Grundlagen der Rechnernetze

Grundlagen der Rechnernetze

Introduction

Overview

• Building blocks and terms

• Basics of communication

• Addressing

• Protocols and Layers

• Performance

• Historical development

Building blocks and terms

Hosts and links

H1 H2

Host

Link

Hosts and links

H1 H2

Host

Link

Host is a computer or

more general a device that communicates with the other host on a network

Link is (in the context of computer networks) connection between two hosts

Point to point connection designates communication connection between two hosts (nodes) or endpoints

Types of communication

• Simplex

• Half-duplex

• Duplex (full duplex)

Medium

• Wired communication

• Wireless communication

• Light(?)

• Sound (ultrasound)

Communication channel between the nodes

• Communication channel – refers to a physical transmission medium (wired or wireless) but it also covers logical connection over

multiplexed medium

Message, stream, packet [1]

H1 ^M H2 H3 S H4

P₁ P₂

…

Payload

Header Trailer

Bytes

Message, stream, packet

Message

Communication primitive, usually consists of multiple packets; usually used in the higher layers of

communication

Stream

A sequence of signals that we use to transmit data

Packet

Formatted unit of data consisting of user data and control data (header and trailer). Essentially a part of a message; several

packets together form a message

Multiple access

H₁ H₂ H₃

…

Collision domain

A network connected by a shared medium; in this network packets may collide with one another when they are sent. A term coming from early versions of

Single hop communication Basically communication within one collision domain; packet reaches destination within one hop

Multiplexing [1]

H₁

H₂

H₃

H₄

H₅

H₆

H₁

H₂

H

H₄

H₅

H₆

…

Multiplexing [2]

• Static multiplexing (predefined)

• Statistical multiplexing (can adapt over time)

• Queueing

• Packet scheduling – the way of controlling packet transmission

Scalability of computer networks

• Scalability

• how networks adapt to the grow of load?

• how networks adapt to the increase of hosts?

• how networks adapt to the increase of links?

Scalability of multiple access networks

H₁ H₂ H₃

…

Assuming that all node pairs communicate the same number of times. What is the share s of the medium per node pair?

Scalability of multiple access networks

H₁ H₂ H₃

…

Assuming that all node pairs communicate the same number of times. What is the share s of the medium per node pair?

𝑠𝑠 = [𝑛𝑛 ∗ (𝑛𝑛 − 1)/2]

= 𝑂𝑂( 1

𝑛𝑛

)

Scalability of fully connected network

H1 H2

H3

H5 H9 H4

H8

H6 H7

H10

H11

number of links l?

Scalability of fully connected network

H1 H2

H3

H5 H9 H4

H8

H6 H7

H10

H11

number of links l?

𝑘𝑘 = 𝑛𝑛 − 1 𝑙𝑙 = 𝑛𝑛 ∗ (𝑛𝑛 − 1)/2

Switched network

H1 H2 H3

H8

H7

H4 S1

S2 S3 S4

S5

Switch – a network device that provides dedicated communication between the hosts

Switched network – computer network that uses network switches

Switched network

H1 H2 H3

H8

H7

H6 H5

H4 S1

S2 S3 S4

S5

Packet switched network – a type of network that uses packets for communication; packet switching is a form of grouping of the data sent over the network; in here network links can be shared

Circuit switched network – a dedicated communication channel (circuit) is established between two hosts; in here network links are dedicated to one specific

communication between the hosts

Switched network

H1 H2 H3

H8

H7

H4 S1

S2 S3 S4

S5

Store and forward – a packet is sent to an intermediate station where it can be either kept or forwarded

Cut through switching – a bigger chunk of the data (frame) is forwarded in smaller pieces even before the whole chunk is received

Multi-hop communication – using multiple stations to transmit data between two hosts

Cloud representation

Internet [1]

N1

N3 N2

R1

H1 H2

H3

R3

R2

H4

H5 H9

H8

Internet [2]

• What is internet?

• Router

• Network interface

• The I nternet and a i nternet

• Physical network

• Intranet

Recursive use of cloud representation

N1

N3 N2

R1

H1 H2

H3

R3

R2

H4

H5 H9

H8

N

Network sizes

• LAN – local-area network

• WAN – wide-area network

• MAN – metropolitan area network ; larger than local area network (LAN) but smaller than the area covered by a wide area network (WAN).

• SAN – storage area network – is a high-speed network of storage devices that also connects those storage devices with servers.

• CAN – Controller Area Network (also known as CAN bus) is a vehicle bus standard designed to allow microcontrollers and devices to communicate with each other in applications without a host computer.

• PAN – personal area network ; network of localized and personalized devices.

• GAN – global area network; connecting everything.

Network sizes

Source: www.cebylon.com/khi1/141-01-GAN-MAN.html

Networks and graphs [1]

N1

N3 N2

R1

H1 H2

H3

R3

R2 H4

H5 H6

H9

H8

H7

H1 H2

H3 N1

R1 R2

N2

H4

H5 H7 H6

H9 H8

N3 R3

Networks and graphs [2]

• Nodes

• Links

• Topology

Formal definition of a network graph:

𝐺𝐺 = 𝑉𝑉, 𝐸𝐸 𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝐸𝐸 ⊆ 𝑉𝑉 × 𝑉𝑉

Topology examples

Bus Tree

Star Ring Mesh

Basics of communication

Types of communication

N1

N3 N2

R1

H1 H2

H3

R3

R2 H4

H5

H6 H9

H8

H7

• Unicast – communication where a piece of information is sent from one point to another point. In this case there is just one sender, and one receiver.

• Multicast – describe communication

where a piece of information is sent from one or more points to a set of other

points. In this case there is may be one or more senders, and the information is distributed to a set of receivers (theer may be no receivers, or any other number of receivers).

• Broadcast – communication where a piece of information is sent from one

point to all other points. In this case there is just one sender, but the information is sent to all connected receivers.

Types of communication

N1

N3 N2

R1

H1 H2

H3

R3

R2 H4

H5 H9

H8

• Forwarding – Packet (frame…) forwarding is the relaying of packets from one

network segment to another by nodes in a computer network. Usually refers to the effective transfer of a packet (frame...)

• Routing – process of selecting a path for traffic in a network, or between or across multiple networks.

• Path – actual path used for transmission between two hosts

Forwarding table

Destination Next hop

4711 3

7893 2

3467 5

2576 2

… …

R

1 2 4 3

5 6

Timeouts and acknowledgements

N1

N3 N2

R1

H1 H2

H3

R3

R2 H4

H5 H9

H8

• Timer

• Timeout

• Acknowledgement ACK

Connection oriented and connectionless communication

N1

N3 N2

R1

H1 H2

H3

R3

R2 H4

H5

H6 H9

H8

H7

Connection oriented

• Telephone, File transfer Connectionless

• VoIP, Post

Client-Server principle

H N S

Client Server

Client-Server principle

H N S

Client Request Server

Client-Server principle

H N S

Client Server

Response

Client-Server principle

H N S

Client Server

• stateful serverremembers client data (state) from one request to the next.

• stateless server does not keep state information. Using a stateless file server, the client must specify

complete file names in each request, specify location for reading or writing.

Adressing

Motivation

N1

N3 N2

R1

H1 H2

H3

R3

R2 H4

H5

H6 H9

H8

H7

How do we transfer message from H8 to H4?

Which path do we use?

Can we always reach the destination?

Physical address – Ethernet example

00001000 00000000 00101011 11100100 10110001 00000010

08 : 00 : 2B : E4 : B1 : 02

Broadcast

11111111 11111111 11111111 11111111 11111111 11111111 FF:FF:FF:FF:FF:FF

Multicast

1XXXXXXX XXXXXXXX XXXXXXXX XXXXXXXX XXXXXXXX XXXXXXXX

Address space

R1

H1 H2 H3

R3 R2

H4

H5 H6

H9 H8

H7

1

2

3 4

1.1 1.2 1.7

2.5

1.10

2.7

2.1

2.4 2.8

3.1

3.2

4.4

4.1 4.2 4.3

Forwarding table and address space

R1

H1 H2 H3

R3 R2

H4

H5 H6

H9 H8

H7

1

2

3 4

1.1 1.2 1.7

2.5

1.10

2.7

2.1

2.4 2.8

3.1

3.2

4.4

4.1 4.2 4.3

Dest Next-Hop H1 After R1 H2 After R1 H3 After R1 H4 Direct H5 Direct H6 Direct H7 After R3

Dest Next-Hop 1.X After R1 2.X Direct 4.X After R3

IP addresses

• 32 bits – approximately 4 billion addresses

• Binary representation 4 groups of 8 bits

• Dot notation – 4 decimal numbers representing 4 groups of bits

• Example:

10101011 01000101 11010010 11110101

171.69.210.245

Classful IP addresses

• What does it mean classful?

• Which are different classes?

• Network address

• Host address

• Broadcast address

Need for additional hierarchical layer

R1

H1 H2 H3

R3 H4 R2

H5 H6

H9 H8

H7

1

2

3 4

1.1 1.2 1.7

2.5

1.10

2.7

2.1

2.4 2.8

3.1

3.2

4.4

4.1 4.2 4.3

Entrance to the University network

Subnetworks

1 0 Network Host

14 16

For example class B Address

11111111 11111111 11111111 (255.255.255.0)

Subnet mask 00000000

Network number

Solution Subnet Host

Subnetting example

R1 H1

Subnet number : 128. 96. 34. 0 = 100000000 01100000 00100010 00000000 Subnet mask : 255.255.255.128 = 111111111 11111111 11111111 10000000

Example: Using one class B network:

128.96.X.X =

= 10000000 01100000 XXXXXXXX XXXXXXXX

128. 96. 34. 15 = 100000000 01100000 00100010 00001111

128. 96. 34. 1 = 100000000 01100000 00100010 00000001

128. 96. 34.130 = 100000000 01100000 00100010 10000010

128. 96. 34.128 = 100000000 01100000 00100010 10000000 255.255.255.128 = 111111111 11111111 11111111 10000000

Changes in forwarding tables

R1 H1

H2

Subnet number : 128. 96. 34. 0 Subnet mask : 255.255.255.128

128. 96. 34. 15

128. 96. 34. 1

128. 96. 34.130

128. 96. 34.128 255.255.255.128

128. 96. 34.139

Subnet number Subnet mask Next hop 128.96.34.0 255.255.255.128 direct (if 1) 128.96.34.128 255.255.255.128 direct (if 2) 128.96.33.0 255.255.255.0 after R2 (if 2)

Interface 1 Interface 2

128. 96. 34.129

Network number Next Hop

128.96 …

R3

Address resolution

H1 R1 H2

128.96.34.15 45:35:FE:36:42:55

128.96.34.1 128.96.34.16 57:FF:AA:36:AB:11 85:48:A4:28:AA:18

IP address Physical address

128.96.34.1 57:FF:AA:36:AB:11 128.96.34.16 85:48:A4:28:AA:18

… …

IP address Physical address

128.96.34.15 ???

128.96.34.16 85:48:A4:28:AA:18

… …

Supernetting – motivation

• Lets assume, for example, the IT department of a university campus, which "autonomously" uses a lot of IP addresses.

• With subnetting, we can efficiently use given set of IP addresses.

• The problem is that the IT department still has to request / manage IP

addressing in the granularities Class-A-, -B-, or -C-network.

Supernetting – motivation

• What happens when for example we need 257 hosts?

1. We can apply for Class B network address. The problem is efficiency

Supernetting – motivation

• What happens when for example we need 257 hosts?

1. We can apply for Class B network address. The problem is efficiency 257

2 ¹⁶ − 1 ≈ 0,39%

Supernetting – motivation

• What happens when for example we need 257 hosts?

1. We can apply for Class B network address. The problem is efficiency 257

2 ¹⁶ − 1 ≈ 0,39%

2. We can also consider 2 class C networks.

Supernetting – motivation

• What happens when for example we need 257 hosts?

1. We can apply for Class B network address. The problem is efficiency 257

2 ¹⁶ − 1 ≈ 0,39%

2. We can also consider 2 class C networks.

This means that we have 2 routing entries in each internet router

Solution: Classless Inter-Domain Routing (CIDR)

• We can aggregate network addresses.

• Example: Lets assume that we have 16 * 256 - 1 hosts.

• We use 16 addresses of Class-C networks.

Not arbitrary addresses, but consecutive, e.g.:

192.4.16

192.4.17

... 192.4.31

Solution: Classless Inter-Domain Routing (CIDR)

• We can aggregate network addresses.

• Example: Lets assume that we have 16 * 256 - 1 hosts.

• We use 16 addresses of Class-C networks.

But not arbitrary addresses, but consecutive, e.g.:

192.4.16 192.4.17 ... 192.4.31

• Now we can observe following:

all addresses begin with the same 20 bits: 11000000 00000100 0001

Solution: Classless Inter-Domain Routing (CIDR)

• Observation: all addresses begin with the same 20 bits:

11000000 00000100 0001

• That means that we need a 20-bit network address

• This is between Class-C (24-bit) and Class-B (16-bit)

• Required output of 2 ^ 4 = 16 Class C addresses

• General question: How many class-C networks requires i-bit network address?

Solution: Classless Inter-Domain Routing (CIDR)

• Observation: all addresses begin with the same 20 bits:

11000000 00000100 0001

• That means that we need a 20-bit network address

• This is between Class-C (24-bit) and Class-B (16-bit)

• Required output of 2 ^ 4 = 16 Class C addresses

• General question: How many class-C networks requires i-bit network address?

2 ^{24−𝑖𝑖}

Solution: Classless Inter-Domain Routing (CIDR)

• We need a notation for the scheme.

• In our example:

192.4.16 192.4.17 ... 192.4.31

• Notation can be summarized as: 192.4.16 / 20

• So this additional number / 20 means network address consists of first 20

bits and summarizes the 2 ^ 4 = 16 successive class-C networks beginning

with 192.4.16.

Quiz

• How to represent the class-C networks from 192.4.0 to 192.4.31 using

/ X notation?

Quiz

• How to represent the class-C networks from 192.4.0 to 192.4.31 using / X notation?

192.4.0 / 19

Quiz

• How to represent the class-C networks from 192.4.0 to 192.4.31 using / X notation?

192.4.0 / 19

• How to represent the single class-C network 192.4.16 in / X notation?

Quiz

• How to represent the class-C networks from 192.4.0 to 192.4.31 using / X notation?

192.4.0 / 19

• How to represent the single class-C network 192.4.16 in / X notation?

192.4.0 / 24

Solution: Classless Inter-Domain Routing (CIDR)

• How are aggregated addresses handled in the router:

• Addresses in the routing tables are pair <length, value>

• This is comparable to the pair <mask, value> in subnetting if the mask

consists of successive 1-bit values

Solution: Classless Inter-Domain Routing (CIDR)

• CIDR allows further route aggregation. For example:

Internet provider Advertise

128.112.128/21

128.112.128/24

128.112.135/24 Client networks

We don’t even need to use 8 consecutive addresses

Solution: Classless Inter-Domain Routing (CIDR)

• What happens with CIDR and routing table entries? Prefixes may overlap.

Lets consider following routing table:

Where do we route the message for 171.69.10.5?

Where do we route the message 171.69.20.5?

Network address Next hop

... ...

171.69/16 if1 171.69.10/24 if2

... ...

Solution: Classless Inter-Domain Routing (CIDR)

• What happens with CIDR and routing table entries? Prefixes may overlap.

Lets consider following routing table:

Where do we route the message for 171.69.10.5?

if2

Where do we route the message 171.69.20.5?

if1

Network address Next hop

... ...

171.69/16 if1 171.69.10/24 if2

... ...

Solution: Classless Inter-Domain Routing (CIDR)

• What happens with CIDR and routing table entries? Prefixes may overlap.

Lets consider following routing table:

Where do we route the message for 171.69.10.5?

if2

Where do we route the message 171.69.20.5?

if1

In general: Longest-Prefix-Match

Network address Next hop

... ...

171.69/16 if1 171.69.10/24 if2

... ...

Subnetting vs CIDR

• Subnetting allows splitting a network address into subnets

• Distribution almost anywhere; everything that can be expressed with the subnet mask

• CIDR is used to aggregate network addresses in a single address

• Aggregation not arbitrary; network addresses must be consecutive; only 2^i sized networks can be aggregated

• Certain flexibility using "dummy networks"

Subnetting and addresses revisited

• Smaller networks using one network address

• Hierarchy

• Better organization

• Better use of resources

• Addresses (network, host, broadcast)

Subnetting example [1]

• We have been given one class C network address: 217.110.20.0 that we want to divide it into two subnets. The questions are:

• How many hosts can we have in each subnet?

• What are the subnet addresses for these two subnets?

• What are corresponding subnet masks?

• What set of IP addresses cover these subnets?

• What are CIDR notations for subnets?

Subnetting example [1]

• We have been given one class C network address: 217.110.20.0 that we want to divide it into two subnets. The questions are:

• How many hosts can we have in each subnet?

2 subnets = address space of 2^7 addresses

Number of hosts = 2^7 – host address – broadcast address = 126

Subnetting example [1]

• We have been given one class C network address: 217.110.20.0 that we want to divide it into two subnets. The questions are:

• What are the subnet addresses for these two subnets?

217.110.20.0

217.110.20.128

Subnetting example [1]

• We have been given one class C network address: 217.110.20.0 that we want to divide it into two subnets. The questions are:

• What are corresponding subnet masks?

255.255.255.128

255.255.255.128

Subnetting example [1]

• We have been given one class C network address: 217.110.20.0 that we want to divide it into two subnets. The questions are:

• What set of IP addresses cover these subnets?

217.110.20.0 - 217.110.20.127

217.110.20.128 - 217.110.20.255

Subnetting example [1]

• We have been given one class C network address: 217.110.20.0 that we want to divide it into two subnets. The questions are:

• What are CIDR notations for subnets?

25 for both networks 217.110.20.0 / 25

217.110.20.128 / 25

Subnetting example [2]

• We have been given one class C network address: 217.110.20.0 that we want to divide it into three subnets, the first having 101 hosts, second 44 and the third 60. The questions are:

• What are the subnet addresses for these three subnets?

• What are corresponding subnet masks?

• What set of IP addresses cover these subnets?

• What are CIDR notations for subnets?

Subnetting example [2]

• We have been given one class C network address: 217.110.20.0 that we want to divide it into three subnets, the first having 101 hosts, second 44 and the third 60. The questions are:

• Actual first question is what are the sizes of those three subnets

• 64 < 101 < 128 => 2

− 1 − 1 = 126

• 32 < 44 < 64 => 2

− 1 − 1 = 62

• 32 < 60 < 64 => 2

− 1 − 1 = 62

Subnetting example [2]

• We have been given one class C network address: 217.110.20.0 that we want to divide it into three subnets, the first having 101 hosts, second 44 and the third 60. The questions are:

• What are the subnet addresses for these three subnets?

217.110.20.0

217.110.20.128

217.110.20.192

Subnetting example [2]

• We have been given one class C network address: 217.110.20.0 that we want to divide it into three subnets, the first having 101 hosts, second 44 and the third 60. The questions are:

• What are corresponding subnet masks?

255.255.255.128

255.255.255.192

255.255.255.192

Subnetting example [2]

• We have been given one class C network address: 217.110.20.0 that we want to divide it into three subnets, the first having 101 hosts, second 44 and the third 60. The questions are:

• What set of IP addresses cover these subnets?

217.110.20.0 - 217.110.20.127

217.110.20.128 - 217.110.20.191

217.110.20.192 - 217.110.20.255

Subnetting example [2]

• We have been given one class C network address: 217.110.20.0 that we want to divide it into three subnets, the first having 101 hosts, second 44 and the third 60. The questions are:

• What are CIDR notations for subnets?

217.110.20.0 / 25

217.110.20.128 / 26

217.110.20.192 / 26

Addresses example

• Given are following IP addresses in CIDR notation. Determine whether these addresses are subnet, host or broadcast addresses.

1. 192.168.0.17 / 24

2. 14.195.1.191 / 26

3. 112.127.0.0 / 14

4. 112.127.0.0 / 16

Addresses example

• Given are following IP addresses in CIDR notation. Determine whether these addresses are subnet, host or broadcast addresses.

1. 192.168.0.17 / 24

=> Host address, host bits are different from 0

Addresses example

• Given are following IP addresses in CIDR notation. Determine whether these addresses are subnet, host or broadcast addresses.

2. 14.195.1.191 / 26

=> Broadcast address, host bits are all equal to 1

Addresses example

• Given are following IP addresses in CIDR notation. Determine whether these addresses are subnet, host or broadcast addresses.

3. 112.127.0.0 / 14

=> Host address, host bits are different from 0

Addresses example

• Given are following IP addresses in CIDR notation. Determine whether these addresses are subnet, host or broadcast addresses.

4. 112.127.0.0 / 16

=> Network address, host bits are equal to 0

Protocols and layers

Protocol and interface

Host 1

Protocol High-Level Object

Host 2

Protocol High-Level Object

Peer-to-peer Interface Service

Interface Service

Interface

Protocol and interface

Host 1

Protocol High-Level Object

Host 2

Protocol High-Level Object

Peer-to-peer Interface Service

Interface Service

Interface

Interoperability

Protocol vs algorithm

Message sequence chart (MSC)

H1 H2

Message sequence chart (MSC)

H1 H2

RTS CTS Data

RTS – request to send

CTS – clear to send

Data – useful data

Protocol state machine

Wait for connection

request

Wait for file request connection request/

connection response

close request

file request/

file response

Example

H N S

Service primitives:

States:

Timeline:

File f GET_FILE(), void ABORT_FILE_RETRIVAL(), ...

CLIENT_IDLE, CLIENT_WAITS_FOR_FILE, ...

if client waits 1000ms then change to state CLIENT_ERROR

Protocol graph

Host 1

Protocol 1

Protocol 3

Protocol 2

Protocol 4

Host 2

Protocol 1

Protocol 3

Protocol 2

Protocol 4

Message encapsulation

Host 1

Application 1

Protocol 1

Protocol 2

Protocol 3

Host 2

Data

Data H1

Data H1

H2

Application 1

Protocol 1

Protokoll 2

Protocol 3

Data

Data H1

Data H1

H2

Multiplexing and demultiplexing

Host 1

Protocol 1

Protocol 3

Protocol 2

Protocol 4

Host 2

Protocol 1

Protocol 3

Protocol 2

Protocol 4

Protocol stack – practical example

OSI model

OSI – Open System Interconnection Communication subnet boundary

OSI model – concepts

Service – set of operations that layer provides to the layer above it Protocol – set of rules that

determine the format and meaning of the packets

(messages) that are exchanged Analogy with OO programming languages: services – abstract data types; protocols implementation of services

OSI model

OSI Principles:

1. A layer should be created where a different abstraction is needed.

2. Each layer should perform a well-defined function.

3. The function of each layer should be chosen with an eye toward defining internationally standardized protocols.

4. The layer boundaries should be chosen to minimize the information flow across the interfaces.

5. The number of layers should be large enough that distinct functions need not be thrown together in the same layer out of necessity and small enough that the

architecture does not become unwieldy.

OSI model

Physical layer– transmitting raw bits (or signal in general) over the communication channel Data link layer – organizes raw data into the data frames (order of 100 – 1000 of bytes) and transmits frames sequentially. Sending back ACK in reliable services. Traffic regulation when we have fast transmitter and slow receiver.

Control of access to the shared medium – separate sublayer media access control

OSI model

Network layer – controls operation of subnet, determines how packets are routed from

source to destination. Possible different types of routes – static tables, routes that can be updated (at the start of each conversation) and highly dynamic routes. Handling congestion (too many packets received).

Transport layer– accepts data above it and splits it in smaller units, passing them to the lower layers. End-to-end layer – carries

communication from source to the destination.

Delivering in the order in which bytes were sent or delivering isolated messages or broadcast to multiple destinations

OSI model

Session layer – allows different machines (users) to establish sessions: dialog control

(whose turn is to transmit), token management (preventing two parties to attempt same

critical operation) and synchronization.

Presentation layer – syntax and semantics of the information. Manages abstract data

structures

Application layer – protocols needed by users

Internet model (TCP/IP reference model)

Internet model (TCP/IP reference model)

Link layer – interface between hosts and transmission (not a layer in standard sense)

Internet layer – corresponds to network layer. “Internet” –

between the networks. Modeled over snail mail, series of packets delivered using one or more gateways. Internet Protocol (IP) and Internet Control Message Protocol (ICMP)

Internet model (TCP/IP reference model)

Transport layer – above internet layer, allows conversation

between the source and

destination. Transport Control Protocol (TCP) – reliable

transmission and User Datagram Protocol (UDP) – connectionless protocol

Application layer – high level protocols. Early examples: Telnet (virtual terminal), FTP (file

transfer), SMTP (electronic mail)

Internet protocols

Comparisons (critique) of OSI and TCP/IP model

• OSI model

• Model made before protocols were made (good and bad)

• Bad technology and implementations (empty layers, big and slow)

• TCP/IP

• Protocols then models

• Not general - it cannot describe anything else but TCP/IP

• Does not distinguish services, services and protocols

• Link layer – not a real layer; no distinction between physical and data layer

• In general a lot of ad-hoc solutions

How do we use TCP (or UDP)

Creating a socket

int socket(int domain, int type, int protocol) domain : PF_INET, PF_UNIX, PF_PACKET, ...

type : SOCK_STREAM, SOCK_DGRAM, ...

protocol : UNSPEC, ...

Passive open on the server side

int bind(int socket, struct sockaddr *address, int len) int listen(int socket, int backlog)

int accept(int socket, struct sockaddr *address, int *len) address : enthält IP-Adresse und Port

backlog : Anzahl erlaubter Pending-Connections

Active open on the client side

int connect(int socket, struct sockaddr *address, int len)

How do we use TCP (or UDP)

Server side: Client side:

Addresses in internet model

Host 2 Host 1

TCP

IP IP

LINK LINK

TCP UDP

UDP

Application Application Application Application

Physical address

IP address Port

Demux-Key

Performance

Bandwidth

1 µs

1 second

0 1 1 0 0 1 1

…

Bandwidth b in this example:

Bandwidth

1 µs

1 second

0 1 1 0 0 1 1

…

Bandwidth b in this example:

𝑏𝑏 = 10 ⁶ 𝑏𝑏𝑏𝑏𝑠𝑠 = 1 𝑀𝑀𝑏𝑏𝑏𝑏𝑠𝑠

Bps and bps

Parameter Order of value Value

KBps 2

Byte/s 1.024

MBps 2

Byte/s 1.048.576

GBps 2

Byte/s 1.073.741.824

TBps 2

Byte/s 1.099.511.627.776

Kbps 10

Bits/s 1.000

Mbps 10

Bits/s 1.000.000

Gbps 10

Bits/s 1.000.000.000

Tbps 10

Bits/s 1.000.000.000.000

Bytes per second vs bits per second

Bps and bps

Parameter Order of value Value

KBps 2

Byte/s 1.024

MBps 2

Byte/s 1.048.576

GBps 2

Byte/s 1.073.741.824

TBps 2

Byte/s 1.099.511.627.776

Kbps 10

Bits/s 1.000

Mbps 10

Bits/s 1.000.000

Gbps 10

Bits/s 1.000.000.000

Tbps 10

Bits/s 1.000.000.000.000

Simplification of surpluses:

Simplification:

𝑛𝑛 𝑀𝑀𝑀𝑀𝑏𝑏𝑠𝑠 = 𝑛𝑛 � 2 ²⁰ 𝑀𝑀𝐵𝐵𝑤𝑤𝐵𝐵

10 ⁶ 𝑀𝑀𝐵𝐵𝑤𝑤𝐵𝐵

= 8𝑛𝑛 � 10 ⁶ 𝑀𝑀𝑤𝑤𝑤𝑤

= 8𝑛𝑛 𝑀𝑀𝑏𝑏𝑏𝑏𝑠𝑠

Propagation delay

H2 H1

d

Time x needed for transmission of one bit at distance d and with signal propagation speed l

x

Propagation delay

H2 H1

d

Time x needed for transmission of one bit at distance d and with signal propagation speed l

𝒙𝒙 = 𝒅𝒅

𝒍𝒍 [ 𝒎𝒎 𝒎𝒎 𝒔𝒔

= 𝒔𝒔]

x

Delay of one 1-hop transmission

H2 H1

d

Time x needed for transmission of n bits at distance d and with signal propagation speed l and bandwidth b:

x

Delay of one 1-hop transmission

H2 H1

d

Time x needed for transmission of n bits at distance d and with signal propagation speed l and bandwidth b:

𝒙𝒙 = 𝒅𝒅

𝒍𝒍 + 𝒏𝒏

𝒃𝒃 [ 𝒃𝒃𝒃𝒃𝒃𝒃𝒔𝒔 𝒃𝒃𝒃𝒃𝒃𝒃𝒔𝒔

𝒔𝒔

= 𝒔𝒔]

x

Delay of one 1-hop transmission

H2 H1

d

Time x needed for transmission of n bits at distance d and with signal propagation speed l and bandwidth b:

𝒙𝒙 = 𝑷𝑷𝑷𝑷 + 𝑻𝑻𝑷𝑷

x

Propagation delay:

𝑷𝑷𝑷𝑷 = 𝒅𝒅

Transmission delay: 𝒍𝒍 𝑻𝑻𝑷𝑷 = 𝒏𝒏

𝒃𝒃 [ 𝒃𝒃𝒃𝒃𝒃𝒃𝒔𝒔 𝒃𝒃𝒃𝒃𝒃𝒃𝒔𝒔

𝒔𝒔

= 𝒔𝒔]

Delay of one multi-hop transmission

d

Time x needed for transmission of n bits at distance d and with signal propagation speed l and bandwidth b and queuing time q:

x H2

H1

Delay of one multi-hop transmission

d

Time x needed for transmission of n bits at distance d and with signal propagation speed l and bandwidth b and queuing time q:

𝒙𝒙 = 𝒅𝒅

+ 𝒏𝒏

+ 𝒒𝒒 [𝒔𝒔]

x H2

H1

Round-trip time (RTT)

d

Round-trip time – time needed for the signal to be sent plus time it takes to get the acknowledgment for that signal

𝑹𝑹𝑻𝑻𝑻𝑻 = 𝟐𝟐 � 𝑷𝑷𝑷𝑷

RTT H2

H1

Bandwidth delay product

Bandbreite

Delay

Definition: Number of the bits n that are contained in one channel with latency of l and bandwidth of b

𝑛𝑛 = 𝑙𝑙 � 𝑏𝑏

Bandwidth delay product

Bandbreite

Delay

Example: Number of the bits n that are contained in one channel with latency of 100ms and bandwidth of 50Mbps

𝑛𝑛 = 50 � 10

𝑏𝑏𝑤𝑤𝑤𝑤𝑠𝑠

𝑠𝑠 � 0,1𝑠𝑠 = 5 � 10

𝑏𝑏𝑤𝑤𝑤𝑤𝑠𝑠

Transfer time and effective throughput

l

x H2

H1

Transfer time and effective throughput

l

x H2

H1

Example: Calculation of transfer time z and effective throughput d and when retrieving a 1MB file over a channel with bandwidth of 1Gbps and RTT of 92ms.

𝑧𝑧 = 1𝑀𝑀𝑀𝑀

1𝐺𝐺𝑏𝑏𝑏𝑏𝑠𝑠 + 0,092 ≈ 8𝑀𝑀𝑏𝑏

1𝐺𝐺𝑏𝑏𝑏𝑏𝑠𝑠 + 0,092 = 8 � 10⁶𝑏𝑏 1 �10⁹𝑏𝑏

𝑠𝑠

+ 0,092 = 0,092 + 0,008 = 0,1𝑠𝑠

𝑑𝑑 = 1𝑀𝑀𝑀𝑀 8𝑀𝑀𝑏𝑏

= 80𝑀𝑀𝑏𝑏𝑏𝑏𝑠𝑠 ≪ 1𝐺𝐺𝑏𝑏𝑏𝑏𝑠𝑠 ‼!

Bit error rate and packet error rate

010100010111100010011101110010110001101

Bit error rate (BER)

Packet error rate (PER)

Packet 1 Packet 2 Packet 3 Packet 4

Bit error rate and packet error rate

010100010111100010011101110010110001101

Bit error

Packet error

Packet 1 Packet 2 Packet 3 Packet 4

Connection between BER and PER for n bit message without correction:

𝑃𝑃𝐸𝐸𝑃𝑃 = 1 − (1 − 𝑀𝑀𝐸𝐸𝑃𝑃 ) ^𝑛𝑛

Additive and bottleneck costs

H1 H2

R1

R2

R3

10ms 5ms 10ms 20ms

1Mbps 1Gbps 1Gbps

1Mbps

e₁ e₂ e₃ e₄

Additive and bottleneck costs

H1 H2

R1

R2

R3

10ms 5ms 10ms 20ms

1Mbps 1Gbps 1Gbps

1Mbps

e₁ e₂ e₃ e₄

Example: What is delay d und bandwidth b between hosts H1 and H2

𝑑𝑑 = �

𝑖𝑖=1 𝑛𝑛

𝐷𝐷𝐵𝐵𝑙𝑙𝐷𝐷𝐵𝐵(𝐵𝐵

) = 45𝑚𝑚𝑠𝑠

𝑏𝑏 = min

𝑀𝑀𝐷𝐷𝑛𝑛𝑑𝑑𝑤𝑤𝑤𝑤𝑑𝑑𝑤𝑤𝑤(𝐵𝐵

) = 1𝑀𝑀𝑏𝑏𝑏𝑏𝑠𝑠

Multiplicative costs

H1 H2

R1

R2

R3

p₁=2/3 p₂=1/3 p₃=1/2 p₄=1/2

e₁ e₂ e₃ e₄

Multiplicative costs

H1 H2

R1

R2

R3

p₁=2/3 p₂=1/3 p₃=1/2 p₄=1/2

e₁ e₂ e₃ e₄

Example: What is the total packet success rate with given packet error rates per link.

𝑟𝑟 = 1 − �

𝑖𝑖=1 𝑛𝑛

(1 − 𝑏𝑏 _𝑖𝑖 )

Performance

example of effective throughput with packet switching

Delay savings

H1 H2

Circuit switching

R1 R2 H1 H2

Message switching

R1 R2 H1 H2

Packet switching

R1 R2

Delay savings

H1 H2

Circuit switching

R1 R2 H1 H2

Message switching

R1 R2 H1 H2

Packet switching

R1 R2

Delay savings

H1 H2

Circuit switching

R1 R2 H1 H2

Message switching

R1 R2 H1 H2

Packet switching

R1 R2

Delay savings

H1 H2

Circuit switching

R1 R2 H1 H2

Message switching

R1 R2 H1 H2

Packet switching

R1 R2

Influence of the packet size

H1

H2

R1 R2

Message size n bits Packet payload k bits Packet header c bits

Bandwidth b bps

Delay per hop d seconds Number of hops h

Effective throughput x

𝒙𝒙 = 𝒏𝒏

𝒃𝒃 ; 𝒃𝒃 = 𝑷𝑷𝑷𝑷 + 𝑻𝑻𝑷𝑷

Influence of the packet size

H1

H2

R1 R2

Message size n bits Packet payload k bits Packet header c bits

Bandwidth b bps

Delay per hop d seconds Number of hops h

Effective throughput x

𝒙𝒙 = 𝒏𝒏

𝒃𝒃 ; 𝒃𝒃 = 𝑷𝑷𝑷𝑷 + 𝑻𝑻𝑷𝑷 = 𝒉𝒉 � 𝒅𝒅 + ( 𝒏𝒏

𝒌𝒌 + 𝒉𝒉 − 𝟏𝟏) � 𝒌𝒌 + 𝒄𝒄

𝒃𝒃

Example plot

EffectiveThroughputin Gbps

Message size 1 GB

Bandwidth 1 Gbps

Header size 64 Byte

Number of hops 10

Delay per hop 10 ms

History and present

Packet switching – the first generation

• The end of 1950s

• Cold war at its peak; DoD (USA – department of defense) looks for command and control center that could survive nuclear attack

• During 1960s

• Contract with RAND corporation (still looking for a solution). Paul Baran develops a distributed and fault tolerant system as a basis for packet switching. AT&T thinks it is not feasible.

Structure of telephone systems

Baran’s distributed switching system

ARPANET

• 1967

• Feasibility of packet switched networks

• Donald Davies (at NPL) independently developed packet switching system as a campus network. They referenced work from Paul Baran

• 1969

• (D)ARPA contracted consulting

company BBN to develop that kind of network and necessary software.

Graduate students from the University of Utah developed host software.

Dec 1969 Jul 1970 Mar 1971

Structure of packet switched subnet according to Clark

ARPANET and NSFNET

• 1974

• First ARPANET protocol (Vinton Cerf and Robert Kahn)

• ARPA pushed usage of TCP/IP;

University of California Berkeley

integrated these protocols in Berkeley Unix

• Late 1970s – end of 1980s

• TCP/IP emerged in its nearly final form

• Associated standards were published in 1981

• Form the 1. January 1983 TCP/IP became the only approved part of ARPANET

NSF backbone 1988

Commercialization of Internet

• During 1980s

•

IP addresses becoming more expensive (scarce); development of hierarchical name structure – DNS – domain name system

•

Further growth of network (universities, research labs, libraries,…); problems with overload;

NSF contract MERIT (consortium from Chicago) to continue operating the network; upgrade of backbone (56kbps -> 448kbps ->1.5Mbps)

•

Merger of ARPANET and NSFNET followed by many other regional networks (Canada, Europe, Pacific)

• 1990

•

First step of commercialization of internet NSFNET donated to nonprofit corporation ANS (Advanced Networks and Services – MERIT, MCI, IBM); further upgrade of the backbone 1.5 Mbps -> 45Mbps

•

NSF ensures fair competition (through the agreements with PacBell, Ameritech, MSF and Sprint)

• 1995

WWW

• During 1990s

• Development in other countries EuropaNET and EBONE (started at 2Mbps then upgraded to 34Mbps)

• Until early 1990s Internet was mainly used in academia.

• Everything changed with development of world wide web (WWW) – Tim Berners-Lee (CERN physicist) and Mosaic Browser Marc Andersen (National Center for Supercomputer Applications in Urbana Illinois)

• Rise of Internet Service Providers – increased number of home computers on

the internet (dial-up service)

Simplified overview of Internet today

Wide area data networks – evolution

• 1970s

• X.25 system – connection-oriented wide area data networks of the first generation. System was used for a decade.

• 1980s

• Frame relay system – mostly used for connections of LANs (even until today)

• 1990s

• Development of ATM (asynchronous transfer mode); the main aim was

transfer of speech, data, cable TV, telegraph using one type of data network.

ATM did not achieved awaited success but it is used for data transport of

Internet traffic

Local area networks

• Early 1970s

• Norman Abrahamson and colleagues from the university of Hawaii developed wireless (short range radio) ALOHANET. They were using computers from

neighboring islands to communicate with main computer in Honolulu

• 1976

• Using previous work from Abrahamson, Bob Metcalfe and David Boggs (Xerox PARC) developed the first LAN, called Ethernet, data rate of 2,94Mbps

• 1978

• Xerox Ethernet is standardized by DEC, Intel and Xerox (10 Mbps Ethernet)

Local area networks

• 1978 onwards

• Bob Metcalfe founded company 3Com which sold over 100 Millions of Ethernet adapters

• Development of Ethernet (100 Mbps and 1000 Mbps, switching, cabling)

• Token bus and token ring were added

• Middle of 1990s

• Standardization of Ethernet compatible wireless communication network WiFi

Standardization communities

Telecommunication

ITU International Telecommunication Union

International Standards

ISO International Standards Organization

IEEE Institute of Electrical and Electronics Engineering

Internet standards ISOC Internet Society

IAB Internet Architecture Board IRTF Internet Research Task Force IETF Internet Engineering Task Force

IEEE 802 Working-Groups

Overview and conclusion

• Definition of a network

• Scalability (hierarchical aggregation)

• Addressing, routing, forwarding

• Multiplexing

• Layering and protocols (separation of concerns)

• OSI model and Internet (TCP/IP) model

• Latency and bandwidth

• Standardization

Literature

[PetersonDavie2007] Larry L. Peterson and Bruce S. Davie, „Computer Networks: A Systems Approach“, Edition 4, 2007.

1.2 Requirements

1.3 Network Architecture

1.4.1 Application Programming Interface (Sockets) 1.5 Performance

4.1.1 What is an Internetwork?

4.1.3 Global Addresses

4.1.4 Datagramm Forwarding in IP 4.3.1 Subnetting

4.3.2 Classless Routing (CIDR)

[Tanenbaum2003] Andrew S. Tanenbaum, „Computer Networks“, Fourth Edition, 2003.

1.5 Example Networks