Data Communications and Networks Unit 9526M Level H MIKE DAWSON

Data Communications and NetworksUnit 9526M Level H

MIKE DAWSON

Introduction

Welcome to the course I am Mike Dawson My email address is

[email protected] You are ? What you expect from the course

mailto:[email protected]

Main Topics covered

The main area’s of study

IP Networking Communication Protocols Routing Switching Transport Security

Course outline topics

Intro to networking and Datacomms OSI 7 Layer model Layer 1 Physical bit level Layer 2 Data Link 802.2 802.3 Ethernet Bridges Routers Hubs Switches Spanning Tree VLANS Layer3 Routing ( IP ) Layer 4 Transport ( TCP/UDP )

Topic 2 IP

IP addressing ClassFull – A, B, C ClassLess (VLSM) PREFIX (CIDR) Subnetting TCP UDP

Topic 3 Routing Fundamentals Overview Classfull routing (RIPv1 IGRP) Classless routing (EIGRP OSPF) Distance Vector routing (RIP) Link State routing (OSPF) Convergence Case study

Topic 4 Protocols

Frame Relay ATM ISDN HDLC PPP

OPEN SHORTEST PATH FIRST (OSPF) Algorithm derivation Dykstra Link State Components AREAS Convergence Configuring

Security

Secure communication PAP CHAP Radius AAA Encryption

Distance Vector routing

Algorithm derivation Bellman-Ford Compare – Contrast Convergence Poison reverse Split Horizon Count to Infinity Hold-downs Metrics

Communication techniques CRC Encoding Error detecting codes Scramblers Manchester code NRZ code RZ code

Real Life

Refer to the network diagram Real life tasks will be examined IP network design IP Subnetting design Overview of applications Windows NT Novell Apple

What This Means

After completing this course you will be familiar with the following.

Networking fundamentals Advanced networking concepts Network design IP Routing using CISCO devices Know things most people don’t

understand. Get a better job .. If you want

Next Steps

Lets start lesson 1

Lesson 1Basic Concepts of Internetworks, Routers and Adressing

Once upon a time Mini computer (8086) Micro computer (Z80,286,386) Decentralisation Sneaker net What does this mean ?

Primary Purpose of a LAN is resource sharing Pooling resources LOCAL AREA NETWORKS Centralisation Sharing resources Printers Files Applications

Shared medium

Ethernet Media

Shared medium Data Link

Ethernet Media

Shared medium Data Link Co-axial cable 10base2 (185m)

Ethernet Media

Shared medium Data Link Co-axial cable 10base2 (185m) 10baseT CAT 3,4,5 UTP Unshielded Twisted Pair

RJ45 terminator upto (100m)

Ethernet Media


RJ45 terminator upto (100m) 100baseTX CAT5,6,7 UTP RJ45 (100m)

Ethernet Media


RJ45 terminator upto (100m) 100baseTX CAT5,6,7 UTP RJ45 (100m) 100baseFX 62.5/125micron multimode fibre 780nm

laser upto (400m)

Ethernet Media


RJ45 terminator upto (100m) 100baseTX CAT5,6,7 UTP RJ45 (100m) 100baseFX 62.5/125micron multimode fibre 780nm

laser upto (400m) 1000baseLX 9-micron core, 1300 nm laser mono-

mode fibre upto (10km)

Ethernet Media

All devices attach to a common data link through some sort of Network Interface Card ( NIC )

Rules must govern communication etiquette Medium Access Control ( MAC ) IEEE 802.3

Data Link

OSI data link identifiers Each devices needs a unique identifier Burned in / Physical / Machine / MAC Data link transport uses Encapsulation Encapsulation is like an electronic envelope

Ethernet IEEE802.2/3 Token Ring IEEE802.5 FDDI ( HDLC encapsulation )

MAC Identifier structure MAC address is not a real address because its fixed

to the device not the area the device lives. Its more like a name than an address!



Ethernet MAC 48 binary bits



Ethernet MAC 48 binary bits 0100 = 4 in Hexadecimal




248 = 281474976710656 individual iDs




248 = 281474976710656 individual iDs The above decimal number can be represented as a

48 bit binary number.





48 bit binary number. 4 Binary bits make up one Hexadecimal number





48 bit binary number. 4 Binary bits make up one Hexadecimal number 8 Binary bits = two Hexadecimal numbers





48 bit binary number. 4 Binary bits make up one Hexadecimal number 8 Binary bits = two Hexadecimal numbers 0100.1111 = 4F

Complete Task 1

Task 1 Answers 11111111 = 255 10000000 = 128 11000000 = 192 11100000 = 224 11110000 = 240 11111000 = 248 11111100 = 252 11111110 = 254

Task 1 Answers 11111111 = FF 10000000 = 80 11000000 = C0 11100000 = E0 11110000 = F0 11111000 = F8 11111100 = FC

Task 1 Answers Convert Hexadecimal to Decimal FE23 = 65059 FFFF = 65535 1010 = 4112 1111 = 4369 00AB = 171 66FA = 26362

Boolean AND function Lets Logically AND the two binary numbers below

1111111111001101 AND11001101 Answer

Network Identifiers 10101100.00010000.00010001.00001010 = 172.16.17.10

The above DOTTED DECIMAL notation is how IP addresses are represented.

More on IP addressing in later lectures.

OSI Layer 2 The Information presented so far may be

summarised.

A Data Network is one or more devices sharing a common transmission medium.

Each and every device must have a unique iD Using iD’s and Encapsulation to send data in a

virtual envelope

A wonderful tool but.. Everybody wants one. As the LAN grows so do the problems

Problem 1 Distance

Three factors

1. Attenuation caused by the length of the cables.

2. Interference as cables get longer there is more chance of external forces causing distortion.

3. Distortion – As the signal gets weaker and interfered with it can become unrecognisable, hence the need for pulse regenerators.

Signal degrading effects

2nd Big Problem

More devices more demand for DataLink capacity.

For Ethernet = more collisions For Token ring = Token rotation

time > We need to do something

How do we reduce the problems.1. Make the LAN segment smaller

Make the LAN segment smaller By reducing the size of the

DataLink you keep local traffic local.

We do this with a BRIDGE Which needs to 1st perform 4 tasks.

Transparent Bridging

has 4 states.

1. BLOCKING

2. LISTENING

3. LEARNING

4. FORWARDING

Transparent BridgingA Transparent bridge must do the following:-

1. Not modify the frames it passes thru.



2. Learn by Listening where each MAC resides




3. Build a table of MAC / Port





4. Forward all Broadcasts out each port





4. Forward all Broadcasts out each port

5. If no destination address is found in the forwarding table then flood the frame out all ports except the port it came in on.

Transparent Bridging

What is a bridge loop

Preventing Bridge Loops The Spanning Tree Protocol was developed to

overcome the problems of redundant links. The Spanning Tree disables redundant links Disabled redundant paths are placed into

Standby mode / Blocking mode

Preventing bridge loops

Bridge Protocol Data Unit The BPDU is a special kind of packet used by all

switches and bridges running the Spanning Tree Algorithm (STA)

BPDU take part in the selection of a ROOT bridge

BPDU’s are sent out every 2 sec of every port in order to maintain a loop free topology.

Root Bridge Selection

“I am the King” At start up each switch assumes it is the root

bridge with the LOWEST iD and King.

The Bridge iD = 2 byte priority + 6 byte MAC

If all bridges have the same priority then the bridge with the Lowest MAC identifier will become the Root Bridge.

Root Association After the root bridge has been Elected by passing

BPDU’s between participating switches.

Each bridge forms an association with the root via BPDU

By receiving BPDU for the root on multiple ports indicates a loop path to the root.

One of the ports must be blocked

Calculating Path Cost

STP Port States

There are FOUR states for a port participating in the STA

1. Blocking

2. Listening

3. Learning

4. Forwarding

Bridge Forwarding Table

Finally Broadcasts

FF.FF.FF.FF.FF.FF Bridges Flood Broadcasts out of all

interfaces except the interface upon which it received the packet.

This can cause a lot of un-necessary traffic

Broadcasts have to be opened by all devices using their time and resources even if its not applicable to them

In order to block broadcasts we need to segment the network logically

Segmentation of LANS logically requires more sophistication.

OSI LAYER 3 ROUTERS can perform this task.

They can also change a DataLink encapsulation type so that Ethernet can communicate with Token Ring

Answers to review questions

1.The primary purpose of a local-area network is to allow resource sharing. The resources may be devices, applications, or information. Examples of shared resources are files, databases, e-mail, modems, and printers


2. A protocol is an agreed-upon set of rules. In data communications, the rules usually govern a procedure or a format.


3. A Media Access Control protocol defines how a given LAN medium is shared, how LAN devices connected to the medium are identified, and how frames transmitted onto the medium are formatted

.


4. A frame is a digital "envelope" that provides the information necessary for the delivery of data across a data link. Typical components of a frame are identifiers (addresses) of the source and destination devices on the data link, an indicator of the type of data enclosed in the frame, and error-checking information.

.


5. A feature common to all frame types is a format for identifying devices on the data link an ID field or MAC address / id.


6. A Media Access Control address or identifier is a means by which individual devices connected to a data link are uniquely identified for the purpose of delivering data .


7. An address specifies a location. A MAC address is not a true address because it is permanently associated with the interface of a specific device and moves whenever the device moves. A MAC identifies the device, not the location of the device.


8. The three sources of signal degradation on a data link are attenuation, interference, and distortion. Attenuation is a function of the resistance of the medium. Interference is a function of noise entering the medium. Distortion is a function of the reactive characteristics of the medium, which react differently to different frequency components of the signal.


9. A repeater is a device that extends the useful range of a physical medium by reading a degraded signal and producing a "clean" copy of the signal


10. A bridge is a device that increases the capacity of a LAN. A bridge divides the data link into segments, forwarding only traffic that is generated on one segment and is destined for another segment. By controlling and limiting the traffic on a data link, more devices may be attached to the LAN.


11. A transparent bridge "listens promiscuously" on each of its ports. That is, it examines all frames on all media to which it is attached. It records the source MAC identifiers of the frames, and the ports on which it learns the identifiers, in a bridging table. It can then refer to the table when deciding whether to filter or forward a frame. The bridge is transparent because it performs this learning function independently of the devices that originate the frames. The end devices themselves have no knowledge of the bridge.


12. Three fundamental differences between local-area and wide-area networks are: LANs are limited to a small geographic area, such as a single building or small campus. WANs cover a large geographic area, from citywide to worldwide. LANs usually consist entirely of privately owned components. Some components of a WAN, such as a packet switching network or point-to-point serial links, are usually leased from a service provider. A LAN provides high bandwidth at a relatively cheap price. The bandwidth across a WAN is significantly more expensive.


13. A broadcast MAC identifier, when used as the destination address of a frame, signifies that the data is for all devices attached to the data link. In binary, the broadcast MAC identifier is all ones. In hex, it is FFFF.FFFF.FFFF


14. The primary similarity between a bridge and a router is that both devices increase the number of hosts that, may be interconnected into a common communications net work.


15. The difference is that a bridge works by interconnecting separate segments of a single network, whereas a router interconnects separate networks


16. A packet is the means by which data is transported from one network to another .


17. The similarity between a frame and a packet is that they both encapsulate data and provide an addressing scheme for delivering the data.


18. The difference between a frame and a packet is that the frame delivers data between two devices sharing a common data link, whereas a packet delivers data across a logical pathway, or route, spanning multiple data links.


19. Neither the source nor the destination address of a packet changes as it progresses from the source of the packet to the destination.


20. Network addresses are the addresses used in packets .


21. Each network address has a network part, which identifies a particular data link, and a host or node part, which identifies a specific device on the data link identified by the network part .


22. A packet identifies a device from the perspective of the entire internetwork. A frame identifies a device from the perspective of a single data link. Because the connection between two devices across an internetwork is a logical path, a network address is a logical address. Because the connection between two devices across a data link is a physical path, a data link identifier is a physical address.

Our Next Lecture will investigate in more detail the layers involved in producing this intelligent switch / routed network

Thank you

Documents

Data Communications and Networks Unit 9526M Level H MIKE DAWSON