Channel Associated Signaling - E1

In my previous post, I have talked about CAS T1 in this post, I will be covering CAS E1. CAS T1 is no more used by service providers. CAS E1 uses 32 channels from DS0 to DS31 but the signalling and framing is carried by different frames. Channel 0 is used for framing and channel 16 is used for signaling. It means channel 1-15 and 17-31 are used for data.
Frame 16 is having 8 bits. First 4 bits are used for 1 to 15 and rest 4 bits are used for 17-31.


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shivlu jain
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How To Maintain S,G For Long Time

Really a awesome command which can help to force S,G entry for a long period.

ip pim sparse sg-expiry-timer


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shivlu jain
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Channel Associated Signaling

CAS is known as channel associated signaling. This is the signaling method used in the digital world.

In digital world, we have two type of signaling method one is CAS and another is CCS. In this post, I will be covering CAS T1 which is also known as channel associated signaling T1. CAS signalling is further classify two parts :-
a) T1 CAS Signaling
b) E1 CAS Signaling

T1 is mainly used in Europe, US and other countries. T1 uses 23B+1D channel. CAS in T1 enviornment is always equal to Robbed Bit Signaling (RBS). T1 is having 24 time slots, each timeslot carries 64Kbps and when we club the whole 24 time slots, a single frame is built up.

Frame1:- |Ist DS0|2nd DS0|3rd DS0|4th DS0|5th DS0|6th DS0|7th DS0|.....|24th DS0|F
Frame2:- |Ist DS0|2nd DS0|3rd DS0|4th DS0|5th DS0|6th DS0|7th DS0|.....|24th DS0|F
Frame3:- |Ist DS0|2nd DS0|3rd DS0|4th DS0|5th DS0|6th DS0|7th DS0|.....|24th DS0|F
Frame4:- |Ist DS0|2nd DS0|3rd DS0|4th DS0|5th DS0|6th DS0|7th DS0|.....|24th DS0|F
Frame5:- |Ist DS0|2nd DS0|3rd DS0|4th DS0|5th DS0|6th DS0|7th DS0|.....|24th DS0|F
Frame6:- |Ist DS0|2nd DS0|3rd DS0|4th DS0|5th DS0|6th DS0|7th DS0|.....|24th DS0|F

1010101 0 --> 0 is signaling bit from ist DS0. Same way get the LSB from every DS0.
The least significant bit in every 6th frame is known as signaling. It means every 6th frame will have only 7 bits instead of 8 bits. Thats true. Lets calculate the length of Frame1. 64Kbps * 24 = 1536 Kbps + 8 (Framing Type) = 1544 kbps or 1.544 Mbps.

T1 is having two typed of framing types:-
a) Super frame:- It sends 12 T1 frame at a time. It means 6 and 12th frame will be sent along and we will be having two RBS.
b) Extended super frame:- It sends 24 T1 frame at a time. It means 6,12,18 and 24th frame will be sent along and we will be having four RBS.


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shivlu jain
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Bandwidth Required By Single Voip Call

Yesterday, One of my collegue walk down to me and had asked a very wonderful question that how much bandwidth is required to deliver a one voip call. I replied that go with www.voip-calculator.com. Then he rasied a point that dusing customer call I would have limited access to internet, so provide me some sort of calculation which could help me to calculate the bandwidth.
To calculate the bandwidth of voip call, calrity of codec, delay and transport media is essential. Depicted is list of Various Overhead

Data Link Overhead

Ethernet 18 Bytes
PPP 6 Bytes
Ethernet With Dot1q 22 Bytes



Network Overhead
IP 20 Bytes
UDP 8 Bytes
RTP 12 Bytes

Payload Size is 20 bytes

Lets calculate how much bandwidth is required for 3 calls, if customer is using G.729 codec and transposrt media is ethernet.

total bandwidth requirement = (total packet size*codec bandwidth requirement)/payload size

((18 Bytes of Ethernet + 20 Bytes Of IP + 8 Bytes Of UDP + 12 Bytes of RTP + 20 Bytes of Payload) * 8 Kbps G.723 Codec Bandwidth)) / 20 Bytes of Payload

(78 Bytes * 8 Kbps ) / 20 Bytes

31 Kbps

31 kbps is the bandwidth required by one call, now multiply it by 3.


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shivlu jain
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Digital Signal Processor

Digital Siginal Processor aka DSP which are used to convert analog to digital and vice versa. It is same like we use the graphic cards in our computers but in case of routers need to use DSP. The DSP is required because router has to do do the routing but when the voice comes in picture the load becomes double, to reduce the load DSP are used. In the old days, cisco used to sell NM-HDV cards which carry 3 or 5 DSP to convert analog to voice signals. But now a days ISR router comes with inbuit DSP processor.


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shivlu jain
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Quantization Techniques In Voip Network

There is mainly two most inportant quanization techniques
a) Wave Codecs
b) Source Codecs

Wave Codecs
G.711 falls in wave codec category and takes bandwidth of 64Kbps approx by using Pulse Code Modulation technique(PCM). It is the same as described by Nyquist. G.726 uses 32,24 or 16 Kbps and uses Adapative Differential Pulse Code Modulation (ADPCM). When G.726 uses 32 Kbps during that time it takes 4 bits for error control and slips the reoccurence bits, for 24 Kbps it uses 3 bit and for 16 Kbps it uses 2 bits. Definately during the less error control bits increases the error rate because you can skips those bits also which need to be used. This is aka quantization error.

Source Codecs
Sorce codecs are designed to work with human voice. Like cisco G.729 takes 8Kbps of bandwidth to deliver voice. Now the question comes in mind why it takes so less? Actually it is having in built codebook in the database and on the basics of frequeny it matches the code and place the code in a packet. Lets assume if you want to say CISCO, it will go like "ccceessccoo" and it is having built in binary for the words. Thats why it eats less bandwidth rather than other protocol.

As move in the less bandwidth protocols the delay need to be considered because it always increases. So one can say bandwidth is inversely proportional to delay.


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shivlu jain
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Why Does EPBX Require 64Kbps

In this post, I will be exploring why EPBX requires 64Kbps of bandwidth to deliver end to end voice. To understand the concept, one should know how analog packets converts into digital packets. To convert analog packets into digital, sampling of packets is required. It can be done with the help of Nyquist theorem.
As per Nyquist theorem, sample the signal with double of its frequency to convert it into digital format. Human ear can understand from 20 - 20000 Hz, human can peak between 200 - 9000 Hz. A telephone channel can generate frequency between 300 - 3400 Hz and Nyquist theorem can convert the packets which fall in between 300 - 4000 Hz. If we want to convert the packets which are flowing in telephone channel, we need to twice the frequency. A telephone channel cannot generate more than 3400Hz and its twice will not cross the nyquest theorem. Lets consider if frequency is 4000Hz, double it which is 8000Hz. So we need to create 8000 samples in single second.
1 Sample = 1/8000 Sec

This is pure analog format. To convert it into digital format we need to calculate in terms of numbers and the process is known as quantization. Every second is divided into 8000 number of samples and each sample represents a scale of 256. As we move higher the difference between the number scale increases. It means 8000 samples per seconds and each sample is of 1 byte. Now calcute

8000 samples * 8 bits (1 Bytes) = 64000bps or 64Kbps



This is the bandwidth required for Epbx to set up end to end voice system.



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shivlu jain
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Zone Based Firewall

Cisco IOS release 12.496)T introduced a new feature called Zone Based Policy Firewall. Prior to this Cisco provided CBAC - Context Based Access Control list, in which the policy was applied to specific interface but in ZFW, policies are configured and mapped to the specific interfaces. In ZFW, interface need to be a part of zone, once it is done then the policies whicha re defined for zone are used.

Rules For ZFW
1. A zone need to be configured before it assigned to any interface.
2. Like firewall, interface cannot participate in multiple zones.
3. By default all the traffoc from one zone to another zone is blocked but within zone it is permitted.
4. If an interface doesnot participate in any ZFS policy, it works as regular port.

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shivlu jain
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Loopback Summarization Is Now Possible

It is always recommended that not to perform the summarization on loopbacks because it does the black holing in the cloud. I have already explained in the post "Why Sumamrization Is Not Recommended On Loopbacks". But the problem of not doing summarization on loopbacks leads to the inconsistency of OSPF hierarchy. As a consequence, RIB & LFIB will contain all the loopbacks as 32 which leads to the memory problem. As per RFC 3036, section 3.5.7.1 LDP recommends that the IP address of the FEC Element should exactly match an entry in the IP Routing Information Base.
To overcome this problem, a new solution has been proposed which is alluded in RFC 5283. As per Proposed solution of Longest-Match Label Mapping Message Procedure which depicts a new label mapping procedure for LDP. Excerpt from the RFC 5283 is given below
"With this new Longest-Match Label Mapping Procedure, an LSR receiving
a Label Mapping message from a neighbor LSR for a Prefix Address FEC
Element FEC1 SHOULD use the label for MPLS forwarding if its routing
table contains an entry that matches the FEC Element FEC1 and the
advertising LSR is a next hop to reach FEC1. If so, it SHOULD
advertise the received FEC Element FEC1 and a label to its LDP peers.

By "matching FEC Element", one should understand an IP longest match.
That is, either the LDP FEC element exactly matches an entry in the
IP RIB or the FEC element is a subset of an IP RIB entry. There is
no match for other cases (i.e., if the FEC element is a superset of a
RIB entry, it is not considered a match).

Note that LDP re-advertises to its peers the specific FEC element
FEC1, and not the aggregated prefix found in the IP RIB during the
longest-match search.

Note that with this Longest-Match Label Mapping Procedure, each LSP
established by LDP still strictly follows the shortest path(s)
defined by the IGP.
"

In future, if vendors implement the solution suggested by RFC 5283, will maintain the OSPF hieracy and save lot of RIB memory.


The main disadvantage of adopting this solution is that service provider need to run BGP on P routers.


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shivlu jain
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Does Different MTU Communicate

Does Different MTU Communicate

A dicey sitution occurs when MTU on different interfaces come in picture. Every is having a thought process that the lowest MTU on interface will negociate each other but in fact, it only works in case of path mtu is enabled. Assume a scenario where a R1 router is connected with R2 router. R1 interface is having MTU of 1500 bytes and latter is having MTU of 1600 bytes. If R1 sends a packets of 1501 bytes to R2 without DF bit set, R1 will fragment the packet and forwards it to R2 reassembles the packet and will forward a single packet of 1501 bytes to R1. R1 interface will drop the packet because the interface MTU is set to 1500 bytes and communiaction will never occur in case of packet size more than 1500 bytes.


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shivlu jain
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Control Word In Pseudowire

Control word plays a vital role in AToM. It is 32 bit field which is inserted between VC label and transport layer in case of AToM. This is added by the ingress PE and removed by the egress PE.

Structure of control Word

Tunnel Lable/IGP Label
VC label
Control Word
Layer 2 Frame

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shivlu jain
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Cisco IOS Upgrade Management

One of my friend GuruPrasad has sent me a white paper on Cisco IOS Upgrade Management. Document describes the full life cycle of planning and upgradation.

Click here to download


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shivlu jain
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Introducing New CCIE R & S Lab

Cisco has revised the certification requirements for CCIE Routing & Switching (CCIE R&S)-the expert level certification for network engineers.
The new certification standards reflect the job skills employers look for at the expert level and are outlined on the Cisco Learning Network at CCIE R&S v4.0 written exam topics and CCIE R&S v4.0 lab exam topics. The revised CCIE R&S v4.0 exams are scheduled for release on October 18, 2009 and will immediately replace the currently available v3.0 exams.
To support the certification changes, the Cisco 360 Learning Program for CCIE R&S is being updated with new lessons on MPLS and Troubleshooting, additions to the instructor-led workshops, new lab exercises for self-paced practice, and new performance assessments. The Program is the only authorized expert training currently aligned to CCIE R&S v4.0. The program is delivered globally by Cisco Learning Partners .

For more please visit
https://cisco.hosted.jivesoftware.com/community/certifications/ccie_routing_switching/written_exam?view=overview

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shivlu jain
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Deploying & Testing Of SSM in Service Provider Cloud

Implementation of SSM is really easy. I have already covered how to implement SSM in service proivder cloud. In this post, a basic test topology is used for a vrf customer which is using multicast at their end. The same stream need to be transported by mpls service proivder.
Implementation is fully covered in the documnet. Click here to download it.


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shivlu jain
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L2TPv3: Transport Header Overhead Used By Various Transport Types

In yesterday's post, I have explained how PMTU help to stop the reassembly of l2tpv3 packets over IP cloud consequence saves router CPU processes. In this post, I have transport header size with respect to the transport type which help to calculate the exact MTU size for CE.

MTU= 1500- (20 Bytes (IPv4 Header) + 4 Bytes (L2TPv3 Overhead) + 8 Bytes (Option Cookie Overhead) + Transport Header Size)


Transport Type Transport Header Size
Ethernet Port 14 Bytes
Ethernet Vlan 18 Bytes
HDLC 2 Bytes
PPP 4 Bytes
Frame Relay DLCI,CISCO 2 Bytes
Frame Relay DLCI,IETF 8 Bytes

E.g. If we want to calcute the overhead occured in case of Ethernet where dot1q is used. The calculation is given below

1500-(20+4+18) = 1462

1462 Bytes is the maximum MTU avail by CE. The same is depicted in previous post.


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shivlu jain
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MTU Problem In L2TPV3

Introduction
Layer 2 VPN is being used by many service providers. L2tpv3 is used to provide layer 2 services to the customer over IP/MPLS cloud. In this document, MTU issue has been simulated with its workaround.


How To Check The Status Of Circuit
R1#show l2tun session

L2TP Session Information Total tunnels 1 sessions 1

LocID RemID TunID Username, Intf/ State Last Chg Uniq ID
Vcid, Circuit
50413 45422 23086 20, Fa0/0 est 00:00:46 4

R2#show l2tun session

L2TP Session Information Total tunnels 1 sessions 1

LocID RemID TunID Username, Intf/ State Last Chg Uniq ID Vcid, Circuit
45423 50414 17633 20, Fa0/0 est 00:00:01 4

Check The Encapsulation Type
L2TP Session Information Total tunnels 1 sessions 1

Session id 20044 is up, tunnel id 59245
Remote session id is 50078, remote tunnel id 25237
Locally initiated session
Call serial number is 4048300002
Remote tunnel name is R2
Internet address is 20.20.20.20
Local tunnel name is R1
Internet address is 10.10.10.10
IP protocol 115
Session is L2TP signaled
Session state is established, time since change 00:00:10
DF bit off, ToS reflect disabled, ToS value 0, TTL value 255
UDP checksums are disabled
FS cached header information:
encap size = 24 bytes
45000014 00000000 FF737F3B 0A0A0A0A
14141414 0000C39E
1 Packets sent, 1 received
60 Bytes sent, 60 received

Encapsulation Type 24 means 20 bytes of IP Header and 4 Bytes of L2tpv3.

Initiate a Ping request of 1500 bytes with df-bit from R0 which is CPE. Before starting a ping check the two given outputs on R2 router.
R2#show interfaces fastEthernet 0/0 switching
FastEthernet0/0

Protocol Other
Switching path Pkts In Chars In Pkts Out Chars Out
Process 0 0 0 40878
Cache misses 0 - - -
Fast 170 52853 158 15680
Auton/SSE 0 0 0 0


R2#sh ip traffic | i fra
0 fragmented, 0 fragments, 0 couldn't fragment
Start the ping from R0 router
R0#ping 192.168.1.3 df-bit size 1500 repeat 10

Now check the outputs again
Show interfaces fa0/0 switching
Protocol Other
Switching path Pkts In Chars In Pkts Out Chars Out
Process 0 0 27 40878
Cache misses 0 - - -
Fast 197 52853 158 15680
Auton/SSE 0 0 0 0

R2#sh ip traffic | i fra
27 fragmented, 54 fragments, 0 couldn't fragment

Nice findings come here. You can R0 is sending packets with DF bit sent and when the packet comes at R1 interface it encapsulate the packet with l2tpv3 but the packet size becomes more than 1500 so the l2tpv3 packet fragments on R1 and sent over l2tpv3 to R2. At R2 again reassemble of fragment packet occurs which can be seen from the above output. Total 27 packets has been sent from R0 but due to fragmentation each packet is divided into two parts and reassemble at R2. This is reason we are getting 54 fragments which is double the sent packets. The main disadvantage of using this is that every time PE tail end router has to reassemble the packet consequence lot of CPU is required. From the “Show interfaces fa0/0 switching”, it is cleared that the packets received as fast switched and processed as process switch.

Solution Of The Problem
The problem can be overcome by using PMTU which is path MTU discovery. Now add ip pmtu command under pseudowire and initiate a ping request from R0 router.
R0#ping
Protocol [ip]:
Target IP address: 192.168.1.3
Repeat count [5]:
Datagram size [100]:
Timeout in seconds [2]:
Extended commands [n]: y
Source address or interface: 192.168.1.1
Type of service [0]:
Set DF bit in IP header? [no]: y
Validate reply data? [no]:
Data pattern [0xABCD]:
Loose, Strict, Record, Timestamp, Verbose[none]: V
Loose, Strict, Record, Timestamp, Verbose[V]:
Sweep range of sizes [n]: y
Sweep min size [36]: 1450
Sweep max size [18024]: 1470
Sweep interval [1]:
Type escape sequence to abort.
Sending 105, [1450..1470]-byte ICMP Echos to 192.168.1.3, timeout is 2 seconds:
Packet sent with a source address of 192.168.1.1
Packet sent with the DF bit set
Reply to request 0 (564 ms) (size 1450)
Reply to request 1 (564 ms) (size 1451)
Reply to request 2 (552 ms) (size 1452)
Reply to request 3 (564 ms) (size 1453)
Reply to request 4 (564 ms) (size 1454)
Reply to request 5 (344 ms) (size 1455)
Reply to request 6 (564 ms) (size 1455)
Reply to request 7 (392 ms) (size 1457)
Reply to request 8 (756 ms) (size 1458)
Reply to request 9 (708 ms) (size 1459)
Reply to request 10 (564 ms) (size 1460)
Reply to request 11 (320 ms) (size 1461)
Reply to request 0 (564 ms) (size 1462)
Unreachable from 192.168.1.3, maximum MTU 1462 (size 1463)
Request 13 timed out (size 1464)

From the above output the packets are being dropped after packet size 1462 and a message is appearing that maximum MTU is 1462.
How MTU becomes 1462? Here is the calculation for MTU
20 Bytes IP Header + 4 Bytes L2tpv3 + 14 Bytes Ethernet = 38 Bytes
1500 Bytes – 38 Bytes = 1462 Bytes
This is in case of Ethernet without dot1q, If dot1q is enabled then we need to add 4 bytes more in this consequence MTU will be reduced to 1458.

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shivlu jain
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