Computer Networks - Network Layer

Network Layer

Unit II Topics

• Services

• Switching Techniques:

  • Datagram Approach

  • Virtual-Circuit Approach

• Internet Protocol: IPv4 & IPv6

• Classful Addressing

• Classless Addressing

• CIDR

• Subnetting, NAT, ICMP

• Routing Algorithms:

  • Distance-Vector (DV) Routing

  • Link State (LS) Routing

• Routing in Internet: RIP, OSPF, BGP

Functions of Network Layer

• The Network Layer is the third layer of the OSI model.

• It handles service requests from the transport layer and forwards them to the data link layer.

• Translates logical addresses into physical addresses.

• Determines the route from source to destination.

• Manages traffic problems like switching and routing.

• Controls data packet congestion.

• Moves packets from sending host to receiving host.

Network Layer Services

• Guaranteed delivery: Ensures packet arrival at the destination.

• Guaranteed delivery with bounded delay: Ensures packet delivery within a specified delay.

• In-Order packets: Ensures packets arrive in the order they were sent.

• Guaranteed max jitter: Maintains equal time between successive transmissions and receipts.

• Security services: Provides security using a session key between source and destination, including encryption and decryption for data integrity and authentication.

Network Layer Services (Detailed)

• Packetizing: Encapsulating data (payload) from upper layers into a network layer packet at the source and decapsulating at the destination.

• Routing: Moving packets to the router's output link after reaching the router's input link.

• Logical Addressing: Implementing logical addressing to distinguish between source and destination systems, adding a header with logical addresses.

• Internetworking: Providing logical connection between different types of networks.

• Fragmentation: Breaking packets into smaller data units for travel through different networks.

Forwarding & Routing

• A router forwards a packet by examining its header field.

• The header field value indexes into the forwarding table.

• The forwarding table's value indicates the router's outgoing interface link for packet forwarding.

Network Layer Design Issues

• Routing: Process of moving data from one device to another, finding the best possible route.

• Forwarding: Action applied by each router when a packet arrives at one of its interfaces.

• Routing involves packet forwarding from an entry interface to an exit interface.

Store and Forward Packet Switching

• The host sends the packet to the nearest router.

• The packet is stored until fully arrived and checksum is verified.

• It is then forwarded to the next router until destination is reached.

Services Provided to Transport Layer

• Services must not depend on router technology.

• The transport layer needs protection from the type, number, and topology of available routers.

• Network addresses for the transport layer should use a uniform numbering pattern at LAN and WAN connections.

Connection-Based Services

• Connectionless: Packets are routed and inserted into the subnet individually without added setup.

• Connection-Oriented: Subnet offers reliable service; all packets are transmitted over a single route.

Switching Techniques

• Circuit Switching

• Packet Switching

• Message Switching

Circuit Switching Techniques

• A dedicated path is established for data transmission between sender and receiver.

• A pre-specified route is needed, and no other data is permitted on it.

• Circuits can be permanent or temporary.

• Three phases:

  • Establish a circuit

  • Transfer the data

  • Disconnect the circuit

Advantages of Circuit Switching

• Once the path is set up, the only delay is in data transmission speed.

• No congestion or garbled message problems.

Disadvantages of Circuit Switching

• Long setup time is required.

• A request token must travel to the receiver and be acknowledged before transmission.

• The line may be held up for a long time.

Message Switching Techniques

• The whole message is treated as a data unit and is switched/transferred in its entirety.

• A switch first receives the whole message and buffers it until resources are available to transfer it to the next hop.

• No dedicated path is established between the sender and receiver.

• If the next hop does not have enough resources, the message is stored, and the switch waits.

• The destination address is appended to the message.

• It is dynamic routing as the message is routed through intermediate nodes based on available information.

• Each node stores the entire message and then forwards it (store and forward network).

Disadvantages of Message Switching

• Every switch in the transit path needs enough storage for the entire message.

• Message switching is very slow because of the store-and-forward technique and waits.

• Message switching was not a solution for streaming media and real-time applications.

Advantages of Message Switching

• Data channels are shared among communicating devices, improving bandwidth efficiency.

• Traffic congestion is reduced because messages are temporarily stored in nodes.

• Message priority can be used to manage the network.

• The size of the message can be varied, supporting data of unlimited size.

Packet Switching Techniques

• The message is divided into smaller pieces (packets) sent individually.

• Packets are given a unique number to identify their order at the receiving end.

• Every packet contains information in its headers, such as source address, destination address, and sequence number.

• Packets travel across the network, taking the shortest path possible.

• All packets are reassembled at the receiving end in the correct order.

• If any packet is missing or corrupted, a message is sent to resend it.

• An acknowledgment message is sent if the correct order of packets is reached.

• Datagram Packet switching

• Virtual Circuit Switching

Advantages of Packet Switching

• Delay in packet delivery is less since packets are sent as soon as available.

• Switching devices don’t require massive storage.

• Data delivery can continue even if some parts of the network face link failure; packets can be routed via other paths.

• It allows simultaneous usage of the same channel by multiple users.

• It ensures better bandwidth usage as packets from multiple sources can be transferred via the same link.

Disadvantages of Packet Switching

• They are unsuitable for applications that cannot afford delays in communication, like high-quality voice calls.

• Packet switching has high installation costs.

• They require complex protocols for delivery.

• Network problems may introduce errors, delays, or loss of packets, which may lead to loss of critical information if not properly handled.

Datagram Packet Switching

• Packets, known as datagrams, are considered independent entities.

• Datagram Packet Switching is also known as connectionless switching.

• Each packet contains information about the destination, which the switch uses to forward the packet.

• Packets are reassembled at the receiving end in the correct order.

• The path is not fixed.

• Intermediate nodes make routing decisions to forward packets.

Virtual Circuit - Packet Switching

• Virtual Circuit Switching is also known as connection-oriented switching.

• A preplanned route is established before messages are sent.

• Call request and call accept packets are used to establish the connection between sender and receiver.

• The path is fixed for the duration of a logical connection.

Comparison of Virtual-Circuit and Datagram

Network Layer: Logical Addressing

• 32 bits in size.

• Logical address is called an IP address.

• Global addresses.

• IPv4 (IP version 4) or IP address are 32 bits long.

• $2^{32}$ addresses can be generated.

• IPv6 (IP version 6) are futuristic/new generation addresses, 128 bits in size.

• $2^{128}$ addresses can be generated, accommodating future needs and addressing IP address depletion issues.

IPv4 Addresses

• Unique and Universal addresses.

• Each address defines one and only one connection to the internet.

• Two devices can never have the same IP address at the same time.

• An address can be assigned to a device for some time, then taken away and assigned to some other device.

IPv4 Addresses - Notations

• Address space is $2^{32}$

• Binary notation: 10000000 00001011 00000011 00011111

• Dotted decimal notation: 128.11.3.31

• Hexadecimal notation: 80 0B 03 1F

Binary and Dotted Decimal Notations

• Example:

  • 1st Octet: 00111111

  • 2nd Octet: 10101011

  • 3rd Octet: 11101010

  • 4th Octet: 10101011

• Dotted Decimal Notation: 63.171.234.171

IPv4: Header

• IPv4 (Internet Protocol Version 4) is the fourth version of the Internet Protocol (IP).

• IP is responsible for delivering data packets from the source host to the destination host.

• This delivery is solely based on the IP Addresses in the packet headers.

• IPv4 is the first major version of IP.

• IPv4 is a connectionless protocol for use on packet-switched networks.

IPv4 Header Fields

• Version: 4 bits, indicates the IP version used (value is 4).

• Type Of Service: 8 bits, used for Quality of Service (QoS).

• Total Length: 16 bits, total length of the datagram in bytes (Header length + Payload length).

  • Minimum total length: 20 bytes.

  • Maximum total length: 65535 bytes.

• Header Length: 4 bits, ranges from 20 to 60 bytes.

  • Minimum length: 5 rows * 4 bytes = 20 bytes.

  • Maximum length: 20 bytes + 40 bytes (Options field) = 60 bytes.

  • Scaling factor of 4 is used: Header length = Header length field value * 4 bytes.

• Identification: 16 bits, used for the identification of the fragments of an original IP datagram; each fragmented datagram is assigned the same identification number.

• DF Bit: Do Not Fragment bit (0 or 1). When set to 0, fragmentation is allowed; when set to 1, fragmentation is not allowed.

• MF Bit: More Fragments bit (0 or 1). When set to 0, the current datagram is the last fragment or the only fragment; when set to 1, more fragments are following.

• Fragment Offset: 13 bits. It indicates the position of a fragmented datagram. Value = Fragment Offset / 8.

• Time To Live: 8 bits. It indicates the maximum number of hops. Prevents datagrams from looping forever in a routing loop, decremented by 1 at each hop. If the value of $TTL$ becomes zero, the datagram is discarded.

• Protocol: 8 bits. It tells the network layer at the destination host to which protocol the IP datagram belongs to (ICMP is 1, IGMP is 2, TCP is 6, and UDP is 17).

• Header Checksum: 16 bits. It contains the checksum value of the entire header; used for error checking of the header. If mismatched, the datagram is discarded.

• Source IP Address: 32 bits. It contains the logical address of the sender.

• Destination IP Address: 32 bits. It contains the logical address of the receiver.

• Options: Variable size (0 to 40 bytes). Used for purposes such as record route, source routing, and padding.

IPv4 Addresses - Classful Addressing

• Address space is divided into classes A, B, C, D, and E.

• Becoming obsolete but important to understand classless addressing.

Classful Addressing

• Class A: 0-127.

• Class B: 128-191.

• Class C: 192-223.

• Class D: 224-239.

• Class E: 240-255.

Host ID and Net ID

• Class A: 1 byte for Net ID, 3 bytes for Host ID.

• Class B: 2 bytes for Net ID, 2 bytes for Host ID.

• Class C: 3 bytes for Net ID, 1 byte for Host ID.

• Class D: Multicast Address.

• Class E: Reserved.

Before 1993:

• 128 networks with 16 million hosts

• 16,384 networks with up to 65,536 hosts

• 2 million networks and 256 hosts

Special IP Addresses

• 0.0.0.0: This host.

• {Network}.0: A host on this Network.

• 255.255.255.255: Broadcast on the local network.

• {Network}.255: Broadcast on a distant network.

• 127.x.x.x: Loopback.

Special IP Addresses: Local Host

• 0.0.0.0/32

• Single address with all 32 bits as zero.

• The host is not connected to the TC/IP network.

• Sends a request packet to the DHCP server to get connected to the internet.

Special IP Addresses: Broadcast Address

• 225.225.225.225/32

• All 32 bits of IPv4 address are $'1'$.

• Last address in IPv4 address space (Limited Broadcast Address).

• If a host wants to send a message to every other host, it uses this address.

• Packets are broadcasted in local network only.

Special IP Addresses: Loopback Addresses

• 127.0.0.0/8

• $2^{32-8} = 2^{24} = 16777216$ addresses.

• Can only be the destination address of a packet.

• The packet never leaves the machine; it returns back to the source.

Special IP Addresses: Multicast Addresses

• 224.0.0.0/4

• $2^{32-4} = 2^{28} = 268435456$ Addresses used for multicast communication.

• Assigned to a group of hosts instead of a single host.

• The packet sent to the multicast address is delivered to all hosts in that group.

Special IP Addresses: Private Addresses

• Never used globally; not routed on the internet.

• Configured by the network administrator.

• Devices on the same network use private IP addresses to communicate with each other.

• They translate private IP addresses into public IP addresses using NAT when communicating outside their network.

Special Addresses Reserved in Each Block

• Network Address: The first address of the block, not allocated to any host.

• Direct Broadcast Address: The last address of the block, used as the destination address in IPv4 packets.

Internet Protocol: IPv4 - Address Space

• When calculating hosts' IP addresses, 2 IP addresses are decreased because they cannot be assigned to hosts.

• The first IP of a network is a network number.

• The last IP is reserved for Broadcast IP.

Classes and Blocks:

Internet Protocol: IPv4 - Classful Addressing

• Address space: 4,294,967,296 addresses

• Class A: 50%, Class B: 25%, Class C: 12.5%, Class D: 6.25%, Class E: 6.25%

• Class Prefixes:

  • Class A: 0, Prefix = 0 to 127

  • Class B: 10, Prefix = 128 to 191

  • Class C: 110, Prefix = 192 to 223

  • Class D: 1110, Multicast addresses, Prefix = 224 to 239

  • Class E: 1111, Reserved for future use, Prefix = 240 to 255

Class A Address

• The first bit of the first octet is always set to 0.

• The first octet ranges from 1–127 (excluding 127.x.x.x, which is reserved for loopback IP addresses).

• The default subnet mask is 255.0.0.0. 126 networks ($2^7$) and 16777214 hosts ($2^{24} - 2$).

  • Format: 0NNNNNNN.HHHHHHHH.HHHHHHHH.HHHHHHHH

Class B Address

• The first two bits in the first octet are set to 10.

• IP Addresses range from 128.0.x.x to 191.255.x.x.

• The default subnet mask for Class B is 255.255.x.x. 16384 Network addresses ($2^{14}$) and 65534 ($2^{16} - 2$) Host addresses.

• Format: 10NNNNNN.NNNNNNNN.HHHHHHHH.HHHHHHHH

Class C Address

• The first 3 bits of the first octet are set to 110.

• IP addresses range from 192.0.0.x to 223.255.255.x.

• The default subnet mask is 255.255.255.x. 2097152 Network addresses ($2^{21}$) and 254 Host addresses ($2^8 - 2$).

• Format: 110NNNNN.NNNNNNNN.NNNNNNNN.HHHHHHHH

Class D Address

• The first four bits of the first octet are set to 1110.

• IP addresses range from 224.0.0.0 to 239.255.255.255.

• Reserved for Multicasting; no need to extract host address, no subnet mask.

Class E Address

• Reserved for experimental purposes only.

• IP addresses range from 240.0.0.0 to 255.255.255.254.

• Like Class D, no subnet mask.

Sample Examples

• Question 1: Given the network address 17.0.0.0, find the class, the block, and the range of the addresses.

• Solution: Class A, block has a netid of 17, addresses range from 17.0.0.0 to 17.255.255.255.

• Question 2: Given the network address 132.21.0.0, find the class, the block, and the range of the addresses.

• Solution: Class B, block has a netid of 132.21, addresses range from 132.21.0.0 to 132.21.255.255.

• Question 3: Given the network address 220.34.76.0, find the class, the block, and the range of the addresses.

• Solution: Class C, block has a netid of 220.34.76, addresses range from 220.34.76.0 to 220.34.76.255.

Mask

• The mask helps to find the netid and hostid.

• In class A, the first 8 bits define the netid; the next 24 bits hostid, hence in this first 8 are 1s.

• /n (e.g., 8, 16, or 24) shows the mask for each class. This /n notation is called Classless Interdomain Routing (CIDR).

Default masks for classful addressing

Classless Addressing

• An improved IP Addressing system, more efficient allocation of IP Addresses.

• Replaces the older classful addressing system.

• Also known as Classless Inter Domain Routing (CIDR).

Why Classless?

• To overcome address depletion.

• Give access to the Internet to more organizations.

• No classes, but addresses are still granted in blocks.

Classless Addressing Details

• The entire address space is partitioned into blocks of varying lengths.

• An address's prefix designates the block (network); its suffix designates the node (device).

• Capable of having a block of $2^0, 2^1, 2^2, …, 2^{32}$ addresses, theoretically.

• A block of addresses must have a power of two addresses and one address block may be given to an organization.

Prefix Lengths

• Prefix lengths that vary from $0$ to $32$ are possible.

• The length of the prefix has an inverse relationship with network size: smaller networks have large prefixes; larger ones have small prefixes.

• Classless addressing is a specific instance of classful addressing. An address in class A can be considered as a classless address with a prefix length of 8. Class B addresses can be viewed as classless addresses with the prefix 16

Classless Address Notation

• The prefix length is to be provided because it is not a property of the address.

• A classless address does not automatically define the block or network to which it belongs.

• The address is inserted, followed by a slash, and the prefix length, $n$.

Calculations for a given address in a block

• the number of addresses in the block,

• the start address in the block,

• the last address.

Formulas

• The block has $N = 2^{32-n}$ addresses.

• The $n$ leftmost bits are kept, and the $(32 - n)$ rightmost bits are all set to zeroes to determine the first address.

• The $n$ leftmost bits are kept, while the $(32 - n)$ rightmost bits are all set to 1s to determine the last address.

Example

• Problem: For the given classless address 167.199.170.82/27 find the number of addresses in the block, the start address in the block, and the last address.

• Solution: There are $2^{32-n} = 2^5 = 32$ addresses in all.

• The first $27$ bits are kept while the remaining bits are converted to $0s$ to determine the first address.

• Keeping the first $27$ bits and turning the remaining bits to $1s$ will allow you to determine the last address.

Network Address and Mask

• Network address:

  • Identifies a network on the internet.

  • Using this, we can find range of addresses in the network and total possible number of hosts in the network.

• Mask:

  • It is a 32-bit binary number that gives the network address in the address block when bitwise AND operation is applied on the mask and any IP address of the block.

• In IPv4 addressing, a block of addresses can be defined as x.y.z.t / n

  • Ex. 192.168.1.1/28

  • Where x.y.z.t defines one of the addresses and /n defines the mask.

Subnetting

• Dividing a large block of addresses into several contiguous sub-blocks and assigning these sub-blocks to different smaller networks.

• A practice that is widely used when classless addressing is done.

Default Mask vs Subnet Mask

• Example: 192 = 11000000

• 72 = 01001000

• Bitwise AND = 01000000 = 64

Subnetting Details

• To reduce the wastage of IP addresses in a block, we use subnetting.

• Some of the Host id bits are used as net id bits of a classful IP address.

• We give the IP address and define the number of bits for the mask along with it (usually followed by a / symbol), like, 192.168.1.1/28.

  • Subnet mask is found by putting the given number of bits out of 32 as 1.

• Ex: In the given address, we need to put 28 out of 32 bits as 1 and the rest as 0, and so, the subnet mask would be 255.255.255.240.

Subnetting Calculations

• Number of subnets : Given bits for mask – No. of bits in default mask

• Subnet address : AND result of subnet mask and the given IP address

• Broadcast address : By putting the host bits as 1 and retaining the network bits as in the IP address

• Number of hosts per subnet : $2^{(32 - Given bits for mask)} - 2$

• First Host ID : Subnet address + 1 (adding one to the binary representation of the subnet address)

• Last Host ID : Subnet address + Number of Hosts

Another Subnetting Method

• The first address in the block can be found out by setting the rightmost 32-$n$ bits to $0s$

• The last address in the block can be found out by setting the rightmost 32-$n$ bits to $1s$

• Ex: A block of addresses is granted to a small organization. One of the address is 205.16.37.39/28. What is first and last address?

• Solution: Binary representation of the given address: 11001101 00010000 00100101 00100111

• Binary representation of mask: 11111111 11111111 11111111 11110000

• First address : 11001101 00010000 00100101 00100000 = 205.16.37.32

• Last address : 11001101 00010000 00100101 00101111 = 205.16.37.47

• No. of addresses = $2^{32 -28} = 2^4 = 16$

Subnetting - Alternate Method

• Ex: A block of addresses is granted to a small organization. One of the addresses is 205.16.37.39/28. What is first and last address?

• Binary representation of given address: 11001101 00010000 00100101 00100111

• Binary representation of mask: 11111111 11111111 11111111 11110000

• First address : Bitwise AND operation of given address and mask = 205.16.37.32

• Last address : Take compliment of the mask and do bitwise OR operation.

• Given address: 11001101 00010000 00100101 00100111

• Compliment of the mask: 000000000 00000000 00000000 00001111 = 205.16.37.47

• No. of addresses : Compliment of the mask + $1 = 000000000 00000000 00000000 00001111 + 1 = 16$

Subnetting Example (Problem)

• An Organization is given a block 17.12.14.0/26, which contains 64 addresses. The organization has 3 offices and needs to divide the addresses into 3 subblocks of $32, 16$ and $16$.

• How will the subnetting be performed?

Subnetting Solution

• Mask for Subnet 1: $2^{(32-n1)} = 32 <br>ewline n1 = 27$

• Mask for Subnet 2: $2^{(32-n2)} = 16 <br>ewline n2 = 28$

• Mask for Subnet 3: $2^{(32-n3)} = 16 <br>ewline n3 = 28$

• We must have masks 27, 28 and 28 with the organization mask being 26.

First and Last Addresses

Subnet 1

• Block 1 = 32 addresses.

• The mask for Subnet 1 is $n1 = 27$

• Mask is /27 = 11111111 11111111 11111111 11100000

• First Address: 00010001 00001100 00001110 00000000 = 17.12.14.0

• Last Address: 00010001 00001100 00001110 00011111 = 17.12.14.31

Subnet 2

• Block 2 = $16$ addresses.

• Given Addresses: 17.12.14.0 = 00010001 00001100 00001110 0