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Chapter 7

Data Communication and Computer Network (old course)

TU BIM past exam questions

Past questions and answers

286 questions set from this chapter, 3 of them more than once; 1 is most repeated (set, or a close variant set, in 3 or more exams). Most repeated first.

  • Most repeated · 3 of 21 exams
  • 2019 AD · 3 marks

Compare Datagram approach and Virtual Circuit approach of Switching.

Similar questions: Why Switching is necessary? Compare between Datagram (2012 AD) · Explain virtual circuit switching approach? How is (2010 AD)

Answer

BasisDatagramVirtual circuit
Connection set-upNot neededNeeded before data transfer
RoutingEach packet routed independentlyRoute chosen once, at set-up, then fixed for all packets
Packet addressFull destination address in every packetShort virtual circuit number
Order of deliveryMay arrive out of order or duplicatedIn order
Failure of a nodePackets re-routed easilyAll circuits through that node are lost
Reliability and congestionHandled by the end hosts, with no guaranteeEasier to control; resources can be reserved
ExampleIP (Internet)X.25, Frame Relay, ATM

Answer checked by Claude Sonnet · Oct 21, 2026

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  • Asked 2 times
  • 2002 AD · 1 mark
  • 2010 AD · 1 mark

What is Piggy backing?

Answer

Piggybacking is the technique in two-way (full duplex) communication where an acknowledgement for a received frame is not sent as a separate frame but is attached (carried in the header) to the next outgoing data frame going the opposite way. It saves bandwidth and reduces the number of frames sent.

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  • Asked 2 times
  • 2005 AD · 3 marks
  • 2010 AD · 3 marks

You would like to send an e-mail to your friend from a computer at your residence. Draw and explain one of the possible models of a communication system that can support your requirement.

Answer

A home computer typically connects to the Internet through a modem or router provided by an ISP, and the mail is delivered through mail servers.

Sender PC --(SMTP)--> Sender's mail server (via ISP) --(SMTP over Internet)--> Receiver's mail server --(POP3/IMAP)--> Friend's PC

Components:

  • User agent (mail client): the program where I compose the message.
  • Access network and ISP: the modem/router and telephone, cable or fibre line link the home computer to the ISP and then to the Internet.
  • SMTP server: the sender's mail server accepts the mail with SMTP (port 25) and relays it. DNS finds the destination mail server using the MX record.
  • Internet backbone: routers forward the TCP/IP packets to the destination server.
  • Receiver's mailbox server: stores the mail until my friend retrieves it with POP3 or IMAP.

This is the basic model: source, transmitter, transmission system, receiver and destination.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • Asked 2 times
  • 2009 AD · 1 mark
  • 2013 AD · 1 mark

Define routing protocol.

Answer

A routing protocol is a protocol that routers use to exchange information about network topology and reachability, so that they can build and update routing tables and choose the best path for packets. Examples are RIP, OSPF and BGP.

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  • 2002 AD · 3 marks

Encode the following bit stream 100101 with
i) Manchester Encoding\nii) ASK

Similar questions: Encode the following bit stream 0001110101 with (2010 AD)

Answer

Bit stream: 1 0 0 1 0 1

(i) Manchester encoding (IEEE 802.3 convention: a 1 is a low-to-high transition in the middle of the bit period, a 0 is a high-to-low transition):

bit:    1     0     0     1     0     1
level: _|-   -|_   -|_   _|-   -|_   _|-

Every bit has a mid-bit transition, which gives clocking; the first half of each bit is the opposite level to the second half.

(ii) ASK (Amplitude Shift Keying): the carrier is sent at high amplitude for bit 1 and at zero (or low) amplitude for bit 0, frequency and phase unchanged:

bit:   1     0     0     1     0     1
wave: /\/\/ ----- ----- /\/\/ ----- /\/\/
      (carrier present for 1, absent for 0)

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  • 2010 AD · 1 mark

List down the various advantages of analog transmission over digital transmission.

Similar questions: List down the various advantages of digital (2013 AD)

Answer

Advantages of analog over digital transmission:

  • It needs less bandwidth for the same information (for example voice), because no encoding to bits is required.
  • It is simpler and cheaper for continuous signals such as voice and audio, since no analog-to-digital conversion is needed.
  • It has no quantization error, so the signal is represented continuously.
  • The existing analog infrastructure (telephone lines, radio and TV) can be used directly.

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  • 2010 AD · 3 marks

Encode the following bit stream 0001110101 with
i) Manchester Encoding\nii) ASK\niii) NRZ-I

Similar questions: Encode the following bit stream 100101 with (2002 AD)

Answer

Bit stream: 0 0 0 1 1 1 0 1 0 1

(i) Manchester encoding (convention: 1 = low-to-high transition in mid-bit, 0 = high-to-low). Each bit becomes two half-bit levels, H = high and L = low:

Bit0001110101
LevelsHLHLHLLHLHLHHLLHHLLH

There is always a transition in the middle of each bit period.

(ii) ASK (amplitude shift keying): the carrier is sent at high amplitude for 1 and at zero (or low) amplitude for 0. The signal is:

0 0 0 carrier off, 1 1 1 carrier on, 0 off, 1 on, 0 off, 1 on.

(iii) NRZ-I: the level inverts for each 1 and stays the same for each 0. Starting from low:

Bit0001110101
LevelLLLHLHHLLH

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  • 2010 AD · 3 marks

Differentiate GO-Back-N and selective repeat ARQ.

Similar questions: Differentiate GO-Back N and selective repeat ARQ (2013 AD)

Answer

Go-Back-N ARQSelective Repeat ARQ
When a frame is lost or damaged, that frame and all later frames are retransmittedOnly the lost or damaged frame is retransmitted
Receiver window size is 1, so out-of-order frames are discardedReceiver window is greater than 1, so out-of-order frames are buffered
No receiver buffer needed for reorderingBuffer and sorting logic needed
Sender window ≤2m−1\le 2^m-1Sender window ≤2m−1\le 2^{m-1}
Simple, but wastes bandwidth on noisy linksMore complex, but efficient
Cumulative ACK or NAKIndividual ACK or NAK for each frame

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  • 2010 AD · 3 marks

Explain virtual circuit switching approach? How is it different from datagram approach?

Similar questions: Compare Datagram approach and Virtual Circuit approach (2019 AD)

Answer

Virtual circuit switching is a packet-switching method that is connection-oriented. A logical path is first set up between source and destination, and every packet follows it.

  1. Setup: a call request establishes the route, and each switch records a virtual circuit identifier (VCI) in its table.
  2. Data transfer: each packet carries only the short VCI, and switches forward it using the table without a routing decision.
  3. Teardown: the circuit is released after the transfer.

Packets arrive in order, and resources can be reserved. The path is not dedicated, since the links are still shared.

Difference from datagram:

Virtual circuitDatagram
Connection-oriented, with setupConnectionless, no setup
All packets follow the same routeEach packet routed independently
Packets arrive in orderMay arrive out of order
Short VCI in headerFull destination address in each header
A switch failure breaks the circuitPackets can be rerouted around failures
Easier QoS and congestion controlHarder

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  • 2012 AD · 3 marks

Why Switching is necessary? Compare between Datagram approach and Virtual circuit approach.

Similar questions: Compare Datagram approach and Virtual Circuit approach (2019 AD)

Answer

Necessity of switching. Connecting every pair of devices by a direct link needs n(n−1)/2n(n-1)/2 links, which is impractical and costly. Switching lets many devices share a smaller set of links: the network connects a sender to a receiver only when needed, making use of the links efficient, scalable and economical.

Datagram vs Virtual circuit

DatagramVirtual circuit
Connectionless; no setup phaseConnection-oriented; a call setup phase is needed
Each packet carries full destination address and is routed independentlyPackets carry only a short VC number; all follow the same path
Packets may arrive out of order or be lost; reordering is done by the end hostPackets arrive in order
Routers keep no per-connection stateRouters keep a table entry per virtual circuit
Robust: a failed node is bypassed by new routesA node failure breaks all circuits through it
Congestion control is hardResources can be reserved, so quality of service is easier
Example: IPExample: X.25, Frame Relay, ATM

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  • 2013 AD · 1 mark

List down the various advantages of digital transmission over analog transmission.

Similar questions: List down the various advantages of analog (2010 AD)

Answer

Advantages of digital over analog transmission:

  • Better noise immunity; signals can be regenerated by repeaters instead of amplified, so noise does not accumulate.
  • Easy error detection and correction by coding.
  • Allows time-division multiplexing and efficient sharing of links.
  • Voice, data and video can be carried in a common form and integrated.
  • Easier encryption and security.
  • Digital hardware is cheaper, smaller and more reliable (VLSI).
  • Easy storage and processing, with higher quality over long distances.

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  • 2013 AD · 3 marks

Differentiate GO-Back N and selective repeat ARQ.

Similar questions: Differentiate GO-Back-N and selective repeat ARQ (2010 AD)

Answer

Go-Back-N ARQSelective Repeat ARQ
On an error, the erroneous frame and all frames after it are retransmittedOnly the erroneous (or lost) frame is retransmitted
Receiver window size is 1; out-of-order frames are discardedReceiver window is greater than 1; out-of-order frames are buffered
Sender window up to 2m−12^m - 1Sender window up to 2m−12^{m-1}
Uses cumulative acknowledgementUses individual acknowledgement (and NAK)
Simple; one timer for the oldest frameMore complex; a timer per frame and sorting logic at the receiver
Wastes bandwidth on noisy linksMore bandwidth efficient
Receiver needs little bufferingReceiver needs larger buffer

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  • 2015 AD · 3 marks

Explain distance vector routing with its working principle.

Similar questions: Explain the working principle of Distance vector (2017 AD)

Answer

In distance vector routing each router maintains a table (vector) giving the least known cost and next hop to every destination.

Working:

  1. At start, each router knows only the cost to its directly connected neighbours; other costs are infinity.
  2. Periodically (and on change), each router sends its whole table to its immediate neighbours.
  3. A router receiving a neighbour's vector computes, for each destination yy,

Dx(y)=min⁡v{c(x,v)+Dv(y)}D_x(y)=\min_v\{c(x,v)+D_v(y)\}

and updates its table if a cheaper route is found, recording vv as the next hop. 4. The process repeats until no table changes (convergence).

Example: if A reaches B at cost 2 and B advertises C at cost 3, A learns C at cost 5 via B.

It is simple, used in RIP, but converges slowly and suffers from count-to-infinity, reduced by split horizon and poison reverse.

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  • 2017 AD · 3 marks

Explain the working principle of Distance vector routing algorithm with an example.

Similar questions: Explain distance vector routing with its working (2015 AD)

Answer

In distance vector routing each router keeps a table (vector) giving the best-known distance to every destination and the next hop to use. It is based on the Bellman-Ford algorithm.

Working:

  1. Each router initialises its table with distance 0 to itself and the cost of directly connected neighbours.
  2. Periodically, each router sends its whole table to its immediate neighbours only.
  3. When a router receives a neighbour's table, for each destination it computes: new distance = cost to neighbour + neighbour's distance to destination. If this is smaller than its current value, it updates the entry and the next hop.
  4. The process repeats until no table changes (convergence).

Example. Take routers A, B, C in a line, with A-B cost 2 and B-C cost 3. At first, A knows B=2 and C=infinity. B sends A its table in which C=3. A calculates C = 2 + 3 = 5 via B, which is better than infinity, and updates its table to C=5, next hop B.

Drawbacks: slow convergence and the count-to-infinity problem (reduced by split horizon and poison reverse). Example protocol: RIP.

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  • 2018 AD · 3 marks

Explain the access control protocol used in Ethernet based LAN Network.

Similar questions: Explain Access Control Protocol used in Ethernet (2019 AD)

Answer

Ethernet LANs use CSMA/CD (Carrier Sense Multiple Access with Collision Detection) to control access to the shared medium.

  1. Carrier sense: A station that has a frame to send first listens to the channel. If the channel is busy, it waits.
  2. Multiple access: If the channel is idle, the station transmits at once. Many stations share the same medium.
  3. Collision detection: While transmitting, the station keeps monitoring the channel. If it detects a collision (two stations transmitted at the same time), it stops sending and transmits a short jam signal so that all stations know.
  4. Backoff: After the collision, each station waits a random time chosen by the binary exponential backoff algorithm, then tries again. After 16 failed attempts the frame is dropped.

The minimum frame size ensures the sender is still transmitting when a collision signal returns, so collisions can be detected.

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  • 2019 AD · 5 marks

Explain Access Control Protocol used in Ethernet based & wireless LAN.

Similar questions: Explain the access control protocol used in (2018 AD)

Answer

Ethernet: CSMA/CD

Ethernet shares a medium, so it uses CSMA/CD.

  1. The station senses the carrier. If the channel is idle, it transmits; if busy, it waits.
  2. While sending, it keeps listening to detect a collision.
  3. On a collision it stops, sends a jam signal, and waits a random time using binary exponential backoff. It then retries, up to 16 attempts.

Detecting collisions works because the cable is short and the signal is strong enough to be sensed by the sender.

Wireless LAN (IEEE 802.11): CSMA/CA

Wireless stations cannot detect collisions while transmitting (a station cannot hear while it sends, and hidden stations exist). Wi-Fi therefore avoids collisions using the DCF mode.

  1. The station senses the channel. If it is idle for a DIFS period, it transmits. Otherwise it defers and counts down a random backoff timer, which freezes whenever the channel is busy.
  2. After receiving a frame correctly, the receiver sends an ACK after a SIFS. If no ACK arrives, the sender assumes a collision, doubles its contention window and retries.
  3. Optionally, RTS/CTS is used to solve the hidden-terminal problem. The sender sends an RTS, the receiver replies with a CTS, and other stations hearing either one set their NAV (virtual carrier sense) and stay silent for the transmission time.

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  • 2001 AD · 5 marks

What is switching?

Answer

Switching is the technique by which a network moves data from an input port to the correct output port so that two end systems can communicate without a direct link between every pair. Instead of a dedicated line between every pair of devices, intermediate nodes (switches) forward the data.

Types of switching:

  • Circuit switching: a dedicated path is set up between sender and receiver for the whole call (set-up, data transfer, tear-down), as in the telephone network. It gives fixed delay and no congestion once connected, but wastes bandwidth when the line is idle.
  • Message switching: the whole message is stored at each node and forwarded when the next link is free (store and forward). There is no set-up, but delays are long and nodes need large buffers.
  • Packet switching: the message is divided into small packets, each routed independently (datagram) or along a virtual circuit, and reassembled at the destination. It uses bandwidth efficiently and is used on the Internet.

Switching therefore provides efficient sharing of links, routing and connection management in large networks.

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  • 2001 AD · 5 marks

What is piggybacking? Write down the merits and demerits of it.

Answer

Piggybacking is a technique in which an acknowledgement for a received frame is not sent as a separate frame but is attached (carried in the ACK field) to the next outgoing data frame going in the opposite direction. It is used in bidirectional data link protocols such as sliding window.

If no data frame is ready within a short timeout, a separate ACK frame is sent.

Merits:

  • Better use of bandwidth, since no separate ACK frame is needed.
  • Fewer frames are sent, so there is less overhead and lower network traffic.
  • Fewer frame arrivals mean fewer interrupts and less processing at the receiver.
  • The ACK field in the header is only a few bits, so the extra cost is tiny.

Demerits:

  • The acknowledgement is delayed until a data frame is available, which may cause the sender's timer to expire and trigger needless retransmission.
  • The data link layer must wait for a data frame and decide how long to wait, which makes the protocol more complex.
  • It works only for two-way (full duplex) traffic; it gives no benefit for one-way flow.

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  • 2001 AD · 5 marks

What is framing? Write down significant of it.

Answer

Framing is the function of the data link layer by which the bit stream received from the network layer is divided into discrete units called frames. Each frame carries a header (addresses, control information) and a trailer (error-detecting code such as CRC) around the data.

Framing methods: character count, flag bytes with byte stuffing, start/end flags with bit stuffing, and physical-layer coding violations.

Significance:

  • It provides frame boundaries so the receiver knows where a frame begins and ends.
  • It allows error detection and correction on a per-frame basis, so only a damaged frame needs to be retransmitted instead of the whole stream.
  • It enables flow control and acknowledgement through frame sequence numbers.
  • Source and destination (MAC) addresses in the header make delivery between nodes on the same link possible and let a shared medium be used.
  • It provides synchronisation between sender and receiver.

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  • 2001 AD · 5 marks

What are services, interface and protocol?

Answer

In a layered network architecture:

Service: the set of operations (primitives) that a layer provides to the layer above it. It defines what the layer does, not how. Services may be connection-oriented or connectionless, e.g. the transport layer offers reliable delivery to the application layer.

Interface: the boundary between two adjacent layers on the same machine. It defines the primitives and the services the lower layer offers to the upper layer, and how the upper layer accesses them (through service access points). A clean interface lets one layer be replaced without affecting the others.

Protocol: the set of rules and conventions for communication between peer layers (same layer) on different machines, covering the format, order and meaning of the messages exchanged, such as TCP, IP and HTTP. Protocols are used to implement the services of a layer.

In short, a layer uses a protocol to talk to its peer on another machine, and provides a service to the layer above through an interface.

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  • 2001 AD · 5 marks

What is topology? Discuss the various types of it.

Answer

Network topology is the geometric arrangement of the nodes and links of a network, i.e. how devices are physically (or logically) connected.

Types:

  1. Bus: all nodes share a single backbone cable with terminators at both ends. It is cheap and easy to install, but a break in the backbone stops the network, performance degrades with traffic, and faults are hard to isolate.
  2. Star: each node connects to a central hub or switch. It is easy to add or remove nodes and a failed node does not affect others, but the central device is a single point of failure and more cable is needed.
  3. Ring: each node connects to exactly two neighbours forming a closed loop, and data travels in one direction (often with token passing). It gives orderly access and equal opportunity, but one node or cable failure can break the ring (a dual ring helps).
  4. Mesh: every node connects to every other (full mesh needs n(n−1)/2n(n-1)/2 links). It is very reliable and has dedicated links, but cabling and ports are expensive and complex.
  5. Tree (hierarchical): star networks arranged in a hierarchy under a root node. It is scalable and easy to manage, but depends on the root and backbone.
  6. Hybrid: a combination of two or more topologies to suit the needs of an organisation.

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  • 2001 AD · 5 marks

What is multiplexing?

Answer

Multiplexing is the technique of combining several signals (data streams) from different sources so that they share one physical link or channel at the same time, and separating them again at the receiver. It makes efficient use of the capacity (bandwidth) of an expensive high-capacity link and reduces cabling cost.

At the sending end a multiplexer (MUX) merges the nn input lines into one high-capacity link; at the receiving end a demultiplexer (DEMUX) splits the stream back into the nn original outputs.

Main types:

  • Frequency Division Multiplexing (FDM): the link bandwidth is divided into frequency bands, each carrying one signal at the same time (used in radio/TV broadcasting, analog telephone trunks). Guard bands separate channels.
  • Time Division Multiplexing (TDM): each source gets the whole bandwidth for a short time slot in turn. It can be synchronous (fixed slots) or statistical/asynchronous (slots given only to active sources).
  • Wavelength Division Multiplexing (WDM): FDM applied to light; different wavelengths share one optical fibre.
  • Code Division Multiplexing (CDM): all users transmit at the same time and frequency, separated by unique spreading codes.

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  • 2001 AD · 5 marks

Write notes on: Error Control and Flow Control

Answer

Error control is the set of methods used by the data link layer to detect and correct errors in frames caused by noise on the channel.

  • Error detection: extra bits are added to each frame, such as parity bit, checksum or CRC (Cyclic Redundancy Check).
  • Error correction: either forward error correction (e.g. Hamming code, which lets the receiver correct the error itself) or retransmission (ARQ).
  • ARQ (Automatic Repeat reQuest): the receiver sends an ACK for a good frame and a NAK (or stays silent so the sender's timer expires) for a bad one, and the sender retransmits. Variants are Stop-and-Wait, Go-Back-N and Selective Repeat ARQ.

Flow control is the mechanism that keeps a fast sender from overwhelming a slow receiver, so that the receiver's buffer does not overflow.

  • Stop-and-Wait: the sender sends one frame and waits for its ACK before sending the next. It is simple but wastes link capacity on long-delay links.
  • Sliding window: the sender may send up to WW frames without waiting for an ACK; the window slides forward as ACKs arrive. Frames are numbered with sequence numbers. It is used by Go-Back-N and Selective Repeat and gives much better link utilisation.

Error control and flow control are normally implemented together in the same protocol (e.g. HDLC), using ACKs, sequence numbers and timers.

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  • 2001 AD · 5 marks

Compare between OSI and TCP/IP.

Answer

BasisOSI modelTCP/IP model
Layers7 (physical, data link, network, transport, session, presentation, application)4 (network access/host-to-network, internet, transport, application)
OriginDeveloped by ISO as a theoretical reference modelDeveloped by DARPA; built from protocols that already worked
ApproachProtocols were designed after the model (model first)Model describes protocols that already existed (protocols first)
Session and presentationSeparate layersMerged into the application layer
Network layer serviceBoth connection-oriented and connectionlessConnectionless only (IP)
Transport layer serviceConnection-oriented onlyBoth: TCP (connection-oriented) and UDP (connectionless)
GeneralityProtocol independent, clear separation of service, interface and protocolProtocol specific, less clear separation
UseMainly a teaching/reference toolThe basis of the Internet in practice

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  • 2002 AD · 3 marks

Describe the communication model along with block diagram.

Answer

A simple communication model has five elements: source, transmitter, transmission system, receiver and destination.

Source -> Transmitter -> Transmission system -> Receiver -> Destination
(data g(t))  (signal s(t))   (noise added)       (signal r(t))   (data g'(t))
  • Source: generates the data to be sent (e.g. a PC or telephone).
  • Transmitter: converts the data into a signal suitable for the transmission medium (e.g. a modem converts digital data to analog signal).
  • Transmission system: the medium (wire, fibre, air) or a whole network between source and destination. Noise and attenuation affect the signal here.
  • Receiver: accepts the received signal and converts it back to data.
  • Destination: takes the incoming data from the receiver and uses it.

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  • 2002 AD · 1 mark

What is the difference between frequency and bandwidth?

Answer

Frequency is the number of cycles of a signal per second, measured in hertz (Hz); it describes a single signal (f=1/Tf = 1/T). Bandwidth is the range of frequencies a signal contains or a channel can pass, i.e. the difference between the highest and lowest frequency (B=fmax−fminB = f_{max} - f_{min}).

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  • 2002 AD · 1 mark

Differentiate between data and signal.

Answer

Data is the entity (information) that conveys meaning, such as text, numbers or voice, and can be analog or digital. A signal is the electrical or electromagnetic representation of data that is actually transmitted over the medium. Data is what is to be sent; signal is how it travels.

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  • 2002 AD · 3 marks

Make OSI/ISO reference model and write down the functions of network layer.

Answer

OSI reference model (7 layers, top to bottom):

7 Application
6 Presentation
5 Session
4 Transport
3 Network
2 Data link
1 Physical

Functions of the network layer:

  • Routing: selects the best path for packets from source to destination across one or more networks (routing algorithms and tables).
  • Logical addressing: gives each host a unique network address (e.g. IP address) so packets can be identified end to end.
  • Packetising and forwarding: carries packets from node to node, with fragmentation and reassembly where link sizes differ.
  • Congestion control: avoids too many packets entering part of the subnet.
  • Internetworking: connects different networks together (gateways/routers).

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  • 2002 AD · 1 mark

What is collision? Define CSMA/CD. What part of 802 project uses CSMA/CD?

Answer

A collision occurs when two or more stations transmit on a shared medium at the same time, so their signals overlap and the frames are corrupted.

CSMA/CD (Carrier Sense Multiple Access with Collision Detection): a station first listens to the channel (carrier sense); if idle it transmits, if busy it waits. While transmitting it keeps monitoring the channel; if it detects a collision it stops, sends a short jam signal, waits a random backoff time (binary exponential backoff) and tries again.

IEEE 802.3 (Ethernet) uses CSMA/CD.

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  • 2002 AD · 2 marks

How 802.4 differ from 802.5?

Answer

BasisIEEE 802.4IEEE 802.5
NameToken BusToken Ring
TopologyPhysical bus/tree, logical ringPhysical and logical ring
MediumCoaxial cable (broadband or carrierband)Twisted pair/shielded cable
TokenToken passed to the next station in the logical ring by addressToken circulates around the physical ring
Station orderLogical order (by address) differs from physical orderSame as physical order
ComplexityMore complex (ring maintenance, joining/leaving, token loss handling)Simpler, uses a monitor station
Frame priorityFour priority classesPriority and reservation bits

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  • 2002 AD · 2 marks

A communication system 'B' receives a transmitted message by another system 'A', which is 11000110111 as generator to calculates FCS, is the received message correct or incorrect and why?

Answer

The question does not state the generator pattern clearly, so I take the common generator G(x)=x4+x+1G(x) = x^4 + x + 1 (bits 10011). The receiver divides the received message by the generator using modulo-2 division: if the remainder (syndrome) is zero the message is accepted as correct, otherwise it contains an error.

Dividing 1100011011111000110111 by 1001110011 gives remainder 11111111.

Because the remainder is not zero, the received message is incorrect: it was corrupted during transmission and the FCS check fails.

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  • 2002 AD · 3 marks

What is framing? What happens when an ACK frame is lost in G-Back-n ARQ?

Answer

Framing is the data link layer function of splitting the stream of bits from the network layer into manageable units called frames, with a header and trailer so the receiver can tell where each frame starts and ends. Methods include character count, flag bytes with byte stuffing, flag bits with bit stuffing, and physical layer coding violations.

Lost ACK in Go-Back-N ARQ: the sender keeps sending frames within its window. The receiver accepts frames in order only. If an ACK is lost, the sender's timer for that frame expires before the ACK arrives. However, ACKs are cumulative, so if a later ACK arrives before the timeout it also acknowledges the earlier frames and nothing needs to be resent. If no later ACK arrives, the timer expires and the sender goes back and retransmits that frame and all frames sent after it. The receiver discards the duplicates (it already has them) and re-acknowledges.

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  • 2002 AD · 3 marks

If all hosts in a class-based network begins with 192.168.0.0 write IP address for followings
i) Network address\nii) Host no 5\niii) This computer\niv) Broadcasting in network 192.168.3.0\nv) Loop back through IP layer\nvi) Class type of this network

Answer

For network 192.168.0.0 (class C):

  • (i) Network address: 192.168.0.0
  • (ii) Host no 5: 192.168.0.5
  • (iii) This computer: 0.0.0.0
  • (iv) Broadcast in network 192.168.3.0: 192.168.3.255 (host part all 1s)
  • (v) Loopback through the IP layer: 127.0.0.1 (any address 127.x.x.x)
  • (vi) Class type: Class C, because the first octet 192 lies in 192-223.

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  • 2002 AD · 3 marks

A TV channel has the signal with bandwidth 6.1 MHz and it is transmitted with power of 2 W. This signal has to travel through a channel, which includes an average noise of 0.1 mW. If this TV signal is digitized and sent, find out the maximum data rate of the transmitted signal.

Answer

Use the Shannon capacity formula for a noisy channel:

C=Blog⁡2(1+SN)C = B\log_2\left(1 + \frac{S}{N}\right)

Given B=6.1 MHzB = 6.1\text{ MHz}, S=2 WS = 2\text{ W}, N=0.1 mW=10−4 WN = 0.1\text{ mW} = 10^{-4}\text{ W}.

SN=210−4=20000\frac{S}{N} = \frac{2}{10^{-4}} = 20000

C=6.1×106×log⁡2(20001)=6.1×106×14.29≈87.2×106 bpsC = 6.1\times10^{6}\times\log_2(20001) = 6.1\times10^{6}\times 14.29 \approx 87.2\times10^{6}\text{ bps}

The maximum data rate is about 87 Mbps.

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  • 2002 AD · 3 marks

What happens when token is lost in token bus?

Answer

In a token bus (IEEE 802.4), the token is passed around a logical ring of stations. If the token is lost (for example the station holding it fails, or noise corrupts it), no station can transmit and the network goes silent.

Recovery:

  • Every station keeps a timer. If a station hears no activity for a long time, it concludes the token is lost.
  • It then starts the claim-token process: it sends a CLAIM_TOKEN frame. If several stations do this at once, contention is resolved by address-based arbitration (the highest address wins, using timeslots).
  • The winner generates a new token and re-initialises the logical ring, so normal operation resumes.
  • If the next station in the ring has failed, the sender fails to get a response after retries and uses a "who follows" frame to find the next live station and removes the failed one from the ring.

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  • 2002 AD · 3 marks

Differentiate between Virtual circuit and Datagram.

Answer

BasisVirtual circuitDatagram
ConnectionConnection-oriented: a path is set up before data transferConnectionless: no set-up
RouteAll packets follow the same routeEach packet routed independently, may take different routes
Packet headerCarries a short virtual circuit numberCarries full destination address
Order of deliveryPackets arrive in orderMay arrive out of order, need reordering
Router stateRouters keep a table entry for each circuitRouters keep no per-connection state
FailureA failed router breaks all circuits through itOnly packets in transit are lost; others reroute
Congestion/QoSEasier to reserve resources and controlHarder
ExampleX.25, Frame Relay, ATMIP

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  • 2002 AD · 3 marks

Difference between TCP and UDP

Answer

BasisTCPUDP
Full nameTransmission Control ProtocolUser Datagram Protocol
ConnectionConnection-oriented (three-way handshake)Connectionless
ReliabilityReliable: acknowledgements, retransmission, sequencingUnreliable: no guarantee of delivery or order
Flow/congestion controlYes (sliding window)No
Header20 bytes minimum8 bytes
SpeedSlower, more overheadFaster, low overhead
Data unitByte stream (segments)Datagrams, message boundaries kept
UseHTTP, FTP, SMTP, TelnetDNS, SNMP, streaming, VoIP, TFTP

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  • 2002 AD · 3 marks

Find the shortest route between A and G in the network given in the figure of the original paper (nodes A-J).

Answer

The figure with the nodes A-J and link weights is not available here, so the numerical route cannot be read off; the method used is Dijkstra's shortest path algorithm:

  1. Label the source A with distance 0 (permanent) and every other node with infinity (tentative).
  2. From the most recently made permanent node, update the tentative distance of each neighbour as d(v)=min⁡(d(v), d(u)+w(u,v))d(v) = \min\big(d(v),\, d(u) + w(u,v)\big), remembering the predecessor.
  3. Make the tentative node with the smallest distance permanent.
  4. Repeat steps 2 and 3 until G becomes permanent.
  5. Read the route by following the predecessors back from G to A; the label on G is the total length of the shortest route.

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  • 2004 AD · 3 marks

Draw the generalized block diagram of wide area network and identify the major components with their function. How is WAN different from LAN?

Answer

Generalised WAN block diagram:

 LAN / Hosts --- Router --- [ Communication subnet ] --- Router --- LAN / Hosts
                              (switching nodes/routers
                               joined by transmission lines)

Major components:

  • Hosts (end systems): computers/LANs that run user applications and generate data.
  • Switching nodes / routers (IMPs): receive packets and forward them towards the destination along the best path.
  • Transmission lines (links): leased lines, fibre, microwave, satellite or telephone circuits carrying bits between nodes.
  • Communication subnet: the collection of nodes and lines that carries data between hosts.
  • Modems/CSU-DSU: connect hosts or routers to the carrier lines.

WAN vs LAN:

BasisLANWAN
AreaBuilding or campusCity, country or world
SpeedHigh (10 Mbps - Gbps)Lower
Error rateLowHigher
OwnershipSingle organisationUsually carriers / many organisations
CostLowHigh
TopologyBus, ring, starMesh, irregular

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  • 2004 AD · 2 marks

Computer generated data is to be represented by analog signal. How many options do you have? List them and draw the waveforms for the data sequence 101101.

Answer

Digital data can be represented by an analog signal (modulating a carrier) in three ways:

  1. ASK - Amplitude Shift Keying
  2. FSK - Frequency Shift Keying
  3. PSK - Phase Shift Keying

(A fourth, QAM, combines ASK and PSK.)

Waveforms for 1 0 1 1 0 1:

bit:  1      0      1      1      0      1
ASK:  big    none   big    big    none   big      (amplitude changes)
FSK:  high f low f  high f high f low f  high f    (frequency changes)
PSK:  0 deg  180    0 deg  0 deg  180    0 deg     (phase changes)

In ASK the carrier is on for 1 and off for 0; in FSK a higher frequency represents 1 and a lower one 0; in PSK the carrier phase is shifted by 180 degrees for a 0 relative to a 1.

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  • 2004 AD · 1 mark

Calculate the percentage overhead for transmitting ASCII character with one stop bit & even parity.

Answer

An asynchronous ASCII character frame has 1 start bit, 7 data bits, 1 even parity bit and 1 stop bit, i.e. 10 bits in total.

Overhead bits =1+1+1=3= 1 + 1 + 1 = 3.

Overhead=310×100=30%\text{Overhead} = \frac{3}{10}\times100 = 30\%

(of the transmitted bits; equivalently 3/7≈43%3/7 \approx 43\% relative to the 7 data bits).

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  • 2004 AD · 1 mark

What is multiplexing? Compare TDM and FDM.

Answer

Multiplexing is combining several signals so they share a single link, with a multiplexer at the sender and a demultiplexer at the receiver.

BasisFDMTDM
SharingBandwidth divided into frequency bandsTime divided into slots
SimultaneityAll signals sent at the same timeSignals sent one after another in turn
Signal typeMainly analogMainly digital
GuardGuard bands between channelsGuard times / synchronisation bits
ProblemsCrosstalk, intermodulation noiseNeeds synchronisation, timing jitter
ExampleRadio/TV, cable TVDigital telephony (T1/E1), SONET

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  • 2004 AD · 3 marks

Draw the layered architecture of the computer communication protocol reference model developed by ISO and lists the major functions (no explanations needed) performed by each layer.

Answer

ISO OSI reference model:

+---------------------+
| 7 Application       |
| 6 Presentation      |
| 5 Session           |
| 4 Transport         |
| 3 Network           |
| 2 Data link         |
| 1 Physical          |
+---------------------+

Major functions:

  • Physical: transmission of raw bits, voltage levels, cables, connectors, bit synchronisation.
  • Data link: framing, error control, flow control, MAC / physical addressing.
  • Network: routing, logical addressing, congestion control, fragmentation.
  • Transport: end-to-end delivery, segmentation, reliability, flow and error control, port addressing.
  • Session: dialog control, synchronisation, checkpoints, session establishment and termination.
  • Presentation: translation, encryption/decryption, compression, data formatting.
  • Application: user services such as file transfer, e-mail, web access, directory services.

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  • 2004 AD · 2 marks

List the different framing techniques used by the data link layer. Explain bit stuffing with an example.

Answer

Framing techniques:

  1. Character (byte) count
  2. Flag bytes with byte stuffing
  3. Flag bits (start/end flag) with bit stuffing
  4. Physical layer coding violations

Bit stuffing: each frame begins and ends with the flag pattern 01111110. To stop the same pattern appearing inside the data, the sender inserts a 0 after every five consecutive 1s in the data. The receiver removes a 0 that follows five consecutive 1s.

Example: data 0111111011111

  • Scan for five consecutive 1s: after 01 11111 a 0 is stuffed, and after the second run 11111 at the end a 0 is stuffed.
  • Stuffed data: 011111010111110

This contains no 01111110 pattern. At the receiver, every 0 following five consecutive 1s is deleted to recover 0111111011111.

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  • 2004 AD · 3 marks

What do you mean by flow control? Illustrate a situation in which a flow control is required. What flow control techniques are used by DLL? Briefly describe the working.

Answer

Flow control is the mechanism that regulates the rate of data transmission so that a fast sender does not overwhelm a slow receiver, whose buffers would otherwise overflow and frames be lost.

Situation requiring it: a powerful server sending at 100 Mbps to a slow device (such as an old PC or printer) that can process only 1 Mbps; without flow control the receiver's buffer fills and frames are dropped.

Techniques used in the data link layer:

  • Stop-and-Wait: the sender transmits one frame and waits for its ACK before sending the next, so the receiver controls the pace. It is simple but gives poor utilisation when propagation delay is long.
  • Sliding window: the sender may have up to WW unacknowledged frames outstanding. The receiver advertises or implies how many it can accept; each ACK slides the window forward. Frames carry sequence numbers. Go-Back-N and Selective Repeat are sliding window protocols and make better use of the link.

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  • 2004 AD · 3 marks

The IEEE 802.x for LAN divides the DLL of the OSI Reference model into two sub layers. What are these sub layers? Why MAC sub layers is required in LAN? Explain the working of CSMA/CD and compare it with token passing protocol. List the IEEE standards for LAN and their corresponding MAC protocols.

Answer

Sub-layers: the data link layer is divided into the LLC (Logical Link Control) sub-layer (IEEE 802.2), which provides flow and error control and interfaces to the network layer, and the MAC (Medium Access Control) sub-layer, which controls access to the shared medium.

Why MAC is needed: in a LAN, many stations share one broadcast channel. Without rules, simultaneous transmissions would collide, so the MAC sub-layer decides who may transmit and when.

CSMA/CD: a station senses the channel; if idle it transmits, otherwise waits. During transmission it listens, and if it detects a collision it stops, sends a jam signal, and retries after a random binary exponential backoff.

CSMA/CD vs token passing:

BasisCSMA/CDToken passing
AccessContention, randomControlled, deterministic
CollisionsPossibleNone
Performance at heavy loadDegradesRemains good
Delay at light loadLowMust wait for token
PrioritiesNot supportedSupported
ComplexitySimpleToken management needed

IEEE standards and MAC protocols:

StandardNameMAC protocol
802.3EthernetCSMA/CD
802.4Token BusToken passing on logical ring
802.5Token RingToken passing on ring
802.11Wireless LANCSMA/CA

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  • 2004 AD · 2 marks

With reference to the following networking hardware, identify their functions, the OSI layer in which they work.
i. Bridge\nii. Repeater\niii. Hub\niv. Router

Answer

DeviceFunctionOSI layer
BridgeConnects two LAN segments, filters and forwards frames using MAC addresses, reduces collision domainsData link (layer 2)
RepeaterRegenerates and amplifies the signal to extend the cable lengthPhysical (layer 1)
HubMulti-port repeater; broadcasts incoming bits to all other portsPhysical (layer 1)
RouterConnects different networks and forwards packets by logical (IP) address using routing tablesNetwork (layer 3)

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  • 2004 AD · 4 marks

List the characteristics and working of network switching techniques that are employed by the Internet.

Answer

The Internet uses packet switching, in two forms: datagram and virtual circuit. (Circuit switching is used by the telephone network.)

Characteristics of packet switching:

  • Messages are divided into small packets, each with a header (addresses, sequence number).
  • No dedicated path is reserved; links are shared by many users (statistical multiplexing), giving efficient use of bandwidth.
  • Store-and-forward: a node receives a whole packet, checks it and forwards it to the next node.
  • Packets may suffer variable delay, and may be lost or reordered.
  • Robust: packets can be rerouted around failures.

Working:

  • Datagram: each packet is routed independently using the destination address and routing tables; packets may take different paths and are reassembled at the destination (IP).
  • Virtual circuit: a logical path is set up first by a call request; all packets then follow it using a virtual circuit number, and the circuit is released at the end.

Circuit switching, for comparison, sets up a dedicated physical path for the whole call, with fixed delay but wasted capacity when idle.

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  • 2004 AD · 3 marks

A packet arrived at a node in the Internet can take many alternative paths to reach its destination. What technique is used to resolve the problem of identifying a path? List its important desirable characteristics and explain the working of shortest path routing.

Answer

The problem of choosing a path is solved by routing (a routing algorithm running in the network layer, which fills the routing tables).

Desirable characteristics: correctness, simplicity, robustness (copes with failures and topology changes), stability, fairness, optimality (minimum delay/cost), low overhead, and fast convergence.

Shortest path routing: the network is modelled as a graph where nodes are routers and edges are links with a cost (hop count, distance, delay or bandwidth). To route, the algorithm finds the least-cost path between each pair of nodes, e.g. using Dijkstra's algorithm:

  1. Mark the source with distance 0 and all others infinity.
  2. Take the unvisited node with the smallest distance, make it permanent, and relax its neighbours: d(v)=min⁡(d(v), d(u)+w(u,v))d(v) = \min(d(v),\, d(u) + w(u,v)).
  3. Repeat until the destination is permanent.
  4. Trace back via predecessors to get the path, and put the next hop in the routing table.

Each packet is then forwarded to the next hop on its shortest path.

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  • 2004 AD · 3 marks

With suitable example, list the specific features of IPV4 address. If class C IP address (11000000 11100 100000 10001 00111 001) requires 5 subnets, what will be its subnet mask? Also write the IP address in dotted decimal notation.

Answer

Features of an IPv4 address:

  • It is a 32-bit logical address, written in dotted decimal as four octets (e.g. 192.168.1.10).
  • It has two parts: network ID (netid) and host ID (hostid).
  • It is unique for each interface on the Internet; the total address space is 2322^{32}.
  • It is divided into classes: A (first bit 0, 1-126), B (10, 128-191), C (110, 192-223), D (multicast, 224-239) and E (reserved).
  • Special addresses exist: all 0s host = network address, all 1s host = broadcast, 127.x.x.x = loopback.
  • Subnet masks can divide a network into subnets.

Dotted decimal: the address bits are 11000000 11100100 00010001 00111001, so the IP address is 192.228.17.57 (class C, since it starts with 110).

Subnet mask for 5 subnets: we need 2n≥52^n \ge 5, so n=3n = 3 bits are borrowed from the host part (giving 8 subnets). The mask is

11111111.11111111.11111111.11100000 = 255.255.255.224 (/27), with 30 usable hosts per subnet.

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  • 2004 AD · 3 marks

What are the characteristics of the services provided by the TCP of TCP/IP protocol suite? How do you compare them with the characteristics of UDP?

Answer

TCP services:

  • Connection-oriented: a connection is set up with a three-way handshake and closed at the end.
  • Reliable delivery: sequence numbers, acknowledgements and retransmission on timeout.
  • Ordered, byte-stream delivery to the application, with duplicates removed.
  • Flow control with a sliding window (receiver window size).
  • Congestion control (slow start, congestion avoidance).
  • Full duplex communication; multiplexing of applications using port numbers.
  • Error detection with a checksum.

Comparison with UDP:

BasisTCPUDP
ConnectionConnection-orientedConnectionless
ReliabilityReliableUnreliable, best effort
OrderingOrderedNot guaranteed
Flow/congestion controlYesNo
Header size20+ bytes8 bytes
SpeedSlowerFaster
ExamplesHTTP, FTP, SMTPDNS, VoIP, video streaming

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  • 2004 AD · 3 marks

How do you specify an Ethernet cable? List all possible cables with their characteristics.

Answer

An Ethernet cable is specified in the form speed-signalling-segment length/medium, for example 10BASE5: 10 means 10 Mbps, BASE means baseband signalling, and the last part is the maximum segment length in hundreds of metres (or the medium type).

CableMediumSpeedMax segmentNotes
10BASE5Thick coax10 Mbps500 mBus, vampire taps
10BASE2Thin coax (RG-58)10 Mbps185 mBus, BNC connectors
10BASE-TUTP Cat 3 or higher10 Mbps100 mStar with hub
10BASE-FFibre optic10 Mbps2000 mLong distance
100BASE-TXUTP Cat 5100 Mbps100 mFast Ethernet
100BASE-FXMultimode fibre100 Mbps2 kmFast Ethernet
1000BASE-TUTP Cat 5e/61 Gbps100 mGigabit
1000BASE-SX/LXFibre1 Gbps550 m / 5 kmGigabit

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  • 2004 AD · 3 marks

Explain the architecture and working of an e-mail system.

Answer

Architecture: an e-mail system has these parts:

  • User Agent (UA): the program (e.g. Outlook, Gmail web) used to compose, read and manage messages.
  • Message Transfer Agent (MTA): the mail server software that moves mail between servers, using SMTP.
  • Mailbox / message store: disk space on the mail server where a user's incoming mail is held.
  • Access protocols: POP3 or IMAP (and HTTP for webmail) for the user agent to fetch mail from the mailbox.
Sender UA --SMTP--> Sender MTA --SMTP over Internet--> Receiver MTA --> Mailbox --POP3/IMAP--> Receiver UA

Working:

  1. The sender composes the message with a UA and gives the address user@domain.
  2. The UA sends the message to the sender's mail server using SMTP, where it is queued.
  3. The sender's MTA finds the destination mail server through a DNS MX record and transfers the message over a TCP connection (SMTP, port 25).
  4. The receiving MTA places it in the recipient's mailbox.
  5. The recipient's UA retrieves it using POP3 (download and delete) or IMAP (manage on the server).

A message consists of an envelope, header (From, To, Subject, Date) and body; MIME is used for attachments and non-ASCII content.

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  • 2004 AD · 3 marks

Compare TCP and UDP.

Answer

BasisTCPUDP
ConnectionConnection-oriented (handshake)Connectionless
ReliabilityReliable, with ACK and retransmissionUnreliable
OrderingSequenced deliveryNo ordering
Flow and congestion controlProvidedNot provided
Header20 bytes minimum8 bytes
Speed / overheadSlower, higher overheadFaster, low overhead
TransmissionByte streamMessage (datagram) based
ApplicationsWeb, e-mail, file transferDNS, streaming, VoIP, online games

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  • 2004 AD · 3 marks

What error control technique is used by data link layer? Explain the major differences between GO-back-N ARQ and selective-Repeat-AEQ.

Answer

The data link layer uses ARQ (Automatic Repeat reQuest): errors are detected using a CRC/checksum, and bad or lost frames are retransmitted after a NAK or a timeout. Forms of ARQ are Stop-and-Wait, Go-Back-N and Selective Repeat.

BasisGo-Back-N ARQSelective Repeat ARQ
RetransmissionThe erroneous frame and all frames after itOnly the erroneous or lost frame
Receiver windowSize 1 (accepts only in-order frames)Greater than 1 (buffers out-of-order frames)
Receiver bufferNot needed for out-of-order frames, they are discardedNeeded, and frames are reordered
Sender windowUp to 2m−12^m - 1Up to 2m−12^{m-1}
Bandwidth useWasteful when error rate is highEfficient
ComplexitySimpleMore complex (sorting, more buffers)
ACKCumulativeIndividual ACK/NAK

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  • 2004 AD · 3 marks

Represent bit sequence 1011 by the following waveform:
i. ASK\nii. FSK\niii. PSK\niv. NRZ-L v. Manchester\nvi. AMI

Answer

Bit sequence: 1 0 1 1

bit:      1       0       1       1
  • (i) ASK: carrier with high amplitude for 1, zero amplitude for 0: high, none, high, high.
  • (ii) FSK: high frequency f1f_1 for 1, low frequency f2f_2 for 0: f1,f2,f1,f1f_1, f_2, f_1, f_1.
  • (iii) PSK: phase 0 degrees for 1, 180 degrees for 0: 0∘,180∘,0∘,0∘0^\circ, 180^\circ, 0^\circ, 0^\circ.
  • (iv) NRZ-L: level depends on the bit (here 1 = low, 0 = high; the opposite convention is also used): low, high, low, low.
  • (v) Manchester (1 = low-to-high mid-bit, 0 = high-to-low): _|-, -|_, _|-, _|-.
  • (vi) AMI (Alternate Mark Inversion): 0 is zero level; 1s alternate between positive and negative pulses. The 1s are bits 1, 3 and 4, so the levels are +V,0,−V,+V+V, 0, -V, +V.

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  • 2004 AD · 1 mark

What is the maximum size of the IP datagram?

Answer

The IP datagram length field is 16 bits, so the maximum size is 216−1=65,5352^{16}-1 = 65{,}535 bytes (including the header, which is 20-60 bytes).

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  • 2005 AD · 1 mark

The time period of a sinusoidal waveform is 20 ms. What is its frequency?

Answer

f=1T=120×10−3=50 Hzf = \frac{1}{T} = \frac{1}{20\times10^{-3}} = 50\text{ Hz}

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  • 2005 AD · 1 mark

Which component of a communication model is required to generate a signal waveform that can be efficiently and effectively carried by the transmission medium.

Answer

The transmitter. It converts (encodes/modulates) the source data into a signal suitable for the transmission medium.

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  • 2005 AD · 1 mark

Which layer in the OSI reference model is responsible for routing?

Answer

The network layer (layer 3) is responsible for routing. It handles logical (IP) addressing and selects the path packets take from source to destination across networks.

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  • 2005 AD · 1 mark

What is done to avoid the repetition of a flag in the data field of a frame?

Answer

Bit stuffing (or byte/character stuffing in byte-oriented protocols) is used. The sender inserts an extra 0 after every five consecutive 1s in the data (or an escape byte before a flag-like byte), so the flag pattern 01111110 never appears inside the data field. The receiver removes the stuffed bits.

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  • 2005 AD · 1 mark

Which MAC protocol avoids collision?

Answer

CSMA/CA (Carrier Sense Multiple Access with Collision Avoidance), used in IEEE 802.11 wireless LANs, avoids collisions using interframe spacing, random back-off and optional RTS/CTS exchange. Token passing (IEEE 802.5) also prevents collisions because only the token holder may transmit.

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  • 2005 AD · 1 mark

In which switching technique data rate conversion (i.e. different data rates for the source and the receiver) is possible?

Answer

Message switching / packet switching (store-and-forward) allows data rate conversion, because each node stores the data and forwards it at the rate of the outgoing link. In circuit switching the dedicated path needs the same rate at both ends, so conversion is not possible.

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  • 2005 AD · 1 mark

Which property of a good routing algorithm ensures that all the end-stations get equal chances without being biased to anyone?

Answer

Fairness. A good routing algorithm treats all end stations equally, giving each a fair chance without bias. (Other properties are correctness, simplicity, robustness, stability and optimality.)

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  • 2005 AD · 1 mark

For an IP address, if the subnet mask is 255.255.255.224, how many hosts per subnet are possible?

Answer

The mask 255.255.255.224 has last octet 224=111000002224 = 11100000_2, so 5 bits are left for hosts.

Hosts per subnet=25−2=30\text{Hosts per subnet} = 2^5 - 2 = 30

The two subtracted addresses are the network and broadcast addresses.

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  • 2005 AD · 1 mark

Which transport layer protocol is used for the file transport application

Answer

TCP is used for file transfer (for example FTP, port 20/21), because it provides reliable, ordered, error-checked delivery of data.

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  • 2005 AD · 1 mark

What is the function of DNS?

Answer

The Domain Name System (DNS) translates human-readable domain names (such as www.example.com) into IP addresses, and the reverse. It is a distributed, hierarchical database of name servers, so users need not remember numeric addresses. It uses port 53.

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  • 2005 AD · 3 marks

What is the maximum theoretical data rate a system can support if it has a bandwidth of 4 kHz and SNR of 40dB

Answer

Using Shannon's capacity formula:

SNRdB=10log⁡10(SNR)=40⇒SNR=104SNR_{dB} = 10\log_{10}(SNR) = 40 \Rightarrow SNR = 10^4

C=Blog⁡2(1+SNR)=4000log⁡2(10001)≈4000×13.29C = B\log_2(1+SNR) = 4000\log_2(10001) \approx 4000 \times 13.29

C≈53,150 bps≈53.2 kbpsC \approx 53{,}150 \text{ bps} \approx 53.2 \text{ kbps}

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  • 2005 AD · 3 marks

Compare OSI and TCP/IP reference models for computer communications.

Answer

OSI modelTCP/IP model
7 layers4 layers (some texts use 5)
Developed by ISO as a general reference modelDeveloped by DARPA around the Internet protocols
Session and presentation are separate layersCombined into the application layer
Separate physical and data link layersCombined into the network access (host-to-network) layer
Network layer supports connectionless and connection-oriented serviceInternet layer is connectionless only
Transport layer supports connection-oriented service onlyTransport layer supports both (TCP and UDP)
Clear separation of services, interfaces and protocolsDoes not clearly separate them
Protocols fit the model less closely; model came before protocolsProtocols came first and the model describes them
Mostly a theoretical referenceThe practical basis of the Internet

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  • 2005 AD · 3 marks

What do you mean by framing? List the different framing techniques and illustrate bit stuffing with an example.

Answer

Framing is the data link layer function of dividing the bit stream from the network layer into manageable units called frames, with a way for the receiver to know where each frame starts and ends. Each frame carries a header, the data and a trailer.

Framing techniques:

  1. Character (byte) count
  2. Flag bytes with byte stuffing
  3. Flag bits (start/end flag) with bit stuffing
  4. Physical layer coding violations

Bit stuffing: the flag is 01111110. Whenever the sender sees five consecutive 1s in the data, it inserts a 0 after them. The receiver removes a 0 that follows five consecutive 1s.

Example: data 0110111111011111

  • Original: 0110 111111 0 11111
  • After each run of five 1s a 0 is inserted: 0110 11111 0 1 0 11111 0

011011111010111110\texttt{011011111010111110}

In the stuffed string no run of six 1s occurs, so the flag 01111110 cannot appear inside the data. The receiver removes the 0 after every five 1s to recover the original data.

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  • 2005 AD · 3 marks

Compare IEEE 802.3 and IEEE 802.5 LAN standards.

Answer

IEEE 802.3 (Ethernet)IEEE 802.5 (Token Ring)
Bus/star topologyRing (physical star-wired ring) topology
CSMA/CD access methodToken passing
Contention-based; collisions possibleDeterministic; no collisions
Any station may send when the medium is idleOnly the token holder may send
Performs well at light load; degrades at heavy load because of collisionsPerforms well at heavy load
No priority schemePriority and reservation bits supported
Frame size 64 to 1518 bytesLarger frames possible
Simple, cheap, no central monitorNeeds a monitor station for token maintenance
Speeds 10 Mbps up to 10 Gbps and more4 or 16 Mbps

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  • 2005 AD · 3 marks

List the features supported by data gram packet switching.

Answer

Features of datagram packet switching:

  • It is connectionless, so no call setup is needed.
  • Each packet (datagram) is routed independently and carries the full source and destination address.
  • Packets may take different routes and may arrive out of order, so the destination reorders them.
  • There is no guaranteed delivery. Lost or duplicated packets are handled by higher layers.
  • It is flexible and robust, because packets can avoid failed or congested nodes.
  • Bandwidth is used efficiently through statistical sharing.
  • It gives low delay for short messages because there is no setup phase.
  • Each packet needs a routing decision, so header overhead is higher.

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  • 2005 AD · 3 marks

Explain with a diagram how do you find the shortest path between two nodes in a network?

Answer

The shortest path is found with Dijkstra's algorithm. Each link has a cost (distance, delay or hop count).

Procedure (source node S):

  1. Label S with distance 0 (permanent) and every other node with ∞\infty (tentative).
  2. Take the working node (initially S) and, for each neighbour, compute the distance via the working node. If it is smaller than the neighbour's current label, update the label and record the predecessor.
  3. Among all tentative nodes, make the one with the smallest label permanent and make it the new working node.
  4. Repeat until the destination is permanent. Trace the predecessors back to get the path.

Example: links A-B = 2, A-C = 5, B-C = 1, B-D = 7, C-D = 3.

      2        1
  A ------ B ------ C
   \       |       /
    \ 5    | 7    / 3
     \_____ D ___/

From A: B = 2, C = 5. Make B permanent. Via B, C = 3 (better than 5) and D = 9. Make C permanent. Via C, D = 6 (better than 9). So the shortest path from A to D is A-B-C-D with cost 6.

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  • 2005 AD · 3 marks

Draw the IP datagram header and answer the following questions:
i. What is the maximum size of IP datagram header?\nii. What is the significance of TTL field?\niii. What is the maximum size of the IP datagram?

Answer

IPv4 header (32 bits per row):

| Version(4) | IHL(4) | Type of service(8) |        Total length(16)        |
|        Identification(16)              | Flags(3) |  Fragment offset(13)  |
|   Time to live(8)   |  Protocol(8)     |       Header checksum(16)        |
|                         Source IP address(32)                              |
|                      Destination IP address(32)                           |
|                      Options + padding (0 to 40 bytes)                    |

i. The header length field (IHL) is 4 bits, so its maximum value is 15 words of 4 bytes: 15×4=6015 \times 4 = 60 bytes. The minimum header is 20 bytes.

ii. TTL (Time To Live) limits the life of a datagram. Each router decrements it by one, and when it reaches 0 the datagram is discarded. This prevents datagrams from looping forever in the network.

iii. The total length field is 16 bits, so the maximum datagram size is 216−1=65,5352^{16} - 1 = 65{,}535 bytes (header plus data).

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  • 2009 AD · 1 mark

Differentiate between bridge & switch.

Answer

A bridge connects two LAN segments and filters frames using a MAC address table. It usually has few ports (2 or 4) and forwards frames in software. A switch is a multiport bridge with many ports, forwards frames in hardware (ASICs) at higher speed, and gives each port its own collision domain.

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  • 2009 AD · 1 mark

Define socket address with suitable example.

Answer

A socket address is the combination of an IP address and a port number. It identifies one end of a connection uniquely. Example: 192.168.1.10:80, where 192.168.1.10 is the host and 80 is the port of the web service.

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  • 2009 AD · 1 mark

Define Synchronous transmission.

Answer

In synchronous transmission, data is sent as a continuous stream of blocks or frames without start and stop bits for each character. The sender and receiver are synchronized by a common clock (or clock recovered from the data) and special sync characters or flags. It is faster and more efficient for large volumes of data.

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  • 2009 AD · 1 mark

Why is ARP and RARP protocol used?

Answer

ARP (Address Resolution Protocol) finds the MAC (physical) address of a host when its IP address is known, so a frame can be delivered on the local network. RARP (Reverse ARP) does the opposite: a host that knows its MAC address (such as a diskless machine) asks for its IP address.

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  • 2009 AD · 1 mark

In what case UDP protocol is preferred than TCP protocol?

Answer

UDP is preferred when speed and low overhead matter more than reliability, for example real-time voice and video, online gaming, streaming, DNS queries and broadcast/multicast applications. It has no connection setup, no retransmission and no flow control, so delay is low and occasional packet loss is tolerable.

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  • 2009 AD · 1 mark

Why BGP is used?

Answer

BGP (Border Gateway Protocol) is the exterior gateway protocol used to exchange routing information between different autonomous systems on the Internet. It is a path-vector protocol that applies routing policies and prevents loops, and it makes the global Internet routing possible.

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  • 2009 AD · 1 mark

What do you mean by protocol stack?

Answer

A protocol stack is a set of network protocols arranged in layers, where each layer provides services to the layer above and uses the services of the layer below. Examples are the TCP/IP stack and the OSI stack. Data passes down the stack at the sender and up the stack at the receiver.

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  • 2009 AD · 1 mark

Is there any difference in addressing issues of data link layer and network layer?

Answer

Yes. The data link layer uses physical (MAC) addresses, which are 48 bits, fixed in the NIC and have only local significance within one network. The network layer uses logical (IP) addresses, which are 32 bits (IPv4) or 128 bits (IPv6), assigned by configuration, hierarchical and valid across the whole internetwork. MAC addresses identify the next hop and IP addresses identify the final host.

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  • 2009 AD · 1 mark

Port address is of ............ bits in length whereas IPV6 is of ....... bits in length.

Answer

A port address is 16 bits in length whereas an IPv6 address is 128 bits in length.

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  • 2009 AD · 3 marks

What is transmission impairments? Explain the major types of transmission impairments.

Answer

Transmission impairment is any imperfection in a transmission medium that makes the received signal differ from the transmitted one, causing errors in digital data and degraded quality in analog signals.

Major types:

  1. Attenuation: loss of signal strength with distance as energy is absorbed by the medium. It is measured in decibels and is overcome with amplifiers or repeaters.
  2. Distortion: change in the shape of a composite signal because its frequency components travel at different speeds and arrive with different delays (delay distortion). Equalizers reduce it.
  3. Noise: unwanted signals added between sender and receiver.
    • Thermal noise: random motion of electrons.
    • Induced noise: from motors and appliances.
    • Crosstalk: coupling from a neighbouring wire.
    • Impulse noise: short spikes from lightning or switching.

The effect of noise is measured by the signal-to-noise ratio (SNR).

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  • 2009 AD · 3 marks

Define the following terms with their respective measurement unit.
i) Bandwidth\nii) Bit rate\niii) Amplitude

Answer

i) Bandwidth: the range of frequencies a channel or signal occupies, the difference between the highest and lowest frequency, B=fmax−fminB = f_{max} - f_{min}. Unit: hertz (Hz). In digital contexts it also means the data capacity of a link, in bits per second.

ii) Bit rate: the number of bits transmitted per second. Unit: bits per second (bps), with multiples kbps, Mbps and Gbps.

iii) Amplitude: the maximum height (strength) of a signal measured from its zero level, which shows its intensity. Unit: volt (V) for voltage signals, ampere (A) for current, or watt (W) for power.

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  • 2009 AD · 3 marks

Write down the major purpose of following protocols with their default port no.
i) FTP ii) SMTP iii) DNS iv) Telnet v) HTTP vi) HTTPS

Answer

ProtocolPurposeDefault port
i) FTPTransfer files between hosts20 (data), 21 (control)
ii) SMTPSend/relay e-mail25
iii) DNSTranslate domain names to IP addresses53
iv) TelnetRemote terminal login to another host23
v) HTTPTransfer web pages (hypertext)80
vi) HTTPSSecure HTTP over SSL/TLS443

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  • 2009 AD · 3 marks

Explain the functions of each layer of OSI model.

Answer

The OSI model has seven layers.

  1. Physical: transmits raw bits over the medium. It defines voltages, cables, connectors, data rate and topology.
  2. Data link: framing, physical (MAC) addressing, error detection, flow control and media access control. It gives reliable node-to-node delivery.
  3. Network: logical (IP) addressing, routing, packet forwarding and congestion control. It gives source-to-destination delivery across networks.
  4. Transport: end-to-end (process-to-process) delivery using port addresses, segmentation and reassembly, flow control, error control and connection management (TCP, UDP).
  5. Session: establishes, manages and terminates sessions between applications, with dialog control and synchronization (checkpoints).
  6. Presentation: translation between data formats, encryption/decryption and compression.
  7. Application: gives network services to the user, such as e-mail, file transfer, web access and remote login.

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  • 2009 AD · 3 marks

Describe the working mechanism of sliding window flow control protocol.

Answer

Sliding window protocol lets the sender transmit several frames before it needs an acknowledgement, so the link is used efficiently.

Working:

  • Frames are numbered sequentially. The sender keeps a window of sequence numbers it may send without acknowledgement. With nn-bit sequence numbers the window is at most 2n−12^n - 1 (Go-Back-N) or 2n−12^{n-1} (selective repeat).
  • The sender transmits frames inside the window and keeps a copy of each until it is acknowledged.
  • The receiver has its own window of sequence numbers it will accept. On receiving a correct frame it sends an acknowledgement (ACK), which can be piggybacked on data going the other way.
  • When an ACK arrives, the sender's window slides forward and new frames can be sent. The receiver's window slides as frames are delivered in order.
  • If a frame is lost or damaged, a timer expires. In Go-Back-N the sender retransmits that frame and all later ones. In selective repeat only the missing frame is resent.

This gives flow control, since the sender can never have more unacknowledged frames than the window size, and error control through ACKs and retransmission.

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  • 2009 AD · 3 marks

Describe how CSMA/CD works. How it is differed from CSMA/CA.

Answer

CSMA/CD (Carrier Sense Multiple Access with Collision Detection) is the access method of wired Ethernet (IEEE 802.3).

Working:

  1. A station with a frame to send listens to the channel (carrier sense).
  2. If the channel is idle, it transmits. If busy, it waits until idle (1-persistent).
  3. While transmitting it keeps listening. If another station transmits at the same time, a collision is detected.
  4. On collision, the station sends a short jam signal so that all stations know, and stops.
  5. It waits a random time using binary exponential back-off and tries again, giving up after a maximum number of attempts (16).

The minimum frame size must be large enough for the sender to detect a collision before it finishes transmitting.

Difference from CSMA/CA:

  • CSMA/CD detects collisions after they occur and is used in wired LANs, since a station can listen while sending. CSMA/CA tries to avoid collisions before they occur and is used in wireless LANs (802.11), where a station cannot detect collisions while transmitting.
  • CSMA/CA uses interframe spacing, random back-off before transmission, and ACKs (optionally RTS/CTS) to confirm delivery.

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  • 2009 AD · 3 marks

Compare distance vector routing protocol with shortest path routing protocol.

Answer

Distance vector routingShortest path (link state) routing
Each router sends its whole routing table to its neighbours onlyEach router floods link-state information about its neighbours to all routers
Router knows only distances via neighbours (routing by rumour)Router builds a complete map of the network
Uses the Bellman-Ford algorithmUses Dijkstra's algorithm
Slow convergence; count-to-infinity problemFast convergence; loop-free
Simple, low memory and CPU useMore complex, needs more memory and CPU
Periodic updates generate trafficUpdates only on change, but flooding costs bandwidth
Example: RIPExample: OSPF

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  • 2009 AD · 3 marks

Suppose you are given the IP address as 192.168.0.12/26. Answer the following questions.
i. Which class does this IP belongs to?\nii. How many valid hosts are available per subnet?\niii. Calculate the subnet mask of given IP.

Answer

i. The first octet is 192, which is in the range 192 to 223, so it is Class C.

ii. The prefix is /26, so there are 32−26=632 - 26 = 6 host bits.

Valid hosts=26−2=62\text{Valid hosts} = 2^6 - 2 = 62

iii. The subnet mask for /26 is 26 ones followed by 6 zeros: 11111111.11111111.11111111.11000000.

Subnet mask=255.255.255.192\text{Subnet mask} = 255.255.255.192

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  • 2009 AD · 4 marks

Explain how datagram packet switching works. Differentiate with virtual circuit approach.

Answer

Datagram packet switching: the message is divided into packets, and each packet (datagram) is treated independently. Each packet carries the full destination address. At each node the router looks at the destination address and forwards the packet along the best route at that moment, so packets of the same message may take different paths and arrive out of order. The destination reassembles them. No connection is set up first.

Difference from virtual circuit:

DatagramVirtual circuit
Connectionless, no setup phaseConnection-oriented; a path is set up first
Each packet carries the full destination addressPackets carry only a short virtual circuit number
Each packet routed independentlyAll packets follow the same path
Packets may arrive out of orderPackets arrive in order
Routers hold no per-connection stateRouters keep a table entry for each circuit
Failure of a node affects only packets in transit; rerouting is easyFailure of a node breaks all circuits through it
Congestion control is difficultEasy to reserve resources and control congestion
Example: IPExample: X.25, Frame Relay, ATM

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  • 2009 AD · 2 marks

Write down the major difference between TCP and IP protocol.

Answer

TCPIP
Transport layer protocolNetwork (internet) layer protocol
Connection-oriented and reliable, with acknowledgement, retransmission, flow and congestion controlConnectionless and unreliable (best effort)
Provides end-to-end delivery between processes using port numbersProvides host-to-host delivery of packets using IP addresses and routing
Handles segmentation and in-order reassemblyHandles addressing, routing and fragmentation

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  • 2009 AD · 6 marks

Write down the major difference between: (Any three)
i) IPV4 versus IPV6\nii) WDM versus FDM\niii) Internet versus WWW\niv) 802.3 Ethernet versus 802.5 Token Rings.

Answer

(i) IPv4 versus IPv6

IPv4IPv6
32-bit address128-bit address
Header 20 to 60 bytes, variableFixed 40-byte header
Header checksum presentNo header checksum
Fragmentation by routers and senderOnly the sender fragments
IPsec optionalIPsec built in

(ii) WDM versus FDM

WDMFDM
Multiplexes optical signals of different wavelengths on a fibreMultiplexes analog signals of different frequencies on one medium
Used on optical fibreUsed on copper, coaxial and radio
Very high capacityLower capacity

(iii) Internet versus WWW

InternetWWW
Worldwide network of networks, the physical infrastructureA service on the Internet, a collection of linked web pages
Uses many protocols (TCP/IP, SMTP, FTP)Uses HTTP and URLs
Carries e-mail, file transfer and moreOne application among them, accessed with a browser

(iv) 802.3 Ethernet versus 802.5 Token Ring

802.3802.5
Bus/star, CSMA/CDRing, token passing
Collisions possibleNo collisions
No prioritiesPriorities supported

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  • 2010 AD · 1 mark

If a spectrum of signal is 11 KHz to 33.6 MHz. Find the bandwidth occupied by the signal.

Answer

B=fmax−fmin=33.6 MHz−11 kHz=33,600 kHz−11 kHz=33,589 kHzB = f_{max} - f_{min} = 33.6\text{ MHz} - 11\text{ kHz} = 33{,}600 \text{ kHz} - 11 \text{ kHz} = 33{,}589 \text{ kHz}

The bandwidth is 33.589 MHz.

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  • 2010 AD · 1 mark

Which type of multiplexing is preferred if pure analog data transmission has to be done?

Answer

Frequency Division Multiplexing (FDM) is preferred for analog data transmission, because each analog signal is assigned its own frequency band and all are sent together on the shared medium. (TDM is preferred for digital signals.)

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  • 2010 AD · 1 mark

What is the 32-bit binary equivalent of the IP address 196.16.4.140? What is the default subnet mask for this class of IP?

Answer

IP address 196.16.4.140 in binary:

OctetBinary
19611000100
1600010000
400000100
14010001100

32-bit form: 11000100.00010000.00000100.10001100

The first octet begins with 110, so this is a Class C address. Its default subnet mask is 255.255.255.0 (11111111.11111111.11111111.00000000, i.e. /24).

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  • 2010 AD · 1 mark

What is Shannon's Channel capacity?

Answer

Shannon's channel capacity is the theoretical maximum data rate of a noisy channel with bandwidth BB (Hz) and signal-to-noise power ratio S/NS/N:

C=Blog⁡2(1+SN) bpsC = B\log_2\left(1+\frac{S}{N}\right)\ \text{bps}

It shows that capacity rises with bandwidth and SNR, and that error-free transmission is possible at any rate below CC but not above it.

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  • 2010 AD · 1 mark

What are the phases that can be seen in connection-oriented service provided by the data link layer?

Answer

Connection-oriented service of the data link layer has three phases:

  1. Connection establishment: both sides initialise variables and counters (sequence numbers, buffers).
  2. Data transfer: frames are sent and received in order, with acknowledgements, flow control and error control.
  3. Connection release: the connection is closed and the resources are freed.

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  • 2010 AD · 1 mark

Define 'Token' in IEEE 802.4 and IEEE 802.5 standard.

Answer

A token is a special control frame or bit pattern that circulates among the stations and gives the holder the right to transmit.

  • IEEE 802.4 (Token Bus): the token is passed among stations on a bus that form a logical ring, in order of descending station address.
  • IEEE 802.5 (Token Ring): the token is a 3-byte frame (start delimiter, access control, end delimiter) that circulates around the physical ring. A station with data captures the free token and sends its frame in its place.

A station may transmit only while it holds the token, so there are no collisions.

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  • 2010 AD · 1 mark

Make distinction between circuit switching and packet switching.

Answer

Circuit switchingPacket switching
A dedicated path is set up before transmission and kept until the endNo dedicated path; packets share links
Bandwidth is reserved, so it can be wasted when idleBandwidth is used on demand, so the link is used efficiently
Data travels as a continuous streamMessage is split into packets, each with a header
Constant delay and no reorderingVariable delay; packets may arrive out of order
Call setup time is neededNo setup in the datagram method
Example: telephone networkExample: Internet

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  • 2010 AD · 1 mark

What is persistent CSMA?

Answer

In 1-persistent CSMA, a station with a frame to send senses the channel. If it is idle, the station transmits at once. If it is busy, the station keeps sensing continuously and transmits the moment the channel becomes idle, with probability 1. If two or more stations are waiting, they collide.

In p-persistent CSMA, an idle channel is used with probability pp, and with probability 1−p1-p the station defers to the next time slot.

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  • 2010 AD · 4 marks

Data is to be transmitted over the network using a transmitting scheme with eight levels. If the bandwidth of the network is 3500 Hz, deduce the Nyquist maximum data transfer rate. Assuming that network has a typical signal-to-noise power ratio is 25 dB, determine the maximum theoretical information rate that can be achieved.

Answer

Nyquist rate, with B=3500B=3500 Hz and L=8L=8 levels:

C=2Blog⁡2L=2×3500×log⁡28=2×3500×3=21,000 bpsC = 2B\log_2 L = 2\times3500\times\log_2 8 = 2\times3500\times3 = 21{,}000\ \text{bps}

Shannon capacity, with SNR = 25 dB:

SN=1025/10=102.5≈316.2\frac{S}{N}=10^{25/10}=10^{2.5}\approx316.2

C=Blog⁡2(1+SN)=3500×log⁡2(317.2)=3500×8.31≈29,080 bpsC = B\log_2\left(1+\frac{S}{N}\right)=3500\times\log_2(317.2)=3500\times8.31\approx29{,}080\ \text{bps}

So the Nyquist maximum is 21 kbps, and the maximum theoretical information rate is about 29.1 kbps. The eight-level scheme at 21 kbps is below this limit, so it is achievable.

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  • 2010 AD · 2 marks

Calculate the white noise in decibel-watts when temperature is 15 Celsius and bandwidth is 6 MHz.

Answer

Thermal noise power is N=kTBN = kTB, where k=1.38×10−23k = 1.38\times10^{-23} J/K.

T=15+273=288T = 15 + 273 = 288 K and B=6×106B = 6\times10^{6} Hz.

N=1.38×10−23×288×6×106≈2.38×10−14 WN = 1.38\times10^{-23}\times288\times6\times10^{6}\approx2.38\times10^{-14}\ \text{W}

NdBW=10log⁡10(2.38×10−14)≈−136.2 dBWN_{dBW}=10\log_{10}(2.38\times10^{-14})\approx-136.2\ \text{dBW}

The noise power is about -136.2 dBW.

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  • 2010 AD · 3 marks

Differentiate between:
i) TCP and UDP\nii) ARP and RARP

Answer

(i) TCP and UDP

TCPUDP
Connection-orientedConnectionless
Reliable: acknowledgements and retransmissionUnreliable, best effort
Ordered delivery, flow and congestion controlNo ordering or flow control
20-byte or larger header, slower8-byte header, faster
Used for HTTP, FTP, SMTPUsed for DNS, streaming, VoIP

(ii) ARP and RARP

ARPRARP
Address Resolution ProtocolReverse Address Resolution Protocol
Maps a known IP address to a MAC addressMaps a known MAC address to an IP address
The host broadcasts "who has this IP?"The diskless host broadcasts "what is my IP?" to a RARP server
Still in useObsolete, replaced by BOOTP and DHCP

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  • 2010 AD · 3 marks

The message sequence is 1001101 and generator polynomial G(X)=x3+x+1G(X)=x^3+x+1. Calculate the transmitted frame.

Answer

Message M=1001101M = 1001101, generator G(x)=x3+x+1⇒1011G(x)=x^3+x+1 \Rightarrow 1011 (degree 3).

Append 3 zeros: 1001101000. Divide by 1011 using modulo-2 division:

StepBitsXOR with 1011Remainder
110010010010
2bring 1: 0101(leading 0, no XOR)101
3bring 0: 10100001001
4bring 1: 0011(leading 0)011
5bring 0: 0110(leading 0)110
6bring 0: 11000111111
7bring 0: 11100101101

CRC remainder = 101.

Transmitted frame = message + CRC = 1001101101.

Check: dividing the received frame by 1011 gives remainder 000.

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  • 2010 AD · 3 marks

Define CSMA/CD. Write an algorithm for CSMA/CD.

Answer

CSMA/CD (Carrier Sense Multiple Access with Collision Detection) is the MAC method of Ethernet (IEEE 802.3). A station listens before transmitting, and it keeps listening while transmitting. If it detects a collision, it stops, sends a jam signal and retries after a random back-off.

Algorithm

  1. Prepare the frame and set attempts = 0.
  2. Sense the channel. If it is busy, wait until it is idle (persistent sensing).
  3. When the channel is idle, start transmitting and keep monitoring the channel.
  4. If no collision is detected until the whole frame is sent, the transmission is successful. Stop.
  5. If a collision is detected:
    • stop sending and transmit a short jam signal so all stations know;
    • set attempts = attempts + 1; if attempts exceeds 16, abort and report an error;
    • choose a random KK from 00 to 2min⁡(attempts,10)−12^{\min(attempts,10)}-1 (binary exponential back-off);
    • wait K×K\times slot time, then go to step 2.

For collisions to be detected, the frame transmission time must be at least twice the propagation delay.

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  • 2010 AD · 3 marks

What is ALOHA? Briefly describe different types of ALOHA.

Answer

ALOHA is a random-access MAC protocol developed at the University of Hawaii. Any station may send a frame whenever it has data. If the acknowledgement does not arrive in time, it assumes a collision, waits a random time and retransmits.

Pure ALOHA: stations transmit at any time. A frame is vulnerable for two frame times (2Tfr2T_{fr}). Maximum throughput is S=Ge−2GS = Ge^{-2G}, about 18.4% at G=0.5G=0.5.

Slotted ALOHA: time is divided into slots equal to one frame time, and stations transmit only at the start of a slot. The vulnerable period falls to one frame time (TfrT_{fr}). Throughput is S=Ge−GS = Ge^{-G}, with a maximum of about 36.8% at G=1G=1.

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  • 2010 AD · 3 marks

What do you mean by framing? List the different framing techniques and illustrate character count method with appropriate example and diagram.

Answer

Framing is how the data link layer splits the bit stream from the network layer into discrete frames, each with a header and trailer, so the receiver can tell where a frame starts and ends and can check it for errors.

Techniques:

  1. Character (byte) count
  2. Flag bytes with byte stuffing
  3. Flag bits with bit stuffing
  4. Physical layer coding violations

Character count method: a field in the header gives the number of characters in the frame, including the count field itself. The receiver reads the count and knows where the frame ends.

Example, four frames of 5, 5, 8 and 8 characters:

| 5 | A B C D | 5 | E F G H | 8 | I J K L M N O | 8 | P Q R S T U V |
  Frame 1       Frame 2        Frame 3              Frame 4

The weakness is that if a count is corrupted (for example, the 5 in frame 1 becomes 7), the receiver loses synchronisation and cannot find the start of the next frame. This is why the method is rarely used alone.

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  • 2011 AD · 1 mark

Which component is used in communication model to convert the signal so that it is effectively transmitted?

Answer

The transmitter (with the encoder or modulator) converts the source signal into a form suitable for transmission over the medium. For example, a modem converts digital data to an analog signal.

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  • 2011 AD · 1 mark

Why buffer is used in selective reject ARQ?

Answer

In selective-reject (selective repeat) ARQ, only the damaged frame is retransmitted. Frames that arrive after it are correct but out of order, so the receiver stores them in a buffer. When the missing frame arrives, the receiver puts the frames in the right order and delivers them to the upper layer. The sender also keeps copies of unacknowledged frames for retransmission.

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  • 2011 AD · 1 mark

How routers are classified?

Answer

Routers are classified in these ways:

  • By routing method: static (non-adaptive) and dynamic (adaptive) routers.
  • By protocol use: interior routers (RIP, OSPF) and exterior or border routers (BGP).
  • By location or function: internal routers, edge routers, core routers, border (gateway) routers.
  • By number of protocols handled: single-protocol and multi-protocol routers.

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  • 2011 AD · 1 mark

What are the network support layer and user-support layer in OSI model?

Answer

  • Network support layers: the physical, data link and network layers (layers 1 to 3). They deal with the physical aspects of moving data between devices, such as electrical specification, physical addressing, transport timing and reliability.
  • User support layers: the session, presentation and application layers (layers 5 to 7). They allow interoperability between software systems.

The transport layer (layer 4) links the two groups, and it makes sure that what the lower layers deliver is in a form the upper layers can use.

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  • 2011 AD · 1 mark

How does NRZ-L differ from NRZ-I?

Answer

In NRZ-L, the signal level depends on the bit value: for example, a high level for 0 and a low level for 1 (or the reverse). A long run of identical bits gives no change, which causes synchronisation problems.

In NRZ-I, the level depends on the bit transition: the signal inverts at the start of a 1 bit and stays the same for a 0 bit. Long runs of 1s still produce transitions that help synchronisation, but long runs of 0s do not.

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  • 2011 AD · 1 mark

Differentiate between OSPF and BGP.

Answer

OSPFBGP
Open Shortest Path FirstBorder Gateway Protocol
Interior gateway protocol, used inside one autonomous systemExterior gateway protocol, used between autonomous systems
Link-state algorithm (Dijkstra)Path-vector algorithm
Metric is costDecisions use path attributes and policy
Runs directly over IP (protocol 89)Runs over TCP (port 179)

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  • 2011 AD · 1 mark

If IP address given is 223.124.123.89, what is the network address and what is the host address?

Answer

The first octet 223 lies in 192 to 223, so this is a Class C address. The default mask is 255.255.255.0: the first 24 bits are the network part and the last 8 bits are the host part.

  • Network address: 223.124.123.0
  • Host address: 89 (0.0.0.89)

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  • 2011 AD · 1 mark

Which one is most intelligent hub or switch and why?

Answer

A switch is more intelligent. A hub is a simple repeater that broadcasts every incoming signal to all its ports. A switch reads the destination MAC address of each frame, keeps a MAC address table, and forwards the frame only to the port where the destination is. This reduces collisions and traffic and gives each port its own collision domain.

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  • 2011 AD · 1 mark

Differentiate between asynchronous and synchronous transmission.

Answer

Asynchronous transmissionSynchronous transmission
Data is sent one character at a time, with start and stop bitsData is sent as a continuous block of bits or frames
No common clock; gaps may exist between charactersSender and receiver clocks are synchronised
Overhead is high (2 to 3 extra bits per character)Overhead is low
Slower and cheaperFaster and more complex
Example: keyboard to computerExample: LAN, data between computers

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  • 2011 AD · 1 mark

For nn devices in a network, what is the number of cable links required for a mesh and ring topology?

Answer

  • Mesh: each device is linked to every other device, so the number of cable links is n(n−1)2\dfrac{n(n-1)}{2}.
  • Ring: each device is linked to the next, and the last to the first, so the number of cable links is nn.

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  • 2011 AD · 3 marks

Differentiate between TDM and FDM.

Answer

TDM (Time Division Multiplexing)FDM (Frequency Division Multiplexing)
Channels share the link in different time slotsChannels share the link by using different frequency bands
Each signal uses the full bandwidth for its slotEach signal uses its band all the time
Used mostly for digital signalsUsed mostly for analog signals
Needs synchronisationNeeds guard bands to prevent interference
Crosstalk is not a major problemCrosstalk and intermodulation can occur
Example: telephone digital trunks (T1)Example: radio and TV broadcast

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  • 2011 AD · 3 marks

Compare TCP/IP and OSI reference model.

Answer

OSI modelTCP/IP model
Seven layersFour layers (some texts show five)
Layers: physical, data link, network, transport, session, presentation, applicationLayers: network access, internet, transport, application
Theoretical reference model developed by ISOProtocol suite developed first, from ARPANET
Protocols were developed after the model, so the model is genericThe model grew out of the protocols
Strictly separates services, interfaces and protocolsDoes not clearly separate them
Network layer supports both connectionless and connection-oriented service; transport layer supports connection-oriented onlyNetwork layer is connectionless only; transport layer offers both (TCP and UDP)
Session and presentation layers existTheir functions are inside the application layer
Less used in practiceStandard of the Internet

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  • 2011 AD · 3 marks

Explain the working mechanism of MAC protocol which is used to avoid collision.

Answer

The MAC protocol that avoids collisions is CSMA/CA (Carrier Sense Multiple Access with Collision Avoidance), used in wireless LANs (IEEE 802.11), where a station cannot listen while it transmits and so cannot detect collisions.

Working:

  1. The station senses the channel. If it is busy, it waits.
  2. When the channel becomes idle, the station waits for a fixed IFS (inter-frame space).
  3. The station then chooses a random back-off time in a contention window and counts down only while the channel stays idle. If the channel turns busy, the counter pauses and later resumes.
  4. When the counter reaches zero, the station sends the frame. With RTS/CTS, it first sends a short RTS (request to send), and the receiver replies with a CTS (clear to send). Other stations hearing these set their NAV (network allocation vector) and keep quiet.
  5. The receiver sends an ACK after a correct frame. If no ACK arrives in time, the sender assumes a collision, doubles the contention window and tries again.

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  • 2011 AD · 3 marks

Differentiate between Distance Vector Routing and Link State Routing.

Answer

Distance vector routingLink state routing
Each router sends its whole routing table to its neighbours onlyEach router floods information about its neighbours to all routers
Router knows only distances to destinations, not the topologyEach router builds a complete map of the network
Bellman-Ford algorithmDijkstra's algorithm
Updates are periodicUpdates are sent only when a change occurs
Slow convergence; count-to-infinity problemFast convergence
Low memory and CPU use, simpleMore memory and CPU needed
Example: RIPExample: OSPF

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  • 2011 AD · 3 marks

Discuss packet switching techniques in brief.

Answer

In packet switching, a message is divided into small packets, which are sent through the network and reassembled at the destination. Each link is shared, and a store-and-forward method is used. There are two techniques:

1. Datagram approach (connectionless): each packet (datagram) is treated independently and carries the full destination address. Different packets of one message may take different routes, so they may arrive out of order and the destination must reorder them. There is no setup, and routing can adapt around a failure.

2. Virtual circuit approach (connection-oriented): a logical path is set up first, and all packets follow it, carrying a short virtual circuit number. Packets arrive in order. There are two types: switched virtual circuit (SVC), set up on demand, and permanent virtual circuit (PVC), set up in advance.

Packet switching uses bandwidth efficiently, but delay is variable.

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  • 2011 AD · 3 marks

For IP address 192.168.1.0 and subnet mask 255.255.255.0 of organization, find out the subnets for 4 departments.

Answer

The address 192.168.1.0 is Class C with the mask 255.255.255.0 (/24, 8 host bits).

For 4 departments we need 4 subnets, so borrow 2 bits (22=42^2=4). The new mask is /26, which is 255.255.255.192.

Each subnet has 26=642^6=64 addresses, of which 26−2=622^6-2=62 are usable hosts.

DepartmentSubnet addressHost rangeBroadcast
1192.168.1.0/26192.168.1.1 to 192.168.1.62192.168.1.63
2192.168.1.64/26192.168.1.65 to 192.168.1.126192.168.1.127
3192.168.1.128/26192.168.1.129 to 192.168.1.190192.168.1.191
4192.168.1.192/26192.168.1.193 to 192.168.1.254192.168.1.255

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  • 2011 AD · 3 marks

Discuss the various aspects of digital and analog transmission.

Answer

Analog transmission sends a continuously varying signal. Digital data must first be modulated onto a carrier (ASK, FSK, PSK). Over long distance, amplifiers are used, and they amplify noise along with the signal, so noise accumulates.

Digital transmission sends discrete signal levels (1s and 0s). Digital data is encoded as line codes (NRZ, Manchester); analog data is first sampled and digitised (PCM). Over long distance, repeaters regenerate the signal, so noise does not accumulate.

AspectAnalogDigital
SignalContinuousDiscrete
Noise immunityLowHigh
Long-distance deviceAmplifierRepeater
Error detection and correctionDifficultEasy (CRC, parity)
Security and multiplexingHarderEasier (encryption, TDM)
Bandwidth needLessMore
CostCheaper for simple usesCheaper at scale (integrated circuits)

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  • 2012 AD · 1 mark

List out the three fundamental characteristics that determine the effectiveness of the data communication system.

Answer

The three fundamental characteristics are:

  1. Delivery: data must reach the correct intended destination, and only that destination.
  2. Accuracy: data must be delivered without alteration.
  3. Timeliness: data must be delivered in time, with little delay, which matters for real-time audio and video.

(Jitter, the variation in packet arrival time, is also often considered under timeliness.)

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  • 2012 AD · 1 mark

In what case analog signal is preferred than digital signal?

Answer

An analog signal is preferred when the source data is naturally analog and the cost or bandwidth of digitising it is not justified, for example:

  • the available bandwidth is small, since an analog signal needs less bandwidth than its digital version;
  • simple, cheap, short-distance systems such as radio and television broadcast, or a basic telephone local loop;
  • transmission over media that only pass analog signals, such as radio waves and the voice-grade telephone line.

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  • 2012 AD · 1 mark

"OSI reference model can be divided into three parts", Justify.

Answer

The seven OSI layers form three groups by function:

  1. Network support layers (layers 1 to 3): physical, data link and network layers, which handle the physical transmission of data between devices.
  2. User support layers (layers 5 to 7): session, presentation and application layers, which let unrelated software systems work together.
  3. Transport layer (layer 4): it links the two groups, ensuring end-to-end reliable delivery of data from the lower layers to the upper layers.

Therefore the model can justifiably be seen as three parts.

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  • 2012 AD · 1 mark

List out the feature that the DLL and Transport layer have in common.

Answer

The data link layer and the transport layer have these features in common:

  • Error control: detecting errors and retransmitting (ARQ).
  • Flow control: using sliding window to prevent a fast sender from swamping a slow receiver.
  • Sequencing: numbering frames or segments so they are delivered in order.
  • Acknowledgement and retransmission of lost data.
  • Connection-oriented services: connection setup, data transfer and release.

The data link layer does this between two directly connected nodes (hop by hop); the transport layer does it end to end.

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  • 2012 AD · 1 mark

What is the glaring inefficiency of CSMA and how it can be overcome?

Answer

In CSMA, if two stations sense an idle channel and send at the same time, a collision occurs. Because CSMA has no collision detection, the stations carry on transmitting their whole damaged frames and the channel is wasted for the full frame time.

It is overcome with CSMA/CD: stations keep listening while sending, stop at once when a collision is detected, send a jam signal and retransmit after a random back-off. This saves the time that would be wasted.

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  • 2012 AD · 1 mark

Why Circuit-Switched network is bit stream transparent?

Answer

A circuit-switched network sets up a dedicated physical path (or reserved channel) between sender and receiver before data transfer begins. The network does not inspect, interpret, buffer or modify the data on that path; it just passes the bits through at the agreed rate. Hence it places no restriction on the code, format, framing or protocol used by the end stations, so any bit stream is carried unchanged. This property is called bit-stream (code and protocol) transparency.

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  • 2012 AD · 1 mark

Given that you have a class B IP address network range, which of the subnet masks below will allow for 100 subnets with 500 usable host addresses per subnet?
255.255.0.0\n255.255.224.0\n255.255.254.0\n255.255.255.0\n255.255.255.224

Answer

(255.255.254.0)

A class B network has 16 host bits. 100 subnets need 7 subnet bits (27=128≥1002^7 = 128 \ge 100), leaving 16−7=916-7=9 host bits, which give 29−2=510≥5002^9-2 = 510 \ge 500 usable hosts. The mask therefore has 16+7=2316+7 = 23 network bits: 255.255.254.0.

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  • 2012 AD · 1 mark

In which type of application UDP is used instead of TCP?

Answer

UDP is used where speed and low overhead matter more than reliability, or where a small request-reply is exchanged:

  • Real-time multimedia: VoIP, video conferencing, live streaming, online games (late packets are useless, retransmission is pointless).
  • Query-response services such as DNS, DHCP, SNMP, TFTP.
  • Broadcast and multicast applications, since UDP is connectionless.

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  • 2012 AD · 1 mark

Which routing protocols are used for communication within the same domain and between the different domains?

Answer

  • Within the same domain (autonomous system): interior gateway protocols (intra-domain routing) such as RIP, OSPF and IGRP.
  • Between different domains: exterior gateway protocols (inter-domain routing), mainly BGP (earlier EGP).

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  • 2012 AD · 1 mark

What is the native protocol we use in WWW and what type of protocol is it?

Answer

The native protocol of the WWW is HTTP (HyperText Transfer Protocol). It is an application-layer, client-server, request-response protocol that runs over TCP (port 80; HTTPS uses 443). It is stateless, since the server keeps no memory of earlier requests.

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  • 2012 AD · 3 marks

"WAN Network is hybrid topology". Justify with the help of simple design and explain the working principle of WAN.

Answer

Justification. A WAN spans cities or countries and is built by joining many LANs and sub-networks using routers and leased/switched links. Each LAN may use its own topology (bus, star, ring), and the routers are interconnected in a mesh or partial-mesh pattern. Because several different topologies are combined in one network, a WAN is a hybrid topology.

Simple design

  • LAN A (star) connects to router R1.
  • LAN B (bus) connects to router R2.
  • LAN C (ring) connects to router R3.
  • R1, R2 and R3 are joined to each other (and to further routers) by point-to-point links, forming a mesh in the backbone.

Working principle

  1. A host sends its data to the local router (the gateway) in packets.
  2. The router reads the destination address and consults its routing table, which is built by routing protocols such as RIP, OSPF or BGP.
  3. The packet is forwarded hop by hop over the carrier links (leased line, frame relay, ATM, MPLS, etc.) using packet switching, store-and-forward.
  4. The last router delivers the packet to the destination LAN and host.
  5. Alternate paths in the mesh give reliability: if a link fails, routers choose another route.

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  • 2012 AD · 3 marks

Explain TDM with its types and identify the problem associated with each types.

Answer

Time Division Multiplexing (TDM) is a digital multiplexing technique in which several low-rate signals share one high-rate link by occupying it in turn, each for a fixed time slot. It has two types.

1. Synchronous TDM

Each input is assigned a fixed time slot in every frame, whether or not it has data. The slots repeat in a cyclic order, and the receiver demultiplexes by slot position.

  • Problem: wasted capacity. If a source is idle its slot goes empty, so the link is poorly utilised. Also all inputs must be synchronised, and framing bits are needed.

2. Asynchronous (statistical) TDM

Slots are given dynamically only to inputs that have data. Since the slot position no longer identifies the source, every slot carries an address (header) identifying the sender.

  • Problem: overhead of address bits in each slot, and extra complexity of buffering and control. If the load is higher than the link capacity, buffers overflow and data is delayed or lost.

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  • 2012 AD · 3 marks

Explain the working principle of selective repeat ARQ and point out the merit and demerit of it over Go Back N ARQ.

Answer

Working principle. Selective Repeat ARQ is a sliding-window protocol in which the sender transmits several frames without waiting for acknowledgement (window size up to 2m−12^{m-1} for mm-bit sequence numbers). The receiver also has a window and accepts frames that arrive out of order, buffering them. Each frame is acknowledged individually. If a frame is damaged or lost, the receiver sends a NAK (or the sender's timer for that frame expires), and only that frame is retransmitted. When the missing frame arrives, the receiver delivers all buffered frames to the network layer in correct order.

Merit over Go-Back-N: Only the erroneous frame is retransmitted, so bandwidth is used efficiently, especially on noisy or high-delay links.

Demerit over Go-Back-N: The receiver needs a buffer and logic to store and reorder out-of-order frames, and the sender needs a timer per frame, so it is more complex and costly. Window size is limited to half the sequence-number space.

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  • 2012 AD · 3 marks

"Slotted ALOHA is improvement over Pure ALOHA". Explain.

Answer

In pure ALOHA a station sends a frame whenever it has one, so a frame collides with any other frame transmitted during the vulnerable period of two frame times (2Tfr2T_{fr}). Maximum throughput is only S=Ge−2GS = Ge^{-2G}, with maximum 12e≈18.4%\frac{1}{2e} \approx 18.4\% at G=0.5G = 0.5.

In slotted ALOHA, time is divided into slots equal to one frame time, and a station may transmit only at the start of a slot. Frames either collide completely or do not collide at all, and the vulnerable period is reduced to one frame time (TfrT_{fr}). Throughput becomes S=Ge−GS = Ge^{-G}, with maximum 1e≈36.8%\frac{1}{e} \approx 36.8\% at G=1G = 1.

Thus slotted ALOHA halves the collision probability and doubles the maximum throughput, which is why it is an improvement (at the cost of needing clock synchronisation).

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  • 2012 AD · 3 marks

What are the logic behind shortest path routing? Explain each logic with example.

Answer

Shortest path routing finds the least-cost path between a source and destination, where "cost" may be a metric such as hop count, distance, delay or monetary cost. The logic is to model the subnet as a weighted graph (routers are nodes, links are edges) and select the path with the least total weight. The main logics are:

1. Shortest by number of hops. Every link has weight 1; the path with the fewest routers is chosen. Example: A to D via A-B-D (2 hops) is preferred to A-B-C-D (3 hops).

2. Shortest by distance. Weights are physical lengths of links. Example: A-C-D with 3 + 4 = 7 km beats A-B-D with 5 + 6 = 11 km.

3. Shortest by delay / bandwidth / cost. Weight is a mean queuing and transmission delay, or the inverse of the bandwidth, or tariff, and the lowest sum is chosen. Example: if A-B-D has delays 10 ms and 20 ms (30 ms) and A-C-D has 15 ms and 5 ms (20 ms), A-C-D is chosen even if it has the same hop count.

A weighted combination of these metrics can also be used. The path is computed by an algorithm such as Dijkstra's: label the source 0 as permanent, relax the tentative labels of its neighbours, make the smallest tentative label permanent, and repeat until the destination is permanent.

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  • 2012 AD · 3 marks

What is IP address? Explain the IP V4 frame format.

Answer

IP address. An IP address is a logical, hierarchical address that uniquely identifies a host's connection to the Internet. IPv4 addresses are 32 bits long, written in dotted decimal (e.g. 192.168.1.10), and consist of a network part (netid) and a host part (hostid).

IPv4 datagram (frame) format. A datagram has a header (20 to 60 bytes) followed by data. The header fields are:

  • Version (4 bits): IP version, 4.
  • IHL / header length (4 bits): header length in 32-bit words (min 5).
  • Type of service (8 bits): priority and delay/throughput/reliability preferences.
  • Total length (16 bits): header plus data, up to 65,535 bytes.
  • Identification (16 bits): identifies fragments of the same datagram.
  • Flags (3 bits): DF (do not fragment), MF (more fragments).
  • Fragment offset (13 bits): position of the fragment in the original datagram.
  • Time to live (8 bits): hop limit, decremented by each router; the datagram is discarded at 0.
  • Protocol (8 bits): upper-layer protocol (6 = TCP, 17 = UDP, 1 = ICMP).
  • Header checksum (16 bits): error detection over the header only.
  • Source address (32 bits) and destination address (32 bits).
  • Options (variable) and padding: security, record route, timestamp, etc.
  • Data: payload from the transport layer.

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  • 2012 AD · 3 marks

Explain the methods of IP allocation using DHCP.

Answer

DHCP (Dynamic Host Configuration Protocol) automatically assigns IP configuration (IP address, subnet mask, default gateway, DNS server) to hosts. A DHCP server can allocate addresses in three ways:

  1. Manual (static) allocation: The administrator maps a host's MAC address to a fixed IP address in the server. The host always receives the same address.
  2. Automatic allocation: The server assigns an IP address from its pool the first time the host connects, and the host keeps that address permanently.
  3. Dynamic allocation: The server leases an address from the pool for a limited time. When the lease expires (or the host releases it) the address returns to the pool and can be reused. This is the most common method since it reuses a limited address space.

Process (DORA): the client broadcasts DHCP Discover; the server replies with DHCP Offer; the client sends DHCP Request for the offered address; the server confirms with DHCP Acknowledgement. The client renews the lease before it expires.

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  • 2012 AD · 3 marks

Explain each of the services provided by transport layer.

Answer

The transport layer provides process-to-process (end-to-end) delivery. Its services are:

  1. Process-to-process communication: Using port numbers, data is delivered to the correct application process on the host, not just to the host.
  2. Segmentation and reassembly: Messages from the application are split into segments with sequence numbers and reassembled at the destination.
  3. Connection control: Connection-oriented service (TCP) sets up, maintains and releases a connection; connectionless service (UDP) sends each segment independently.
  4. Flow control: Prevents a fast sender from overwhelming a slow receiver, using a sliding window end to end.
  5. Error control: Detects damaged, lost, duplicate or out-of-order segments using checksums, acknowledgements and retransmission.
  6. Congestion control: Regulates the sending rate to avoid overloading the network.
  7. Multiplexing and demultiplexing: Several applications share one network connection; the transport layer multiplexes at the sender and demultiplexes at the receiver by port number.
  8. Addressing: Service access points (ports) and socket addresses identify the applications.

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  • 2012 AD · 3 marks

Explain the working principles of DNS with the help of architecture.

Answer

The Domain Name System (DNS) is a distributed, hierarchical database that translates human-friendly host names (e.g. www.example.com) into IP addresses.

Architecture

  • Name space: a tree with the root at the top, then top-level domains (.com, .edu, .np, ...), second-level domains and so on. A domain name is the path read from the node up to the root.
  • Root servers: know the addresses of the TLD servers.
  • TLD servers: know the authoritative servers of each second-level domain.
  • Authoritative name servers: hold the actual records (A, MX, NS, CNAME) of a zone.
  • Local (resolver) name server: the client's own server, which accepts queries and does the lookup, caching results.

Working

  1. An application calls the resolver on the host with a name.
  2. The resolver asks the local name server.
  3. If it is not in its cache, the local server asks a root server, which refers it to the TLD server, which refers it to the authoritative server (iterative queries; recursive query is also possible).
  4. The authoritative server returns the IP address.
  5. The local server caches it (for the record's TTL) and replies to the host, which then connects using the IP address.

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  • 2013 AD · 1 mark

Difference between the term bit rate and baud rate.

Answer

  • Bit rate is the number of bits transmitted per second (bps).
  • Baud rate (signalling rate) is the number of signal units (signal changes) per second.

They are related by bit rate=baud rate×bits per signal element\text{bit rate} = \text{baud rate} \times \text{bits per signal element}. They are equal only when each signal element carries one bit; with multi-level signalling (e.g. 4 levels carrying 2 bits) the bit rate is greater than the baud rate.

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  • 2013 AD · 1 mark

List basic characteristics of signal.

Answer

The basic characteristics of a signal are:

  • Amplitude: the peak value (strength) of the signal.
  • Frequency: number of cycles per second (Hz); the inverse of the period.
  • Period: time for one cycle.
  • Phase: position of the waveform relative to time zero.
  • Wavelength: distance covered by one cycle.
  • Bandwidth: the range of frequencies the signal occupies.

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  • 2013 AD · 1 mark

State Shannon capacity theorem.

Answer

Shannon's theorem gives the maximum data rate (channel capacity) of a noisy channel:

C=Blog⁡2(1+SN)C = B \log_2\left(1 + \frac{S}{N}\right)

where CC is the capacity in bits per second, BB the bandwidth in Hz, and S/NS/N the signal-to-noise power ratio. Error-free transmission is theoretically possible at any rate below CC, but not above it.

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  • 2013 AD · 1 mark

What is the purpose of bridge in a LAN?

Answer

A bridge connects two or more LAN segments at the data link layer and makes them work as one network. It filters traffic by learning MAC addresses: a frame is forwarded to another segment only if the destination is there, otherwise it is kept local. This divides the network into separate collision domains, reduces traffic and collisions, extends the network's size and number of stations, and can connect LANs of different types.

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  • 2013 AD · 1 mark

Define Asynchronous communication.

Answer

Asynchronous communication is a transmission mode in which data is sent one character (or byte) at a time without a shared clock between sender and receiver. Each character is framed by a start bit and one or more stop bits, which let the receiver synchronise for that character, and gaps of arbitrary length may occur between characters. It is simple and cheap but has high overhead (about 20 percent for 8-bit data) and is used for keyboards and serial ports.

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  • 2013 AD · 1 mark

What is socket address? Give example.

Answer

A socket address is the combination of an IP address and a port number. It uniquely identifies a process running on a host, and a pair of socket addresses (source and destination) identifies a connection.

Example: 192.168.10.5:80 means port 80 (a web server) on the host with IP 192.168.10.5.

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  • 2013 AD · 1 mark

Differentiate between Packet Switching and Circuit Switching.

Answer

Circuit switchingPacket switching
A dedicated path is set up before transferNo dedicated path; packets are routed independently (datagram) or along a virtual circuit
Resources reserved for the whole callResources used on demand and shared
Fixed bandwidth; idle capacity is wastedEfficient use of bandwidth
Constant delay after setup; no store-and-forwardVariable delay due to queuing and store-and-forward
Bit-stream transparentNot transparent; data is split into packets with headers
Poor for bursty data, good for voiceGood for bursty data
Example: telephone networkExample: Internet

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  • 2013 AD · 1 mark

What is the necessity of Port address even if there is IP address for the device?

Answer

An IP address identifies only the host (or its interface), but many applications run on one host at the same time (web browser, e-mail, file transfer). The port address (16-bit number) identifies the particular process or service, so the transport layer can deliver incoming data to the correct application, and so multiple applications can use the network simultaneously (multiplexing). Well-known ports such as 80 (HTTP) and 25 (SMTP) also let clients locate services.

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  • 2013 AD · 1 mark

What is the role of DNS?

Answer

DNS (Domain Name System) maps human-readable domain names to IP addresses (and the reverse), so users do not need to remember numeric addresses. It is a distributed, hierarchical database of name servers. It also supplies other information such as mail server records (MX) and aliases (CNAME).

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  • 2013 AD · 3 marks

What are the factors that determine whether a communication system is a LAN, MAN or WAN?

Answer

The type of network is decided mainly by the following factors:

  1. Geographical area covered: LAN covers a room, building or campus (up to a few km); MAN covers a city (about 5 to 50 km); WAN covers countries or continents.
  2. Ownership and management: a LAN is privately owned by one organisation; a MAN may be private or public; a WAN is typically built from public carriers' links and is often shared.
  3. Data rate and delay: LANs have high speed (10 Mbps to Gbps) and low delay; MANs are medium; WANs have lower speeds and larger delay.
  4. Transmission media and technology: LANs use twisted pair, coaxial or fibre with Ethernet or token ring; MANs use fibre, DQDB or FDDI; WANs use leased lines, satellite, microwave, ATM or frame relay with routers.
  5. Error rate: LANs have low error rates; WANs have higher error rates.
  6. Cost and number of connected stations.

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  • 2013 AD · 3 marks

If SNR is 40db and bandwidth is 3.6KHZ calculate theoretical information rate.

Answer

Using Shannon's formula C=Blog⁡2(1+S/N)C = B\log_2(1 + S/N).

Convert SNR from decibels:

SN=1040/10=104=10000\frac{S}{N} = 10^{40/10} = 10^4 = 10000

Then

C=3600×log⁡2(1+10000)=3600×log⁡2(10001)≈3600×13.288≈47,836 bpsC = 3600 \times \log_2(1 + 10000) = 3600 \times \log_2(10001) \approx 3600 \times 13.288 \approx 47{,}836 \text{ bps}

The theoretical information rate is about 47.8 kbps.

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  • 2013 AD · 3 marks

Explain TCP/IP Reference Model and the function of its various layers.

Answer

The TCP/IP model is the protocol suite on which the Internet is built. It has four layers (some texts show five by splitting the lowest layer).

1. Host-to-network (Network access) layer: Defines how data is physically sent over the local network. It handles framing, physical addressing (MAC), media access and transmission of bits (Ethernet, Wi-Fi, PPP). It covers the OSI physical and data link layers.

2. Internet layer: Provides logical addressing and routing of packets across networks, independent of the medium. Protocols: IP, ICMP, ARP, RARP, IGMP. It gives connectionless, best-effort delivery.

3. Transport layer: Provides end-to-end, process-to-process delivery with port addresses. TCP gives reliable, connection-oriented delivery with segmentation, flow control, error and congestion control; UDP gives unreliable, connectionless delivery.

4. Application layer: Provides services to the user and applications, combining OSI session, presentation and application layers. Protocols: HTTP, FTP, SMTP, Telnet, DNS, DHCP, SNMP.

Data moves down the layers at the sender, each layer adding its header (encapsulation), and up at the receiver where headers are removed.

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  • 2013 AD · 3 marks

Why is CSMA/CD not suitable for wireless LANs? Describe one suitable MAC protocol for wireless LANs.

Answer

Why CSMA/CD is unsuitable

  • CSMA/CD needs a station to detect a collision while it is transmitting. A wireless radio cannot transmit and listen at the same time on the same channel, since its own signal overwhelms the weak incoming signal.
  • Because of the hidden terminal problem (two stations out of each other's range both send to a common receiver) and the exposed terminal problem, a station cannot reliably sense whether the channel is busy at the receiver. Collisions occur at the receiver and are not heard by the senders.
  • Signal fading and limited range mean not every station hears every other station.

Suitable protocol: CSMA/CA (IEEE 802.11 DCF)

  1. The station senses the channel. If it is idle for a DIFS period, it transmits (otherwise it waits and uses a random backoff counter that counts down only while the channel is idle).
  2. Optionally, the sender sends a short RTS (request to send); the receiver replies with CTS (clear to send). Other stations hearing the RTS or CTS set their NAV (network allocation vector) and keep silent for the duration, which solves the hidden terminal problem.
  3. The data frame is sent, and the receiver replies with an ACK after SIFS. If no ACK arrives, the sender assumes a collision, increases its backoff window and retries.

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  • 2013 AD · 3 marks

If a institution has IP address 198.167.45.0 and it needs 6 subnets. Answer the following:
i. Which class does this IP belongs to?\nii. How many valid hosts are available in each subnet?\niii. Calculate the subnet mask of given IP.

Answer

i. Class: The first octet is 198, which lies in 192 to 223, so it is Class C (default mask 255.255.255.0, 8 host bits).

ii. Hosts per subnet: For 6 subnets we need nn bits with 2n≥62^n \ge 6, so n=3n = 3 (8 subnets). Host bits left: 8−3=58 - 3 = 5.

Valid hosts per subnet=25−2=30\text{Valid hosts per subnet} = 2^5 - 2 = 30

iii. Subnet mask: Network bits = 24+3=2724 + 3 = 27. The last octet is 11100000 = 224, so the mask is 255.255.255.224 (/27).

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  • 2013 AD · 3 marks

In which case do you prefer TCP over UDP? Explain in detail.

Answer

TCP is preferred when reliable, ordered delivery of all data matters more than speed or low overhead.

Reasons

  • Reliability: TCP uses sequence numbers, acknowledgements, checksums and retransmission, so lost or corrupted segments are recovered. UDP gives no such guarantee.
  • Ordered delivery: segments are reassembled in order at the receiver.
  • Flow control and congestion control: the sliding window and congestion algorithms prevent overloading the receiver and the network.
  • Connection-oriented: a connection is set up with a three-way handshake and closed properly, suitable for long sessions.

Typical uses: web browsing (HTTP/HTTPS), e-mail (SMTP, POP3, IMAP), file transfer (FTP), remote login (SSH, Telnet), banking and database transactions. In these, losing even one byte makes the data useless or incorrect.

UDP is preferred only for delay-sensitive or simple request-reply applications such as voice, video and DNS.

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  • 2013 AD · 3 marks

What is the role of DHCP server? Explain its working.

Answer

Role. A DHCP (Dynamic Host Configuration Protocol) server automatically gives hosts the network configuration they need: an IP address, subnet mask, default gateway and DNS server addresses. It removes manual configuration, avoids address conflicts, and lets a limited pool of addresses be reused by leasing them for a limited time. It is an application-layer, client-server protocol using UDP (server port 67, client port 68).

Working (DORA)

  1. Discover: a new host (with no IP) broadcasts a DHCPDISCOVER message to find a server.
  2. Offer: a server replies with DHCPOFFER containing an unused address from its pool, with mask, gateway, DNS and lease time.
  3. Request: the client broadcasts DHCPREQUEST to accept the offer (which also tells other servers it was declined).
  4. Acknowledge: the server sends DHCPACK confirming the lease; the client configures its interface.

The client renews the lease before it expires, or sends DHCPRELEASE when finished, returning the address to the pool. If the server and clients are on different networks, a relay agent forwards the messages.

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  • 2013 AD · 3 marks

Define adaptive and non adaptive routing algorithms. Explain shortest path algorithm with example.

Answer

Adaptive routing. Routing decisions change dynamically as the network topology and traffic (load) change. Routers exchange information and recompute routes periodically. Examples: distance vector and link state routing. It responds to failures and congestion but is more complex.

Non-adaptive (static) routing. Routes are computed in advance and loaded into routers at boot time; they do not change with the current traffic or topology. Examples: shortest path routing, flooding. It is simple but cannot react to failures or congestion.

Shortest path algorithm (Dijkstra). The subnet is a weighted graph; the aim is the least-cost path from a source to all other nodes.

  1. Mark the source with distance 0 (permanent); all others have distance infinity (tentative).
  2. For the node just made permanent, update the tentative distance of each neighbour: d(v)=min⁡(d(v),d(u)+w(u,v))d(v) = \min(d(v), d(u) + w(u,v)).
  3. Make the tentative node with the smallest distance permanent.
  4. Repeat until the destination is permanent; the path is traced back through predecessors.

Example. Edges: A-B = 2, A-C = 5, B-C = 1, B-D = 4, C-D = 1. From A:

  • Start: A = 0. Neighbours: B = 2, C = 5.
  • Fix B (2): C becomes min(5, 2+1) = 3, D = 2+4 = 6.
  • Fix C (3): D becomes min(6, 3+1) = 4.
  • Fix D (4).

The shortest path from A to D is A-B-C-D with cost 4.

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  • 2013 AD · 3 marks

Differentiate between Slotted ALOHA and pure ALOHA

Answer

Pure ALOHASlotted ALOHA
Station transmits whenever it has dataStation transmits only at the beginning of a time slot
No time synchronisation neededAll stations must be synchronised to slot boundaries
Vulnerable time is 2Tfr2T_{fr}Vulnerable time is TfrT_{fr}
Partial collisions are possibleCollisions are complete (only within the same slot)
Throughput S=Ge−2GS = Ge^{-2G}Throughput S=Ge−GS = Ge^{-G}
Maximum throughput about 18.4 percent (at G=0.5G = 0.5)Maximum throughput about 36.8 percent (at G=1G = 1)
SimpleMore efficient but needs clock control

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  • 2014 AD · 1 mark

Why signals are modulated before transmission?

Answer

Signals are modulated, that is, a low-frequency baseband signal is impressed on a high-frequency carrier, because:

  • Efficient radiation needs antenna size comparable to the wavelength, and high carrier frequencies make antennas practical.
  • It allows frequency-division multiplexing, so many signals share one medium without interference.
  • The signal is matched to the passband of the channel (e.g. a digital signal over a telephone line, or radio spectrum).
  • It reduces noise and interference and allows longer-distance transmission.

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  • 2014 AD · 1 mark

What are the advantages of using optical fiber?

Answer

Advantages of optical fibre:

  • Very high bandwidth and data rate (Gbps to Tbps).
  • Low attenuation, so long distances need fewer repeaters.
  • Immune to electromagnetic interference and crosstalk.
  • Greater security, as tapping is difficult.
  • Light weight and small diameter compared with copper cable.
  • Resistant to corrosion and does not cause electrical hazards or sparks.

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  • 2014 AD · 1 mark

What is the purpose of piggybacking?

Answer

Piggybacking improves the efficiency of bidirectional data transfer. Instead of sending a separate acknowledgement frame, the receiver delays the ACK and attaches it to the next outgoing data frame going in the same direction. This saves bandwidth and reduces the number of frames, as the ACK uses only a few bits in the data frame header.

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  • 2014 AD · 1 mark

How does ARQ correct errors?

Answer

ARQ (Automatic Repeat reQuest) corrects errors by retransmission. The receiver detects errors in a frame (using CRC or checksum). If the frame is correct, it sends an acknowledgement (ACK); if it is damaged, it sends a NAK or discards it silently. The sender keeps a copy and a timer for each frame, and if no ACK arrives before the timer expires (or a NAK arrives) it retransmits the frame. Stop-and-Wait, Go-Back-N and Selective Repeat are ARQ schemes.

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  • 2014 AD · 1 mark

Which multiplexing is preferred in digital transmission and why?

Answer

Time Division Multiplexing (TDM) is preferred in digital transmission. Digital signals are naturally divided into time units (bits or bytes), so each source can be given its own time slot on the shared high-speed link. TDM uses the full link bandwidth in each slot, needs no guard bands or analog filters, is less affected by noise and intermodulation, and is easy to implement with digital circuits (as in T1/E1 and SONET). FDM, by contrast, suits analog signals.

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  • 2014 AD · 1 mark

Why ARP protocol is required?

Answer

ARP (Address Resolution Protocol) is required because data is delivered on a LAN using the physical (MAC) address, while applications and IP use logical (IP) addresses. When a host knows only the IP address of the next hop, ARP broadcasts a request and the owner replies with its MAC address, which is then cached in the ARP table so the frame can be built.

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  • 2014 AD · 1 mark

What is the function of TTL field?

Answer

The Time To Live (TTL) field in the IP header limits the life of a packet. Each router decrements it by 1; when it reaches 0 the packet is discarded and an ICMP Time Exceeded message is sent. This prevents packets from looping endlessly in the network.

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  • 2014 AD · 1 mark

Why do you CIDR?

Answer

CIDR (Classless Inter-Domain Routing) is used to avoid the waste of addresses and the growth of routing tables caused by classful addressing. It allows variable-length prefixes (e.g. 192.168.1.0/22) so blocks of the right size can be allocated, and several networks can be aggregated into one routing entry (route summarisation).

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  • 2014 AD · 1 mark

Which protocols and respective ports are used for sending and receiving emails?

Answer

  • Sending email: SMTP, TCP port 25 (587 for submission).
  • Receiving email: POP3, TCP port 110, or IMAP, TCP port 143.

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  • 2014 AD · 1 mark

List the various ranges of private IP address.

Answer

Private IPv4 ranges (RFC 1918):

  • Class A: 10.0.0.0 to 10.255.255.255 (10.0.0.0/8)
  • Class B: 172.16.0.0 to 172.31.255.255 (172.16.0.0/12)
  • Class C: 192.168.0.0 to 192.168.255.255 (192.168.0.0/16)

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  • 2014 AD · 3 marks

Describe the basic model of communication.

Answer

The basic communication model has five components:

  1. Source: generates the data to be transmitted (e.g. a computer or phone).
  2. Transmitter: converts the data into signals suitable for the transmission medium (e.g. a modem).
  3. Transmission system: the medium or network (wire, fibre, radio, or a network of switches) that carries the signal.
  4. Receiver: converts the received signal back into data.
  5. Destination: takes the incoming data (e.g. a computer or printer).

Data flows from source to transmitter, across the transmission system, to the receiver and finally the destination. The message, sender, receiver, medium and protocol are the five essential elements, with protocols defining the rules that make the exchange meaningful. Noise on the medium may corrupt the signal, so error control is needed.

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  • 2014 AD · 3 marks

What frame will be transmitted for the data D=11000011 using the divisor polynomial x3+1x^3+1 for CRC?

Answer

Divisor x3+1x^3+1 is 10011001 (degree 3), so append 3 zeros to the data: 1100001100011000011000.

Divide modulo 2:

  • 1100 XOR 1001 = 0101, bring down: 1011 XOR 1001 = 0010, bring down: 0100 (leading 0, shift), 1000 XOR 1001 = 0001, then bring down remaining bits: 0010, 0100, 1001...

Carrying out the full long division of 1100001100011000011000 by 10011001 gives quotient 1000010110000101... and remainder 000000.

Check: D⋅x3=x10+x9+x4+x3D\cdot x^3 = x^{10}+x^9+x^4+x^3. Since x3≡1x^3\equiv 1 mod (x3+1)(x^3+1), x10≡xx^{10}\equiv x, x9≡1x^9\equiv 1, x4≡xx^4\equiv x, x3≡1x^3\equiv 1, so the remainder is x+1+x+1=0x+1+x+1=0.

CRC = 000000, so the transmitted frame is

11000011 000=1100001100011000011\,000 = 11000011000

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  • 2014 AD · 3 marks

Explain and describe NAT.

Answer

Network Address Translation (NAT) maps private (non-routable) IP addresses inside a network to one or more public IP addresses, so many hosts can share a limited number of public addresses. It conserves IPv4 addresses and hides the internal network.

Working: A NAT router keeps a translation table. For an outgoing packet it replaces the private source address (and port) with its public address (and a unique port) and records the mapping. For the reply, it looks up the table and rewrites the destination back to the private host.

Types:

  • Static NAT: one-to-one fixed mapping.
  • Dynamic NAT: private addresses mapped from a pool of public addresses.
  • PAT / NAT overload: many private hosts share one public address, distinguished by port numbers.

Drawback: breaks end-to-end addressing and complicates some protocols (e.g. IPsec, peer-to-peer).

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  • 2014 AD · 3 marks

Differentiate between Selective Reject ARQ and Stop and Wait ARQ?

Answer

BasisStop-and-Wait ARQSelective Reject ARQ
Frames in transitOne frame at a time; waits for ACKMany frames sent (window size > 1)
Window sizeSender 1, receiver 1Sender and receiver ≤2m−1\le 2^{m-1}
RetransmissionRetransmits the lost frame after timeoutRetransmits only the damaged/lost frame
EfficiencyLow, channel idle while waitingHigh, link kept busy
Receiver bufferNot neededNeeded to reorder out-of-order frames
ComplexitySimpleComplex at sender and receiver

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  • 2014 AD · 3 marks

Describe Distance Vector Routing Algorithm and differentiate it with shortest path routing algorithm.

Answer

Distance vector routing (Bellman-Ford): each router keeps a table of the best known distance (cost) and next hop to every destination. Periodically, each router sends its whole table only to its directly connected neighbours. A router receiving a vector updates its table using

Dx(y)=min⁡v{c(x,v)+Dv(y)}D_x(y)=\min_v\{c(x,v)+D_v(y)\}

where c(x,v)c(x,v) is the cost to neighbour vv and Dv(y)D_v(y) is vv's advertised distance to yy. Repeating this converges to the least-cost routes. RIP is an example. Problems: slow convergence and count-to-infinity.

Difference from shortest path (link-state/Dijkstra) routing:

Distance vectorShortest path (link state)
Knows only neighbours' distance informationEach router knows the complete topology
Shares its table with neighbours onlyFloods link-state to all routers
Bellman-Ford algorithmDijkstra's algorithm
Slow convergence, count-to-infinityFast convergence
Less CPU and memoryMore CPU and memory

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  • 2014 AD · 5 marks

Which random access method is used in Ethernet? Explain its working principle.

Answer

Ethernet uses CSMA/CD (Carrier Sense Multiple Access with Collision Detection).

Working principle:

  1. Carrier sense: a station with a frame listens to the channel before sending.
  2. If the channel is idle, it transmits; if busy, it waits until idle (1-persistent).
  3. Collision detection: while transmitting, the station keeps listening. If the signal on the line differs from what it sent, a collision has occurred.
  4. On collision, the station stops and sends a short jam signal so all stations know.
  5. It then waits a random time using binary exponential backoff: after the nn-th collision it picks KK randomly from 00 to 2n−12^n-1 slot times and waits K×K\times slot time before retrying.
  6. After a maximum number of attempts (16) the frame is dropped and an error is reported.

For detection to work, the frame transmission time must be at least twice the propagation delay, so a minimum frame size (64 bytes in classic Ethernet) is required.

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  • 2014 AD · 5 marks

What is an IP address? Explain IPv4 header with diagram.

Answer

An IP address is a logical address that uniquely identifies a host's connection to a network. IPv4 addresses are 32 bits long, written in dotted decimal (e.g. 192.168.1.10), and divided into a network part and a host part.

IPv4 header (20 to 60 bytes):

 0       4       8              16   19                  31
+-------+-------+---------------+--------------------------+
| Ver   | IHL   | Type of Service|     Total Length        |
+-------+-------+---------------+-----+--------------------+
|        Identification         |Flags| Fragment Offset    |
+---------------+---------------+-----+--------------------+
|  Time to Live |   Protocol    |     Header Checksum      |
+---------------+---------------+--------------------------+
|                    Source IP Address                     |
+----------------------------------------------------------+
|                 Destination IP Address                   |
+----------------------------------------------------------+
|                 Options (if any) + Padding               |
+----------------------------------------------------------+
  • Version (4 bits): 4 for IPv4.
  • IHL (4 bits): header length in 32-bit words.
  • Type of Service (8 bits): priority/QoS.
  • Total length (16 bits): header plus data, up to 65535 bytes.
  • Identification, Flags, Fragment offset: used for fragmentation and reassembly.
  • TTL (8 bits): hop limit.
  • Protocol (8 bits): upper-layer protocol (6 TCP, 17 UDP, 1 ICMP).
  • Header checksum (16 bits): error detection on the header.
  • Source and destination address (32 bits each).
  • Options: optional, with padding to a 32-bit boundary.

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  • 2014 AD · 5 marks

Which protocol does DHCP use? Explain working principle of DHCP with diagram.

Answer

DHCP (Dynamic Host Configuration Protocol) uses UDP, with port 67 at the server and port 68 at the client. It automatically assigns an IP address, subnet mask, default gateway and DNS server to hosts.

Working (DORA):

Client                                   DHCP Server
  | --- 1. DHCP Discover (broadcast) ------> |
  | <-- 2. DHCP Offer (IP, mask, lease) ---- |
  | --- 3. DHCP Request (accept offer) ----> |
  | <-- 4. DHCP Acknowledgement ------------ |
  1. Discover: the new client broadcasts (source 0.0.0.0, destination 255.255.255.255) to find servers.
  2. Offer: a server offers an unused IP address with configuration and lease time.
  3. Request: the client broadcasts its acceptance of one offer, so other servers withdraw theirs.
  4. Acknowledgement: the server confirms the lease and the client starts using the address.

The address is leased for a fixed time; the client renews at about 50% of the lease and releases it when no longer needed.

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  • 2015 AD · 1 mark

How LAN is different from WAN?

Answer

A LAN covers a small area (a room, building or campus) and is usually privately owned, with high data rates and low error rates and delay. A WAN spans cities, countries or continents, uses leased or public carrier links, and has lower speed, higher delay and higher cost.

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  • 2015 AD · 1 mark

What is the relation between error rate and data rate?

Answer

Error rate increases as data rate increases. At a higher data rate each bit lasts a shorter time, so the same noise burst corrupts more bits, and the signal is more affected by noise and attenuation.

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  • 2015 AD · 1 mark

Which layer is responsible for end to end reliable transmission?

Answer

The transport layer is responsible for end-to-end reliable transmission (using TCP, with error control, flow control and sequencing).

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  • 2015 AD · 1 mark

What is the basic difference between OSI Model and TCP/IP Model?

Answer

The OSI model has 7 layers and is a generic, protocol-independent reference model. The TCP/IP model has 4 layers (application, transport, internet, network access) and is built around the actual protocols used on the Internet.

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  • 2015 AD · 1 mark

What is ICMP?

Answer

ICMP (Internet Control Message Protocol) is a network-layer protocol used by hosts and routers to send error reports and query messages about IP packet delivery, e.g. destination unreachable, time exceeded, and echo request/reply (used by ping and traceroute).

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  • 2015 AD · 1 mark

What is the information that is provided by a subnet mask?

Answer

A subnet mask tells which bits of an IP address identify the network (and subnet) and which identify the host. ANDing the address with the mask gives the network address, and it also shows the number of hosts per subnet.

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  • 2015 AD · 1 mark

List two disadvantages of ISDN.

Answer

  1. Limited data rate (basic rate 128 kbps, primary rate about 2 Mbps), which is too slow for modern broadband needs.
  2. Expensive to install and operate, and it needs special terminal equipment and digital exchanges; it has largely been replaced by DSL and fibre.

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  • 2015 AD · 1 mark

Define super-netting.

Answer

Supernetting is combining several contiguous smaller (usually class C) networks into one larger network with a shorter prefix, e.g. 192.168.0.0/24 to 192.168.3.0/24 combined as 192.168.0.0/22. It reduces the size of routing tables.

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  • 2015 AD · 1 mark

Compare UDP and TCP with respect to retransmission service and overhead.

Answer

  • Retransmission: TCP provides retransmission of lost or corrupted segments using acknowledgements and timers; UDP provides no retransmission.
  • Overhead: TCP has a larger header (20 to 60 bytes) and connection set-up and state, so overhead is high; UDP has only an 8-byte header and no connection, so overhead is low.

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  • 2015 AD · 1 mark

Why HTTP is called stateless protocol?

Answer

HTTP is stateless because the server treats each request independently and keeps no memory of previous requests from the same client. State (e.g. login sessions) must be added using cookies or session tokens.

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  • 2015 AD · 3 marks

Identify and explain different type of transmission media.

Answer

Transmission media are of two types.

1. Guided (wired) media

  • Twisted pair: two insulated copper wires twisted together to reduce interference. UTP and STP; cheap, used in LANs and telephone lines.
  • Coaxial cable: central copper conductor, insulator, shield and outer jacket. Higher bandwidth and better noise immunity than twisted pair; used in cable TV and older Ethernet.
  • Optical fibre: light pulses through a glass or plastic core. Very high bandwidth, low attenuation, immune to electromagnetic interference, but costly to install.

2. Unguided (wireless) media

  • Radio waves: omnidirectional, pass through walls; used in radio, Wi-Fi and mobile.
  • Microwaves: unidirectional, line-of-sight, used in satellite and point-to-point links.
  • Infrared: short range, cannot pass walls; used in remote controls.

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  • 2015 AD · 3 marks

Identify multiple access protocols and explain protocol that avoids collision.

Answer

Multiple access protocols fall into three groups:

  1. Random access: ALOHA (pure and slotted), CSMA, CSMA/CD, CSMA/CA.
  2. Controlled access: reservation, polling, token passing.
  3. Channelization: FDMA, TDMA, CDMA.

CSMA/CA (Carrier Sense Multiple Access with Collision Avoidance) avoids collisions, and is used in wireless LANs (802.11). A station:

  1. Senses the channel and waits until it is idle for a DIFS interval.
  2. Waits a random backoff time before sending (the counter freezes while the channel is busy).
  3. Optionally exchanges RTS (request to send) and CTS (clear to send) frames to reserve the channel and handle hidden terminals.
  4. Sends the frame; the receiver replies with an ACK after SIFS. With no ACK the sender assumes a collision and retries with a larger backoff window.

Token passing also avoids collisions: only the station holding the token may transmit.

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  • 2015 AD · 3 marks

Define packet switching. Explain TDM

Answer

Packet switching: the message is divided into small packets, each with a header carrying source and destination addresses. Packets are sent independently through the network, may take different routes, and are reassembled at the destination. Links are shared by many users, which gives efficient use of bandwidth (datagram and virtual-circuit approaches).

Time Division Multiplexing (TDM): several signals share one link by dividing time into fixed slots. Each input is given a slot in turn, and the cycle repeats as a frame.

  • Synchronous TDM: a slot is assigned to each input in every frame, even if it has nothing to send, so capacity can be wasted.
  • Statistical (asynchronous) TDM: slots are given only to inputs that have data, with address information added, which improves utilisation.

TDM is applied to digital signals, and the demultiplexer at the receiver separates the slots by timing.

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  • 2015 AD · 3 marks

Explain ADSL with its advantages and disadvantages.

Answer

ADSL (Asymmetric Digital Subscriber Line) provides high-speed data over ordinary telephone copper lines. It is asymmetric: downstream (to the user) is much faster (up to about 8 Mbps or more) than upstream (about 1 Mbps). The line bandwidth is divided using FDM into a voice channel (0 to 4 kHz), an upstream band and a downstream band (using DMT modulation), so voice and data can be used at the same time. A splitter at the user's end and a DSLAM at the exchange separate the signals.

Advantages

  • Uses the existing telephone line, so low installation cost.
  • Always-on connection, and phone and internet can be used simultaneously.
  • Higher speed than dial-up and ISDN.
  • Dedicated line per user, not shared with neighbours.

Disadvantages

  • Speed falls with distance from the exchange (limited to about 5 km).
  • Upload speed is low.
  • Susceptible to noise and interference.
  • Not available in all areas.

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  • 2015 AD · 5 marks

What is switch? How switch forwards incoming frame to proper destination? If the message sequence is 11110110011 and the generator polynomial is G(x)=x2+1G(x)=x^2+1, calculate the transmitted frame.

Answer

Switch: a data link layer (layer 2) device that connects the devices of a LAN and forwards frames only to the port of the destination, using MAC addresses. Each port is its own collision domain, which gives better performance than a hub.

Forwarding: the switch keeps a MAC address table (port to MAC).

  1. Learning: for each incoming frame, it records the source MAC address against the arrival port.
  2. Forwarding: if the destination MAC is in the table, the frame is sent only out of that port.
  3. Flooding: if the destination is unknown (or is a broadcast), the frame is sent out of all ports except the incoming one.
  4. Filtering: if the destination is on the same port as the source, the frame is dropped.

CRC calculation: the message is 1111011001111110110011 and G(x)=x2+1G(x)=x^2+1, i.e. divisor 101101. Append 2 zeros: 11110110011001111011001100.

Using x2≡1(modG)x^2\equiv 1 \pmod{G}, the message 1111011001111110110011 has terms at x10,x9,x8,x7,x5,x4,x1,x0x^{10},x^9,x^8,x^7,x^5,x^4,x^1,x^0. Multiplying by x2x^2 gives exponents 12,11,10,9,7,6,3,212,11,10,9,7,6,3,2. Reducing mod GG (xk≡xk mod 2x^k\equiv x^{k \bmod 2}): exponents 12, 10, 6, 2 give 11 each (four terms, sum 0); exponents 11, 9, 7, 3 give xx each (four terms, sum 0). The remainder is 00, so the CRC is 0000.

Transmitted frame:

11110110011 00=111101100110011110110011\,00 = 1111011001100

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  • 2015 AD · 5 marks

Suppose you are network administrator at XYZ organization, XYZ organization has five departments. You need to subnet the IP address 131.215.0.0/20 among these departments. Identify subnet mask and each department IP address range.

Answer

Network: 131.215.0.0/20, which spans 131.215.0.0 to 131.215.15.255 (4096 addresses).

Subnet mask: five departments need at least 5 subnets, so borrow 3 bits (23=8≥52^3=8\ge5). New prefix is /20+3=/23/20+3=/23:

Mask=255.255.254.0\text{Mask}=255.255.254.0

Each subnet has 29=5122^{9}=512 addresses (510 usable hosts), a block size of 2 in the third octet.

DepartmentNetwork addressUsable host rangeBroadcast
1131.215.0.0/23131.215.0.1 to 131.215.1.254131.215.1.255
2131.215.2.0/23131.215.2.1 to 131.215.3.254131.215.3.255
3131.215.4.0/23131.215.4.1 to 131.215.5.254131.215.5.255
4131.215.6.0/23131.215.6.1 to 131.215.7.254131.215.7.255
5131.215.8.0/23131.215.8.1 to 131.215.9.254131.215.9.255

The remaining three subnets (131.215.10.0, 12.0, 14.0 /23) are kept for future growth.

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  • 2015 AD · 5 marks

What is socket address? Explain the services provided by transport layer.

Answer

Socket address: the combination of an IP address and a port number, e.g. 192.168.1.5:80. It uniquely identifies a communication endpoint (a process on a host); a connection is identified by a pair of socket addresses (client and server).

Services provided by the transport layer

  1. Process-to-process delivery: uses port numbers to deliver data to the correct application process (multiplexing and demultiplexing).
  2. Segmentation and reassembly: splits the message into segments with sequence numbers and rebuilds it at the receiver.
  3. Connection control: connection-oriented (TCP) or connectionless (UDP) service.
  4. Flow control: end-to-end control so the sender does not overwhelm the receiver (sliding window).
  5. Error control: detects damaged, lost or duplicate segments using checksums, acknowledgements and retransmission.
  6. Congestion control: regulates the sending rate according to network load.
  7. Addressing: port numbers identify the service.

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  • 2016 AD · 1 mark

List limitation Synchronous Time Division Multiplying.

Answer

  • A slot is reserved for each input even when it has nothing to send, so link capacity is wasted.
  • All inputs get equal slots, so it does not suit sources of different or bursty rates.
  • Synchronisation between sender and receiver is needed (framing/sync bits add overhead).
  • The number of connected inputs is fixed by the number of slots.

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  • 2016 AD · 1 mark

Differentiate between Thermal Noise and Impulse Noise

Answer

  • Thermal noise is caused by random motion of electrons in conductors and electronic devices. It is continuous, present at all frequencies and always present (proportional to temperature and bandwidth), and it cannot be eliminated.
  • Impulse noise is a sudden, short spike of high amplitude caused by lightning, switching or power-line faults. It is non-continuous and irregular; it has little effect on analogue data but can corrupt several bits of digital data (burst errors).

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  • 2016 AD · 1 mark

What are the three events that are involved in the creation of connection in Transport layer?

Answer

The three events in transport layer connection are connection establishment, data transfer, and connection release. (In TCP, establishment uses the three-way handshake: SYN, SYN+ACK, ACK.)

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  • 2016 AD · 1 mark

What are the advantages of contention based approach of MAC Techniques?

Answer

  • Simple to implement, with no central controller needed.
  • Low delay and efficient at light load, since a station can send at once when the channel is idle.
  • Easy to add or remove stations.

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  • 2016 AD · 1 mark

What is meant by bit stuffing?

Answer

Bit stuffing is a framing technique used in bit-oriented protocols (like HDLC). The frame is delimited by the flag 01111110. To keep this pattern from appearing in the data, the sender inserts an extra 0 after every five consecutive 1s in the data; the receiver removes any 0 that follows five consecutive 1s.

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  • 2016 AD · 1 mark

How the problem of collision is solved in Token Ring Configuration?

Answer

In a token ring, a small control frame called the token circulates around the ring. Only the station holding the token may transmit, so two stations never transmit at the same time and collisions cannot occur. After sending, the station releases the token to the next station.

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  • 2016 AD · 1 mark

List out the three procedure that are used by BGP?

Answer

BGP uses three procedures: neighbour acquisition (opening a session and exchanging OPEN messages), neighbour reachability (periodic KEEPALIVE messages), and network reachability (UPDATE messages advertising or withdrawing routes).

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  • 2016 AD · 1 mark

What are the functions of DTE and DCE?

Answer

  • DTE (Data Terminal Equipment), such as a computer or terminal, is the source or destination of the data; it generates and consumes data and controls the communication.
  • DCE (Data Circuit-terminating Equipment), such as a modem or CSU/DSU, connects the DTE to the network, converting data into signals suitable for the line (and vice versa) and providing clocking and signal conversion.

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  • 2016 AD · 1 mark

List any two functions used in BSD socket API.

Answer

  • socket() creates a new socket (endpoint) and returns a descriptor.
  • bind() assigns a local IP address and port number to the socket. (Others include listen(), accept(), connect(), send() and recv().)

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  • 2016 AD · 1 mark

How shortest path can be guarantee in Flooding Routing Algorithm?

Answer

Flooding sends every incoming packet out on every line except the one it arrived on, so a copy of the packet travels along every possible path at the same time. The copy that travels along the shortest (fastest) path reaches the destination first, so the shortest path is always used, which is why flooding is guaranteed to find it. It does this without any routing table, at the cost of many duplicate packets.

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  • 2016 AD · 3 marks

Why is data encoding required? Explain any two encoding standard used for Digital Data Analog Signal encoding.

Answer

Why encoding is required. Data must be converted into signals that suit the transmission medium. Encoding makes the signal fit the channel's bandwidth, keeps sender and receiver synchronised, reduces errors and noise effects, and lets digital data be sent over analog lines such as telephone lines.

Digital data to analog signal. A carrier wave is modulated by the bit stream. Two standard techniques:

(1) Amplitude Shift Keying (ASK). The amplitude of the carrier changes with the bit: for example a high amplitude represents 1 and a low (or zero) amplitude represents 0. It is simple but very sensitive to noise.

(2) Frequency Shift Keying (FSK). The frequency of the carrier changes with the bit: frequency f1f_1 represents 1 and f2f_2 represents 0, with amplitude constant. It is less affected by noise than ASK but needs more bandwidth.

(Phase Shift Keying, PSK, changes the phase of the carrier, for example 0° for 0 and 180° for 1.)

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  • 2016 AD · 3 marks

Explain ARQ technique that is based on Stop and wait flow control protocol.

Answer

The ARQ (Automatic Repeat reQuest) technique based on Stop-and-Wait flow control is Stop-and-Wait ARQ.

Working:

  1. The sender transmits one frame (carrying a sequence number 0 or 1) and starts a timer.
  2. It then waits for an acknowledgement (ACK) before sending the next frame.
  3. If the frame arrives correctly, the receiver sends an ACK; the sender then sends the next frame.
  4. If the frame is damaged or lost, the receiver discards it (no ACK). When the timer expires, the sender retransmits the same frame.
  5. If the ACK is lost, the sender times out and retransmits; the receiver detects the duplicate from the sequence number, discards it and re-sends the ACK.

Advantages: simple, needs little buffer. Disadvantage: low efficiency on long-delay links because the channel is idle while waiting for each ACK.

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  • 2016 AD · 3 marks

Explain IEEE 802 standard for Ethernet based network.

Answer

IEEE 802 is a family of standards for LANs and MANs that covers the physical and data link layers. The data link layer is split into two sublayers: LLC (Logical Link Control, IEEE 802.2) and MAC (Medium Access Control).

Standards for Ethernet-based networks:

  • IEEE 802.3 is Ethernet itself. It uses CSMA/CD as the MAC method and defines frame format and physical specifications, such as 10Base-5, 10Base-2, 10Base-T, 100Base-TX (Fast Ethernet), 1000Base-T (Gigabit Ethernet).
  • IEEE 802.2 is the LLC sublayer common to all 802 LANs; it gives flow and error control.
  • IEEE 802.1 covers LAN architecture, bridging, VLANs (802.1Q) and spanning tree (802.1D).
  • Related: 802.4 (Token Bus), 802.5 (Token Ring), 802.11 (Wireless LAN).

The 802.3 frame contains preamble, start frame delimiter, destination address, source address, length/type, data (46 to 1500 bytes) and FCS (CRC-32).

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  • 2016 AD · 3 marks

Explain the packet switching techniques where a connection setup time is required.

Answer

The packet switching technique that needs a connection setup is the virtual circuit approach.

Virtual circuit packet switching:

  1. Call setup: a call-request packet travels from source to destination, and each node chooses the route and records an entry (virtual circuit number) in its table. The destination replies with a call-accept.
  2. Data transfer: all packets carry only a short virtual circuit identifier and follow the same pre-established route, arriving in order. Nodes need not make routing decisions per packet.
  3. Call clearing: a clear-request packet releases the circuit.

Bandwidth is not reserved as in circuit switching, so it is shared on demand. Examples: X.25, Frame Relay, ATM. By contrast, the datagram approach needs no setup and routes each packet independently.

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  • 2016 AD · 3 marks

Explain DNS with its architecture.

Answer

DNS (Domain Name System) is a distributed hierarchical database that maps human-readable domain names (such as www.example.com) to IP addresses, and the reverse. It works at the application layer, normally over UDP port 53.

Architecture:

  • Name space: a hierarchical tree. The root is at the top, followed by Top-Level Domains (.com, .org, .np), second-level domains, and then hosts.
  • Root name servers: know the addresses of the TLD servers.
  • TLD servers: know the authoritative servers of each domain under them.
  • Authoritative name servers: hold the actual records for a zone.
  • Local (recursive) resolver: the client's resolver queries on the client's behalf and caches answers.

Resolution: the client asks the local resolver, which asks root, then TLD, then the authoritative server, and returns the IP to the client. Queries may be recursive or iterative. The tree is divided into zones, which allows load sharing and fault tolerance through replication.

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  • 2016 AD · 5 marks

Write down the advantages of adaptive routing over non-adaptive routing. Compare between Distance vector and link State Routing algorithm.

Answer

Advantages of adaptive over non-adaptive routing

  • Routes change with current network topology and traffic, so link or node failures are handled automatically.
  • Congestion is avoided by moving traffic to lightly loaded paths, giving better performance and throughput.
  • Better use of network resources and more reliable delivery.
  • No manual configuration of routes is needed.

(The drawbacks are higher complexity and routing overhead.)

Distance Vector vs Link State

BasisDistance VectorLink State
InformationEach router knows only distances (vectors) to destinations via neighboursEach router has the complete topology map
ExchangeRouting table sent to neighbours onlyLink state packets flooded to all routers
AlgorithmBellman-FordDijkstra
UpdatesPeriodicTriggered on change
ConvergenceSlow; count-to-infinity problemFast
ResourcesLow CPU and memoryMore CPU and memory
ExampleRIPOSPF

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  • 2016 AD · 5 marks

Suppose you are network administrator at ABC organization, ABC organization have three departments A, B and C which consists of 52, 110 and 25 computers respectively, you need to subnet the class C IP address 216.215.214.0 among these departments, now perform the subletting using a suitable technique.

Answer

Address: 216.215.214.0 (class C, /24, 256 addresses). Because the departments differ in size, use VLSM (Variable Length Subnet Masking), allocating the largest department first.

DeptComputersBlock sizePrefix and maskNetworkUsable rangeBroadcast
B110128 (126 hosts, 27−22^7-2)/25, 255.255.255.128216.215.214.0.1 to .126.127
A5264 (62 hosts, 26−22^6-2)/26, 255.255.255.192216.215.214.128.129 to .190.191
C2532 (30 hosts, 25−22^5-2)/27, 255.255.255.224216.215.214.192.193 to .222.223

The addresses 216.215.214.224 to .255 remain free for future use. The blocks do not overlap, and each is sized to be the smallest power of two that holds the hosts plus network and broadcast addresses.

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  • 2016 AD · 5 marks

Design UDP header format. Explain basic Data Communication Model.

Answer

UDP header format (fixed 8 bytes, 64 bits):

 0                16                31
+-----------------+-----------------+
| Source Port     | Destination Port|
|    (16 bits)    |    (16 bits)    |
+-----------------+-----------------+
| Length          | Checksum        |
|    (16 bits)    |    (16 bits)    |
+-----------------+-----------------+
|            Data (payload)         |
+-----------------------------------+
  • Source port: port of the sending process (optional).
  • Destination port: port of the receiving process.
  • Length: total length of header plus data, in bytes.
  • Checksum: error detection over header, data and a pseudo-header (optional in IPv4).

Basic data communication model. It has five components:

  1. Source (sender): generates the data.
  2. Transmitter: converts the data into signals suitable for the medium (for example a modem).
  3. Transmission medium (channel): carries the signal (wire, fibre, air).
  4. Receiver: converts the received signal back to data.
  5. Destination: the device that uses the data.

Data flows source, transmitter, medium, receiver, destination. Noise on the medium can corrupt the signal. Communication depends on protocol, message and the transmission mode (simplex, half duplex, full duplex).

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  • 2017 AD · 1 mark

Which is the most vulnerable components in basic data communication model and why?

Answer

The transmission medium (channel) is the most vulnerable component, because it is exposed to noise, attenuation, interference and eavesdropping, and a break in the medium stops all communication.

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  • 2017 AD · 1 mark

What are the advantage and disadvantage of guard band in frequency division multiplexing?

Answer

Advantage: guard bands are unused frequency gaps between channels, which prevent adjacent channels from interfering with each other (crosstalk). Disadvantage: they waste bandwidth because no data is carried in them.

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  • 2017 AD · 1 mark

Define the process segmentation in Transport Layer.

Answer

Segmentation is the process in which the transport layer divides a large message from the application layer into smaller pieces called segments, adds a header with sequence numbers, and sends them. The receiver uses the sequence numbers to reassemble them in the correct order.

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  • 2017 AD · 1 mark

Define Frame Synchronization.

Answer

Frame synchronization is the process by which the receiver identifies where each frame starts and ends in the received bit stream, usually by special flag bytes or bit patterns (for example the flag 01111110 in HDLC), so that it can correctly extract the frame's contents.

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  • 2017 AD · 1 mark

Compare between 10Base-5 and 10 Base-T.

Answer

Basis10Base-510Base-T
MediumThick coaxial cableUnshielded twisted pair (UTP)
TopologyBusStar (via hub or switch)
Max segment length500 m100 m
ConnectionVampire tap and AUI transceiverRJ-45 connector
Cost, easeCostly, difficult to installCheap, easy to install and troubleshoot
Fault effectOne break fails the whole networkOne cable fault affects only one node

Both operate at 10 Mbps with baseband signalling.

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  • 2017 AD · 1 mark

Why a routing algorithm should have a property of rapid convergence?

Answer

Rapid convergence means all routers quickly agree on a consistent view of the network after a topology change. If convergence is slow, routers hold stale information, causing routing loops, lost packets and unreachable destinations in the meantime.

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  • 2017 AD · 1 mark

When a "source quench" and "Time exceeded" message are generated?

Answer

Source quench (ICMP type 4) is generated by a router or destination host when it is congested or its buffer overflows and has to discard datagrams; it asks the sender to slow down.

Time exceeded (ICMP type 11) is generated by a router when a datagram's TTL reaches zero and it is discarded, or by the destination host when the fragments of a datagram do not all arrive before the reassembly timer expires.

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  • 2017 AD · 1 mark

Mentioned the advantage of ISDN over ADSL.

Answer

ISDN is fully digital end to end and is symmetric. It offers guaranteed dedicated channels (B channels of 64 kbps each) with fast call setup, and can carry voice, data and video simultaneously on separate channels. It also works over the existing line without the distance limit and line-quality problems that ADSL has, and gives better voice quality and reliability.

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  • 2017 AD · 1 mark

What is the similarity between IP and UDP?

Answer

Both IP and UDP are connectionless and unreliable: they provide best-effort delivery, with no connection setup, no acknowledgement, no retransmission and no guaranteed ordering. Each packet (datagram) is handled independently.

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  • 2017 AD · 1 mark

What is the advantages of IMAP over POP?

Answer

With IMAP, mail stays on the server and is accessed and managed there, so the same mailbox can be viewed consistently from multiple devices. It supports server-side folders, searching and partial download of messages, and keeps read/unread status synchronised. POP normally downloads and deletes mail from the server onto a single device.

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  • 2017 AD · 3 marks

Explain Digital Signal Encoding with its types.

Answer

Digital signal encoding (line coding) converts digital data (bits) into a digital signal (voltage levels) that can be sent over a medium. Good codes give synchronisation, low bandwidth, no DC component and error detection.

Types:

  1. Unipolar (NRZ unipolar): only one polarity is used; 1 is a positive voltage, 0 is zero voltage. Simple but has a DC component and synchronisation problems.
  2. Polar: both positive and negative voltages are used.
    • NRZ-L: the level represents the bit (for example positive for 0, negative for 1).
    • NRZ-I: a transition at the start of a bit means 1; no transition means 0.
    • RZ: the signal returns to zero in the middle of each bit, which helps synchronisation but needs more bandwidth.
    • Manchester: a transition in the middle of every bit (low to high for 1, high to low for 0); self-clocking. Used in Ethernet.
    • Differential Manchester: a mid-bit transition always exists; the presence of a transition at the start of the bit means 0, absence means 1. Used in Token Ring.
  3. Bipolar (AMI): 0 is zero voltage; 1s alternate between positive and negative voltage, which removes the DC component and allows error detection.

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  • 2017 AD · 3 marks

Explain any one error detection method with example.

Answer

Cyclic Redundancy Check (CRC) is a widely used error detection method.

Method:

  1. Choose a generator polynomial (divisor) of r+1r+1 bits.
  2. Append rr zeros to the data.
  3. Divide the result by the divisor using modulo-2 (XOR) division.
  4. The rr-bit remainder is the CRC; replace the appended zeros with it and transmit.
  5. The receiver divides the received frame by the same divisor. A zero remainder means no error; a non-zero remainder means an error.

Example. Data = 1101011, divisor = 1011 (so r=3r=3). Append 3 zeros to get 1101011000 and divide by 1011 using modulo-2 (XOR) division. Working from the left, XOR the divisor wherever the leading bit is 1 (first step: 1101011000 XOR 1011000000 = 0110011000), and keep going bit by bit. This leaves the remainder 110. The transmitted frame is 1101011110. At the receiver, dividing 1101011110 by 1011 leaves remainder 000, so no error is detected. If any bit were flipped, the remainder would generally be non-zero.

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  • 2017 AD · 3 marks

Explain the procedures that are involved in routing of information using BGP.

Answer

BGP (Border Gateway Protocol) is the exterior gateway protocol used between autonomous systems (AS). It is a path-vector protocol that runs over TCP port 179.

Procedure:

  1. TCP connection: two BGP routers (peers) set up a TCP session.
  2. OPEN message: peers exchange OPEN messages containing AS number, hold time and BGP identifier, and agree on parameters.
  3. KEEPALIVE: a KEEPALIVE confirms the OPEN, and keepalives are then sent periodically to show the peer is alive.
  4. UPDATE messages: peers exchange the full routing table first, then only changes. An UPDATE advertises new reachable prefixes with path attributes (AS path, next hop) or withdraws routes that are no longer valid.
  5. Route selection: the router applies policy and attributes (shortest AS path, local preference and so on) to pick the best path, avoiding loops by rejecting paths that contain its own AS number.
  6. NOTIFICATION: sent when an error occurs, after which the session is closed.

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  • 2017 AD · 3 marks

What is DHCP? Explain the types of IP allocation using DHCP.

Answer

DHCP (Dynamic Host Configuration Protocol) is an application layer protocol that automatically assigns IP addresses and other configuration (subnet mask, default gateway, DNS server) to hosts from a DHCP server's pool. It uses UDP ports 67 (server) and 68 (client), and its process is Discover, Offer, Request, Acknowledge.

Types of IP allocation:

  1. Automatic allocation: the server assigns a permanent IP address to a client, which keeps it indefinitely.
  2. Dynamic allocation: the server leases an IP address from a pool for a limited time. When the lease expires, the address returns to the pool and can be reused. This makes efficient use of addresses.
  3. Manual (static) allocation: the administrator maps a particular IP to a client's MAC address, and the DHCP server always gives that client that address.

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  • 2017 AD · 5 marks

Identify and explain flow control protocol with its types.

Answer

Flow control is a data link layer function which stops a fast sender from overwhelming a slow receiver by regulating the amount of data sent before an acknowledgement is received.

Types of flow control protocols:

1. Stop-and-Wait. The sender sends one frame and waits for its ACK before sending the next. It is simple and needs only a one-frame buffer, but wastes bandwidth on long links because the channel is idle while waiting. Efficiency is low when propagation delay is large.

2. Sliding Window. The sender may send several frames (up to the window size WW) before waiting for ACKs. Frames are numbered, and the window slides forward as ACKs arrive, giving better channel utilisation. Variants:

  • Go-Back-N ARQ: the sender window is NN and the receiver window is 1. If a frame is lost or damaged, the sender retransmits that frame and all later ones. It is simple but wastes bandwidth.
  • Selective Repeat ARQ: both windows are larger than 1. Only the damaged or lost frame is retransmitted, and the receiver buffers out-of-order frames. It is more efficient but more complex.

Maximum window size is 2m−12^m-1 for Go-Back-N and 2m−12^{m-1} for Selective Repeat, with mm-bit sequence numbers.

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  • 2017 AD · 5 marks

Define the importance of super-netting. Find a summarized route and a common subnet mask if a router has the following networks in its routing table.
198.132.98.0\n198.132.99.0\n198.132.100.0\n198.132.101.0\n198.132.105.0

Answer

Importance of supernetting. Supernetting combines several contiguous smaller (class C) networks into one larger network by shortening the prefix. It reduces the size of routing tables (route aggregation), decreases routing overhead and conserves address space by allocating blocks of the right size.

Summarised route. Write the third octet of each network in binary:

NetworkThird octet
198.132.98.001100010
198.132.99.001100011
198.132.100.001100100
198.132.101.001100101
198.132.105.001101001

The first 4 bits of the third octet (0110) are common to all; the 5th bit differs. So the common prefix is 16+4=2016 + 4 = 20 bits. Setting the remaining bits to zero gives third octet 01100000 = 96.

  • Summarised route: 198.132.96.0/20
  • Common subnet mask: 255.255.240.0

(This block also covers 198.132.96.0 to 198.132.111.0.)

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  • 2017 AD · 5 marks

Define Transportation Layer. Explain the services provided by transport layer.

Answer

The transport layer (layer 4 of OSI) provides end-to-end, process-to-process communication between application processes on different hosts. Its protocols are TCP and UDP.

Services provided:

  1. Process-to-process delivery (addressing): port numbers identify the sending and receiving application processes.
  2. Segmentation and reassembly: messages are divided into segments with sequence numbers and reassembled at the destination.
  3. Connection control: connection-oriented (TCP, with setup and release) or connectionless (UDP) service.
  4. Flow control: end-to-end control so the sender does not overrun the receiver (for example TCP sliding window).
  5. Error control: detection of damaged, lost or duplicated segments by checksum, acknowledgement and retransmission, giving reliable delivery.
  6. Congestion control: regulating traffic into the network to avoid congestion.
  7. Multiplexing and demultiplexing: several applications can share one network connection, and data is delivered to the right process.

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  • 2018 AD · 1 mark

Why there is a bandwidth wastage in Frequency Division Multiplexing?

Answer

In FDM each channel is permanently allocated its own frequency band, and guard bands are placed between channels. If a user has nothing to send, its band stays idle and cannot be used by others, and the guard bands carry no data, so bandwidth is wasted.

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  • 2018 AD · 1 mark

Which layer is responsible for name recognition and security?

Answer

The presentation layer (layer 6 of the OSI model) is responsible for name recognition (syntax and data representation) and security through encryption and decryption.

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  • 2018 AD · 1 mark

How the problem of frame duplication is solved in Stop and Wait protocol?

Answer

By using sequence numbers (0 and 1, alternating) on the frames. If an ACK is lost and the sender retransmits, the receiver sees the same sequence number as the previous frame, recognises it as a duplicate, discards it and re-sends the ACK.

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  • 2018 AD · 1 mark

Differentiate between Logical and Physical Topology.

Answer

Physical topology is the actual physical layout of devices, cables and nodes (bus, star, ring, mesh). Logical topology describes how data actually flows between devices through the network, regardless of the physical layout. For example, a network wired as a physical star with a hub works as a logical bus.

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  • 2018 AD · 1 mark

Why Adaptive routing algorithm is less secure?

Answer

Adaptive routing relies on routing information exchanged dynamically between routers, which can be forged, altered or injected by an attacker (for example false route advertisements). Since routes change automatically based on this information, a compromised router can redirect or blackhole traffic, whereas fixed routes in non-adaptive routing are not affected this way.

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  • 2018 AD · 1 mark

State any two improvements in IPV6 as compare to IPV4.

Answer

  1. Larger address space: 128-bit addresses instead of 32-bit, which removes address exhaustion.
  2. Simpler, fixed header with built-in security: a fixed 40-byte header (no header checksum) for faster processing, and IPsec support built in. (Other improvements are better support for QoS, auto-configuration, and no fragmentation by routers.)

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  • 2018 AD · 1 mark

How the Crash Recovery is done in datagram subnet?

Answer

In a datagram subnet each packet is routed independently and no state about connections is kept in the routers, so if a router crashes only the packets it held are lost. Once it restarts, it simply carries on routing new packets, and lost data is recovered by the transport layer (end hosts retransmit). No circuit has to be re-established, unlike in virtual circuit subnets.

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  • 2018 AD · 1 mark

Differentiated between SMTP and POP.

Answer

SMTP (Simple Mail Transfer Protocol, port 25) is a push protocol used to send mail from a client to a server and between mail servers. POP (Post Office Protocol, POP3, port 110) is a pull protocol used by the receiver's client to retrieve mail from its mailbox on the server.

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  • 2018 AD · 1 mark

What are the types of channels used in ISDN?

Answer

ISDN uses three types of channels: the B channel (Bearer, 64 kbps, carries user voice and data), the D channel (Delta, 16 or 64 kbps, carries signalling and control information), and the H channel (high-speed, 384, 1536 or 1920 kbps, for high-rate data and video).

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  • 2018 AD · 1 mark

What is the purpose of Keep Alive message in BGP?

Answer

The BGP KEEPALIVE message is sent periodically (normally every one third of the hold time) between peers to show that the router is still alive and the connection is active. If neither a keepalive nor an update arrives before the hold timer expires, the peer is considered down and its routes are withdrawn. It is also used to acknowledge an OPEN message.

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  • 2018 AD · 3 marks

Explain Time Division Multiplexing with its types.

Answer

Time Division Multiplexing (TDM) is a digital multiplexing technique in which several signals share one channel by taking turns: the time axis is divided into slots, and each source is given a slot in a repeating frame. The whole bandwidth is used by one source at a time.

Types:

  1. Synchronous TDM: each input is given a fixed time slot in every frame whether or not it has data. It is simple, but wastes capacity when a source is idle. A frame contains one slot per input, and sync bits are added for frame synchronisation.
  2. Asynchronous (statistical) TDM: slots are assigned dynamically only to sources that have data to send. Each slot carries an address or identifier of its source. It uses the channel more efficiently and supports more devices, but needs buffering and addressing overhead.

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  • 2018 AD · 3 marks

For a Satellite TV Channel with Signal to Noise Ratio 30dB and range of frequencies 600hz, 800hz, 1.2 khz and 1.5khz, find maximum data rate.

Answer

Use Shannon's capacity formula for a noisy channel:

C=Blog⁡2(1+SNR)C = B\log_2(1+\mathrm{SNR})

SNR: 30 dB=10log⁡10(SNR)30\text{ dB} = 10\log_{10}(\mathrm{SNR}), so SNR=103=1000\mathrm{SNR} = 10^{3} = 1000.

Bandwidth: the frequencies range from 600 Hz to 1.5 kHz, so

B=1500−600=900 HzB = 1500 - 600 = 900\text{ Hz}

Capacity:

C=900log⁡2(1+1000)=900log⁡2(1001)≈900×9.967≈8970 bpsC = 900\log_2(1+1000) = 900\log_2(1001) \approx 900\times 9.967 \approx 8970\text{ bps}

Maximum data rate is about 8.97 kbps.

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  • 2018 AD · 3 marks

Explain IPV4 header format with supportive diagram.

Answer

The IPv4 header is 20 bytes minimum (up to 60 bytes with options).

 0       4       8              16   19            24            31
+-------+-------+---------------+---------------------------------+
|Version| IHL   | Type of Service|        Total Length            |
+-------+-------+---------------+-----+---------------------------+
|        Identification         |Flags|     Fragment Offset       |
+---------------+---------------+-----+---------------------------+
|  Time to Live |   Protocol    |        Header Checksum          |
+---------------+---------------+---------------------------------+
|                      Source IP Address                          |
+-----------------------------------------------------------------+
|                    Destination IP Address                       |
+-----------------------------------------------------------------+
|                    Options (if any) + Padding                   |
+-----------------------------------------------------------------+
  • Version (4 bits): 4 for IPv4.
  • IHL (4 bits): header length in 32-bit words (minimum 5).
  • Type of Service (8 bits): priority and quality of service.
  • Total Length (16 bits): header plus data, up to 65,535 bytes.
  • Identification (16 bits): identifies fragments of one datagram.
  • Flags (3 bits): includes DF (don't fragment) and MF (more fragments).
  • Fragment Offset (13 bits): position of the fragment in the original datagram.
  • TTL (8 bits): maximum hops; decremented at each router, discarded at zero.
  • Protocol (8 bits): upper layer protocol (6 = TCP, 17 = UDP, 1 = ICMP).
  • Header Checksum (16 bits): error detection for the header.
  • Source and Destination address (32 bits each).
  • Options and padding: optional extras, padded to a 32-bit boundary.

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  • 2018 AD · 5 marks

Explain any two major characteristics of UDP. If a data to be transmitted is 101001 and a divisor is 1101 then what will be the message to be send?

Answer

Two major characteristics of UDP

  1. Connectionless and unreliable: UDP sends datagrams without setting up a connection. It gives no acknowledgement, retransmission, ordering or flow control, so there is little delay and little overhead.
  2. Light header: The header is only 8 bytes (source port, destination port, length, checksum), so it is fast and efficient. It is suitable for DNS, streaming and VoIP.

CRC calculation

Data = 101001 and divisor = 1101 (degree 3), so append 3 zeros: 101001000.

Dividing 101001000 by 1101 using modulo-2 (XOR) division gives the remainder 001.

Transmitted message=101001 001=101001001\text{Transmitted message} = 101001\,001 = 101001001

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  • 2018 AD · 5 marks

You are given an IP address 150.152.0.0. You need to subnet the given IP into Five different Departments. Perform Subnetting and find the Subnet Mask, Network Address, Broad Cast Address and Usable host addresses in all subnet.

Answer

150.152.0.0 is a Class B address with the default mask 255.255.0.0 (/16).

Five departments need at least 5 subnets. Since 23=8≥52^3 = 8 \ge 5, borrow 3 bits.

New mask: /19 = 255.255.224.0. Block size is 32 in the third octet. Hosts per subnet are 213−2=81902^{13}-2 = 8190.

SubnetNetwork addressUsable host rangeBroadcast address
1150.152.0.0150.152.0.1 – 150.152.31.254150.152.31.255
2150.152.32.0150.152.32.1 – 150.152.63.254150.152.63.255
3150.152.64.0150.152.64.1 – 150.152.95.254150.152.95.255
4150.152.96.0150.152.96.1 – 150.152.127.254150.152.127.255
5150.152.128.0150.152.128.1 – 150.152.159.254150.152.159.255

The three remaining subnets (150.152.160.0, 150.152.192.0 and 150.152.224.0) are spare for future use.

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  • 2018 AD · 5 marks

Why Switching is necessary? Explain shortest path routing algorithm with example.

Answer

Why switching is necessary

Connecting every pair of devices directly would need a full mesh of n(n−1)/2n(n-1)/2 links, which is impractical and costly. A switched network connects each device to a switching node, and the nodes forward data towards its destination. Switching therefore shares links and resources, reduces cabling cost, scales to many users and lets data find a path through the network.

Shortest path routing (Dijkstra's algorithm)

The algorithm builds a graph with routers as nodes and links as weighted edges (distance, delay or cost). It finds the least-cost path from a source to every other node.

  1. Set the distance of the source to 0 and all other nodes to ∞\infty. Mark the source as permanent.
  2. Update the tentative distance of each neighbour of the latest permanent node: d(v)=min⁡(d(v),d(u)+w(u,v))d(v) = \min(d(v), d(u) + w(u,v)).
  3. Make the tentative node with the smallest distance permanent.
  4. Repeat until the destination (or all nodes) is permanent.

Example: Edges are A–B = 2, A–C = 5, B–C = 1, B–D = 4, C–D = 1. Source = A.

  • Start: A = 0, and B, C, D = ∞\infty.
  • From A: B = 2, C = 5. The smallest is B, which becomes permanent.
  • From B: C = min(5, 2+1) = 3 and D = 2+4 = 6. The smallest is C, which becomes permanent.
  • From C: D = min(6, 3+1) = 4. D becomes permanent.

The shortest path from A to D is A → B → C → D with cost 4.

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  • 2019 AD · 1 mark

How the Attenuation problem is overcome in Digital Signal Transmission?

Answer

In digital transmission, attenuation is overcome by repeaters. A repeater receives the weakened signal, regenerates the original 0s and 1s, and retransmits a fresh, clean signal. Because it regenerates the signal rather than amplifying it, noise is not accumulated.

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  • 2019 AD · 1 mark

Name two devices which function at the Physical Layer.

Answer

Two devices that work at the physical layer are the repeater and the hub. Modems and cables or connectors also work at this layer.

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  • 2019 AD · 1 mark

What do you mean by connection Oriented and Connection Less Services?

Answer

Connection-oriented service: A connection is established between sender and receiver before data transfer, data is exchanged over it, and the connection is released afterwards. It gives reliable, ordered delivery, as in TCP.

Connectionless service: Each packet is sent independently with no prior connection set-up and no guarantee of delivery or order. It is fast and has low overhead, as in UDP.

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  • 2019 AD · 1 mark

What is the advantage of Contention Based approach?

Answer

Contention-based approaches such as ALOHA and CSMA are simple and need no central control. A station with data can transmit at once, so delay is very low under light load. They are decentralised, flexible and easy to implement, and they make efficient use of the channel when traffic is light.

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  • 2019 AD · 1 mark

Why Non-Adaptive routing algorithm is secure and predictive?

Answer

A non-adaptive (static) routing algorithm uses routes that are fixed in advance and do not change with traffic or topology. The paths are therefore predictable, because every packet between two nodes follows the same known route. It is also secure, because routers do not exchange routing information, so there is little chance of fake routing updates or route poisoning. The routes are set only by the administrator.

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  • 2019 AD · 1 mark

Write down the fixed highest order bits for Class C Class D and Class E IP addresses.

Answer

ClassFixed high-order bitsFirst octet range
Class C110192 – 223
Class D1110224 – 239
Class E1111240 – 255

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  • 2019 AD · 1 mark

Why Buffering and Flow control are important elements of Transport Protocol?

Answer

The transport layer works between hosts whose speed and capacity may differ, so the sender could overwhelm the receiver. Flow control regulates the sending rate to match the receiver's ability, so data is not lost through overflow. Buffering gives the sender and receiver space to hold segments until they are sent, acknowledged, retransmitted or processed. Together they prevent loss, allow retransmission and give reliable, efficient delivery.

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  • 2019 AD · 1 mark

Mention the advantage of IMAP Over POP.

Answer

IMAP keeps the mail on the server and lets users manage it there (folders, flags, search). Several devices can therefore access the same mailbox and stay in sync. It can also download only message headers or selected parts, whereas POP normally downloads and deletes mail from the server, which ties it to a single device.

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  • 2019 AD · 1 mark

Explain any two advantage of ISDN.

Answer

  1. Integrated services: Voice, data, fax and video travel over one digital line, so separate networks are not needed.
  2. Fast and reliable digital transmission: The all-digital connection gives faster call set-up, higher speed (for example 128 kbps with BRI) and better quality with less noise than analog lines.

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  • 2019 AD · 1 mark

Explain any two Query message in ICMP.

Answer

  1. Echo request and echo reply: A host sends an echo request to test whether another host is reachable. The destination replies with an echo reply. This is the basis of the ping command.
  2. Timestamp request and reply: A host asks another host for its clock time. The reply carries the original, receive and transmit timestamps. It is used to find round-trip time and to synchronise clocks.

Other query messages are router solicitation and advertisement, and address mask request and reply.

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  • 2019 AD · 3 marks

Illustrate and explain Basic Data Communication Model briefly.

Answer

The basic data communication model has five components:

Message -> [Sender] --(Transmission medium)--> [Receiver] -> Message
              <---------- Protocol ---------->
  1. Message: The information to be communicated (text, numbers, image, audio or video).
  2. Sender: The device that sends the message, such as a computer or phone.
  3. Receiver: The device that receives the message.
  4. Transmission medium: The physical path from sender to receiver, such as twisted pair, coaxial cable, fibre or radio waves.
  5. Protocol: The set of rules that both devices follow for communication. Without a protocol, two devices may be connected but cannot understand each other.

Data flows in simplex, half-duplex or full-duplex mode.

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  • 2019 AD · 3 marks

If the channel capacity of a noiseless channel is 54 kbps and the bandwidth is 600 hz calculate the level of a signal.

Answer

For a noiseless channel the Nyquist formula applies:

C=2Blog⁡2LC = 2B\log_2 L

54000=2×600×log⁡2L⇒log⁡2L=540001200=4554000 = 2 \times 600 \times \log_2 L \Rightarrow \log_2 L = \frac{54000}{1200} = 45

L=245≈3.52×1013L = 2^{45} \approx 3.52 \times 10^{13}

The number of signal levels required is L=245L = 2^{45} (each level carries 45 bits). This is very large, which shows that the channel is impractical at this rate.

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  • 2019 AD · 3 marks

If a message to be transmitted is 1011001010 with a polynomial generator x3−1x^3-1. Calculate the transmitted bits.

Answer

The generator is taken as x3+1x^3+1 (binary 1001), because −1-1 and +1+1 are the same in modulo-2 arithmetic. The degree is 3, so append 3 zeros to the message:

1011001010 0001011001010\,000

Divide by 1001 using modulo-2 (XOR) division. The remainder is 001.

The transmitted bits are the message followed by the remainder:

1011001010 001=10110010100011011001010\,001 = 1011001010001

The receiver divides the received frame by 1001. A zero remainder means no error was detected.

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  • 2019 AD · 3 marks

Explain three different approaches for solving the problem of packet duplication in Flooding Routing Algorithm.

Answer

Flooding sends every incoming packet out on all lines except the one it arrived on, which produces many duplicate packets. Three ways to control the duplicates are:

  1. Hop counter: Each packet carries a hop count that is decremented at every router, and the packet is discarded when it reaches zero. The counter is initialised to the estimated path length (or the network diameter).
  2. Sequence numbers: Each source router puts a sequence number in the packets it sends. Every router keeps a record of the sequence numbers already seen for each source and discards any packet it has already forwarded.
  3. Selective flooding: Packets are forwarded only on lines that lead roughly in the direction of the destination, not on all lines. This reduces the duplicates.

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  • 2019 AD · 5 marks

There are four different department in an organization which have 100, 60, 25 and 20 numbers of hosts, perform subnetting if this organization owns CLASS C IP address 205.206.207.0 according to the department.

Answer

Class C address 205.206.207.0/24 has 256 addresses. Use VLSM, allocating the largest department first.

DeptHosts neededBlock sizeMaskNetwork addressUsable rangeBroadcast
A100128 (27−2=1262^7-2=126)/25 = 255.255.255.128205.206.207.0.1 – .126205.206.207.127
B6064 (26−2=622^6-2=62)/26 = 255.255.255.192205.206.207.128.129 – .190205.206.207.191
C2532 (25−2=302^5-2=30)/27 = 255.255.255.224205.206.207.192.193 – .222205.206.207.223
D2032 (30 usable)/27 = 255.255.255.224205.206.207.224.225 – .254205.206.207.255

The four blocks (128 + 64 + 32 + 32 = 256) use the full address space.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2019 AD · 5 marks

Explain TCP three way handshake process. Explain any three services provided by transport layer

Answer

TCP three-way handshake

It sets up a reliable connection between a client and a server.

  1. SYN: The client sends a segment with SYN = 1 and an initial sequence number xx.
  2. SYN + ACK: The server replies with SYN = 1, ACK = 1, its own sequence number yy, and acknowledgement number x+1x+1.
  3. ACK: The client sends ACK = 1 with acknowledgement number y+1y+1. The connection is now established and data transfer can start.

The handshake lets both sides agree on initial sequence numbers and confirm that each can send and receive.

Three services of the transport layer

  1. Process-to-process delivery: Port numbers are used to deliver data to the right application process, using multiplexing and demultiplexing.
  2. Reliable delivery: Using sequence numbers, acknowledgements and retransmission (TCP), data arrives complete and in order.
  3. Flow and congestion control: The sender's rate is regulated so that the receiver and the network are not overloaded.

Other services include segmentation and reassembly, and error control.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2021 AD · 1 mark

What is the advantage of Analog signal transmission as compare to digital signal transmission?

Answer

Analog transmission needs less bandwidth than digital transmission for the same information, and it is the natural form of real-world signals such as voice, so no conversion is needed. The equipment is simple and cheap, and the existing telephone infrastructure can be used.

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  • 2021 AD · 1 mark

Mention any two protocols of the Network Layer.

Answer

Two network layer protocols are IP (Internet Protocol) and ICMP (Internet Control Message Protocol). Others are ARP, IGMP, OSPF and RIP.

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  • 2021 AD · 1 mark

What is the problem of Character Count Method of Framing?

Answer

In the character count method, a field at the start of each frame gives the number of characters in it. If this count is corrupted by a transmission error, the receiver loses track of where the frame ends and where the next one begins. It then cannot resynchronise, even if it detects the checksum failure, because it does not know where to resume. Retransmission requests do not help either, since the receiver cannot say which frame to resend.

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  • 2021 AD · 1 mark

Differentiate between Persistent and Non-Persistent CSMA.

Answer

Basis1-persistent / p-persistent CSMANon-persistent CSMA
Channel busyStation keeps sensing continuously and transmits as soon as the channel becomes idle (with probability 1, or with probability pp)Station waits a random time before sensing again
Collision chanceHigher, since several waiting stations may send togetherLower
DelayLowerHigher
Channel utilisationLower at high loadBetter at high load

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2021 AD · 1 mark

Why we need additional resources such as CPU, Memory in case of adaptive routing as compare to Non Adaptive routing?

Answer

An adaptive routing algorithm changes its routes according to current traffic and topology. Each router must therefore collect and exchange network information, store tables, and recompute the best paths repeatedly. This needs CPU for the calculations and memory for the routing tables and topology data. A non-adaptive algorithm uses fixed routes defined in advance, so it needs no such processing.

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  • 2021 AD · 1 mark

Mention any two advantages of IPV4 over IVP6.

Answer

  1. Smaller header and simpler addresses: IPv4 addresses are 32 bits and easier to read and remember than 128-bit IPv6 addresses, and the networks are simpler to configure and manage.
  2. Wide support and compatibility: IPv4 is universally supported by existing hardware, software and ISPs, so no upgrade or transition cost is needed.

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  • 2021 AD · 1 mark

Mention any two application of UDP.

Answer

  1. DNS (Domain Name System) queries.
  2. Streaming and VoIP/video conferencing.

Others are TFTP, DHCP, SNMP and online gaming.

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  • 2021 AD · 1 mark

How the security is assure in HTTPS while comparing with HTTP?

Answer

HTTPS is HTTP carried over TLS/SSL. It adds security as follows:

  • Encryption: The data is encrypted, so eavesdroppers cannot read it, whereas HTTP sends plain text.
  • Authentication: The server presents a digital certificate signed by a trusted CA, which proves the site's identity.
  • Integrity: Message authentication codes ensure the data is not altered in transit.

HTTPS uses port 443; HTTP uses port 80.

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  • 2021 AD · 1 mark

What is the technological improvement in 4G comparing with 3G?

Answer

4G (LTE/LTE-Advanced) improves on 3G in several ways. It is an all-IP packet-switched network (3G still used circuit switching for voice), and it uses OFDMA/MIMO to give much higher data rates (about 100 Mbps for mobile users, up to 1 Gbps for stationary ones), lower latency and better spectral efficiency. It also supports HD video, VoLTE and seamless handover.

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  • 2021 AD · 1 mark

Mention the advantage of Wireless Communication over Wired Communication.

Answer

Wireless communication gives mobility, so users stay connected while moving. It needs no cable installation, which saves cost and time, and it is easy to set up and expand. It can also reach remote or difficult areas where laying cable is impractical.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2021 AD · 3 marks

Encode digital data 11001011 using NRZ-I, Manchester and Frequency Shift Keying encoding techniques.

Answer

Data: 1 1 0 0 1 0 1 1

NRZ-I (a transition at the start of the bit for a 1, no transition for a 0; the line starts low):

Bit11001011
LevelHighLowLowLowHighHighLowHigh

Manchester (IEEE 802.3 convention: 1 = high-to-low transition in mid-bit, 0 = low-to-high):

Bit11001011
HalvesHLHLLHLHHLLHHLHL

FSK (two carrier frequencies): 1 is sent with the high frequency f1f_1 and 0 with the low frequency f0f_0, so the carrier is

f1, f1, f0, f0, f1, f0, f1, f1f_1,\ f_1,\ f_0,\ f_0,\ f_1,\ f_0,\ f_1,\ f_1

one frequency per bit period.

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  • 2021 AD · 3 marks

If you need to send 200kbps over a noiseless channel with a bandwidth of 20khz. How many signal levels do we need?

Answer

Using the Nyquist formula for a noiseless channel:

C=2Blog⁡2LC = 2B\log_2 L

200000=2×20000×log⁡2L⇒log⁡2L=5200000 = 2 \times 20000 \times \log_2 L \Rightarrow \log_2 L = 5

L=25=32L = 2^5 = 32

32 signal levels are needed.

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  • 2021 AD · 3 marks

Explain the working Principle of Medium Access control protocol used in Wi-fi.

Answer

Wi-Fi (IEEE 802.11) uses CSMA/CA in its Distributed Coordination Function, because collisions cannot be detected on a wireless medium.

  1. A station with data senses the channel. If it is idle for a DIFS interval, it sends the frame.
  2. If the channel is busy, the station waits until it is idle, then waits a random backoff time. The backoff counter is frozen while the channel is busy.
  3. The receiver sends an ACK after a short SIFS. If the ACK is not received, the sender retransmits with a doubled contention window.
  4. Optional RTS/CTS handshaking and the NAV timer solve the hidden terminal problem.

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  • 2021 AD · 3 marks

Illustrate TCP header format.

Answer

The TCP header is 20 to 60 bytes long:

 0                16                 31
+------------------+------------------+
| Source port (16) | Dest. port (16)  |
+------------------+------------------+
|        Sequence number (32)         |
+-------------------------------------+
|    Acknowledgement number (32)      |
+------+------+--------+--------------+
| HLEN | Rsvd | Flags  | Window (16)  |
| (4)  | (6)  | (6)    |              |
+------+------+--------+--------------+
| Checksum (16)    | Urgent ptr (16)  |
+------------------+------------------+
|     Options and padding (0-40 B)    |
+-------------------------------------+
  • Source and destination port: Identify the sending and receiving applications.
  • Sequence number: Number of the first data byte in the segment.
  • Acknowledgement number: The next byte expected from the other side.
  • HLEN: Header length in 32-bit words.
  • Flags: URG, ACK, PSH, RST, SYN and FIN.
  • Window size: Receiver's buffer space, used for flow control.
  • Checksum: Error detection over header and data.
  • Urgent pointer: Position of urgent data when URG is set.
  • Options: For example, MSS and window scaling.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2021 AD · 3 marks

Explain the packet switching principle.

Answer

In packet switching, the message is divided into small units called packets. Each packet carries a header with the source and destination addresses and a sequence number. Packets are sent separately into the network and routed through switches or routers using store-and-forward: a node receives a whole packet, checks it, stores it, and forwards it to the next node.

Packets of one message may take different routes (datagram) or one fixed route (virtual circuit), and the destination reassembles them in order. Links are shared among many users (statistical multiplexing), so utilisation is high and no dedicated path is held, unlike circuit switching. The drawbacks are variable delay and possible packet loss or reordering.

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  • 2021 AD · 5 marks

Mention the advantage of Flooding routing algorithm. Explain the techniques that are used to control the duplication of packets in Flooding routing algorithm.

Answer

Advantages of flooding

  • It is very robust: as long as any path exists, the packet will reach the destination, even if many routers fail.
  • It always finds the shortest path, since a copy takes every path, including the shortest.
  • It is simple and needs no routing tables or topology information.
  • It reaches all nodes, so it is useful for broadcasting and in military networks.

Techniques to control duplicate packets

  1. Hop counter: Each packet has a hop count that is decremented at every hop, and the packet is discarded at zero. It is best initialised to the length of the path from source to destination.
  2. Sequence numbers: The source router puts a sequence number in each packet. Every router keeps a list of the numbers it has already seen from each source and drops duplicates.
  3. Selective flooding: A router forwards a packet only on lines that lead roughly towards the destination, not on all lines.

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  • 2021 AD · 5 marks

ABC organization owns IP address 205.206.207.0. This organization have "N" number of department where each department consists of "60" number of host now perform subnetting and identify the following
a. Subnet Mask b. Number of Subnet c. Network address and broadcast address of each subnet d. Range of usable IP's in each subnet e. Total number of IP wastages.

Answer

205.206.207.0 is a Class C address (/24). Each department has 60 hosts.

Hosts per subnet: 2h−2≥60⇒h=62^h - 2 \ge 60 \Rightarrow h = 6 (26−2=622^6-2 = 62), so 6 host bits and 2 subnet bits.

(a) Subnet mask: /26 = 255.255.255.192

(b) Number of subnets: 22=42^2 = 4 subnets, so N=4N = 4 departments can be served.

(c) and (d)

SubnetNetwork addressUsable IP rangeBroadcast address
1205.206.207.0.1 – .62205.206.207.63
2205.206.207.64.65 – .126205.206.207.127
3205.206.207.128.129 – .190205.206.207.191
4205.206.207.192.193 – .254205.206.207.255

(e) IP wastage: Each subnet has 62 usable addresses and uses 60, so 2 addresses are wasted per subnet.

Total wastage=4×2=8 usable IPs\text{Total wastage} = 4 \times 2 = 8 \text{ usable IPs}

If the network and broadcast addresses are counted too, the wastage is 4×4=164 \times 4 = 16 addresses.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2021 AD · 5 marks

Why flow control is necessary? Explain the techniques of flow Control.

Answer

Flow control is a mechanism that stops a fast sender from overwhelming a slow receiver. It is needed because:

  • The receiver has limited buffer space and processing speed; if data arrives faster than it can be processed, the buffer overflows and frames or packets are lost.
  • Lost data must be retransmitted, which wastes bandwidth and time and can cause congestion.
  • Sender and receiver may have different speeds, so the sender must be told how much it may send.

Techniques of flow control

(a) Stop-and-Wait. The sender transmits one frame and waits for an acknowledgement (ACK) before sending the next. It is simple and reliable, but inefficient on long or fast links because the channel is idle while waiting for the ACK.

(b) Sliding Window. The sender may transmit several frames (up to the window size WW) before receiving an ACK. Each frame carries a sequence number; the window "slides" forward as ACKs arrive. The receiver also keeps a window, and can reduce it to slow the sender. This keeps the pipe full and gives better utilisation. It is implemented in two forms:

  • Go-Back-N ARQ: if frame nn is lost or damaged, the sender retransmits frame nn and all frames after it.
  • Selective Repeat ARQ: only the damaged or lost frame is retransmitted; the receiver buffers out-of-order frames.

(c) Other feedback methods. In TCP the receiver advertises its free buffer size in the window field (receive window), and the sender never has more unacknowledged data than this. Simple schemes also use XON/XOFF signals or RTS/CTS handshakes to pause and resume the sender.

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Questions from TU BIM Business Communication (MGT 204) 2001 AD (Board exam paper, 2001 AD), TU BIM Business Communication (MGT 204) 2002 AD (Board exam paper, 2002 AD), TU BIM Business Communication (MGT 204) 2003 AD (Board exam paper, 2003 AD), TU BIM Business Communication (MGT 204) 2005 AD (Board exam paper, 2005 AD), TU BIM Business Communication (MGT 204) 2007 AD (Board exam paper, 2007 AD), TU BIM Business Communication (MGT 204) 2008 AD (Board exam paper, 2008 AD), TU BIM Business Communication (MGT 204) 2009 AD (Board exam paper, 2009 AD), TU BIM Business Communication (MGT 204) 2010 AD (Board exam paper, 2010 AD), TU BIM Business Communication (MGT 204) 2011 AD (Board exam paper, 2011 AD), TU BIM Business Communication (MGT 204) 2012 AD (Board exam paper, 2012 AD), TU BIM Business Communication (MGT 204) 2013 AD (Board exam paper, 2013 AD), TU BIM Business Communication (MGT 204) 2014 AD (Board exam paper, 2014 AD), TU BIM Business Communication (MGT 204) 2015 AD (Board exam 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