Group A — Short Answer Questions (1 Mark Each)

Q1Define Computer Network.

Ans: A computer network is a collection of autonomous computing devices (hosts) interconnected by a transmission medium so that they can exchange data under a common set of protocols. "Autonomous" means no node can forcibly start, stop or control another, which distinguishes a network from a multiprocessor system. Its purpose is resource sharing, reliability and communication, e.g. a college LAN sharing a printer and a file server.

Q2What is OSI Reference Model?

Ans: The OSI (Open Systems Interconnection) Reference Model is a seven-layer conceptual framework standardised by ISO (ISO 7498, 1984) that partitions network communication into layers, each providing a defined service to the layer above and using the services of the layer below. It is a reference model only, not a protocol suite — real Internet traffic runs the TCP/IP protocols — but it remains the standard vocabulary for describing layer functions.

Q3Name all 7 layers of OSI Model in order.

Ans: From bottom to top the seven OSI layers are: (1) Physical, (2) Data Link, (3) Network, (4) Transport, (5) Session, (6) Presentation, (7) Application. Layers 1–3 are the network-support (hardware) layers, layers 5–7 the user-support layers, and layer 4 (Transport) links the two. Mnemonic: "Please Do Not Throw Sausage Pizza Away".

Q4What is TCP/IP Model?

Ans: The TCP/IP model is the four-layer protocol suite on which the Internet actually runs: Network Access (Host-to-Network), Internet (IP, ICMP, ARP), Transport (TCP, UDP) and Application (HTTP, FTP, DNS, SMTP). It collapses OSI layers 1–2 into one layer and OSI layers 5–7 into the Application layer; unlike OSI it was defined after its protocols existed, so the model fits the protocols rather than the reverse.

Q5Define Physical Topology.

Ans: Physical topology is the geometric arrangement of the physical links and nodes of a network, i.e. how the cabling actually runs; the basic types are bus, star, ring, mesh and hybrid. It must be distinguished from logical topology, which describes how data actually flows: a 10BASE-T Ethernet is physically a star (all nodes wired to a hub) but logically a bus.

Q6What is Star Topology?

Ans: In a star topology every node has a dedicated point-to-point link to a central controller (hub or switch) and all traffic passes through that centre; no two nodes are directly linked. For \( n \) nodes it needs \( n \) links and \( n \) ports on the hub, is cheap to install and easy to fault-isolate, but the central device is a single point of failure. It is the topology of modern switched Ethernet LANs.

Q7What is Mesh Topology?

Ans: In a mesh topology every node has a dedicated point-to-point link to every other node, so a fully connected mesh of \( n \) nodes needs \( \frac{n(n-1)}{2} \) duplex links and \( n-1 \) I/O ports per node. It gives maximum reliability, privacy and no traffic contention, but the \( O(n^2) \) cabling and port cost restricts it to small or backbone networks.

Q8Define Bandwidth.

Ans: Bandwidth is the capacity of a channel, used in two senses: bandwidth in hertz is the width of the band of frequencies the medium can pass, \( B = f_{max} - f_{min} \); bandwidth in bits per second is the maximum theoretical data rate the channel can carry. The two are linked by Nyquist, \( C = 2B\log_2 L \), and Shannon, \( C = B\log_2(1+\mathrm{SNR}) \). Bandwidth is a theoretical ceiling, not the rate actually achieved.

Q9Define Throughput in networking.

Ans: Throughput is the actual measured rate at which data is successfully delivered across a link or path, in bits per second. It differs from bandwidth: bandwidth is the theoretical maximum capacity of the medium, whereas throughput is what is really obtained after protocol overhead, collisions, retransmissions and congestion, so always \( \text{Throughput} \le \text{Bandwidth} \). A 100 Mbps link may deliver only 20 Mbps of throughput under load.

Q10Define Propagation Delay.

Ans: Propagation delay is the time a single bit takes to travel from sender to receiver across the physical medium: \( T_p = \frac{\text{Distance}}{\text{Propagation speed}} \), the speed being about \( 2 \times 10^8 \) m/s in copper or fibre. It depends only on distance and medium — never on packet size or data rate.

Q11Define Transmission Delay.

Ans: Transmission delay (store time) is the time the sender needs to push all the bits of a packet onto the link: \( T_t = \frac{L}{R} \), where \( L \) is the packet length in bits and \( R \) the bandwidth in bps. It depends on packet size and data rate but not on distance — the exact opposite of propagation delay.

Q12What is Bandwidth-Delay Product?

Ans: The bandwidth–delay product is \( \text{BDP} = \text{Bandwidth} \times \text{Delay} \), and it gives the number of bits that can be "in flight" on the link at any instant, i.e. the volume of the pipe. Taken with the round-trip time it is the minimum sender window needed to keep the link fully utilised; a sliding window smaller than the BDP leaves the channel idle.

Q13Define Simplex, Half-Duplex, and Full-Duplex.

Ans: Simplex transmission is unidirectional only, one device sending and the other only receiving (keyboard to CPU). Half-duplex is bidirectional but only one direction at a time, the entire channel capacity going to whichever station is transmitting (walkie-talkie, Ethernet on a hub). Full-duplex is bidirectional simultaneously, the capacity being shared between the two directions or carried on two separate paths (telephone, switched Ethernet).

Q14What is Guided Transmission Media?

Ans: Guided (bounded) transmission media conduct the signal along a solid physical conductor that directs the wave from one device to another. The three types are twisted-pair cable (UTP/STP), coaxial cable and optical fibre. Because the signal is confined to the conductor, guided media give higher security, lower interference and higher data rates than wireless.

Q15What is Unguided Transmission Media?

Ans: Unguided (unbounded) transmission media carry electromagnetic waves through free space with no physical conductor, so the signal is broadcast and available to anyone with a suitable antenna. The three bands are radio waves (3 kHz–1 GHz, omnidirectional, penetrate walls), microwaves (1–300 GHz, line-of-sight, unidirectional dish antennas) and infrared (300 GHz–400 THz, short range, blocked by walls).

Q16Define Optical Fiber communication.

Ans: Optical fibre communication transmits data as pulses of light through a glass or plastic core surrounded by cladding of lower refractive index, so that rays striking the boundary at more than the critical angle undergo total internal reflection and remain inside the core. It offers very high bandwidth, extremely low attenuation, and complete immunity to electromagnetic interference and crosstalk; propagation modes are single-mode and multimode (step-index or graded-index).

Q17What is Bit Rate and Baud Rate?

Ans: Bit rate is the number of bits transmitted per second, whereas baud rate (signal rate) is the number of signal elements or symbol changes per second. They are related by \( \text{Bit rate} = \text{Baud rate} \times \log_2 L \), where \( L \) is the number of distinct signal levels. Baud rate determines the bandwidth required while bit rate measures the data carried, so 16-QAM at 1000 baud yields 4000 bps.

Q18Define Pulse Code Modulation (PCM).

Ans: Pulse Code Modulation (PCM) is the standard technique for converting an analog signal into a digital bit stream in three steps: sampling (PAM) at the Nyquist rate \( f_s \ge 2 f_{max} \), quantization of each sample into one of \( L \) levels, and encoding of each level into \( n = \log_2 L \) bits. Voice band sampled at 8000 Hz with 8-bit codes therefore gives the 64 kbps digital telephone channel.

Q19What is Framing in Data Link Layer?

Ans: Framing is the Data Link layer (OSI layer 2) function of dividing the continuous bit stream received from the Physical layer into distinguishable units called frames, each carrying a header with addresses and a trailer with a checksum, so the receiver knows exactly where a message begins and ends. Frame delimitation is achieved by character count, byte (character) stuffing with FLAG and ESC bytes, bit stuffing (insert a 0 after five consecutive 1s, as in HDLC), or physical-layer coding violations.

Q20Define Error Detection and Error Correction.

Ans: Error detection only determines whether the received bits differ from the transmitted bits — the receiver knows an error exists but not where, so it must ask for retransmission (ARQ); parity, checksum and CRC are detection codes. Error correction additionally locates and repairs the corrupted bits at the receiver with no retransmission (Forward Error Correction), as in Hamming code, and therefore needs far more redundant bits.

Q21What is Parity Check?

Ans: A parity check appends one redundant bit to the data unit so that the total number of 1s becomes even (even parity) or odd (odd parity). It is the simplest detection scheme, with minimum Hamming distance \( d_{min} = 2 \), so it detects all single-bit errors and any odd number of bit errors, but fails on every even number of errors and can never correct. Two-dimensional (LRC + VRC) parity improves this and can correct a single-bit error.

Q22What is Cyclic Redundancy Check (CRC)?

Ans: Cyclic Redundancy Check (CRC) is a polynomial-based error-detecting code used in data link networks. The sender appends a checksum \(R(x)\) derived from modulo-2 division of data polynomial \(M(x) \cdot 2^r\) by generator polynomial \(G(x)\).

Q23What is Hamming Code?

Ans: Hamming code is a linear block error-correcting code that inserts \( r \) parity bits at the bit positions that are powers of two \( (1, 2, 4, 8, \dots) \) among \( m \) data bits, where \( r \) is the smallest integer satisfying \( 2^r \ge m + r + 1 \). Each parity bit checks a distinct set of positions, so the parity results read together form a syndrome whose binary value is the address of the erroneous bit. With \( d_{min} = 3 \) it corrects any single-bit error (and with one extra bit becomes SEC-DED).

Q24Define Hamming Distance.

Ans: The Hamming distance \( d(x,y) \) between two codewords of equal length is the number of bit positions in which they differ, obtained as the number of 1s in \( x \oplus y \). The minimum Hamming distance \( d_{min} \) of a code fixes its power: the code detects up to \( d_{min}-1 \) bit errors and corrects up to \( \left\lfloor \frac{d_{min}-1}{2} \right\rfloor \) bit errors.

Q25What is Flow Control?

Ans: Flow control is the set of procedures that prevents a fast sender from overwhelming a slow receiver, by limiting the amount of data that may be sent before an acknowledgement is received. It is a receiver-driven mechanism operating between two endpoints (stop-and-wait, sliding window, TCP's advertised receive window). It must not be confused with congestion control, which protects the routers inside the network rather than the end receiver's buffer.

Q26What is Stop-and-Wait Protocol?

Ans: In the Stop-and-Wait protocol the sender transmits one frame and then halts until the acknowledgement for that frame arrives, so the window size is 1 in each direction and a single sequence-number bit suffices. With \( a = \frac{T_p}{T_t} \) its link utilisation is \( \text{Efficiency} = \frac{1}{1+2a} \), which becomes very poor on long or high-speed links and motivates the sliding-window protocols.

Q27What is Sliding Window Protocol?

Ans: A sliding window protocol lets the sender transmit several frames before any acknowledgement arrives, by maintaining a sender window of outstanding unacknowledged frames and a receiver window of frames it is prepared to accept; the windows "slide" forward as acknowledgements are received. With \( m \)-bit sequence numbers the numbering is modulo \( 2^m \), and a window of \( N \) raises efficiency to \( \frac{N}{1+2a} \), capped at 1. Go-Back-N and Selective Repeat are its two ARQ forms.

Q28What is Go-Back-N ARQ?

Ans: Go-Back-N ARQ is a sliding-window protocol with a sender window of \( N \) but a receiver window of 1: the receiver accepts frames strictly in order, silently discards every frame arriving after a lost one, and returns cumulative acknowledgements. On timeout the sender retransmits the lost frame and all frames sent after it, so it needs no receiver buffering but wastes bandwidth. To keep sequence numbers unambiguous, \( N \le 2^m - 1 \).

Q29What is Selective Repeat ARQ?

Ans: Selective Repeat ARQ retransmits only the frame that was actually lost or damaged, because the receiver window is also \( N \): the receiver buffers out-of-order frames, acknowledges them individually and sends a NAK for the missing one. It is far more bandwidth-efficient than Go-Back-N but requires receiver buffering and resequencing logic, and the tighter window constraint \( N \le 2^{m-1} \).

Q30Define Piggybacking.

Ans: Piggybacking is the technique of carrying the acknowledgement of received frames inside the header of an outgoing data frame travelling in the reverse direction, instead of sending a separate ACK frame. On a full-duplex link this saves bandwidth and reduces the number of frames, at the cost of a small added delay while the data link layer waits for a data frame to attach the ACK to.

Q31What is MAC Address?

Ans: A MAC (Media Access Control) address is the 48-bit (6-byte) physical hardware address burned into the NIC and used by the Data Link layer (OSI layer 2) to identify a station within one LAN. It is written as 12 hexadecimal digits (e.g. 00:1A:2B:3C:4D:5E), the first 24 bits being the OUI assigned to the manufacturer and the last 24 bits the card number. Unlike an IP address it is flat, permanent and globally unique, and it does not change when the host joins a different network.

Q32Define CSMA/CD.

Ans: CSMA/CD (Carrier Sense Multiple Access with Collision Detection) is the wired Ethernet (IEEE 802.3) MAC protocol in which a station senses the channel, transmits when it is idle, and keeps listening while transmitting; on detecting a collision it aborts at once, sends a 32-bit jam signal and retries after a binary exponential backoff of \( R \times \text{slot time} \) with \( R \) chosen randomly from \( [0,\, 2^k - 1] \). Detection requires \( T_t \ge 2T_p \), which fixes the 51.2 µs slot time and 64-byte minimum frame at 10 Mbps.

Q33Define CSMA/CA.

Ans: CSMA/CA (Carrier Sense Multiple Access with Collision Avoidance) is the wireless (IEEE 802.11) MAC protocol, used because a radio NIC is half-duplex and suffers the hidden-terminal problem, so collisions cannot be detected and must instead be avoided. Avoidance uses interframe spacing (IFS), a random backoff drawn from a contention window before transmitting, a compulsory ACK for every frame, and optional RTS/CTS handshaking with a NAV timer that silences other stations.

Q34What is Pure ALOHA?

Ans: Pure ALOHA is the original random-access protocol (Abramson, 1970) in which a station transmits a frame whenever it has data, with no carrier sensing at all, and retransmits after a random backoff time if no acknowledgement arrives. Since a frame is destroyed by any overlap beginning either one frame time before or after it, the vulnerable period is \( 2T_{fr} \), giving throughput \( S = G e^{-2G} \) with maximum \( S_{max} = \frac{1}{2e} \approx 0.184 \) (18.4%) at \( G = 0.5 \).

Q35What is Slotted ALOHA?

Ans: Slotted ALOHA divides time into slots of exactly one frame time and forces every station to begin transmitting only at the start of a slot. This halves the vulnerable period to \( T_{fr} \), so throughput becomes \( S = G e^{-G} \) with maximum \( S_{max} = \frac{1}{e} \approx 0.368 \) (36.8%) at \( G = 1 \) — exactly double Pure ALOHA, at the cost of requiring global clock synchronisation.

Q36What is IP Address?

Ans: An IP address is the logical, hierarchical address assigned at the Network layer (OSI layer 3) that uniquely identifies the connection of a host or router interface to the Internet. It is universal, software-configured, and changes when the host moves to another network — unlike the flat, permanent 48-bit MAC address — and it is the address routers use to forward datagrams end-to-end across different networks.

Q37Define IPv4 Address structure.

Ans: An IPv4 address is 32 bits long, written in dotted-decimal notation as four 8-bit octets separated by dots (e.g. 192.168.1.10), giving an address space of \( 2^{32} \approx 4.3 \) billion addresses. It is hierarchically divided into a network ID (prefix) and a host ID (suffix); classful addressing defined class A (/8), B (/16), C (/24), D (multicast) and E (reserved), now superseded by CIDR.

Q38Define IPv6 Address structure.

Ans: An IPv6 address is 128 bits long, written in colon-hexadecimal notation as eight groups of four hexadecimal digits (e.g. 2001:0DB8:0000:0000:0000:0000:1428:57AB, abbreviable to 2001:DB8::1428:57AB), giving \( 2^{128} \approx 3.4 \times 10^{38} \) addresses. Beyond the enlarged space it uses a fixed 40-byte base header with optional extension headers, drops the header checksum and router fragmentation, and replaces broadcast with anycast.

Q39What is Subnetting?

Ans: Subnetting is the process of logically partitioning a single IP network into multiple smaller sub-networks (subnets) using a custom subnet mask, reducing broadcast domain size and optimizing IP allocation.

Q40What is CIDR?

Ans: CIDR (Classless Inter-Domain Routing) abolishes the fixed class boundaries by writing an address as a.b.c.d/n, where the slash notation \( /n \) states how many leading bits form the network prefix; the mask is therefore \( n \) ones followed by \( 32-n \) zeros and the block contains \( 2^{32-n} \) addresses. Freed from the rigid /8, /16 and /24 boundaries it supports variable-length subnet masking and route aggregation (supernetting), which conserves IPv4 addresses and shrinks backbone routing tables.

Q41What is Network Address Translation (NAT)?

Ans: Network Address Translation (NAT) is the technique by which a border router rewrites the private source IP addresses (RFC 1918 blocks 10.0.0.0/8, 172.16.0.0/12, 192.168.0.0/16) of outgoing datagrams into one or a few globally routable public addresses, and reverses the mapping on the replies using a translation table. PAT (NAT overload) also translates port numbers so that many internal hosts can share a single public IP; NAT conserves scarce IPv4 addresses and hides the internal topology.

Q42What is ARP?

Ans: ARP (Address Resolution Protocol) maps a known 32-bit IPv4 address to the unknown 48-bit MAC address of a host on the same physical network. The requesting host broadcasts an ARP request frame; only the owner of that IP answers with a unicast ARP reply carrying its MAC address, which is then stored in an ARP cache so repeat broadcasts are avoided. ARP is the glue between the Network and Data Link layers.

Q43What is RARP?

Ans: RARP (Reverse Address Resolution Protocol) is the inverse of ARP: a station that knows only its own 48-bit MAC address broadcasts a request so that a RARP server can tell it its 32-bit IP address. It was used by diskless workstations at boot time, but because it works only within one physical network and returns nothing but the IP address it has been replaced by BOOTP and DHCP.

Q44What is ICMP?

Ans: ICMP (Internet Control Message Protocol) is the Network layer companion to IP that reports errors and carries diagnostic queries, since IP itself is an unreliable, connectionless best-effort protocol with no error-reporting mechanism of its own. ICMP messages (IP protocol number 1) are encapsulated inside IP datagrams and include error types such as Destination Unreachable, Time Exceeded, Source Quench, Parameter Problem and Redirect, plus query types Echo Request/Reply used by ping and traceroute. It only reports errors, never corrects them.

Q45What is DHCP?

Ans: DHCP (Dynamic Host Configuration Protocol) is an Application layer client–server protocol running over UDP (server port 67, client port 68) that automatically leases a host its IP address, subnet mask, default gateway and DNS server addresses for a limited lease time. The client acquires them by the four-step DORA exchange — DISCOVER (broadcast), OFFER, REQUEST, ACK — and DHCP is a backward-compatible extension of BOOTP.

Q46What is Distance Vector Routing?

Ans: Distance vector routing is a decentralised algorithm based on Bellman–Ford in which every router keeps a vector of (destination, least cost, next hop) entries and periodically sends its entire routing table to its directly connected neighbours only, updating its own table whenever a neighbour advertises a cheaper path. No router ever learns the full topology; convergence is slow and the scheme suffers the count-to-infinity problem, mitigated by split horizon, poison reverse and an infinity of 16 hops. RIP is the classic example.

Q47What is Link State Routing?

Ans: Link state routing is an algorithm in which each router measures the cost of its own directly attached links, floods this Link State Packet to every router in the area, and so builds an identical and complete topology map (the link state database); it then runs Dijkstra's shortest-path-first algorithm locally to compute its routing table. Compared with distance vector it converges much faster and is free of count-to-infinity, at the cost of higher memory and CPU use. OSPF is the classic example.

Q48Define Congestion Control.

Ans: Congestion control is the set of mechanisms that keeps the total load offered to the network below its carrying capacity, so that router queues do not overflow and throughput does not collapse. It is global and network-oriented — open-loop (prevention: traffic shaping, admission control) or closed-loop (feedback: choke packets, ECN, TCP's congestion window with slow start and AIMD) — in contrast to flow control, which protects only an individual receiver's buffer.

Q49What is Leaky Bucket Algorithm?

Ans: The leaky bucket algorithm is a traffic-shaping mechanism that models the host as a bucket with a hole in it: however bursty the arriving traffic may be, the bucket drains onto the network at a fixed constant rate, and packets that arrive when the finite queue is full are discarded. It therefore converts a bursty flow into a perfectly smooth average-rate flow, but it permits no burstiness at all and can discard data from a well-behaved bursty source.

Q50What is Token Bucket Algorithm?

Ans: The token bucket algorithm adds tokens at a constant rate \( r \) to a bucket of capacity \( b \), and a packet may be sent only by consuming a token; while the host is idle tokens accumulate, so a later burst can be sent at full line speed. Over an interval \( t \) at most \( rt + b \) bytes may leave, so unlike the leaky bucket it preserves burstiness up to the bucket size and never discards data merely for arriving in a burst.

Q51What is DNS?

Ans: DNS (Domain Name System) is the Application layer distributed, hierarchical database that maps human-readable domain names to IP addresses and back, using port 53 over UDP for ordinary queries and TCP for zone transfers or replies longer than 512 bytes. Its name space is an inverted tree of root, top-level (.com, .in) and second-level domains, resolved recursively or iteratively through root, TLD and authoritative servers, with caching at every step.

Q52What is HTTP and HTTPS?

Ans: HTTP (HyperText Transfer Protocol) is the stateless Application layer request–response protocol used to transfer Web pages over TCP port 80, with methods GET, POST, PUT, HEAD and DELETE. HTTPS is the same protocol tunnelled through TLS/SSL on TCP port 443, adding encryption, server authentication by digital certificate and message integrity — so HTTPS is secure HTTP, not a different application protocol.

Q53What is FTP?

Ans: FTP (File Transfer Protocol) is the Application layer TCP-based standard for copying files between hosts, and it is distinctive in using two parallel connections: a persistent control connection on port 21 that carries commands and replies for the whole session, and a separate data connection on port 20 opened and closed for each individual transfer. This out-of-band control lets it cope with differing file types and structures, and it operates in active or passive mode.