chapter 4 optimal channel allocation algorithm...

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CHAPTER 4

OPTIMAL CHANNEL ALLOCATION ALGORITHM WITH EFFICIENT BANDWIDTH RESERVATION FOR CELLULAR

NETWORKS

4.1 Introduction

In cellular wireless networks, the challenges are increasing more and more as

there are more applications being developed in the technology. Optimization leads to

many more challenges to dig out more from the scientific growth [11].

The enormous growth of mobile users and applications in the present scenario

with limited bandwidth is a challenging task today. The bandwidth should be utilized

in an optimal way so that more number of users may be serviced. One of the important

factors to improve the quality of cellular service is to make handoffs nearly invisible to

the user and successful [5]. Unsuccessful, handoff requests are one of the main causes

of end to end delay.

In spite of the last several years of research on wireless networks, massive real-

life deployments of networks still remain a challenge. Although the freedom of

networks from utilizing fixed infrastructure for offering wireless communication

services makes them attractive for fast deployment in application domains such as the

military and emergency services, they are limited by their ability to efficiently offer

global accessibility and web-based services such as file sharing, messenger services

and voice-over-IP [12]. Nodes of traditional cellular wireless networks are maintained

by a base station manager (BSM) or server for routing. On the other hand, nodes of

purely networks behave as routers by relaying messages in order to improve the

performance of the network. One of the most important issues in providing ubiquitous

communication is mobility management [13], which primarily concern effectively

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tracking the locations of the nodes. In case of hybrid networks, BSM can be used for

effective mobility management, which can be otherwise more challenging in cellular

networks, because of their lack in using a dedicated router/server having a network-

wide knowledge of the location of the nodes. In the case of pure cellular wireless

networks, the goal of adaptive call admission control is to ensure that there is

sufficient bandwidth reservation for handoff, i.e., for transferring an ongoing call in a

cell to another. The reserved bandwidth in a target cell is proportional to the traffic

intensity in the surrounding cells [3]. In the absence of sufficient bandwidth for

handoff, new connections are subject to getting dropped. One common approach used

to reduce the connection dropping rate (CDR) is to reserve some bandwidth solely for

handoff use [4]. [1]

In a cellular network, a mobile user may visit different cells in his lifetime. In

each of these cells, resources must be made available to support the mobile user else

the user will suffer a forced termination of his call in progress. Therefore, careful

resource allocation along with call admission control is required to mitigate the

chances of forced termination or dropping of a call. Keeping the probability of a user

getting dropped (Pdrop) below a pre-specified target value is considered as a practical

design goal of any resource allocation scheme. Achieving the above goal provides the

probabilistic quality of service (QoS) guarantee as desired by a mobile user. [9]

Early work in handoff prioritization proposes the static reservation of

bandwidth at each BS as a solution [14], in which a fixed portion of the radio capacity

is permanently reserved for handoffs. However, such a static approach is unable to

handle variable traffic load and mobility [15]. [10]

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Fig. 4.1 Cellular Network

4.2 System Model

In the Bandwidth reservation based on user mobility scheme, we assume that

base stations are equipped with road-map information and that mobile stations are

equipped with global positioning systems (GPS) devices. Mobile stations periodically

report their GPS location information to their base stations. Based on the location

information of the mobile stations at two consecutive epochs, the base stations

estimate the speed and moving direction of the mobile stations. Furthermore, the base

stations estimate the probability that the mobile stations will enter the neighbouring

cells based on their velocity and the road-map information stored in the base stations.

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The base stations then compute the amount of bandwidth to be reserved, based on such

estimation. With the road-map information, the base stations can make a more-

accurate prediction on the user’s mobility and, hence, reduce unnecessary bandwidth

reservation. The cellular network architecture is shown in Fig. 4.1 and the structure of

the cellular network estimated is shown in Fig. 4.2

Indicates road

MH Mobile Host

Fig. 4.2 Structure of a Cellular Network

Mobile user’s motion pattern is mostly restricted by man-made constructions,

such as roads. Obviously, mobile users who are making handoff requests are in

motion; most are either walking or riding in automobiles on the roads. That is to say,

mobility is always restricted by the layout of the roads. A mobile user’s mobility is

restricted in the sense that mobile users can only move on the roads. We propose a

new mobility prediction and bandwidth-reservation method called “bandwidth

reservation based on user mobility”. This scheme makes use of a mobile user’s

moving speed, direction, and the road information stored in the base stations to predict

the handoff probabilities to neighbouring cells. The amount of reserved bandwidth is

MH3

MH4

MH1

MH5

MH2 . .

.

.

.

.

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dynamically adjusted according to the handoff probability and the traffic load in each

cell.

4.2.1 Measuring Traffic Intensity to Know How Many Channels to Reserve at

Each Base Station

The intensity of the traffic varies during the day. It is normally high in morning

business hours and less in afternoon during lunchtime and again goes high in evening.

The traffic in busiest periods is called busy-hour traffic. The intensity of the busy-hour

traffic also varies depending on the day of the week. During weekends traffic in

morning time may be less and evening time will be high. The network operator needs

to meet the demands of the average busy-hour traffic.

Traffic intensity can be measured using two dimensionless units

1.) Erlang

2.) Circuit Centum Seconds (CCS)

One Erlang is equivalent to number of calls (made in one hour) multiplied by

the duration of these calls (in hours). Each call has a different duration or a different

call holding time: for traffic intensity measurements the average call holding time is

taken into account [7] [8]. The typical values for average call holding time vary

between 120 and 180 seconds. Therefore, the traffic intensity in Erlangs can be

defined as:

Number of call in an hour * average call holding time in sec T (in Erlangs) = (1) 3600

If a call attempt is made when all channels in cellular networks are serving

other calls, the call attempt will be blocked. The probability of call blocking in a

telecommunication network is called Grade of Service (GoS). The Grade of Service of

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a telecommunication network varies between zero and one. A GoS of 0.02 is normally

taken as acceptable for communication systems.

4.2.2 Estimating the Number of Subscribers in the Cellular System

The number of subscribers in the system can be estimated assuming the

relation between the number of subscribers in the busy hour (n) and the number of

calls per hour per cell. The maximum number of calls per hour that a cell can take

depends on the number of channels allocated for that cell based on traffic conditions

under its geographic area. The estimated number of subscribers in the system M is

∑ maximum number of calls per cell M= (2) n

In this chapter, we consider the flows to be as originating and handoff flows.

The originating flows are the flows which originated in a particular cell and the

handoff flows are generated when a mobile user moves from one cell location to

another. The cross layer architecture proposed in [6] is used for bandwidth reservation

scheme proposed in this paper. We also use the architecture proposed in [1]. In this

architecture, the call admission control estimates the available bandwidth and reserves

the same for handoff flows.

Fig. 4.3 System Model

BWO - represents the bandwidth reserved for originating flows

BWC - represents the bandwidth which can be utilized by both originating and

handoff flows

BWH - represents the bandwidth which can be utilized by only handoff flows.

BWO BWC BWH

SR

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In this paper, we propose the bandwidth reservation scheme which reserves

some amount of bandwidth for handoff flows and some amount of bandwidth for

originating flows and keeps some amount of bandwidth which can be used by both

handoff flows and originating flows [2].The bandwidth procedure defined in this paper

is fair-fair reservation which maintains the minimum delay and provide high

throughput. To allocate the bandwidth for originating flow, the BWO region is verified.

If bandwidth can be allocated, it is done otherwise the BWC is checked and bandwidth

is allocated. Finally if the bandwidth cannot be allocated the originating flow is

blocked. To allocate the bandwidth for handoff flow, the BWH region is verified. If

bandwidth can be allocated, it is done otherwise the BWC is checked and bandwidth is

allocated. Finally if the bandwidth cannot be allocated the handoff flow is dropped.

This procedure is explained below.

4.3 Bandwidth Reservation Procedure

1. For originating flow

a. if bandwidth can be allocated from BWO then

i. bandwidth is allocated from BWO

b. else

i. if bandwidth can be allocated from BWC then

1. bandwidth is allocated from BWC

ii. else

1. call is blocked

2. else

3. For handoff flow

a. if bandwidth can be allocated from BWH then

i. bandwidth is allocated from BWH

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b. else

i. if bandwidth can be allocated from BWC then

1. bandwidth is allocated from BWC

ii. else

1. call is dropped

The flow diagram of the above given procedure is shown in Fig. 4.4

Fig. 4.4 Flow diagram for the Bandwidth reservation procedure

Originating Handoff Flow

Drop the Call

Start

Is BWO range free?

Is BWH

range free?

Is BWC range free?

Is BWC range free?

Allocate bandwidth

Block the call

Stop

•

Yes Yes

Yes Yes

No No

No No

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4.4 Sequence Flow Diagrams

4.4.1 Mobile Originated Wireless Calls

Fig. 4.5 Sequence Flow Diagram for Mobile Originated Wireless Calls

In the above sequence flow diagram Fig. 4.5, when the mobile station or user

dials a number and starts a call the setup message is sent to the Mobile Switching

Center (MSC). It will identify from the visitor location register (VLR) whether the

user is valid user or not. If the user is valid, call is processed and will inform the base

station (BS) to provide a channel to the user. Then the base station will provide the

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channel to the user and a channel assignment complete message is received by both

BS and MSC. Upon receiving the number of mobile station 1 has dialed MSC sends an

Initial Address Message to the mobile station 2 to call setup. The mobile station 2 then

sets up the call and send an address complete message. When the mobile station 2

answers the call, connection will be established between two mobiles stations and

connect acknowledge message is sent to the mobile switching center.

4.4.2 Mobile Terminated Wireless Calls

Fig 4.6 Sequence Flow Diagram for Mobile Terminated Wireless calls

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The above Fig. 4.6 shows about mobile terminated wireless calls. At the end of

the call the mobile station 1 sends call disconnect message to MSC. The MSC then

sends a release message to the other mobile station 2 and also to the mobile station 1

to start the clearing procedure. After the clearing procedure, the mobile station 1 sends

release complete message, the mobile station 2 also sends release complete message to

MSC. After receiving that the MSC sends a message to base station to release all the

allocated dedicated resources using clear command. The base station then sends

channel release message to release the allocated channel to the mobile station 1. After

the release of channel the base station informs it to the MSC by clear complete

message.

4.4.3 Call Handoff

The Fig. 4.7 shows how the call flow scenario takes place when the mobile

host takes handoff, moving from one base station to another base station when the

serving and target base stations do not reside within the same BSS. Here MSC will

coordinate the handover and performs switching tasks between serving and target base

stations. When the mobile station determines that a handover is required, it sends the

measurement report message to serving base station. The base station sends a

handover request message to MSC. The MSC then checks which target base station it

is intended to go and sends handover request message to it. The handover is performed

and the target base station should be able to provide the channel for it. If the channel is

free the target base station sends an acknowledgement message to MSC. Then the

handover message is send to mobile host through the serving base station. The mobile

host responses with handover access message to target base station. After the handover

is completed the target base station sends handover complete message to MSC. Then

MSC will send message to serving base station to release the old channel.

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Fig. 4.7 Sequence Flow Diagram for handoff calls

4.5 Analytical Analysis

The no of channels in a system are divided into three groups. Why here it has

taken three groups is, it is the minimum number where it can have the reusability of

channels available. If it takes four or five and so on… the time taken in finding the

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group and the time taken in allocating the free channels in it increases. So it has

considered only three groups namely Gp, Gq, Gr. Here it has considered seven cells in

a cluster. It can also consider 3 or 5 or 9 or 12 so on... cells in a cluster. Since in most

of the cases it was taken as seven cells in a cluster it has also considered seven as the

number of cells in a cluster. The cluster after dividing them into three groups it

appears as shown below in one particular case.

In this dynamic channel allocation with reusing concept which one will yield

better the grouping of channels into equally or unequally? It is going to consider all

cases like grouping channels equally and reserved channels for both unequal and

equal. In other case like grouping channels unequally and reserved channels for both

equal and unequal. How the maximum number of users it supports without blocking or

dropping will differ it is going to show. Another problem is after reserving some

channels initially in a system whether done equally or unequally from the above

groups, how many more channels to reserve after that is a difficult problem. When

almost all the cells in a given system has occurring the same number of handoff calls

then it can reserve the channels in the system later by choosing a constant K value.

Where K ranges from 0 to 0.5 .It will give the number of channels to be reserved for a

given cell for handoff calls, when it is multiplied with the no of channels in that

group. But if there are both hot cells and cold cells in a system where in hot cell the

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handoff calls occurs more and in cold cells less number of handoff occurs. By

choosing a same K value for both the hot cells as well as cold cells will result in more

dropping as well as blocking of calls. So In that case it is better to give a high K value

for hot cells and low K value for cold cells. The mathematical formulas for finding the

maximum no of users the system supports without blocking and dropping of calls for

the above said cases are shown below.

4.5.1 Formulas to find the maximum no of users the cellular network can handle

without blocking

Case 1: When there are C channels in a system and g is the number of groups, and the

C channels are divided equally among these groups if R channels are the no of

reserved from each group then the total no of users the system can support without

blocking them is given by:

Proof:

N * [(C / g) – R]

Since we have divided the groups equally each group will have channels of

C/g. since each group reserved R channels , the no of channels without blocking each

group supports is C/g-R. Since each cell are having the same number of channels and

if there are N cells in a network then the maximum calls it will support without

blocking will be N * [(C/g)-R].

Here in my algorithm I am dividing in to 3 groups so g value is 3 then

N * [ ( C - 3R ) / 3 ]

Case 1a: If the number of the cells in the system are in the multiples of clusters and

for a maximum of seven clusters i.e. 49 cells..When there are C channels in a system

and g is the number of groups, and these C channels are divided equally among these

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groups, if the channels are reserved unequally from each group then the total no of

users the system can support without blocking them is given by:

For even no. of clusters

(2M + 2) * [ (C/g) – Rp ] + 2M * [ (C/g) – Rq ] + (3M-2) *[(C/g) – Rr]

For odd no. of clusters

(2M + 1) * [ (C/g) – Rp ] + (2M+1) * [ (C/g) – Rq ] + (3M-2) *[(C/g) – Rr]

Proof:

Since we have divided the groups equally each group will have channels of

C/g. Since each group reserves channels unequally and like the channels reserved by

group Gp be Rp and by group Gq be Rq and by group Gr be Rr. Then the no of

channels now each group is having for without blocking is (C/g) – Rp, (C/g) – Rq and

(C/g) – Rr respectively

In this for cluster one there 3 Gp, 3Gq, and 1Gr. If another cluster added then

For 2nd cluster 4 Gp, 6 Gq and 4 Gr,

For 3rd cluster 7 Gp, 7 Gq and 7 Gr,

For 4th cluster 8 Gp, 10 Gq and 10 Gr,


For 6thcluster 12 Gp, 14 Gq and 16 Gr,

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For 7th cluster 15 Gp, 15 Gq and 19 Gr.

Since even clusters are in form 4, 8, 12 for Gp and 6, 10, 14 for Gq and odd

clusters are in form 3,7,11,15 for Gp and Gq and last Gr is increasing by 3 from each

previous one we can write it as 2m+2 for Gp in even and 2m+1 for Gp in odd and 2m

for Gq in even and 2m+1 for Gq in odd and 3m-2 for Gr in both.

Hence,


(2M + 2) * [ (C/g) – Rp ] + 2M * [ (C/g) – Rq ] + (3M-2) *[(C/g) – Rr]


(2M + 1) * [ (C/g) – Rp ] + (2M+1) * [ (C/g) – Rq ] + (3M-2) *[(C/g) – Rr]

Will give the maximum no of calls system supports without blocking.

Case 2: Here we have considered seven cells as a single cluster and the total no of

cells in the system are in the multiples of cluster, and if C channels in the system are

divided unequally in to three groups and the if the reservation of channels are done

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equally, then the no of users the system can support without blocking them is given

by:

For even number of clusters

(2M + 2) * [ Gp – R ] + 2M * [ Gq – R ] + (3M-2) *[Gr – R]

For odd no of clusters

(2M + 1) * [ Gp – R ] + (2M+1 ) * [ Gq – R ] + (3M-2) *[Gr – R]

M- No. of clusters

Gp, Gq and Gr denote the three groups in cellular network

Proof:

Since the channels are divided in to three groups unequally each will be having

channels like Gp, Gq and Gr respectively. If the Reserved channels are done equally

then the no. of channels without reservation are Gp-R and Gq-R and Gr-R respectively.

As said above for different clusters we get the form as

For even number of clusters

(2M + 2) * [ Gp – R ] + (2M ) * [ Gq – R ] + (3M-2) *[Gr – R]

For odd no of clusters

(2M + 1) * [ Gp – R ] + (2M+1 ) * [ Gq – R ] + (3M-2) *[Gr – R]

Will give maximum no of users the system supports without blocking.

To find which group number the value r in Gr belongs, the following condition is

being used

If the count of Gr = M then count of Gp = count of Gq = 3M

This condition is repeated from 1 to the no. of groups present. Here in the

given algorithm we are taking as 3 groups so from 1 to 3 times the condition will be

checked at maximum.

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Case 2a: If there are C channels in a system, and these channels are divided

unequally in to three groups and if the reservation of channels are also done unequally,

then the no of users the cellular network can support without blocking them is given

by:


(2M + 2) * [ Gp – Rp ] + (2M) * [ Gq – Rq ] + (3M-2) *[Gr – Rr]


(2M + 1) * [ Gp – Rp ] + (2M + 1) * [ Gq – Rq ] + (3M-2) *[Gr – Rr]

Proof

Since the channels are divided in to three groups unequally each will be having

channels like Gp, Gq and Gr respectively. If the Reserved channels are done unequally

then the no. of channels without reservation are Gp-Rp and Gq-Rq and Gr-Rr

respectively. As said above for different clusters we get the form as


(2M + 2) * [ Gp – Rp ] + (2M) * [ Gq – Rq ] + (3M-2) *[Gr – Rr]


(2M + 1) * [ Gp – Rp ] + (2M + 1) * [ Gq – Rq ] + (3M-2) *[Gr – Rr]

Will give maximum no of users the system supports without blocking

4.5.2 Formula for calculating the total number of handoff calls the cellular

network system can support

Case 1: If the channels reserved are equal from each group and if the traffic

distribution of handoff calls is uniform then the formula is

N * R + N * │[(C/g) – R] * k │

Where k ranges from 0 to 0.5

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k – is a parameter which ranges from 0 to 0.5 which gives you how many channels to

reserve in a group of a given cell for both uniform and non uniform traffic distribution

of handoff calls.

Proof:

Let R be the no of channels reserved in each group initially. So each cell can

have R channels reserved for handoff calls. If there are N numbers of cells then N * R

give the total no of handoff calls the system supports initially. Now each group

contains (C/g) – R channels where C is total number of channels and g is the number

of groups. If the initial reserved channels are not sufficient they may acquire more

channels from the remaining one. The parameter “k” gives you how many channels

still we are reserving in that group. This k value is assigned for each cell.

Suppose, Let N =7 cells, C=36 channels, g=3 groups, No. of channels available

in each group are 12, 12, 12. Let R be 2 initially then each group will have 10, 10, 10

channels. Now if the reserved channels are not sufficient they may take from the group

to some extent by seeing not giving too many channels for reservation as it may

increase in blocking probability. I am considering we can reserve channels at most half

the channels in the group. If you don’t want any reserved channels k value will be 0.If

you don’t reserve any channels initially and want to reserve later, then k value can be

at most 0.5.

Here we have reserved 2 channels initially, for suppose we may want to

reserve channels up to 5 then three more channels we require. By giving K value as

0.3 we are reserving three channels from each group. Since there are N cells N * │

[(C/g) – R] * k │ gives the no of channels we have reserved later. I have [(C/g) – R] *

k in mod to get the value in whole number. Therefore total no of reserved channels for

handoff calls in the system now will be

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N * R + N * │[(C/g) – R] * k │

Case 2: If the channels reserved are equal from each group and if the traffic

distribution of handoff calls is non uniform then the formula is

N * R + │ [(C/g) – R] * k1 │+ │ [( C/g ) – R] * k2 │+ …. + │ [(C/g) – R] *KN │

Proof:

Let R be the no of channels reserved in each group initially. So each cell can

have R channels reserved for handoff calls. If there is N number of cells then N * R

gives the total no of handoff calls the system supports initially. Now each group

contains (C/g) – R channels where C is total number of channels and g is the number

of groups. If the initial reserved channels are not sufficient they may acquire more

channels from the remaining one. The parameters “k1, k2…kN” give you how many

channels still we want to reserve for handoff calls. These k values are assigned for

each and every cells. Since the system may have some cells with high number handoff

calls and in some cells having very less. By giving low k value in the cells which are

having low number of handoff calls and high k value to cells having high number of

handoff calls will reduce dropping probability. When there are N cells then there will

be k1, k2, k3….kN values. Since each group are now having (C/g)-R channels

multiplied with this k values gives you how many channels we are reserving in that

cell. Since there are N cells the no. of reserved channels we are reserving later will be.

│ [( C/g ) – R] * k1 │+ │ [( C/g ) – R] * k2 │+ …. + │ [( C/g ) – R] *KN │

Hence total no of reserved channels for handoff calls in the system will be

N * R + │ [( C/g ) – R] * k1 │+ │ [( C/g ) – R] * k2 │+ …. + │ [( C/g ) – R] *KN │

Case 3: If the channels reserved are unequal from each group and if the traffic

distribution is uniform then the formula is

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(2M + 1) * Rp + (2M + 1) * Rq + (3M – 2)* Rr +

(2M + 1) * │ k* (C/g-Rp) │ + (2M + 1) *│ k* (C/g-Rq)│ + (3M – 2)* │k * (C/g-Rr ) │


(2M + 2) * Rp + 2M * Rq + (3M – 2)* Rr +

(2M + 2) * │ k* (C/g-Rp) │ + 2M * │ k* (C/g-Rq)│ + (3M – 2)* │k * (C/g-Rr) │

Proof:

The channels reserved are unequal in this case and let each group Gp,Gq,Gr

reserved channels Rp,Rq,Rr respectively. For one cluster there are 3Gp, 3Gq, Gr.



For 4th cluster 10Gp, 8 Gq and 10 Gr,




We are taking into assumption that after forming one cluster the remaining

clusters are formed at an angle of 30o from the center and in the anti clock wise

direction.

We are calling as 1st cluster as level 0 and from 2 to 7 clusters as level 1 as

after arranging clusters up to 7 the whole thing will look like a single cluster.

Here there is a sequence in even cluster as 6,10,14 and in odd clusters as

3,7,11,15 for Gp, even clusters 4,8,12 for Gq and odd clusters 3,7,11,15 for Gq and

same sequence for Gr for both odd and even 1,4,7,10,13,16,19. by seeing the sequence

we can write a generalized formula as:

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(2M + 1) for both Gp and Gq for even clusters,

(2M + 2) for Gp in odd cluster, 2M for Gq in odd cluster and

(3M – 2) for Gr in both even and odd clusters

Form that we can write the no of reserved channels initially as

(2M + 1) * Rp + (2M + 1) * Rq + (3M – 2)* Rr

for odd no of clusters and

(2M + 2) * Rp + 2M * Rq + (3M – 2)* Rr

for even number of clusters:

Now each group will be having C/g-Rp, C/g-Rq and C/g-Rr for Gp,Gq, and Gr

respectively. As said above in case 1 and by using the “k” value and above said

sequence we can find the no of channels we have reserved later by

(2M + 1) * k* (C/g-Rp) │+ (2M + 1) *│ k* (C/g-Rq)│+ (3M – 2)* │k * (C/g-Rr ) │

For odd no of cluster and

(2M + 2) * │ k* (C/g-Rp) │ + 2M * │ k* (C/g-Rq)│ + (3M – 2)* │k * (C/g-Rr) │

For even number of clusters.

Hence the total number of handoff calls it supports is given as


(2M + 1) * Rp + (2M + 1) * Rq + (3M – 2)* Rr +(2M + 1) * │ k* (C/g-Rp) │ +

(2M + 1) *│ k* (C/g-Rq)│ + (3M – 2)* │k * (C/g-Rr)│


(2M + 2) * Rp + 2M * Rq + (3M – 2)* Rr +(2M + 2) * │ k* (C/g-Rp) │ + 2M * │

k* (C/g-Rq)│ + (3M – 2)* │k * (C/g-Rr) │

Case 4: If the channels reserved are unequal from each group and if the traffic

distribution is non uniform then the formula is

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For odd no. of clusters:

(2M + 1) * Rp + (2M + 1) * Rq + (3M – 2)* Rr +

2M+1 2(2M+1) N

∑ │ ki * (C/g-Rp) │ + ∑ │ ki * (C/g-Rq) │ + ∑ │ ki * (C/g-Rr)│

i=1 i=2(M+1) i= 2(2M+1)+1

For even no. of clusters:

(2M + 2) * Rp + 2M * Rq + (3M – 2)* Rr +

2M+2 2(2M+2) N


i=1 i=2(M+1)+1 i= 2(2M+2)+1

Proof:

The channels reserved are unequal in this case and let each group Gp,Gq,Gr

reserved channels Rp,Rq,Rr respectively. For one cluster there are 3Gp,3Gq,Gr.



For 4th cluster 10Gp, 8 Gq and 10 Gr,




We are taking into assumption that after forming one cluster the remaining

clusters are formed at an angle of 30o from the center and in the anti clock wise

direction.

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We are calling as 1st cluster as level 0 and from 2 to 7 clusters as level 1 as

after arranging clusters up to 7 the whole thing will look like a single cluster.

Here there is a sequence in even cluster as 6,10,14 and in odd clusters as

3,7,11,15 for Gp, even clusters 4,8,12 for Gq and odd clusters 3,7,11,15 for Gq and

same sequence for Gr for both odd and even 1,4,7,10,13,16,19. by seeing the sequence

we can write a generalized formula as:

(2M + 1) for both Gp and Gq for even clusters,

(2M + 2) for Gp in odd cluster, 2M for Gq in odd cluster and

(3M – 2) for Gr in both even and odd clusters

Form that we can write the no of reserved channels initially as

(2M + 1) * Rp + (2M + 1) * Rq + (3M – 2)* Rr for odd no of clusters and

(2M + 2) * Rp + 2M * Rq + (3M – 2)* Rr for even number of clusters.

Now each group will be having C/g-Rp, C/g-Rq and C/g-Rr for Gp,Gq,and Gr

respectively. As said above in case 2 and by using the different k values for each cell

depending on the number of handoff calls occurring into that cell and by using above

said sequence we can find the no of channels we have reserved later by

2M+1 2(2M+1) N


i=1 i=2(M+1) i= 2(2M+1)+1

for odd number of clusters

2M+2 2(2M+2) N


i=1 i=2(M+1)+1 i= 2(2M+2)+1

for even number of clusters

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Hence the total number of handoff calls it supports is given as


(2M + 1) * Rp + (2M + 1) * Rq + (3M – 2)* Rr +

2M+1 2(2M+1) N


i=1 i=2(M+1) i= 2(2M+1)+1

for even no. of clusters:

(2M + 2) * Rp + 2M * Rq + (3M – 2)* Rr +

2M+2 2(2M+2) N


i=1 i=2(M+1)+1 i= 2(2M+2)+1

With the above methods for dividing the channels among the groups will help

us in handling the calls without blocking and handling the handoff calls more

efficiently by reserving the channels among groups.

4.6 Performance Analysis

The performance of the proposed system is estimated using the following

metrics:

• Packet Delivery Throughput:

The ratio of the number of packets received by the destinations to the number

of packets sent by the CBR sources is defined as packet delivery throughput.

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• End-to-End Delay of Data Packets:

The difference between the time at which the packet is received by the

destination and the time at which the packet is sent by the source is defined as end to

end delay of data packets. The packets lost in the journey from source to destination

are not considered. The delay metric do not includes the delay related to the route

discovery, queuing and retransmissions.

The proposed system performance is compared with the RSVP and EDCF

reservation policies. Fig. 4.8 and Fig. 4.9 shows the performance results of

comparison in terms of throughput and delay respectively and can be observed that the

performance is improved by the proposed system with respect to the legacy systems.

The QoS parameters specified above are examined by simulation. Analytically they

are examined by using the blocking and dropping probabilities formulae [2] given

below. The blocking probability for an originating call is given as

1

( )c

s

oi s

B p i= +

= ∑

The dropping probability for a handoff call is given as

( )(0)

!

C O CS S S SO H H

H SB p

S

λ λ λµ

− −+=

Where,

S = Total Bandwidth

SR = Bandwidth reserved for handoff flows.

SO = Bandwidth reserved for only originating calls.

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The steady state probability P(i) can be obtained:

Where,

4.7 Results and Discussion

The performance of the proposed system is evaluated by comparing the

performance of the proposed system with the performance of RSVP and EDCF

reservation policies. Fig. 4.8 shows the comparison of the performance of the

proposed system to the RSVP and EDCF in terms of throughput and can be observed

that the throughput is high for the proposed system. The experiment is carried out for

35 runs for various flows between the number of flows and the number of packets

transmitted successfully. The Packet Delivery Throughput for the proposed system

when the number of flows 4 is 0.97 compared with the RSVP and EDCF which are

0.91 and 0.69 respectively. When the number of flows increases then the proposed

system throughput will increases to 23% more than the existing systems. It is observed

that the throughput of the proposed system improves when compared with the

66

performance of RSVP, EDCF methods. Fig. 4.9 shows the comparison of the

performance of the proposed system to the RSVP and EDCF in terms of end to end

delay of data packets and can be observed that the delay is low for the proposed

system. The experiment is carried out for 35 runs for various flows between the

number of flows and the time taken to successfully deliver the total flows. When the

number of flows increases the delay of the proposed system will be reduced for

approximately about 14% compared with the existing systems. It is observed that the

delay of the proposed system is low when compared with the performance of RSVP,

EDCF methods.

Fig. 4.8 Comparison of packet delivery throughput

Packet Delivery Throughput

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

2 4 6 8 10 12 14Number of Flows

Th

rou

gh

pu

t

RSVP

EDCF

Proposed System

67

Fig. 4.9 Comparison of End to End Delay

4.8 Conclusion

A new efficient cross layer based bandwidth reservation scheme is proposed in

this paper. The amount of bandwidth to be reserved at each base station was calculated

dynamically based on the user mobility and the traffic intensity of mobile users. This

paper assumes several probabilities that the user may move left, right, straight and

backwards. The analysis of the QoS performance metrics like packet throughput and

end to end packet delay is carried out and observed that the proposed system is better

in performance in terms of delay and throughput metrics.

chapter 4 optimal channel allocation algorithm...

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