2026 question paper

Computer Networks

28 questions

1(a). A fully connected mesh network contains 8 nodes. Every pair of nodes requires a dedicated point-to-point link. How many links are required? - (i) 16 - (ii) 28 - (iii) 32 - (iv) 5620262m
Network Topologies
A fully connected mesh network contains 8 nodes. Every pair of nodes requires a dedicated point-to-point link. How many links are required?
- (i) 16
- (ii) 28
- (iii) 32
- (iv) 56
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Worked Solution
Answer: (ii) 28

In a fully connected mesh, every pair of nodes has a dedicated link:

$\text{Links} = \frac{n(n-1)}{2} = \frac{8 \times 7}{2} = 28$

(Each node also needs $n-1 = 7$ I/O ports.)
1(b). Wavelength Division Multiplexing (WDM) is primarily associated with: - (i) Copper cables - (ii) Radio communication only - (iii) Optical fiber communication - (iv) Coaxial cable only20262m
Multiplexing
Wavelength Division Multiplexing (WDM) is primarily associated with:
- (i) Copper cables
- (ii) Radio communication only
- (iii) Optical fiber communication
- (iv) Coaxial cable only
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Worked Solution
Answer: (iii) Optical fiber communication

WDM sends several optical signals of different wavelengths (colors) simultaneously over a single fiber. It is the optical equivalent of FDM, so it applies to fiber, not copper or coaxial cable.
1(c). A communication code has minimum Hamming distance d_{min} = 5. What is the maximum number of errors that can be detected? - (i) 2 - (ii) 3 - (iii) 5 - (iv) 420262m
Error Detection
A communication code has minimum Hamming distance $d_{min} = 5$ . What is the maximum number of errors that can be detected?
- (i) 2
- (ii) 3
- (iii) 5
- (iv) 4
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Worked Solution
Answer: (iv) 4

A code with minimum Hamming distance $d_{min}$ can detect up to $d_{min} - 1$ errors and correct up to $\lfloor (d_{min}-1)/2 \rfloor$ errors.

$d_{min} - 1 = 5 - 1 = 4$
1(d). A sender transmits frame 5 but does not receive its acknowledgement within the timeout interval. Assume the frame or its acknowledgement has been lost. What should the sender do? - (i) Retransmit frame 5 - (ii) Move directly to frame 6 - (iii) Discard all previously transmitted frames - (iv) Restart the entire connection20262m
Data Link Layer / ARQ
A sender transmits frame 5 but does not receive its acknowledgement within the timeout interval. Assume the frame or its acknowledgement has been lost. What should the sender do?
- (i) Retransmit frame 5
- (ii) Move directly to frame 6
- (iii) Discard all previously transmitted frames
- (iv) Restart the entire connection
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Worked Solution
Answer: (i) Retransmit frame 5

In ARQ protocols, when the timer expires without an ACK, the sender assumes the frame (or its ACK) was lost and retransmits the same frame. It does not move on, discard earlier frames, or restart the connection.
1(e). Which statement correctly distinguishes IPv6 from IPv4? - (i) IPv6 uses 32-bit addresses, while IPv4 uses 128-bit addresses - (ii) Both use 32-bit addresses - (iii) IPv6 uses 128-bit addresses, while IPv4 uses 32-bit addresses - (iv) Both use 128-bit addresses20262m
IP Addressing
Which statement correctly distinguishes IPv6 from IPv4?
- (i) IPv6 uses 32-bit addresses, while IPv4 uses 128-bit addresses
- (ii) Both use 32-bit addresses
- (iii) IPv6 uses 128-bit addresses, while IPv4 uses 32-bit addresses
- (iv) Both use 128-bit addresses
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Worked Solution
Answer: (iii) IPv6 uses 128-bit addresses, while IPv4 uses 32-bit addresses

Feature IPv4 IPv6

Address size 32 bits 128 bits

Address space ~ $2^{32}$ ~ $2^{128}$

Notation Dotted decimal Hexadecimal, colon-separated
1(f). Host A knows the IP address of Host B, and both hosts are on the same LAN. Host A does not have Host B's MAC address in its ARP cache. What happens first? - (i) Host A broadcasts an ARP request asking for the MAC address corresponding to Host B's IP address - (ii) Host A unicasts the IP packet directly to Host B - (iii) Host B broadcasts its MAC address - (iv) The router assigns a new IP address to Host A20262m
ARP
Host A knows the IP address of Host B, and both hosts are on the same LAN. Host A does not have Host B's MAC address in its ARP cache. What happens first?
- (i) Host A broadcasts an ARP request asking for the MAC address corresponding to Host B's IP address
- (ii) Host A unicasts the IP packet directly to Host B
- (iii) Host B broadcasts its MAC address
- (iv) The router assigns a new IP address to Host A
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Worked Solution
Answer: (i) Host A broadcasts an ARP request asking for the MAC address corresponding to Host B's IP address

ARP resolves IP to MAC. With no cache entry, Host A broadcasts an ARP request (destination MAC FF:FF:FF:FF:FF:FF) on the LAN. Only Host B replies, with a unicast ARP reply carrying its MAC address.
1(g). An application requires reliable, ordered delivery with retransmission of lost data. Which transport protocol is more appropriate? - (i) UDP - (ii) TCP - (iii) ARP - (iv) DHCP20262m
Transport Layer
An application requires reliable, ordered delivery with retransmission of lost data. Which transport protocol is more appropriate?
- (i) UDP
- (ii) TCP
- (iii) ARP
- (iv) DHCP
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Worked Solution
Answer: (ii) TCP

TCP is connection-oriented and provides reliable, in-order delivery using sequence numbers, acknowledgements and retransmission. UDP is unreliable; ARP and DHCP are not transport protocols.
1(h). A client wants to access www.example.com but does not know its IP address. Which application-layer service is responsible for resolving the domain name? - (i) FTP - (ii) TELNET - (iii) DNS - (iv) SMTP20262m
DNS
A client wants to access www.example.com but does not know its IP address. Which application-layer service is responsible for resolving the domain name?
- (i) FTP
- (ii) TELNET
- (iii) DNS
- (iv) SMTP
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Worked Solution
Answer: (iii) DNS

The Domain Name System translates human-readable domain names (e.g. www.example.com) into IP addresses. FTP is for file transfer, TELNET for remote login and SMTP for email.
1(i). Consider the following TCP exchange: 1. Client → Server: SYN 2. Server → Client: SYN + ACK 3. Client → Server: ACK What is the purpose of this exchange? - (i) IP fragmentation - (ii) TCP connection termination - (iii) UDP connection establishment - (iv) TCP connection establishment20262m
TCP
Consider the following TCP exchange:
1. Client → Server: SYN
2. Server → Client: SYN + ACK
3. Client → Server: ACK
What is the purpose of this exchange?
- (i) IP fragmentation
- (ii) TCP connection termination
- (iii) UDP connection establishment
- (iv) TCP connection establishment
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Worked Solution
Answer: (iv) TCP connection establishment

This is the three-way handshake:

Client → Server: SYN (client's initial sequence number)

Server → Client: SYN + ACK

Client → Server: ACK

After this, both sides are synchronized and the connection is open.
1(j). A web browser sends a request to a web server and receives a response containing the requested resource. Which communication model is primarily illustrated? - (i) Request-response - (ii) Token passing - (iii) Broadcast - (iv) Error correction20262m
Application Layer
A web browser sends a request to a web server and receives a response containing the requested resource. Which communication model is primarily illustrated?
- (i) Request-response
- (ii) Token passing
- (iii) Broadcast
- (iv) Error correction
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Worked Solution
Answer: (i) Request-response

The client sends a request (HTTP GET) and the server returns a response with the resource. This is the client-server request-response model used by the web.
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Feature	IPv4	IPv6
Address size	32 bits	128 bits
Address space	~ $2^{32}$	~ $2^{128}$
Notation	Dotted decimal	Hexadecimal, colon-separated

2(a). Explain the layered architecture of the OSI Reference Model with a neat block diagram. Describe the: - primary functions - data units - responsibilities of each of the seven layers.20267m

OSI Model

Explain the layered architecture of the OSI Reference Model with a neat block diagram. Describe the:

primary functions
data units
responsibilities

of each of the seven layers.

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Worked Solution

OSI Reference Model

The OSI model (ISO) divides network communication into seven layers. Each layer provides services to the layer above and uses services of the layer below.

SVGVector Architecture Diagram (SVG)

Layer	Data unit	Main functions
7 Application	Data / message	Interface to user applications; email, file transfer, web (HTTP, FTP, SMTP, DNS)
6 Presentation	Data	Translation, encryption/decryption, compression, character encoding
5 Session	Data	Establishes, manages and terminates sessions; dialog control; synchronization/checkpoints
4 Transport	Segment (TCP) / Datagram (UDP)	End-to-end delivery, segmentation and reassembly, flow control, error control, port addressing
3 Network	Packet	Logical addressing (IP), routing, path selection, fragmentation
2 Data Link	Frame	Framing, physical (MAC) addressing, error detection, flow control, media access control
1 Physical	Bit	Transmission of raw bits, signal encoding, voltage levels, data rate, topology, transmission medium

Key points

Encapsulation: each layer adds its own header (data link adds a trailer too) as data moves down; the receiver strips them going up.
Layers 1-3 are network-support layers; layers 5-7 are user-support layers; layer 4 links the two.
Peer layers communicate logically using protocols; real data flows vertically through the stack.

2(b). Explain mesh and star network topologies with suitable diagrams. Compare their advantages and disadvantages.20267m

Network Topologies

Explain mesh and star network topologies with suitable diagrams. Compare their advantages and disadvantages.

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Worked Solution

Mesh Topology

Every device has a dedicated point-to-point link to every other device.

SVGVector Architecture Diagram (SVG)

Links for $n$ nodes: $n(n-1)/2$ ; each node needs $n-1$ ports.

Advantages

No traffic problems; dedicated links
Robust: failure of one link does not affect others
Privacy and security (point-to-point)
Easy fault identification

Disadvantages

Very high cabling and port cost
Complex installation and reconfiguration
Poor scalability

Star Topology

All devices connect to a central hub/switch; no direct links between devices.

SVGVector Architecture Diagram (SVG)

Links for $n$ nodes: $n$ ; each node needs one port.

Advantages

Cheap, easy to install and extend
Failure of one node/link does not affect others
Easy fault detection and management

Disadvantages

Central hub is a single point of failure
Needs more cable than bus; hub capacity limits network size

Comparison

Basis	Mesh	Star
Links (n nodes)	n(n-1)/2	n
Cost	Very high	Low
Reliability	Very high	Depends on central device
Scalability	Poor	Good
Ports per device	n-1	1
Use	Backbone/critical networks	LANs (Ethernet with switch)

3(a). Explain unguided transmission media used in computer networks with suitable examples. Discuss their: - characteristics - advantages - disadvantages20267m

Transmission Media

Explain unguided transmission media used in computer networks with suitable examples. Discuss their:

characteristics
advantages
disadvantages

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Worked Solution

Unguided (Wireless) Transmission Media

Signals travel through free space (air, vacuum, water) as electromagnetic waves, with no physical conductor.

1. Radio Waves (3 kHz - 1 GHz)

Omnidirectional, propagate in all directions; penetrate walls
Examples: AM/FM radio, TV broadcast, cordless phones, paging
Pros: long range, penetrates buildings, simple
Cons: interference, susceptible to eavesdropping, limited bandwidth

2. Microwaves (1 - 300 GHz)

Unidirectional, line-of-sight; need aligned antennas
Examples: cellular networks, satellite links, Wi-Fi, point-to-point towers
Pros: high data rate, no cabling cost
Cons: line of sight needed, affected by rain/weather, cannot penetrate buildings well

3. Infrared (300 GHz - 400 THz)

Short range, line-of-sight; cannot pass through walls
Examples: TV remotes, wireless keyboards/mice, IrDA
Pros: secure within a room, no interference with other rooms
Cons: very short range, blocked by obstacles, affected by sunlight

4. Satellite Communication

Microwave relay through orbiting satellites (GEO, MEO, LEO)
Examples: GPS, DTH TV, VSAT
Pros: wide coverage, remote areas
Cons: propagation delay, high cost

Summary

Medium	Range	Directionality	Penetration	Use
Radio	Long	Omni	Good	Broadcast
Microwave	Medium-long	Unidirectional	Poor	Cellular, satellite
Infrared	Very short	Directional	None	Remote controls

General advantages

Mobility, no cable installation, easy deployment in difficult terrain.

General disadvantages

Less secure, interference and noise, signal attenuation, weather dependence, lower reliability than fiber.

3(b). Explain wired LAN and wireless LAN. Compare their characteristics, advantages and limitations.20267m

LAN

Explain wired LAN and wireless LAN. Compare their characteristics, advantages and limitations.

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Worked Solution

Wired LAN

Devices are connected through physical media (twisted pair, coaxial, fiber) using switches/hubs. The standard is IEEE 802.3 (Ethernet) with CSMA/CD (legacy) or full-duplex switched operation.

Wireless LAN (WLAN)

Devices connect through radio waves via an access point (AP). The standard is IEEE 802.11 (Wi-Fi) using CSMA/CA.

SVGVector Architecture Diagram (SVG)

Comparison

Basis	Wired LAN	Wireless LAN
Standard	IEEE 802.3	IEEE 802.11
Medium	Cables / fiber	Radio waves
Access method	CSMA/CD	CSMA/CA
Speed	Very high (1/10/100 Gbps)	Lower, shared (up to a few Gbps)
Reliability	High, little interference	Affected by interference and obstacles
Security	Physical access needed	Needs encryption (WPA2/WPA3)
Mobility	Fixed	High
Installation	Costly, cabling work	Easy, flexible
Latency	Low	Higher, variable

Advantages and limitations

Wired: + fast, stable, secure; - cabling cost, no mobility, hard to reconfigure.

Wireless: + mobility, easy setup, scalable; - interference, limited range, security risks, lower throughput.

4(a). Explain Time Division Multiplexing (TDM) with a suitable diagram. Discuss its working and advantages.20267m
Multiplexing
Explain Time Division Multiplexing (TDM) with a suitable diagram. Discuss its working and advantages.

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Worked Solution
Time Division Multiplexing (TDM)

TDM is a digital technique where multiple signals share one channel by occupying it in different time slots. The full bandwidth is used by one signal at a time.

SVGVector Architecture Diagram (SVG)
Working

Each input gets a fixed time slot.

The multiplexer scans inputs in round-robin order, taking a small unit (bit/byte) from each.

One round of slots forms a frame; frames repeat continuously.

The demultiplexer, synchronized using framing bits, routes each slot to the correct output.

Types

Synchronous TDM: slots are pre-assigned; an idle input wastes its slot.

Statistical TDM: slots are assigned dynamically, only to active inputs; better utilization (needs addressing info).

Advantages

Efficient use of link capacity

No crosstalk or intermodulation distortion (unlike FDM)

Simple, suited to digital data and voice (PCM, T1/E1)

Easy to add or remove channels

Disadvantages

Needs synchronization

Wasted slots in synchronous TDM

4(b). Compare Frequency Division Multiplexing (FDM), Time Division Multiplexing (TDM) and Wavelength Division Multiplexing (WDM) based on: - signal type - medium - operation - applications20267m

Multiplexing

Compare Frequency Division Multiplexing (FDM), Time Division Multiplexing (TDM) and Wavelength Division Multiplexing (WDM) based on:

signal type
medium
operation
applications

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Worked Solution

FDM vs TDM vs WDM

Basis	FDM	TDM	WDM
Full form	Frequency Division Multiplexing	Time Division Multiplexing	Wavelength Division Multiplexing
Signal type	Analog	Digital (also analog after sampling)	Optical (light) signals
Medium	Coaxial, radio, twisted pair	Copper, microwave, fiber	Optical fiber only
Operation	Channels share the link by occupying different frequency bands simultaneously	Channels share the link in different time slots, one at a time	Signals of different wavelengths travel simultaneously in one fiber
Sharing	Frequency	Time	Wavelength (color)
Guard needed	Guard bands	Guard times / sync bits	Guard wavelengths
Interference	Crosstalk, intermodulation noise	Needs synchronization	Minimal
Capacity	Moderate	Moderate to high	Very high (DWDM)
Applications	Radio/TV broadcasting, cable TV, 1G cellular	Telephone networks (T1/E1), GSM, SONET	Backbone and long-haul fiber networks

5(a). Explain the principle of error detection using Cyclic Redundancy Check (CRC). Given: - Dataword: 1101011011 - Generator polynomial: G(x) = x^4 + x + 1 Determine the transmitted codeword using binary modulo-2 division. Verify how the receiver detects errors if no bit is altered.20267m

Error Detection / CRC

Explain the principle of error detection using Cyclic Redundancy Check (CRC).

Given:

Dataword: 1101011011
Generator polynomial: $G(x) = x^4 + x + 1$

Determine the transmitted codeword using binary modulo-2 division. Verify how the receiver detects errors if no bit is altered.

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Worked Solution

Principle of CRC

CRC is a polynomial-based error detection technique. The sender appends $r$ redundant bits (the CRC) to the dataword so that the resulting codeword is exactly divisible by a predetermined generator (divisor) using modulo-2 arithmetic (XOR, no carries). The receiver divides the received codeword by the same generator: remainder 0 means no error detected; nonzero means error.

Steps (sender):

Append $r$ zeros to the dataword ( $r$ = degree of generator).
Divide by the generator using modulo-2 division.
Replace the appended zeros with the remainder; this is the codeword.

Given

Dataword: 1101011011
$G(x) = x^4 + x + 1 \Rightarrow$ divisor 10011 (degree 4, so append 4 zeros)
Dividend: 1101011011 + 0000 = 11010110110000

Modulo-2 division

Step	Current bits	XOR with	Result (remainder)
1	11010	10011	01001 -> 1001
2	1001+1 = 10011	10011	00000 -> 0000
3	0000+1 = 00001	00000	0001
4	0001+0 = 00010	00000	0010
5	0010+1 = 00101	00000	0101
6	0101+1 = 01011	00000	1011
7	1011+0 = 10110	10011	00101 -> 0101
8	0101+0 = 01010	00000	1010
9	1010+0 = 10100	10011	00111 -> 0111
10	0111+0 = 01110	00000	1110

CRC (remainder) = 1110

Transmitted codeword

Dataword + CRC = 1101011011 | 1110  ->  11010110111110

Receiver verification (no bit altered)

The receiver divides 11010110111110 by 10011. Since the codeword equals dataword-with-zeros XOR remainder, the division leaves remainder 0000. A zero remainder means no error is detected and the dataword 1101011011 is accepted. If any bit flipped, the remainder would be nonzero (unless the error pattern is itself a multiple of $G(x)$ ), and the frame would be discarded.

5(b). Explain the Stop-and-Wait ARQ protocol with a suitable diagram. Discuss its working.20267m
ARQ
Explain the Stop-and-Wait ARQ protocol with a suitable diagram. Discuss its working.

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Worked Solution
Stop-and-Wait ARQ

The simplest ARQ (Automatic Repeat reQuest) flow and error control protocol. The sender transmits one frame at a time and waits for its acknowledgement before sending the next.

Working

Sender sends frame 0 and starts a timer; keeps a copy.

Receiver checks the frame (CRC). If correct, it sends ACK; the sender then sends the next frame (1), alternating sequence numbers 0 and 1.

If the frame is corrupted, the receiver discards it silently; the sender's timer expires and it retransmits.

If the ACK is lost, the timer expires, the sender retransmits, and the receiver recognizes a duplicate by the sequence number, discards it and re-sends the ACK.

SVGVector Architecture Diagram (SVG)
Features

Window size: sender = 1, receiver = 1

Sequence numbers: 1 bit (0, 1)

Uses timer, ACKs and retransmission

Drawback

Low efficiency on long or high-bandwidth links since the channel sits idle while waiting for the ACK. Efficiency $= \dfrac{1}{1 + 2a}$ where $a = T_p / T_t$ . Sliding window protocols (Go-Back-N, Selective Repeat) overcome this.

6(a). Describe the IPv4 header format with a suitable diagram. Explain the functions of the following fields: - TTL - Header Checksum - Flags - Fragmentation Offset20267m

IPv4

Describe the IPv4 header format with a suitable diagram. Explain the functions of the following fields:

TTL
Header Checksum
Flags
Fragmentation Offset

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Worked Solution

IPv4 Header Format

The IPv4 header is 20 to 60 bytes (20-byte base header plus up to 40 bytes of options).

SVGVector Architecture Diagram (SVG)

Field	Size	Purpose
Version	4 bits	IP version (4)
IHL	4 bits	Header length in 32-bit words
Type of Service	8 bits	DSCP/ECN; QoS
Total Length	16 bits	Header + data, max 65,535 bytes
Identification	16 bits	Identifies fragments of one datagram
Flags	3 bits	Reserved, DF, MF
Fragment Offset	13 bits	Position of fragment
TTL	8 bits	Hop limit
Protocol	8 bits	Upper layer (TCP=6, UDP=17, ICMP=1)
Header Checksum	16 bits	Header error detection
Source / Destination IP	32 bits each	Addresses

Functions of key fields

TTL: set by the source and decremented by each router; when it reaches 0 the packet is discarded (ICMP Time Exceeded). Prevents infinite looping.
Header Checksum: 16-bit one's complement checksum over the header only. Recomputed at every router since TTL changes. Packets with a bad checksum are dropped.
Flags: 3 bits: bit 0 reserved; DF (Don't Fragment): 1 means do not fragment; MF (More Fragments): 1 means more fragments follow, 0 means last fragment.
Fragmentation Offset: gives the position of the fragment's data in the original datagram, in units of 8 bytes, allowing the receiver to reassemble fragments in order.

6(b). Explain the working of ARP in mapping an IP address to a MAC address. Describe the ARP request and ARP reply process with a neat diagram.20267m
ARP
Explain the working of ARP in mapping an IP address to a MAC address. Describe the ARP request and ARP reply process with a neat diagram.

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Worked Solution
Address Resolution Protocol (ARP)

ARP maps a known IP address (logical, network layer) to the corresponding MAC address (physical, data link layer) on the same LAN. A host needs the destination MAC to build an Ethernet frame.

Process

Host A wants to send to Host B (IP known) and checks its ARP cache.

If there is no entry, A broadcasts an ARP Request (destination MAC FF:FF:FF:FF:FF:FF) containing A's IP/MAC and B's IP, asking "Who has IP B?"

All hosts receive it; only B recognizes its own IP.

B sends a unicast ARP Reply with its MAC address to A (and may cache A's mapping).

A stores the mapping in its ARP cache (with a timeout) and sends the frame.

SVGVector Architecture Diagram (SVG)
Notes

Request: broadcast; Reply: unicast.

If the destination is on another network, the host ARPs for the default gateway's MAC instead.

ARP packet fields: hardware type, protocol type, hardware/protocol lengths, operation (1=request, 2=reply), sender MAC/IP, target MAC/IP.
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7(a). Describe the connection establishment and connection termination mechanisms in TCP with sequence timing diagrams. - Explain the role of the SYN, ACK and FIN control flags - Explain the significance of the TIME_WAIT state during connection teardown20267m
TCP
Describe the connection establishment and connection termination mechanisms in TCP with sequence timing diagrams.
- Explain the role of the SYN, ACK and FIN control flags
- Explain the significance of the TIME_WAIT state during connection teardown
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Worked Solution
TCP Control Flags

SYN: synchronize sequence numbers; used to open a connection

ACK: acknowledgement field is valid

FIN: sender has no more data; used to close a connection

Connection Establishment: Three-Way Handshake

SVGVector Architecture Diagram (SVG)

Client sends SYN with its initial sequence number $x$ .

Server replies SYN + ACK with its own sequence number $y$ and ack $x+1$ .

Client sends ACK with ack $y+1$ . Connection is established.

Connection Termination (four-way)

SVGVector Architecture Diagram (SVG)

Side A sends FIN.

Side B ACKs it (it may still send remaining data: half-close).

Side B sends its own FIN when done.

Side A ACKs and enters TIME_WAIT.

Significance of TIME_WAIT

The side that closes first waits 2 x MSL (Maximum Segment Lifetime) before fully closing. Reasons:

Reliable termination: if the final ACK is lost, the peer retransmits its FIN, and the waiting side can re-send the ACK.

Avoid stale segments: lets old duplicate segments from the previous connection expire so they are not mistaken for data of a new connection using the same address/port pair.
7(b). Explain the Leaky Bucket algorithm with a suitable diagram. Discuss its working and applications.20267m
Congestion Control
Explain the Leaky Bucket algorithm with a suitable diagram. Discuss its working and applications.

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Worked Solution
Leaky Bucket Algorithm

A traffic shaping / congestion control algorithm that converts bursty traffic into a constant-rate output, like water leaking from a hole at the bottom of a bucket at a fixed rate.

SVGVector Architecture Diagram (SVG)
Working

Incoming packets are placed in a finite-size queue (bucket).

Packets leave the bucket at a fixed rate $\rho$ , regardless of the input rate.

If the bucket is full, arriving packets are discarded.

For variable-size packets, a fixed number of bytes per tick is released.

Characteristics

Output rate is constant and smooth

Bursts are absorbed up to the bucket size

Excess traffic is dropped

Does not allow idle-time credit (unlike Token Bucket, which permits bursts)

Applications

Traffic shaping at network edges (ISP-customer)

Enforcing a Service Level Agreement/rate limit

QoS in ATM and multimedia traffic

Congestion prevention in routers
8(a). Explain the hierarchical architecture of the Domain Name Space. Describe the domain name resolution process using recursive and iterative queries with a suitable sequence diagram.20267m
DNS
Explain the hierarchical architecture of the Domain Name Space. Describe the domain name resolution process using recursive and iterative queries with a suitable sequence diagram.

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Worked Solution
Domain Name Space

DNS is a distributed, hierarchical database mapping names to IP addresses. The name space is an inverted tree with up to 128 levels; each node has a label (max 63 characters), and a domain name is the sequence of labels from the node up to the root, separated by dots.

SVGVector Architecture Diagram (SVG)

Root servers know the TLD servers.

TLD servers (.com, .org, country codes) know authoritative servers of domains.

Authoritative servers hold the actual records for a zone.

Local/recursive resolver (from ISP) queries on behalf of clients and caches answers.

Name Resolution

Recursive query

The client asks its local resolver, which takes full responsibility for returning the final answer (or an error).

Iterative query

The queried server returns either the answer or a referral to another server; the requester contacts the next server itself.

SVGVector Architecture Diagram (SVG)
Client -> local resolver is recursive; local resolver -> root/TLD/authoritative uses iterative queries. Results are cached according to their TTL to reduce traffic.
8(b). Describe the working of File Transfer Protocol (FTP) with a neat diagram. Describe the control connection and data connection used for file transfer.20267m
FTP
Describe the working of File Transfer Protocol (FTP) with a neat diagram. Describe the control connection and data connection used for file transfer.

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Worked Solution
File Transfer Protocol (FTP)

FTP is an application-layer protocol (RFC 959) for transferring files between a client and a server over TCP. It uses two separate connections.

SVGVector Architecture Diagram (SVG)
Control Connection

Uses server port 21; opened first and stays open for the whole session

Carries commands (USER, PASS, LIST, RETR, STOR, QUIT) and replies (status codes)

Text (NVT ASCII) based; handles login/authentication

Data Connection

Uses server port 20 (active mode)

Opened for each file transfer and closed afterward

Carries the actual file or directory listing

Modes

Active: server initiates the data connection to the client.

Passive: client initiates the data connection (works through firewalls/NAT).

Working

Client opens the control connection to port 21 and logs in.

Client sends a command (e.g. RETR file).

A data connection is set up and the file is transferred.

The data connection closes; the control connection remains for further commands until QUIT.

Using separate connections (out-of-band control) allows commands to be sent even during a transfer.
9(a). Write a short note on Virtual LAN (VLAN). > Question 9 asks candidates to write short notes on any two listed topics.20267m
VLAN
Write a short note on Virtual LAN (VLAN).

Question 9 asks candidates to write short notes on any two listed topics.

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Worked Solution
Virtual LAN (VLAN)

A VLAN is a logical grouping of devices into a separate broadcast domain, independent of their physical location, configured on switches.

SVGVector Architecture Diagram (SVG)
Features

Each VLAN is its own broadcast domain; inter-VLAN traffic needs a router / Layer-3 switch.

Membership by port, MAC address, protocol or IP subnet.

IEEE 802.1Q tags frames (4-byte tag with a 12-bit VLAN ID) on trunk links.

Advantages

Reduced broadcast traffic and better performance

Improved security through logical isolation

Flexibility: users can be grouped by function, not location

Lower cost and easier network management
9(b). Write a short note on Piggybacking. > Question 9 asks candidates to write short notes on any two listed topics.20267m
Data Link Layer
Write a short note on Piggybacking.

Question 9 asks candidates to write short notes on any two listed topics.

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Worked Solution
Piggybacking

Piggybacking is a data link layer technique in bidirectional (full-duplex) communication where an acknowledgement is attached to an outgoing data frame instead of being sent as a separate ACK frame.

SVGVector Architecture Diagram (SVG)
Working

Each frame carries both a data field and an acknowledgement field.

When a node has data to send, it includes the pending ACK in that frame.

If no data is ready within a short time, the node sends a separate ACK to avoid sender timeout (ACK timer).

Advantages

Better bandwidth utilization

Fewer frames, lower overhead

Disadvantage

ACK may be delayed while waiting for outgoing data, risking timeouts.
9(c). Write a short note on Quality of Service (QoS). > Question 9 asks candidates to write short notes on any two listed topics.20267m
Quality of Service
Write a short note on Quality of Service (QoS).

Question 9 asks candidates to write short notes on any two listed topics.

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Worked Solution
Quality of Service (QoS)

QoS is the set of techniques to guarantee a required level of network performance for different types of traffic.

QoS parameters

Bandwidth (throughput)

Delay (latency)

Jitter (variation in delay)

Reliability / packet loss

Techniques to achieve QoS

Scheduling: FIFO, priority queuing, weighted fair queuing

Traffic shaping: leaky bucket, token bucket

Admission control: accept a flow only if resources are available

Resource reservation: RSVP

Models

Integrated Services (IntServ): per-flow reservation

Differentiated Services (DiffServ): per-class marking (DSCP) and treatment

Applications

VoIP, video conferencing, streaming and online gaming, which are delay- and jitter-sensitive, unlike email or file transfer.
9(d). Write a short note on Bluetooth. > Question 9 asks candidates to write short notes on any two listed topics.20267m
Wireless Networks
Write a short note on Bluetooth.

Question 9 asks candidates to write short notes on any two listed topics.

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Worked Solution
Bluetooth

Bluetooth (IEEE 802.15.1) is a short-range wireless technology for connecting devices to form a Personal Area Network (PAN).

Features

Operates in the 2.4 GHz ISM band

Uses FHSS (frequency hopping, 1600 hops/s) to avoid interference

Range about 10 m (Class 2), up to 100 m (Class 1)

Low power and low cost

Architecture

Piconet: one master and up to 7 active slaves

Scatternet: multiple interconnected piconets

Uses TDD with time slots of 625 microseconds

Applications

Wireless headsets, keyboards/mice, file transfer between phones, IoT devices and car hands-free systems.

Limitations

Short range, lower data rate than Wi-Fi, security concerns (pairing attacks).

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Computer Networks

Pocketly on Android

OSI Reference Model

Key points

Mesh Topology

Star Topology

Comparison

Unguided (Wireless) Transmission Media

1. Radio Waves (3 kHz - 1 GHz)

2. Microwaves (1 - 300 GHz)

3. Infrared (300 GHz - 400 THz)

4. Satellite Communication

Summary

General advantages

General disadvantages

Wired LAN

Wireless LAN (WLAN)

Comparison

Advantages and limitations

Time Division Multiplexing (TDM)

Working

Types

Advantages

Disadvantages

FDM vs TDM vs WDM

Principle of CRC

Given

Modulo-2 division

Transmitted codeword

Receiver verification (no bit altered)

Stop-and-Wait ARQ

Working

Features

Drawback

IPv4 Header Format

Functions of key fields

Address Resolution Protocol (ARP)

Process

Notes

Pocketly on Android

TCP Control Flags

Connection Establishment: Three-Way Handshake

Connection Termination (four-way)

Significance of TIME_WAIT

Leaky Bucket Algorithm

Working

Characteristics

Applications

Domain Name Space

Name Resolution

Recursive query

Iterative query

File Transfer Protocol (FTP)

Control Connection

Data Connection

Modes

Working

Virtual LAN (VLAN)

Features

Advantages

Piggybacking

Working

Advantages

Disadvantage

Quality of Service (QoS)

QoS parameters

Techniques to achieve QoS

Models

Applications

Bluetooth

Features

Architecture

Applications

Limitations

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