Tech report

1
Fair Channel Allocation and Access Design for
Cognitive Ad Hoc Networks
Le Thanh Tan and Long Bao Le
Abstract—We investigate the fair channel assignment and
access design problem for cognitive radio ad hoc network in
this paper. In particular, we consider a scenario where ad hoc
network nodes have hardware constraints which allow them
to access at most one channel at any time. We investigate a
fair channel allocation problem where each node is allocated a
subset of channels which are sensed and accessed periodically
by their owners by using a MAC protocol. Toward this end,
we analyze the complexity of the optimal brute-force search
algorithm which finds the optimal solution for this NP-hard
problem. We then develop low-complexity algorithms that can
work efficiently with a MAC protocol algorithm, which resolve
the access contention from neighboring secondary nodes. Also,
we develop an throughput analytical model, which is used in
the proposed channel allocation algorithm and for performance
evaluation of its performance. Finally, we present extensive
numerical results to demonstrate the efficacy of the proposed
algorithms in achieving fair spectrum sharing among traffic flows
in the network.
Index Terms—Channel assignment, MAC protocol, cognitive
ad hoc network, fair resource allocation.
I. INTRODUCTION
Cognitive radio has recently emerged as an important re-
search field, which promises to fundamentally enhance wire-
less network capacity in future wireless system. To exploit
spectrum opportunities on a given set of channels of interest,
each cognitive radio node must typically rely on spectrum
sensing and access mechanisms. In particular, an efficient
spectrum sensing scheme aims at discovering spectrum holes
in a timely and accurate manner while a spectrum access
strategy coordinates the spectrum access of different cognitive
nodes so that high spectrum utilization can be achieved. These
research themes have been extensively investigated in many
researchers in recent years [1]-[9]. In [1], a survey of recent
advances in spectrum sensing for cognitive radios has been
reported.
There is also a rich literature on MAC protocol design
and analysis under different network and QoS provisioning
objectives. In [2], a joint spectrum sensing and scheduling
scheme is proposed where each cognitive user is assumed to
possess two radios. A beacon-based cognitive MAC protocol
is proposed in [4] to mitigate the hidden terminal problem
while effectively exploiting spectrum holes. Synchronized and
channel-hopping based MAC protocols are proposed in [5] and
[6], respectively. Other multi-channel MAC protocols [7], [8]
are developed for cognitive multihop networks. However, these
existing papers do not consider the setting where cognitive
radios have access constraints that we investigate in this paper.
The authors are with INRS-EMT, University of Quebec, Montréal, Québec,
Canada. Emails: {lethanh,long.le}@emt.inrs.ca.
In [9], we have investigated the channel allocation problem
considering this access constraint for a collocated cognitive
network where each cognitive node can hear transmissions
from other cognitive nodes (i.e., there is a single contention
domain). In this paper, we make several fundamental contribu-
tions beyond [9]. First, we consider the large-scale cognitive
ad hoc network setting in this paper where there can be
many contention domains. In addition, the conflict constraints
become much more complicated since each secondary node
may conflict with several neighboring primary nodes and vice
versa. These complex constraints indeed make the channel
assignment and the throughput analysis very difficult. Second,
we consider a fair channel allocation problem under the max-
min fairness criterion [13] while throughput maximization is
investigated in [9]. Third, we propose optimal brute-force
search and low-complexity channel assignment algorithms and
analyze their complexity. Finally, we develop a throughput
analytical model, which is used in the proposed channel
allocation algorithms and for performance analysis.
II. SYSTEM MODEL AND PROBLEM FORMULATION
A. System Model
We consider a cognitive ad-hoc network where there are
Ms flows exploiting spectrum opportunities in N channels for
their transmissions. Each secondary flow corresponds to one
cognitive transmitter and receiver and we refer to secondary
flows as secondary users (SU) in the following. We assume
there are Mp primary users (PU) each of which can transmit
their own data on these N channels. We assume that each SU
can use at most one channel for his/her data transmission. In
addition, time is divided fixed-size cycle where SUs perform
sensing on assigned channels at the beginning of each cycle to
explore available channels for communications. For simplicity,
we assume that there is no sensing error although the analysis
presented in this paper can be extended to consider sensing
errors. It is assumed that SUs transmit at a constant rate which
is normalized to 1 for throughput calculation purposes.
To model the interference among SUs in the secondary
network, we form a contention graph G = {N, L}, where
N = {1, 2, . . . , Ms} is the set of nodes (SUs) representing
SUs and the set of links L = {1, 2, . . . , L} representing
contention relationship among SUs. In particular, there is a
link between two SUs in L if these SUs cannot transmit
packet data on the same channel at the same time, which is
illustrated in Fig. 1. To model the activity of PUs on each
channel, let us define pp
ij as the probability that PU i does
not transmit on channel j. We stack these probabilities and
define Pi = (pp
i1, . . . , pp
iN ), i ∈ [1, Mp], which captures the
activity of PU i on all channels. In addition, let us define

2
Fig. 1. The contention graph.
Pp
=
(
P1, . . . , PMp
)
where Pi is the vector representing
activities of PU i.
We now model the contention relationship among SUs and
between PUs and SUs. Specifically, we assume that Un
i be the
set of neighboring SUs that conflict with SU i (i.e., there is
a link connecting each SU in Un
i to SU i in the contention
graph). Also, assume that SU k has a set of neighboring PUs
denoted as Up
k , which is the subset of 1, . . . , Mp so that if
any PU in the set Up
k transmit on a particular channel then
SU k is not allowed to transmit on this channel to protect the
primary transmission. Assuming that the activities of different
PUs on any channel are independent then the probability that
channel j is available for SU k indicates can be written as
pkj =
∏
i∈Up
k
pp
ij since channel j is available for SU k if all
conflicting PUs in Up
k do not use channel j.
B. Problem Formulation
We are interested in performing channel assignment that
maximizes the minimum throughput among all SUs (i.e., max-
min fairness [13]). Let Ti denote the throughput achieved by
SU i. Let xij describe the channel assignment decision where
xij = 1 if channel j is assigned to SU i and xij = 0,
otherwise. Then, the max-min channel assignment problem
can be written as
max
x
min
i
Ti (1)
where x is the channel assignment vector whose elements are
xij. For the case where each SU is allocated a distinct set of
channels, i.e., we have
Ms∑
i=1
xij = 1, for all j. Under this non-
overlapping channel assignments, let Si be the set of channels
assigned to SU i. Recall that pij is the probability that channel
j is available at SU i. Then, Ti can be calculated as Ti =
1 −
∏
j∈Si
pij = 1 −
N∏
j=1
(¯pij)
xij
where pij = 1 − pij is the
probability that channel j is not available for SU i [9]. In
fact, 1 −
∏
j∈Si
pij is the probability that there is at least one
channel available for SU i. Because each SU can use at most
one available channel, its maximum throughput is 1.
In general, it would be beneficial if each channel is allocated
to several SUs in a common neighborhood to exploit the multi-
user diversity. Under both non-overlapping and overlapping
channel assignments, it can be observed that the channel
assignment problem with the objective defined in (1) is a non-
linear integer program, which is a NP-hard problem (interest
readers can refer to [12] for detailed treatment of this hardness
result).
C. Optimal Algorithm and Its Complexity
We describe a brute-force search (i.e., exhaustive search) to
determine the optimal channel assignment solution. Specifi-
cally, we can enumerate all possible channel assignment solu-
tions then determine the best one by comparing their achieved
throughput. While throughput can be calculated quite easily for
the non-overlapping channel assignments as being presented in
Section II-B, developing a throughput analytical model for an
overlapping channel assignment solution is indeed challenging
task, which is performed in Section III-B2 of this paper.
We now quantify the complexity of the optimal brute-force
search algorithm. Let us consider SU i (i.e., i ∈ {1, . . . , Ms}).
Suppose we assign it k channels where k ∈ {1, . . . , N}).
Then, there are Ck
N ways to do so. Since k can take any
values in k ∈ {1, . . . , N}, the total number of ways to assign
channels to SU i is
N∑
k=1
Ck
N ≈ 2N
. Hence, the total number
of ways to assign channels to all SUs is
(
2N
)Ms
= 2NMs
.
Recall that we need to calculate the throughputs achieved
by Ms SUs for each potential assignment to determine the
best one. Therefore, the complexity of the optimal brute-
force search algorithm is O(2NMs
). Given the exponentially
large complexity of this brute-force search, we will develop
low-complexity channel assignment algorithms, namely non-
overlapping and overlapping assignment algorithms.
III. CHANNEL ALLOCATION AND ACCESS DESIGN
A. Non-overlapping Channel Assignment
We develop a low-complexity algorithm for non-overlapping
channel assignment in this section. Recall that Si is the set of
channels assigned for secondary user i. In the non-overlapping
channel assignment scheme, we have Si ∩ Sj = ∅, i ̸= j
where SUs i and j are neighbors of each other (i.e., there
is a link connecting them in the contention graph G). Note
that one particular channel can be assigned to SUs who are
not neighbors of each other. This aspect makes the channel
assignment different from the collocated network setting con-
sidered in [9]. Specifically all channels assigned for different
SUs should be different in [9] under non-overlapping channel
assignment since there is only one contention domain for the
collocated network investigated in [9].
The greedy channel assignment algorithm iteratively allo-
cates channels to one of the minimum-throughput SUs so that
we can achieve maximum increase in the throughput for the
chosen SU. Detailed description of the proposed algorithm is
presented in Algorithm 1. In each channel allocation iteration,
each minimum-throughput SU i calculates its increase in
throughput if the best available channel (i.e., channel j∗
i =
arg max
j∈Sa
pij) is allocated. This increase in throughput can be
calculated as ∆Ti = Ta
i − Tb
i = pij∗
i
∏
j∈Si
(1 − pij) [9].
In step 4, there may be several SUs achieving the minimum
throughput. We denote this set of minimum-throughput SUs

3
Algorithm 1 NON-OVERLAPPING CHANNEL ASSIGNMENT
1: Initialize SU i’s set of available channels, Sa
i :=
{1, 2, . . . , N} and Si := ∅ for i = 1, 2, . . . , Ms where
Si denotes the set of channels assigned for SU i.
2: continue := 1
3: while continue = 1 do
4: Find the set of SUs who currently achieve the min-
imum throughput Smin
= argmin
i
Tb
i where Smin
=
{i1, . . . , im} ⊂ {1, . . . , Ms} is the set of minimum-
throughput SUs.
5: if OR
il∈Smin
(
Sa
il
̸= ∅
)
then
6: For each SU il ∈ Smin
and channel jil
∈ Sa
il
,
find ∆Til
= Ta
il
− Tb
il
where Ta
il
and Tb
il
are the
throughputs after and before assigning channel jil
;
and we set ∆Til
= 0 if Sa
il
= ∅
7:
{
i∗
l , j∗
i∗
l
}
= argmax
il∈Smin,jil
∈Sa
il
∆Til
(jil
)
8: Assign channel j∗
i∗
l
to SU i∗
l .
9: Update Si∗
l
= Si∗
l
∪ j∗
i∗
l
and Sa
k = Sa
k j∗
i∗
l
for all
k ∈ Un
i∗
l
.
10: else
11: Set continue := 0
12: end if
13: end while
as Smin
. Then, we assign the best channel that results in
the maximum increase of throughput among all SUs in the
set Smin
. We update the set of available channels for each
SU after each allocation. Note that only neighboring SUs
compete for the same channel; hence, the update of available
channels for the chosen minimum-throughput SU is only
performed for its neighbors. This means that we can exploit
spatial reuse in a large cognitive ad hoc network. It can be
verified that if the number of channels is sufficiently large (i.e.,
N>>maxi |Un
i |), then the proposed non-overlapping channel
assignment achieves throughput close to 1 for all SUs.
B. Overlapping Channel Assignment
1) MAC Protocol: Overlapping channel assignment can
improve the minimum throughput but we need to design a
MAC protocol to resolve access contention among different
SUs. Note that a channel assignment solution needs to be
determined only once while the MAC protocol operates re-
peatedly using the chosen channel assignment solution in each
cycle. Let Si be the set of channels solely assigned for SU i
and Scom
i be the set of channels assigned for SU i and some
other SUs. These two sets are referred to as separate set and
common set in the following. Let denote Stot
i = Si ∪ Scom
i ,
which is the set of all channels assigned to SU i.
Assume that there is one control channel, which is always
available and used for access contention resolution. We con-
sider the following MAC protocol run by any particular SU
i, which belongs the class of synchronized MAC protocol
[11].1
The MAC protocol operates a cyclic manner where
1Since we focus on the channel assignment issue in this paper, we do not
attempt different alternative MAC protocol designs. Interest readers can refer
to [11] for detailed treatment of this issue.
synchronization and sensing phases are employed before the
channel contention and transmission phase in each cycle.
After sensing the assigned channels in the sensing phase, if
a particular SU i finds at least one channel in Si available,
then it chooses one of these available channels randomly for
communication. If this is not the case, SU i will choose
one available channel in Scom
i randomly (if any). Then, it
chooses a random backoff value which is uniformly distributed
in [0, W − 1] (i.e., W is the contention window) and starts
decreasing its backoff counter while listening on the control
channel.
If it overhears transmissions of RTS/CTS from any other
SUs, it will freeze from decreasing its backoff counter until
the control channel is free again. As soon as a SU’s backoff
counter reaches zero, its transmitter and receiver exchange
RTS/CTS messages containing the chosen available channel
for communication. If the RTS/CTS message exchange fails
due to collisions, the corresponding SU will quit the contention
and wait until the next cycle. In addition, by overhearing
RTS/CTS messages of neighboring SUs, which convey infor-
mation about the channels chosen for communications, other
SUs compared these channels with their chosen ones. Any SU
who has his/her chosen channel coincides with the overheard
channels quits the contention and waits until the next cycle.
Note that in the considered cognitive ad hoc setting each SU
i only competes with its neighbors in the set Un
i , which is
different from the setting investigated in [9].
2) Throughput Analysis: To analyze the throughput
achieved by one particular SU i, we consider all possible
sensing outcomes for the considered SU i on its assigned
channels. We will consider the following cases.
• Case 1: If there is at least one channel in Si available,
then SU i will exploit this available channel and achieve
the throughput of one. Here, we have
Ti {Case 1} = Pr {Case 1} = 1 −
∏
j∈Si
¯pij.
• Case 2: We consider scenarios where all channels in Si
are not available; there is at least one channel in S com
i
available, and SU i chooses the available channel j for
transmission. Suppose that channel j is shared by SU i
and MSj neighboring SUs (i.e., MSj = |Uj| where
Uj denotes the set of these MSj neighboring SUs).
Recall that all MSj SUs conflict with SU i (i.e., they
are not allowed to transmit data on the same channel
with SU i). There are four possible groups of SUs ik,
k = 1, . . . , MSj sharing channel j, which are described
in the following
– Group I: channel j is not available for SU ik.
– Group II: channel j is available for SU ik and SU
ik has at least 1 channel in Sik
available.
– Group III: channel j is available for SU ik, all
channels in Sik
are not available and there is another
channel j′
in Scom
ik
available for SU ik. In addition,
SU ik chooses channel j′
̸= j for transmission in
the contention stage.
– Group IV: channel j is available for SU ik, all chan-
nels in Sik
are not available. Also, SU ik chooses

4
channel j for transmission in the contention stage.
Hence, SU ik competes with SU i for channel j.
Let Up
j,i be the set of PUs who are neighbors of SUs in Uj.
Then, the throughput achieved by SU i can be written as
Ti ( Case 3) = (1 − δ)Θi
MSj
∑
A1=0
MSj −A1
∑
A2=0
MSj −A1−A2
∑
A3=0
1
1 + A4
C
A1
MSj
∑
c1=1
C
A2
MSj −A1
∑
c2=1
C
A3
MSj −A1−A2
∑
c3=1
ΘjΦ1(A1)Φ2(A2)Φ3(A3)
where A4 = MSj − A1 − A2 − A3 and δ denotes the MAC
protocol overhead, which will be derived in Section III-B4. In
this derivation, we consider all possible cases where SUs in
Uj are divided into four groups defined above with sizes A1,
A2, A3, and A4, respectively. For one such particular case,
let Up,1
j,i be the set of PUs who are only neighbors of SUs
in group I with size A1 and Up,2
j,i = Up
j,iUp,1
j,i be the set of
remaining PUs in Up
j,i. In addition, let Up,3
j,i be the set of PUs
who are neighbors of SUs in group III and IV with sizes A3
and A4, respectively. The terms Θi , Θj, Φ1(A1), Φ2(A2),
and Φ3(A3) in the above derivation are
• Θi is the probability that all channels in Si are not
available and SU i chooses an available channel j in Scom
i
for transmission.
• Θj is the probability that all PUs in Up,2
j,i do not use
channel j.
• Φ1(A1) denotes the total probability of all cases for PUs
in Up,1
j,i such that channel j is not available for all A1
SUs in group I.
• Φ2(A2) represents the probability that there is at least
one available channel in the separate set for each of the
A2 SUs in Group II.
• Φ3(A3) describes the total probability of all cases for
PUs in Up,3
j,i such that each SU in group III chooses
other available channel j
′
̸= j for transmission, each SU
in group IV chooses channel j for transmission; and all
channels in the separate sets of users in group III and
group IV are not available.
In this formula, we have considered all possible events and
combinations that can happen for neighboring SUs of the
underlying SU i. Note that only A4 SUs in Group IV compete
with SU i for channel j by using the proposed MAC pro-
tocol. Therefore, SU i wins this contention with probability
1/(1 + A4). In addition, the throughput is reduced by a factor
1 − δ where δ is the MAC protocol overhead.
We derive these probabilities in the following. First, we can
calculate the probability Θi as
Θi =
[
∏
k∈Si
pik
]
.
Hi∑
Bi=1
C
Bi
Hi∑
h=1
∑
j∈Ψh
i
1
Bi
∏
j1∈Ψh
i
pij1
∏
j2∈Scom
i Ψh
i
pij2
(2)
where Hi denotes the number of channels in Scom
i . The first
product term in (2) represents the probability that all channels
in Si are not available for SU i. The second term in (2) de-
scribes the probability that SU i chooses an available channel j
among Bi available channels in Scom
i for transmission. Here,
we consider all possible cases where there are Bi available
channels in Scom
i and Ψh
i describes one particular set of Bi
available channels. Next, we can express the probability Θj
as
Θj =
∏
l∈Up,2
j,i
pp
lj. (3)
To calculate Φ1(A1), let Ωp
1 be an index set capturing all
possible cases for which channel j is not available for all A1
SUs in group I. Then, the probability Φ1(A1) can be expressed
as
Φ1(A1) =
∑
h∈Ωp
1
∏
i1∈Λp
h
pp
i1j
∏
i2∈Up,1
j,i Λp
h
pp
i2j (4)
where pp
ij = 1 − pp
ij; and Λp
h ⊆ Up,1
j,i denotes the set of PUs
in Up,1
j,i , which do not transmit on channel j for the particular
case h ∈ Ωp
1. Furthermore, the probability Φ2(A2) can be
calculated as
Φ2(A2) =
∏
i∈A2
Li∑
Ci=1
C
Ci
Li∑
h=1
∑
k∈Ωh
i
∏
j1∈Λh
k
pij1
∏
j2∈SiΛh
k
pij2
(5)
where Li denotes the total number of channels in the separate
set Si of SU i. In this equation, we consider all possible cases
for which each SU i ∈ A2 has Ci ≥ 1 channels available in
their separate set. In addition, Λh
k denotes a particular set (i.e.,
a subset of Si) with Ci available channels. Finally, we can
express the probability Φ3(A3) as
Φ3(A3) =
[
∏
i∈A3∪A4
∏
n∈Si
pin
]
.
∑
i3∈A3
|Si3 |−1
∑
ki3 =1
∑
i4∈A4
|Si4 |−1
∑
ki4 =0
(6)
∏
i3∈A3
1
1 + ki3
∏
i4∈A4
1
1 + ki4
C
ki3
|Si3
|−1
∑
h3=1
C
ki4
|Si4
|−1
∑
h4=1
(7)
∏
j∈Scom
A3∪A4
∏
i1∈Λp,j
h3,h4
pp
i1j
∏
i2∈Up,1
j,i Λp,j
h3,h4
pp
i2j (8)
where A3 and A4 denote the sets of SUs belonging to groups
III and IV described above, respectively. The product term
inside [.] in (6) captures the probability that all SUs in A3 ∪A4
have no channel available in their separate sets. In addition,
we consider all possible cases where each SU i3 ∈ A3 chooses
channel j′
̸= j for transmission while each SU i4 ∈ A4
chooses channel j among available channels for transmission.
For each particular case represented by indices h3 and h4,
there are a set of PUs Λp,j
h3,h4
which do not use channel
j ∈ SA3∪A4 where Scom
A3∪A4
=
∪
i∈A3∪A4
Scom
i .
Summarizing all considered cases, the throughput achieved
by SU i is given as
Ti = Ti {Case 1} + Ti {Case 2} . (9)
This throughput derivation is used for channel assignment and
performance evaluation of the proposed algorithms.
3) Configuration of Contention Window: We show how to
calculate contention window W so that collision probabilities
among contending SUs are sufficiently small. Note that the
probability of the first collision among potential collisions is
largest because the number of contending SUs decreases for

5
successive potential collisions. Derivation of these collision
probabilities for the cognitive ad-hoc networks is more com-
plicated than that for collocated networks considered in [9]
since the interference constraints are more complicated.
We calculate contention window Wk for each SU k consid-
ering the contention with its neighbors. Let us calculate Pc,k
as a function of Wk assuming that there are m secondary SUs
in the contention phase. Without loss of generality, assume
that the random backoff times of m SUs are ordered as
r1 ≤ r2 ≤ . . . ≤ rm. The conditional probability of the first
collision if there are m SUs in the contention stage can be
written as
P
(m)
c,k =
m∑
j=2
Pr (j users collide)
=
m∑
j=2
Wk−2∑
l=0
Cj
m
(
1
Wk
)j (
Wk − l − 1
Wk
)m−j
(10)
where each term in the double-sum represents the probability
that j users collide when they choose the same backoff value
equal to l. Hence, the probability of the first collision can be
calculated as
Pc,k =
Mn
k∑
m=2
P
(m)
c,k × Pr {m users contend} , (11)
where Mn
k = |Un
k | + 1 is the total number of SUs (in-
cluding SU k and its neighbors), P
(m)
c,k is given in (10) and
Pr {m users contend} is the probability that m SUs contend
with SU k in the contention phase. To compute Pc,k, we now
derive Pr {m users contend}.
We can divide the set of neighbors of SU k into two groups.
In particular, there are m SUs contending with SU k while
the remaining Mn
k − m SUs do not join the contention phase.
There are Cm
Mn
k
such combinations for a particular value of m
where it happens with the following probability
Pr {m users contend} =
Cm
Mn
k∑
n=1
P(n)
con (12)
where P
(n)
con is the probability of one particular case where m
SUs contend with SU k. We can divide the set of remaining
Mn
k − m SUs who do not join the contention into two
subgroups, namely SUs who could not find any available
channels in their allocated channels Stot
i2
(first subgroup) and
SUs who find some available channels in their separate sets
Si1
(second subgroup).
Now, let Λn be one particular set of m SUs in the first
group and A1 denote the number of SUs in the first subgroup
of the remaining Mn
k − m SUs. Then, we can calculate P
(n)
con
as follows:
P(n)
con =
∏
i1∈Λn


∏
l1∈Si1
pi1l1

 (13)
Mn
k −m
∑
A1=0
C
A1
Mn
k
−m
∑
c1=1
∏
i2∈Ω
(1)
c1
∏
l2∈Si2
pi2l2
∏
i3∈Ω
(2)
c1

1 −
∏
l3∈Si3
pi3l3

 (14)
β(1)
∑
n(1)=1
Cn(1)
β(1)
∑
q(1)=1
. . .
β(m)
∑
n(m)=1
Cn(m)
β(m)
∑
q(m)=1
∏
i4∈Up
c1
∏
l4∈Λ
(1)
c1
pp
i4l4
∏
l5∈Λ
(2)
c1
pp
i4l5
. (15)
The term inside [.] in (13) represents the probability that all
channels in the separate sets Si1 for all SUs i1 ∈ Λn are
not available so that these SUs contend to access available
channels in Scom
i1
. The term in (14) denotes the probability
that each of A1 SUs in the first subgroup (i.e., in the set Ω
(1)
c1 )
find no available channels in their separate sets and each of
the Mn
k −m−A1 SUs in the second subgroup (i.e., in the set
Ω
(2)
c1 ) find at least one available channel in their separate sets
(therefore, these SUs will not perform contention). Here, c1 is
the index of one particular case where there are A1 SUs in the
first subgroup and a particular set Λn. The last term in (15)
denotes the probability of the event representing the status of
all PUs who are neighbors of SUs in the set Un
k (i.e., neighbors
of SU k) so that there are exactly m contending SUs in the
set Λn and A1 SUs in the first subgroup. In (15) we consider
all possible scenarios where for each SU i ∈ Λn, there are
n(i)
available channels among β(i)
= |Scom
i | channels in the
set Scom
i where q(i)
represents the index of one such particular
case. Corresponding to such (n(i)
, q(i)
), Up
c1
denotes the set
of PUs who are neighbors of SUs in Un
k so that indeed m
underlying SUs perform contention.
By substituting P
(n)
con calculated above into (12), we can
calculate the collision probability in Pc,k in (11). From this,
we can determine Wk as follows:
Wk = min {Wk such that Pc,k(Wk) ≤ ϵPk
} (16)
where ϵPk
controls the collision probability and overhead
tradeoff and for clarity we denote Pc,k(Wk), which is given
in (11) as a function of Wk. Then, we will determine the
contention window for all SUs as W = maxk Wk.
4) Calculation of MAC Protocol Overhead: Let r be the
average value of the backoff value chosen by any SU. Then,
we have r = (W − 1)/2 because the backoff counter value
is uniformly chosen in the interval [0, W − 1]. As a result,
average overhead can be calculated as follows:
δ (W) =
[W − 1] θ/2 + tRTS + tCTS + 3tSIFS + tSEN + tSYN
Tcycle
where θ is the time corresponding to one backoff unit; tRTS,
tCTS, tSIFS are the corresponding time of RTS, CTS and SIFS
(i.e., short inter-frame space) messages; tSEN is the sensing
time; tSYN is the transmission time of the synchronization
message; and Tcycle is the cycle time.
5) Overlapping Channel Assignment Algorithm: In the
overlapping channel assignment algorithm described in Algo-
rithm 2, we run Algorithm 1 to obtain the non-overlapping
channel assignment solution in the first phase and perform
overlapping channel assignments by allocating channels that
have been assigned to a particular SU to other SUs in the sec-
ond phase. We calculate the increase-of-throughput metric for
all potential channel assignments that can improve the through-
put of minimum-throughput SUs. To calculate the increase-
of-throughput, we use the throughput analytical model in
Subsection III-B2, where the MAC protocol overhead, δ < 1
is derived from III-B4. After running Algorithm 1 in the first

6
Algorithm 2 OVERLAPPING CHANNEL ASSIGNMENT
1: After running Algorithm 1, each SU i has Si, Scom
i = ∅
and Sn
i , i = 1, . . . , Ms.
2: continue := 1.
3: while continue = 1 do
4: Find Tmin and i∗
= argmin
i∈{1,...,Ms}
Tb
i .
5: SUni
i∗ =
∪
l∈Un
i∗
Stot
l .
6: SSep
i∗ = SETXOR
l∈Un
i∗
(Stot
l ).
7: SInt
i∗ = SUni
i∗ SSep
i∗ Scom
i∗ .
8: Find all minimum-throughput SUs and find the best
channels from either SSep
i∗ or SInt
i∗ for these minimum-
throughput SUs to improve the overall minimum
throughput.
9: if
∪
i∈{1,...,Ms}
Scom,temp
i ̸= ∅ then
10: Assign Scom
i = Scom,temp
i and Si = Stemp
i .
11: else
12: Set continue := 0.
13: end if
14: end while
phase, each SU i has the set of assigned non-overlapping
channels, Si, and it initiates the set of overlapping channels
as Scom
i = ∅, i = 1, . . . , Ms. Recall that the set of all assigned
channels for SU i is Stot
i = Si ∪ Scom
i . Let SUni
i∗ is the set of
all channels that have been assigned for SU i∗
’s neighboring
SUs. Also, let SSep
i∗ be the set of all channels assigned solely
for each individual neighbor of SU i∗
(i.e., each channel in
SSep
i∗ is allocated for only one particular SU in Un
i∗ ). Therefore,
SInt
i∗ defined in step 7 of Algorithm 2 is the set of “intersecting
channels”, which are shared by at least two neighbors of SU
i∗
. Here, SETXOR(A,B) would return the set of all elements
in A or B but not the common elements of both A and B.
In each iteration, we determine the set of SUs which achieve
the minimum throughput. Then, we need to search over two
sets SSep
i∗ or SInt
i∗ to find the best channel for each of these
minimum-throughput SUs. Note that allocation of channels
in SInt
i∗ to minimum-throughput SUs can indeed decrease
the achievable throughput of their owners (i.e., SUs which
own these channels before the allocation). Therefore, channel
allocations in step 8 are only performed if the minimum
throughput can be improved. In step 9, Scom,temp
i is the
potential set of channels for SU i. Algorithm 2 terminates
when there is no assignment that can improve the minimum
throughput. Detailed description of step 8 of Algorithm 2 is
given in Algorithm 3.
6) Complexity Analysis: In each iteration of Algorithm 1,
the number of minimum-throughput SUs is at most Ms and
there are at most N channel candidates which can be allocated
for each of them. Therefore, the complexity involved in each
iteration is upper bounded by MsN. We can also determine
an upper bound for the number of iterations, which is MsN.
This is simple because each SU can be allocated at most N
channels and there are Ms SUs. Therefore, the complexity of
Algorithm 1 is upper bounded by M2
s N2
. In Algorithm 2, we
run Algorithm 1 in the first phase and perform overlapping
Algorithm 3 SEARCH POTENTIAL CHANNEL ASSIGNMENT
1: (*Search potential channel assignment from separating
sets*)
2: for j ∈ SSep
i∗ do
3: Find SU i′
where j ∈ Stot
i′ and i′
∈ Sn
i∗ (j ∈ Si′ or
Scom
i′ ).
4: for l = 0 to |Sn
i∗ | − 1 do
5: for k = 1 to Cl
|Sn
i∗ |−1
do
6: Find Ta
i∗ , Ta
i′ and Ta
m m∈Ul
j
, where Ul
j is the set
of l new SUs sharing channel j.
7: if min
(
Ta
i∗ , Ta
i′ , Ta
m m∈Ul
j
)
> Tmin then
8: Temporarily assign channel j to SU i∗
, all SUs
m and i′
: Scom,temp
i∗ = Scom
i∗ ∪ j, Scom,temp
m =
Scom
m ∪j; and if j ∈ Si′ then Scom,temp
i′ = Scom
i′ ∪j,
Stemp
i′ = Si′ j.
9: Update Tmin = min
(
Ta
i∗ , Ta
i′ , Ta
m m∈Ul
j
)
.
10: Reset all temporary sets of other SUs to be
empty.
11: end if
12: end for
13: end for
14: end for
(*Search potential channel assignment from separating
sets*)
15: for j ∈ SInt
i∗ do
16: Find the subset of SUs i′
, SUse
who use channel j as a
non-overlapping or overlapping channel (i.e., j ∈ Stot
i′ );
and SUse
⊂ Sn
i∗ .
17: for l = 0 to |Sn
i∗ | − SUse
do
18: for k = 1 to Cl
|Sn
i∗ |−|SUse|
do
19: Find Ta
i∗ , Ta
i′ |i′∈SUse and Ta
m m∈Ul
j
, where Ul
j is
the set of l new SUs sharing channel j.
20: if min
(
Ta
i∗ , Ta
i′ |i′∈SUse , Ta
m m∈Ul
j
)
> Tmin then
21: Temporarily assign channel j to SU i∗
, all
SUs m and all SUs i′
: Scom,temp
i∗ = Scom
i∗ ∪ j,
Scom,temp
m = Scom
m ∪ j; and if j ∈ Si′ , then
Scom,temp
i′ = Scom
i′ ∪ j, Stemp
i′ = Si′ j.
22: Update Tmin =
min
(
Ta
i∗ , Ta
i′ |i′∈SUse , Ta
m m∈Ul
j
)
.
23: Reset all temporary sets of other SUs to be
empty.
24: end if
25: end for
26: end for
27: end for
channel assignments in the second phase. The complexity
of this second phase can also be upper-bounded by M2
s N2
.
Therefore, the complexity of both Algorithms 1 and 2 can be
upper-bounded by O
(
M2
s N2
)
, which is much lower than that
of the brute-force search algorithm presented in Section II-C.
IV. NUMERICAL RESULTS
To obtain numerical results, we choose the length of control
packets as follows: RTS including PHY header 288 bits, CTS

7
Fig. 2. The scenario with 3 SUs and 2 PUs.
2 3 4 5 6
0.4
0.5
0.6
0.7
0.8
0.9
1
Number of channels (N)
Throughput(T)
Non−The
Non−Sim
Ove−The
Ove−Sim
Opt−The
Opt−Sim
p
p
ij
= 0.8
p
p
ij
= 0.6
(a)
5 6 7 8 9 10
0
0.2
0.4
0.6
0.8
1
Number of channels (N)
Throughput(T)
Non−The
Non−Sim
Ove−The
Ove−Sim
p
p
ij
= 0.6
p
p
ij
= 0.8
(b)
Fig. 3. Throughput versus the number of channels, pp
ij = 0.6 and 0.8,
Non: Non-overlapping, Ove: Overlapping, The: Theory, Sim: Simulation,
Opt:Optimal.(a) Mp = 2, Ms = 3 (b) Mp = 5, Ms = 8
0.3 0.4 0.5 0.6 0.7 0.8 0.9
0
0.2
0.4
0.6
0.8
0.9
pp
ij
Throughput(T)
Non−The, N = 7
Non−Sim, N = 7
Ove−The, N = 7
Ove−Sim, N = 7
Non−The, N = 9
Non−Sim, N = 9
Ove−The, N = 9
Ove−Sim, N = 9
(a)
1 2 3 4 5 6 7 8
0
0.2
0.4
0.6
0.8
1
SU
Throughput(Ti)
Non−The
Non−Sim
Ove−The
Ove−Sim
(b)
Fig. 4. (a) Throughput versus pp
ij, N = 7 and 9. (b) Throughput achieved
by each SU, Mp = 5, Ms = 8, pp
ij = 0.8, N = 8.
including PHY header 240 bits, which correspond to tRTS =
48µs, tCTS = 40µs for transmission rate of 6 Mbps, which is
the basic rate of 802.11a/g standards [14]. Other parameters
are chosen as follows: cycle time Tcycle = 3ms; θ = 20µs,
tSIFS = 28µs, target collision probability ϵP = 0.03; tSEN and
tSYN are assumed to be negligible so they are ignored. Note
that these values of θ and tSIFS are typical (e.g., see [10]).
To compare the performance of optimal brute-force search
and our proposed algorithms, we consider a small network
shown in Fig. 2 where we choose Ms = 3 SUs, Mp = 2 PUs
and pp
ij = 0.6 and 0.8. Fig. 3(a) shows that the minimum
throughputs achieved by Algs 2 are very close to that obtained
the optimal search, which confirms the merit of this low-
complexity algorithm. Also, the simulation results match the
analytical results very well, which validates the proposed
throughput analytical model. Figs. 3(b), 4(a), and 4(b) il-
lustrate the minimum throughputs achieved by our proposed
algorithms for a larger network shown in Fig. 1. In particular,
Fig. 3(b) shows the minimum throughput versus the number
of channels for pp
ij equal to 0.6 and 0.8. This figure confirms
that Alg. 2 achieves significantly larger throughput than that
due to Alg. 1 thanks to overlapping channel assignments.
Fig. 4(a) illustrates the minimum throughput versus pp
ij. It
can be observed that as pp
ij increases, the minimum achievable
throughput indeed increases. This figure also shows that the
minimum throughput for N = 9 is greater than that for
N = 7. This means our proposed algorithms can efficiently
exploit available spectrum holes. In Fig. 4(b), we illustrate
the throughputs achieved by different SUs to demonstrate the
fairness performance. It can be observed that the differences
between the maximum and minimum throughputs under Alg.
2 is much smaller than that due to Alg. 1. This result implies
that Alg. 2 not only achieves better throughput but also results
in improved fairness compared to Alg. 1.
V. CONCLUSION
We have investigated the fair channel allocation problem
in cognitive ad hoc networks. Specifically, we have pre-
sented both optimal brute-force search and low-complexity
algorithms and analyzed their complexity and throughput
performance through analytical and numerical studies.
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