BLOCK 1 UNIT 1
INTRODUCTION TO COMPUTER NETWORKS
Q1. What is computer network?
Ans: A computer network consists of two or more autonomous computers that are linked or connected together in order to
(a) Share resources (files, printers, modems, fax machines)
(b) Share application software like MS Office
(c) Allow electronic communications
(d) Increase productivity and makes it easier to share data among users.
Q2. How do you classify computer network?
Ans: Computer networks are generally classified according to their structure and the area they are localized in as follows:
(i) Local Area Network (LAN)
The network that spans a relatively small area that is in the single building or campus in known as Local Area Network.
(ii) Metropolitan Area Network (MAN)
The type of computer network that is designed for a city or town is known as Metropolitan Area Network.
(iii) Wide Area Network (WAN)
A network that covers a large geographical area and covers different cities, states and sometimes even countries, is known as Wide Area Network.
Q3. What are the characteristics of network?
Ans: The characteristics of network are explained below:
(a) Topology
Topology is the graphical arrangement of computer systems in a network. Common topologies include a bus, star, ring and mesh.
(b) Protocol
The protocol defines a common set of rules which are used by computers on the network that communicate between hardware and software entities. One of the most popular protocols for LANs is the Ethernet. Another popular LAN protocol for PCs is the token ring network.
(c) Architecture
Networks can be broadly classified as using either peer to peer or client or server architecture.
Q4. What are the basic goals and motivations of computer network?
Ans: Some of the basic goals that a computer network should satisfy are:
1) Cost reduction by sharing hardware and software resources.
2) Provide high reliability by having multiple sources of supply.
3) Provide an efficient means of transport for large volumes of data among various locations.
4) Provide inter process communication among users and processors.
5) Reduction in delay driving data transport.
6) Increase productivity by making it easier to share data amongst users.
7) Repairs, upgrades, expansions, and changes to the network should be performed with minimal impact on the majority of network users.
8) Standards and protocols should be supported to allow many types of equipment from different vendors to share the network.
9) Provide centralized or distributed management and allocation of network resources like host processors, transmission facilities, etc.
Q5. Explain the classification of networks depending on the transmission technology.
Ans: Depending on the transmission technology, networks are classified as follows:
(a) Broadcast Networks
Broadcast networks have a single communication channel that is shared by all the machines on the network. In this type of network, short messages sent by any machine are received by all the machines on the network. The packet contains an address field, which specifies for whom the packet is intended. All the machines, upon receiving a packet check for the address field, if the packet intended for itself, it processes it and if not the packet is just ignored.
(b) Point to Point or Switched Networks
Point to point or switched, networks are those in which there are many connections between individual pairs of machines. In these networks, when a packet travels from source to destination it may have to first visit one or more intermediate machines. Routing algorithms play an important role in point to point or switched networks because often multiple routes of different lengths are available.
Q6. What are the typical characteristics of LAN?
Ans: The typical characteristics of LAN are:
(1) Confined to small area, i.e. it connects several devices over a distance of 5 to 10km.
(2) High speed
(3) Most inexpensive equipment.
(4) Low error rates.
(5) Data and hardware sharing between users owned by the user.
(6) Operates at speeds ranging from 10Mbps to 100Mbps. Now-a-days 1000Mbps are available.
Q7. Explain the different types of switched network.
Ans: The different types of switched network are explained below:
(i) Circuit Switched Networks
Circuit switched networks use a networking technology that provides a temporary, but dedicated connection between two stations no matter how many switching devices are used in the data transfer route. Circuit switching was originally developed for the analog based telephone system in order to guarantee steady and consistent service for two people engaged in a phone conversation. Analog circuit switching has given way to digital circuit switching and the digital counterpart still maintains the connection until broken. This means bandwidth is continuously reserved and silence is transmitted just the same as digital audio in voice conversation.
(ii) Packet Switched Networks
Packet switched networks use a networking technology that breaks up a message into smaller packets for transmission and switches them to their required destination. Unlike circuit switching, which requires a constant point to point circuit to be established each packet in a packet – switched network contains a destination address. Thus all packets in a single message do not have to travel the same path.
Q8. What is bus topology? Give its advantages and disadvantages.
Ans: In Bus topology, all devices are connected to a central cable, called the bus or backbone. The bus topology connects workstations using a single cable. Each workstation is connected to the next workstation in a point to point fashion. All workstation connect to the same cable.
Advantages of Bus Topology
a) Installation is easy and cheap when compared to other topologies.
b) Connections are simple and this topology is easy to use.
c) Less cabling is required.
Disadvantages of Bus Topology
(a) Used only in comparatively small networks.
(b) As all computers share the same bus, the performance of the network deteriorates when we increase the number of computers beyond a certain limit.
(c) Faults identification is difficult.
(d) A single fault in the cable stops all transmission.
Q9. What is star topology? Give its advantages and disadvantages.
Ans: Star topology uses a central hub through which, all components are connected. In a star topology, the central hub is the host computer, and at the end of each connection is a terminal. Nodes communicate across the network by passing data through the hub. A star network uses a significant amount of cable as each terminal is wired back to the central hub even if two terminals are side by side but several hundred meters away from the host. The central hub makes all routing decisions and all other workstations can be simple.
Advantages of Star Topology
1) Installation and configuration of network is easy.
2) Less expensive when compared to mesh topology.
3) Faults in the network can be easily traced.
4) Expansion and modification of star network is easy.
5) Single computer failure does not affect the network.
6) Supports multiple cable types like shielded twisted pair cable, unshielded twisted pair cable, ordinary telephone cable etc.
Disadvantages of Star Topology
(i) Failure in the central hub brings the entire network to a halt.
(ii) More cabling is required in comparison to tree or bus topology because each node is connected to the central hub.
Q10. What is ring topology? Give its advantages and disadvantages.
Ans: In ring topology, all devices are connected to one another in the shape of a closed loop, so that each device is connected directly to two other devices, one on either side of it, i.e. the ring topology connects workstations in a closed loop. Each terminal is connected to two other terminals with the last terminal being connected to the first. Data is transmitted around the ring in one direction only; each station passing on the data to the next station till it reaches its destination.
Advantages of Ring Topology
(i) Easy to install and modify the network.
(ii) Fault isolation is simplified.
(iii) Unlike bus topology, there is no signal loss in Ring topology because the tokens are data packets that are re-generated at each node.
Disadvantages of Ring Topology
(1) Adding or removing computers disrupts the entire network.
(2) A break in the ring can stop the transmission in the entire network.
(3) Finding fault is difficult.
(4) Expensive when compared to other topologies.
Q11. What is tree topology? Give its advantages and disadvantages.
Ans: Tree topology is a hybrid topology, it is similar to the star topology but the nodes are connected to the secondary hub, which in turn is connected to the central hub. In this topology group of star configured networks are connected to a linear bus backbone.
Advantages of Tree Topology
(i) Installation and configuration of network is easy.
(ii) Less expensive when compared to mesh topology.
(iii) Faults in the network can be detected traced.
(iv) The addition of the secondary hub allows more devices to be attached to the central hub.
(v) Supports multiple cable types like shielded twisted pair table, unshielded twisted pair cable, ordinary telephone cable, etc.
Disadvantages of Tree Topology
(1) Failure in the central hub brings the entire network to a halt.
(2) More cabling is required when compared to bus topology because each node is connected to the central hub.
Q12. What is mesh topology? Give its advantages and disadvantages.
Ans: In mesh topology, devices are connected with many redundant interconnections between network nodes. In a well connected topology, every node has a connection to every other node in the network. The cable requirements are high, but there are redundant built in. Failure in one of the computers does not cause the network to break down, as they have alternative paths to other computers.
Advantages of Mesh Topology
(i) Use of dedicated links eliminates traffic problems.
(ii) Failure in one of the computers does not affect the entire network.
(iii) Point to point links makes fault isolation easy.
(iv) It is robust.
(v) Privacy between computers is maintained as messages travel along dedicated path.
Disadvantages of Mesh Topology
(i) The amount of cabling required is high.
(ii) A large number of I/O ports are required.
Q13. What is cellular topology? Give its advantages and disadvantages.
Ans: Cellular topology, divides the area being serviced into cells. In wireless media each point transmits in a certain geographical area called a cell, each cell represents a portion of the total network area. Devices that are present within the cell, communicate through a central hub. Hubs in different cells are interconnected and hubs are responsible for routing data across the network.
Advantages of Cellular Topology
(a) If the hubs maintain a point to point link with devices, trouble shooting is easy.
(b) Hub to hub fault tracking is more complicated, but allows simple fault isolation.
Disadvantages of Cellular Topology
(i) When a hub fails, all devices serviced by the hub lose service (are affected).
Q14. Explain the applications of network.
Ans: The applications of network are explained below:
(1) Share Resources and Information
We can share expensive resources such as laser printers, DVD-ROM drives, fax machines, etc. using computer network. We can also share information and many persons can work together on projects and tasks that require co-ordination and communication, even though these users may not be physically close.
(2) Access to Remote Information
Access to remote information involves interaction between a person and a remote database. Financial institutions allow access to their information so that people can pay their bills, handle their investments and manage their bank accounts electronically. Online shopping also allows people access to product information before purchasing the product.
(3) Person to Person Communication
Person to person communication is mostly carried out using email. Email is a simple, yet potent facility. Email is more valuable and useful than the telephone because by using email we can convey information that is difficult or impossible to read over the telephone, like reports, tables, charts, images, etc. Using a computer network, it is also possible to organize virtual meeting among people who are far away from each other and this is called video conferencing.
(4) Interactive Entertainment
Computer networks such as internet offer a wide variety of entertainment; there are many companies online, which offer video on demand. A large variety of multi person real time simulation games, like hide and seek in a virtual dungeon and flight simulators with the players in one team trying to shoot down the players in the opposite team and many such games are available online.
Q15. What is OSI reference model?
Ans: The Open System Interconnection (OSI) model is a set of protocols that attempt to define and standardize the data communication process; we can say that it is a concept that describes how data communications should take place. It was set up by the International Standards Organization (ISO) in 1984. The OSI reference model describes how information from a software application in one computer moves through a network medium to a software application in another computer. The OSI reference model is a conceptual model composed of seven layers. These layers are fitted the protocol standards developed by the ISO and other standards bodies.
Q16. Explain the different layers of OSI reference model.
Ans: There are seven layers in Open System Interconnection (OSI) reference model which are explained below:
(1) Application Layer (Layer 7)
The application layer is the top layer and it defines the language and syntax that program use to communicate with other programs. The application layer represents the purpose of communicating in the first place. For example, a program in a client workstation uses commands to request data from a program in the server. Common functions at this layer are opening, closing, reading and writing files, transferring files and email messages, executing remote jobs and obtaining directory information about network resources, etc.
(2) Presentation Layer (Layer 6)
The presentation layer performs code conversion and data reformatting (syntax translation). It is the translator of the network; it makes sure the data is in the correct form for the receiving application. When data are transmitted between different types of computer systems, the presentation layer negotiates and manages the way data are represented and encoded. This layer is also used for encryption and decryption. It also provides security features through encryption and decryption.
(3) Session Layer (Layer 5)
The session layer decides when to turn communication on and off between two computers. It provides the mechanism that controls the data exchange process and coordinates the interaction (communication) between them in an orderly manner. It sets up and clears communication channels between two communicating components. It determines one way or two way communications and manages the dialogue between both parties; for example, making sure that the previous request has been fulfilled before the next one is sent. It also marks significant parts of the transmitted data with checkpoints to allow for fast recovery in the event of a connection failure.
(4) Transport Layer (Layer 4)
The transport layer is responsible for overall end to end validity and integrity of the transmission i.e. it ensures that data is successfully sent and received between two computers. The lower data link layer is only responsible for delivering packets from one node to another. Thus if a packet gets lost in a router somewhere in the enterprise internet, the transport layer will detect that. This layer acts as an interface between the bottom and top three layers.
(5) Network Layer (Layer 3)
This network layer establishes the route between the sending and receiving stations. The unit of data at the network layer is called a packet. It provides network routing and flow and congestion functions across computer network interface. The network layer establishes, maintains, and terminates logical and physical connections. The network layer is responsible for translating logical address, or names into physical address. The main device found at the network layer is a router.
(6) Data Link Layer (Layer 2)
The data link layer groups the bits that we see on the physical layer into frames. It is primarily responsible for error free delivery of data on a hop. The data link layer is split into two sub-layers i.e. the logical link control and media access control. The data link layer handles the physical transfer, framing, flow control and error control functions over a single transmission link. The main network device found at the data link layer is a bridge.
(7) Physical Layer (Layer 1)
The data units on this layer are called bits. This layer defines the mechanical and electrical definition of the network medium or cable and network hardware. This includes how data is impressed onto the cable and retrieved from it. The physical layer is responsible for passing bits onto and receiving them from the connecting medium. This layer gives the data link layer its ability to transport a stream of serial data bits between two communicating systems; it conveys the bits that move along the cable. The main network device found in physical layer is a repeater.
Q17. What is TCP/IP?
Ans: TCP/IP stands for Transmission Control Protocol / Internet Protocol. It is a protocol suite used by most communications software. TCP/IP is a robust and proven technology that was first tested in the early 1980s on ARPA Net, the US military’s Advanced Research Projects Agency Network and the world’s first packet switched network; TCP/IP was designed as an open protocol that would enable all types of computers to transmit data to each other via a common communication language.
Q18. Explain the different layers of TCP/IP reference model.
Ans: The different layers of TCP/IP reference model are explained below:
a) Application Layer (Layer 4)
The top layer of the protocol stack is the application layer. It refers to the programs that initiate communication in the first place. TCP/IP includes several application layer protocols for mail, file transfer, remote access, authentication and name resolution. These protocols are embodied in programs that operate at the top layer just as any custom made or packaged client or server application would. Examples include HTTP, FTP, SMTP, etc.
b) Transport Layer (Layer 3)
The Transport Layer also known as the Host to Host Transport Layer is responsible for providing the application layer with session and datagram communication services. The TCP/IP does not contain presentation and session layers, the services are performed if required, but they are not part of the formal TCP/IP stack. For example, Layer 6 or Presentation is where data conversion, floating point to binary, etc. and encryption or decryption is performed. Examples include TCP, UDP, etc.
c) Internet Layer (Layer 2)
The internet layer handles the transfer of information across multiple networks through the use of gateways and routers. The internet layer corresponds to the part of the OSI network layer that is concerned with the transfer of packets between machines that are connected to different networks. It deals with the routing of packets across these networks as well as with the control of congestion. A key aspect of the internet layer is the definition of globally unique addresses for machines that are attached to the internet. Examples include IP, ARP, ICMP, IGMP, etc.
d) Link or Physical Layer (Layer 1)
The link / physical layer also called the network access layer is responsible for placing TCP/IP packets on the network medium and receiving TCP/IP packets of the network medium. TCP/IP was designed to be independent of the network access method, frame format, and medium. In this way, TCP/IP can be used to connect differing network types. This includes LAN technologies such as Ethernet or Token Ring and WAN technologies such as X.25 or Frame Relay. Independence from any specific network technology gives TCP/IP the ability to be adapted to new technologies such as Asynchronous Transfer Mode.
Q19. Explain the difference between OSI and TCP/IP Model.
Ans: The difference between OSI and TCP/IP model are given below:
1) OSI contains 7 layers while TCP / IP contain 4 layers.
2) OSI uses strict layering resulting in vertical layers but TCP/IP uses loose layering resulting in horizontal layers.
3) OSI supports both connectionless and connection oriented communication in the network layer, but only connection oriented communication in transport layer. TCP supports only connectionless communication in the network layer, but both connectionless and connection oriented communication in Transport Layer.
4) OSI distinguishes between service, interface and protocol. But TCP/IP does not clearly distinguish between service, interface and protocol.
5) Protocols are better hidden and can be replaced relatively easily as technology changes in OSI model. But in TCP/IP model, protocols are not hidden and thus cannot be replaced easily. Replacing IP by a substantially different protocol would be virtually impossible.
6) OSI reference model was devised before the corresponding protocols were designed but the protocols came first and the model was a description of the existing protocols in TCP/IP.
Q20. What are the drawbacks of OSI reference model?
Ans: The drawbacks of OSI reference model are given below:
1) All layers are not roughly of equal size and complexity. In practice, the session layer and presentation layer are absent from many existing architectures.
2) Some functions like address, flow control, retransmission are duplicated at each layer, resulting in deteriorated performance.
3) The initial specification of the OSI model ignored the connectionless model, thus leaving much of the LANs behind.
Q21. What are the drawbacks of TCP/IP model?
Ans: The drawbacks of TCP/IP model are given below:
1) TCP/IP model does not clearly distinguish between the concepts of service, interface and protocol.
2) TCP/IP model is not a general model and therefore it cannot be used to describe any protocol other than TCP/IP.
3) TCP/IP model does not distinguish or even mention the physical or the data link layer. A proper model should include both these layers as separate.
Q22. Explain Client / Server Architecture.
Ans: Client Server architecture is one in which the client personal computer or workstation is the requesting machine and the server is the supplying machine, both of which are connected via LAN or WAN. Since the early 1990s client server has been the buzzword for building application on LANs in contrast to centralized minis and mainframes with dedicated terminals. A client server network is called centralized or server based network.
Q23. Explain the different types of client or server architecture.
Ans: The different types of client / server architecture are explained below:
a) Non – Client / Server
In non client / server architecture, the server is nothing more than a remote disk drive. The user machine does all the processing. If many users routinely perform lengthy searches, this can bog down the network, because each client has to pass the entire database over the net. At 1000 bytes per record, a 10000 record database requires 10MB of data be transmitted.
b) Two Tier Client / Server
Two tier client / server is really the foundation of client / server. The database processing is done in the server. An SQL request is generated in the client and transmitted to the server. The DBMS searches locally and returns only matching records. If 50 records met the criteria, only 50K would be transmitted. This reduces traffic in the LAN.
c) Three Tier Client / Server
Many applications lend themselves to centralized processing. If they contain propriety algorithms, security is improved, upgrading is also simpler. Sometimes programs are just too demanding to be place into every client PC. In three tier client or server, application processing is performed in one or more servers.
Q24. What is peer to peer network?
Ans: A type of network in which ends workstation has equal capabilities and responsibilities is called peer to peer network. Here each workstation acts as both a client and a server. There is no central repository for information and there is no central server to maintain. Data and resources are distributed throughout the network, and each user is responsible for sharing data and resources connected to their system. This differs from client or server architecture in which some computers are dedicated to serving the others. Peer to peer networks are generally simpler and less expensive but they usually do not offer the same performance under heavy loads. A peer to peer network is also known as a distributed network.
Q25. What is Novell Network?
Ans: Novell NetWare is the most popular network system in the PC world. Novell NetWare contains the protocols that are necessary to allow communication between different types of PCs and devices. The most recent version NetWare 4.X can probably run on almost any type of machine.
Q26. What is ARPANET?
Ans: ARPRANET stands for Advanced Research Projects Agency (ARPA) Network. The network was developed in 1969 by ARPA and funded by the Department of Defense (DoD). In the mid 1960s at the height of the cold war, the DoD wanted a command and control network, which could survive the nuclear war. The Traditional circuit switched telephone networks were considered too vulnerable, since the loss of one line would certainly terminates all conversation using them and might even partition the network.
Q27. What is ATM Network?
Ans: Asynchronous Transfer Mode (ATM) is a network technology adopted by the telecommunication sector. It is a high performance, cell-oriented switching and multiplexing technology that utilizes fixed length packets to carry different types of traffic. The data transfer takes place in the form of cells or packets of a fixed size (53 bytes). The cell used with ATM is relatively small compared to units used with older technologies. The small constant cell size allows ATM equipment to transmit video, audio and computer data over the same network, and assures that no single type of data hogs the line.
The services provided by ATM Network are given below:
1) Constant Bit Rate (CBR) guarantees bandwidth for real-time voice and video.
2) Real Time variable Bit Rate (rt-VRB) supports interactive multimedia that requires minimal delays.
3) Non – Real time variable bit rate (nrt-VBR) is used for bursty transaction traffic.
4) Available Bit Rate (ABR) adjusts bandwidth according to congestion levels for LAN traffic.
5) Unspecified Bit Rate (UBR) provides the best effort for non-critical data such as file transfers.
Q28. What are the advantages and disadvantages of ATM?
Ans: The advantages and disadvantages of ATM are given below:
Advantages of ATM
1) Flexible bandwidth allocation.
2) Simple routing due to connection oriented technology.
3) High bandwidth utilization due to statistical multiplexing.
4) Potential quality of service guarantees.
Disadvantages of ATM
i) Overhead of cell header (5 bytes per cell)
ii) Complex mechanisms for achieving quality of service.
iii) Congestion may cause cell losses.
iv) It is costly compared to IP.
Q29. Give the typical characteristics of MAN.
Ans: The typical characteristics of a MAN are:
1) Confined to a larger area than a LAN and can range from 10km to a few 100km in length.
2) Slower than a LAN but faster than a WAN.
3) Operates at a speed of 1.5 to 150 Mbps.
4) Expensive equipment.
5) Moderate error rates.
Q30. Give the typical characteristics of WAN.
Ans: The typical characteristics of a WAN are:
i) A WAN can range from 100km to 1000km and the speed between cities can vary from 1.5Mbps to 2.4Gbps.
ii) WAN supports large number of computers and multiple host machines.
iii) Various segments of network are interconnected using sophisticated support devices like routers and gateways.
iv) Usually the speed is much slower than LAN speed.
v) Highest possible error rate compared to LAN and MAN.
Q31. Explain the advantage of Networks.
Ans: The advantages of networks are explained below:
a) Resource Sharing
The main aim of a computer network is to make all programs, equipment and data available to anyone on the network without regard to the physical location of the resource and the user. Users need to share resources other than files as well. A common example is being printers.
b) High Reliability
Computer networks provide high reliability by having alternative sources of supply. For example, all files could be replicated on two or three machines, so if one of them is unavailable due to hardware failure, the other copies could be used.
c) Saving Money
Small computers have a much better price and performance ratio than larger ones. Mainframes are roughly a factor of ten faster than personal computers but they cost much more. This imbalance has caused many systems designers to build systems consisting of personal computers; one per user, with data kept on one or more share files server machines.
d) Scalability
The ability to increase the system performance gradually as the workload grows just by adding more processors. With centralized mainframes, when a system is full, it must be replaced by a larger one, usually at great expense and even greater disruption to the users.
e) Communication Medium
A computer network can provide a powerful communication medium among widely separated users. Using a computer network, it is easy for two or more people who are working on the same project and who live far apart to write a report together.
f) Increased Productivity
Networks increase productivity as several people can enter data at the same time, but they can also evaluate and process the shared data. So one person can handle accounts receivable, and someone else processes the profit and loss statements.
Q32. Explain the difference between Client / Server and Peer to Peer.
Ans: The difference between Client / Server and Peer to Peer architecture are:
Client / Server
i) Also known as centralized or server based network.
ii) Some computers in the network are dedicated to a particular task and are called servers.
iii) Only some computers are dedicated to serve others.
iv) These networks are more expensive when compare to peer to peer network.
Peer to Peer
i) Also know as distributed network.
ii) Each workstation has equal rights and responsibilities i.e. each work station acts as both a client and a server.
iii) Data and resources are distributed throughout the network and each user is responsible for sharing data and resources connected to their system.
iv) These networks are generally simpler and less expensive.
Q33. Differentiate between TCP and UDP.
Ans: Difference between TCP and UDP are:
Transmission Control Protocol (TCP)
i) Transmission control protocol is a highly reliable, connection oriented, end to end transport layer protocol.
ii) TCP provides message fragmentation and reassembly and can accept message of any length from upper layer protocols.
iii) TCP can maintain multiple conversations with upper layer protocols and can improve use of network bandwidth by combining multiple messages into the same segment.
iv) User Datagram protocol is a unreliable connectionless transport host to host layer protocol.
v) UDP does not provide message acknowledgements, it simply transports datagrams.
vi) UDP is used when the amount of data to be transferred is small such as the data that would fit into a single packet, when the overhead of establishing a TCP connection is not desired, or when the applications or upper layer protocols provide reliable delivery.
vii) Like TCP, UDP utilizes port addresses to deliver datagrams.
BLOCK 1 UNIT 2
DATA TRANSMISSION
Q1. Define Data Communication.
Ans: The transfer of data from one machine to another machine such that the sender and the receiver both interpret the data correctly is known as Data Communication. All communication between devices requires that the devices agree on the format of the data. The set of rules defining a format is known as a protocol. At the very least, a communication protocol must define the following:
1. Transmission media used.
2. Rate of transmission in baud or bps.
3. Whether transmission is to be synchronous or asynchronous.
4. Whether data is to be transmitted in half duplex or full duplex mode.
Q2. Define the following terms:
(a) Channel
The term channel refers to a path of communications between two computers or devices. A communication channel provides everything that is needed for the transfer of electronic information from one location to another. It may refer to the physical medium such as coaxial cable, or to a specific carrier frequency within a larger channel or a wireless medium.
(b) Baud
Baud is the number of signaling elements that occur each second. The term is named after JME Baudot, the inventory of the Baudot telegraph code. At slow speeds, only one bit of information or signaling element is encoded in each electrical change. The baud, therefore indicates the number of bits per second that are transmitted. For example, 300 bits are transmitted each second.
(c) Bandwidth
The amount of data or signals that the transmission media can carry in a fixed amount of time is called Bandwidth. The bandwidth depends upon the length, media and signaling technique used. A high bandwidth allows increased throughput and better performance. A medium that has a high capacity has a high bandwidth. A medium that has limited capacity has a low bandwidth.
(d) Frequency
Frequency is the number of cycles or periods a signal completes within one second. The unit of measuring frequency is called Hertz name after a German mathematician Hienrich Hertz. One Hz is one cycle or second. We use one Kilohertz or one kHz to mean 1000Hz and one mega hertz or one MHz to mean 1000 kHz or 1000000Hz.
Q3. Explain the difference between Serial and Parallel Communication.
Ans: The differences between Serial and Parallel communications are:
(a) Serial Communication
In serial data transmission, bits are transmitted serially, one after the other. The least significant bit (LSB) is usually transmitted first. While sending data serially, characters or bytes have to be separated and sent bit by bit. Thus, some hardware is required to convert the data from parallel to serial. At the destination, all the bits are collected, measured and put together as bytes in the memory of the destination. This requires conversion from serial to parallel.
(b) Parallel Communication
In parallel transmission, all the bits of a byte are transmitted simultaneously on separate wires. Here multiple connections between the two devices are therefore required. This is a very fast method of transmitting data from one place to another. The disadvantage of parallel transmission is that it is very expensive, as it requires several wires for both sending as well as receiving equipment. Secondly, it demands extraordinary accuracy that cannot be guaranteed over long distances.
Q4. What is Asynchronous communication? What are its advantages and disadvantages?
Ans: Asynchronous communication sends individual characters one at a time framed by a start bit and 1 or 2 stops bits. Each frame begins with a start bits that enables the receiving device to adjust to the timing of the transmitted signal. The message can begin at any time. Here, messages are kept as short as possible because the sending and receiving devices should not draft out of synchronization, when the message is being transferred. Asynchronous communication is most frequently used to transmit character data and is ideally suited for characters that are transmitted at irregular intervals such as when users are typing in character data from the keyboard.
Advantages of Asynchronous Communications
(a) Asynchronous transmission is simple, inexpensive and suited for transmitting small frames at irregular.
(b) As each individual character is complete in itself, if a character is corrupted during transmission, its successor and predecessor will not be affected.
Disadvantages of Asynchronous Communications
(a) As start, stop and parity bits must be added to each character that is to be transmitted; this adds a high overhead to transmission.
(b) Successful transmission inevitably depends on the recognition of the start bits, hence, as these bits can be easily missed or occasionally spurious, as start bits can be generated by line interference, the transmission may be unsuccessful.
(c) Due to the effects of distortion the speed of asynchronous transmission is limited.
Q5. What is Synchronous communication? What are its advantages and disadvantages?
Ans: In synchronous communication the whole block of data bits is transferred at once, instead of one character at a time. Here, transmission begins at a predetermined regular time instant. A sync signal is used to tell the receiving station that a new frame is arriving and to synchronize the receiving station. Sync signals; generally utilize a bit pattern that cannot appear elsewhere in the messages, ensuring that they will always be distinct and easy for the receiver to recognize. As the transmitter and receiver remain in synchronization for the duration of the transmission, frames can be of longer length.
Advantages of Synchronous Communications
i) Synchronous transmission is more efficient because only 4 additional bytes are required to transmit upto 64k bits.
ii) Synchronous transmission is not really prone to distortion as a result, it can be used at high speeds.
Disadvantages of Synchronous Communications
i) Synchronous transmission is expensive as complex circuitry is required and it is difficult to implement.
ii) If an error occurs during transmission, rather than just a single character the whole block of data is lost.
iii) The sender cannot transmit characters simply, as they occur, but has to store them until it has built up a block. Thus this is not suitable where characters are generated at irregular intervals.
Q6. What is Isochronous communication? What are its advantages and disadvantages?
Ans: This method combines the approaches of asynchronous and synchronous communications. As in the asynchronous method, each character has both the start and stop bits. The idle period between the two characters is not random but an exact multiple of one character time interval. If the time to transmit a character is t, the time interval between characters cannot be random as in the asynchronous method.
Advantages of Isochronous Communications
a) Isochronous transmission guarantees transmission rates and it is almost deterministic.
b) It has low overheads.
c) It has high speed.
Disadvantages of Isochronous Communications
i) Isochronous transmission it is necessary to ensure that the clocking device is fault tolerant.
Q7. Explain following types of communications.
(a) Simplex
This simplest signal flow technique is the simplex configuration. In simplex transmission, one of the communicating devices can only send data, whereas the other can only receive it. Here communication is only in one direction where one party is the transmitter and the other is the receiver. Simple radio and public broadcast television are examples of simplex.
(b) Half Duplex
Half duplex refers to two way communication where only one party can transmit data at a time. Unlike the simplex mode here, both devices can transmit data through not at the same time, that is half duplex provides simplex communication in both directions in a single channel. When one device is sending data the other device must only receive it and vice versa. Half duplex refers to two way communication where one party can transmit data at a time. Simplex refers to one way communication where one party is the transmitter and the other is the receiver.
(c) Full Duplex
Full duplex refers to the transmission of data in two directions simultaneously. Here both the devices are capable of sending as well as receiving data at the same time that simultaneously bi-directional communication is possible. As a result, this configuration requires full and independent transmitting and receiving capabilities at both ends of the communication channel. For example, a telephone is a full duplex device because both parties can talk to each other simultaneously. In contrast, a walkie talkie is a half duplex device because only one party can transmit at a time.
Q8. Explain Analog Signal Transmission.
Ans: Analog signals vary constantly in one or more values; these changes in value can be used to represent data. An analog is continuous and can be represented by using sine waves. Human voice, video and music are all examples of analog signals which vary in amplitude (volume) and frequency (pitch). Amplifiers are used to overcome the attenuation that the signal suffers on its way. The drawback is that amplifiers amplify noise along with the original signal and hence, if the signal gets distorted, it cannot be reconstructed and it is a permanent loss. Due to this reason, this type of transmission is not used where a high level of accuracy is needed. This is used in telephony where a slight distortion in human communication does not matter. By converting analog signals into digital, the original audio or video data can be preserved indefinitely within the specified error bounds and copied over and over without deterioration.
Q9. Explain Digital Data Transmission.
Ans: Digital data transmission describes any system based on discontinuous data or events. Computers are digital machines because at their most basic level they can distinguish between just two values, 0 and 1 or off and on. There is no simple way to represent all the values in between, such as 0.25. All data that a computer processes must be encoded digitally as a series of zeroes and ones.
Q10. Explain the transmission impairments.
Ans: The transmission impairments are explained below:
ii) Attenuation
Attenuation is the loss of energy as the signal propagates outwards. On guided media, the signal falls off logarithmically with the distance. Attenuation is very small at short distances; therefore, the original signal can be recognized without too much distortion. Attenuation increase with the distance as some of the signal energy is absorbed by the medium. The loss is expressed in decibels per kilometer.
iii) Delay Distortion
Delay distortion is caused by the fact that the signals of varying frequencies travel at different speeds along the medium. Any complex signal can be decomposed into different sinusoidal signals of different frequencies, resulting in a frequency bandwidth for every signal.
iv) Noise
Noise is unwanted energy from sources other than the transmitter. Thermal noise is caused by the random motion of the electrons in a wire and is unavoidable. Cross talk is caused by inductive coupling between two wires that are close to each other. Sometimes when talking on the telephone, you can hear another conversation in the background. That is crosstalk.
Q11. Explain the concept of delays.
Ans: The average delay required delivering a packet from source or origin to destination has a large impact on the performance of the data network. A large delay is disastrous for data transfer. The total delay can be categorized into two types which are explained below:
1. Transmission Delay
Transmission delay is the delay, which is present due to link capacities. When resource reservation methods are supported in routers, transmission delays can probably be kept low enough to satisfy the overall delay constraint of 200 ms.
2. Propagation Delay
Propagation delay is the time between the last bit transmitted at the head node of the link and the time that last bit is received at the tail node. This is proportional to the physical distance between the transmitter and the receiver; it can be relatively substantial particularly for a satellite link or a very high speed link. The propagation delay depends on the physical characteristics of the link and is independent of the traffic carried by the link.
Q12. Explain the different transmission media.
i) Magnetic Media
One of the most common ways to transport data from one computer to another is to write them onto magnetic tapes or floppy disks physically transport the tape or disks to the destination machine, read them back in again.
ii) Twisted Pair
A twisted pair consists of two insulated copper wires, typically about 1mm thick. The purpose of twisting the wires is to reduce electrical interference from similar pairs that may be close by. The most common application of the twisted pair is in the telephone system. Nearly all telephones are connected using twisted pair. But the disadvantage is that adequate performance and lowest cost per meter as compare to other cables types, these are expensive to install.
iii) Baseband Coaxial Cable
A coaxial cable consists of a stiff copper wire as the core, surrounded by an insulating material. It has better shielding than twisted pairs so it can span longer distance at higher speeds. The construction and shielding of the coaxial cable gives it a good combination of high bandwidth and excellent noise immunity.
iv) Broadband Coaxial Cable
The term broadband comes from the telephone world, where it refers to anything wider than 1MHz, in the computer networking world, broadband cable means any cable network using analog transmission. Since broadband networks use standard cable television technology, the cable can be used up to 300 MHz and can run for nearly 100km due to the analog signaling, which is much less critical than digital signaling. To transmit digital signals on an analog network, each interface must contain electronics to convert the outgoing bit stream to an analog signal, and the incoming analog signal to a bit stream.
v) Optical Fiber
Fiber optic cable is made of fine strands of silicon glass fiber thinner than a human hair, and is coated with a refractive surface. The signals are converted to light pulses before being sent. It is used for long haul telecommunications links, for providing high speed data communications links for computers and information services to homes.
Q13. What are the advantages and disadvantages of fiber optic?
Ans: Advantages and disadvantages of fiber optic cable are:
Advantages of Fiber optic cable
i) High capacity or laser bandwidth
ii) Immune to interference or electromagnetic
iii) Can go long distances or low attenuation.
Disadvantages of Fiber optic cable
i) It is costly.
ii) It is difficult to join
iii) It is expensive to install and greater skill is required.
Q14. Compare and contrast between Optical Fiber and Copper Wire.
Ans: Optical fiber has the following advantages over copper wire:
i) Optical fiber can handle much higher bandwidths than copper wire. Due to low attenuation, repeaters are needed only about every 30km on long lines whereas copper wires require repeaters every 5km, which is substantial saving in cost.
ii) Optical fibers are not affected by power surges, electromagnetic interference, or power failure. Neither is it affected by corrosive chemicals in the air making it ideal for harsh factory environments.
iii) Optical fibers is lighter than copper. One thousand twisted pairs 1km long weight 8000kg. Two fibers have more capacity and weight only 100kg, which greatly reduces the need for expensive mechanical support systems that must be maintained.
iv) Optical fibers do not leak light and are quite difficult to tap. This gives them excellent security against potential wire trappers.
The disadvantages of optical fibers are
i) Optical is a complex technology requiring skills that most engineers do not posses.
ii) Since optical transmission is inherently unidirectional, two way communications requires either two fibers or two frequency bands on one fiber.
iii) Fiber interfaces cost more than electrical interfaces.
Q15. What is wireless transmission?
Ans: Wireless transmission is also called unbounded media. Here there are no physical connectors between the two communicating devices. Usually the transmission is sent through the atmosphere, but sometimes it can be just across a room. Wireless media is used when a physical obstruction or distance blocks the use of normal cable media.
Q16. What is microwave transmission? What are the advantages and disadvantages of microwave?
Ans: Microwave is a radio system, which uses very high frequencies to send and receive data. Because of the high frequencies involved, microwave stations are located about 30 kilometers apart and in line of sight. Microwave system has sufficient bandwidth capacity to support a large number of voice channels and one or two TV channels.
The advantages of microwave are medium capacity, medium cost and can go long distances. But the disadvantages are nose interference, geographical problems due to line of sight requirements and becoming outdated.
Q17. Give a comparison between microwave and optical fiber.
Ans: The microwave has several significant advantages over optical fiber cable. The main one is that no right of way is needed, and by buying a small plot of ground every 50km and putting a microwave tower on it, one can bypass the telephone system and communicate directly. Microwave is also relatively inexpensive. Putting up two simple towers and putting antennas on each one may be cheaper than burying 50km of fiber through a congested urban area or up over a mountain, and it may also be cheaper than leasing optical fiber from a telephone company.
Q18. Explain the different types of microwave.
Ans: The different types of microwave are:
a) Terrestrial Microwave
Terrestrial microwave employs earth based transmitters and receivers. The frequencies used are in the low gigahertz range, hence limiting all communication to line of sight. Microwave transmissions are carried out using a parabolic antenna that produces a narrow, highly directional signal. At the receiving site a similar antenna which is sensitive to signals, only within a narrow focus, is placed. As both the transmitter and receiver are highly focused, they must be carefully adjusted so that the transmitted signal is aligned with the receiver.
b) Satellite Microwave
Satellite microwave systems relay transmission through communication satellites that operate in geosynchronous orbit 36000 – 40000km above the earth. Satellite microwave communication is extremely expensive as a result of which organizations share the cost or they purchase services from a commercial provider.
Q19. What is radio transmission?
Ans: The radio portion of the electromagnetic spectrum extends from 10KHz to 1GHz. Within this range, there are numerous bands or ranges of frequencies that are designated for specific purposes. Some of the familiar frequency bands are:
a) Short wave,
b) Very high frequency used for television and FM radio.
c) Ultra high frequency used for television.
Radio waves are easy to generate, can travel long distances and generate building easily so they are widely used for communications.
Q20. What is infrared and millimeter waves?
Ans: Unguided infrared and millimeter waves are widely used for short range communication. The remote control transmits pulses of infrared light that carry coded instruction to the receiver on the TV. They are relatively directional, cheap, and easy to build but they have a major drawback that is they do not pass through solid objects. Infrared waves do not pass through solid walls well, as a result of which an infrared system in one room of a building will not interfere with a similar system in adjacent rooms which is a great advantages.
Q21. What is point to point infrared?
Ans: The point to point networks operate by relaying infrared signals from one device to the next. Hence, transmissions are focused on a narrow beam, and the transmitter and receiver must be align carefully. Point to point infrared is not suitable for use with devices that moves frequently because here, the devices must be carefully aligned and setup.
Q22. What is broadcast infrared?
Ans: Broadcast infrared disperse transmissions so that they are visible to several receivers. The two possible approaches to broadcast infrared networking are:
a) The active transmitter can be located at a high point so that it can transmit to all devices.
b) A reflective material can be placed on the ceiling. Devices transmit toward the ceiling from where the light signals are dispersed to other devices in the room.
Installation is simple as devices alignment is not critical. Hence it is essential that each device has clear transmission and reception pathways.
Q23. What is wirelss LAN? What are the disadvantages of wireless LAN?
Ans: Wireless LAN’s are a system of portable computers that communicate using radio transmission. Wireless LANs make use of the omni-directional wireless communication. The main objective of wireless LAN is to provide high speed communication among computers that are located in relatively close proximity.
The disadvantages of using wireless LAN are:
a) When the receiver is within the range of two active transmitters, the resulting signal will usually be grabbled and hence useless.
b) In indoor wireless LANs, the presence of walls between the stations has great impact on the effective range of each station.
Q24. Differentiate between Parallel Transmission and Serial Transmission.
Ans: The difference between parallel transmission and serial transmission are:
Parallel Transmission
1. All the data bits are transmitted simultaneously on separate wires.
2. Here multiple circuits interconnecting the two devices are required.
3. It is a very fast method of data transmission.
4. Expensive as it requires several wires for sending as well as receiving equipment.
5. Suitable for short distance data transmission where accuracy is of prime importance.
Serial Transmission
1. Data bits are transmitted one bit at a time serially.
2. Requires only one circuit interconnecting the two devices.
3. It is slow compared to parallel transmission.
4. Comparatively cheaper.
5. Suitable for long distance data transmission.p
BLOCK 1 UNIT 3
DATA ENCODING AND COMMUNICATION TECHNIQUE
Q1. What is the need of encoding?
Ans: To send a signal over a physical medium, we need to encode or transform the signal in some way so that the transmission medium can transmit it. The sender and the receiver must agree on what kind of transformation has been done so that, it can be reversed at the other end and the actual signal or information can be recovered. The information content in a signal is based upon having some changes in it.
Q2. What is modulation?
Ans: The act of changing or encoding the information in the signal is known as modulation. The information content of a signal is dependent on the variation in the signal. After sending one cycle of a sine wave there is no point in sending another cycle because we will not be able to convey anything more than what we have already done.
Q3. Explain the different ways to encode analog signal to analog information.
Ans: There are three different ways in which analog to analog encoding can be done which are explained below:
(a) Amplitude Modulation
In this type of modulation the frequency and phase of the carrier or base signal are not altered. Only the amplitude changes and we can see that the information is contained in the envelope of the carrier signal. It can be proud / demonstrated that, the bandwidth of the composite signal is twice that of the highest frequency in the information signal that modulates the carrier.
(b) Frequency Modulation
Here it is the frequency of the base signal that is altered depending on the information that is to be sent. The amplitude and phase of the carrier signal are not changed at all. It can be shown that this results in a bandwidth requirement of 10 times the modulating signal, centered on the carrier frequency. This method is less susceptible to noise and gives the best performance of all data encoding types as far as the quality of the transmitted signal is concerned. Although digital encoding methods may give better performance over multiple hops because in their case, the original signal can be accurately reconstructed at each hop, frequency modulation is the best as far as single hop transmission goes.
(c) Phase Modulation
In phase modulation (PM) the modulating signal leaves the frequency and amplitude of the carrier signal unchanged but alter its phase. The characteristics of this encoding technique are similar to FM, but the advantage is that of reduce complexity of equipment. However, this method is not used in commercial broadcasting.
Q4. Explain how to convert analog to digital signal?
Ans: Pulse Code Modulation (PCM) is the method used for converting analog to digital. This method gives very good quality and when sued in conjunction with error correction techniques, can be regenerated at every hop on the way to its final destination. This is the big advantage of digital signals. The first step in PCM is to convert the analog signal into a series of pulses. This is called Pulse Amplitude Modulation (PAM). To do this the analog signal is sampled at fixed intervals and the amplitude at each sample decides the amplitude of that pulse. You can see that at this step, the resulting signal is still actually an analog signal because the pulse can have any amplitude; equal to the amplitude of the original signal at the time of the sample was taken. In PAM, the sampled value is held for a small time so that the pulse has a finite width. In the original signal the value occurs only for the instant at which it was sampled. PCM is commonly used to transmit voice signals over a telephone line. It gives good voice quality but is a fairly complex method of encoding the signal.
Q5. State the Nyqist’s theorem.
Ans: Nyqist’s theorem, states that to be able to get back the original signal, we must sample at a rate that is at least twice that of the highest frequency contained in the original signal.
Q6. Explain the different ways to convert digital to analog.
Ans: There are three different ways to convert digital to analog which are explained below:
(a) Amplitude Shift Keying
In this kind of encoding, abbreviated as ASK, the amplitude of the analog signal is changed to represent the bits, while the phase and the frequency are kept the same. The bigger the difference between the two values, the lower the possibility of noise causing errors. One method could be to represent the high voltage as a 1 and the negative high voltage as a 0 although the reverse convention could also be used. Sometimes, to reduce the energy needed for transmission, we can choose to represent a 0 and a 1 by a zero voltage and the other bit by the high voltage. This scheme is called on-off-keying.
(b) Frequency Shift Keying
When a bit is encoded, the frequency changes and remains constant for a particular duration. The phase and the amplitude of the carrier are unaltered by FSK. Just as in FM, noise in the line has little effect on the frequency and so this method is less susceptible to noise. It is possible to show that an FSK signal can be analyzed as the sum of two ASK signals. One of these is centered on the frequency used for a 0 bit and the other for the frequency used for a 1 bit. From this, it is easy to show that the bandwidth needed for an FSK signal is equal to the sum of the baud rate and the difference between the two frequencies. Thus, the requirement of FSK is a higher bandwidth that helps us to avoid the noise problem of ASK.
(c) Phase Shift Keying
We can also encode by varying the phase of the carrier signal. This is called phase shift keying and here the frequency and amplitude of the carrier are not altered. To send a 1 we could use a phase of 0 while we could change it to 180 degrees to represent a 0. Such an arrangement is not affected by the noise in the line, because that affects amplitude rather than the phase of the signal. This quality of PSK can be used to achieve more efficient encoding.
Q7. Explain the concept of Quadrature Amplitude Modulation (QAM).
Ans: But it can be quite fruitful to combine ASK and PSK together and reap the advantages. This technique is called Quadrature Amplitude Modulation (QAM). There are a large number of schemes possible in the QAM method. For example, 1 amplitude and 4 phases is called 4-QAM. It is the same as 4 PSK because there is no change in the amplitude. With 2 amplitudes and 4 phases we can send 3 bits at a time and so this scheme is called 8 QAM.
Q8. Explain the concept of digital to digital encoding.
Ans: There are different techniques for converting digital to digital encoding which are explained below:
(a) Unipolar Encoding
In this method, there is only one polarity used in the transmission of the pulses. A positive pulse may be taken as a 1 and a 0 voltage can be taken as 0, or the other way round depending on the convention we are using. Unipolar encoding uses less energy and also has a low cost.
(b) Polar
The polar methods use both a positive as well as a negative voltage level to represent the bits. So, a positive voltage may represent a 11 and a negative voltage may represent a 0 or the other way round. Because positive and negative voltages are present, the average voltage is much lower than in the unipolar case, mitigating the problem of the DC component in the signal.
(c) Bipolar
This encoding method uses three levels, with a zero voltage representing a 0 and a 1 being represented by positive and negative voltages respectively. This way the DC component is done away with because, every 1 cancels the DC component introduced by the previous 1. This kind of transmission also ensures that every 1 bit is synchronized. The simplest bipolar method is called Alternate Mark Inversion. The problem with Bipolar Ami is that synchronization can be lost in a long string of 0s.
Q9. What are the different methods of Polar encoding?
Ans: The different methods of polar encoding are explained below:
(a) Non Return to Zero (NRZ)
Non return is zero (NRZ) is a polar encoding scheme which uses a positive as well as a negative voltage to represent the bits. A zero voltage will indicate the absence of any communication at the time. Here again there are two variants based on level and inversion. In NRZ-L for level encoding, it is the levels themselves at indicate the bits, for example positive voltage for a 1 and a negative voltage for a 0. On the other hand, in NRZ-I (Inversion) encoding, an inversion of the current level indicates a 1, while a continuation of the level indicates a 0.
(b) Return to Zero (RZ)
The return to zero is a method that allows for synchronization after each bit. This is achieved by going to the zero level midway through each bit. So a 1 bit could be represented by the positive to zero transition and a 0 bit by a negative to zero transition.
Q10. Explain the difference between Manchester encoding and Differential Manchester encoding.
Ans: In Manchester encoding, there is a transition at the middle of each bit that achieves synchronization. A 0 is represented by a negative to positive transition while a 1 is represented by the opposite. In Differential Manchester method, the transition halfway through the bit is used for synchronization as usual, but a further transition at the beginning of the bit represents the bits itself. There is no transition for a 1 and there is a transition for a 0 so that a 0 is actually represented by two transitions.
Q11. Explain High Density Bipolar 3 and 8 Zero Substitution (B8ZS).
Ans: HDB3 changes the pattern of four 0’s depending on the number of 1’s since the last such substitution and the polarity of the last 1. If the number of 1’s since the last substitution is odd, the pattern is violated at the fourth 0. So if he previous 1 was positive the 0 is also made positive to avoid confusion with a 1. If the number of 1’s since the last substitution is even, the first and fourth zeroes have a pattern violation.
In 8 Zero Substitution (B8ZS), a string of 8 zeroes is encoded depending on the popularity of the previous actual 1 bit that occurred. If the previous 1 was sent as a negative voltage, the zeroes are sent as three 0’s followed by a negative, positive, zero, positive, negative voltage sequence. A similar but opposite pattern violation is performed for the case where the last 1 was positive. This way a deliberate pattern violation is used to achieve synchronization.
BLOCK 1 UNIT 4
MULTIPLEXING AND SWITCHING
Q1. Define Frequency Division Multiplexing (FDM).
Ans: Frequency Division Multiplexing is the process of dividing up the channel from 300 kHz upto 312 kHz into different frequencies for sending data and all the different transmissions are happening at the same time. The transmission is received at the other end, the destination and there, it has to be separated into its original components by de-multiplexing.
Q2. What is time division multiplexing? Explain the different types of time division multiplexing.
Ans: In time division multiplexing, there are four different transmissions occurring with each being one-fourth of the time slice. This slice should be small enough so that the slicing is not apparent to the application. So, for a voice transmission a cycle of 100ms could be sufficient as we would not be able to detect the fact that there are delays. In that case, each transmission could be allotted a slice of 25ms. At the receiving end, the transmission has to be reconstructed by dividing up the cycle into the different slices, taking into account the transmission delays.
There are two types of time division multiplexing which are explained below:
(a) Synchronous Time Division Multiplexing
In this kind of Time Division Multiplexing, the simpler situation is where the time slots are reserved for each transmission, irrespective of whether it has any data to transit or not. There, this method can be inefficient because many time slots may have only silence. But it is a simpler method to implement because the act of multiplexing and de-multiplexing and de-multiplexing is easier. This kind of TDM is known as Synchronous TDM.
(b) Statistical Time Division Multiplexing
This is a method where there are more devices than the number of time slots available. Each input data stream has a buffer associated with it. The multiplexer looks at the buffer of each device that provides the input data and checks to see if it has enough to fill a frame. If there are enough the multiplexer sends the frame. Otherwise, it goes to the next, device in line and checks its input buffer. This cycle is continued indefinitely. At the receiving end the de-multiplexer decomposes the signal into the different data streams and sends them to the corresponding output line.
Q3. Explain the different types of switching.
Ans: The different types of switching are explained below:
(a) Circuit Switching
In circuit switching, we create a link or circuit between the devices that need to communicate for the duration of the transmission only. It entails setting up the circuit, doing the actual transmission that can be simplex, half duplex or full duplex, and then dismantling the circuit for use by another pair of devices. These are divided into two groups of 3 on the left and 4 on the right. To ensure complete connectivity between them at all times, we would need 12 physical links.
(b) Packet Switching
It is based on the concept of data packets, called datagrams. These are self contained units that include the address of the destination, the actual data and other control information. It takes care of the sporadic nature of data communication where transmission tends to occur in bursts. So we do not waste transmission capacity by keeping circuits connected but idle while there are periods of silence. The packet is stored for verification and then transmitted further to the next router along in ways until it reaches for fixed destination machine. This mechanism is called packet switching.
Q4. What do you meant by Time Slot Interchange (TSI)?
Ans: It consists of a control unit that does the actual reordering of the input. For this, it has to first buffer the input it gets, in the order it gets it. This would be stored in some kind of volatile memory. The control unit then sends out the data from the buffer in the order in which it is desired. Usually the size of each buffer would be that of the data that the inputs generate in one time slice.
Q5. What are the limitations of circuit switching?
Ans: The limitations of circuit switching are given below:
(a) It is not possible to prioritize a transmission. In a circuit switching mechanism, all the data will be sent in the order in which it is generated, irrespective of its importance or urgency. This is fine for voice transmission, in which we want to hear things in the sequence, in which they are spoken at the other end, but this does not work for data where some packets may be more important than others and the order of delivery does not have to be sequential.
(b) A circuit once set up, defines the route that will be taken by the data until it is dismantled and set up again. Sometimes, that circuit may have been set up via a less advantages set of links because that was the best route available at the quality or other parameters. Now subsequently, even if another, better route is released by other devices, we cannot change over to this better route without disconnecting the previous circuit and forcing the participants to set up the call again.
Q6. What is virtual switching?
Ans: A virtual circuit is et up at the time a pair of devices begins to communicate using packet switching. Unlike circuit switching, we can change the path or route used if there is a failure in the route. Virtual circuits can also be permanent, that is dedicated to a pair of devices, or can be set up for a limited time between different devices. Another difference is that in virtual circuits multiplexing can happen at the level of individual switches as well.
BLOCK 2 UNIT 1
DATA LINK LAYER FUNDAMENTALS
Q1. What are the functions of data link layer?
Ans: The functions of the data link layer are framing, frame synchronization, error handling, flow regulation, addressing and access control. The data link layer is divided into two sub-layers; the Media Access Control (MAC) layer and the Logical Link Control (LLC) layer. The MAC sub layer controls how a computer on the network gains access to the link resources and grants permission to transmit it. The LLC layer controls frame synchronization, flow control and error checking.
Q2. What do you understand by framing?
Ans: The raw data coming from the physical layer is converted into frames for forwarding to the network layer. This is done to ensure that the transmission of data is error free. Error detection and correction is done by a data link layer. This is called framing.
Q3. Explain the different types of errors which occur during transmission over the network.
Ans: The different types of errors which occur during transmission over the network are given below:
(a) 1 bit error
1-bit error or single bit error means that only one bit is changed in the data during transmission from the source to the destination node i.e. either 0 is changed to 1 or 1 is change to 0. This error will not appear generally in case of serial transmission. But it might appear in case of parallel transmission.
(b) Burst Error
Burst error means that 2 or more bits of data are altered during transmission from the source to the destination node. But it is not necessary that error will appear in consecutive bits. Size of burst error is from the first corrupted bit to the last corrupted bit.
(c) Lost Message or Frame
The sender has sent the frame but that is not received properly, this is known as loss of frame during transmission. To deal with this type of error, a retransmission of the sent frame is required by the sender.
Q4. What is redundancy?
Ans: Sending every data unit twice increases the transmission time as well as overhead in comparison. Hence the basic strategy, for dealing with errors is to include group of bits as additional information in each transmitted frame, so that the receiver can detect the presence of error. This method is called Redundancy as extra bit appended in each frame are redundant.
Q5. Explain the different error detection methods.
Ans: The different error detection methods are given below:
(a) Parity Check
This is the most common method used for detecting errors when the number of bits in the data is small. A parity bit is an extra binary digit added to the group of data bits, so that the total number of one in the group is even or odd. Data bits in each frame is inspected prior to transmission and an extra bit is computed and appended to the bit string to ensure even or odd parity, depending on the protocol being used. If odd parity is being used, the receiver expects to receive a block of data with an odd number of 1s. For even parity, the number of 1s should be even.
(b) Block Sum Check Method
This is the extension to the single parity bit method. This can be used to detect up to two bits errors in a block of characters. Each byte in the frame is assigned a parity bit. An extra bit is computed for each bit position. The resulting set of parity bits for each column is called block sum check. Each bit that makes up the character is the modulo2 sum of all the bits in the corresponding column.
(c) Cyclic Redundancy Check
In this method, the sender divides frame or data string by a predetermined Generator. Polynomial and then appends the remainder called checksum onto the frame before starting the process of transmission. At the receiver end, the receiver divides the received frame by the same Generator polynomial. If the remainder obtained after the division is zero, it ensure that he data received at the receiver end is error free. All operations are done modulo – 2.
(d) Checksum
In this method, the checksum generator divides the given input data into equal segments of k bits. The addition of these segments using ones complement arithmetic is complimented. This result is know as the checksum and it is appended the transmission media. At the receiver end add all received segments. If the addition of segments at the receiver end is all 1’s then the data received is error free as complement of the same will be all 0s. Then the data can be accepted, otherwise data can be discarded.
Q6. What is forward error correction?
Ans: Forward Error Correction (FEC) is a type of error correction method which improves on simple error detection schemes by enabling the receiver to correct errors once they are detected. This reduces the need for retransmissions of error frames.
Q7. What is Hamming Code?
Ans: In Hamming code, each redundant bit is the combination of data bits where each data bit can be included in more than one combination.
Q8. Give the algorithm for computing the checksum.
Ans: Let D(x) be the data and G(x) be the generating polynomial. Let r be the degree of generator polynomial G(x).
Step 1: Multiple the data D(x) by X(to the power)r giving r zeroes in the low order end of the frame.
Setp 2: Divide the result obtained in step 1 by G(x) using modulo-2 division.
Step 3: Append the remainder from step 2 to D(x) thus placing r terms in the r low order positions.
Q9. What is flow control?
Ans: Flow control means using some feedback mechanism by which the sender can be aware of when to the send next frame and not at the usual speed of the sender. If the frame is accepted or processed by the receiver then only with the sender send the next frame. It may be said that the speed of the sender and the receiver should be compatible with each other, so that the receiver will receive or process all frames sent by the sender as every receiver has a limited block of memory called the buffer reserved for storing incoming frames.
Q10. What is stop and wait method? What is the problem with this method?
Ans: Stop and wait flow control is the simplest form of flow control. In this method, the receiver indicates its readiness to receive data for each frame; the message is broken into multiple frames. The sender waits for an ACK or acknowledgement after ever frame for a specified time called time out. It is sent to ensure that the receiver has received the frame correctly. It will then send the next frame only after the ACK has been received. If the frame or ACK is lost during transmission, then it has to be transmitted again by the sender. This retransmission process is known as Automatic Repeat Request (ARQ). Stop and wait ARQ is one of the simplest methods of flow control.
The problem with stop and wait is that only one frame can be transmitted at a time and that often leads to inefficient transmission channel till we get the acknowledgement the sender can not transmit any new packet. During this time, both the sender and the channel are unutilized.
Q11. Explain Sliding Window Protocol.
Ans: In this flow control method, the receiver allocates buffer space for n frames in advances and allows transmission of multiple frames. This method allows the sender to transmit n frames without an ACK. A k-bit sequence number is assigned to each frame. The range of sequence number assigned to each frame. The range of sequence number uses modulo 2 arithmetic. To keep track of the frames, that have been acknowledged, each ACK has a sequence number. The receiver acknowledges a frame by sending an ACK that includes the sequence number of the next expected frame. The sender sends the next n frames starting with the last received sequence number that has been transmitted by the receiver (ACK). Hence, a single ACK can acknowledge multiple frames.
BLOCK 2 UNIT 2
RETRANSMISSION STRATEGIES
Q1. What is the response of the Stop and Wait ARQ to:
(a) Normal Operation
If the sender is sending frame 0, then it will wait for ack 1 which will be transmitted by the receiver with the expectation of the next frame numbered frame1. As it receives ACK1 in time, it will send frame 1. This process will be continuous till computer data transmission takes place. This will be successful transmission if ack for all frames ent it received in time.
(b) When ACK is lost
Here, the sender will receive corrupted ACK1 for frame sent frame 0. It will simply discard corrupted ACK1 and as the time expires for this ACK, it will retransmit frame 0. The receiver has already received frame 0 and is expecting frame1, hence it will discard duplicate copy of frame 0. In this way, the numbering mechanism solves the problem of duplicate copy of frames. Finally, the receiver has only one correct copy of one frame.
(c) When frame is lost
If the receiver receives corrupted damaged frame1, it will simply discard it and assumes that the frame was lost on the way. And correspondingly, the sender will not get ACK0 as frame has been received by the receiver. The sender will be in waiting stage for ACK0 till its time out occurs in the system.
(d) When ACK time out occurs
In this operation, the receiver is not able to send ACK1 for received frame 0 in time due to some problem at the receiver’s end or network communication. The sender retransmits frame0 as ACK1 is not received in time. Then, the sender retransmits frame0 as ACK2 time expires. At the receiver end, the receiver discards this frame0 as the duplicate copy is expecting frame1 but sends the ACK1 once gain corresponding to the copy received for frame0.
Q2. What is Piggybacking?
Ans: In bidirectional transmission both the sender and the receiver can send frames as well as acknowledgement. Hence both will maintain S and R variables. To have an efficient use of bandwidth the ACK can be appended with the data frame during transmission. The process of combining ACK with data is known as Piggybacking.
Piggybacking can also be defined as the process which appends acknowledgement of frame with the data frame. Piggybacking process can be used if Sender and Receiver both have some data to transmit. This will increase the overall efficiency of transmission. While data is to be transmitted by both sender and receiver, both will maintain control variable S and R.
Q3. Explain Go Back N ARQ method.
Ans: In this method, many frames can be transmitted during the process without waiting for acknowledgement. In this, we can send n frames without making the sender wait for acknowledgement. At the same time, the sender will maintain a copy of each frame till acknowledgement reaches it safely. Each frame will have a sequence of number that will be added with the frame. If the frame can have a k bit sequence number then the number will range between 0 to 2(to the power)k – 1.
The sender can send 4 frames continuously without waiting for acknowledgement. But the receiver will look forward to only one frame that must be in order. If the frame received is not in order, it will simply keep on discarding the frame till, it reaches the desired sequence number frame. The receiver is not bound to send an individual acknowledgement for all frames received; it can send a cumulative acknowledgement also. The receiver will send a positive acknowledgement if, the received frame is the desired sequence number frame. Otherwise, it will keep on waiting if, the frame received is corrupted or out of order. As soon as the timer expires for the frame sent by the sender, the sender will GoBack and retransmit all frames including the frame for which the time expired, till the last sent frame. Hence, it is named as Go-Back-N ARQ.
Q4. Explain Selective Repeat ARQ method.
Ans: This method increases the efficiency of the use of bandwidth. In this method, the receiver has a window with the buffer that can hold multiple frames sent by the sender. The sender will retransmit only that frame which has some error and not all the frames as Go-Back-N ARQ. Hence, it is name as selective repeat ARQ. Here the size of the sender and the receiver window will be same. The receiver will not look forward only to one frame as in Go-Back-N ARQ but it will look forward to continuous range of frames. The receiver also sends NAK for the frame which had error and required to be retransmitted by the sender before the time out event fires. If the window is 2 and acknowledgement for frame0 and frame1 both gets lost during transmission then, after timer expires the sender will retransmit it frame0 though the receiver is expecting frame2 after frame1 was received without any error.
Q5. What is pipelining?
Ans: Pipelining in the network is one task that starts before the previous one is completed. We might also say that the number of tasks is buffered in line, to be processed as this is called pipelining. Go-Back-N ARQ and Selective Repeat ARQ as both methods can send multiple frames without holding the sender for receiving the acknowledgement for frame sent earlier. This process of pipelining improves the efficiency of bandwidth utilization. Here frame0 is received by the receiver and without waiting for acknowledgement of frame0 sent at sender site; the sender is allowed to send frame1. This process is known as pipelining.
BLOCK 2 UNIT 3
CONTENTION BASED MEDIA ACCESS PROTOCOLS
Q1. Explain the concept of Pure Aloha.
Ans: In an ALOHA network one station will work as the central controller and the other station will be connected to the central station. If any of stations want to transmit data among themselves, then the station sends the data first to the central station which broadcast it to all the station. Here the medium is shared between the stations. So if two stations transmit a frame at overlapping time then, collision will occur in the system. Here, no station is constrained; any station that has data or frame to transmit can transmit at any time. Once one station sends a frame (when it receives its own frame and assumes that the destination has received it) after 2 times the maximum propagation time. If the sender station does not receive the its own frame during this time limit then, it retransmit this frame by using back off algorithm that we will discuss later on. As in the case of Pure ALOHA protocol frame can be sent any time so, the probability of collision will be very high. Hence to present a frame from colliding no other frame should be sent within its transmission time.
Q2. Explain the concept of slotted ALOHA.
Ans: In this, we can improve the performance by reducing the probability of collision. In the slotted ALOHA stations are allowed to transmit frames in slots only. If more than one station transmits in the same slot, it will lead to collision. This reduces the occurrence of collision in the network system. Here every station has to maintain the record of tie slot. The process of transmission will be initiated by any station at the beginning of the time slot only. Here also, frames are assumed to be of constant length and with the same transmission time. Here the frame will collide with the reference frame only if, it arrives in the interval t0 to tc. Hence here the vulnerable period is reduced that is to t seconds long.
Q3. Explain the different protocols used in Carrier Sense Multiple Access (CSMA).
Ans: The different protocols used in CSMA are:
(a) 1 – Persistent CSMA
In this protocol a station i.e. who wants to transmit some frame will sense the channel first, if it is found busy than that some transmission is going on the medium, then this station will continuously keep sensing that the channel.
(b) Non-Persistent CSMA
To reduce the frequency of the occurrences of collision in the system then, another version of CSMA that is non-persistent CSMA can be used. Here the station that has frames to transmit first sense whether the channel is busy or free. If the station finds that channel to be free it simply transmits its frame. Otherwise it will wait for a random amount of time and repeat the process after that time span is over.
(c) P-Persistent CSMA
This category of CSMA combines features of the above version of CSMA that is 1-persistent CSMA and non-persistent CSMA. This version is applicable for the slotted channel. The station that has frames to transmit senses the channel and if found free then simply transmits the frame with p probability and with probability 1-p it, defers the process.
Q4. Explain CSM with Collision Detection (CSMA/CD) method.
Ans: In CSMA/CD the station aborts the process of transmission as soon as they detect some collision in the system. If two stations sense that the channel is free at the same time, then both start transmission process immediately. And after that both stations get information that collision has occurred in the system. Here after the station detecting the collision, the system aborts the process of transmission. In this way, time is saved and utilization of bandwidth is optimized. This protocol is known as CSMA/CD and this scheme is commonly used in LAN.
In CSMA/CD a station with a frame ready to begin transmission senses the channel and starts transmission if it finds that the channel is idle. Otherwise, if it finds it busy the station can persist and use back off algorithm.
Q5. Explain back off algorithm.
Ans: After a collision is sensed by the channel, time is divided up into discrete slots. For example, if first collision identified then each station waits for either 0 or 1 slot time. Similarly, if third collision occurs then random interval will be 0 to 7 and for ith collision random number interval will be 0 to 2(to power i) – 1. Subsequently, these many numbers of slots will be skipped before re-transmission. This is called a binary exponential back off algorithm. CSMA/CD uses back off algorithm.
BLOCK 2 UNIT 4
WIRELESS LAN AND DATALINK LAYER SWITCHING
Q1. Explain the wireless LAN architecture (IEEE 802.11).
Ans: The wireless LAN is based on a cellular architecture where the system is sub-divided into cells. Each cell is controlled by a base station. Wireless LAN may be formed by a single cell, with a single access point most stations will be formed by several cells, where the APs are connected through some kind of backbone. This backbone may be the Ethernet and in some cases, it can be the wireless system. The DS appears to upper level protocols as a single 802 network, in much the same way as a bridge in wire 802.3. The Ethernet network appears as a single 802 network to upper layer protocols.
Q2. What are the problems associated with wireless network?
Ans: There are two fundamental problems associated with a wireless network. Assume that there are four nodes, A, B, C and D. B and C are in the radio range of each other. Similarly A and B are in the radio range of each other. But C is not in the radio range of A. Now suppose that there is a transmission going on between A and B. If C also wants to transmit to B, first it will sense the medium but will not listen to A’s transmission to B because, A is outside its range. Thus, C will create garbage for the frame coming from A if it transmits to B. This is called the Hidden Station Problem.
There is another problem called exposed station problem. In this case, B is transmitting to A. Both are within radio range of each other. Now C wants to transmit to D. As usual, it senses the channel and hears an ongoing transmission and falsely concludes that it cannot transmit to D. But the fact is transmission between C and D would not have caused any problem because, the intended receivers C and D are in different range. This is called exposed station problem.
Q3. What are the features of MACAW?
Ans: The features of MACAW are explained below:
(a) Data Link Layer Acknowledgement
It was noticed that without data link layer acknowledgements, lost frames were not retransmitted until the transport layer noticed their absence, much later. They solved this problem by introducing an ACK frame after each successful data frame.
(b) Addition of Carrier Sensing
They also observed that CSMA has some use, namely to keep a station from transmitting a RTS while at the same time another nearby station is also doing so to the same destination, so carrier sensing was added.
(c) An improved Back Off Mechanism
It runs the back off algorithm separately for each data stream (source and destination pair) rather than for each station. This change improves the fairness of the protocol.
(d) Data Sending Message
Say a neighbor of the sender overhears an RTS but not CTS from the receiver. In this case it can tell if RTS-CTS were successful or not. When it overhears DS, it realizes that the RTS-CTS were successful and it defers its own transmission.
Q4. What is physical channel sensing?
Ans: In physical channel sensing method, before the transmission, it senses the channel. If the channel is sensed idle, it just wants and then transmitting. But it does not sense the channel while transmitting until it goes idle and then starts transmitting. If a collision occurs, the colliding stations wait for a random time, using the Ethernet binary exponential back off algorithm and then may try again later.
Q5. What are the six reasons pointed out by Tanenbaun for multiple LANs?
Ans: The six reasons pointed out by Tanenbaun for multiple LANs are explained below:
1) Multiple LANs in organization
Since the goals of the various departments differ, different departments choose different LANs. But there is a need for interaction so bridges are needed.
2) Geographical Difference
The organization may be geographically spread over several buildings separated by considerable distances. It may be cheaper to have separate LANs in each building and connect them with bridges and later link them to run a single cable over the entire site.
3) Load Distribution
It may be necessary to split what is logically a single LAN into separate LANs to accommodate the load.
4) Long Round Trip Delay
In some situations, a single LAN would be adequate in terms of the load, but the physical distance between the most distant machines can be great. Even if laying the cable is easy to do, the network would not work due to the excessively long round trip delay. The only solution is to partition the LAN and install bridges between the segments.
5) Reliability
By inserting bridges at critical points reliability can be enforced in the network by isolating a defective mode. Unlike a repeater, which just copies whatever it sees, a bridge can be programmed to exercise some discretion reading what forwards and what it does not forward.
6) Security
Security is a very important feature in bridges today, and can control the movement of sensitive traffic by isolating the different parts of the network.
Q6. What are the characteristics of bridges?
Ans: The characteristics of bridges are:
(a) Fully Transparent
It means that it should allow the movement of a machine from one cable segment to another cable segment without change of hardware and software or configuration tools.
(b) Interpretability
It should allow a machine on one LAN segment to talk to another machine on another LAN segment.
Q7. What are the difficulties in building bridges between the various 802 LANs?
Ans: The difficulties in building bridges between the various 802 LANs are:
(a) Different Frame Format
Each of the LANs uses a different frame format. As a result, any copying between different LANs requires reformatting, which takes CPU time, requires a new checksum calculation and introduces the possibility of undetected errors due to bad bits in the bridge memory.
(b) Different Data Rates
When forwarding a frame from a fast LAN to a slower one, the bridge will not be able to get rid of the frames as fast as they come in. Therefore, it has to be buffered.
(c) Different Frame Lengths
When a long frame forwarded onto a LAN, it cannot be accepted due to different frame length. This is the most serious problem. The only solution is that the frame must be split or discarded.
(d) Security
Both 802.11 and 802.16 support encryption in the data link layer, but the Ethernet does not do so. This means that the various encryption services available to the wireless networks are lost when traffic passes over the Ethernet.
(e) Quality of Services
Ethernet has no concept of quality of service, so traffic from other LANs will lose its quality of service when passing over the Ethernet.
Q8. What are the features of transparent bridge?
Ans: Transparent Bridge is a plug and play device. There are no additional requirement of hardware and software changes, no setting of address switches, no downloading of routing table, in case, it is used to connect LANs. It does not affect the operation of LANs. The operation of existing LANs would not be affected by the bridges at all. The algorithm used by the transparent bridges is called backward learning.
Q9. Explain the Spanning Tree Bridges
Ans: To prevent networking loops in networking, multiple bridges are used. The bridges communicate with each other and establish a map of the network to derive what is called a spanning tree for all the networks. A spanning tree consists of a single path between source and destination nodes that does not include any loops. Thus, a spanning tree can be considered to be a loop free subset of a network’s topology. The spanning tree algorithm, specified in IEEE 802.1d describes how bridges or switches can communicate to avoid network loops.
Q10. Explain Source Routing Bridges.
Ans: In Source Routing Bridges, the sending machine is responsible for determining whether, a frame is destined for a different network, and then the source machine designates this by setting the high order bit of the group address bit of the source address to 1. It also includes in the frame header the path the frame is to be followed from source to destination. Source routing bridges are based on the assumption that a sending machine will provide routing information for messages destined for different networks. By making the sending machine responsible for this task, a source routing bridge can ignore frames that have not been marked and forward only those frames with their high order destination bit set to 1.
BLOCK 3 UNIT 1
INTRODUCTION TO LAYER FUNCTIONALITY AND DESIGN ISSUES
Q1. What is connection oriented services?
Ans: Connection oriented services define a way of transmitting data between a sender and a receiver, in which an end to end connection is established before sending any data. After establishing a connection, a sequence of packets, can be sent one after another. All the packets belonging to a message are sent from the same connection. When all packets of a message have been delivered, the connection is terminated. In connection oriented services, the devices at both the endpoints use a protocol to establish an end to end connection before sending any data.
Q2. What are the characteristics of connection oriented services?
Ans: The characteristics of connection oriented services are:
1) The network guarantees that all packets will be delivered in order without loss or duplication of data.
2) Only a single path is established for the call, and all the data follows that path.
3) The network guarantees a minimal amount of bandwidth and this bandwidth is reserved for the duration of the call.
4) If the network is over utilized, future call requests are refused.
Q3. Why connection oriented service is called reliable?
Ans: Connection oriented service is called reliable because:
1) It guarantees that data will arrive in the proper sequence.
2) Single connection for entire message facilities acknowledgement process and retransmission of damaged and lost frames.
Q4. What are the different stages of connection oriented transmission?
Ans: The different stages of connection oriented transmission are:
(i) Connection Establishment
In connection oriented services, before transmitting data, the sending devices must first determine the availability of the other to exchange data and a connection must be established by which data can be sent.
(ii) Data Transfer
After the connection gets established, the sender starts sending data packets to the receiver.
(iii) Connection Termination
After all the data get transferred, the connection has to be terminated. Connection termination also requires a three way handshake.
Q5. Explain the concept of Connection Less Services.
Ans: Connection less service is a service that allows the transfer of information among subscribers without the need for end to end connection establishment procedures. Connection less services define a way of communication between two network end points in which a message can be sent from one end point to another without prior arrangement. The sender simply starts sending packets, addressed to the intended recipient.
Q6. What is datagram subnet?
Ans: In connectionless service, packets are individual injected into the subnet and their routing decisions are not dependent on each other. Therefore, no advance setup is needed. In this context, the packets are frequently called datagrams and the subnet is called a datagram subnet.
Q7. Compare and contrast between virtual circuit and datagram subnet.
Ans: Comparison between virtual circuit and datagram subnet are:
(a) Address Machine
Each datagram in datagram subnet contains the full source and destination address. But in virtual circuit subnet, each datagram contains a small VC number.
(b) Referencing of Circuit Setup
Datagram subnet does not require referencing of circuit setup. But virtual circuit subnet requires referencing of circuit setup.
(c) State information by a router
In datagram subnet, routers do not hold state information about connection. But in virtual circuit subnet, each VC requires router table space per connection.
(d) Routing Procedure
In datagram subnet, each datagram is routed independently. But in virtual circuit subnet, route is selected when VC is set up and all the packets follow all routes.
(e) Effect of Router Failures
In datagram subnet, there is no effect of router failure except the datagram lost during the crash. But in virtual circuit subnet, all VCs that passed through the failed router are terminated and the new virtual circuit is established.
(f) Quality of Service
Datagram subnet is difficult while virtual circuit subnet is easy.
(g) Congestion Control Mechanism
Datagram subnet is difficult while virtual circuit subnet is easy.
Q8. Explain the different types of network address.
Ans: The different types of network address are explained below:
(a) Data Link Layer Addresses
Data link layer addresses sometimes are referred to as physical or hardware addresses, uniquely identify each physical network connection of a network device. Usually data link addresses have a pre-established and fixed relationship to a specific device.
(b) Media Access Control (MAC) Addresses
Media Access Control (MAC) addresses are used to identify network entities in LANs that implement the IEEE MAC addresses of the data link layer. These addresses are 48 bits in length and are expressed as 12 hexadecimal digits.
(c) Network Layer Addresses
Network addresses are sometimes called virtual or logical addresses. These addresses are used to identify an entity at the network layer of the OSI model. Network addresses are usually hierarchical addresses.
Q9. Differentiate between Hierarchical address and Flat Address.
Ans: Hierarchical addresses are organized into a number of subgroups, each successively narrowing an address until it points to a single device as a house address.
A flat address space is organized into a single group such as your enrolment no. Hierarchical addressing offers certain advantages over flat addressing schemes. In hierarchical addressing, address sorting and recalling is simplified using the comparison operation.
Q10. Differentiate between static and dynamic address.
Ans: The different between static and dynamic address are:
(a) Static Address Assignment
Static addresses are assigned by a network administrator according to a preconceived inter-network addressing plan. A static address does not change until the network administrator changes it manually.
(b) Dynamic Addresses
Dynamic addresses are obtained by devices when they are attached to a network, by means of some protocol specific process. A device using dynamic address often has a different address each time it connects to the network.
Q11. What is IP Address?
Ans: IP Address is a unique address i.e. no two machines on the internet can have same IP address. It encodes its network number and host number. Every host and router, in a inter-network has an IP Address. The format of an IP address is 32 bit numeric address written as four numbers separated by periods. Each number can be zero to 255. For example, 1.160.10.240 could be an IP address. These numbers defines three fields:
(a) Class Type
Indicate the IP class, to which the packing belongs.
(b) Network Identifier (NETID)
Network Identifiers indicates the network (a group of computers). Networks with different identifiers can be interconnected with routers.
(c) Host Identifier (HOSTID)
Host identifier indicates a specific computer on the network.
Q12. Explain the concept of congestion.
Ans: In the network layer, when the number of packets sent to the network is greater than the number of packets the network can handle, a problem occurs that is known as congestion. Congestion occurs on shared networks when multiple users contend for access to the same resources.
Q13. What is congestion control? What are the techniques used to manage congestion?
Ans: Congestion control refers to the network mechanism and techniques used to control congestion and keep the load below the network capacity.
Some of the basic techniques to manage congestion are:
(a) End System Flow Control
This is not a congestion control scheme. It is a way of preventing the sender from overrunning the buffers of the receiver.
(b) Network Congestion Control
In this scheme, end systems throttle back in order to avoid congesting the network. The mechanism is similar to end to end flow controls, but the intention is to reduce congestion in the network not at the receiver’s end.
(c) Network Based Congestion Avoidance
In this scheme, a router detects that congestion may occur and attempts to slow down senders before queues become full.
(d) Resource Allocation
This technique involves scheduling the use of physical circuit or other resources, perhaps for a specific period of time. A virtual circuit built across a series of switches with a guaranteed bandwidth is a form of resource allocation.
Q14. What is routing? What are the various activities performed by a router?
Ans: Routing is the act of moving data packets in packet switched network, from source to a destination. A router is a device used to handle the job of routing. It identifies the address on data passing through it, determines which route the transmission should take and collects data in packets which are hen sent to their destinations.
Q15. Differentiate between adaptive routing and non-adaptive routing.
Ans: In adaptive routing, routing decisions are taken on each packet separately i.e. for the packets belonging to the same destination, the router may select a new route for each packet.
In non-adaptive routing, routing decisions are not taken again and again i.e. once the router decides a route to the destination. It sends all packets fro that destination on the same route.
BLOCK 3 UNIT 2
ROUTING ALGORITHMS
Q1. Explain the Dijkstra Algorithm.
Ans: Dijkstra algorithm is the algorithm for finding the shortest path routing. In this algorithm, each node has a label which represents its distance from the source node along the best known path. On the basis of these labels, the algorithm divides the node into two sets i.e. tentative and permanent. The algorithm works in the following manner:
i) First mark source node as current node (T node).
ii) Find all the neighbors of the T-node and make them tentative.
iii) Now examine all these neighbors.
iv) Then among all these node, label one node as permanent i.e. node with the lowest weight would be labeled as permanent and mark it as the t-node.
v) If, the destination node is reached or tentative list is empty then stop else go to step 2.
Q2. What is Bellman Ford Algorithm.
Ans: The Bellman Ford algorithm can be states find the shortest paths from given source node subject keeping in mind the constraint that the paths contain at most one link; then, find the shortest paths, keeping in mind a constraint of paths of at most two links and so on. This algorithm also proceeds in stages.
Q3. What is flooding?
Ans: This is also a very simple method of broadcasting. In this method every incoming packet is sent out on every outgoing line except the line from which it arrived. This algorithm is very simple to implement. But the major advantage of this algorithm is that it generates lots of redundant packets, thus consumes too much bandwidth.
Q4. What is reverse path forwarding?
Ans: Reverse path forwarding is a method of broadcasting. In this method, when a broadcast packet arrives at a router, the router checks whether the packet has arrived on the line that is normally used for sending packets to the source of the broadcast or not. If the packet has arrived on the line that is normally used for sending packets to the source of the broadcast then the router forwards copies of it onto all lines except the one it has arrived on. The router discards the packet as a like duplicate.
Q5. Define pruning.
Ans: In multicasting, pruning is the task of removing all lines from the spanning tree of the router that do not lead to hosts that are members of a particular group.
BLOCK 3 UNIT 3
CONGESTION CONTROL IN PUBLIC SWITCHED NETWORK
Q1. Differentiate between congestion control and flow control.
Ans: The differences between congestion control and flow control are:
Congestion Control
i) Congestion control is needed when buffers in packet switches overflow or cause congestion.
ii) Congestion is end to end, it includes all hosts links and routers. It is global issue.
iii) If viewed as a congested buffer then the switched tells the source of the data stream to slow down using congestion control modifications. When output buffers at a switch fill up and packets are dropped this lead to congestion control actions.
Flow Control
i) Flow control is needed when; the buffers at the receiver are not depleted as fast as the data arrives.
ii) Flow is between one data sender and one receiver. It can be done on link to link or end to end basis. It is a local issue.
iii) If there is a queue on the input side of a switch, and link by link flow control is used, then as a flow control action the switch tells its immediate neighbor to slow down if the input queue fills up.
Q2. Differentiate between Leaky Bucket and Token Bucket.
Ans: The different between Leaky Bucket and Token Bucket are:
Leaky Bucket
i) Leaky bucket discards packets.
ii) With LB, a packet can be transmitted if the bucket is not full.
iii) LB sends the packets at an average rate.
iv) LB does not allow saving, a constant rate is maintained.
Token Bucket
i) Token bucket discard tokens.
ii) With TB, a packet can only be transmitted if there are enough tokens to cover its length in bytes.
iii) TB allows for large bursts to be sent faster by speeding up the output.
iv) TB allows saving up of tokens to send large bursts.