3em}{0ex}}\left({O}_{2}\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}{O}_{1}{O}_{2}\phantom{\rule{0.3em}{0ex}}+\phantom{\rule{0.3em}{0ex}}{O}_{1}\right)\\ \phantom{\rule{1em}{0ex}}\times \left(\right)close="]">{O}_{1}^{M-2}+{O}_{1}^{M-1}& ,\end{array}\n

(34)

and

{P}^{\text{SR}}=1-{\left[\left(1-{O}_{1}\right){O}_{2}+{O}_{1}\right]}^{M}.

(35)

It is clear that *P*^{SR} → 1 and *P*^{FR} → 0 when *M* → *∞*; hence, the upper bound is obtained. □

### Throughput for primary system in access mode

In access mode, SR transmits an INF signal to corrupt the primary ACK/NACK of the first transmission, and the failure to receive the ACK/NACK is treated as an implicit NACK. Since the secondary transmitter utilizes the retransmission of primary packets to achieve spectrum access, the retransmission of the primary packets will fail due to the strong interference from secondary access. Thus, whether the primary packets can be successfully transmitted or not in access mode is determined by the first transmission slot. There are two mutually exclusive events for primary system in access mode as shown in Figure 2:

Event 1: *A*_{1} = {the primary packet is successfully received at PR in the first transmission, but the corresponding ACK is corrupted by SR; PT then retransmits this packet and an ACK is received from PR}.

Event 2: *A*_{2} = {the primary packet is not successfully received at PR in the first transmission and the corresponding NACK is corrupted by SR; PT then retransmits this packet but still fails due to the strong interference from the secondary access and a NACK is received at PR, i.e., a packet loss}.

Thus, the probability of the above two events are respectively given by

\begin{array}{l}Pr\left\{{A}_{1}\right\}=1-{O}_{0}.\end{array}

(36)

\begin{array}{l}Pr\left\{{A}_{2}\right\}={O}_{0}.\end{array}

(37)

The throughput for primary system under access mode thus given by

\begin{array}{ll}{\eta}_{{A}_{\mathrm{P}}}& =\frac{Pr\left\{{A}_{1}\right\}}{2Pr\left\{{A}_{1}\right\}+2Pr\left\{{A}_{2}\right\}}\\ =\frac{1-{O}_{0}}{2}.\end{array}

(38)

### Credits and penalties

From the above analysis, it is apparent that the throughput for the primary system decreases severely under access mode, but improves significantly under cooperation mode. The performance loss in access mode incurred by the secondary system is regarded as penalties and the performance gain collected by the secondary system in cooperation mode is regarded as credits. Thus we propose an ARQ-based spectrum sharing protocol to regulate the credits and the penalties such that secondary spectrum access can be achieved without degrading the primary performance.

In cooperation mode, ‘credits’ is defined as

{\mathrm{\Gamma}}_{\mathrm{C}}\stackrel{\Delta}{=}{\eta}_{{\mathrm{C}}_{\mathrm{P}}}-{\eta}_{\mathrm{P}}\phantom{\rule{1em}{0ex}}(\mathrm{\text{credits}}/\mathrm{\text{slot}})

(39)

In access mode, ‘penalties’ is denoted as

{\mathrm{\Gamma}}_{\mathrm{P}}\stackrel{\Delta}{=}{\eta}_{\mathrm{P}}-{\eta}_{{\mathrm{A}}_{\mathrm{P}}}\phantom{\rule{1em}{0ex}}(\mathrm{\text{penalties}}/\mathrm{\text{slot}})

(40)

It is apparent that the throughput for the conventional system *η*_{P} is taken as a benchmark in the definition of credits and penalties. For ease of exposition, we define the ratio between penalties and credits as

{\stackrel{\u0304}{N}}_{\mathrm{C}}=\frac{{\mathrm{\Gamma}}_{\mathrm{P}}}{{\mathrm{\Gamma}}_{\mathrm{C}}}.

(41)

To achieve spectrum sharing without degrading the performance of the primary system, every {\stackrel{\u0304}{N}}_{\mathrm{C}} (or l{\stackrel{\u0304}{N}}_{\mathrm{C}}) complete transmission for a packet in cooperation mode allows one (or *l*) access opportunity for the secondary system, where l\in {\mathbb{Z}}^{+}. Thus, the overall average throughput for primary system in the proposed spectrum sharing protocol is given by

{\eta}_{{\mathrm{T}}_{\mathrm{P}}}={\eta}_{{\mathrm{C}}_{\mathrm{P}}}\frac{{\stackrel{\u0304}{N}}_{\mathrm{C}}}{{\stackrel{\u0304}{N}}_{\mathrm{C}}+1}+{\eta}_{{A}_{\mathrm{P}}}\frac{1}{{\stackrel{\u0304}{N}}_{\mathrm{C}}+1}.

(42)

### Throughput for secondary system in access mode

For the secondary system, if the primary packet is successfully received at SR in the first transmission, the interference from PT’s retransmission can be perfectly canceled out at SR. Otherwise, we assume that the secondary packet fails to be received at SR due to the interference from the primary retransmission. Thus in access mode, there are three mutually exclusive events for the secondary system:

Event 1: *S*_{1} = {the primary packet is successfully received at SR in the first transmission slot, and the secondary packet is also successfully received at SR in the second transmission slot}

Event 2: *S*_{2} = {the primary packet is successfully received at SR in the first transmission slot, but the secondary packet fails to be received at SR in the second transmission slot}

Event 3: *S*_{3} = {the primary packet failed to be received at SR in the first transmission slot}

Thus, the probability of the above three events can be derived as

\begin{array}{l}Pr\left\{{S}_{1}\right\}=(1-{O}_{4})\left(1-{{O}_{3}}^{M}\right),\end{array}

(43)

\begin{array}{l}\phantom{\rule{4em}{0ex}}Pr\left\{{S}_{2}\right\}=(1-{O}_{4}){{O}_{3}}^{M},\end{array}

(44)

\begin{array}{l}\phantom{\rule{8em}{0ex}}Pr\left\{{S}_{3}\right\}={O}_{4}.\end{array}

(45)

The throughput for the secondary system under access mode is thus given by

\begin{array}{ll}{\eta}_{\mathrm{S}}& =\frac{Pr\left\{{S}_{1}\right\}}{2Pr\left\{{S}_{1}\right\}+2Pr\left\{{S}_{2}\right\}+2Pr\left\{{S}_{3}\right\}}\\ =\frac{(1-{O}_{4})\left(1-{{O}_{3}}^{M}\right)}{2}.\end{array}

(46)

#### Lemma 2

An upper bound of the secondary throughput under access mode with an increasing *M* is given by

\begin{array}{ll}{\eta}_{\mathrm{S}}^{\ast}& =\underset{M\to \infty}{\text{lim}}{\eta}_{\mathrm{S}}\\ =\frac{1-{O}_{4}}{2}.\end{array}

(47)

#### Proof

It is clear that {{O}_{3}}^{M}\to 0 when *M* → *∞*; hence, the upper bound is obtained. □

Through the proposed credit system, the secondary system switches between cooperation and access modes, according to the accumulated credits, continuously. Thus, the overall average throughput for secondary system in the proposed spectrum sharing protocol is given by

{\eta}_{{\mathrm{T}}_{\mathrm{S}}}=\frac{1}{{\stackrel{\u0304}{N}}_{\mathrm{C}}+1}{\eta}_{\mathrm{S}}.

(48)

#### Lemma 3

An upper bound of the overall average throughput for secondary system with an increasing *M* is given by

\begin{array}{ll}{\eta}_{{\mathrm{T}}_{\mathrm{S}}}^{\ast}& =\underset{M\to \infty}{\text{lim}}{\eta}_{{\mathrm{T}}_{\mathrm{S}}}\\ =\frac{1-{O}_{4}}{2({\stackrel{\u0304}{N}}_{\mathrm{C}}+1)}.\end{array}

(49)