An Efficient Algorithm for Channel Assignment in Cellular Mobile Networks

Goutam K. Audhya1, Koushik Sinha2, Kalikinkar Mandal3,

Rana Dattagupta4, Sasthi C. Ghosh5 and Bhabani P Sinha3 ∗†‡§¶

Abstract

This paper deals with the channel assignment problem

in a hexagonal cellular network, having non-homogeneous

demands in a 2-band buffering system. We first partition

the given problem into smaller subproblems each of which

is constituted by a homogeneous demand vector on a simple

subgraph of the original network graph. The subgraphs and

the demands on each node of these subgraphs are chosen in

such a way that all the required channels can be assigned

to these nodes with the minimum bandwidth by repeating

an appropriate sequence of channels in a regular and sys-

tematic manner, without causing any interference. Based on

this idea, we next present an algorithm for solving the chan-

nel assignment problem with non-homogeneous demands.

When tested on the Philadelphia benchmark problems, the

proposed algorithm assigns the channels with a bandwidth

always within 5% more than the optimal bandwidth, requir-

ing a very small execution time (less than 50 mSec on a

HPxw8400 workstation). In contrast to this, the best known

algorithm [24] generates optimal assignments with about

10-20 seconds on a comparable workstation. This makes

our proposed algorithm attractive for very fast assignments

of channels in real-life situations, with a marginally small

deviation from the optimal bandwidth.

1. Introduction

Radio Frequency Spectrum is a scarce natural resource

and it must be utilized with the objective of increasing net-

work capacity and minimizing interference. Any wireless

network is a spectrum limited system and hence, frequen-

cies are reused to enhance the capacity. In a cellular mobile

network, each cell is assigned a set of frequency channels

to provide services to the individual calls of that cell. The

∗1 BSNL, Kolkata 700001, India, email : gkaudhya@gmail.com†2 Honeywell Technology Solutions Lab, Bangalore 560076, India,

email : sinha kou@yahoo.com‡3 Indian Statistical Institute, Kolkata 700108, India, email : bha-

bani@isical.ac.in§4 Jadavpur University, Kolkata - 700032¶5 Indian Statistical Institute, Bangalore, email : sasthi@isibang.ac.in

Channel Assignment Problem (CAP) in such a network is

the task of assigning frequency channels to the calls satis-

fying some frequency separation constraints with a view to

avoiding channel interference and using as small bandwidth

as possible.

A lot of work has already been done on the optimal as-

signment of channels [1, 4, 7, 10, 12, 19, 15, 16, 17, 18, 21,

22, 23]. The available radio frequency spectrum is divided

into non-overlapping frequency bands termed as channels.

The frequency bands are assumed to be of equal length and

are numbered as 0, 1, 2, 3, · · ·, from the lower end. The

highest numbered channel required in an assignment prob-

lem is termed as the required bandwidth. The same fre-

quency channel may be assigned to different cells (reused)

if they are at a sufficient distance, without causing any inter-

ference. For avoiding interference, the assignment of chan-

nels should satisfy certain constraints : i) co-channel con-

straint, due to which the same channel is not allowed to be

assigned to certain pair of calls simultaneously, ii) adjacent

channel constraint, for which adjacent channels are not al-

lowed to be assigned to certain pair of calls simultaneously,

and iii) co-site constraint, which implies that any pair of

channels assigned to calls in the same cell must be sepa-

rated by a certain number [22].

The cellular network is often modelled as a graph and

the problem of channel assignment is mapped to the prob-

lem of graph coloring. In its most general form, the chan-

nel assignment problem (CAP) is NP-complete [2]. As a

result, researchers attempted to develop approximation al-

gorithms or heuristic approaches using genetic algorithms

[22, 23, 18, 19, 14], neural networks [20] or simulated an-

nealing to solve the problem. They considered hexagonal

cellular networks, where every cell has a demand for one

channel of only one type with 2-band buffering, i.e., the

channel interference does not extend beyond two cells, with

s0, s1 and s2 as the minimum frequency separation between

the calls in the same cell, and in cells at distance one and

two apart, respectively. Resent work on approximation al-

gorithm presented an approach by selecting a small subset

of the network, on which they applied genetic algorithm to

find its assignment, and next repeating the assignment for

the whole network. Their technique shows faster rate of

1

convergence and improvement in the minimum bandwidth

requirement for channel assignment problems with homo-

geneous demand.

In general, nodes may not have uniform demand, lead-

ing to a CAP with a non-homogeneous demand vector

W = (wi), where wi is the demand from cell i. In or-

der to evaluate the performance of the channel assignment

algorithms, certain well-known benchmark problems, com-

monly known as Philadelphia benchmarks, are widely used

in the literature [23]. These benchmark problems are de-

fined on a 21-node cellular network, as shown in Fig. 10,

where each node represents a cell and two nodes have an

edge between them when the corresponding cells are adja-

cent to each other. The non-homogeneous demand vector

W = (wi), of the cells are represented by either of the de-

mand vectors D1 and D2, as shown in Table 1. The column-

i of Table 1 refers to the channel demand from cell i corre-

sponding to the demands D1 or D2. Table 2 shows the spec-

ifications of these eight problems (problem 1 through 8) in

terms of the specific values of the frequency separation con-

straints s0, s1 and s2, respectively, for a 2-band buffering

system.

For the problems 1, 3, 4, 5, 7 and 8, out of the eight

benchmark instances, the required number of channels for

the assignments is primarily limited by the co-site interfer-

ence s0 only. The results from these problems, following

earlier approaches, give optimal solution within a reason-

able computational time. However, problem 2 and 6 are re-

garded as most difficult ones in the literature, and even with

the latest approach available, the computational times of the

optimal solution for these problems take 60 seconds and 72

seconds, respectively on a standard workstation. In real-life

situation, a channel may be required to be assigned to a cell

within 1-2 seconds to provide services to an individual call

of that cell.

In this paper, we propose a novel method for near-

optimal channel assignment in a hexagonal cellular network

with non-homogeneous demands, with a very small execu-

tion time. We partition the given problem into smaller sub-

problems each of which is constituted by a homogeneous

demand vector on a simple subgraph of the original network

graph, with the objective that each such subproblem can be

solved very quickly by repeated application of a simple as-

signment scheme. Stated differently, these subgraphs and

the demands on each node of these subgraphs, will be cho-

sen in such a way that we can assign all the required chan-

nels to these nodes with a minimum bandwidth requirement

by repeating an appropriate sequence of channels in a reg-

ular and systematic manner, without causing any interfer-

ence. We next propose our assignment algorithm based on

the above idea so that finally we come up with an assign-

ment which is marginally away from the optimal one (need-

ing a bandwidth less than 5% more than the optimal one),

but requires very small execution time (less than 50 mil-

liseconds on an HP xw 6400 workstation). Thus, the pro-

posed algorithm is very suitable in real-life scenarios where

fast channel allocation is demanded, although a marginal

deviation (say, less than 5%) from the optimal bandwidth

may be tolerated.

The paper is organized as follows. In section 2, we

present the system model for formulating the channel as-

signment problem with non-homogeneous demands in a 2-

band buffering system. Section 3 describes the basic con-

cepts. Section 4 presents the assignment scheme for the dif-

ferent sub-networks. The assignment algorithm is presented

in section 5, followed by conclusions in section 6.

2. System Model

We follow the same system model as in [23] for repre-

senting the channel assignment problem. The system model

is described by the following components:

(1) A set X of n distinct cells, with labels 0, 1, 2, · · ·n−1.

(2) A set of distinct channels numbered as 0, 1, 2, · · ·.

(3) A demand vector W = (wi), (0 ≤ i ≤ n − 1), where

wi represents the channel demand from cell i.

(4) A channel assignment matrix Φ = (φij), where φij

represents the channel assigned to call j in cell i (0 ≤i ≤ n − 1, 1 ≤ j ≤ wi). The assigned frequencies

φij’s are assumed to be evenly spaced, and can be rep-

resented by integers ≥ 0.

(5) A frequency separation matrix C = (cij), where

cij represents the minimum frequency separation re-

quirement between a call in cell i and a call in cell j(0 ≤ i, j ≤ n − 1).

(6) A set of frequency separation constraints specified by

the frequency separation matrix :

|φik − φjl| ≥ cij for all i, j, k, l (except when both

i = j and k = l).

3. Basic Concepts

The above model represents the channel assignment

problem in the most general form. However, considering

the symmetric nature of the hexagonal cellular network, we

define the following simpler sub-networks of the original

network which may be used in our assignment algorithm

discussed later for a 2-band buffering system.

Definition 1 : A path is defined as an alternative se-

quence of nodes and edges (both starting and ending with

nodes) like a1e1a2e2a3e3 · · · an−1en−1an, where ai’s, 1 ≤

a1 a2 a3 a4 a5 an

Figure 1. A Path

a1 a2 ai ai+1ai+2ai+3 an−ian

aj aj+1

Figure 2. A Path augmented with multiple tri­

angles

i ≤ n are the nodes and ei’s, 1 ≤ i ≤ n − 1 are the edges

with ei connecting the node ai to ai+1. Alternatively, for

simplicity, we would also represent a path only as a se-

quence of the nodes appearing successively on the path.

Thus, the path a1e1a2e2a3e3 · · · an−1en−1an in Fig. 1 will

be equivalently represented by a1a2a3 · · · an.

Remark 1 : Given only a path, it may be possible to

judiciously select a finite sequence of frequency channels

which may be assigned to the successive nodes in the path

and the same sequence may be repeated along the path with-

out causing any interference.

Definition 2 :set of sparse paths A set of paths will form

a set of sparse paths if, for any two nodes u and v taken on

any pair of paths P1 and P2 belonging to that set such that

u ∈ P1 and v ∈ P2, the distance between u and v is always

greater than or equal to two.

Remark 2 : Given any pair of sparse paths P1 and P2,

we can judiciously repeat the same sequence of frequency

channels to both P1 and P2 so that the same frequency chan-

nel can be assigned to two nodes u ∈ P1 and v ∈ P2 which

are at distance three apart (satisfying the re-use distance of

three under 2-band buffering scheme).

Definition 3 : (Path augmented with a triangle) Given a

path P = a0a1a2 · · · aiai+1ai+2 · · · an−1, if there is some

node aj /∈ P but aj is adjacent to nodes ai and ai+1 both ∈P , (for some i, 0 ≤ i ≤ n−1), then the union of path P and

the triangle formed by the nodes aiajai+1ai will constitute

a path augmented with a triangle.

Remark 3 : Instead of only aj , if we also have nodes

aj+1, aj+2 · · ·, where aj , aj+1, aj+2 · · ·⋂

P 6= ∅, and

aj+k (k ≥ 0) is adjacent to ai+2k to ai+2k+1, then the union

of the nodes in P and aj , aj+1, aj+2 · · · will form a path

augmented with multiple triangle, as shown in Fig. 2. The

assignment of a triangle requires higher bandwidth com-

pared to a path consisting of three nodes. Hence, it is eco-

nomical to assign multiple triangles with a sequence of fre-

quency channels, where most of the frequency integers can

be used for the assignment.

Definition 4 : Path augmented with a quadrilateral

We can have a path augmented with a quadrilateral in

two possible forms as shown in Fig. 3 and explained below

a1 a2 ai ai+1ai+2ai+3 an−ian

aj ak al

am

x

Figure 3. A path augmented with rhombuses

a1 a2 ai ai+1ai+2ai+3 an−ian

aj ak al

Figure 4. A path augmented with a 7­node

trapezoid

:

i) Given a path P = a0a1a2 · · · aiai+1ai+2 · · · an−1, if

there exist two nodes aj , ak /∈ P and aj is adjacent to both

ai and ak, ak is adjacent to both aj and ai+1, then the union

of path P and the rhombus aiajakai+1ai formed by the

nodes ai, aj , ak, ai+1 will constitute a path augmented with

a rhombus.

ii) Given a path P = a0a1a2 · · · aiai+1ai+2 · · · an−1, if

there are two nodes al, am /∈ P and al is adjacent to both

ai and ai+1 forming a triangle aialai+1ai, am is also ad-

jacent to both ai and ai+1, but situated on the other side

of P , forming another triangle aiamai+1ai, then these two

triangles together form a rhombus aialai+1amai by the

nodes ai, al, ai+1, am. The union of path P and the rhom-

bus aialai+1amai will constitute a path augmented with a

rhombus.

Remark 4 : Assignment of a quadrilateral requires more

bandwidth compared to a triangle or a path having four con-

secutive nodes. Hence, for any channel assignment prob-

lem, it is economical to modify a quadrilateral suitably so

that most of the integers in a sequence of frequency chan-

nels can be utilized for an assignment. If we consider one

more nodes al /∈ P , such that, al is adjacent to ai+2, ai+3

and aj , respectively, as shown in Fig. 4, then the seven

nodes ai, ai+1, ai+2, ai+3, al, ak and aj will form a 7-node

trapezoid aiai+1ai+2ai+3alakaj augmented with a straight

path aiai+1ai+2ai+3.

Quadrilateral augmented with nodes and triangles:

a1 a2 ai ai+1ai+2ai+3 an−ian

aj ak al

ap

Figure 5. A 7­node trapezoid augmented with

a triangle

a1 a2 ai ai+1ai+2ai+3 an−ian

ap

aj ak

Figure 6. A 5­node trapezoid augmented with

a triangle

a1 a2 an−ian

ap

aj ak al

ai ai+1

ai+2

Figure 7. A 5­node trapezoid augmented with

nodes

Given a trapezoid aiai+1ai+2ai+3alakaj , if there is a

node ap /∈ P , ap is adjacent to both ai and ai+1 forming a

triangle aiapai+1(as shown in Fig.5), then the union of the

7-node trapezoid and the triangle will constitute a 7-node

trapezoid augmented with a triangle.

Note: If, during an assignment of a 7-node trapezoid

augmented with a triangle, any one of the nodes (say, al)

of the quadrilateral is not considered for the assignment,

then the sub-network forms a 5-node trapezoid augmented

with a triangle as shown in Fig. 6. Also, there could be

other possibilies where a quadrilaterals can be augmented

either with a node or a triangle or both. A 5-node trapezoid

aiakalai+1apai augmented with two nodes aj and ai+2,

where aj is adjacent to ak and ai+2 is adjacent to ai+1, is

shown in Fig. 7. Similarly, Fig. 8 shows a 5-node trapezoid

augmented with a triangle and a node.

4. Proposed Assignment Scheme

For a hexagonal cellular network with demand vector

W = (wi), where wi is the non-homogeneous demand

from cell i and w = max(wi), the rough estimation of the

minimum bandwidth requirement is ws0 [23]. That is, the

required number of channels in each round of assignment

a1 a2 an−ian

ap

ai ai+1ai+2

ai+3

aj ak

Figure 8. A 5­node trapezoid augmented with

a triangle and a node

is primarily limited by the co-site interference constraint s0

only. Also, the minimum value of s0 is restricted by the

values of s1 and s2. When a second frequency channel is

assigned to a node in the network, it may conflict not only

with the channel assigned to the same node but may also

conflict with the channels assigned to other nodes, that are

within distance two, in the previous rounds of assignment.

We now state the following lemma:

Lemma 1 To avoid co-site interference, s0 must be greater

than or equal to 2s1 + s2.

Proof : Let the frequency integer 0 be assigned to one

of the nodes u in a hexagonal cellular network. A second

node v, at a distance one apart from u, can be assigned

the minimum frequency 0 + s1, to avoid interference. A

third node w, at a distance one apart from v and two apart

from u, can be assigned the minimum frequency 0 + 2s1,

to avoid any conflict with the channels assigned to u and

v. If we now, assign a second channel to u to meet another

demand, then that channel has to satisfy the frequency sep-

aration constraints with all the three channels 0, s1 and 2s1

already assigned to u, v and w, respectively. As the highest

frequency channel, out of the three, is 2s1 at w, which is

at a distance two apart from u, the minimum frequency that

can be assigned for the second demand from u is given by

s0 = 2s1 + s2. ✷

We now, propose an assignment scheme, for assigning

s0 integers (0 through s0 − 1) following a definite sequence

to a set of s0 consecutive nodes in a path such that there is

no interference within the set. This assignment scheme is

based on the following lemmas:

Lemma 2 Given a path P = a0a1a2 · · · an−1, we can as-

sign a frequency channel ks1(mods0), k = 0, 1, 2, · · ·, to

the node ak ∈ P , satisfying the frequency separation con-

straints among all the nodes in P .

Proof : We first note that s1 ≥ 1 and s2 ≥ 1. Thus,

s0 ≥ 2 ∗ 1 + 1 = 3. Hence, if we assign ks1(mods0) chan-

nel number to the node ak in the path P = a0a1a2 · · · an−1,

the same channel number may be assigned to two nodes

(reused) which are at least distance three apart. Further, any

two consecutive nodes in the path will be assigned channels

separated by s1. Also, for any three consecutive nodes ak,

ak+1 and ak+2 in the path P , ak and ak+2 will be assigned

channels which are separated by 2s1(mods0) = 2s1 > s2

(since s0 ≥ 2s1 + s2). Thus, the frequency separation con-

straints among all nodes in the path P are satisfied by this

assignment scheme. ✷

Lemma 3 If s0 and s1 are co-prime, then by assigning a

frequency ks1(mods0), k = 0, 1, 2, · · ·, to the node ak in

a path P = a0a1a2 · · · an−1, we can use all the frequency

channels 0, 1, 2, · · · , s0−1 repeatedly on every sequence of

nodes of length s0 occurring in the path P .

Proof : Since, s0 and s1 are co-prime, ks1(mods0),k = 0, 1, 2, 4, 5, · · · · · · will generate s0 distinct numbers

< 0, s1, 2s1 · · · (s0−1)s1 > followed by s0s1(mods0) = 0again. That is, all the s0 numbers 0, 1, 2, · · · , s0 − 1 will be

generated by the above assignment scheme. ✷

Example 1 : Let us assume the values s0 = 7, s1 =2 and s2 = 1. Then by following lemmas 2 and 3, the

sequence of channels < 0, 2, 4, 6, 1, 3, 5 > can be assigned

to the seven nodes a0a1a2a3a4a5a6 in a path, without any

conflict. Now, the sequence can also be assigned repeatedly

to the sets of s0 nodes along the path, without any conflict.

Remark 5 : If s0 and s1 are not co-prime, then the fre-

quency channel 0 will be repeated within a sequence length

less than s0, i.e., all the frequency channels will not be uti-

lized in this case.

We next consider assigning channels to the nodes in a

path augmented with consecutive triangles. We take a set

of triangles and derive a sequence of frequency channels to

assign the nodes of that set, without causing any interfer-

ence. We next repeat the sequence of frequency channels to

the consecutive set of triangles along the path, without any

conflict. Depending upon the different relative values of the

frequency separation constraints, we now state the follow-

ing lemmas:

Lemma 4 For s1 < 2s2, for a subgraph consisting

of a path P = a0a1a2 · · · agai+1ai+2 · · · an−1 aug-

mented with two consecutive triangles aiajai+1ai and

ai+2aj+1ai+3ai+2, aj , aj+1 /∈ P , we can assign channels

to all nodes in such a subgraph with a bandwidth of 6s2.

Proof :

We start with assigning channel 0 to the node ai as shown

in Fig. 2. We next assign the channel s2 to the node ai+2

which is at distance two from ai. Now, node aj is at dis-

tance two from ai+2. Thus, we successively choose the

nodes aj , ai+3, ai+1 and aj+1, where any two consecutive

nodes in this order are distance two apart from each other,

and assign the channels 2s2, 3s2, 4s2 and 5s2, respectively,

without any interference (as s1 < 2s2). For the second

round of assignment, we have to start with assigning chan-

nel 6s2 to the node ai. Thus, the channels can be assigned

to all nodes of this structure with a bandwidth of 6s2. ✷

Lemma 5 For s1 ≥ 2s2, for a subgraph consisting

of a path P = a0a1a2 · · · aiai+1ai+2 · · · an−1 aug-

mented with two consecutive triangles aiajai+1ai and

ai+2aj+1ai+3ai+2, aj , aj+1 /∈ P , we can assign channels

to all nodes in such a subgraph with a bandwidth of 3s1.

Proof : We proceed similar to Lemma 4 by choosing

successive nodes ai, ai+2, aj , ai+3, ai+1 and aj+1, where

any two consecutive nodes in this order are distance two

apart from each other (as shown in Fig.2). We assign the

frequency channels 0 and s2 to the nodes ai and ai+2, re-

spectively, without any interference. The next node aj is at

a distance two apart from ai+2 but one apart from ai. As

s1 ≥ 2s2, the frequency channel s1 is assigned to aj to

meet the frequency separation requirements. Following the

same logic, we next assign the channels s1 + s2, 2s1 and

2s1 +s2 to the nodes ai+3, ai+1 and aj+1, respectively, sat-

isfying the frequency separation requirements with all the

previously assigned nodes. To assign channels in the sec-

ond round (to meet the second demand on each node), we

have to start with assigning the channel 3s1 to the node ai.

Hence, the channels can be assigned to all nodes of this

structure with a bandwidth of 3s1. ✷

Remark 6 : The sequence of channels derived in lem-

mas 4 and 5 can be repeatedly used if the path is augmented

with further sets of consecutive two triangles.

We now consider assignment of a path augmented with a

7-node trapezoid, as shown in Fig. 4. It is seen that the se-

quence of channels derived for assigning a path augmented

with a 7-node trapezoid can be repeated conveniently to as-

sign all the nodes in two adjacent paths, without any con-

flict. We now state the following lemmas.

Lemma 6 For s1 < 2s2, for a subgraph consisting of a

path P = a0a1a2 · · · aiai+1ai+2 · · · an−1 augmented with

a 7-node trapezoid aiai+1ai+2ai+3alakaj , aj , ak, al /∈ P ,

we can assign channels to all nodes in such a subgraph with

a bandwidth of 7s2.

Proof :

We start with assigning the channel 0 to the node ai,

as shown in Fig. 4. We next choose successive nodes

ak, ai+3, ai+1, al, aj and ai+2, where any two consecutive

nodes in this order are distance two apart from each other,

and assign channels s2, 2s2, 3s2, 4s2, 5s2 and 6s2, respec-

tively, without any conflict (as s1 < 2s2). For the second

round of assignment (to meet the second demand on each

node), we have to start with assigning the channel 7s2 to the

node ai. Hence, the channels can be assigned to all nodes

of this structure with a bandwidth of 7s2. ✷

Lemma 7 For s1 ≥ 2s2, for a subgraph consisting of a

path P = a0a1a2 · · · aiai+1ai+2 · · · an−1 augmented with

a 7-node trapezoid aiai+1ai+2ai+3alakaj , aj , ak, al /∈ P ,

we can assign channels to all nodes in such a subgraph with

a bandwidth of 3s1 + s2.

Proof :

We proceed similar to lemma 6, and choose succes-

sive nodes ai, ak, ai+3, ai+1, al, aj and ai+2, where any

two consecutive nodes in this order are distance two apart

from each other. We assign the channels 0, s2 and 2s2 to

the nodes ai, ak and ai+3, respectively, without any in-

terference. The next node ai+1 is at a distance two apart

ai ai+1ai+2ai+3

aj aj+1aj+2(1) (4)(5)

aj+3aj+4aj+5aj+6

ai+4ai+5ai+6

(5) (1) (4)

(0) (3) (6) (2)

(0) (3) (6) (2)

Figure 9. Assignment of two adjacent paths

from ai+3 but one apart from ak. As s1 ≥ 2s2, the fre-

quency channel s1 + s2 is assigned to ai+1 to meet the fre-

quency separation requirements. We next assign the chan-

nels s1 + 2s2, 2s1 + s2 and 2s1 + 2s2 to the nodes al, aj

and ai+2, respectively, to avoid interference. To assign the

second round of channels, we have to start with assigning

the channel 3s1 + s2 to the node ai. Thus, we can assign

channels to all nodes in such a subgraph with a bandwidth

of 3s1 + s2. ✷)

Example 2 : For s0 = 5, s1 = 2 and s2 = 1, we have

s1 = 2s2. Then by lemma 7, the channels 0, 1, 2, 3, 4, 5

and 6 can be assigned to the nodes ai, ak, ai+3, ai+1, al, aj

and ai+2, respectively, of the 7-node trapezoid, as shown in

Fig. 9. It can be seen that the 7-node trapezoid can now be

inverted along with its assigned frequencies and augmented

with the path, without any interference. Thus, the sequence

of channels can be repeated with such a pair of 7-node trape-

zoid to cover all nodes in two adjacent paths of the network,

without any interference.

4.1 Assignment of a quadrilateral Augmentedwith Nodes and Triangles

When s1 < 2s2:

i) We consider the assignment of the 7-node trapezoid

augmented with a triangle, as shown in Fig. 5. We

start with assigning channel 0 to the node ai. Node ak

is at a distance two apart from ai. Proceeding similarly,

we choose next successive nodes ak, ai+3, ai+1al, aj , ap

and ai+2, where any two consecutive nodes in the order

are distance two apart from each other, and assign chan-

nels s2, 2s2, 3s2, 4s2, 5s2, 6s2 and 7s2, respectively, with-

out any conflict (as s1 < 2s2). To assign the second round

of channels, we have to start with assigning channel 8s2 to

the node ai. Hence, we can assign channels to all nodes in

such a subgraph with a bandwidth of 8s2.

ii) We next consider the assignment of the 5-node

trapezoid augmented with a triangle, as shown in Fig.

6. We follow similar strategy as above and assign

the channels 0, s2, 2s2, 3s2, 4s2, 5s2, 6s2 to the nodes

ai, ak, ai+3, ai+1ap, aj , ai+2, respectively. For the second

round of assignment, we have to start with assigning fre-

quency channel 7s2 to the node ai. Hence, we can assign

channels to all nodes in such a subgraph with a bandwidth

of 7s2.

0 1 2 3 4

5 6 7 8 9 10 11

12 13 14 15 16 17

18 19 20

Figure 10. The benchmark cellular network

0 1 2 3 4

5 6 7 8 9 10 11

12 13 14 15 16 17

18 19 20

Figure 11. A pair of sparse paths selected in

the first phase of assignment

iii) For the 5-node trapezoid augmented with two nodes,

as shown in Fig. 7, we proceed similar to the case above and

assign the channel 0, s2, 2s2, 3s2, 4s2, 5s2, 6s2 to the nodes

ai, ak, ai+3, ai+1ap, aj , ai+2. Considering the assignment

for the second demand from the node ai, the required band-

width will be 7s2.

iv) For the 5-node trapezoid augmented with a tri-

angle and a node, as shown in Fig. 8, we assign

the channels 0, s2, 2s2, 3s2, 4s2, 5s2, 6s2 to the nodes

ai+2, aj , ap, ak, ai, ai+3, ai+1. Considering the second de-

mand from node ai+2, the required bandwidth will be 8s2.

When s1 ≥ 2s2, it can be seen that the assignments fol-

lowing the similar strategy can be done with the correspond-

ing bandwidth requirements of 3s1+2s2, 3s1+s2, 3s1+s2

and 3s1 +2s2 for the above four cases i), ii), iii) and iv), re-

spectively.

5. Assignment Algorithm

We now discuss below our proposed assignment algo-

rithm stepwise , illustrating each step with the help of the

benchmark problem 6 (as shown in tables 1 and 2) as an

example. The problem is defined on a hexagonal cellular

network of 21 nodes (node 0 through 20), as shown in Fig.

10, with non-homogeneous demand vector D2, as given

in Table 1. We first partition the CAP into a sequence of

subproblems, each consisting of a specific subgraph with

a uniform demand to be satisfied at each node of this sub-

graph. Solution to such a subproblem will constitute an as-

signment phase, and the whole assignment process for the

Table 1. Two different demand vectors for the benchmark problemsCell nos. 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20D1 8 25 8 8 8 15 18 52 77 28 13 15 31 15 36 57 28 8 10 13 8D2 5 5 5 8 12 25 30 25 30 40 40 45 20 30 25 15 15 30 20 20 25

Table 2. Specification of the benchmark problemsProblems 1 2 3 4 5 6 7 8Frequency s0 5 5 7 7 5 5 7 7separation s1 1 2 1 2 1 2 1 2constraints s2 1 1 1 1 1 1 1 1Demand vector D1 D1 D1 D1 D2 D2 D2 D2

original given problem will thus consist of all such assign-

ment phases. When we move from one assignment phase,

say phase i, to the next phase i + 1, we must ensure that to

start with, every node u in the subgraph of phase i + 1 is

assigned a channel number that does not cause co-site inter-

ference with the channel last assigned to u in phase i. This

may be satisfied if i) either u was not included in the sub-

graph at phase i or, ii) u was last assigned with a channel

number f in phase i and then a channel f ′ is assigned to uin phase i + 1 with f ′ − f ≥ s0.

In our algorithm, we select first the simplest subgraph

(e.g., a set of sparse paths) of the network conveniently so

that it includes at least all the highest demand nodes, and

then gradually try for more complicated structure as men-

tioned in section 3. The criterion for selecting such a sub-

graph in each phase is to include the maximum number of

nodes along with the highest and the relatively higher de-

mand nodes for the assignment. The assignment algorithm

is now described in the following steps:

Step 1: Find a subgraph with a set of sparse paths such

that the node(s) with the highest demand is(are) included in

this subgraph. Set w′ = w.

Step 2 : Assign a channel to each node in the chosen

set of sparse paths following the scheme given in lemmas

2 and 3, and repeat this for H1 successive rounds where

H1 ≤ w′, and H1 is the maximum possible value such

that after these H1 rounds of assignment, no node in the

whole network has still now an unsatisfied demand more

than that of the node(s) having the maximum demand in the

subgraph. The residual demands of all nodes in the chosen

set of sparse paths are reduced uniformly by H1, and the

maximum residual demand amongst all nodes in the net-

work is now set to w′ = w′ − H1.

Example 3 :

Let the simplest subgraph of the example network (Fig.

10) be a pair of sparse paths formed by the paths 5 6 7 8

9 10 11 and 18 19 20 for the first phase of assignment, as

0 1 2 3 4

5 6 7 8 9 10 11

12 13 14 15 16 17

18 19 20

Figure 12. Pairs of sparse paths selected in

the second phase of assignment

shown by solid lines in Fig. 11. The subgraph not only

cover the node 11, having the highest demand but also the

nodes 9 and 10, having relatively higher demand along with

seven other nodes of the network. The nodes outside the

subgraph have relatively lower demands and left out of this

phase of assignment. From Table 1 and 2, we have w =45, s0 = 5, s1 = 2 and s2 = 1. We follow lemmas 2

and 3, and generate a sequence 0, 2, 4, 1, 3 of five integers

for assigning the nodes of the subgraph. We next, choose

the homogeneous demand vector H1 = 15, H1 ≤ w′ and

repeat the assignment for fifteen successive rounds. After

the fifteenth round of assignment, it is seen from Table 3

that the maximum unsatisfied demand in the whole network

becomes same as that of the sub-graph and is equal to 30.

The last row of Table 3 denotes the residual demands RD21

for all the nodes in the whole network after the first phase

of assignment.

Step 3: Look for any other set of sparse paths such that

the node(s) with the highest demand is(are) included in this

subgraph. If such a subgraph is found, then repeat step 1

with this subgraph; otherwise, go to step 4.

Example 4 :

From RD21of Table 3, it is seen that nodes 11 and 17

have highest demand of 30, for the example network. It

0 1 2 3 4

5 6 7 8 9 10 11

12 13 14 15 16 17

18 19 20

Figure 13. Pairs of sparse paths augmented

with triangles selected in the third phase of

assignment

is again possible to include both the highest demand nodes

by pairs of sparse paths 5 13, 0 7 15 19, 2 9 17 and 4 11,

as shown by solid lines in Fig. 12, in the second phase of

the assignment along with nine more nodes. We proceed

similar to example 3 and repeat the finite sequence of five

integers along the pairs of sparse paths to cover the sub-

network, maintaining all the frequency separation criteria

with the last round of assignment in the first phase. We next,

choose the homogeneous demand vector H1 = 5, H1 ≤ w′,

for the sub-graph and repeat the assignment for five succes-

sive rounds. After the fifth round of assignment, it is seen

from Table 4 that the maximum unsatisfied demand in the

whole network becomes same as that of the subgraph and

is equal to 25. The last row of Table 4 denotes the residual

demands RD22for all the nodes after the second phase of

assignment.

/* Now the next phase begins with the next difficult

structure, such as a set of sparse paths augmented with tri-

angles */

Step 4: Find a subgraph with a set of sparse paths aug-

mented with triangles such that the node(s) with the highest

demand is(are) included in this subgraph (as it was not pos-

sible to cover all the highest demand nodes with just a set

of sparse paths).

Step 5 : Assign a channel to each node in the chosen

subgraph according to the schemes mentioned in lemmas 4

and 5 and also taking care of inter-phase frequency sepa-

ration requirements as discussed above, and repeat this for

H2 successive rounds where H2 ≤ w′, and H2 is the max-

imum possible value such that after these H2 rounds of as-

signment, no node in the whole network has still now an

unsatisfied demand more than that of the node having the

maximum demand in this subgraph. Reduce the residual

demand of each node in this subgraph uniformly by H2 and

then set w′ = w′ − H2.

Example 5 :

From RD22of Table 4, it is seen that nodes 10, 11, 13,

14 and 17 have highest demand of 25. We now select a set

of sparse paths augmented with triangles, as shown by solid

0 1 2 3 4

5 6 7 8 9 10 11

12 13 14 15 16 17

18 19 20

Figure 14. Pairs of sparse paths augmented

with triangles selected in the forth phase of

assignment

lines in Fig. 13, in the third phase of the assignment, such

that all the highest demand nodes are included. Path 12 13

14 15 16 17 10 11 is augmented with the triangles formed

by the nodes 4 10 11, 5 12 13, 14 15 18 and 16 17 20, re-

spectively. The node 1 is also included, considering it to

be a vertex of a triangle augmented with a virtual path not

accommodated in the network. By lemma 5 and for s2 = 1,

we follow the assignment scheme where a sequence of six

integers 0, 4, 2, 1, 3 and 5 is assigned to the each node of

two consecutive triangles augmented with a path. We repeat

the sequences in the same order of on every two consecu-

tive triangles augmented with the path, without any conflict.

We now choose the homogeneous demand vector H2 = 5,

H2 ≤ w′, for the sub-graph and repeat the assignment for

five successive rounds. After the fifth round of assignment,

it is seen from Table 5 that the maximum unsatisfied de-

mand in the whole network becomes same as that of the

sub-graph and is equal to 20. The last row of Table 5 de-

notes the residual demands RD23for all the nodes after the

third phase of assignment.

Step 6: Look for any other set of sparse paths augmented

with triangles such that the node(s) with the highest de-

mand is(are) included in this subgraph. If such a subgraph

is found, then repeat step 5 with this subgraph; otherwise,

go to step 7.

Example 6 :

From RD23of Table 5, nodes 9, 10, 11, 13, 14 and 17

have highest demand of 20. It is once again possible to

include all the highest demand nodes by paths augmented

with triangle, as shown by solid lines in Fig. 14, in the forth

phase of the assignment. The path 8 9 10 11 is augmented

with triangle 9 10 17, and the path 12 13 14 15 is augmented

with triangle 6 13 14. The node 20 is also included, consid-

ering it to be a vertex of a triangle augmented with a virtual

path not accommodated in the network. Proceeding simi-

lar to the example 5, we follow lemma 5 and assign all the

nodes within the selected subgraph. We now choose the ho-

mogeneous demand vector H2 = 10, H2 ≤ w′, for the sub-

graph and repeat the assignment for ten successive rounds.

Table 3. Residual demands after the first phase of assignment

Cell nos. 0 1 2 3 4 (5) (6) (7) (8) (9) (10) (11) 12 13 14 15 16 17 (18) (19) (20)D2 5 5 5 8 12 (25) (30) (25) (30) (40) (40) (45) 20 30 25 15 15 30 (20) (20) (25)RD21

5 5 5 8 12 (10) (15) (10) (15) (25) (25) (30) 20 30 25 15 15 30 (5) (5) (10)

Table 4. Residual demands after the second phase of assignment

Cell nos. (0) 1 (2) 3 (4) (5) 6 (7) 8 (9) 10 (11) 12 (13) 14 (15) 16 (17) 18 (19) 20RD21

(5) 5 (5) 8 (12) (10) 15 (10) 15 (25) 25 (30) 20 (30) 25 (15) 15 (30) 5 (5) 10RD22

5 8 (7) (5) 15 (5) 15 (20) 25 (25) 20 (25) 25 (10) 15 (25) 5 10

Table 5. Residual demands after the third phase of assignment

Cell nos. 0 (1) 2 3 (4) (5) 6 7 8 9 (10) (11) (12) (13) (14) (15) (16) (17) (18) 19 (20)RD22

(5) 8 (7) (5) 15 5 15 20 (25) (25) (20) (25) (25) (10) (15) (25) (5) (10)RD23

8 (2) 15 5 15 20 (20) (20) (15) (20) (20) (5) (10) (20) (5)

Table 6. Residual demands after the forth phase of assignment

Cell nos. 0 1 2 3 4 5 (6) 7 (8) (9) (10) (11) (12) (13) (14) (15) 16 (17) 18 19 (20)RD23

8 2 (15) 5 (15) (20) (20) (20) (15) (20) (20) (5) 10 (20) (5)RD24

8 2 (5) 5 (5) (10) (10) (10) (5) (10) (10) 10 (10)

Table 7. Residual demands after the fifth phase of assignment

Cell nos. 0 1 2 (3) 4 5 (6) (7) 8 (9) (10) (11) (12) (13) (14) 15 (16) (17) 18 19 20RD24

(8) 2 (5) (5) 5 (10) (10) (10) (5) (10) (10) (10) (10)RD25

(3) 2 (5) (5) (5) (5) (5) (5) (5) (5)

Table 8. Residual demands after the sixth phase of assignmentCell nos. 0 1 2 3 (4) 5 6 7 (8) (9) (10) (11) 12 (13) (14) 15 (16) (17) 18 19 20RD25

3 (2) (5) (5) (5) (5) (5) (5) (5) (5)RD26

3 (3) (3) (3) (3) (3) (3) (3) (3)

Table 9. Residual demands after the seventh phase of assignmentCell nos. 0 1 2 (3) 4 5 6 7 (8) (9) (10) (11) 12 (13) (14) 15 (16) (17) 18 19 20RD26

(3) (3) (3) (3) (3) (3) (3) (3) (3)RD27

(0) (0) (0) (0) (0) (0) (0) (0) (0)

0 1 2 3 4

5 6 7 8 9 10 11

12 13 14 15 16 17

18 19 20

Figure 15. Trapezoids augmented with trian­

gles selected in the fifth phase of assignment

After the tenth round of assignment, it is seen from Table 6

that the maximum unsatisfied demand in the whole network

becomes same as that of the sub-graph and is equal to 10.

The last row of Table 6 denotes the residual demands RD24

for all the nodes after the forth phase of assignment.

Step 7: Find a subgraph with paths augmented with

quadrilaterals such that the node(s) with the highest demand

is(are) included in this sub-graph (as, it was not possible

to cover all the highest demand nodes with just the set of

sparse paths or paths augmented with triangles).

Step 8 : Assign a channel to each node in this subgraph

following the schemes mentioned in lemmas 6 and 7, and

also taking care of inter-phase frequency separation require-

ments as discussed above, and repeat this for H3 successive

rounds where H3 ≤ w′, and H3 is the maximum possi-

ble value such that after these H3 rounds of assignment,

no node in the whole network has still now an unsatisfied

demand more than that of the node having the maximum

demand in this subgraph. Reduce the residual demand of

each node in this subgraph uniformly by H3, and then set

w′ = w′ − H3.

Example 7 : From RD24of Table 6, nodes 9, 10, 11, 13,

14, 16 and 17 have highest demand of 10. It is now pos-

sible to include all the highest demand nodes by a 7-node

trapezoid and a 5-node trapezoid augmented with a nodes,

as shown by solid lines in Fig. 15, in the fifth phase of

assignment. Nodes 3, 6, 7 and 12 are also included along

with the highest demand nodes. For s1 ≥ 2s2 and s2 = 1,

we assign the subgraph by following lemma 7 and the as-

signment scheme described in sub-section 4.1. We choose

the homogeneous demand vector H3 = 5, H3 ≤ w′, for

the sub-graph and repeat the assignment for five successive

rounds. After fifth round of assignment, it is seen from Ta-

ble 7 that the maximum unsatisfied demand in the whole

network becomes same as that of the sub-graph and is equal

to 5. The last row of Table 7 denotes the residual demands

RD25for all the nodes after the fifth phase of assignment.

Step 9: Look for any other set of sparse paths augmented

with quadrilaterals such that the node(s) with the highest de-

mand is(are) included in this subgraph. If such a subgraph

0 1 2 3 4

5 6 7 8 9 10 11

12 13 14 15 16 17

18 19 20

Figure 16. Trapezoids augmented with trian­

gles selected in the fifth phase of assignment

0 1 2 3 4

5 6 7 8 9 10 11

12 13 14 15 16 17

18 19 20

Figure 17. Trapezoids augmented with trian­

gles selected in the fifth phase of assignment

is found, then repeat step 8 with this subgraph; otherwise

stop.

Example 8 :

From RD25of Table 7, nodes 8, 9, 10, 11, 13, 14, 16 and

17 have highest demand of 5. It is possible to include all

the highest demand nodes by a 5-node trapezoid augmented

with a triangle and a path consisting of only two nodes. as

shown by solid lines in Fig. 16, in the sixth phase of assign-

ment. For s1 ≥ 2s2 and s2 = 1, we assign the subgraph by

following the assignment scheme described in sub-section

4.1. Proceeding similarly, we choose the homogeneous de-

mand vector H3 = 2, H3 ≤ w′, for the sub-graph and

repeat the assignment for two successive rounds. After the

second round of assignment, it is seen from Table 8 that

the maximum unsatisfied demand in the whole network be-

comes same as that of the sub-graph and is equal to 3. From

the last row of Table 8, it is seen that the residual demands

RD26for all the nodes is same after the sixth phase of as-

signment and is equal to 3.

Example 9 :

From Table 8, nodes 3, 8, 9, 10, 11, 13, 14, 16 and 17

have same same of 3. It is possible to include all the high-

est demand nodes by a 5-node trapezoid augmented with

a triangle and a node, and a path consisting of only two

nodes. as shown by solid lines in Fig. 17, in the last phase

of the assignment. For s1 ≥ 2s2 and s2 = 1, we assign the

subgraph by following the assignment scheme described in

sub-section 4.1, and after repeating the assignment for three

successive round the assignment of the example network is

complete for the non-homogeneous demand vector D2.

Remark 7 : With the above algorithm, the assignment of

the benchmark problem 6 requires 264 channels with an ex-

ecution time of less than 50 milliseconds on an HP xw8400

workstation. The optimal bandwidth required for this prob-

lem is 253 [23]. Thus, the deviation from optimality in the

required bandwidth with our algorithm is less than 5%. For

other benchmark problems as well, we have tested our algo-

rithm to generate assignments with bandwidth always less

than 5% away from the optimal one, and requiring execu-

tion time much less than 50 milliseconds. These results may

be contrasted with the so far best known results reported in

[24] which assigns the channels with an optimal bandwidth

but needs an execution time of 10-20 seconds on a compa-

rable HP workstation. Thus, our proposed algorithm will

be very attractive in real-life situations where very fast as-

signment of channel is demanded, although the assignment

bandwidth may be marginally away from the optimal one.

6. Conclusion

We have proposed a novel technique for optimal chan-

nel assignment in a hexagonal cellular network with non-

homogeneous demands in a 2-band buffering system. We

have partitioned the channel assignment problem into a se-

quence of very simple subproblems, each consisting of a

subgraph with a specific structure and having a uniform de-

mand at each node of this subgraph, so that each such sub-

problem can be solved very quickly by repeated application

of a simple assignment scheme. We have then presented

an algorithm based on our novel technique for solving the

channel assignment problem with non-homogeneous de-

mands. We have illustrated our algorithm with the help of

the Philadelphia benchmark problem 6, which is regarded

as the most difficult one in the literature. When tested on

these benchmark problems, our proposed algorithm pro-

vides near-optimal assignment (with a bandwidth never re-

quiring more than 5% from the optimal requirement) in

each case, with an execution time always less than 50 mil-

liseconds. These features may be contrasted with the exist-

ing best known results [24] which generate the assignments

with optimal bandwidth, but with a lot of execution time

(10-20 seconds). Thus, the proposed algorithm is very suit-

able for real-life applications where very fast assignment of

channel is necessary, but a bandwidth marginally away from

the optimal one may be tolerable.

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