On the Design of Relay--Assisted Primary--Secondary Networks

0.44 MB·

by  Ahmed El Shafie

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On the Design of Relay–Assisted
Primary–Secondary Networks
AhmedEl Shafie,Member, IEEE, Tamer Khattab,Member, IEEE,Ahmed Sultan, Member, IEEE and H.
VincentPoor, Fellow,IEEE
✦
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Abstract—The use of N cognitive relays to assist primary and sec- channel only when the source nodes are idle, i.e. not
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ondarytransmissionsinatime-slottedcognitivesettingwithoneprimary utilizing the spectrum.
y user(PU)andonesecondaryuser(SU)isinvestigated.Anoverlapped
a spectrumsensingstrategyisproposedforchannelsensing,wherethe
M [4]considersacognitiverelaythataidsmultiplenodes
SU senses the channel for τ seconds from the beginning of the time
in transmitting their data to a common receiver. The
slot and the cognitive relays sense the channel for 2τ seconds from
0 the beginning of the time slot, thus providing the SU with an intrinsic proposed protocol exploits source burstiness to enable
2 priorityovertherelays.Therelayssensethechannelovertheinterval cooperation during silence periods of different nodes
[0,τ] to detect primary activity and over the interval [τ,2τ] to detect in a time-division multiple access (TDMA) network. In
T] secondaryactivity.TherelayshelpboththePUandSUtodelivertheir [5], Krikidis et al. proposed to deploy a dumb relay
undeliveredpacketsandtransmitwhenbothareidle.Twooptimization-
I node in cognitive radio networks to improve network
. basedformulationswithqualityofserviceconstraintsinvolvingqueueing
s spectrumefficiency.Therelayaidsboththeprimaryand
delayarestudied.Bothcasesofperfectandimperfectspectrumsensing
c the secondary users. The proposed protocol is analyzed
are investigated. These results show the benefits of relaying and its
[
abilitytoenhancebothprimaryandsecondaryperformance,especially and optimized for a network model consisting of a

2 in the case of no direct link between the PU and the SU transmitters pair of primary users (PUs) and a pair of secondary
v andtheirrespectivereceivers.Threepacketdecodingstrategiesatthe users (SUs). In [6], multiple relays serve multiple PUs
8 relaysarealsoinvestigatedandtheirperformanceiscompared.
during their silence periods. A number of secondary
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links coexist with the system and two secondary access
1 IndexTerms—CognitiveRadio,QueueingDelay,Relaying,Cooperative
3 Communications,StabilityAnalysis. scenarios are investigated. Under the first, the SUs are
. able to sense the activity of both the PUsand the relays,
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0 1 INTRODUCTION thereby remaining silent when any of them is active. In
4 the second scenario, the relays and the SUs randomly
1 In the quest for efficient usage of radio spectrum, high access the channel and their transmissions may collide.
: reliability and high speed wireless transmission, cogni-
v
i tive radio and cooperative communications emerge as Relays with buffers are also considered in [7]–[12].
X
two of the most promising technologies. In coopera- The max-max relay selection policy is considered in [9].
ar tive communications [1]–[3], a portion of the channel Buffered relays enable the selection of the relay with
resources are assigned to one or more relays for co- the best source-relay channel for reception and the best
operation. These relays cooperate with a source node relay-destination channel for transmission. The scheme
to help in forwarding its data to a destination. This relies on a two-slot protocol where the schedule for
enhancescommunicationreliability,reducestherequired the source and relay transmission is fixed a priori. This
transmitted power and achieves spatial diversity. The limitation is relaxed in [10] where each slot is allocated
use of relays may, however, result in some bandwidth dynamically to the source or relay transmission accord-
efficiencyloss becauseofthe channel resourcesassigned ingtotheinstantaneousqualityofthelinksandthestate
to the relays to perform their task. A cognitive relay is of the buffers. In [11] and [12], the authors considered
a viable solution to this problemasthe relayutilizes the two-hop communication, where the SU exploits periods
of silence of the PU to transmit its packets to a set of
A. El Shafie is with Wireless Intelligent Networks Center (WINC), Nile
relays. Moreover, the relays can transmit even when the
University, Giza, Egypt. He is also with Electrical Engineering, Qatar
University,Doha,Qatar(e-mail:ahmed.salahelshafi[email protected]). PU is busy because they can act together and create a
T. Khattab is with Electrical Engineering, Qatar University, Doha, Qatar beamformer to suppress or even null the interference at
(email:[email protected]).
the primary receiver. The instantaneous channel gains
A.SultaniswithElectricalEngineering,AlexandriaUniversity,Alexandria,
Egypt(email:[email protected]). are assumed to be known at the relay stations.
H. V. Poor is with the Department of Electrical Engineering, Princeton
University,Princeton,NJ08544USA(email:[email protected]).
In this work, we consider buffered relays with cog-
This researchwork is supportedbyQatar National ResearchFund(QNRF)
undergrantnumberNPRP09-1168-2-455. nitive capabilities. The relays serve two users with dif-
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ferent priorities: a PU and an SU.1 The relays accept 4. We provide some numerical results in Section 5 and
a fraction of the undelivered primary and secondary conclude the paper in Section 6.
packets into their buffers and forward these packets
to the primary and secondary destinations. We do not
2 SYSTEM MODEL
assume instantaneous channel knowledge and, hence,
our protocol does not involve relay selection on the Thenetworkconsists ofone primarytransmitter‘p’, one
basis of instantaneous channel quality. We propose a secondary transmitter ‘s’, one primary destination ‘pd,
particularoverlappedspectrumsensingschemeinorder one secondary destination ‘sd’, and a set of N relays
to regulate the operation of the PU, the SU and the labeled as 1,2,3,...,N as shown in Fig. 1. The relays
relays. arehalf-duplex,whichmeansthattheyeithertransmitor
We can summarize the contributions in this paper receivebutcannotdobothatthesametime.Weconsider
as follows. We consider one PU and one SU in the a wireless collision channel model where concurrent
presence of N cognitive relays. The relays are used to transmissions by two or more nodes are assumed to be
help both the PU and the SU in communicating their lost. Eachofthe PUandthe SUhasaninfinite bufferfor
data packets to their respective receivers. We propose a storing fixed-length packets. Each terminal operates as
novel overlapped spectrum sensing technique to coor- a discrete-time Geo/Geo/1 queue [13].2 The arrivals at
dinate channel access. More specifically, the SU senses the primaryand secondaryqueues areindependentand
the channel for τ seconds from the beginning of the identicallydistributed(i.i.d.)Bernoullirandomvariables
time slot to detect possible activity of the PU, while from slot to slot with means λp ∈ [0,1] and λs ∈ [0,1]
the relays sense the channel for 2τ seconds from the packets per time slot, respectively. Arrival processes
beginning ofthe timeslot. Eachrelaysensesthe channel at the primary and secondary buffers are statistically
over the interval [0,τ] to detect possible activity of the independent of one another. Each relayhas two queues:
PU and over the interval [τ,2τ] to detect the activity a queue for relaying the primary packets denoted by
of the SU. The SU transmits a packet from its queue Qp,k, and a queue for relaying the secondary packets
if the PU is sensed to be idle. For the relays to transmit, denoted by Qs,k, where k ∈ {1,2,...,N}. The relays
they must sense both the PU and the SU to be inactive. help both the PU and the SU to deliver their packets
We investigate three strategies for the decoding of the in the periods of silence of both of them. If a terminal
primary and the secondary transmissions at the relays. transmits during a time slot, it sends exactly one packet
We propose an ordered acceptance strategy, denoted to its respective receiver.
by S , in which the relays are ordered in terms of We propose an overlapped spectrum sensing scheme
OD
accepting the undelivered packets of the PU and the asdepictedinFig.2.TheSUsensesthechannelfromthe
SU into their queues. To simplify the decoding process, beginningofthetime slotuptoτ secondsrelativetothe
we propose random assignment decoding, denoted by beginning of the time slot, while all the relays sense the
S , and round robin decoding, denoted by S , in channelover the interval[0,τ] to detectprimaryactivity
RD RR
which each relay is assigned to the decoding role for and over the interval [τ,2τ] to detect secondary activity.
a fraction of the time slots. We study the optimal sec- If the channel is sensed to be free over both intervals,
ondaryaverageservicerategivencertainaveragearrival thenalltherelaysremainidleduringtherestofthetime
rates to the primary and the secondary queues. Also, slot except the relay that is scheduled for transmission
we investigate the minimum number of relays needed provided that its queues are nonempty. If either the PU
to achieve a specific level of quality of service (QoS) or the SU is sensed to be active, then the relays may
for the users. We study the case of sensing errors at the switchtothereceivingmodedependingonthedecoding
relays’ spectrum sensors. In contrast with many works strategy as explained later in Subsection 2.1. There is a
involving automatic repeat request (ARQ) feedback, we feedbackphaseattheendofthetime slottoindicatethe
take into account the cost of the feedback duration, status of packet delivery.
whichisathroughputlossasthetimeallowedforactual
data transmission is reduced. Finally, in Appendix A,
2.1 Mediumaccesscontrol(MAC)Layer
we provide a proof of the advantage of S over S
OD RD
The PU transmits the packet at the head of its queue
and S , in terms of the service rates, for the case of a
RR
starting at the beginning of the time slot. If the PU is
negligible feedback duration per relay.
sensed to be idle by the SU, the SU transmits the packet
Therestofthepaperisorganizedasfollows.InSection
at the head of its queue after τ seconds. A relay with
2, we describe the system model adopted in this paper.
nonempty queues transmits during a time slot after 2τ
The problem formulations are presented in Section 3.
seconds if it is scheduled to transmit and it senses the
ThesystemwithsensingerrorsisinvestigatedinSection
PU and the SU to be idle. The probability that relay
k is scheduled to transmit during a time slot is ω .
1.Theproposedcognitivecooperationprotocolsandthetheoretical k
development in this paper can be generalized to cognitive radio
networks with more PUs and more SUs, in which the PUs and the 2.Thenotionofdiscrete-timeGeo/Geo/1queueisusedtodescribe
SUsareoperatingunderTDMAorfrequency-divisionmultipleaccess aqueueingsystemwithaBernoulliarrivalprocessandgeometrically
(FDMA). distributedservicetimes.
3
Secondary user starts
transmission after Random decoding and
PD se coisn ddesc ilfa trheed cidhlaen.n(cid:2028)el round robin decoding (cid:3)
(cid:2028)(cid:2028)(cid:2028)(cid:2028)(cid:3033)(cid:3033) (cid:2028)(cid:2028)(cid:2028)(cid:2028)(cid:3033)(cid:3033)(cid:3033)
Time slot

PU (cid:4666)(cid:1846)(cid:4667)
(cid:884)(cid:2028)(cid:3033)

Secondary user (cid:2028) Rela(cid:2028)ys start transmmiissssiioonnFF eeee dddbbuaarcca(cid:1846)kkt (cid:2890)ipporrnoocceessssss (cid:2028) (cid:3033) (cid:2028)(cid:3033) (cid:2028)(cid:2028)(cid:2028)(cid:2028) (cid:3033)(cid:3033) (cid:2028)(cid:3033)
and relays start after 2 seconds if the
sensing process. channel is declared idle.
(cid:2028) Or(cid:4666)(cid:4666)d(cid:1840)(cid:1840)ere(cid:3397)(cid:3397)d(cid:883)(cid:883) a(cid:4667)(cid:4667)c(cid:2028)(cid:2028)(cid:2028)(cid:2028)c(cid:3033)(cid:3033)eptance
k
threlay
Fig. 2. Time slot structure.The explanation for the feed-
SU
back process and its duration is provided at the end of
Section2.
destination (SD) acknowledges the correct reception of
the transmitted packet by sending an acknowledgment
SD
(ACK) message, the relays discard what they have re-
ceived from the PU or the SU. If the PD or SD declares
Fig. 1. System model: N relays assist primary and itsfailuretodecodethereceivedpacketcorrectlybygen-
secondary transmissions from the PU and the SU to the erating a negative acknowledgment (NACK) message,
primarydestination(PD)andsecondarydestination(SD), the relays attempt to decode the received packet and
respectively. determine its origin. If the received packet is correctly
decoded and, hence, its origin is identified by the first-
ranked relay, it decides whether to accept the packet.
This means that over a large number of time slots relay If the packet is admitted, an ACK is transmitted by
k is assigned to transmit during a fraction ωk of the the relay to inform the PU or SU to drop the packet
N
total time slots. It is clear that ω = 1. We define from its queue and to notify the other relays that the
k=1 k
a vector ω = [ω1,ω2,...,ωN] Pto indicate the fraction packet has already been accepted. If the first-ranked
of time slots allocated to each relay for transmission. relay receives the packet in error or does not accept it,
If relay k is scheduled for transmission, which occurs it remains silent and the second-ranked relay makes the
with probability ωk, it chooses a packet from Qp,k with acceptancedecisionincaseithasreceivedthepacketcor-
probabilityα andfromQ withprobability1−α .We rectly.Generally,arelay,dependingonitsdecodingrank,
k s,k k
define the vector α=[α1,α2,...,αN]. decides whether to accept a correctly decoded packet
If a relay receives during a time slot, it distinguishes provided that all the preceding relays do not admit the
between the primary and the secondary transmissions packet. We assume perfect decoding of the feedback
through an identifier contained in each transmitted messages at all nodes. This assumption is reasonable
packet.3 If a relay correctly receives a packet, it decides whenstrongchannelcodeswithlowmodulationindices
to accept it with a certain probability. The acceptance are employed for the feedback channel [4].
probability vector of the undelivered primary packets The relays’ acceptance order is the N-tuple m =
n
is fp, where the element fp,k is the probability that (m1,m2,...,mi,...,mN), where mi ∈ {1,2,...,N} and
relay k admits a correctly received primary packet to m 6= m ,∀i,k : i 6= k. The N-tuple (m ,m ,...,m )
i k 1 2 N
Qp,k. Similarly, the vector fs has N elements with fs,k meansthatrelay1isassignedthem1thacceptancerank,
being the probability of admitting a correctly received relay2isassigned them thrankandsoon.Itisevident
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secondary packet to Qs,k. that mn is a permutation, Πn, over the set {1,2,...,N}
and there are N! such permutations, where N! indicates
2.1.1 OrderedAcceptanceStrategy the factorial of N. We define the probability ρ(np) as the
Under the ordered acceptance strategy, SOD, if a relay probability of the jth permutation, Πn, resulting in the
senses either the PU or SU to be busy, it operates in the acceptanceorder(m ,m ,...,m )ifthereceivedpacket
1 2 N
receivingmodetillthetransmissiontimewithinthetime comesfromthePU.Thisprobabilitydenotesthefraction
slotisover.Iftheprimarydestination(PD)orsecondary of time slots with this ranking order. Probability ρ(ns) is
defined in a similar fashion if the packet is transmitted
3.The reason the origin of a packet is identified by a certain
embedded identifier is that spectrum sensing may be erroneous. If bytheSU.Vectorsρ(p) andρ(s) havetheaforementioned
sensingwereperfect,therelaycouldidentifytheoriginoftransmission probabilities as elements and both have N! elements.
dependingon whetherit has proceeded at the beginningof the time
The medium access control (MAC) operation can be
slotorafterτ seconds.Inthispaper,althoughwestartwiththeperfect
sensingcase,welateraddresstheissueofspectrumsensingerrors. summarized as follows:
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• At the beginning of a time slot, the PU transmits • If the primary packet is not received correctly, a
the packet at the head of its queue to the primary NACK message is fed back from the primary re-
receiver.Due tothe broadcastnatureof thewireless ceiver.Therelaythatisassignedforpacketdecoding
channel, the SU and the relays can listen to the tries to decode the undelivered primary packet.
transmitted primary packet. If the packet is decoded correctly, it is accepted
• The SU senses the channel over the first τ seconds with a certain probability and an ACK message is
of the time slot. If the SU detects the channel to transmitted, thereby inducing the PU to drop the
be free from primary activity, it transmits from its packet. If the cognitive relay assigned for decoding
queue ifitisnonempty. Therelayscanoverhearthe failstodecodetheprimarypacketordoesnotaccept
secondary transmission. it, the packet is kept in the primary queue for
• If the PU is active and the transmitted packet is retransmission.
received correctly by the primary receiver, an ACK • If the PU is sensed by the SU to be idle, the SU,
message is fedbackfromthe receiver.The packetis if its queue is not empty, starts transmission and
then dropped from the primary queue. The relays the relays repeat the same operation as described
also discard what they have received. earlierforthePU’srelayingscenario.Recallthatthe
• If the primary packet is not received correctly, a origin of the received packets at the assigned relay
NACK message is fed back from the primary re- canbe known fromthe packet’sidentifier.Basedon
ceiver. The relays then attempt to decode the re- the identifier, assigned relay k accepts the packet
ceived packet and determine its origin. Based on into Q with probability f , or into Q with
p,k p,k s,k
their primary packet acceptance ranking, the first- probability f .
s,k
ranked relay decides whether or not to accept the
Randomroundrobindecoding,S ,isasimplification
RR
primary packet if it is decoded correctly. If the
of S in which the decoding assignment probability
RD
packet is accepted, an ACK message is transmitted,
is equal for all relays. That is, β = 1/N, where k ∈
k
thereby inducing the primary transmitter to drop
{1,2,...,N}.
the packet.Ifthe first-rankedcognitive relayfailsto
decodetheprimarypacketordoesnotacceptit,the
second-ranked relaytries to do so. This relay issues
2.2 Physical(PHY)Layer
an ACK signal if it decodes the packet successfully
and decides to accept it. This operation continues The channel outage event for the relays and the SU
inrankingordertillarelaydecodesandacceptsthe can be calculated as follows. The transmitters adjust
packet. If no relay accepts the packet, it is kept in their transmission rates depending on when they start
the PU’s queue for retransmission. transmission during the time slot. Assuming that the
• In the case of secondary transmission, the relays numberofbitsinapacketisbandthetimeslotduration
perform the same operation as described for pri- is T, the transmission rate is
mary transmission. The ranking of relays to accept
b
the secondary packets differs from the ranking of ri = T 1− TSF . (1)
accepting the primary packets. T
(cid:0) (cid:1)
• If both the PU and the SU are found to be idle, with TSF = iτ +TF < T, where TF is the time needed
therelaysstarttransmittingthepacketsattheheads to execute the feedback process. The parameter i = 0
of their queues. The fraction of time slots in which if the transmitter is the PU as transmission proceeds at
relay k is scheduled to transmit is ω . the very beginning of the time slot, i = 1 for the SU
k
as its transmission is preceded by a spectrum sensing
2.1.2 Random Assignment Decoding and Random period of τ seconds, and i = 2 for all relays because
RoundRobinStrategies their transmission is preceded by a spectrum sensing
period of 2τ seconds relative to the beginning of the
The difference between random assignment decoding,
timeslot. Outage ofalinkoccurswhenthetransmission
S , and S is that inS only one relayis scheduled
RD OD RD rate exceeds the channel capacity. Hence, the outage
todecode,and possibly accept,the undeliveredprimary
probability of the link between node j and node k is
or secondary packet at any slot. The probability that
given by [4]
relay k is assigned the decoding role in a time slot is
denotedasβk.Wedefinethevectorβ =(cid:20)β1,β2,...,βN(cid:21) Pj,k =Pr(cid:26)ri >W log2(1+γj,khj,k)(cid:27) (2)
with the constraint Nk=1βk = 1. The vectors α, ω, fp where W is the bandwidth of the channel, γj,k is the
and fs are similar toPthose in SOD. The operation of the received signal to noise ratio (SNR) when the channel
relays can be summarized as follows: gain is equal to unity, and h is the channel power
j,k
• At the beginning of each time slot, the index, k, of gain, which is exponentially distributed in the case of
the randomly selected relay is generated according Rayleigh fading. The channel gain, h , is assumed to
j,k
to the discrete distribution β. be independent from slot to slot and link to link. The
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outage probability can be written as The secondary queue can be analyzed in a similar
fashion. The probability of the primary queue being
P =Pr h < 2Wri −1 (3) empty is [4], [14], [15]
j,k j,k
γ
n j,k o
λ
p
Assuming that the mean value of hj,k is σj,k, πp,◦ =1− µ . (7)
p
2Wri −1 When the primary queue is empty, a packet in the
P =1−exp − . (4)
j,k
(cid:18) γj,kσj,k(cid:19) secondary queue can be served in either one of the
following events: the channel between the SU and SD is
LetP =1−P betheprobabilityofcorrectreception.
j,k j,k notinoutage;orthechannelbetweentheSUandtheSD
It is therefore given by
is in outage and one of the relays decodes and decides
b to accept the packet. Therefore, the average service rate
2TW(1−TTSF) −1 of the secondary queue can be written as
P =exp − . (5)
j,k
(cid:18) γ σ (cid:19)
j,k j,k
N
Note that the duration of the feedback process, TF, µs =πp,◦ Ps,sd+Ps,sd Ps,kfs,k
varies according to the strategy in the decoding that the (cid:20) Xk=1(cid:18)
N (8)
relays adopt. In the case of S , the relays are ordered
in terms of sending the ACKOmDessages if they accept a × ρ(ns) Ps,vfs,v .
correctlyreceivedpacket.If eachrelayneedsτ seconds, XΠn vvY6==k1 (cid:19)(cid:21)
f mv<mk
then the overall feedback duration is T = (N +1)τ
F f
The probability that the secondary queue is empty is
given that the PD also needs τ seconds to acknowledge
f
given by
the reception of a data packet. On the other hand, S
RD λ
and S need only T = 2τ for the feedback process π =1− s. (9)
RR F f s,◦
µ
to be executed. The increase in the feedback duration is s
interpretedasanincreaseintheoutageprobabilityofthe Let λp,k and λs,k be the arrival rates at the queues
channels. This fact can be seen easily from (5). It should Qp,k and Qs,k of relay k, respectively. Note that for
be mentioned that the decoding process, taking into an arrival event to occur at Qp,k, the primary queue
account the feedbackduration, will cause a reduction in should be nonempty. For an arrival event to happen
theallowabledatatransmissiontimeofboththeprimary at Qs,k, the primary queue must be empty to preclude
and secondary transmissions, i.e., the total transmission primary transmission and the secondary queue should
time will be reduced to T −T . benonempty.Theexpressionsforthearrivalratesfollow
SF
directly from (6) and (8) and are given by
3 PROBLEM FORMULATIONS N
We startthe performanceanalysis of the differentproto- λp,k=πp,◦Pp,pd Pp,kfp,k ρ(np) Pp,vfp,v . (10)
cols with the case of perfect spectrum sensing; then we (cid:20) XΠn vvY6==k1 (cid:21)
mv<mk
consider spectrum sensing errors in the next section.
and
N
3.1 AverageArrivalandServiceRates
3.1.1 OrderedAcceptance λs,k=πs,◦πp,◦Ps,sd(cid:20)Ps,kfs,kXΠn ρ(ns) vvY6==k1 Ps,vfs,v(cid:21). (11)
Consider the primary queue first. A packet can be mv<mk
servedin either one of the following events: the channel For a relay to transmit, both the primary and sec-
between PU and PD is not in outage; or the primary ondaryqueues should be empty. Relay k transmits from
channel is in outage and one of the relays decodes Q with probability α and from Q with probability
p,k k s,k
correctly and accepts the packet. Note that for relay 1−α . The average service rates, µ and µ , of Q
k p,k s,k p,k
k to get the primary packet, all the relays having a and Q at relay k, respectively, are given by
s,k
higher priority in accepting the packetshould either fail
to receive the packet correctly due to channel outage, µp,k=ωkπp,◦πs,◦αkPk,pd, (12)
or reject the packet. Hereinafter, we adopt the notation µs,k=ωkπp,◦πs,◦ 1−αk Pk,sd
x=1−x.The averageservicerate ofthe primaryqueue We can upper bound the (cid:0)mean(cid:1)service rate of the
is given by
primaryqueueasfollows.Themaximumservicerateoc-
N curswhenallrelaysdecidetoacceptthe primarypacket
µp =Pp,pd+Pp,pd Pp,kfp,k eachtimeslot,i.e.,fp,k =1∀k,regardlessofthedecoding
Xk=1(cid:20) orderdistribution.4 In this case, the meanservice rate of
N (6)
× ρ(np) Pp,vfp,v . 4.This is because, in any arbitrary slot, each relay, whatever its
XΠn mvvvY<6==km1k (cid:21) dit,ecifotdhienglorwaenrkr,awniklledatrteelmaypstftaoildinecdoedceodthinegpirtimduaerytopcahckanetnealnsdouatdamgeit.
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theprimarynodebecomestheprobabilitythatoneofthe N P β ≤max P ,P ,...,P , and
receivingnodes’channelsisnotinoutage.Therefore,the k=1 p,k k (cid:26) p,1 p,2 p,N(cid:27)
P
maximummeanservicerateoftheprimaryqueueunder N
strategy SOD is µp ≤Pp,pd+Pp,pd Pp,kβk
N Xk=1
µmpax =1−Pp,pd Pp,v (13) ≤Pp,pd+Pp,pdmax Pp,1,Pp,2,...,Pp,N (18)
v=1 (cid:26) (cid:27)
Y
where 1−Pp,pd Nv=1Pp,v is the probability that either =1−Pp,pdmin Pp,1,Pp,2,...,Pp,N =µmpax
the PD or one ofQthe relays decodes the primary packet (cid:26) (cid:27)
where max{·} and min{·} return the maximum and the
correctly. Similarly, the maximum mean secondary ser-
minimumofallthevaluespresentintheirarguments,re-
vice rate under strategy S is
OD
spectively. The maximum mean primary and secondary
N
service rates are
µms ax = 1−Ps,sd Ps,v π˜p,◦
(cid:20) vY=1 (cid:21) (14) µmpax=1−Pp,pdmin Pp,1,Pp,2,...,Pp,N (19)
N
=(cid:20)1−Ps,sdv=1Ps,v(cid:21)"1−1−Pp,pdλpNv=1Pp,v# and (cid:26) (cid:27)
Y
wthheeprerimπ˜pa,r◦y=qu1eu−e1b−ePinp,gpdλeQpmNv=p1tyPp,wvhiesnQthµe =proµbmaabxilwithyicohf µms ax=(cid:20)1−Ps,sdmin(cid:26)Ps,1,Ps,2,...,Ps,N(cid:27)(cid:21)
upper bounds π . p p × 1− λp (20)
p,◦
(cid:20) 1−Pp,pdmin Pp,1,Pp,2,...,Pp,N (cid:21)
(cid:26) (cid:27)
3.1.3 RoundRobinDecoding
In S , each relay is assigned the decoding role with
RR
3.1.2 RandomAssignmentDecoding equal probability, i.e., 1/N, in a cyclic manner. The
expressionsarethussimilartoS withthesubstitution
RD
β =1/N.Asintheprevioussubsectionandwithsetting
k
In SRD, the kth relay is scheduled to decode the trans- βk = 1/N, we can obtain the maximum mean service
mitted packet with probability βk. Hence, the average rates of the primary and secondary queues under strat-
service rates of the PU and the SU are given by egy S . The maximum mean primary and secondary
RR
service rates are
N
µp =Pp,pd+Pp,pd Pp,kfp,kβk, µmax =1−P 1− Nv=1Pp,v (21)
Xk=1 (15) p p,pd N
N (cid:18) P (cid:19)
µs =πp,◦ Ps,sd+Ps,sd Ps,kfs,kβk . and
(cid:18) Xk=1 (cid:19) N P
Theaveragearrivalratestotherelayingqueuesaregiven µms ax = 1−Ps,sd 1− v=N1 s,v
(cid:20) (cid:18) P (cid:19)(cid:21)
by
λ (22)
× 1− p .
λλps,,kk ==PPsp,,spddPPs,kp,fksf,kpβ,kkβπksπ,◦pπ,◦p,,◦. (16) " 1−Pp,pd(cid:18)1− PNv=N1Pp,v(cid:19)#
The average service rates of the relaying queues are Theorem 1. The queue service rates of SOD always outper-
the same as in the ordered acceptance case. The mean form the queue service rates of SRD and SRR for a network
service rate of the primary queue is upper bounded as with N relays if the feedback duration per relay is negligible.
follows. The mean service rates of the primary queue
Proof:Theproofforthistheoremunderperfectsens-
under strategy S can be upper bounded as follows.
RD ing and imperfect sensing (sensing errors) is presented
N in Appendix A.
µ =P +P P f β
p p,pd p,pd p,k p,k k Proposition 1. The SU’s maximum mean service rate, µ∗,
Xk=1 (17) s
N for an arbitrary decoding strategy is given by
≤E P +P P β ,
p,pd p,pdXk=1 p,k k µ∗s =1−λp. (23)
wheretheinequalityE holdswithequalitywhenfp,k =1 Proof: Regardless of decoding strategy, the sec-
for all relays. Since Pp,k belongs to the convex set [0,1] ondary average service rate can be always upper
N
and βk ∈ [0,1] is a convex set with k=1βk = 1, bounded by the probability of the PU’s queue being
N
then k=1Pp,kβk isaconvex hullwithmPaximumvalue empty assuming that when the PU is idle due to its
locatePd at the edges, i.e., at βk ∈ {0,1}. Accordingly emptyqueue,theSUcansuccessfullytransmititspacket
7
withprobabilityone. Thiscanbe expressedasµ ≤π . relays’ queues to be stable. The total number of opti-
s p,◦
Since the probability of the PU being empty is π = mization parameters in case of ordered acceptance is
p,◦
1− λp ≤1−λ , then µ ≤1−λ . 2N!+4N.
µp p s p
It is worth noting that the optimization problems are
solved at a controller which then supplies the required
3.2 AverageQueueingDelayAnalysis
informationtotherelaystations.Theoptimalparameters
Since all network queues are decoupled, the jth queue
are functions of many parameters such as the channels
Q queueing delaywhen itis stable, is given by [4], [16]
j outage between all nodes in the network (based on
1−λ the expression in (3), the channel outage between any
j
D = (24)
j µ −λ two nodes is a function of the packet length, channel
j j
bandwidth, SNR, time slot duration, and many other
where j ∈ p,s,(p,k),(s,k) and µ > λ . The end- parameters), primary and secondary arrivals rate, delay
j j
(cid:26) (cid:27)
constraints, number of relays, misdetection probability,
to-end mean queueing delay is the average delay that
and false alarm probability at each relay. Thus, we
any packet experiences from its arrival at the source
note that for a given system’s parameters, the optimal
queue till it arrives at the destination. In our system,
parameters are fixed as far as these parameters remain
each packet arriving at Q experiences on the average
ℓ constant. Once the optimal parameters are obtained,
delayofD timeslots,whereℓ∈{p,s}.Further,apacket
ℓ the controller generates a long sequence of decoding
has an additional delayD if it reachesthe destination
ℓ,k orders and time slot accessing distribution over time
through relay k. Since, on the average, the probability
slots to be supplied to the relay stations during the
that a packet serviced from Q is buffered at the kth
ℓ whole operational time of the system. This occurs all
relay before reaching its destination is λℓ,k, the average
λℓ at once before the actual operation of the system. The
queueing delays of the primary and secondary packets
optimal acceptance probabilities of users’ packets at the
are given by
relay stations and the probability of selecting one of the
N N relaying queues over the other for a given time slot are
λ D λ D
Dp(T)=Dp+Pk=1λp,k p,k, Ds(T)=Ds+Pk=1λs,k s,k. all generated locally at each relay station. However, the
p s
values of the probabilities are also supplied to the relay
(25)
stations by the controller all at once before the actual
A similar approach for computing the end-to-end delay
operation of the system.
is found in [17]–[19].
This optimization problem and the others presented
in this work are solved numerically Specifically, we
3.3 OptimizationProblems
use Matlab’s fmincon as in [20]–[26] and the references
3.3.1 SecondaryThroughputMaximization
therein.
Our first optimization problem is concerned with the Now, we investigate the case in which all relays are
constrained maximization of the secondary average ser- set to accept the users’ undelivered packets every time
vice rate given λp, λs and N subject to predefined slot. Precisely, fp,k = fs,k = 1 for all k. Moreover, we
tolerableend–to–endmeanqueuingdelayconstraintsfor assumethattheprobabilityofselectingarelayingqueue
the primary and secondary packets. Under the ordered fortransmissionis1/(2N)where2N isthe totalnumber
acceptancestrategy, SOD, the maximum secondaryaver- of possibilities. According to the above case, µp = µmpax
ageserviceratecanbeobtainedbysolvingthefollowing in (13), µ = µmax in (14), π = 1 −λ /µmax, π =
s s p,◦ p p s,◦
problem: 1−λ /µmax,
s s
max µs N
α,fp,fs,ω,ρ(p),ρ(s) λp,k=πp,◦Pp,pd Pp,k ρ(np) Pp,v , (27)
s.t. Ds(T) ≤Ds(T),Dp(T) ≤Dp(T), (cid:20) XΠn vvY6==k1 (cid:21)
λ <µ , λ <µ , λ <µ , λ <µ mv<mk
s s p p p,k p,k s,k s,k
and
0≤α,f ,f ≤1,
p s
N
0≤ω,ρ(p),ρ(s), λs,k=πs,◦πp,◦Ps,sd Ps,k ρ(ns) Ps,v . (28)
kωk1,kρ(p)k1,kρ(s)k1 =1 (cid:20) XΠn vvY6==k1 (cid:21)
(26) mv<mk
whereD(T) <∞isthemaximumtolerableprimaryend– The relaying queues’ mean service rates become con-
p
to–endmeanqueueingdelay,D(T) <∞isthemaximum stants. That is,
s
tolerable secondary end–to–end mean queueing delay, 1
the notation a ≤ x is an element wise condition on µp,k=2Nπp,◦πs,◦Pk,pd,
(29)
vector x implying that a ≤ x and kxk is the l –norm 1
of the vector x defined as kxkk = 1|x| . The1 delay µs,k=2Nπp,◦πs,◦Pk,sd.
1 k k
constraintsimplicitlyrequiretheprimPary,secondaryand Theoptimizationproblemisaconvexfeasibilityproblem
8
which can be solved efficiently [27]. We note that the It can be solved as follows:
objective function is constant. Moreover, Dp and Ds N
1
are constants. We need to prove the convexity of the max π P +P P =constant
p,◦ s,sd s,sd s,k
constraints Ds(T) ≤ Ds(T), Dp(T) ≤ Dp(T), λp,k < µp,k, and zk,yk∀k (cid:18) N kX=1 (cid:19)
λs,k < µs,k. From (27), (28) and (29), λp,k < µp,k and s.t. λp,k <zkπp,◦πs,◦Pk,pd, λs,k <ykπp,◦πs,◦Pk,sd
λs,k < µs,k are linear in ρ(np) and ρ(ns).The second term 0≤zk,yk ≤1,
ofthe queueingdelaysDp(T) andDs(T), PNk=1λλpp,kDp,k and 0≤zk,yk,
PNk=1λs,kDs,k,respectively,areconvexifeachoftheterms N
insideλsthesummationisconvex.Thus,we needtoprove (zk+yk)=1
the convexity of λjDj, j ∈ {(p,k),(s,k)}. The second Xk=1
derivative of D = λ 1−λj with respect to λ is given (33)
j jµj−λj j with λs <µs and λp <µp. The feasible values of zk and
by
y are
k
∂2Dj =2µj(1−µj) ≥0. (30) λ λ N
∂λ2j (µj−λj)3 p,k <zk, s,k <yk, (zk+yk)=1
π π P π π P
Since the second derivative is positive for µ ≥ λ p,◦ s,◦ k,pd p,◦ s,◦ k,sd kX=1
j j (34)
(stability constraint) and µ ≤ 1, the delay constraint is
j
convex over λ .
j If the usersaredelaysensitive, the optimization prob-
lem can be shown to be a convex feasibility problem.
ForS ,themaximumsecondaryaverageservicerate
RD We note that λ , λ , µ , and µ are constants with
p,k s,k s p
can be obtained by solving the following optimization
respect to the optimization variables, ω and α . The
k k
probmleamx: µs itnersmidePthNk=e1sλλuss,mkDms,katiisoncoinsvceoxnvoevxeroµvjerifµeja.cThhoufs,thweetenremeds
α,f ,f ,ω,β
p s toprovetheconvexityofλ D .Thesecondderivative
s,k s,k
s.t. Ds(T) ≤Ds(T),Dp(T) ≤Dp(T), of Dj =λjµ1j−−λλjj with respect to µj is given by
λ <µ , λ <µ , λ <µ , λ <µ
s s p p p,k p,k s,k s,k
0≤α,f ,f ≤1, ∂Dj =2λj(1−λj) ≥0. (35)
p s ∂µj (µj−λj)3
0≤ω,β,
Since the second derivative is positive, the delay con-
kωk1,kβk1 =1. straint is convex over µj. We solve the problem with
(31)
respect to z and y then we get the values of ω and
k k k
The totalnumber ofoptimization parametersinthe case
α .
k
of random decoding is 5N.
It should be noticed that the total number of opti-
ForSRR,themaximumsecondaryaverageservicerate mization parameters is a reflection of both the degrees
can be obtained by solving an optimization problem of freedom and the degree of complexity of the system.
similar to (31) with all elements of β equal to 1/N. The Therefore, the ordered acceptance is considered as the
optimization problem is stated as follows: strategy with the highest degrees of freedom and the
highestcomplexityamongtheproposedstrategiesinthis
paper. On the other hand, round robin is the simplest
max µs strategyamongtheproposedstrategiesanditneedsless
α,f ,f ,ω
p s cooperation between the relays than other strategies; it
s.t. Ds(T) ≤Ds(T),Dp(T) ≤Dp(T), is a cyclic switching operation shared among relays.
λ <µ , λ <µ , λ <µ , λ <µ
s s p p p,k p,k s,k s,k
0≤α,f ,f ≤1,
p s
0≤ω,
kωk =1.
1
(32)
The total number of optimization variables is equal to 3.3.2 NumberofRelaysMinimization
4N.
Consider the case f = f = 1. Let z = α ω and
s,k k,p k k k
yk = (1 − αk)ωk with zk + yk = ωk. If the queueing Our second formulation is to minimize the number of
delay requirements are large, i.e., Dp(T) = Ds(T) = ∞, relays,N,neededtoachievecertaindelayorservicerate
which means that the users are delay insensitive, then requirements for the users. Given λ and λ and under
p s
theoptimizationproblemisaconvexfeasibilityproblem. the ordered acceptance strategy, S , the optimization
OD
9
problem is given by P(pk). Sensing false alarms have probability Pk . All
MD FA
relays are adjusted on the receiving mode and attempt
min N
to decode the transmitter packet. The relay scheduled
α,f ,f ,ω,ρ(p),ρ(s)
p s
for transmission is the only relay that decides after 2τ
s.t. D(T) ≤D(T),D(T) ≤D(T),
s s p p seconds relative to the beginning of the time slot about
λ <µ ,λ <µ ,λ <µ ,λ <µ
s s p p p,k p,k s,k s,k the state of the time slot: busy or free. If the slot is
0≤α,f ,f ≤1, sensed tobe free,thatrelayswitches tothe transmission
p s
0≤ω,ρ(p),ρ(s), mode and start retransmission of one of the packets in
its relaying queues. If the channel is sensed to be busy
kωk ,kρ(p)k ,kρ(s)k =1.
1 1 1 over either interval, the relay continues in the receiving
(36)
mode. Upon decoding, the relay will be able to identify
In case of S , the minimum number of relays required
RD the packet’s origin from the identifier attached to the
in the network is given by the following optimization
packet and will use the appropriate decoding order in
problem:
case of order decoding. In case of random decoding
min N or round robin decoding, one of the relay stations is
α,fp,fs,ω,β assigned the decoding task in each time slot. Based on
s.t. D(T) ≤D(T),D(T) ≤D(T), the above, the service rates of the users’ queues and the
s s p p
arrivalratesof the relayingqueues under sensing errors
λ <µ , λ <µ , λ <µ , λ <µ
s s p p p,k p,k s,k s,k
are only affected by the activity of the relay scheduled
0≤α,f ,f ≤1,
p s for transmission. The reduction of the mean service and
0≤ω,β, arrival rates is equal to ω B(ℓr), where B(ℓr) denotes
r r
kωk1,kβk1 =1. thecomplementoftheprPobabilitythatrelayr,scheduled
(37) for transmission, erroneously finds the time slot free
ForSRR,weconstructanoptimizationproblemsimilar given there is an active transmission from user ℓ.
to (37) with all elements in β being set to 1/N. Now we compute B(ℓr) for both users. The relay
r scheduled for transmission disrupts the primary if
4 THE CASE OF SENSING ERRORS it fails to detect the activity of the PU during both
sensing intervals. That is, the probability that the rth
We address here the specific scenario of a strong sens-
relay detects the time slot as a busy slot due to activity
ing channel between the PU and the SU and consider
of PU is B(pr) = (1−P(pr)P(pr)).6 The probability that
sensing errors at the relay stations. In other words, we MD MD
the relay r scheduled for transmission does not disrupt
assume that the sensing errors at the SU are negligible,
the secondary activity is equal to the probability that
whereas spectrum sensing at the relays may generate
the relay either detects the secondary transmission, or
erroneous sensing results that should be accounted for.
falsely finds the PU to be active while it is not. In either
To render the problem tractable and avoid the difficulty
case, it will abstain from transmission, thereby avoiding
of queue interaction due to sensing errors, we impose
collision with the secondary transmission. Thus, the
the assumption that Q and Q are never empty.
s,k p,k
Specifically, when either Qs,k or Qp,k is empty, the kth probability is given by B(sr) = 1−PM(sDr) 1−PFrA .
relay sends dummy packets.5 The dummy packets do Accordingly, we have the followin(cid:16)g set of (cid:2)arrival a(cid:3)n(cid:17)d
not contribute to the service rates of Q and Q service rates:
s,k p,k
but cause interference during concurrent transmission
with the primary and secondary terminals. Based on
N
this assumption, the relay scheduled for transmission µ(ℓSE)=µℓ ωr B(ℓr), ℓ∈{p,s} (38)
couldcauseinterferencewiththeprimaryandsecondary r=1
X
transmissions, when it misdetects their transmissions,
even if it is empty in the original system. Accordingly, N
the service rates of the primary and secondary queues λ(ℓS,kE) =λℓ,k ωr B(ℓr), ℓ∈{p,s} (39)
are reduced, and the probability of having any of them r=1
X
emptyisreducedaswell.Consequently,theservicerates
of the relays are reduced. Therefore, our results provide µ(SE)=µ 1−Pk 2, ℓ∈{p,s} (40)
lower bounds on the primary, secondary and relays ℓ,k ℓ,k FA
service rates. where µ(SE), λ(SE) and(cid:0)µ(SE) ar(cid:1)e the rates of users and
ℓ ℓ,k ℓ,k
The kth relay scheduled for transmission at a slot relays in the case of sensing error and µ , λ and
ℓ ℓ,k
misdetects the SU’s transmission with probability P(sk) µ are the rates of users and relays in the case of
MD ℓ,k
and misdetects the PU’s transmission with probability
5.The assumption of a node sending dummy packets when it is 6.As mentioned in Section 2, we assume in this paper that if two
emptyhasbeenconsideredinmanyworks(see,forexample,[4],[6], terminals transmit simultaneously, their packets cannot be decoded
[15],[28]andreferencestherein). correctlyattherespectivereceivers.
10
TABLE1
N=0
Relayschannelsoutageprobabilities whereN =5and 0.9 Ordered acceptance
τ =0. Upper bound
f
Random assignment decoding
0.8
Round robin decoding
Relay-SD Relay-PD SU-relay PU-relay
P1,SD =0.1 P1,PD =0.1 Ps,1 =0.1 Pp,1 =0.1 0.7
P =0.1 P =0.1 P =0.1 P =0.02
2,SD 2,PD s,2 p,2
P3,SD =0.2 P3,PD =0.2 Ps,3 =0.02 Pp,3 =0.2 µs0.6
P =0.1 P =0.01 P =0.1 P =0.1
4,SD 4,PD s,4 p,4
P =0.01 P =0.01 P =0.01 P =0.01
5,SD 5,PD s,5 p,5 0.5
0.4
perfect sensing.7 We note that the term 1 − Pk 2 in
FA
(40) represents the probability that the k(cid:0)th relay (cid:1)finds 0.30. 1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5
the time slot free from transmissions. This equals the λ
p
probability that the sensor of relay k does not generate
false alarm over both sensing intervals. Fig. 3. Optimal secondary average service rate versus
To obtain the optimal secondary average service rate theprimaryaveragearrivalrate,λp.
in case of sensing errors, we construct an optimization
problemsimilarto(26)and(31).Fortheminimumnum-
ber of relays needed to achieve certain QoS constraints, 1
weconstructanoptimizationproblemsimilarto(36)and
(37).
0.998
5 NUMERICAL RESULTS 0.996
p
In this section, we provide some numerical results for µ
theoptimizationproblemsconsideredinthispaper.Figs. 0.994
3 and 4 demonstrate the case of negligible feedback
Ordered acceptance
duration, i.e., τ = 0, and low outage probabilities for
f 0.992 Random assignment decoding
the PU-PD and the SU-SD direct links: P = 0.2 and
s,sd Round robin decoding
P = 0.1. The figures are generated using N = 2,
p,pd
λs = 0.1 packets per time slot, Dp(T) ≤ 1.6 time slots, 0.990 0.1 0.2 0.3 0.4 0.5
(T) λ
Ds ≤ 3 time slots, λs = 0.2 packets per time slot, and p
the outage probabilities given in the first two lines of
Table 1. As evident from Fig. 3, the ordered acceptance Fig. 4. Primary average service rate versus λ for the
p
strategy with two relays almost achieves the upper sameparametersusedtogenerateFig.3.
bound on the secondary average service rate, which is
equal to 1−λ . Random assignment and round robin
p
decoding give almost the same performance for the
averageservicerateoftheSU(maximumµ )approaches
s
parametersused inthe simulation. Theprimaryaverage
the upper bound.
service rate, as shown in Fig. 4, is constant and almost
Figs. 6 and 7 also show the case of τ = 0, but this
unity for the proposed decoding strategies compared to f
time there are no direct links between the PU and the
1−P = 0.9 when no relays are used. The primary
p,pd SU and their respective receivers. That is, P = 1
mean service rate is constant because the solution of the s,sd
and P = 1. The parameters used to generate these
optimization problemmakes µ =µmax (see expressions p,pd
p p figures are: τ = 0, N = 2, λ = 0.1 packets per time
(13), (19), and (21)). f s
slot, D(T) ≤ 25, D(T) ≤ 25, and the outage probabilities
Fig. 5 reveals the impact of increasing the number p s
given in the first three lines of Table 1. Note that in this
of relays on the optimal secondary average service rate
case relaying is essential since without cooperation (no
for S with τ = 0. This figure is generated using
RD f
relays),theprimaryservicerateisequalto1−P =0
P = 0.3 and P = 0.4, λ = 0.2 packets per time p,PD
s,sd p,pd s
slot, Dp(T) ≤ 5 time slots, Ds(T) ≤ 10 time slots, and abnacdkbloogtghetdheanpdrimuanrsytaabnledasnecdonpdaacrkyetqsuaerueesneavreeralbweainygs
the outage probabilities given in Table 1. As shown in
served.Hence,thequeueingdelayofeachuserisinfinity.
the figure, when the number of relays, N, increases, the
Fig. 8 represents the solution of the optimization
problems (36) and (37), which is the number of relays
7.Thesevaluesdependonthedecodingstrategyusedasexplained
earlier. requiredtoachievespecificQoSrequirementsforthePU