7. Computer Networks — Computer Science | CSS Solved Past Papers

FPSC CSS Computer Science past papers (2016–2026) — 50 questions: 6 MCQs, 44 subjective parts. Newest first.

Part A — Subjective Questions

2026 · Paper-I · Q2(b) (6 marks)

Question

You receive an email asking for your banking details. What kind of risk or security issue this might be? How to respond?

Solution

This is a classic phishing attack — a social-engineering technique where an attacker sends a fraudulent email impersonating a trustworthy entity (e.g., a bank) to trick the recipient into revealing sensitive information such as account numbers, passwords, or card details. It may also be an example of identity theft groundwork or a precursor to financial fraud, and the email itself may contain malicious links (leading to fake login pages) or attachments carrying malware.

How to respond:

Do NOT click any links or reply with the requested information.
Verify the sender's actual email address and check for spelling/domain inconsistencies, urgency/threat language, and generic greetings — common phishing red flags.
Contact the bank directly using its official, independently verified phone number or website (never using contact details given in the suspicious email) to confirm whether the request is genuine.
Report the email as phishing/spam to your email provider and to the bank's fraud department, then delete it.
Never share OTPs, passwords, PINs, or card details via email — legitimate banks never ask for such details through email.

2025 · Paper-II · Q3(a) (10 marks)

Question

Answer the Following Questions: (i) Is it possible for an organization's Web server and mail server to have exactly the same alias for a hostname (for example, foo.com)? What would be the type for the RR that contains the hostname of the mail server? (2) (ii) Draw the structure of Internet in terms of ISPs, also show the concept in terms of Edge and Core networks. (3) (iii) A packet switch receives a packet to be forwarded. When the packet arrives, one other packet is halfway done being transmitted on this outbound link and four other packets are waiting to be transmitted in FIFO manner. Suppose all packets are 1,500 bytes and the link rate is 2 Mbps. What is the queuing delay for the packet? (3) (iv) Why do HTTP, FTP, SMTP, and POP3 run on top of TCP rather than on UDP? (2)

Solution

(i) Yes — it is possible. DNS allows both the Web server and the mail server to be reached via the same alias hostname (e.g., foo.com) by using CNAME (canonical name) records that map the alias to the actual canonical hostnames of each server. The resource record (RR) type that specifically identifies the hostname of the mail server for a domain is the MX (Mail eXchange) record.

(ii) The Internet's structure is a rough hierarchy of Internet Service Providers (ISPs): at the 'core' are a small number of highly-interconnected Tier-1 ISPs (forming the global backbone, all directly connected to each other), below them regional/Tier-2 ISPs connect to one or more Tier-1s, and at the 'edge' are access/Tier-3 ISPs and end-user access networks (home, enterprise, mobile) that connect end hosts to the core. Traffic from an edge host typically travels up through access and regional ISPs into the core backbone, then back down to the destination edge network — the 'network edge' comprises end systems and their access links, while the 'network core' comprises the mesh of interconnected routers and ISPs that switch/forward packets between edges.

(iii) Queuing delay = transmission time of the currently-in-progress packet remaining + transmission time of all fully-queued packets ahead of the arriving one. Transmission time per packet = (1500 bytes x 8 bits/byte) / 2 Mbps = 12000 bits / 2,000,000 bps = 6 ms. The in-progress packet is halfway done, so its remaining transmission time = 6/2 = 3 ms. Four more full packets are waiting ahead in FIFO order = 4 x 6 ms = 24 ms. Total queuing delay = 3 + 24 = 27 ms.

(iv) HTTP, FTP, SMTP and POP3 all need to reliably deliver complete, ordered, and error-free data (web pages, files, emails) — a single dropped or reordered segment would corrupt a downloaded file or garble an email. TCP provides exactly this: reliable, in-order, connection-oriented, byte-stream delivery with automatic retransmission, flow control and congestion control, whereas UDP is unreliable, connectionless and does not guarantee ordering — acceptable for real-time media where an occasional lost packet is tolerable, but not for document/file/mail transfer where every byte must arrive intact.

2025 · Paper-II · Q3(b) (10 marks)

Question

(i) Why do we need conditional GET when we have regular GET with respect to HTTP? What is its difference in comparison to GET? (4) (ii) With respect to web caching, if an average object size is 100K bits and the average request rate from browsers to origin servers is 15/sec. Suppose RTT from an institutional router to any origin server is 2 sec. If LAN speed is 100Mbps and access link speed is 1.54 Mbps then with a cache hit of 50%, how much total delay will be there? (6)

Solution

(i) A conditional GET is needed to support efficient caching: instead of a browser/proxy re-downloading a full object from the origin server every time, it sends a GET request with an If-Modified-Since (or If-None-Match) header carrying the cached copy's last-known modification time/ETag. The difference from a regular GET: if the object has NOT changed since that time, the server replies with a lightweight 304 Not Modified response (no body), letting the client reuse its cached copy and saving bandwidth and time; if it HAS changed, the server returns the full object with 200 OK just like a normal GET. A plain GET always returns the full object regardless of whether the cached copy is still valid, wasting bandwidth when nothing changed.

(ii) Set up: object size = 100,000 bits; request rate = 15/sec; RTT institution<->origin = 2 sec; LAN = 100 Mbps (fast, so LAN delay is negligible, ~0); access link = 1.54 Mbps; cache hit ratio = 50%, so only the 50% of requests that MISS the cache actually go out over the access link to the origin server.

Traffic reaching the access link = 15 x 0.5 = 7.5 requests/sec. Traffic intensity on the access link, β = (7.5 x 100,000) / 1,540,000 = 750,000 / 1,540,000 ≈ 0.487.

Transmission delay of one object over the access link, Δ = object size / access-link rate = 100,000 / 1,540,000 ≈ 0.0649 s ≈ 64.9 ms.

Using the standard queuing approximation for average access-link delay: Access delay ≈ Δ / (1 - β) = 0.0649 / (1 - 0.487) = 0.0649 / 0.513 ≈ 0.1266 s ≈ 126.6 ms.

Internet delay (origin round trip, dominated by the given RTT) ≈ 2 s.

For a cache MISS, total delay ≈ LAN delay(~0) + access delay(126.6 ms) + Internet delay(2 s) ≈ 2.127 s. For a cache HIT, the object is served from the local cache on the fast LAN, so delay ≈ negligible (~0).

Average total delay = (hit ratio x hit delay) + (miss ratio x miss delay) = 0.5 x (~0) + 0.5 x (2.127 s) ≈ 1.06 seconds.

So the average total delay per request, with a 50% cache hit ratio, is approximately 1.06 seconds — much better than if every request had to go all the way to the origin server (~2.13 s), demonstrating the benefit of web caching.

2024 · Paper-II · Q4(a) (7 marks)

Question

Compare the OSI and TCP/IP models in terms of their simplicity and practicality. Why is a layered approach beneficial in network design?

Solution

The OSI model has 7 layers (Physical, Data Link, Network, Transport, Session, Presentation, Application) and was designed as a comprehensive, general-purpose theoretical reference model, giving very fine-grained separation of concerns, but it was developed somewhat independently of actual working protocols and is more complex than necessary for practical implementation.

The TCP/IP model has only 4 layers (Link, Internet, Transport, Application), combining the OSI's session, presentation, and application layers into a single application layer. It was developed alongside the actual protocols that built the modern Internet (IP, TCP, UDP, HTTP, etc.), making it simpler and directly practical/implementable - it is the model actually used in real-world networking.

A layered approach is beneficial in network design because it provides: modularity (each layer can be designed, implemented, and upgraded independently, as long as its interface to adjacent layers stays consistent); simplified troubleshooting (problems can be isolated to a specific layer); interoperability (standardized interfaces let different vendors' implementations of a layer work together); and reusability (a lower layer, e.g., the physical/link layer, can support many different upper-layer protocols).

2024 · Paper-II · Q4(b) (7 marks)

Question

Explain how overlay networks and content distribution networks enhance the performance and scalability of internet services? Provide a numerical example to illustrate their impact on content delivery.

Solution

An overlay network is a virtual network built on top of the existing physical network infrastructure (e.g., peer-to-peer networks, VPNs), enabling new routing paths, services, or logical topologies without requiring changes to the underlying hardware/routers.

A Content Distribution Network (CDN) is a geographically distributed set of edge/proxy servers that cache copies of content (web pages, videos, images) physically close to end users, rather than serving every request from one distant origin server. This reduces the physical distance data must travel, cutting latency, and offloads traffic from the origin server, improving scalability under high demand.

Numerical example: suppose a user in Asia requests content from an origin server located in the US, incurring roughly 200ms round-trip latency. If a CDN edge server located within the user's own region serves a cached copy instead, latency might drop to roughly 20ms - a 10x improvement. Simultaneously, if a popular resource would otherwise be requested from the origin server 1,000,000 times per day, but 80% of those requests are now served by CDN edge caches instead, the origin server's load drops to only 200,000 requests per day (an 80% reduction), letting it scale to serve far more total users without becoming a bottleneck.

2024 · Paper-II · Q4(c) (6 marks)

Question

If the internet were a city, and each device had its own unique street address, how does IP addressing work in this scenario? Explain the purpose of subnetting using a neighborhood analogy.

Solution

In this analogy, every device connected to the internet has a unique "street address" - its IP address - which lets data (like mail) be routed to exactly the right destination. Routers act like postal sorting centers/post offices: each one reads the destination address on a packet and forwards it toward the next router closer to that address, hop by hop, until it reaches the correct device, just as a letter passes through several sorting offices before final delivery.

Subnetting is like dividing the city into distinct neighborhoods, each with its own reserved block of "street addresses" (a contiguous IP address range). Instead of every router in the city needing to know the exact location of every individual house, it only needs to know which neighborhood block to forward mail toward; once the mail reaches a router local to that neighborhood, that local router (rather than a distant, central one) handles final delivery to the specific house. This hierarchical grouping dramatically reduces the size of routing tables and the amount of routing information that needs to be exchanged network-wide, while also letting each neighborhood manage its own internal addressing independently - exactly what subnetting achieves for real IP networks.

2023 · Paper-II · Q3(a) (7 marks)

Question

The CPU in a router can process 2 million packets/sec. The load offered to it is 1.5 million packets/sec. If a route from source to destination contains 10 routers, how much time is spent being queued and serviced by the CPUs?

Solution

Modeling each router's CPU as an M/M/1 queue with service rate mu = 2,000,000 packets/sec and arrival rate lambda = 1,500,000 packets/sec, the average time a packet spends in the system (queueing + service) at one router is: T = 1 / (mu - lambda) = 1 / (2,000,000 - 1,500,000) = 1 / 500,000 seconds = 2 microseconds.

For a path passing through 10 such routers, the total time spent being queued and serviced = 10 x 2 microseconds = 20 microseconds.

2023 · Paper-II · Q3(b) (7 marks)

Question

An IP datagram using the strict source routing option has to be fragmented. Do you think the option is copied into each fragment, or is it sufficient to just put it in the first fragment? Explain your answer.

Solution

The strict source routing option must be copied into EVERY fragment, not just the first one. This is because each fragment of a datagram is routed through the network as an independent unit once fragmentation occurs, and strict source routing dictates the exact sequence of routers/hops that MUST be followed to reach the destination. If the routing option were present only in the first fragment, later fragments would be routed using ordinary (non-source) routing and could take a different path, or fail to be delivered along the mandated route entirely, breaking the very guarantee that strict source routing is meant to provide. This is reflected in the IP option format itself, which has a "copy" flag bit that, for the strict source route option, is set to 1, mandating that it be copied into the header of every fragment produced.

2023 · Paper-II · Q3(c) (6 marks)

Question

Give two examples of computer applications for which connection-oriented service is appropriate and also give two examples for which connection-less service is best.

Solution

Connection-oriented service (reliable, ordered delivery, e.g., TCP) is appropriate for:

File transfer (FTP) - requires every byte to arrive correctly and in order; any loss/corruption would produce a broken file.
Remote login/terminal access (Telnet/SSH) - requires reliable, in-order delivery of every keystroke and response for correct interactive behavior.

Connection-less service (best-effort, no session setup, e.g., UDP) is best for:

DNS lookups - a single small request/response exchange where the overhead of setting up a connection is not worth it, and the client can simply retry if no reply arrives.
Real-time voice/video streaming or online gaming - timeliness matters more than guaranteed delivery of every packet; occasional loss is tolerable and retransmitting late data is often worse than just skipping it.

2023 · Paper-II · Q5(b) (5 marks)

Question

Suppose that instead of using 16-bits for the network part of a class B address originally, 20 bits had been used. How many class B networks would there have been?

Solution

A class B address is 32 bits, with the fixed prefix "10" identifying it as class B; originally, the 16-bit network field (including this 2-bit prefix) leaves 14 bits actually distinguishing network numbers, giving 2^14 = 16,384 class B networks.

If instead 20 bits were used for the network part (still including the fixed 2-bit class-B prefix "10"), then the number of bits actually distinguishing networks becomes 20 - 2 = 18 bits, giving: Number of class B networks = 2^18 = 262,144.

2022 · Paper-I · Q6(b) (8 marks)

Question

Compare and contrast the DES, AES, and RSA encryption algorithms.

Solution

DES (Data Encryption Standard): a symmetric block cipher using a 56-bit key and 64-bit blocks, based on a 16-round Feistel structure. It is now considered insecure because its small key space is vulnerable to brute-force attack; largely superseded, though 3DES (applying DES three times) extended its life.
AES (Advanced Encryption Standard): the modern symmetric block cipher, using 128-bit blocks with key sizes of 128, 192, or 256 bits, based on substitution-permutation network rounds. It is fast, efficient in both hardware and software, and currently considered secure - the de facto standard for bulk data encryption.
RSA: an asymmetric (public-key) algorithm based on the mathematical difficulty of factoring large numbers. It uses a public key for encryption and a private key for decryption (or vice versa for signatures), with key sizes typically 2048 bits or more.

Key differences: DES and AES are symmetric (same key encrypts and decrypts, must be kept secret and shared securely), whereas RSA is asymmetric (public key can be shared openly, only the private key is secret). Symmetric ciphers (DES/AES) are much faster and used for encrypting bulk data; RSA is computationally expensive and typically used only for key exchange and digital signatures, or to encrypt a symmetric session key which then encrypts the actual data (hybrid encryption).

2022 · Paper-II · Q5(a) (8 marks)

Question

Describe any one of the routing algorithms in detail, which is used to resolve the conflict between path selections.

Solution

Dijkstra's Shortest Path (Link-State Routing) Algorithm: Each router first discovers its directly connected neighbors and the cost (link metric) to reach them, then floods this link-state information to every other router in the network, so that every router builds an identical, complete map (topology graph) of the network.

Each router then independently runs Dijkstra's algorithm on this graph to compute the shortest-cost path from itself to every other node:

Initialize: distance to self = 0, distance to all other nodes = infinity; mark all nodes as unvisited.
Repeatedly select the unvisited node with the smallest known distance (initially the source itself).
For each neighbor of that node, calculate the distance through the current node; if it is smaller than the neighbor's current known distance, update it (relaxation) and record the current node as the predecessor.
Mark the current node as visited (finalized) and repeat until all nodes are visited.
Trace back predecessor pointers from any destination to the source to obtain the actual shortest path.

This resolves conflicts between multiple possible paths by always choosing the path with minimum cumulative cost/metric, and each router uses the results to build its own forwarding table, forwarding each packet to the next hop on the shortest path to its destination. This is the basis of link-state protocols like OSPF.

2022 · Paper-II · Q5(b) (6 marks)

Question

A company occupies four buildings on a campus. Each of these building has a 100 base T network running to all the floors. The buildings form a square 100 meter on each side. The network must support 100 Mbps data transfer rate. What will be the proposed design solution under the given condition?

Solution

100BASE-T (Fast Ethernet over UTP/copper) has a maximum cable segment length of 100 meters. Since the buildings sit at the corners of a 100m x 100m square, adjacent buildings are 100m apart (borderline usable) but diagonal buildings are about 141m apart (100*sqrt(2)), which exceeds the UTP distance limit and would not reliably support 100 Mbps.

Proposed design: use a hierarchical, two-tier design.

Horizontal cabling (within each building): keep 100BASE-T UTP cabling from each floor's wiring closet to the desktops, since floor-to-closet distances are well within the 100m UTP limit.
Backbone cabling (between buildings): connect a central switch/router in each building to the others using fiber-optic links (e.g., 100BASE-FX or Gigabit fiber), since fiber supports distances of up to 2 km (multimode) or more, comfortably covering the inter-building distances regardless of exact layout, and is immune to the electromagnetic interference outdoor cable runs are exposed to.

This creates a star (or ring, for redundancy) backbone connecting the four buildings' main distribution frames via fiber, while each building's internal floors use standard 100BASE-T, together guaranteeing the required 100 Mbps end-to-end within the distance limitations of each medium.

2022 · Paper-II · Q5(c) (6 marks)

Question

What is the main fault in TCP 3-way handshake which makes it a boon for attackers? Many operating systems generally associate a "Backlog" parameter with a listening socket that sets cap on the number of TCBs simultaneously in SYN-RECEIVED state. Does this parameter solve the problem in TCP 3-way handshake?

Solution

The main fault: when a server receives a SYN, it allocates resources (a Transmission Control Block, sequence numbers, buffers) and replies with SYN-ACK, then waits in the SYN-RECEIVED state for the final ACK - before the connection is actually established. An attacker can exploit this by sending a flood of SYN packets, often with spoofed/unreachable source addresses, and never completing the handshake (never sending the final ACK). Each such half-open connection ties up server resources until it times out, and enough of them exhaust the server's connection table, denying service to legitimate clients - this is the classic SYN flood attack.

Does the Backlog parameter solve it? No. The backlog only sets an upper bound on how many half-open (SYN-RECEIVED) connections the OS will track for a listening socket; it does not fix the underlying vulnerability. Once an attacker's spoofed SYNs fill up the backlog queue, the server simply starts dropping/rejecting new legitimate SYN requests as well, meaning the backlog limit merely bounds resource consumption rather than preventing the denial-of-service - it converts unbounded resource exhaustion into a hard-but-still-effective flood target. Real mitigation requires techniques like SYN cookies, which avoid storing state for a connection until the handshake actually completes.

2021 · Paper-II · Q5(a) (6 marks)

Question

Compare Transmission Control Protocol and User Datagram Protocol. How they are used in wireless networks?

Solution

TCP: connection-oriented (three-way handshake), reliable (acknowledgements and automatic retransmission of lost segments), provides in-order delivery, includes flow control and congestion control, and has higher overhead/latency. Suited to applications needing guaranteed data integrity: file transfer, web browsing (HTTP), email. UDP: connectionless, unreliable (no acknowledgement or retransmission), no ordering or congestion control guarantees, minimal header overhead, and lower latency. Suited to real-time applications that tolerate some loss: video/voice streaming, DNS queries, online gaming. In wireless networks specifically: TCP's congestion-control algorithm was designed assuming packet loss signals network congestion, but wireless links naturally suffer higher packet loss from interference, fading, and handoffs unrelated to congestion — causing TCP to unnecessarily shrink its sending rate and under-utilise available bandwidth. UDP, by not reacting to loss at all, performs better for real-time wireless multimedia (VoIP calls, video calls over WiFi/cellular), tolerating occasional lost packets as minor glitches rather than triggering costly retransmission delays — though applications using UDP over wireless must add their own error-concealment or selective-retransmission logic if needed.

2021 · Paper-II · Q5(b) (6 marks)

Question

What mechanism is used by TCP to protect itself from miss delivery?

Solution

TCP protects against mis-delivery through several mechanisms working together: (1) each TCP connection/segment is uniquely identified by a 4-tuple — source IP, source port, destination IP, and destination port — so the operating system's demultiplexing logic always routes incoming segments to the exact correct socket/application process, even when many simultaneous connections share the same destination port; (2) sequence numbers on every byte of data let the receiver detect and discard duplicate segments and reorder segments arriving out of order, so data cannot be delivered to the wrong position in the stream; and (3) a checksum field in the TCP header detects corrupted segments in transit, which are simply discarded (triggering retransmission via timeout/duplicate ACK) rather than being delivered incorrectly to the application.

2021 · Paper-II · Q5(c) (8 marks)

Question

Calculate the TCP window size to fully utilize the line on which WAN link is 2 Mbps and RTT between source and destination is 300 msec.

Solution

To fully utilise (keep 'full') a link, the sender must have enough data in flight to cover one full round-trip time before needing an acknowledgement — this required amount is the Bandwidth-Delay Product (BDP): BDP = Bandwidth × RTT = 2 Mbps × 300 ms = (2 × 10⁶ bits/sec) × (0.3 sec) = 600,000 bits = 600,000 / 8 bytes = 75,000 bytes ≈ 73.2 KB So the TCP window size should be at least 600,000 bits (75,000 bytes) to keep the WAN link fully utilised without the sender stalling while waiting for acknowledgements. (Note: the standard TCP window field is only 16 bits, capping it at 65,535 bytes; achieving a ~75 KB window in practice requires the TCP Window Scaling option, RFC 1323.)

2020 · Paper-I · Q3(a) (4 marks)

Question

Write standard ports for following services HTTP, FTP, SMTP, HTTPS, DNS.

Solution

HTTP: port 80. FTP: port 21 (control connection; port 20 is used for the data connection in active mode). SMTP: port 25. HTTPS: port 443. DNS: port 53 (used for both UDP queries and TCP zone transfers/large responses).

2020 · Paper-I · Q3(c) (4 marks)

Question

If you are transferring a file over the Internet, would you prefer TCP or UDP as the underlying protocol. Explain

Solution

TCP is preferred for file transfer. File transfer requires every single byte to arrive intact and in the correct order — a single lost or corrupted segment can render the whole file unusable (e.g. a corrupted byte in an executable or archive). TCP provides exactly this guarantee through acknowledgements, automatic retransmission of lost segments, sequencing, and flow/congestion control, at the cost of somewhat higher overhead and latency, which is an acceptable trade-off since file transfer is not typically time-critical to the millisecond.

2020 · Paper-I · Q3(d) (4 marks)

Question

If you are transferring live audio in real-time over the Internet, would you prefer TCP or UDP as the underlying protocol. Explain.

Solution

UDP is preferred for live real-time audio. Real-time audio tolerates occasional lost packets (perceived as a brief glitch/click) far better than it tolerates delay, whereas TCP's retransmission of a lost segment plus its strict in-order delivery would stall playback and introduce noticeable, unacceptable lag/jitter while waiting for the missing data to be resent. UDP's low overhead, connectionless, no-retransmission delivery keeps latency minimal, which matters far more than occasional imperfection for live audio/VoIP.

2020 · Paper-II · Q4(a) (8 marks)

Question

Compare IPv4 and IPv6 headers. Explain the use of NAT technology to overcome IPv4 scarcity.

Solution

IPv4 header: variable length (20-60 bytes with options), 32-bit source/destination addresses, includes a header checksum and fragmentation fields (Identification, Flags, Fragment Offset). IPv6 header: fixed 40-byte length, 128-bit source/destination addresses, no header checksum (relies on other layers), no fragmentation fields in the base header (handled via extension headers only at the source), adds Traffic Class and Flow Label fields for QoS, and replaces 'Options' with a chainable set of Extension Headers, keeping the base header simpler and faster for routers to process. NAT (Network Address Translation): since IPv4's 32-bit address space (~4.3 billion addresses) is insufficient for the number of devices worldwide, NAT lets an entire private network (using non-routable private addresses like 192.168.x.x) share a single public IPv4 address for external internet access. The router/gateway rewrites the source IP and port of outgoing packets to its own public address (with a unique port per internal connection), and maintains a translation table to correctly route the corresponding return traffic back to the right internal host — effectively multiplexing many internal devices behind one scarce public address, which greatly extended the usable life of IPv4.

2020 · Paper-II · Q4(b) (6 marks)

Question

Find the maximum number of valid subnets and usable hosts per subnet that you can get from the network 172.23.0.0/23.

Solution

A /23 mask leaves 32 − 23 = 9 host bits. Total addresses in this block = 2⁹ = 512. As a single network block, 172.23.0.0/23 itself represents exactly 1 valid subnet (network address 172.23.0.0, broadcast address 172.23.1.255), with usable hosts = 512 − 2 (network + broadcast addresses reserved) = 510 usable host addresses. (If this /23 block were further subnetted down to the smallest practical subnet size — /30, which gives 2 usable hosts each, commonly used for point-to-point router links — it could be divided into 2^(30−23) = 128 subnets of 2 usable hosts each; but taken as given, 172.23.0.0/23 is itself the single subnet with 510 usable hosts.)

2020 · Paper-II · Q5(a) (8 marks)

Question

What is signal encoding? Explain different encoding techniques used in data communication.

Solution

Signal encoding is the process of converting data (digital or analog) into an electrical, optical, or electromagnetic signal suitable for transmission over a communication medium, balancing bandwidth efficiency, clock synchronisation, and noise immunity. Digital-to-digital encoding techniques: NRZ (Non-Return-to-Zero: NRZ-L represents bits directly as voltage levels; NRZ-I represents a 1 as a transition and 0 as no transition); Manchester encoding (a transition occurs in the middle of every bit period, making it self-clocking; used in classic Ethernet); Differential Manchester; Bipolar-AMI (alternates polarity for successive 1s, 0 stays at zero level); and multilevel codes like 4B/5B and MLT-3 used in modern high-speed LANs. Digital-to-analog encoding (modulation): ASK (Amplitude Shift Keying), FSK (Frequency Shift Keying), PSK (Phase Shift Keying), and QAM (Quadrature Amplitude Modulation, combining amplitude and phase for higher data rates), used in modems and wireless transmission. Analog-to-digital encoding: PCM (Pulse Code Modulation) samples and quantizes an analog signal (e.g. voice) into a digital bit stream.

2020 · Paper-II · Q5(b) (5 marks)

Question

Explain the functions and needs of ARP and RARP protocols in computer networks.

Solution

ARP (Address Resolution Protocol) resolves a known IP (logical, layer-3) address into its corresponding MAC (physical, layer-2) address on a local network. It is needed because IP routing operates using logical IP addresses, but actual frame delivery on a LAN requires the physical MAC address of the destination network interface — ARP bridges this gap by broadcasting a request ('who has this IP?') and receiving a unicast reply with the MAC address. RARP (Reverse Address Resolution Protocol) performs the opposite mapping — given a known MAC address, it discovers the corresponding IP address. It was historically used by diskless workstations at boot time (which know their own hardware/MAC address but have no stored IP configuration) to query a RARP server for their assigned IP address. RARP is now largely obsolete, superseded by BOOTP and DHCP, which provide richer configuration information.

2020 · Paper-II · Q5(c) (7 marks)

Question

Explain multiplexing and demultiplexing at the transport layer. Explain in the context of TCP/IP protocol.

Solution

Multiplexing (at the sender): the transport layer (TCP or UDP) gathers data from multiple application processes running on the same host — each identified by a unique port number — and combines/wraps each into segments handed down to the single shared network (IP) layer for transmission, allowing one host to run many network applications simultaneously over its one network connection. Demultiplexing (at the receiver): the transport layer examines each arriving segment's header fields — destination port number for UDP, or the full 4-tuple (source IP, source port, destination IP, destination port) for TCP — and uses this to deliver the payload to the correct socket/application process, distinguishing simultaneous connections correctly. In the TCP/IP context: every TCP connection is uniquely identified by its 4-tuple, so a web server listening on port 80 can simultaneously serve thousands of different clients — even though they all connect to the SAME destination port 80 — because each client has a distinct source IP/port combination, letting the OS's demultiplexing table route each segment to the correct per-connection socket buffer.

2019 · Paper-I · Q5(a) (10 marks)

Question

Discuss the security issues associated with the cloud computing.

Solution

Key cloud security issues:

Data breaches — sensitive data stored off-premises on shared multi-tenant infrastructure is an attractive target; a breach at the provider can expose many customers' data at once.
Insecure APIs/interfaces — cloud services are managed and consumed through APIs; weak authentication or poor API design can be exploited.
Account/credential hijacking — phishing or weak credentials let attackers take over cloud accounts and resources.
Multi-tenancy and shared technology vulnerabilities — a flaw in the hypervisor or shared component can allow one tenant to affect or access another's resources ('VM escape').
Insider threats — malicious or careless staff at the cloud provider have privileged access.
Loss of control/visibility — customers cannot fully audit the provider's physical security, patch status, or internal processes.
Data loss — accidental deletion, provider outage, or ransomware without adequate backup/DR strategy.
Compliance and jurisdiction — data may be stored in a different legal jurisdiction, creating regulatory/data-sovereignty problems.
Denial of Service — shared infrastructure can be a target for large-scale DDoS affecting many tenants. Mitigations include encryption at rest/in transit, strong IAM with MFA, regular audits, contractual SLAs, and a shared-responsibility security model between provider and customer.

2019 · Paper-II · Q3(a) (8 marks)

Question

Show field by field comparison for IPv4 and IPv6 packets.

Solution

IPv4 header (20-60 bytes, variable due to Options): Version(4b), IHL/Header Length(4b), Type of Service(8b), Total Length(16b), Identification(16b), Flags(3b), Fragment Offset(13b), TTL(8b), Protocol(8b), Header Checksum(16b), Source Address(32b), Destination Address(32b), Options(variable). IPv6 header (fixed 40 bytes): Version(4b), Traffic Class(8b), Flow Label(20b), Payload Length(16b), Next Header(8b), Hop Limit(8b), Source Address(128b), Destination Address(128b). Key differences: address size grows from 32-bit to 128-bit; IPv6 has NO header checksum (relies on link/transport layer checks, saving per-hop recomputation) and NO fragmentation fields in the base header (fragmentation is handled end-to-end via an extension header, not by routers); IPv6 replaces 'Type of Service' with 'Traffic Class' and adds a 'Flow Label' for QoS flow identification; IPv6 replaces 'Protocol' with 'Next Header' (which can chain to extension headers) and 'TTL' with 'Hop Limit'; IPv6 drops the variable-length Options field in favour of extension headers, making its base header fixed-length and faster for routers to process.

2019 · Paper-II · Q3(b) (6 marks)

Question

Explain the following routing techniques using suitable examples. (i) Link State Routing (ii) Distance Vector Routing

Solution

Distance Vector Routing: each router only knows the distance (metric/hop count) to each destination via each of its directly connected neighbours, and periodically broadcasts its ENTIRE routing table to those neighbours. Routers update their tables using the Bellman-Ford algorithm ('routing by rumour'). Example: RIP (Routing Information Protocol). It converges slowly and is prone to routing loops (the 'count-to-infinity' problem). Link State Routing: each router discovers its directly connected links/neighbours and floods this information (Link State Advertisements) to EVERY router in the network, so every router builds an identical, complete map (topology) of the network. Each router then independently computes shortest paths using Dijkstra's algorithm. Example: OSPF (Open Shortest Path First). It converges faster and is more robust than distance-vector, at the cost of higher memory/processing overhead for maintaining the full topology.

2019 · Paper-II · Q3(c) (6 marks)

Question

Show step by step procedure of error detection using cyclic redundancy check method for a 7 bit code block "1001001". Assume appropriate generator polynomial.

Solution

Take generator G = 1011 (degree 3, representing x³+x+1), so 3 zero bits are appended to the message. Message M = 1001001 → Augmented dividend = 1001001000 (10 bits) Perform modulo-2 (XOR) long division of 1001001000 by 1011: 1001 XOR 1011 = 0010 → bring down 0 → 0100 0100 (leading bit 0, no XOR) → bring down 0 → 1000 1000 XOR 1011 = 0011 → bring down 1 → 0111 0111 (leading bit 0, no XOR) → bring down 0 → 1110 1110 XOR 1011 = 0101 → bring down 0 → 1010 1010 XOR 1011 = 0001 → bring down 0 → 0010 No bits remain; remainder (CRC) = last 3 bits = 010. Transmitted codeword = message + CRC = 1001001 010 = 1001001010. Error detection at the receiver: divide the received codeword by the same generator G = 1011 using the same XOR division. If the remainder is exactly 000, the block is accepted as error-free; any non-zero remainder indicates the block was corrupted in transit.

2019 · Paper-II · Q5(a) (8 marks)

Question

Demonstrate various types of multiplexing techniques in the context of computer networks using suitable examples.

Solution

Multiplexing lets multiple signals share one physical transmission medium.

FDM (Frequency Division Multiplexing): the available bandwidth is divided into separate, non-overlapping frequency bands, one per channel, all transmitted simultaneously. Example: cable television, where each channel occupies its own frequency band on the same coaxial cable.
TDM (Time Division Multiplexing): the transmission time is divided into fixed time slots, and each channel is given its own slot in a repeating cycle. Synchronous TDM example: digital telephone T1 lines carrying 24 voice channels; Statistical/Asynchronous TDM dynamically allocates slots only to channels with data to send.
WDM (Wavelength Division Multiplexing): used in fiber-optic links — different data streams are carried on different wavelengths (colours) of light through the same fiber simultaneously.
CDM/CDMA (Code Division Multiplexing): every transmitter is assigned a unique orthogonal code; all signals share the same time and frequency but are separated at the receiver by their code. Example: CDMA cellular networks.

2019 · Paper-II · Q5(b) (6 marks)

Question

Show step by step procedure to find MAC address of a node in a network using Address Resolution Protocol.

Solution

Host A wants to send a packet to Host B on the same LAN, and knows B's IP address but not its MAC address.
Host A checks its local ARP cache for an existing IP→MAC mapping for B; if found, it uses it directly (no broadcast needed).
If not cached, Host A broadcasts an ARP Request frame on the LAN (destination MAC = FF:FF:FF:FF:FF:FF) containing: 'Who has IP address X? Tell A (A's IP and MAC)'.
Every host on the LAN receives the broadcast, but only the host whose own IP matches (Host B) responds.
Host B sends a unicast ARP Reply directly to Host A's MAC address, containing B's own MAC address.
Host A receives the reply and caches the (IP,MAC) mapping for B in its ARP table for future use (with a timeout).
Host A can now encapsulate and send the actual data frame addressed directly to Host B's MAC address.

2019 · Paper-II · Q5(c) (6 marks)

Question

For transmission of voice signal in real time over the network, select a suitable switching technique. Justify your answer using an example.

Solution

Circuit switching is the most suitable technique for strict real-time voice, because it establishes a single dedicated, fixed-bandwidth end-to-end path for the entire duration of the call before transmission begins. This guarantees a constant, predictable delay and no queuing jitter, both essential for intelligible voice communication. Example: a traditional PSTN telephone call, where the exchange sets up a dedicated circuit between caller and callee for the whole call. By contrast, packet switching (as used by the modern Internet/VoIP) shares the network among many flows, so packets can experience variable queuing delay and jitter at each hop, which can cause choppy or delayed audio unless heavily supplemented with QoS prioritisation, jitter buffers, and using UDP/RTP to avoid retransmission delay — which is why real VoIP systems must add these mechanisms to approximate the guarantees that circuit switching provides natively.

2018 · Paper-I · Q2(a) (4 marks)

Question

How many layers are in the TCP/IP stack? What are the names?

Solution

The TCP/IP model has 4 layers:

Network Access (Link) Layer - handles physical addressing and transmission of data over the local network medium (e.g. Ethernet, Wi-Fi).
Internet Layer - handles logical addressing and routing of packets across networks (IP, ICMP, ARP).
Transport Layer - provides end-to-end communication, reliability, and flow control (TCP, UDP).
Application Layer - provides network services directly to user applications (HTTP, FTP, SMTP, DNS).

(Note: some textbooks present a 5-layer version that separates Physical and Data Link, mapping more closely to the OSI model, but the classical, original TCP/IP reference model has exactly these 4 layers.)

2018 · Paper-II · Q3(A) (8 marks)

Question

Differentiate between hub, bridge, switch and router.

Solution

Hub: a simple, unintelligent Layer-1 (Physical layer) networking device that broadcasts (repeats) every incoming signal out to ALL other connected ports, regardless of the intended destination; it has no addressing awareness, causing unnecessary traffic and collisions on the shared medium.

Bridge: a Layer-2 (Data Link layer) device that connects two network segments and forwards frames between them based on MAC addresses, learning which devices are on which segment; it reduces unnecessary traffic and collisions compared to a hub, but typically has only a small number of ports.

Switch: also a Layer-2 device, essentially a multi-port bridge; it examines the destination MAC address of each incoming frame and forwards it ONLY to the specific port where the destination device is connected (using a MAC address table it learns dynamically), rather than broadcasting to all ports - this dramatically reduces collisions and improves efficiency compared to a hub, and supports many more ports than a bridge typically does.

Router: a Layer-3 (Network layer) device that connects different networks (e.g. a LAN to the Internet, or two different subnets) and forwards packets between them based on IP (logical) addresses, using routing tables and routing protocols/algorithms to determine the best path. Unlike hubs/bridges/switches (which typically operate within a single network/broadcast domain), routers connect and separate distinct networks/broadcast domains.

Summary progression: Hub (Layer 1, broadcasts everything) -> Bridge/Switch (Layer 2, MAC-address-based selective forwarding within one network) -> Router (Layer 3, IP-address-based forwarding between different networks).

2018 · Paper-II · Q3(B) (6 marks)

Question

Discuss how Network Address Translation (NAT) works and why is it useful?

Solution

NAT (Network Address Translation) is a technique, typically implemented on a router/gateway, that translates private (internal) IP addresses used within a local network into a single (or small pool of) public IP address(es) when communicating with the outside Internet, and translates responses back to the correct internal device.

How it works: when an internal device (e.g. 192.168.1.5) sends a packet to the Internet, the NAT router replaces the packet's private source IP address with its own public IP address, and also typically replaces the source port number with a unique port it chooses, recording this mapping (private IP:port <-> public IP:port) in a NAT translation table. When a reply packet arrives from the Internet addressed to the router's public IP and that specific port, the router looks up the table and rewrites the destination back to the correct internal private IP:port, then forwards it to the correct internal device.

Why it's useful:

IPv4 address conservation: allows an entire private network (potentially hundreds/thousands of devices) to share a single public IPv4 address, significantly easing the shortage of available public IPv4 addresses.
Security/obscurity: internal (private) IP addresses are hidden from the external Internet, since outsiders only see the router's single public address, making it harder for external attackers to directly target internal devices.
Flexibility: internal network addressing can be changed freely without needing to obtain/renumber public IP addresses.

2018 · Paper-II · Q3(C) (6 marks)

Question

Elaborate the working of multiplexing/de-multiplexing at the transport layer.

Solution

Multiplexing at the sender's transport layer: since a single host may run multiple applications/processes simultaneously (e.g. a web browser, an email client, a chat app), each communicating over the network at the same time, the transport layer (TCP/UDP) must combine (multiplex) data from these multiple application-layer processes into a single stream of segments to be handed down to the network layer (IP) for transmission. Each segment is tagged with a source port number identifying which local application process it originated from, and a destination port number identifying the intended application process on the receiving host.

De-multiplexing at the receiver's transport layer: when segments arrive at the receiving host from the network layer, the transport layer must correctly deliver each segment's data to the correct application process among potentially many running processes. It does this by examining the header fields of each segment: for UDP, de-multiplexing uses only the destination port number (and destination IP) to identify the socket; for TCP, de-multiplexing uses the full 4-tuple (source IP, source port, destination IP, destination port) to identify the specific connection/socket, since TCP is connection-oriented and a server may have many simultaneous connections on the same destination port from different clients, each needing to be distinguished.

In short: multiplexing gathers data from many processes into one outgoing stream (using port numbers as identifiers), and de-multiplexing uses those same port numbers (and, for TCP, the full connection tuple) to route incoming segments back to the correct process.

2017 · Paper-I · Q2(a) (8 marks)

Question

The internet era has given rise to the problem of cybercrimes. Given the need to maintain privacy which is an ethical responsibility of the government, what technical means would you suggest to curb this problem?

Solution

Technical measures to curb cybercrime while respecting privacy:

Strong encryption: use end-to-end encryption (e.g. TLS/SSL for web traffic, encrypted messaging) so intercepted data is unreadable, protecting citizens from criminals while still allowing lawful, targeted, warrant-based access rather than blanket surveillance.
Firewalls and Intrusion Detection/Prevention Systems (IDS/IPS): monitor and block malicious traffic at network boundaries (ISP and organizational level).
Multi-factor authentication and strong access control: reduces account takeover and identity theft.
Digital forensics and cyber-crime units: equip law-enforcement with tools to trace and prosecute cybercriminals without needing mass surveillance of ordinary citizens.
Content filtering and monitoring at ISP level targeted narrowly (e.g. blocking known malware/phishing domains) rather than deep packet inspection of all citizens' traffic.
Public Key Infrastructure (PKI) and digital signatures: ensure authenticity and non-repudiation of online transactions/communication.
Regular security audits, vulnerability patching, and cybersecurity awareness/education campaigns.
Anonymization/pseudonymization techniques for data the government does collect, and strict data-retention/access-logging policies (audited access) so any lawful interception is accountable and narrowly scoped, balancing crime prevention against the ethical duty to protect citizens' privacy.

2017 · Paper-II · Q3(a) (8 marks)

Question

Explain the following network protocols: (i) HTTP and SIP (ii) TCP and UDP

Solution

HTTP (HyperText Transfer Protocol): an application-layer protocol used for transferring web content (HTML pages, images, etc.) between web browsers (clients) and web servers, using a request-response model (GET, POST, PUT, DELETE methods); runs over TCP.

SIP (Session Initiation Protocol): an application-layer signalling protocol used to establish, modify, and terminate real-time multimedia sessions (VoIP calls, video conferencing, instant messaging) between participants; it handles call setup/teardown but not the actual media transport (which is typically handled by RTP).

TCP (Transmission Control Protocol): a connection-oriented, reliable transport-layer protocol - it establishes a connection (three-way handshake), guarantees ordered, error-checked, and complete delivery of data via acknowledgments and retransmission of lost segments, and performs flow/congestion control. Used where reliability matters more than speed (web browsing via HTTP, email, file transfer).

UDP (User Datagram Protocol): a connectionless, unreliable (best-effort) transport-layer protocol - no handshake, no guarantee of delivery, ordering, or duplicate protection, but has very low overhead and latency. Used where speed matters more than perfect reliability (video/audio streaming, VoIP, DNS queries, online gaming), since SIP-initiated media (RTP) commonly runs over UDP for this reason.

2017 · Paper-II · Q3(b) (6 marks)

Question

What is the transmission time of a packet sent by a station if the length of the packet is 1 million bytes and the bandwidth of the channel is 200 Kbps?

Solution

Transmission time = Packet size (in bits) / Bandwidth (in bits per second).

Packet size = 1,000,000 bytes = 1,000,000 x 8 bits = 8,000,000 bits. Bandwidth = 200 Kbps = 200,000 bits per second.

Transmission time = 8,000,000 / 200,000 = 40 seconds.

So the packet takes 40 seconds to be transmitted onto the channel at this bandwidth.

2017 · Paper-II · Q3(c) (6 marks)

Question

Given the IP address 10.5.118.3 and the network mask 255.255.240.0, what are the assignable IP addresses and broadcast in this subnet?

Solution

Mask 255.255.240.0 = /20, meaning the block size in the third octet is 256 - 240 = 16.

The subnet boundaries in the third octet are multiples of 16: 0, 16, 32, ..., 112, 128, ... The given third octet is 118, which falls between 112 and 127 (112 = 16 x 7), so this address belongs to the subnet block 112-127.

Network address = 10.5.112.0 Broadcast address = 10.5.127.255 (the last address in that block)

Assignable (usable host) IP address range = 10.5.112.1 to 10.5.127.254

(10.5.112.0 is reserved as the network address and 10.5.127.255 is reserved as the broadcast address, so they cannot be assigned to hosts.)

2017 · Paper-II · Q5(b) (6 marks)

Question

Discuss the functionality of Ethernet LAN and its types.

Solution

Ethernet is the most widely used LAN (Local Area Network) technology, defined by IEEE 802.3, providing wired connectivity between devices within a limited geographic area (office, building, campus). It uses frame-based communication with MAC (Media Access Control) addressing for identifying devices, and historically used CSMA/CD (Carrier Sense Multiple Access with Collision Detection) to manage shared-medium access, though modern switched Ethernet largely avoids collisions via full-duplex point-to-point links through switches.

Types of Ethernet (by speed/generation):

Standard Ethernet (10 Mbps): the original 10BASE-T, 10BASE2, 10BASE5 variants over twisted pair, coaxial cable.
Fast Ethernet (100 Mbps): 100BASE-TX, 100BASE-FX, introduced to increase speed over twisted-pair/fiber.
Gigabit Ethernet (1000 Mbps / 1 Gbps): 1000BASE-T, 1000BASE-SX/LX (fiber), widely used in modern LANs and data centers.
10 Gigabit Ethernet and beyond (10GbE, 40GbE, 100GbE): used in high-performance backbones and data centers.

Ethernet also comes in different topological/physical implementations - originally a shared bus (coax) or hub-based star, now almost universally switch-based star topology - providing dedicated collision-free bandwidth per link.

2016 · Paper-II · Q2(a) (6 marks)

Question

Suppose you add two new devices to an existing five-device network. If you have a fully connected mesh topology, how many new cable lines are needed? If, however, the devices are arranged in a ring, how many new cable lines are needed?

Solution

Mesh topology: a fully connected mesh with n devices needs n(n-1)/2 links. Original (5 devices): 5x4/2 = 10 links. New total (7 devices): 7x6/2 = 21 links. New cable lines needed = 21 - 10 = 11 new links.

Ring topology: a ring with n devices needs exactly n links (each device connects to exactly two neighbours, forming one cycle). Original (5 devices): 5 links. New total (7 devices): 7 links. New cable lines needed = 7 - 5 = 2 new links.

This illustrates why mesh topology, despite being the most robust/fault-tolerant, scales very poorly (O(n^2) links) compared to a ring (O(n) links) as the network grows.

2016 · Paper-II · Q2(b) (7 marks)

Question

Transmission media are not perfect because of imperfections and impairments in the signal sent through the medium. Signals at the beginning and at the end of the medium are not the same. Discuss in detail the impairments in the transmission medium.

Solution

Three main transmission impairments:

Attenuation: the signal loses strength/energy as it travels along the medium, due to the medium's resistance; longer distances cause greater signal loss. Amplifiers/repeaters are used to boost the signal periodically to compensate. Measured in decibels (dB).
Distortion: occurs when a composite signal (made of multiple frequency components) travels through a medium, and each frequency component experiences a different propagation speed/delay, so the components arrive at the destination with different phase shifts, changing the shape of the composite signal - most significant for signals of differing frequency composition rather than a single-frequency signal.
Noise: unwanted signal energy that gets added to the transmitted signal from an external source. Types include thermal noise (random electron motion, present in all media), induced noise (from motors/appliances), crosstalk (signal from an adjacent wire/cable bleeding in), and impulse noise (sudden spikes from power lines or lightning). Noise corrupts the signal, and is commonly measured via signal-to-noise ratio (SNR).

Together, these impairments mean the received signal is a degraded/altered version of what was sent, which is why real communication systems must use error detection/correction, amplification/regeneration and shielding techniques.

2016 · Paper-II · Q2(c) (7 marks)

Question

Whenever multiple devices are used in a network, the problem arises that how to connect them to make one-on-one communication possible. Switching is the best solution for this kind of problem. A switched network consists of a series of inter-linked nodes called switches. Explain briefly the methods of switching used by computer networks.

Solution

Three principal switching methods:

Circuit Switching: a dedicated end-to-end physical/logical path (circuit) is established between sender and receiver before data transfer begins, and reserved for the entire duration of the communication (e.g. traditional telephone networks). Advantage: guaranteed, constant bandwidth and delay; disadvantage: inefficient use of resources if the link is idle during the connection.
Packet Switching: data is broken into small packets, each carrying addressing information, and routed independently through the network; packets may take different paths and arrive out of order (reassembled at the destination). Two forms:
- Datagram approach: each packet routed independently (connectionless), e.g. IP.
- Virtual-circuit approach: a logical path is set up first, and all packets follow the same path (connection-oriented), e.g. ATM, Frame Relay. Advantage: efficient bandwidth sharing among many users; disadvantage: variable delay (jitter).
Message Switching: an entire message is sent from node to node, with each intermediate node storing the whole message and forwarding it when the next link is free ('store-and-forward'); mostly of historical interest today (e.g. early telegraph/email-relay systems).

Modern data networks (including the Internet) predominantly use packet switching for its efficiency and resilience.

Part B — MCQs

2025 · Paper-I · Q9 (MCQ)

What is the main purpose of using Trojans?

A. To get passwords
B. To delete data
C. To corrupt files
D. None of these

Answer: A

Explanation: A Trojan horse disguises itself as legitimate software to trick users into installing it, and is primarily used to gain unauthorized access and steal sensitive information such as passwords.

2025 · Paper-II · Q18 (MCQ)

ICMP is primarily used for:

A. Error and diagnostic functions
B. Addressing
C. Forwarding
D. Routing

Answer: A

Explanation: The Internet Control Message Protocol (ICMP) is used by network devices to report errors and provide diagnostic information (e.g., 'destination unreachable', ping/traceroute), not for addressing, forwarding or routing decisions themselves.

2025 · Paper-II · Q19 (MCQ)

Which multiplexing technique is used to transmit digital signals?

A. FDM
B. TDM
C. WDM
D. FDM & WDM

Answer: B

Explanation: Time Division Multiplexing (TDM) allocates fixed time slots to each channel and is the natural fit for digital signals, whereas FDM/WDM (frequency/wavelength division) are used mainly for analog and optical signal multiplexing respectively.

2025 · Paper-II · Q20 (MCQ)

DHCP (dynamic host configuration protocol) provides __________ to the client.

A. IP address
B. MAC address
C. Url
D. None of these

Answer: A

Explanation: DHCP automatically and dynamically assigns an IP address (plus subnet mask, gateway, DNS server, etc.) to a client device joining the network.

2024 · Paper-II · Q4 (MCQ)

Network traffic estimation is:

A. Impossible
B. Easily computable via linear equations
C. Can only be solved using AI techniques
D. None of these

Answer: B

Explanation: Classical traffic engineering models network flow/traffic estimation using linear equations and queuing theory (e.g., traffic matrices, flow conservation equations), not requiring AI.

2024 · Paper-II · Q6 (MCQ)

Which of the following is the most efficient encoding to send data via networks? Also consider the reasoning in your answer.

A. 3-Excess codes because it is not weighted
B. Binary, because its representation can be done simply with zeros and ones
C. Decimal, because a decimal requires lesser space to represent the same number
D. None of these

Answer: B

Explanation: Binary encoding maps directly and efficiently onto the two-voltage-level digital signaling used by network hardware, making it the natural and most efficient choice for data transmission, unlike decimal or excess-N codes which need extra encoding overhead.