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
| Basis | Datagram | Virtual circuit |
|---|---|---|
| Connection set-up | Not needed | Needed before data transfer |
| Routing | Each packet routed independently | Route chosen once, at set-up, then fixed for all packets |
| Packet address | Full destination address in every packet | Short virtual circuit number |
| Order of delivery | May arrive out of order or duplicated | In order |
| Failure of a node | Packets re-routed easily | All circuits through that node are lost |
| Reliability and congestion | Handled by the end hosts, with no guarantee | Easier to control; resources can be reserved |
| Example | IP (Internet) | X.25, Frame Relay, ATM |
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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.
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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:
| Bit | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 1 | 0 | 1 |
|---|---|---|---|---|---|---|---|---|---|---|
| Levels | HL | HL | HL | LH | LH | LH | HL | LH | HL | LH |
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:
| Bit | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 1 | 0 | 1 |
|---|---|---|---|---|---|---|---|---|---|---|
| Level | L | L | L | H | L | H | H | L | L | H |
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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 ARQ | Selective Repeat ARQ |
|---|---|
| When a frame is lost or damaged, that frame and all later frames are retransmitted | Only the lost or damaged frame is retransmitted |
| Receiver window size is 1, so out-of-order frames are discarded | Receiver window is greater than 1, so out-of-order frames are buffered |
| No receiver buffer needed for reordering | Buffer and sorting logic needed |
| Sender window | Sender window |
| Simple, but wastes bandwidth on noisy links | More complex, but efficient |
| Cumulative ACK or NAK | Individual 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.
- Setup: a call request establishes the route, and each switch records a virtual circuit identifier (VCI) in its table.
- Data transfer: each packet carries only the short VCI, and switches forward it using the table without a routing decision.
- 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 circuit | Datagram |
|---|---|
| Connection-oriented, with setup | Connectionless, no setup |
| All packets follow the same route | Each packet routed independently |
| Packets arrive in order | May arrive out of order |
| Short VCI in header | Full destination address in each header |
| A switch failure breaks the circuit | Packets can be rerouted around failures |
| Easier QoS and congestion control | Harder |
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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 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
| Datagram | Virtual circuit |
|---|---|
| Connectionless; no setup phase | Connection-oriented; a call setup phase is needed |
| Each packet carries full destination address and is routed independently | Packets 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 host | Packets arrive in order |
| Routers keep no per-connection state | Routers keep a table entry per virtual circuit |
| Robust: a failed node is bypassed by new routes | A node failure breaks all circuits through it |
| Congestion control is hard | Resources can be reserved, so quality of service is easier |
| Example: IP | Example: 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 ARQ | Selective Repeat ARQ |
|---|---|
| On an error, the erroneous frame and all frames after it are retransmitted | Only the erroneous (or lost) frame is retransmitted |
| Receiver window size is 1; out-of-order frames are discarded | Receiver window is greater than 1; out-of-order frames are buffered |
| Sender window up to | Sender window up to |
| Uses cumulative acknowledgement | Uses individual acknowledgement (and NAK) |
| Simple; one timer for the oldest frame | More complex; a timer per frame and sorting logic at the receiver |
| Wastes bandwidth on noisy links | More bandwidth efficient |
| Receiver needs little buffering | Receiver 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:
- At start, each router knows only the cost to its directly connected neighbours; other costs are infinity.
- Periodically (and on change), each router sends its whole table to its immediate neighbours.
- A router receiving a neighbour's vector computes, for each destination ,
and updates its table if a cheaper route is found, recording 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:
- Each router initialises its table with distance 0 to itself and the cost of directly connected neighbours.
- Periodically, each router sends its whole table to its immediate neighbours only.
- 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.
- 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.
- Carrier sense: A station that has a frame to send first listens to the channel. If the channel is busy, it waits.
- Multiple access: If the channel is idle, the station transmits at once. Many stations share the same medium.
- 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.
- 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.
- The station senses the carrier. If the channel is idle, it transmits; if busy, it waits.
- While sending, it keeps listening to detect a collision.
- 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.
- 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.
- 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.
- 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:
- 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.
- 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.
- 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).
- Mesh: every node connects to every other (full mesh needs links). It is very reliable and has dedicated links, but cabling and ports are expensive and complex.
- 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.
- 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 input lines into one high-capacity link; at the receiving end a demultiplexer (DEMUX) splits the stream back into the 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 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
| Basis | OSI model | TCP/IP model |
|---|---|---|
| Layers | 7 (physical, data link, network, transport, session, presentation, application) | 4 (network access/host-to-network, internet, transport, application) |
| Origin | Developed by ISO as a theoretical reference model | Developed by DARPA; built from protocols that already worked |
| Approach | Protocols were designed after the model (model first) | Model describes protocols that already existed (protocols first) |
| Session and presentation | Separate layers | Merged into the application layer |
| Network layer service | Both connection-oriented and connectionless | Connectionless only (IP) |
| Transport layer service | Connection-oriented only | Both: TCP (connection-oriented) and UDP (connectionless) |
| Generality | Protocol independent, clear separation of service, interface and protocol | Protocol specific, less clear separation |
| Use | Mainly a teaching/reference tool | The 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 (). Bandwidth is the range of frequencies a signal contains or a channel can pass, i.e. the difference between the highest and lowest frequency ().
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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
| Basis | IEEE 802.4 | IEEE 802.5 |
|---|---|---|
| Name | Token Bus | Token Ring |
| Topology | Physical bus/tree, logical ring | Physical and logical ring |
| Medium | Coaxial cable (broadband or carrierband) | Twisted pair/shielded cable |
| Token | Token passed to the next station in the logical ring by address | Token circulates around the physical ring |
| Station order | Logical order (by address) differs from physical order | Same as physical order |
| Complexity | More complex (ring maintenance, joining/leaving, token loss handling) | Simpler, uses a monitor station |
| Frame priority | Four priority classes | Priority 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 (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 by gives remainder .
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:
Given , , .
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
| Basis | Virtual circuit | Datagram |
|---|---|---|
| Connection | Connection-oriented: a path is set up before data transfer | Connectionless: no set-up |
| Route | All packets follow the same route | Each packet routed independently, may take different routes |
| Packet header | Carries a short virtual circuit number | Carries full destination address |
| Order of delivery | Packets arrive in order | May arrive out of order, need reordering |
| Router state | Routers keep a table entry for each circuit | Routers keep no per-connection state |
| Failure | A failed router breaks all circuits through it | Only packets in transit are lost; others reroute |
| Congestion/QoS | Easier to reserve resources and control | Harder |
| Example | X.25, Frame Relay, ATM | IP |
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- 2002 AD · 3 marks
Difference between TCP and UDP
Answer
| Basis | TCP | UDP |
|---|---|---|
| Full name | Transmission Control Protocol | User Datagram Protocol |
| Connection | Connection-oriented (three-way handshake) | Connectionless |
| Reliability | Reliable: acknowledgements, retransmission, sequencing | Unreliable: no guarantee of delivery or order |
| Flow/congestion control | Yes (sliding window) | No |
| Header | 20 bytes minimum | 8 bytes |
| Speed | Slower, more overhead | Faster, low overhead |
| Data unit | Byte stream (segments) | Datagrams, message boundaries kept |
| Use | HTTP, FTP, SMTP, Telnet | DNS, 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:
- Label the source A with distance 0 (permanent) and every other node with infinity (tentative).
- From the most recently made permanent node, update the tentative distance of each neighbour as , remembering the predecessor.
- Make the tentative node with the smallest distance permanent.
- Repeat steps 2 and 3 until G becomes permanent.
- 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:
| Basis | LAN | WAN |
|---|---|---|
| Area | Building or campus | City, country or world |
| Speed | High (10 Mbps - Gbps) | Lower |
| Error rate | Low | Higher |
| Ownership | Single organisation | Usually carriers / many organisations |
| Cost | Low | High |
| Topology | Bus, ring, star | Mesh, 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:
- ASK - Amplitude Shift Keying
- FSK - Frequency Shift Keying
- 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 .
(of the transmitted bits; equivalently 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.
| Basis | FDM | TDM |
|---|---|---|
| Sharing | Bandwidth divided into frequency bands | Time divided into slots |
| Simultaneity | All signals sent at the same time | Signals sent one after another in turn |
| Signal type | Mainly analog | Mainly digital |
| Guard | Guard bands between channels | Guard times / synchronisation bits |
| Problems | Crosstalk, intermodulation noise | Needs synchronisation, timing jitter |
| Example | Radio/TV, cable TV | Digital 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:
- Character (byte) count
- Flag bytes with byte stuffing
- Flag bits (start/end flag) with bit stuffing
- 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 11111a 0 is stuffed, and after the second run11111at 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 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:
| Basis | CSMA/CD | Token passing |
|---|---|---|
| Access | Contention, random | Controlled, deterministic |
| Collisions | Possible | None |
| Performance at heavy load | Degrades | Remains good |
| Delay at light load | Low | Must wait for token |
| Priorities | Not supported | Supported |
| Complexity | Simple | Token management needed |
IEEE standards and MAC protocols:
| Standard | Name | MAC protocol |
|---|---|---|
| 802.3 | Ethernet | CSMA/CD |
| 802.4 | Token Bus | Token passing on logical ring |
| 802.5 | Token Ring | Token passing on ring |
| 802.11 | Wireless LAN | CSMA/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
| Device | Function | OSI layer |
|---|---|---|
| Bridge | Connects two LAN segments, filters and forwards frames using MAC addresses, reduces collision domains | Data link (layer 2) |
| Repeater | Regenerates and amplifies the signal to extend the cable length | Physical (layer 1) |
| Hub | Multi-port repeater; broadcasts incoming bits to all other ports | Physical (layer 1) |
| Router | Connects different networks and forwards packets by logical (IP) address using routing tables | Network (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:
- Mark the source with distance 0 and all others infinity.
- Take the unvisited node with the smallest distance, make it permanent, and relax its neighbours: .
- Repeat until the destination is permanent.
- 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 .
- 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 , so 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:
| Basis | TCP | UDP |
|---|---|---|
| Connection | Connection-oriented | Connectionless |
| Reliability | Reliable | Unreliable, best effort |
| Ordering | Ordered | Not guaranteed |
| Flow/congestion control | Yes | No |
| Header size | 20+ bytes | 8 bytes |
| Speed | Slower | Faster |
| Examples | HTTP, FTP, SMTP | DNS, 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).
| Cable | Medium | Speed | Max segment | Notes |
|---|---|---|---|---|
| 10BASE5 | Thick coax | 10 Mbps | 500 m | Bus, vampire taps |
| 10BASE2 | Thin coax (RG-58) | 10 Mbps | 185 m | Bus, BNC connectors |
| 10BASE-T | UTP Cat 3 or higher | 10 Mbps | 100 m | Star with hub |
| 10BASE-F | Fibre optic | 10 Mbps | 2000 m | Long distance |
| 100BASE-TX | UTP Cat 5 | 100 Mbps | 100 m | Fast Ethernet |
| 100BASE-FX | Multimode fibre | 100 Mbps | 2 km | Fast Ethernet |
| 1000BASE-T | UTP Cat 5e/6 | 1 Gbps | 100 m | Gigabit |
| 1000BASE-SX/LX | Fibre | 1 Gbps | 550 m / 5 km | Gigabit |
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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:
- The sender composes the message with a UA and gives the address
user@domain. - The UA sends the message to the sender's mail server using SMTP, where it is queued.
- 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).
- The receiving MTA places it in the recipient's mailbox.
- 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
| Basis | TCP | UDP |
|---|---|---|
| Connection | Connection-oriented (handshake) | Connectionless |
| Reliability | Reliable, with ACK and retransmission | Unreliable |
| Ordering | Sequenced delivery | No ordering |
| Flow and congestion control | Provided | Not provided |
| Header | 20 bytes minimum | 8 bytes |
| Speed / overhead | Slower, higher overhead | Faster, low overhead |
| Transmission | Byte stream | Message (datagram) based |
| Applications | Web, e-mail, file transfer | DNS, 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.
| Basis | Go-Back-N ARQ | Selective Repeat ARQ |
|---|---|---|
| Retransmission | The erroneous frame and all frames after it | Only the erroneous or lost frame |
| Receiver window | Size 1 (accepts only in-order frames) | Greater than 1 (buffers out-of-order frames) |
| Receiver buffer | Not needed for out-of-order frames, they are discarded | Needed, and frames are reordered |
| Sender window | Up to | Up to |
| Bandwidth use | Wasteful when error rate is high | Efficient |
| Complexity | Simple | More complex (sorting, more buffers) |
| ACK | Cumulative | Individual 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 for 1, low frequency for 0: .
- (iii) PSK: phase 0 degrees for 1, 180 degrees for 0: .
- (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 .
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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 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
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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 , so 5 bits are left for hosts.
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:
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- 2005 AD · 3 marks
Compare OSI and TCP/IP reference models for computer communications.
Answer
| OSI model | TCP/IP model |
|---|---|
| 7 layers | 4 layers (some texts use 5) |
| Developed by ISO as a general reference model | Developed by DARPA around the Internet protocols |
| Session and presentation are separate layers | Combined into the application layer |
| Separate physical and data link layers | Combined into the network access (host-to-network) layer |
| Network layer supports connectionless and connection-oriented service | Internet layer is connectionless only |
| Transport layer supports connection-oriented service only | Transport layer supports both (TCP and UDP) |
| Clear separation of services, interfaces and protocols | Does not clearly separate them |
| Protocols fit the model less closely; model came before protocols | Protocols came first and the model describes them |
| Mostly a theoretical reference | The 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:
- Character (byte) count
- Flag bytes with byte stuffing
- Flag bits (start/end flag) with bit stuffing
- 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:
0110111111011111 - After each run of five 1s a 0 is inserted:
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 topology | Ring (physical star-wired ring) topology |
| CSMA/CD access method | Token passing |
| Contention-based; collisions possible | Deterministic; no collisions |
| Any station may send when the medium is idle | Only the token holder may send |
| Performs well at light load; degrades at heavy load because of collisions | Performs well at heavy load |
| No priority scheme | Priority and reservation bits supported |
| Frame size 64 to 1518 bytes | Larger frames possible |
| Simple, cheap, no central monitor | Needs a monitor station for token maintenance |
| Speeds 10 Mbps up to 10 Gbps and more | 4 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):
- Label S with distance 0 (permanent) and every other node with (tentative).
- 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.
- Among all tentative nodes, make the one with the smallest label permanent and make it the new working node.
- 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: 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 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:
- 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.
- 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.
- 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, . 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
| Protocol | Purpose | Default port |
|---|---|---|
| i) FTP | Transfer files between hosts | 20 (data), 21 (control) |
| ii) SMTP | Send/relay e-mail | 25 |
| iii) DNS | Translate domain names to IP addresses | 53 |
| iv) Telnet | Remote terminal login to another host | 23 |
| v) HTTP | Transfer web pages (hypertext) | 80 |
| vi) HTTPS | Secure HTTP over SSL/TLS | 443 |
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- 2009 AD · 3 marks
Explain the functions of each layer of OSI model.
Answer
The OSI model has seven layers.
- Physical: transmits raw bits over the medium. It defines voltages, cables, connectors, data rate and topology.
- Data link: framing, physical (MAC) addressing, error detection, flow control and media access control. It gives reliable node-to-node delivery.
- Network: logical (IP) addressing, routing, packet forwarding and congestion control. It gives source-to-destination delivery across networks.
- Transport: end-to-end (process-to-process) delivery using port addresses, segmentation and reassembly, flow control, error control and connection management (TCP, UDP).
- Session: establishes, manages and terminates sessions between applications, with dialog control and synchronization (checkpoints).
- Presentation: translation between data formats, encryption/decryption and compression.
- 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 -bit sequence numbers the window is at most (Go-Back-N) or (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:
- A station with a frame to send listens to the channel (carrier sense).
- If the channel is idle, it transmits. If busy, it waits until idle (1-persistent).
- While transmitting it keeps listening. If another station transmits at the same time, a collision is detected.
- On collision, the station sends a short jam signal so that all stations know, and stops.
- 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 routing | Shortest path (link state) routing |
|---|---|
| Each router sends its whole routing table to its neighbours only | Each 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 algorithm | Uses Dijkstra's algorithm |
| Slow convergence; count-to-infinity problem | Fast convergence; loop-free |
| Simple, low memory and CPU use | More complex, needs more memory and CPU |
| Periodic updates generate traffic | Updates only on change, but flooding costs bandwidth |
| Example: RIP | Example: 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 host bits.
iii. The subnet mask for /26 is 26 ones followed by 6 zeros: 11111111.11111111.11111111.11000000.
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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:
| Datagram | Virtual circuit |
|---|---|
| Connectionless, no setup phase | Connection-oriented; a path is set up first |
| Each packet carries the full destination address | Packets carry only a short virtual circuit number |
| Each packet routed independently | All packets follow the same path |
| Packets may arrive out of order | Packets arrive in order |
| Routers hold no per-connection state | Routers keep a table entry for each circuit |
| Failure of a node affects only packets in transit; rerouting is easy | Failure of a node breaks all circuits through it |
| Congestion control is difficult | Easy to reserve resources and control congestion |
| Example: IP | Example: X.25, Frame Relay, ATM |
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- 2009 AD · 2 marks
Write down the major difference between TCP and IP protocol.
Answer
| TCP | IP |
|---|---|
| Transport layer protocol | Network (internet) layer protocol |
| Connection-oriented and reliable, with acknowledgement, retransmission, flow and congestion control | Connectionless and unreliable (best effort) |
| Provides end-to-end delivery between processes using port numbers | Provides host-to-host delivery of packets using IP addresses and routing |
| Handles segmentation and in-order reassembly | Handles 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
| IPv4 | IPv6 |
|---|---|
| 32-bit address | 128-bit address |
| Header 20 to 60 bytes, variable | Fixed 40-byte header |
| Header checksum present | No header checksum |
| Fragmentation by routers and sender | Only the sender fragments |
| IPsec optional | IPsec built in |
(ii) WDM versus FDM
| WDM | FDM |
|---|---|
| Multiplexes optical signals of different wavelengths on a fibre | Multiplexes analog signals of different frequencies on one medium |
| Used on optical fibre | Used on copper, coaxial and radio |
| Very high capacity | Lower capacity |
(iii) Internet versus WWW
| Internet | WWW |
|---|---|
| Worldwide network of networks, the physical infrastructure | A 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 more | One application among them, accessed with a browser |
(iv) 802.3 Ethernet versus 802.5 Token Ring
| 802.3 | 802.5 |
|---|---|
| Bus/star, CSMA/CD | Ring, token passing |
| Collisions possible | No collisions |
| No priorities | Priorities 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
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:
| Octet | Binary |
|---|---|
| 196 | 11000100 |
| 16 | 00010000 |
| 4 | 00000100 |
| 140 | 10001100 |
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 (Hz) and signal-to-noise power ratio :
It shows that capacity rises with bandwidth and SNR, and that error-free transmission is possible at any rate below 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:
- Connection establishment: both sides initialise variables and counters (sequence numbers, buffers).
- Data transfer: frames are sent and received in order, with acknowledgements, flow control and error control.
- 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 switching | Packet switching |
|---|---|
| A dedicated path is set up before transmission and kept until the end | No dedicated path; packets share links |
| Bandwidth is reserved, so it can be wasted when idle | Bandwidth is used on demand, so the link is used efficiently |
| Data travels as a continuous stream | Message is split into packets, each with a header |
| Constant delay and no reordering | Variable delay; packets may arrive out of order |
| Call setup time is needed | No setup in the datagram method |
| Example: telephone network | Example: 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 , and with probability 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 Hz and levels:
Shannon capacity, with SNR = 25 dB:
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 , where J/K.
K and Hz.
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
| TCP | UDP |
|---|---|
| Connection-oriented | Connectionless |
| Reliable: acknowledgements and retransmission | Unreliable, best effort |
| Ordered delivery, flow and congestion control | No ordering or flow control |
| 20-byte or larger header, slower | 8-byte header, faster |
| Used for HTTP, FTP, SMTP | Used for DNS, streaming, VoIP |
(ii) ARP and RARP
| ARP | RARP |
|---|---|
| Address Resolution Protocol | Reverse Address Resolution Protocol |
| Maps a known IP address to a MAC address | Maps 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 use | Obsolete, replaced by BOOTP and DHCP |
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- 2010 AD · 3 marks
The message sequence is 1001101 and generator polynomial . Calculate the transmitted frame.
Answer
Message , generator (degree 3).
Append 3 zeros: 1001101000. Divide by 1011 using modulo-2 division:
| Step | Bits | XOR with 1011 | Remainder |
|---|---|---|---|
| 1 | 1001 | 0010 | 010 |
| 2 | bring 1: 0101 | (leading 0, no XOR) | 101 |
| 3 | bring 0: 1010 | 0001 | 001 |
| 4 | bring 1: 0011 | (leading 0) | 011 |
| 5 | bring 0: 0110 | (leading 0) | 110 |
| 6 | bring 0: 1100 | 0111 | 111 |
| 7 | bring 0: 1110 | 0101 | 101 |
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
- Prepare the frame and set
attempts = 0. - Sense the channel. If it is busy, wait until it is idle (persistent sensing).
- When the channel is idle, start transmitting and keep monitoring the channel.
- If no collision is detected until the whole frame is sent, the transmission is successful. Stop.
- If a collision is detected:
- stop sending and transmit a short jam signal so all stations know;
- set
attempts = attempts + 1; ifattemptsexceeds 16, abort and report an error; - choose a random from to (binary exponential back-off);
- wait 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 (). Maximum throughput is , about 18.4% at .
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 (). Throughput is , with a maximum of about 36.8% at .
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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:
- Character (byte) count
- Flag bytes with byte stuffing
- Flag bits with bit stuffing
- 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
| OSPF | BGP |
|---|---|
| Open Shortest Path First | Border Gateway Protocol |
| Interior gateway protocol, used inside one autonomous system | Exterior gateway protocol, used between autonomous systems |
| Link-state algorithm (Dijkstra) | Path-vector algorithm |
| Metric is cost | Decisions 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 transmission | Synchronous transmission |
|---|---|
| Data is sent one character at a time, with start and stop bits | Data is sent as a continuous block of bits or frames |
| No common clock; gaps may exist between characters | Sender and receiver clocks are synchronised |
| Overhead is high (2 to 3 extra bits per character) | Overhead is low |
| Slower and cheaper | Faster and more complex |
| Example: keyboard to computer | Example: LAN, data between computers |
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- 2011 AD · 1 mark
For 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 .
- Ring: each device is linked to the next, and the last to the first, so the number of cable links is .
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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 slots | Channels share the link by using different frequency bands |
| Each signal uses the full bandwidth for its slot | Each signal uses its band all the time |
| Used mostly for digital signals | Used mostly for analog signals |
| Needs synchronisation | Needs guard bands to prevent interference |
| Crosstalk is not a major problem | Crosstalk 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 model | TCP/IP model |
|---|---|
| Seven layers | Four layers (some texts show five) |
| Layers: physical, data link, network, transport, session, presentation, application | Layers: network access, internet, transport, application |
| Theoretical reference model developed by ISO | Protocol suite developed first, from ARPANET |
| Protocols were developed after the model, so the model is generic | The model grew out of the protocols |
| Strictly separates services, interfaces and protocols | Does not clearly separate them |
| Network layer supports both connectionless and connection-oriented service; transport layer supports connection-oriented only | Network layer is connectionless only; transport layer offers both (TCP and UDP) |
| Session and presentation layers exist | Their functions are inside the application layer |
| Less used in practice | Standard of the Internet |
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- 2011 AD · 3 marks
Make distinction between link state routing and flooding protocols.
Answer
| Link state routing | Flooding |
|---|---|
| Each router learns the full topology and calculates the shortest path (Dijkstra) | A router sends each incoming packet out on every line except the one it arrived on |
| Link-state packets, flooded to all routers, only carry information about neighbours | Data packets themselves are flooded |
| Routing tables are built, so a packet follows one best path | No routing table or topology knowledge needed |
| Efficient in traffic | Creates many duplicate packets, so it uses a lot of bandwidth (limited with hop counters) |
| Used in OSPF and IS-IS | Used in military networks and as a part of link-state distribution |
| Gives a good adaptive route | Always finds the shortest path, because it tries all paths |
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- 2011 AD · 3 marks
List the services provided by Data Link Layer to Network layer.
Answer
The data link layer provides these services to the network layer:
- Framing: splitting the bit stream into frames.
- Physical (MAC) addressing: identifying the sender and receiver on the link.
- Unacknowledged connectionless service: frames are sent without any acknowledgement.
- Acknowledged connectionless service: each frame is acknowledged.
- Acknowledged connection-oriented service: connection setup, ordered and reliable delivery, release.
- Error control: detecting and correcting errors with CRC, parity and ARQ.
- Flow control: stopping a fast sender from overwhelming a slow receiver.
- Media access control: deciding who may use a shared link.
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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:
- The station senses the channel. If it is busy, it waits.
- When the channel becomes idle, the station waits for a fixed IFS (inter-frame space).
- 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.
- 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.
- 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 routing | Link state routing |
|---|---|
| Each router sends its whole routing table to its neighbours only | Each router floods information about its neighbours to all routers |
| Router knows only distances to destinations, not the topology | Each router builds a complete map of the network |
| Bellman-Ford algorithm | Dijkstra's algorithm |
| Updates are periodic | Updates are sent only when a change occurs |
| Slow convergence; count-to-infinity problem | Fast convergence |
| Low memory and CPU use, simple | More memory and CPU needed |
| Example: RIP | Example: 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 (). The new mask is /26, which is 255.255.255.192.
Each subnet has addresses, of which are usable hosts.
| Department | Subnet address | Host range | Broadcast |
|---|---|---|---|
| 1 | 192.168.1.0/26 | 192.168.1.1 to 192.168.1.62 | 192.168.1.63 |
| 2 | 192.168.1.64/26 | 192.168.1.65 to 192.168.1.126 | 192.168.1.127 |
| 3 | 192.168.1.128/26 | 192.168.1.129 to 192.168.1.190 | 192.168.1.191 |
| 4 | 192.168.1.192/26 | 192.168.1.193 to 192.168.1.254 | 192.168.1.255 |
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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.
| Aspect | Analog | Digital |
|---|---|---|
| Signal | Continuous | Discrete |
| Noise immunity | Low | High |
| Long-distance device | Amplifier | Repeater |
| Error detection and correction | Difficult | Easy (CRC, parity) |
| Security and multiplexing | Harder | Easier (encryption, TDM) |
| Bandwidth need | Less | More |
| Cost | Cheaper for simple uses | Cheaper 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:
- Delivery: data must reach the correct intended destination, and only that destination.
- Accuracy: data must be delivered without alteration.
- 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:
- Network support layers (layers 1 to 3): physical, data link and network layers, which handle the physical transmission of data between devices.
- User support layers (layers 5 to 7): session, presentation and application layers, which let unrelated software systems work together.
- 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 (), leaving host bits, which give usable hosts. The mask therefore has 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
- A host sends its data to the local router (the gateway) in packets.
- The router reads the destination address and consults its routing table, which is built by routing protocols such as RIP, OSPF or BGP.
- The packet is forwarded hop by hop over the carrier links (leased line, frame relay, ATM, MPLS, etc.) using packet switching, store-and-forward.
- The last router delivers the packet to the destination LAN and host.
- 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 for -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 (). Maximum throughput is only , with maximum at .
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 (). Throughput becomes , with maximum at .
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:
- 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.
- Automatic allocation: The server assigns an IP address from its pool the first time the host connects, and the host keeps that address permanently.
- 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:
- Process-to-process communication: Using port numbers, data is delivered to the correct application process on the host, not just to the host.
- Segmentation and reassembly: Messages from the application are split into segments with sequence numbers and reassembled at the destination.
- Connection control: Connection-oriented service (TCP) sets up, maintains and releases a connection; connectionless service (UDP) sends each segment independently.
- Flow control: Prevents a fast sender from overwhelming a slow receiver, using a sliding window end to end.
- Error control: Detects damaged, lost, duplicate or out-of-order segments using checksums, acknowledgements and retransmission.
- Congestion control: Regulates the sending rate to avoid overloading the network.
- Multiplexing and demultiplexing: Several applications share one network connection; the transport layer multiplexes at the sender and demultiplexes at the receiver by port number.
- 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
- An application calls the resolver on the host with a name.
- The resolver asks the local name server.
- 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).
- The authoritative server returns the IP address.
- 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 . 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:
where is the capacity in bits per second, the bandwidth in Hz, and the signal-to-noise power ratio. Error-free transmission is theoretically possible at any rate below , 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 switching | Packet switching |
|---|---|
| A dedicated path is set up before transfer | No dedicated path; packets are routed independently (datagram) or along a virtual circuit |
| Resources reserved for the whole call | Resources used on demand and shared |
| Fixed bandwidth; idle capacity is wasted | Efficient use of bandwidth |
| Constant delay after setup; no store-and-forward | Variable delay due to queuing and store-and-forward |
| Bit-stream transparent | Not transparent; data is split into packets with headers |
| Poor for bursty data, good for voice | Good for bursty data |
| Example: telephone network | Example: 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:
- 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.
- 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.
- 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.
- 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.
- Error rate: LANs have low error rates; WANs have higher error rates.
- 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 .
Convert SNR from decibels:
Then
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)
- 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).
- 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.
- 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 bits with , so (8 subnets). Host bits left: .
iii. Subnet mask: Network bits = . 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)
- Discover: a new host (with no IP) broadcasts a DHCPDISCOVER message to find a server.
- Offer: a server replies with DHCPOFFER containing an unused address from its pool, with mask, gateway, DNS and lease time.
- Request: the client broadcasts DHCPREQUEST to accept the offer (which also tells other servers it was declined).
- 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.
- Mark the source with distance 0 (permanent); all others have distance infinity (tentative).
- For the node just made permanent, update the tentative distance of each neighbour: .
- Make the tentative node with the smallest distance permanent.
- 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 ALOHA | Slotted ALOHA |
|---|---|
| Station transmits whenever it has data | Station transmits only at the beginning of a time slot |
| No time synchronisation needed | All stations must be synchronised to slot boundaries |
| Vulnerable time is | Vulnerable time is |
| Partial collisions are possible | Collisions are complete (only within the same slot) |
| Throughput | Throughput |
| Maximum throughput about 18.4 percent (at ) | Maximum throughput about 36.8 percent (at ) |
| Simple | More 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:
- Source: generates the data to be transmitted (e.g. a computer or phone).
- Transmitter: converts the data into signals suitable for the transmission medium (e.g. a modem).
- Transmission system: the medium or network (wire, fibre, radio, or a network of switches) that carries the signal.
- Receiver: converts the received signal back into data.
- 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 for CRC?
Answer
Divisor is (degree 3), so append 3 zeros to the data: .
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 by gives quotient ... and remainder .
Check: . Since mod , , , , , so the remainder is .
CRC = , so the transmitted frame is
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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
| Basis | Stop-and-Wait ARQ | Selective Reject ARQ |
|---|---|---|
| Frames in transit | One frame at a time; waits for ACK | Many frames sent (window size > 1) |
| Window size | Sender 1, receiver 1 | Sender and receiver |
| Retransmission | Retransmits the lost frame after timeout | Retransmits only the damaged/lost frame |
| Efficiency | Low, channel idle while waiting | High, link kept busy |
| Receiver buffer | Not needed | Needed to reorder out-of-order frames |
| Complexity | Simple | Complex 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
where is the cost to neighbour and is 's advertised distance to . 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 vector | Shortest path (link state) |
|---|---|
| Knows only neighbours' distance information | Each router knows the complete topology |
| Shares its table with neighbours only | Floods link-state to all routers |
| Bellman-Ford algorithm | Dijkstra's algorithm |
| Slow convergence, count-to-infinity | Fast convergence |
| Less CPU and memory | More 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:
- Carrier sense: a station with a frame listens to the channel before sending.
- If the channel is idle, it transmits; if busy, it waits until idle (1-persistent).
- Collision detection: while transmitting, the station keeps listening. If the signal on the line differs from what it sent, a collision has occurred.
- On collision, the station stops and sends a short jam signal so all stations know.
- It then waits a random time using binary exponential backoff: after the -th collision it picks randomly from to slot times and waits slot time before retrying.
- 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 ------------ |
- Discover: the new client broadcasts (source 0.0.0.0, destination 255.255.255.255) to find servers.
- Offer: a server offers an unused IP address with configuration and lease time.
- Request: the client broadcasts its acceptance of one offer, so other servers withdraw theirs.
- 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
- Limited data rate (basic rate 128 kbps, primary rate about 2 Mbps), which is too slow for modern broadband needs.
- 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:
- Random access: ALOHA (pure and slotted), CSMA, CSMA/CD, CSMA/CA.
- Controlled access: reservation, polling, token passing.
- 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:
- Senses the channel and waits until it is idle for a DIFS interval.
- Waits a random backoff time before sending (the counter freezes while the channel is busy).
- Optionally exchanges RTS (request to send) and CTS (clear to send) frames to reserve the channel and handle hidden terminals.
- 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 , 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).
- Learning: for each incoming frame, it records the source MAC address against the arrival port.
- Forwarding: if the destination MAC is in the table, the frame is sent only out of that port.
- Flooding: if the destination is unknown (or is a broadcast), the frame is sent out of all ports except the incoming one.
- Filtering: if the destination is on the same port as the source, the frame is dropped.
CRC calculation: the message is and , i.e. divisor . Append 2 zeros: .
Using , the message has terms at . Multiplying by gives exponents . Reducing mod (): exponents 12, 10, 6, 2 give each (four terms, sum 0); exponents 11, 9, 7, 3 give each (four terms, sum 0). The remainder is , so the CRC is .
Transmitted frame:
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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 (). New prefix is :
Each subnet has addresses (510 usable hosts), a block size of 2 in the third octet.
| Department | Network address | Usable host range | Broadcast |
|---|---|---|---|
| 1 | 131.215.0.0/23 | 131.215.0.1 to 131.215.1.254 | 131.215.1.255 |
| 2 | 131.215.2.0/23 | 131.215.2.1 to 131.215.3.254 | 131.215.3.255 |
| 3 | 131.215.4.0/23 | 131.215.4.1 to 131.215.5.254 | 131.215.5.255 |
| 4 | 131.215.6.0/23 | 131.215.6.1 to 131.215.7.254 | 131.215.7.255 |
| 5 | 131.215.8.0/23 | 131.215.8.1 to 131.215.9.254 | 131.215.9.255 |
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
- Process-to-process delivery: uses port numbers to deliver data to the correct application process (multiplexing and demultiplexing).
- Segmentation and reassembly: splits the message into segments with sequence numbers and rebuilds it at the receiver.
- Connection control: connection-oriented (TCP) or connectionless (UDP) service.
- Flow control: end-to-end control so the sender does not overwhelm the receiver (sliding window).
- Error control: detects damaged, lost or duplicate segments using checksums, acknowledgements and retransmission.
- Congestion control: regulates the sending rate according to network load.
- 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 includelisten(),accept(),connect(),send()andrecv().)
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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 represents 1 and 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:
- The sender transmits one frame (carrying a sequence number 0 or 1) and starts a timer.
- It then waits for an acknowledgement (ACK) before sending the next frame.
- If the frame arrives correctly, the receiver sends an ACK; the sender then sends the next frame.
- If the frame is damaged or lost, the receiver discards it (no ACK). When the timer expires, the sender retransmits the same frame.
- 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:
- 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.
- 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.
- 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
| Basis | Distance Vector | Link State |
|---|---|---|
| Information | Each router knows only distances (vectors) to destinations via neighbours | Each router has the complete topology map |
| Exchange | Routing table sent to neighbours only | Link state packets flooded to all routers |
| Algorithm | Bellman-Ford | Dijkstra |
| Updates | Periodic | Triggered on change |
| Convergence | Slow; count-to-infinity problem | Fast |
| Resources | Low CPU and memory | More CPU and memory |
| Example | RIP | OSPF |
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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.
| Dept | Computers | Block size | Prefix and mask | Network | Usable range | Broadcast |
|---|---|---|---|---|---|---|
| B | 110 | 128 (126 hosts, ) | /25, 255.255.255.128 | 216.215.214.0 | .1 to .126 | .127 |
| A | 52 | 64 (62 hosts, ) | /26, 255.255.255.192 | 216.215.214.128 | .129 to .190 | .191 |
| C | 25 | 32 (30 hosts, ) | /27, 255.255.255.224 | 216.215.214.192 | .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:
- Source (sender): generates the data.
- Transmitter: converts the data into signals suitable for the medium (for example a modem).
- Transmission medium (channel): carries the signal (wire, fibre, air).
- Receiver: converts the received signal back to data.
- 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
| Basis | 10Base-5 | 10Base-T |
|---|---|---|
| Medium | Thick coaxial cable | Unshielded twisted pair (UTP) |
| Topology | Bus | Star (via hub or switch) |
| Max segment length | 500 m | 100 m |
| Connection | Vampire tap and AUI transceiver | RJ-45 connector |
| Cost, ease | Costly, difficult to install | Cheap, easy to install and troubleshoot |
| Fault effect | One break fails the whole network | One 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:
- 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.
- 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.
- 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:
- Choose a generator polynomial (divisor) of bits.
- Append zeros to the data.
- Divide the result by the divisor using modulo-2 (XOR) division.
- The -bit remainder is the CRC; replace the appended zeros with it and transmit.
- 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 ). 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:
- TCP connection: two BGP routers (peers) set up a TCP session.
- OPEN message: peers exchange OPEN messages containing AS number, hold time and BGP identifier, and agree on parameters.
- KEEPALIVE: a KEEPALIVE confirms the OPEN, and keepalives are then sent periodically to show the peer is alive.
- 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.
- 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.
- 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:
- Automatic allocation: the server assigns a permanent IP address to a client, which keeps it indefinitely.
- 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.
- 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 ) 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 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 for Go-Back-N and for Selective Repeat, with -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:
| Network | Third octet |
|---|---|
| 198.132.98.0 | 01100010 |
| 198.132.99.0 | 01100011 |
| 198.132.100.0 | 01100100 |
| 198.132.101.0 | 01100101 |
| 198.132.105.0 | 01101001 |
The first 4 bits of the third octet (0110) are common to all; the 5th bit differs. So the common prefix is 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:
- Process-to-process delivery (addressing): port numbers identify the sending and receiving application processes.
- Segmentation and reassembly: messages are divided into segments with sequence numbers and reassembled at the destination.
- Connection control: connection-oriented (TCP, with setup and release) or connectionless (UDP) service.
- Flow control: end-to-end control so the sender does not overrun the receiver (for example TCP sliding window).
- Error control: detection of damaged, lost or duplicated segments by checksum, acknowledgement and retransmission, giving reliable delivery.
- Congestion control: regulating traffic into the network to avoid congestion.
- 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
- Larger address space: 128-bit addresses instead of 32-bit, which removes address exhaustion.
- 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:
- 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.
- 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:
SNR: , so .
Bandwidth: the frequencies range from 600 Hz to 1.5 kHz, so
Capacity:
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 · 3 marks
Explain the types of Datalink layer services provided to Network Layer.
Answer
The data link layer offers three main kinds of service to the network layer:
- Unacknowledged connectionless service: The sender transmits independent frames and the receiver does not acknowledge them. No connection is set up and lost frames are not recovered at this layer. It suits low-error media such as Ethernet and real-time traffic such as voice.
- Acknowledged connectionless service: No connection is set up, but each frame is individually acknowledged. If the ACK does not arrive within a timeout, the sender retransmits the frame. It suits unreliable channels such as wireless links.
- Acknowledged connection-oriented service: A connection is established first, and every frame is numbered and acknowledged. The data link layer guarantees that each frame is received exactly once and in the correct order. The connection is released after the transfer, as in WAN links and satellite channels.
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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
- 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.
- 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.
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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 , borrow 3 bits.
New mask: /19 = 255.255.224.0. Block size is 32 in the third octet. Hosts per subnet are .
| Subnet | Network address | Usable host range | Broadcast address |
|---|---|---|---|
| 1 | 150.152.0.0 | 150.152.0.1 – 150.152.31.254 | 150.152.31.255 |
| 2 | 150.152.32.0 | 150.152.32.1 – 150.152.63.254 | 150.152.63.255 |
| 3 | 150.152.64.0 | 150.152.64.1 – 150.152.95.254 | 150.152.95.255 |
| 4 | 150.152.96.0 | 150.152.96.1 – 150.152.127.254 | 150.152.127.255 |
| 5 | 150.152.128.0 | 150.152.128.1 – 150.152.159.254 | 150.152.159.255 |
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 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.
- Set the distance of the source to 0 and all other nodes to . Mark the source as permanent.
- Update the tentative distance of each neighbour of the latest permanent node: .
- Make the tentative node with the smallest distance permanent.
- 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 = .
- 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
| Class | Fixed high-order bits | First octet range |
|---|---|---|
| Class C | 110 | 192 – 223 |
| Class D | 1110 | 224 – 239 |
| Class E | 1111 | 240 – 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
- Integrated services: Voice, data, fax and video travel over one digital line, so separate networks are not needed.
- 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
- 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
pingcommand. - 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 ---------->
- Message: The information to be communicated (text, numbers, image, audio or video).
- Sender: The device that sends the message, such as a computer or phone.
- Receiver: The device that receives the message.
- Transmission medium: The physical path from sender to receiver, such as twisted pair, coaxial cable, fibre or radio waves.
- 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:
The number of signal levels required is (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 . Calculate the transmitted bits.
Answer
The generator is taken as (binary 1001), because and are the same in modulo-2 arithmetic. The degree is 3, so append 3 zeros to the message:
Divide by 1001 using modulo-2 (XOR) division. The remainder is 001.
The transmitted bits are the message followed by the remainder:
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:
- 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).
- 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.
- 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.
| Dept | Hosts needed | Block size | Mask | Network address | Usable range | Broadcast |
|---|---|---|---|---|---|---|
| A | 100 | 128 () | /25 = 255.255.255.128 | 205.206.207.0 | .1 – .126 | 205.206.207.127 |
| B | 60 | 64 () | /26 = 255.255.255.192 | 205.206.207.128 | .129 – .190 | 205.206.207.191 |
| C | 25 | 32 () | /27 = 255.255.255.224 | 205.206.207.192 | .193 – .222 | 205.206.207.223 |
| D | 20 | 32 (30 usable) | /27 = 255.255.255.224 | 205.206.207.224 | .225 – .254 | 205.206.207.255 |
The four blocks (128 + 64 + 32 + 32 = 256) use the full address space.
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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.
- SYN: The client sends a segment with SYN = 1 and an initial sequence number .
- SYN + ACK: The server replies with SYN = 1, ACK = 1, its own sequence number , and acknowledgement number .
- ACK: The client sends ACK = 1 with acknowledgement number . 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
- Process-to-process delivery: Port numbers are used to deliver data to the right application process, using multiplexing and demultiplexing.
- Reliable delivery: Using sequence numbers, acknowledgements and retransmission (TCP), data arrives complete and in order.
- 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.
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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
| Basis | 1-persistent / p-persistent CSMA | Non-persistent CSMA |
|---|---|---|
| Channel busy | Station keeps sensing continuously and transmits as soon as the channel becomes idle (with probability 1, or with probability ) | Station waits a random time before sensing again |
| Collision chance | Higher, since several waiting stations may send together | Lower |
| Delay | Lower | Higher |
| Channel utilisation | Lower at high load | Better at high load |
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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
- 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.
- 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
- DNS (Domain Name System) queries.
- 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.
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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):
| Bit | 1 | 1 | 0 | 0 | 1 | 0 | 1 | 1 |
|---|---|---|---|---|---|---|---|---|
| Level | High | Low | Low | Low | High | High | Low | High |
Manchester (IEEE 802.3 convention: 1 = high-to-low transition in mid-bit, 0 = low-to-high):
| Bit | 1 | 1 | 0 | 0 | 1 | 0 | 1 | 1 |
|---|---|---|---|---|---|---|---|---|
| Halves | HL | HL | LH | LH | HL | LH | HL | HL |
FSK (two carrier frequencies): 1 is sent with the high frequency and 0 with the low frequency , so the carrier is
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:
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.
- A station with data senses the channel. If it is idle for a DIFS interval, it sends the frame.
- 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.
- The receiver sends an ACK after a short SIFS. If the ACK is not received, the sender retransmits with a doubled contention window.
- Optional RTS/CTS handshaking and the NAV timer solve the hidden terminal problem.
Answer checked by Claude Sonnet · Oct 21, 2026
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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.
Answer checked by Claude Sonnet · Oct 21, 2026
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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
- 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.
- 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.
- Selective flooding: A router forwards a packet only on lines that lead roughly towards the destination, not on all lines.
Answer checked by Claude Sonnet · Oct 21, 2026
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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: (), so 6 host bits and 2 subnet bits.
(a) Subnet mask: /26 = 255.255.255.192
(b) Number of subnets: subnets, so departments can be served.
(c) and (d)
| Subnet | Network address | Usable IP range | Broadcast address |
|---|---|---|---|
| 1 | 205.206.207.0 | .1 – .62 | 205.206.207.63 |
| 2 | 205.206.207.64 | .65 – .126 | 205.206.207.127 |
| 3 | 205.206.207.128 | .129 – .190 | 205.206.207.191 |
| 4 | 205.206.207.192 | .193 – .254 | 205.206.207.255 |
(e) IP wastage: Each subnet has 62 usable addresses and uses 60, so 2 addresses are wasted per subnet.
If the network and broadcast addresses are counted too, the wastage is 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 ) 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 is lost or damaged, the sender retransmits frame 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.
Answer checked by Claude Sonnet · Oct 21, 2026
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