GSM-To-UMTS Training Series 03_WCDMA Radio Network Capacity Dimensioning_V1.0

8/13/2019 GSM-To-UMTS Training Series 03_WCDMA Radio Network Capacity Dimensioning_V1.0

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HUAWEI TECHNOLOGIES CO., LTD.

www.huawei.com

HUAWEI Confidential

Security Level:

WCDMA Radio Network

Capacity Dimensioning

GSM-to-UMTS Training Series_V1.0


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HUAWEI TECHNOLOGIES CO., LTD. HUAWEI Confidential Page 2

Revision Record

Date Revision

Version

Description Author

2008-10-31 1.0 Draft completed. Zang Liang

2008-12-31 1.1 Added comparison of the capacity dimensioning of

the GPRS and WCDMA to Page 3.

Explained the formula for calculation of the data

volume on Page 23.

Explained BHSA on Page 24.Explained the related units on Page 34.

Explained the formula and parameters on Page 35.

Described the downlink interference on Page 57.

Zhang Bibo

2009-01-15 1.2 Added remarks on Pages 3, 16, and 79 to compare

capacity dimensioning of a GSM network with that

of a WCDMA network.

Hou Chong


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Capacity Features: WCDMA vs GSM

WCDMA is a self-interference system.

The WCDMA system capacity is closely related to coverage.

The WCDMA network capacity has the soft capacity feature.

The capacity dimensioning of a WCDMA network is based on

a certain traffic model.

Interference in the GSM network is mainly from the cells that use the same

frequency. There is negligible interference from the subscribers in the same cell.

In the GSM system, the capacity is independent of the coverage.

The GSM system capacity is relatively fixed and can be estimated based on the

frequency and the number of timeslots.

The uplink capacity of the GSM network is the same as the downlink capacity.

The capacity dimensioning of a GSM network is based on a certain traffic model.


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Foreword

At the early stage of radio network planning, we must aim to know the

scale of the network. For example, we must know how many devices

are required and how to configure the devices. Based on coverage

estimation, we can calculate the cell radius under a specific load by

using the path loss, and then obtain the number of subscribers under

the cell coverage in combination with the subscriber density.

In actual planning, can we obtain the cell radius only by considering

the coverage? Should other aspects be taken into consideration?


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Objectives

Understand the parameters of 3G traffic model

Understand the factors that restrict the

WCDMA network capacity

Understand the methods and procedures for

estimating multi-service capacity

Understand the key technologies for

enhancing network capacity

Upon completion of this course, you will be able to:


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Contents1. Traffic Model

2. Uplink capacity analysis

3. Downlink capacity analysis

4. Multi-service capacity estimation

5. Network estimation procedure

6. Capacity enhancement technologies


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1.TrafficModel

1.1 Overview o f Traff ic Model

1.2 CS Traffic Model

1.3 PS Service Model

1.4 PS Traffic Model Parameters


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Service Overview

The WCDMA system supports multiple services.

Variable-rate services

Hybrid services

High-speed data packet services

Asymmetrical services

Large-capacity and flexible service bearing


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QoSType

Real-time

category

Session

It is necessary to maintain the time relationship between

the information entities in the stream. Low delay toleranceand symmetric data rate are required.

Voice

service,video

phone,

video

game

Streaming

It is necessary to maintain the time relationship between

the information entities in the stream. Unidirectionalservices, high error tolerance, and high data rate are

required.

Streaming

multimedia

Non real-time

category

Interactive

Data integrity must be maintained in request-response

mode. High error tolerance and low delay tolerance are

required.

Web page

browsing,

online

game

Background

The information receiving end does not expect the data

arrival within a period of time. Data integrity should be

maintained. Small delay restriction and error-free

transmission are required.

Backgroun

d download

of Email


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Objectives of Establishing Traffic Models

To determine the system configuration, calculate the call capacity of air

interfaces.

To determine the system capacity. In data applications, different

transmission models lead to different system capacities.

-To plan the network properly by establishing a data transmission model

that meets the customers expectations.

Traffic models are provided by telecom operators. If a telecom operator

cannot provide the traffic model, a traffic model for similar scenarios or

a traffic model recommended by Huawei can be used as the basis for the

planning upon the approval of the telecom operator.


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TrafficModel

A traffic model deals with the capacity features of each service

type and the QoS expected by the subscribers who are using

the service from the perspective of data transmission.

In data applications, the subscriber behavior research predicts

the service types available on a 3G network, number of

subscribers of each service type, usage frequency of the

service, and distribution of subscribers in different regions.


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System Configuration

User Behavior

Service Model

Traffic Model

Results

Contents of Traffic Model


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Typical Service Features

Typical service features include the following featureparameters:

Subscriber type (such as indoor, outdoor, and inside a

vehicle)

Average motion speed of a subscriber

Service type

Uplink and downlink service rates

Spreading factor (SF)

Time delay requirements of the service

QoS requirements of the service


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1. Traffic Model

1.1 Overview of Traffic Model

1.2 CS Traff ic Mo del




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CS Traffic Model

A typical CS service is the voice service. The arrival of the CS

subscribers takes on a Poisson distribution, while the interval

takes on a negative exponential distribution.

Related parameters of the model:Penetration rate

BHCA Mean busy-hour call attempts

Mean call duration (s)

Activation factor

Mean rate of service (kbps)


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CS Traffic Model Parameters

Mean busy-hour traffic per user (Erlang) = (BHCA x Mean call

duration)/3600 (Erl)

Mean busy-hour throughput per user (kbit) (G) = BHCA x Mean call

duration x activation factor x Mean rate of service (kbps)

Mean busy-hour throughput per user (bps) (H) = (Mean busy-hour

throughput per user x 1000)/3600


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1. Traffic Model






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PS Traffic Model

The most frequently used model is the packet service session process

model described in ETSI UMTS30.03.

PS Services


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PS Services


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1.TrafficModel




1.4 PS Traff ic Model Parameters


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Service Model

PSServiceModelParameters

Packet Call Num/Session

Packet Num/Packet Call

Packet Size (bytes)

Reading Time (sec)

Typical Bearer Rate (kbps)

BLER


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Determining Parameters

The basic parameters in the traffic model are determined in the

following ways:

Obtain numerous basic parameter sample data from the

existing network.

Obtain the probability distribution of the parameters by

processing the sample data.

Take the distribution most proximate to the standard probability

as the corresponding parameter distribution through comparison

between the obtained distribution and the standard distribution

function.


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Typical Bearer Rate (kbit/s):

The bearer rate varies during the actual transmission.

BLock Error Rate (BLER):

For the PS domain service, we need to consider the retransmission

caused by error blocks when calculating the data transmission time.

Assume that the data volume of a service source is N and that the

block error rate at the air interface is BLER. The total data volume to be

transmitted on the air interface is as follows:

PSTrafficModelParameters


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User behavior

Penetration Rate

BHSA

User Distribution (High, Medium,

Low end)

PS User Behavior Parameters


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PS User Behavior Parameters

Penetration Rate

The percentage of the users that activates this service to all the

users registered in the network.

BHSA The number of sessions of single-user in busy hour

specific to a service

User Distribution(High, Medium, Low end)

The users are divided into high-end, mid-end and low-end users. Different

operators and different application scenarios have different user

distributions.


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Session traffic volume (Byte):Average traffic of single session of the

service

Session traffic volume = Packet size x Packet number per packet call x

Packet call number per session

Data transmission time (s):Transmission time for a single session of

service

Data transmission time = (1/(1-BLER)) x (((Session traffic volume x

8)/1000)/Typical bearer rate)

Holding time (s):Average duration of a single session of service

Holding time = (Packet call number per session -1) x Reading time + Data

transmission time

Derivative Parameters of the PS Traffic Model


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Activation factor: The proportion of the time when services transmits

at full-rate in the duration of a single session.

Busy hour throughput per user (Kbit):

Derivative Parameters of the PS Traffic Model

timeHolding

timeontransmissiDatafactorActivation

1000/)8**(/ VolumeTrafficSessionBHSAuserThroughputHourBusy


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Traffic Model Sample

Mobile

Video

(Stream) on

Demand

Penetration

RateBHSA

Busy Hour

Throughput/use

r (kbit)

Typical

Bearer

Rate

(kbps)

BLERActivation

Factor

Uplink 22.0% 0.100 2.304 8 10% 0.1798

Downlink 22.0% 0.100 102.528 64 10% 1.0000

Mobile

Video

(Stream) on

Demand

Packet Call

Number/Se

ssion

Packet

Number/P

acket Call

Packet Size

(Bytes)

Reading

Time

(sec)

Session

Traffic

Volume

(Byte)

Holding

Time

Uplink 2 3 480 14.6000 2880 17.8000

Downlink 1 267 480 0.0000 128160 17.8000

Data_Erlang = ((Proportion of subscribers at each level in typical application

environment x Penetration rate x Busy throughput per user in typical application

environment)/(Bearer rate x 3600 x Activation factor of this service))


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Questions

1. What are the two parts of the traffic model?

2. What are the main parameters of the CS traffic

model?

3. What are the main parameters of the PS service

model?

4. What is the formula for calculating the equivalent

Erlang of the PS service?


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Summary

Content of the traffic model

Main parameters of CS service traffic model

Structure, main parameters, and derivative parameters ofthe PS service model

Calculation of the equivalent Erlang of the PS services


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Basic Principles


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Basic Principles

The radio system capacity depends on the uplink and downlink

capacities. When planning the capacity, analyze the uplink anddownlink.

In the WCDMA system, all the cells may share a same spectrum,

which greatly helps improve the system capacity. However, co-frequency reuse causes interference between subscribers. Such

multi-access interference restricts the system capacity.


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NotherownTOT PIII

Iown: interference from subscribers of this cell

Iother: interference from subscribers of adjacent

cells

PN: receiver noise floor

Uplink Interference AnalysisComposition of

Uplink Interference

Uplink Interference Analysis Composition of


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Receiver noise floor PNPN= 10lg (KTW) + NF

K: Boltzmann constant, 1.38 x 10-23 J/K

T: Kelvin temperature. The normal temperature is 290 K.

W: signal bandwidth. The WCDMA signal bandwidth is 3.84

MHz.

NF: noise factor of a receiver

10lg (KTW) = -108 dBm/3.84 MHz

NF = 3 dB (typical value of macro-cellular NodeB)

PN= 10lg (KTW) + NF = -105 dBm/3.84 MHz


Uplink Interference



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Iown: interference from subscribers of this cell

Interference to be overcome by each subscriber: ITOT- Pj

Pjis the receiving power of subscriber j

Under ideal power control:

Hence:

The interference from subscribers of this cell are the sum

of powers of all the subscribers arriving at the receiver:

jjjTOT

j

jvR

W

PI

PNoEb

1/

jjj

TOTj

vR

W

NoEb

IP1

/

11

N

jown PI1

p y p

Uplink Interference


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Iother: interference from subscribers of neighbor cells

It is difficult to theoretically analyze the interference from

subscribers of neighbor cells, because the interference is

related to subscriber distribution, cell layout, and antenna

pattern.

Interference factor of a neighbor cell:

When the subscribers are distributed evenly:

For an omni-directional cell, the typical interference factor

of neighbor cells is 0.55.

For a three-sector directional cell, the typical interferencefactor of neighbor cells is 0.65.

own

other

I

I

i


Uplink Interference

Uplink Interference Analysis


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N

N

jjj

TOT

NotherownTOT

P

vR

W

NoEb

Ii

PIII

1

1

/

11

1

Define:

jjj

j

vR

W

NoEb

L1

/

11

1

Then, NN

jTOTTOT PLiII 1

1




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Obtain

N

j

NTOT

Li

PI

1

11

1

Assume that:

All the subscribers are 12.2kbit/s voice subscribers,

and the demodulation

threshold is Eb/No = 5 dB.

Voice activation factor vj=

0.67.

Interference factor of

neighbor cells i = 0.55.



U li k L d F t


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Uplink Load Factor

Define the uplink load factor

When the load factor is 1, ITOTis infinite and the corresponding

capacity is called threshold capacity.

Under this assumption, the threshold capacity is approximately 96

subscribers.

N

jjj

N

jUL

vR

W

Ebv sNo

iLi11

111

111



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Load Factor and Interference

Based on the previous formulas,

we can obtain the noise rise:

50% Load3 dB

60% Load4 dB

75% Load6 dB


Li it ti f th C t M th d


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In the preceding theoretical analysis, there are certain limitations as some

factors have not been taken into considerations and some assumptions have

been made.

No consideration of the influence of soft handover

The interference generated by the subscribers in soft handover state

are slightly lower than those generated by ordinary subscribers.

No consideration of the influence of AMRC and hybrid service

AMRC reduces the voice service rate of certain subscribers, reduces

the interference generated by these subscribers, and increases the

number of subscribers supported by the system at the cost of lowering

the call quality of these subscribers.

Different services have different data rates and demodulation thresholds.

The preceding method can still be used for analysis, but the calculation

process is complicated.

Due to the time-variable feature of the mobile transmission environment,

the demodulation threshold even for a same service is time-variable.

Limitations of the Current Method


Li i i f h C M h d


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Assumption of ideal power control

The power control commands of an actual system have certain error

codes. As a result, the power control process is not ideal, and the

system capacity is reduced.

Assumption that the subscribers are evenly distributed and the neighbor

cell interference is constant.

In consideration of the preceding factors, the system emulation is a more

accurate method:

Static emulation: Monte Carlo method

Dynamic emulation

Limitations of the Current Method


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Downlink Interference Analysis

C iti f th D li k I t f


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NotherownTOT PIII Iown: interference from subscribers of this cell

Iother: interference from subscribers of adjacent cells

PN: receiver noise floor

Composition of the Downlink Interference




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Receiver noise floor PN

PN= 10lg (KTW) + NF

K: Boltzmann constant, 1.38 x 10-23 J/K

T: Kelvin temperature. The normal temperature is 290 K.

W: signal bandwidth. The WCDMA signal bandwidth is

3.84 MHz.

NF: noise factor of a receiver

10lg (KTW) = -108 dBm/3.84 MHz

NF = 7 dB (typical value of UE)

PN= 10lg (KTW) + NF = -105 dBm/3.84 MHz



Composition of Downlink Interference


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Iown: interference from this cell

Downlink subscribers are differentiated with the mutually

orthogonal OVSF codes. In static propagation conditions

without multi-path, no mutual interference exists.

In the case of multi-path propagation, certain energies that

cannot be detected by the RAKE receiver become interference

signals. The orthogonal factor

is defined to describe thisphenomenon.

In the formula, PT

indicates the total transmit power of a

NodeB, including the dedicated-channel transmit power

and the common-channel transmit power.

1 Town jjj

PI

PL

N

jCCHT PPP1





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Iother: interference from neighbor cells

The transmitting signals of neighbor cells cause interference to

the subscribers in the current cell. As the scrambling codes are

different, such interference is non-orthogonal.

Assume that the service is evenly distributed and the transmit

powers of all NodeBs are equal. The system consists of K

NodeBs in neighbor cells, and the path loss from NodeB k to

subscriber j is PLk,j. The interference from neighbor cells forsubscriber j is calculated as follows:

K

jk

TjotherPL

PI

1 ,

1




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NK

jk

T

j

Tj

NotherownTOT

PPL

PPLP

PIII

1 ,

11

Assume the power control is ideal, then:

jjjTOT

j

j

jvR

W

I

PLP

EbvsNo 1

The following formula can be obtained:

jjTOTjj

jj PLIvW

REbvsNoP



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Based on this

formula: N

jCCHT PPP

1

The following formula

can be obtained:

jN

K

jk

j

TTj

N

j

j

jCCH

N

K

jk

T

j

Tj

N

jj

j

jCCH

N

jjTOTj

j

jCCHT

PLP

PL

PLPPv

W

REbv sNoP

PPL

PPL

PPLv

W

REbv sNoP

PLIv

W

REbv sNoPP

1 ,1

1 ,1

1

1

11



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Calculate PTas follows:

N

j

j

jjj

N

jj

j

jNCCH

T

v

W

REbvsNoi

PLvW

REbvsNoPP

P

1

1

11

Where ijis the interference factor of neighbor cells of J,

assume that the following formula is true:

K

jk

jj

PLPLi

1 ,



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According to the preceding analysis, we can define the downlink load factor:

When the downlink load factor reaches 100%, the transmit power of the

NodeB is infinite, and the corresponding capacity is called threshold capacity.

Different from the theoretic calculation of uplink capacity, j and ijin the

downlink capacity formula are variables related to subscriber positions. That

is, the downlink capacity is related to the spatial distribution of subscribers,

and can be determined only through system emulation.

N

j

j

jjjDL vW

REbvsNoi

1

1


Simulation Result


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Simulation Result


Si l ti R lt A l i


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When the transmitting power of the base station is 43 dBm(20 W), a maximum of about 114 subscribers are supported.

In order to ensure system stability, the mean transmitting

power of the base station cannot exceed 80% of the

maximum transmitting power, namely, 42 dBm. Thus, about

112 subscribers are supported.

Simulation Result Analysis

How to Control the Interference


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How to Control the Interference

Impact of the interference to the network

Handover success rate

Access efficiency

Call drop rate

Voice quality

Method for interference control

Improve the precision of the power control

Improve the Rake receiving efficiency

Proper network planning

Contents


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4. Multi-service Capacity Estimation


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p y

4.1 Facto rs th at restr ict the radio

network c apaci ty

4.2 Commonly used capacity design

method

4.3 Capacity estimation example based

on the Campbellstheorem

Factors that Restrict the Radio Network Capacity


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p y

The WCDMA network capacity is

restricted by the following factors in the

radio network part:

Uplink interference

Downlink power

Downlink channel code resources

Channel Element (CE)

Iub interface capacity

Downlink Transmit Power


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The downlink transmit power is divided into

two parts: One part is used for common

channels, and the other part for dedicated(traffic) channels.

The transmit power that is allocated by a

cell to subscribers varies with the service

demodulation threshold, propagation path

loss, and interference received by asubscriber.

The downlink transmit power of a cell is

shared by all the subscribers in the cell.

The emulation method is often used to

analyze downlink interference.

N

jCCHT PPP1

Downlink Channel Code Resources


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The WCDMA network uses the code

words with the SF ranging from 4 to

512. The smaller the SF is, the higherthe supported data rate is.

In a code tree, the allocable codes

must meet the following conditions:

No codes are allocated on the

path from this code to the rootnode.

No codes that take this code

as the root node are allocated

in this sub-tree.

Principle for code allocation:

Try to reserve the code words

with small SFs to improve the

utilization rate.

1

1 -1

1 1

1 1 1 1

1 1 -1 -1

1 -1 1 -1

1 -1 -1 1

C1,0

C2,0

C2,1

C4,0

C4,1

C4,2

C4,3

SF = 1 SF = 2 SF = 4

1

1 -1

1 1

1 1 1 1

1 1 -1 -1

1 -1 1 -1

1 -1 -1 1

C1,0

C2,0

C2,1

C4,0

C4,1

C4,2

C4,3

SF = 1 SF = 2 SF = 4

Downlink Channel Code Resources


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Following is an example of code resources allocation

SF 4 8 16 32 64 128 256 512

C(256,0):PCPICH 2

0

C(256,1):PCCPCH 3

0

C(256,2): AICH 6

1

C(256,3): PICH 10

0

C(64,1):SCCPCH 8

0

C(64,2):SCCPCH 9

1

3

0

1

0

1

1

2

3



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The CE is the quantitative data that logically measures the resources occupied for

service processing.

The resources occupied for service processing are mainly related to the SF of the

service. The smaller the SF is, the greater the data traffic is, and the more the

occupied resources are.

The SFs of typical services are: AMR12.2 kbps SF = 128

CS64 kbps SF = 32

PS64 kbps SF = 32

PS144 kbps SF = 16

PS384 kbps SF = 8



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HUAWEI TECHNOLOGIES CO., LTD.HUAWEI Confidential

Page 62

If we define the resources required for processing AMR services of 12.2 kbpsas a CE, the number of CEs occupied by other services are as follows:

AMR12.2 kbps 1

CS64 kbps 4

CS144 kbps 8

CS384 kbps 16

PS64 kbps 4

PS144 kbps 8

PS384 kbps 16

4. Multi-service Capacity Estimation


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Page 63

5.1 Factors that restrict the radio

network capacity

5.2 Commonly u sed capaci ty design

method

5.3 Capacity estimation example based

on the Campbells theorem

Erlang-B Formula (I)


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Page 64

Erlang-B formula is used for

estimating the peak traffic that meets

certain call loss rate when the average

traffic (Erlang) is given

Erlang-B formula is only used for

Circuit switched services

Single service

The WCDMA system provides CS

and PS domain multi-services

Erlang B Formula (I)

Erlang-B Formula (II)


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Page 65

The prerequisite of the Erlang-B is that the requests for resources

should be based on Poisson distribution, that is, the resource request

variance should be equal to its mean value.

If, when a service establishes a connection, the service requests for

resources more than the unit resources, the resource request is no

longer equal to its mean value. The Erlang-B formula is not applicable in

this case.

Comparison of the methods for multi-service capacity estimation:

Post Erlang-B

Equivalent Erlangs

Campbells Theorem

g ( )

Post Erlang-B (I)


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Page 66

g ( )

By summing up the capacities

required for different services,

we obtain the capacities

required for the combined

services.

The resource efficiency ofdifferent services is not taken

into account.

Post Erlang-B (II)


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Assume that two types of services share resources

Service 1: one unit resource/connection.12 Erlang

Service 2: three unit resources/connection.6 Erlang

Calculate capacity required for each service

Service 1: 12 Erlangs require 19 unit resources, meeting the 2%blocking rate

Service 2: 6 Erlangs require 12 connections (equivalent to the 36

unit resources of service 1), meeting the 2% blocking rate

Total 55 unit resources

g ( )

Assume that two services use the same resources

Post Erlang-B (III)


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Post Erlang-B overestimates

the capacity requirements!

Assume that two services use the same resources.

Service 1: one unit resource/connection, 12 Erlangs


Calculate the capacity required for each service.

Service 1: A 12-Erlang traffic volume requires 19 unit resources to meet the

blocking rate of 2%.

Service 2: A 6-Erlang traffic volume requires 12 unit resources to meet the blocking

rate of 2%.

In total, 31 unit resources are required.

However, the reasonable result should be as follows: A 18-Erlang traffic volume

requires 26 unit resources to meet the blocking rate of 2%.

Equivalent Erlangs (I)


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Combine different services

by converting the bandwidth

from one service to another,

and then calculate the

required capacity.

Selecting different services

as the measurement

benchmark may lead to

different capacity

requirements.

A th t t i h

Equivalent Erlangs (II)


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Assume that two services share resources.


Service 2: three unit resources/connection, 6 Erlangs

If service 1 is taken as the measurement benchmark, the total traffic

volume for the two services is equivalent to 30 Erlangs.

A 30-Erlang traffic volume requires 39 unit resources to meet the

blocking rate of 2%.

If service 2 is taken as the measurement benchmark, the total traffic

volume for the two services is equivalent to 10 Erlangs.

A 10-Erlang traffic volume requires 17 unit resources (equivalent to

51 unit resources of service 1) to meet the blocking rate of 2%.

The forecast results

do not match!

Campbells Theorem (I)


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The Campbells Theorem sets up a kind of combined distribution.

Where:

indicates the service amplitude, that is, the channel resources

required for a single link of the service.

indicates the mean value, and v indicates the variance.

ia

cff icOfferedTra

c

aCCapacity ii

)(

ii

i

i

aErlangs

aErlangsv

c

2

A th t t t f i h

Campbells Theorem (II)


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Assume that two types of services share resources

Service 1: one unit resource/connection.12 Erlang

Service 2: three unit resources/connection.6 Erlang

The system mean value is

The system variance is

The capacity factor c is 1

6636112 222

iaErlangsv

3063121 iaErlangs

2.23066

vc

Campbells Theorem (III)

Combined traffic is calculated as follows:


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Combined traffic is calculated as follows:

A capacity of 21 resources is required to meet the blocking rate of 2%.

For the target services that meet a same Grade of Service (GoS), the

capacity required is as follows (based on the unit resource of service 1):

Target service 1: C1 = (2.2 x 21) + 1 = 47

Target service 2: C2 = (2.2 x 21) + 3 = 49

For a same GoS, different services require different capacities.For a given capacity, the GoSs of different services differ slightly.

63.132.2

30

cf f icOfferedTra

Comparison Between Common Capacity DesignMethods


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Post Erlang-B

Service 1 (one unit resource/connection, 12 Erlangs) and service 2(three unit resources/connection, 6 Erlangs) require 55 unit

resources in total.

Equivalent Erlangs

As calculated based on service 1 (one unit resource/connection, 12

Erlangs), a total of 39 unit resources are required.

As calculated based on service 2 (three unit resources/connection,

6 Erlangs), a total of 51 unit resources are required.

Campbells Theorem

Under the same conditions, 47 to 49 unit resources are required in

total.

Questions


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1. What are the factors that restrict the network

capacity?

2. What are the common methods for estimation of the

multi-service capacity?

Summary


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This chapter describes three methods for the estimation of the multi-

service capacity.

This chapter details the process of calculating the capacity with the

Campbells theorem.



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a c ode






5 Network Estimation Procedure


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Capacity dimensioning ideas

Determining the Service Model

Determining the QoS

B d t ffi di t ib ti d hi l f t di id th

Capacity Dimensioning Ideas


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Based on traffic distribution and geographical features, divide the

areas to be planned into dense urban areas, ordinary urban areas,

suburban areas, and rural areas.

Perform traffic model analysis for all target areas.

Based on the traffic models in different target areas, determine

the single-TRX planning capacity of each target area.

Determine the NodeB quantity and TRX quantity in the target

areas that meet the capacity requirements.

Compare the NodeB quantity and TRX quantity determined

based on the capacity requirements with those based on the

coverage requirements. Select the larger NodeB quantity and

TRX quantity to meet the capacity and coverage requirements at

the same time.

Determining the Service Model


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Subscriber distribution data

Subscribers Distributed in Different Scenarios

(Number of Subscribers/km2)

Application

Scenario

2003 2004 2005

Dense Urban 11128 12060 18683Common Urban 462 499 676

Suburban 246 266 341

Rural 15 16 18

Road/Main Road 23 35 48

Determining the ServiceModel

P ti f b ib t diff t l l


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Proportion of subscribers at different levels

Proportion of Subscribers at Different Levels

(Consider This in Combination with the Time)

Application

Scenario

High-end Mid-end Low-end

Denseurban 40% 40% 20%

Common

urban

15% 25% 60%

Suburban 5% 25% 70%

Rural 1% 10% 89%Road/Main

road

1% 10% 89%

Determining the Service ModelCS Service Model


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ServiceType

PenetrationRate

BHCA AHT (s) ActivationFactor

Mean

Rate

(kbit/s)

AMR

speech100 % 1 90 0.5 8

Video

phone100 % 0.1 54 1 64

Determining the Service ModelPS ServiceModel of Low-end Subscribers


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Service

Type

Penetr

ation

Rate

BHSA

Packet Call

Number/Se

ssion

Packet

Number/Pa

cket Call

Packet

Size

(bytes)

Inter-Arrival

Time Between

Packet Calls(sec)

Email 10 % 0.10 2 32 480 320

Internet 30 % 0.18 5 25 480 412

Onlinegame, ICQ

25 % 0.10 2 3 480 8

Picture

and ring

downloadi

ng, FTP

25 % 0.10 2 62 480 5

Real-timevideo

0 % 0.00 1 267 1500 0

SMS 50 % 0.50 1 1 160 0

EMS /

MMS50 % 0.50 2 32 480 320

Determining the Service ModelPS Service

Model of Mid-end SubscribersInter-


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Service

Type

Penetrati

on Rate BHSA

Packet

Call

Number/S

ession

Packet

Number/Packet Call

Packet

Size(bytes)

Inter-

Arrival

Time

Between

Packet

Calls (sec)

Email 20 % 0.20 2 32 480 320

Internet 30 % 0.24 5 25 480 412

Online

game,ICQ

15 % 0.20 2 3 480 8

Picture

and ring

download

ing, FTP

15 % 0.20 2 62 480 5

Real-time

video 10 % 0.10 1 267 1500 0

SMS 100 % 0.80 1 1 160 0

EMS /

MMS100 % 0.80 2 32 480 320

Determining the Service ModelPS Service

Model of High-end SubscribersInter-


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Service

Type

Penetrati

on Rate BHSA

Packet

Call

Number/S

ession

Packet

Number/Packet Call

Packet

Size(bytes)

Inter-

Arrival

Time

Between

Packet

Calls (sec)

Email 30 % 0.30 2 32 480 320

Internet 20 % 0.30 5 25 480 412

Online

game,ICQ

5 % 0.30 2 3 480 8

Picture

and ring

download

ing, FTP

10 % 0.30 2 62 480 5

Real-time

video 20 % 0.20 1 267 1500 0

SMS 100 % 0.60 1 1 160 0

EMS /

MMS100 % 0.60 2 32 480 320

With the data service models of various user types we can obtain the

Determining the Service ModelPS Traffic ModelParameters


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With the data service models of various user types, we can obtain the

traffic model parameters for calculation.

Busy hour throughput per subscriber (kbit) = Penetration rate of low-

end subscribers x Busy hour throughput per low-end subscriber x

Proportion of low-end subscribers + Penetration rate of mid-range

subscribers x Busy hour throughput per mid-range subscriber x

Proportion of mid-range subscribers + Penetration rate of high-endsubscribers x Busy hour throughput per high-end subscriber x

Proportion of high-end subscribers

Theoretical length of a session (bytes) = Packet Call Num/Session x

Packet Num/Packet Call x Packet Size (bytes)

Average reading time (s) = (Packet Call Num/Session-1) x Inter-Arrival

Time Between Packet Calls (s)

Determining the QoS

Capacity dimensioning determines the capacity under certain QoS


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p y g p y

conditions. CS services generally take the call loss or blocking rate toevaluate the GoS. PS services generally take the acceptable time delay or

acceptable minimum throughput to evaluate the GoS. In the tender

documents of operators, the GoS of PS services is sometimes described

as call loss.

Determining the QoS

Requirements for the blocking rate


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q g

BlockingRate

VoiceService

SMS

MMS

VideoStream

AudioStream

VideoPhone

VideoConference

Internet Email

Dense

urban

Common

urban

Suburban

Rural

Road/mai

n road



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Transmit Diversity


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The transmit diversity can increase the downlink capacity

and coverage.

Conclusion on capacity improvement through the transmit

diversity

STTD mode: capacity increase by 17% to 24%

TxAA(1) mode: capacity increase by 16% to 23%

TxAA(2) mode: capacity increase by 31% to 37%

Sectorization

In dense urban areas and ordinary urban areas with high traffic,


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y g ,

increasing the sectors in a NodeB helps increase the capacity.

A six-sector NodeB generally uses the antennas with the horizontal lobe

33.

The capacity of a 6-sector NodeB is 1.67 times as large as that of a three-

sector NodeB.

Capacity enhancement technologies of GSM

Division of cells

Closer frequency reuse

Adding micro-cellular devices

Extended frequency bands

CoBCCH

Half rate

Comparison Between GSM Capacity Expansion

and WCDMA Capacity Expansion

GSM capacity expansion


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GSM capacity expansion

If capacity expansion is required for the network in future, you only need toallocate new channels to the corresponding cells. Other changes are not needed

on the network as long as the expansion does not exceed the maximum capacity

(determined by the frequency resources and frequency reuse) of cells. Otherwise,

new NodeBs or sectors must be added with frequency planning.

WCDMA capacity expansion

The handover between different frequencies (or between the TRXs in a samesector) in the WCDMA system is hard handover, which requires you to enable the

compression mode and occupies huge system resources. Therefore, the capacity

expansion of the WCDMA system is not as simple as that of the GSM system,

which can be implemented through additional TRXs to the corresponding cells.

The method of dividing cells for capacity expansion also costs a lot. In the early

phase of capacity dimensioning, a signal margin must be determined to serve as

a compensation for interference in the case of traffic increase during thecalculation of the cell area.


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Thank You!

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GSM-To-UMTS Training Series 03_WCDMA Radio Network Capacity Dimensioning_V1.0

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