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1、手机异常位置更新导致的立即指配成功率降低问题分析内部公开2018-3-12华为机密，未经许可不得扩散第 1 页, 共 16 页手机异常位置更新导致的立即指配成功率降低问题分析 1 问题现象描述问题现象描述部分少量手机异常位置更新导致立即指配成功率降低，网络中业务（主被叫、数据业务及附加业务等）均正常。问题手机位置更新未成功的情况下，会不断地向其他小区发起位置更新直到成功为止。因此问题手机也不会发生掉网等情况，业务可正常进行。图1.1 BSC下出现立即指配成功率小区话统手机异常位置更新导致的立即指配成功率降低问题分析内部公开2018-3-12华为机密，未经许可不得扩散第 2 页, 共 16 页图
2、1.2 立即指配成功率低小区话统变化情况2 问题原因问题原因部分手机不符合协议，导致无理由向网络发送位置更新请求，但是本身又不按照协议完成流程，造成立即指配命令下发后，空口无响应，立即指配流程失败。3 问题分析问题分析问题的关键在空口交互。3.1位置更新（立即指配）流程正常位置更新流程：手机异常位置更新导致的立即指配成功率降低问题分析内部公开2018-3-12华为机密，未经许可不得扩散第 3 页, 共 16 页图1.3 正常位置更新流程问题发生时，异常信令流程(立即指配命令下发后没有SABM接入)：图1.4 异常位置更新流程异常信令流程实例：图1.5 异常信令流程实例3.2信道请求 Chann
3、el request（Acess Burst）3.2.1 Channel request（Aces Burst）协议规范手机异常位置更新导致的立即指配成功率降低问题分析内部公开2018-3-12华为机密，未经许可不得扩散第 4 页, 共 16 页空口手机MS以发送随机接入帧的方式，发起信道请求，协议对此有明确的结构（0502）和信道编码方式（0503）统一规定（位置更新的信令请求和其他的SDCCH信道请求）。1、0502协议对Acess Burst 结构的定义Access Burst主要由两部分组成：（1）41比特长度的synch. sequence bits（2）36比特长度的encrypt
4、ed bits（实际是信道编码后比特数据，非加密过的比特数据）Bit Number(BN)Lengthof fieldContentsof fieldDefinition0 - 78extended tail bits(below)8 - 4841synch. sequence bits(below)49 - 8436encrypted bits (e0.e35)05.0385 - 873tail bits(below)(88 - 15668,25extended guard period (bits)subclause 5.2.8)where the “extended tail bits“
5、 are defined as modulating bits with the following states:(BN0, BN1, BN2 . BN7)= (0, 0, 1, 1, 1, 0, 1, 0)-where the “tail bits“ are defined as modulating bits with the following states:(BN85, BN86, BN87) = (0, 0, 0)-where the “synch. sequence bits“ are defined as modulating bits with the following s
6、tates, unless explicitly stated otherwise (see GSM 04.60):(BN8, BN9 . BN48)=(0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0,手机异常位置更新导致的立即指配成功率降低问题分析内部公开2018-3-12华为机密，未经许可不得扩散第 5 页, 共 16 页1, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0)in case alternative training (synchroni
7、zation) sequence “TS1“ is used, the “synch. sequence bits“ shall be defined as modulating bits with the following states:(BN8, BN9 . BN48) =(0, 1, 0, 1, 0, 1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 1, 1, 0, 0, 1, 0, 0, 1, 1, 0, 1)in case alternative training (synchroniz
8、ation) sequence “TS2“ is used, the “synch. sequence bits“ shall be defined as modulating bits with the following states:(BN8, BN9 . BN48) =(1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 1, 0, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1)2、0503协议中关于Access Burst encrypted bits
9、 信道编码方式的描述编码过程如下：（1）8比特的接入信息先增加6比特的CRC校验比特（2）将目标基站的BSIC码和6比特的CRC校验比特异或（3）进行1/2的卷积编码，输出36比特encrypted bitsThe burst carrying the random access uplink message has a different structure. It contains 8 information bits d(0),d(1),.,d(7).Six parity bits p(0),p(1),.,p(5) are defined in such a way that in GF
10、(2) the binary polynomial:d(0)D13 +.+ d(7)D6 + p(0)D5 +.+ p(5), when divided by D6 + D5 + D3 + D2 + D + 1 yields a remainder equal to 0.The six bits of the BSIC, B(0),B(1),.,B(5), of the BS to which the 手机异常位置更新导致的立即指配成功率降低问题分析内部公开2018-3-12华为机密，未经许可不得扩散第 6 页, 共 16 页Random Access is intended, are add
11、ed bitwise modulo 2 to the six parity bits, p(0),p(1),.,p(5). This results in six colour bits, C(0) to C(5) defined as C(k) = b(k) + p(k) (k = 0 to 5) where:b(0) = MSB of PLMN colour codeb(5) = LSB of BS colour code.This defines u(0),u(1),., u(17) by:u(k)= d(k)for k = 0,1,.,7u(k)= C(k-8)for k = 8,9,
12、.,13u(k)= 0for k = 14,15,16,17 (tail bits)The bits e(0),e(1),., e(35) are obtained by the same convolutional code of rate 1/2 as for TCH/FS, defined by the polynomials:G0 = 1 + D3 + D4G1 = 1 + D + D3 + D4and with:e(2k)= u(k) + u(k-3) + u(k-4)e(2k+1)= u(k) + u(k-1) + u(k-3) + u(k-4)for k = 0,1,.,17 ;
13、 u(k) = 0 for k 03、GSM 4.08协议信道请求具体含义：The message is only one octet long, coded as shown in figure 9.1/GSM 4.08 and table 9.9/GSM 04.08.手机异常位置更新导致的立即指配成功率降低问题分析内部公开2018-3-12华为机密，未经许可不得扩散第 7 页, 共 16 页8 7 6 5 4 3 2 1 ESTABLISHMENT RANDOM octet 1 CAUSE REFERENCE FigureFigure 9.1/GSM9.1/GSM 04.08:04.08:
14、 CHANNELCHANNEL REQUESTREQUEST messagemessage contentcontent位置更新相关bit定义:TableTable 9.9/GSM9.9/GSM 04.08:04.08: CHANNELCHANNEL REQUESTREQUEST messagemessage contentcontentMS codesAccording to Establishment cause:bits8 . 1101xxxxxEmergency call110xxxxxCall re-establishment; TCH/F was in use, orTCH/H w
15、as in use but the network does notset NECI bit to 1011010xxCall re-establishment; TCH/H was in use and thenetwork sets NECI bit to 1011011xxCall re-establishment; TCH/H + TCH/H was in useand the network sets NECI bit to 1手机异常位置更新导致的立即指配成功率降低问题分析内部公开2018-3-12华为机密，未经许可不得扩散第 8 页, 共 16 页100xxxxxAnswer to paging0010xxxx0011xxxxSee table 9.9a/GSM 04.080001xxxx111xxxxxOriginating call and TCH/F is needed, or originating calland the network does not set NECI bit to 1, orprocedures that can be completed wi
