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Channel element dimensioning

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A 2000-04-04 Approved at RT System meeting

1 IntroductionThis document describes how to dimension the required amount of channelelements in the BTS. This document will later be incorporated in the RFGuidelines documentation.

This document is intended as a support for engineers performingdimensioning for RFI/RFQ. This is not customer documentation.

The channel element dimensioning presented in this document covers onlythe UL RAX boards in the RBS. The method described in this document doesnot take into consideration requirements on the DL base band board. It isassumed that the capacity will be sufficient in the DL especially for P2+versions of the hardware. An additional base band board can be included tointroduce redundancy or increase capacity.

2 Calculating max simultaneous usersIn order to calculate the number of hardware channel elements a number ofmaximum simultaneous users must be derived. This is done based on theexpected traffic to be carried in the cell.

2.1 Circuit services (symmetric traffic)For circuit services we dimension in the air interface for a given grade ofservice. Naturally, this has to be done in the hardware dimensioning as well.

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The GoS chosen must be equal or less than that chosen in the air interface tootherwise hardware congestion will limit the system. If nothing else is chosen2% GoS is a good starting point.

The hardware in the RBS can be treated as a pooled resource. This meansthat all cells connected to one RBS use the same hardware resources forhandling calls and we get trunking efficiency within the RBS. Therefore, trafficshould be considered per site and not per cell as in the air interfacedimensioning.Table 1, Calculation of simultaneous users, circuit services

Basic parameters

3-sector site

2% GoS

Service Erlang/cell Erlang/site Max #simultaneoususers/site

Speech 10 30 39

C64 1 3 8

C128 1 3 8

Erlang per site indicates that on average during busy hour there is a givenamount of users active. By including a grade of service measure a peak valuemust be used to fulfil the grade of service. This can be found in an ordinaryErlang table.

The number simultaneous users per cell in the table above are givenassuming a 2% blocking probability.

2.2 Packet services

In contrast to circuit services a packet service does not allocate the wholehardware resource during the session. This enables more users to share thesame resources. However, when a user is active and transmitting informationin UL or DL hardware resources are allocated similar to a circuit switchedservice.

Typical traffic profile for packet users are asymmetric, with different trafficvolume in UL and DL when the user is active. This must be considered whenthe number of simultaneous users is calculated.

To calculate the number of simultaneous users consider the average totaluser bitrate/service and divide with the bearer user bitrate for that link. Theresult is an equivalent number of simultaneous users in order to fulfil the totalbitrate/service. This calculation must be done for UL and DL.

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When hardware dimensioning is concerned we must consider the maximumof the UL & DL to ensure that there is enough hardware available. If not thereis a risk for hardware congestion. For example: if the DL traffic volume is suchthat it requires 8 simultaneous users and the UL traffic volume requires 3simultaneous users then we can expect that 8 users will be active in worstcase. If we dimensioned for the 3 simultaneous users this would mean that 5users would not have enough hardware to transfer data. When a user isactive both UL and DL hardware resources are used.

Packet data services can be treated a little differently since we assume besteffort services. This means that we do not have to consider any grade ofservice with a peak value to cope with a given blocking probability.

Table 2, number of simultaneous users, packet services

Total userbitrate/service

#simultaneoususers

Service UL [kbit/s] DL [kbit/s] UL DL

Max #simultaneoususers

P64/64 180 500 2,8 7,8 8

P384/384 180 500 0,47 1,3 2

P64/384 180 500 2,8 1,3 3

The number of simultaneous users are discrete and in order to cope with theworst case the number of simultaneous users always have to be rounded upto the nearest whole number. It is not possible to allocate a fraction of achannel element.

For example, a P384/384 service may require 1,3 simultaneous users onaverage. This would mean that it is quite probable that 2 users are activemomentarily and we have to cater for that possibility. If hardware resourcesfor 2 simultaneous users were not allocated then it is not possible to reach upto 1,3 simultaneous users on average.

3 Calculating channel elements3.1 Definition of channel element

A channel element is a resource that has the ability to handle one voiceequivalent on the receiver (RAX) board in the RBS.

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3.2 Mapping of user data rate to channel elements

The number of hardware resources needed on the RAX board is dependenton the spreading factor defined in the radio access bearer. A lower SF meanshigher user data rate and therefore more resources are required. The channelelements are therefore not dependent if a circuit or packet service is used,only the physical data rate is relevant. The following table maps typical userdata rates with required channel elements. This information is taken from [1].

Table 3, Mapping of user data to channel elements

User data rate Uu Bitrate [kbps] SF #channel elements

Speech 12.2 kbps 60 64 1

32 kbps 240 16 2

64 kbps 240 16 2

128 kbps 480 8 4

144 kbps 480 8 4

384 kbps 960 4 8

3.3 Channel elements per site

With the mapping in Table 3 it is possible to calculate the required amount ofchannel elements per site if the number of simultaneous users are known persite.

Step 1: taking #simultaneous users into account

The number of simultaneous users per service is multiplied with the requirednumber of channel elements per service. All CE per service is added up into asubtotal.

Step 2: compensating for soft handover

The next step is to compensate for users in soft handover since that willrequire additional hardware. Typically it is estimated that 30% of the users arein soft hand over. Therefore the subtotal obtained is calculated by 1.3 tocompensate for soft handover resulting in the total required amount ofchannel elements per site.

Step 3: calculating the required amount of hardware kits for pricing

In the first phase the hardware (RAX board) will be able to handle 16 CE. Inorder to calculate the required amount of hardware kits the total amount of CEcalculated in step 2 is divided by 16 and the result rounded upwards to thenearest whole number.

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For future releases hardware capacity per card will increase but software keyswill be implemented such that capacity will be unlocked in increments of 16CE.

The three steps above are exemplified in the table below.Table 4, Calculating channel elements

Service UL bearermax physicalrate

#simultaneoususersper/service

#CE/service #CE

Speech 60 39 1 39

Circuit 64 240 8 2 16

Circuit 128 480 8 4 32

Packet 64/64 240 8 2 16

Packet 64/384 240 3 2 6

Packet 384/384 960 2 8 16

Subtotal before SHO 125

Users in SHO 30%

Total CE 125x1.3=163

#CE per kit 16

Total #kits 163/16=11

3.4 Hardware limitations

In the base station there is a limit to 18 RAX boards where 2 of these areshared with transmission. This means that it is possible to guarantee 16boards for capacity but the remaining 2 board slots will depend on thetransmission solution chosen for the site.

With 16 CE per board this means it will be possible to have 256-288 CE permacro BTS.

4 References[1] RBS HW Capacity, 9/0363-10/FCP1032565, Rev pB3