UNIT 4

No. Questions Marks

1What is Hidden Station and Exposed Station Problem? How does RTS and CTS solve these problems?    10.0

2 Describe in detail IEEE 802.11architecture with its frame format.    10.03 Briefly explain Bluetooth architecture. Give examples of some Bluetooth devices    10.04 Explain the MAC sub layer of IEEE 802.11 Wireless Ethernet standard    10.05 Explain Bluetooth layers in detail.    10.0

6Write a detailed note on Fast Ethernet. Provide the summary table along with figure and explanation of each type.    10.0

7What is ARP Packet? Provide the frame format of ARP Packet. How encapsulation of ARP Packet is done? Discuss four cases of using ARP.    10.0

8 How is a repeater different from an amplifier?    2.09 Why address mapping required?     2.010 Define Fast Ethernet.    2.011 Define Ethernet.    2.012 What is ARP?    2.013 What is unicast address?     2.014 Name any two unicast routing protocols    2.015 Write a note on address space    2.016 What is logical address?     2.017 What is static mapping?    2.018 Discuss about Internet Protocol.    6.019 What are the 4 different cases in which the services of ARP can be used?    6.020 Give an account on the classes of IP address with their range    6.021 Explain briefly IEEE 802.11.    6.022 Differentiate between IPv4 and IPv6. Also draw the datagram format of IPv4 and IPv6.    6.023 List the advantages of IPv6 over IPv4.    6.0

Q1. What is Hidden Station and Exposed Station Problem? How does RTS and CTS solve these problems?

Sol.

A significant difference between wired and wireless LANs is the fact that, in general a fully connected topology between the WLAN nodes cannot b assumed. This problem gives rise to ‘hidden’ and ‘exposed’ station problems.

A significant difference between wired and wireless LANs is the fact that, in general a fully connected topology between the WLAN nodes cannot be assumed. This problem gives rise to ‘hidden’ and ‘exposed’ station problems.Hidden Terminal: • As seen in the above problem, the transmission range of A reaches B but not C. Similarly, the range of C reaches B but not A. Also, the range of B reaches both A and C. • Now, the node A starts to send something to B and C doesn’t receive this transmission. • Now C also wants to send data to B and senses the carrier. As it senses it to be free, it also starts sending to B. • Hidden terminal problem occurs when two nodes that are outside each other’s range performs simultaneous transmission to a node that is within the range of each of them resulting in a collision. • That means the data from both parties A and C will be lost during the collision. • Hidden nodes mean increased probability of collision at receiver end. • One solution to avoid this is to have the channel sensing range much greater than the receiving range. Another solution is to use the Multiple Access with Collision Avoidance (MACA).

Exposed Terminal: • Consider the same above diagram. Here imagines a situation wherein the B node is currently sending some data to node A. • Now the other node C which is right now free want to send data to some node D (not in diag) which is outside the range of A and B. • Now before starting transmission it senses the carrier and realizes that the carrier is busy (due to interference of B’s signal). • Hence, the C node postpones the transmission to D until it detects the medium to be idle.• However, such a wait was un-necessary as A was outside the interference range of C. • Also, a collision at B will be a weak enough to be unable to penetrate into C

• Exposed terminal problem occurs when the node is within the range of a node that is transmitting and it cannot be transmitted to any node. • Exposed node means denied channel access unnecessarily which ultimately results in under-utilization of bandwidth resources. • It also results in wastage of time-resource.

RTS (Request to Send) & CTS (Clear to Send) was used to avoid frame collisions in the hidden node problem and was used by the 802.11 networking protocol as an optional mechanism. Modern RTS & CTS includes acknowledgments and does not solves exposed node problem which was solved earlier.

Hidden Station problem:

Exposed Station Problem:

The RTS frame contains five fields, which are: • Frame Control • Duration • RA (Receiver Address) • TA (Transmitter Address) • FCS

The CTS frame contains four fields, which are: • Frame Control • Duration

• RA (Receiver Address) • FCS

Q2. Describe in detail IEEE 802.11architecture with its frame format. Sol.

IEEE has defined the specifications for a wireless LAN, called IEEE 802.11, which covers the physical and data-link layers. It is sometimes called wireless Ethernet. In some countries, including the United States, the public uses the term Wi-Fi (short for wireless fidelity) as a synonym for wireless LAN.The IEEE protocol promotes fragmentation, which is the division of large frames into smaller ones. It is more efficient to resend a small frame than a large frame. The MAC Layer consists of nine fields as shown below.

1) Frame control (FC) - The FC field is 2 bytes long and defines the type of frame and some control information. Subfields in the FC Field –

a) D - This field defines the duration of the transmission that is used to set the value of NAV. In one control frame, it defines the ID of the frame.

b) Addresses - There are four address fields, each 6 bytes long. The meaning of each address field depends on the value of the To DS and From DS subfields and will be discussed later.

c) Sequence control - This field, often called the SC field, defines a 16-bit value. The first four bits define the fragment number; the last 12 bits define the sequence number, which is the same in all fragments.

d) Frame body - This field, which can be between 0 and 2312 bytes, contains information based on the type and the subtype defined in the FC field.

e) FCS - The FCS field is 4 bytes long and contains a CRC-32 error-detection sequence.

Q3. Briefly explain Bluetooth architecture. Give examples of some Bluetooth devices.

Sol.

Bluetooth Architecture: -

The Bluetooth architecture, showing all the major layers in the Bluetooth system. The layers below can be considered to be different hurdles in an obstacle course. This is because all the layers function one after the other. One layer comes into play only after the data has been through the previous layer.

• Radio: The Radio layer defines the requirements for a Bluetooth transceiver operating in the 2.4 GHz ISM band. • Baseband: The Baseband layer describes the specification of the Bluetooth Link Controller (LC), which carries out the baseband protocols and other low-level link routines. It specifies Piconet/Channel definition, “Low-level” packet definition, Channel sharing • LMP: The Link Manager Protocol (LMP) is used by the Link Managers (on either side) for link set-up and control. • HCI: The Host Controller Interface (HCI) provides a command interface to the Baseband Link Controller and Link Manager, and access to hardware status and control registers.

• L2CAP: Logical Link Control and Adaptation Protocol (L2CAP) supports higher level protocol multiplexing, packet segmentation and reassembly, and the conveying of quality of service information. • RFCOMM: The RFCOMM protocol provides emulation of serial ports over the L2CAP protocol. The protocol is based on the ETSI standard TS 07.10. • SDP: The Service Discovery Protocol (SDP) provides a means for applications to discover, which services are provided by or available through a Bluetooth device. It also allows applications to determine the characteristics of those available services.

Q4. Explain the MAC sub layer of IEEE 802.11 Wireless Ethernet standard

Sol.

The Media Access Control (MAC) data communication Networks protocol sub-layer, also known as the Medium Access Control, is a sub-layer of the data link layer specified in the seven-layer OSI model. The medium access layer was made necessary by systems that share a common communications medium. Typically, these are local area networks. The MAC layer is the "low" part of the second OSI layer, the layer of the "data link". In fact, the IEEE divided this layer into two layers "above" is the control layer the logical connection (Logical Link Control, LLC) and "down" the control layer the medium access (MAC).

The LLC layer is standardized by the IEEE as the 802.2 since the beginning 1980 Its purpose is to allow level 3 network protocols (for eg IP) to be based on a single layer (the LLC layer) regardless underlying protocol used, including WiFi, Ethernet or Token Ring, for example. All WiFi data packets so carry a pack LLC, which contains itself packets from the upper network layers. The header of a packet LLC indicates the type of layer 3 protocol in it: most of the time, it is IP protocol, but it could be another protocol, such as IPX (Internet Packet Exchange) for example. Thanks to the LLC layer, it is possible to have at the same time, on the same network, multiple Layer 3 protocols.

In LAN nodes uses the same communication channel for transmission. The MAC sub-layer has two primary responsibilities:

Data encapsulation, including frame assembly before transmission, and frame parsing/error detection during and after reception. Media access control, including initiation of frame transmission and recovery from transmission failure.

Q5. Explain Bluetooth layers in detail.

Sol.

1. L2CAP The Logical Link Control and Adaptation Protocol, or L2CAP (L2 here means LL), is roughly equivalent to the LLC sublayer in LANs. It is used for data exchange on an ACL link; SCO channels do not use L2CAP. Figure 15.20 shows the format of the data packet at this level.

The L2CAP has specific duties: multiplexing, segmentation and reassembly, quality of service (QoS), and group management.

a. Multiplexing The L2CAP can do multiplexing. At the sender site, it accepts data from one of the upper-layer protocols, frames them, and delivers them to the baseband layer. At the receiver site, it accepts a frame from the baseband layer, extracts the data, and delivers them to the appropriate protocol layer. It creates a kind of virtual channel that we willdiscuss in later chapters on higher-level protocols.

b. Segmentation and Reassembly The maximum size of the payload field in the baseband layer is 2774 bits, or 343 bytes. This includes 4 bytes to define the packet and packet length. Therefore, the size of the packet that can arrive from an upper layer can only be 339 bytes. However, application layers sometimes need to send a data packet that can be up to 65,535 bytes (an Internet packet, for example). The L2CAP divides these large packets into segments and adds extra information to define the location of the segments in the original packet. The L2CAP segments the packets at the source and reassembles them at the destination.

c. QoS Bluetooth allows the stations to define a quality-of-service level. We discuss quality of service in Chapter 30. For the moment, it is sufficient to know that if no quality-of-service level is defined, Bluetooth defaults to what is called best-effort service; it will do its best under the circumstances.

d. Group Management Another functionality of L2CAP is to allow devices to create a type of logical addressing between themselves. This is similar to multicasting. For example, two or three secondary devices can be part of a multicast group to receive data from theprimary.

2. Baseband Layer The baseband layer is roughly equivalent to the MAC sublayer in LANs. The access method is TDMA (discussed later). The primary and secondary stations communicate with each other using time slots. The length of a time slot is exactly the same as the dwell time. This means that during the time that one frequency is used, a primary sends a frame to a secondary, or a secondary sends a frame to the primary. Note that the communication is only between the primary and a secondary; secondaries can-not communicate directly with one another.

a. TDMA Bluetooth uses a form of TDMA that is called TDD-TDMA (time-division duplex TDMA). TDD-TDMA is a kind of half-duplex communication in which the sender and receiver send and receive data, but not at the same time (half-duplex); however, the communication for each direction uses different hops. This is similar to walkie-talkies using different carrier frequencies.

b. Links Two types of links can be created between a primary and a secondary: SCO links andACL links.

c. Frame Format

A frame in the baseband layer can be one of three types: one-slot, three-slot, or five-slot. A slot, as we said before, is 625 μs. However, in a one-slot frame exchange, 259 μs is needed for hopping and control mechanisms.

1. Radio Layer The radio layer is roughly equivalent to the physical layer of the Internet model. Bluetooth devices are low-power and have a range of 10 m.

a. Band Bluetooth uses a 2.4-GHz ISM band divided into 79 channels of 1 MHz each.

b. FHSS Bluetooth uses the frequency-hopping spread spectrum (FHSS) method in the physical layer to avoid interference from other devices or other networks. Bluetooth hops 1600 times per second, which means that each device changes its modulation frequency 1600 times per second. A device uses a frequency for only 625 μs (1/1600 s) before it hops to another frequency; the dwell time is 625 μs.

c. Modulation To transform bits to a signal, Bluetooth uses a sophisticated version of FSK, called GFSK (FSK with Gaussian bandwidth filtering; a discussion of this topic is beyond the scope of this book). GFSK has a carrier frequency. Bit 1 is represented by a frequency deviation above the carrier; bit 0 is represented by a frequency deviation below the carrier. The frequencies, in megahertz, are defined according to the following formula for each channel.

Q6. Write a detailed note on Fast Ethernet. Provide the summary table along with figure and explanation of each type.

Sol.

• Fast Ethernet was designed to compare with LAN protocols such as FDDI or a fibre channel.• IEEE channel fast Ethernet under the name 802.3a.• Fast Ethernet is backward-compatible with standard Ethernet, but it can transmit data 10 times faster at a rate of 100 Mbps.• The goals of fast Ethernet can be summarised as follows:

1. Upgrade the data rate to 100 mbps2. Make it compatible with standard Ethernet3. Keep the same 48- bit address4. Keep the same frame format5. Keep the same minimum and maximum frame lengths

PHYSICAL LAYER:

• The physical layer in fast Ethernet is more complicated than the one in standardEthernet.• Some features of this layer i.e. topology

Topology:

• Fast Ethernet is designed to two or more stations together.• If there are only two stations, they can be connected point-to-point.

• Three or more stations need to be connected in a star topology with a hub or a switch at the centre.

Fast Ethernet topology:

Implementation:• Fast Ethernet implementation at the physical layer can be categorized as either twowire or four wire• The two-wire implementation can be either category 5 UTP (100Base-TX) or fibreoptic cable (100Base – FX).• The four-wire implementation in designed only for category 3 UTP (100Base-T4).

Four Ethernet implementation:

Encoding:• Manchester encoding needs a 200 Mbaud bandwidth for a data rate of 100Mbps,which makes it unsuitable for a medium such as twisted-pair cable.• For this reason, the fast Ethernet designers sought some alternative encoding/decodingscheme. however, it was found that one scheme would not perform equally well for allthree implementations.• Therefore, three different encoding schemes were chosen.

Encoding for fast Ethernet implementations:

100Base-TX• It uses two-pairs of twisted pair cables (either category 5 UTP or STP).• For this implementation, the MLT-3 scheme was selected since it has good bandwidthperformance.100Base-FX• It uses two-pairs of twisted pair cables• Optical fibre can easily handle high bandwidth requirements by using simpleencoding schemes.100Base-T4• It was designed to use category 3 or higher UTP.• The implementation uses four pairs of UTP for transmitting 100 Mbps.• Encoding/decoding in 100Base—T4 is more complicated.

Summary Table:

Q7. What is ARP Packet? Provide the frame format of ARP Packet. How encapsulation of ARP Packet is done? Discuss four cases of using ARP.

Sol.

The Address Resolution Protocol uses a simple message format containing one address resolution request or response. The size of the ARP message depends on the upper layer and lower layer address sizes, which are given by the type of networking protocol (usually IPv4) in use and the type of hardware or virtual link layer that the upper layer protocol is running on. The message header specifies these types, as well as the size of addresses of each. The message header is completed with the operation code for request (1) and reply (2). The payload of the packet consists of four addresses, the hardware and protocol address of the sender and receiver hosts.

FORMAT OF ARP PACKET

The ARP Packet is Encapsulated within an Ethernet packet. In Ethernet, the Frame Type for ARP is x0806.

4 cases of using ARP

Q8. How is a repeater different from an amplifier?

Sol.

A repeater is an electronic device which regenerates or replicates the attenuated or weakened signal due to transmission over long distances. Signals also get distorted when it passes through areas with high level of electromagnetic interference. In this sense, the noise attached with signal doesn’t get amplified and the regenerated signal is passed further.

An amplifier is an electronic device that increases the power of a signal. It is the circuit which magnifies the weak signal to a signal with more power. An amplifier can be equipped with an electrical circuit or can be a separate device. It increases power, current or voltage of the carrier signal. Amplifier just increases the voltage, signal or power irrespective of attenuation in the signal. It is possible that after amplifying the signal, there is noise added in the signal by the amplifier.

Q9. Why address mapping required? 

Sol.

Address Mapping (also known as pin mapping or geocoding) is the process of assigning map coordinate locations to addresses in a database. A GIS is capable of doing this by comparing the elements of each address in the database (e.g., the number, prefix direction, street name, suffix direction, and type) to the attributes of each line segment in a street centreline layer to find a match.

The starting and ending address of the matched line segment is used to locate the precise coordinate position. The output of address mapping is a point layer attributed with all of the data from the input database.

Q10. Define Fast Ethernet.

Sol.

Fast Ethernet is a local area network (LAN) transmission standard that provides a data rate of 100 megabits per second The MAC sublayer was left unchanged, which meant the frame format and the maximum and minimum size could also remain unchanged. By increasing the transmission rate, features of the Standard Ethernet that depend on the transmission rate, access method, and implementation had to be reconsidered. The goals of Fast Ether- net can be summarized as follows:

a. Upgrade the data rate to 100 Mbps.b. Make it compatible with Standard Ethernet. c. Keep the same 48-bit address.d. Keep the same frame format.

Q11. Define Ethernet.

Sol.

Ethernet is an array of networking technologies and systems used in local area networks (LAN), where computers are connected within a primary physical space.

Systems using Ethernet communication divide data streams into packets, which are known as frames. Frames include source and destination address information, as well as mechanisms used to detect errors in transmitted data and retransmission requests.

Q12. What is ARP?

Sol.

Address Resolution Protocol (ARP) is a protocol for mapping an Internet Protocol address (IP address) to a physical machine address that is recognised in the local network. For example, in IPv4, an address is 32 bits long. In an Ethernet local area network, however, addresses for attached devices are 48 bits long. A table, usually called the ARP cache, is used to maintain a correlation between each MAC address and its corresponding IP address. ARP

provides the protocol rules for making this correlation and providing address conversion in both directions.

Q13. What is unicast address? 

Sol.A unicast address is an address that identifies a unique node on a network. Unicast addressingis available in IPv4 and IPv6 and typically refers to a single sender or a single receiver,although it can be used in both sending and receiving. A unicast address packet is transferredto a network node, which includes an interface address. The unicast address is then insertedinto the destination's packet header, which is sent to the network device destination. It is themost common form of IP addressing.

Q14. Name any two unicast routing protocols

Sol.

1. Routing Information Protocol (RIP) - Routing Information Protocol (RIP) is a distance vector protocol that uses hop count as its primary metric. RIP defines how routers should share information when moving traffic among an interconnected group of local area networks (LANs).2. Open Shortest Path First (OSPF) - Routers connect networks using the Internet Protocol (IP), and OSPF (Open Shortest Path First) is a router protocol used to find the best path for packets as they pass through a set of connected networks.

Q15. Write a note on address space

Sol.

•Address space is the amount of memory allocated for all possible addresses for a computational entity, such as a device, a file, a server, or a networked computer. • Address space may refer to a range of either physical or virtual addresses accessible to a processor or reserved for a process.• Address space may be differentiated as either flat, in which addresses are expressed as incrementally increasing integers starting at zero, or segmented, in which addresses are expressed as separate segments augmented by offsets (values added to produce secondary addresses).Q16. What is logical address?

Sol.

A logical address is the address at which an item (memory cell, storage element, network host) appears to reside from the perspective of an executing application program. A logical address may be different from the physical address due to the operation of an address translator or mapping function.

Q17. What is static mapping?

Sol.

Static NAT also called inbound mapping, is the process of mapping an unregistered IP address to a registered IP address on a one-to-one basis. The unregistered or mapped IP address is assigned with the same registered IP address each time the request comes through. This process is particularly useful for web servers or hosts that must have a consistent address that is accessible from the Internet.Simply, Static NAT enables a PC on a stub domain to maintain an assigned IP address when communicating with other devices outside its network or the Internet.

Q18. Discuss about Internet Protocol.

Sol.

• The IP (Internet Protocol) is a protocol that uses datagrams to communicate over a packet-switched network, such as the Internet. • The IP protocol operates at the network layer protocol of the OSI reference model and is a part of a suite of protocols known as TCP/IP. • The network layer at the source layer at a source is responsible for creating a packet from the data coming from another protocol (such as a transport layer protocol or a routing protocol).

TCP/IP: -

The original TCP/IP protocol suite was defined as having four layers: host-to-network, internet, transport, and application. There are 4 layers in the TCP/IP protocols

1. Physical and data link layers 2. Network layer 3. Transport layer 4. Application layer

• They did not clearly distinguish between service, interface and protocol. • Protocols in TCP/IP model are not hidden and tough to replace if technology changes • The protocols came first, and the model was really just a description of the existing protocols. • Designers have much experience with the subject and have clear idea of which functionality to put in which layer. • Network Layer supports only connectionless communication.

Q19. What are the 4 different cases in which the services of ARP can be used?

Sol.

Q20. Give an account on the classes of IP address with their range.

Sol.

Internet Protocol hierarchy contains several classes of IP Addresses to be used efficiently in various situations as per the requirement of hosts per network. Broadly, the IPv4 Addressing system is divided into five classes of IP Addresses. All the five classes are identified by the first octet of IP Address.

Class A Address

The first bit of the first octet is always set to 0 (zero). Thus the first octet ranges from 1 – 127, i.e.

Class A addresses only include IP starting from 1.x.x.x to 126.x.x.x only. The IP range 127.x.x.x is reserved for loopback IP addresses.

The default subnet mask for Class A IP address is 255.0.0.0 which implies that Class A addressing can have 126 networks (27-2) and 16777214 hosts (224-2).

Class B Address

An IP address which belongs to class B has the first two bits in the first octet set to 10, i.e.

Class B 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.

Class B has 16384 (214) Network addresses and 65534 (216-2) Host addresses.

Class C Address

The first octet of Class C IP address has its first 3 bits set to 110, that is:

Class C IP addresses range from 192.0.0.x to 223.255.255.x. The default subnet mask for Class C is 255.255.255.x.

Class C gives 2097152 (221) Network addresses and 254 (28-2) Host addresses.

Class D Address

Very first four bits of the first octet in Class D IP addresses are set to 1110, giving a range of:

Class D has IP address range from 224.0.0.0 to 239.255.255.255. Class D is reserved for Multicasting. In multicasting data is not destined for a particular host, that is why there is no need to extract host address from the IP address, and Class D does not have any subnet mask.

Class E Address

This IP Class is reserved for experimental purposes only for R&D or Study. IP addresses in this class ranges from 240.0.0.0 to 255.255.255.254. Like Class D, this class too is not equipped with any subnet mask.

Q21. Explain briefly IEEE 802.11.

Sol.

• IEEE 802.11:• IEEE 802.11 is a set of media access control (MAC) and physical layer (PHY)

specifications for implementing wireless local area network (WLAN) computer communication in the 900 MHz and 2.4, 3.6, 5, and 60 GHz frequency bands

• The IEEE developed an international standard for WLANs. The 802.11 standard focuses on the bottom two layers of the OSI model, the physical layer (PHY) and data link layer (DLL).

• The objective of the IEEE 802.11 standard was to define a medium access control (MAC) sublayer, MAC management protocols and services, and three PHYs for wireless connectivity of fixed, portable, and moving devices within a local area.

• The three physical layers are an IR baseband PHY, an FHSS radio in the 2.4 GHz band, and a DSSS radio in the 2.4 GHz.

• 802.11 Features:• CSMA/CA based MAC protocol• DCF (Distributed Coordination Function)• Support for both time-critical• PCF (Point Coordination Function) and non-critical traffic (DCF)• Support multiple priority levels• Spread spectrum technology (no licensing) power management allows a node to

doze offIEEE 802.11 Architecture:

• The architecture of the IEEE 802.11 WLAN is designed to support a network where most decision making is distributed to mobile stations. Two network architectures are defined in the IEEE 802.11 standard:

• Infrastructure network: An infrastructure network is the network architecture for providing communication between wireless clients and wired network resources. The transition of data from the wireless to wired medium occurs via an AP. An AP and its associated wireless clients define the coverage area. Together all the devices form a basic service set (see Figure 21.5).

• Point-to-point (ad hoc) network: An ad hoc network is the architecture that is used to support mutual communication between wireless clients. Typically, an ad hoc network is created

spontaneously and does not support access to wired networks. An ad hoc network does not require an AP.

Q22. Differentiate between IPv4 and IPv6. Also draw the datagram format of IPv4 and IPv6.

Sol.

a) An IPv6 address consists of 128 bits, while an IPv4 address consists of only 32.b) IPv6 has a lot of more usable addresses compared to IPv4.c) IPv6 makes the router’s task simpler compared to IPv4.d) IPv6 is better suited to mobile networks than IPv4.e) IPv6 addresses are represented in a hexadecimal, colon-separated notation, while IPv4

address use the dot-decimal notation.f) IPv6 allows for bigger payloads than what is allowed in IPv4.g) IPv6 is used by less than 1% of the networks, while IPv4 is still in use by the

remaining 99%.

Q23. List the advantages of IPv6 over IPv4.

Sol.

ADVANTAGES OF IPv6 OVER IPv4: -

• LARGER ADDRESS SPACE: An IPv6 address is 128 bits long. Compared with the 32-bit address of IPv4, this is a huge (2^96) increase in the address space. • BETTER HEADER FORMAT: IPv6 uses a new header format in which options are separated from the base header and inserted, when needed, between the base header and the upper-layer data. This simplifies and speeds up the routing process because most of the options do not need to be checked by routers. • NEW OPTIONS: IPv6 has new options to allow for additional functionalities. • ALLOWANCE FOR EXTENSION: IPv6 is designed to allow the extension of the protocol if required by new technologies or applications. • SUPPORT FOR RESOURCE ALLOCATION: In IPv6, the type of service field has been removed, but a mechanism (called flow label) has been added to enable the source to request special handling of the packet. This mechanism can be used to support traffic such as real time audio and video. • SUPPORT FOR MORE SECURITY: The encryption and authentication options in IPv6 provide confidentially and integrity of the packet.