2019 question paper
Computer Networks
22 questions
Q1a. This layer is an addition to OSI model when compared with TCP/IP model: (i) Application layer (ii) Presentation layer (iii) Session layer (iv) Both session layer and presentation layer20192m
Module 1: Physical Layer & Data Communication
View this question on its own page →This layer is an addition to OSI model when compared with TCP/IP model:
(i) Application layer
(ii) Presentation layer
(iii) Session layer
(iv) Both session layer and presentation layerWorked Solution(iv) Both session layer and presentation layer
The OSI model has 7 layers while TCP/IP has 4 (or sometimes described as 5). The Application layer of TCP/IP absorbs the responsibilities of OSI's top three layers — Application, Presentation, and Session. So the Session and Presentation layers exist as distinct layers only in the OSI model; TCP/IP does not define them separately.
Q1b. Application layer is implemented on: (i) end systems (ii) NIC (iii) ethernet (iv) route20192m
Module 5: Application Layer & Security
View this question on its own page →Application layer is implemented on:
(i) end systems
(ii) NIC
(iii) ethernet
(iv) routeWorked SolutionAnswer: (i) end systems
Why: The application layer (HTTP, DNS, FTP, SMTP, …) is implemented by the communicating end hosts — the sender and receiver processes — not by network-core devices. NIC is a data-link/physical-layer device, Ethernet is a link-layer technology, and routers forward packets at the network layer, never processing application-layer content.
Q1c. Which of the following is not applicable for IP? (i) Error reporting (ii) Handle addressing conventions (iii) Datagram format (iv) Packet handling conventions20192m
Module 3: Network Layer & Routing
View this question on its own page →Which of the following is not applicable for IP?
(i) Error reporting
(ii) Handle addressing conventions
(iii) Datagram format
(iv) Packet handling conventionsWorked SolutionAnswer: (i) Error reporting
Why: IP is a connectionless, best-effort service that only handles addressing conventions (source/destination address fields), the datagram format, and packet-handling conventions such as fragmentation, TTL and reassembly. Error reporting and diagnostic notification are done by ICMP, which sits on top of IP, so it is not an IP function.
Q1d. Which one of the following is not a function of network layer? (i) Routing (ii) Inter-networking (iii) Congestion control (iv) None of the above20192m
Module 3: Network Layer & Routing
View this question on its own page →Which one of the following is not a function of network layer?
(i) Routing
(ii) Inter-networking
(iii) Congestion control
(iv) None of the aboveWorked SolutionAnswer: (iv) None of the above
Why: Routing (choosing the best path), inter-networking (delivering a packet across multiple interconnected networks) and congestion control (regulating traffic so buffers/queues do not overflow) are all recognised functions of the network layer. Since options (i), (ii) and (iii) are each genuine network-layer functions, none of them is a "non-function", making (iv) the correct choice.
Q1e. For a 10 Mbps Ethernet link, if the length of the packet is 32 bits, the transmission delay is (in milliseconds): (i) 3.2 (ii) 32 (iii) 0.32 (iv) None of the above20192m
Module 1: Physical Layer & Data Communication
View this question on its own page →For a 10 Mbps Ethernet link, if the length of the packet is 32 bits, the transmission delay is (in milliseconds):
(i) 3.2
(ii) 32
(iii) 0.32
(iv) None of the aboveWorked Solution(i) 3.2
Transmission delay = Packet size / Bandwidth
Packet size = 32 bits Bandwidth = 10 Mbps = 10 × 10^6 bits/sec Delay = 32 / (10 × 10^6) s = 3.2 × 10^-6 s = 3.2 microsecondsNote: the numeric answer works out to 3.2 microseconds, not milliseconds as the question states — this is a common typo in this question across question banks. The intended/keyed answer is still (i) 3.2, matching the computed value.
Q1f. Transmission delay does not depend on: (i) packet length (ii) distance between the routers (iii) transmission rate (iv) None of the above20192m
Module 1: Physical Layer & Data Communication
View this question on its own page →Transmission delay does not depend on:
(i) packet length
(ii) distance between the routers
(iii) transmission rate
(iv) None of the aboveWorked Solution(ii) distance between the routers
Transmission delay = Packet size / Bandwidth — it depends only on packet length and transmission rate. Distance between routers affects propagation delay (= distance / signal speed), which is a separate delay component. So transmission delay does not depend on distance.
Q1g. In TCP, sending and receiving data is done as: (i) stream of bytes (ii) sequence of characters (iii) lines of data (iv) packets20192m
Module 4: Transport Layer & Congestion Control
View this question on its own page →In TCP, sending and receiving data is done as:
(i) stream of bytes
(ii) sequence of characters
(iii) lines of data
(iv) packetsWorked SolutionAnswer: (i) stream of bytes
Why: TCP delivers an ordered, reliable byte stream to the application — it has no notion of message boundaries (characters or lines) and exposes no record framing. "Packets" is the network-layer term (IP datagrams); TCP segments are simply chunks of the byte stream that it numbers, sends and acknowledges by byte offset.
Q1h. Suppose a TCP connection is transferring a file of 1000 bytes. The first byte is numbered 10001. What is the sequence number of the segment if all data is sent in only one segment? (i) 10000 (ii) 10001 (iii) 12001 (iv) 1100120192m
Module 4: Transport Layer & Congestion Control
View this question on its own page →Suppose a TCP connection is transferring a file of 1000 bytes. The first byte is numbered 10001. What is the sequence number of the segment if all data is sent in only one segment?
(i) 10000
(ii) 10001
(iii) 12001
(iv) 11001Worked SolutionAnswer: (ii) 10001
Why: A TCP segment's sequence number is the number of the first byte of data it carries, not the last byte or a byte count. Since the file's first byte is numbered 10001 and the whole file goes in one segment, the segment carries bytes 10001 through 11000 with sequence number 10001; the next byte's number (11001) would become the ACK number, while 10000/12001 are off by one block.
Q1i. The protocol on which DNS application run is: (i) Telnet (ii) UDP (iii) HTTP (iv) None of the above20192m
Module 5: Application Layer & Security
View this question on its own page →The protocol on which DNS application run is:
(i) Telnet
(ii) UDP
(iii) HTTP
(iv) None of the aboveWorked SolutionAnswer: (ii) UDP
Why: DNS primarily uses UDP on port 53 for ordinary queries because its request/response exchanges are short and the low overhead of connectionless UDP gives fast answers. It falls back to TCP only for responses larger than 512 bytes (zone transfers, AXFR); Telnet (port 23) and HTTP (port 80) are unrelated application-layer protocols.
Q1j. __________ allows you to connect and login to a remote computer. (i) Telnet (ii) FTP (iii) HTTP (iv) None of the above20192m
Module 5: Application Layer & Security
View this question on its own page →__________ allows you to connect and login to a remote computer.
(i) Telnet
(ii) FTP
(iii) HTTP
(iv) None of the aboveWorked SolutionAnswer: (i) Telnet
Why: Telnet (port 23) establishes a TCP connection that gives a text-command interface to log in and work on a remote host. FTP transfers files (not a login session), and HTTP retrieves web pages — neither provides remote terminal login.
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Q2a. Draw a neat diagram of TCP/IP protocol stack with the protocols and functionality implemented on each layer. Also explain, why there is a need for having a layered architecture (like OSI and TCP/IP) in networks.201914m
Module 1: Physical Layer & Data Communication
View this question on its own page →Draw a neat diagram of TCP/IP protocol stack with the protocols and functionality implemented on each layer. Also explain, why there is a need for having a layered architecture (like OSI and TCP/IP) in networks.
Worked SolutionTCP/IP Protocol Stack Diagram
+-----------------------------------------------------+ | Application Layer | | HTTP | FTP | SMTP | DNS | Telnet | SNMP | +-----------------------------------------------------+ | Transport Layer | | TCP | UDP | +-----------------------------------------------------+ | Internet Layer | | IP | ICMP | ARP | IGMP | +-----------------------------------------------------+ | Network Access Layer | | Ethernet | Wi-Fi (802.11) | PPP | Frame Relay | +-----------------------------------------------------+ | physical medium (copper/fiber/air) |Data flows top-down at the sender (each layer adds its own header — encapsulation) and bottom-up at the receiver (each layer strips its header — decapsulation).
Why layered architecture is needed
Modularity — each layer solves one well-defined problem (e.g. routing, reliable delivery) independently of the others, making the system easier to design, understand, and reason about.
Independent evolution — a layer's internal implementation can change (e.g. Ethernet replaced by Wi-Fi) without requiring changes to the layers above it, as long as the interface between layers stays the same.
Interoperability — standardized interfaces between layers let hardware and software from different vendors work together, since each only needs to honor the layer boundary contract.
Simplified troubleshooting — problems can be isolated to a specific layer (e.g. a routing issue is a Network-layer problem, not an Application-layer one), speeding up diagnosis.
Reusability — lower layers (e.g. IP, Ethernet) can be reused by many different upper-layer applications, avoiding duplicated effort.
Without layering, every application would need to handle addressing, routing, framing, and physical transmission itself — an unmanageable, tightly-coupled design.
Q3a. We know that there are 2 types of address schemes (i.e. IP address and MAC address) in computer networks. Explain what the need of having two types of addresses. What will happen if we use only IP address in computer networks?201914m
Module 3: Network Layer & Routing
View this question on its own page →We know that there are 2 types of address schemes (i.e. IP address and MAC address) in computer networks. Explain what the need of having two types of addresses. What will happen if we use only IP address in computer networks?
Worked SolutionNeed for Two Address Schemes: IP Address vs MAC Address
A host reachable on a network needs two fundamentally different kinds of identification: a logical (software-assigned) address for locating it across the whole internet and a physical (hardware-burned) address for delivering a frame to the correct interface on the local link. The network layer uses the IP address; the data link layer uses the MAC address. They solve different problems at different scopes, and neither can replace the other.
1. What each address is
Property IP address (logical) MAC address (physical) Assigned by Network administrator / DHCP, by software Manufacturer, burned into NIC ROM Scope Global — meaningful across the entire internetwork Local — meaningful only within the LAN segment Structure Network part + host part (hierarchical, e.g. /24) Flat, no hierarchy: 48-bit OUI + device ID Changes when Host moves to another network Never (fixed to the NIC) Layer Network layer (L3) Data link layer (L2) Delivery goal Find a route to the destination network Deliver the frame to one interface on the link 2. Why both are required
- Hierarchy vs flat identification. Routing tables would explode if every one of billions of devices had its own globally-routable entry. IP addresses are hierarchical (network + host), so a router only stores network-level routes. MAC addresses are flat and unique but carry no topological information, so they cannot be used for global routing.
- Separation of duties across layers. IP addresses are used end-to-end, from source host to destination host, by routers along the path. MAC addresses are used hop-to-hop: each router strips the incoming frame and re-frames the packet with the MAC addresses of the next hop. This layering keeps L2 LAN technologies (Ethernet, Wi-Fi) independent of L3 internetworking.
- Interoperability of heterogeneous networks. An IP datagram can cross Ethernet, token ring, ATM or Wi-Fi links because the logical address stays constant while the link-layer (MAC) address is rewritten at every hop.
- Dynamic reconfiguration. A laptop moving from Wi-Fi to Ethernet keeps its IP configuration (via DHCP) but gets a new link-layer context; the ARP mechanism binds the stable logical address to the current physical one on demand.
3. What happens if only the IP address is used
- No local delivery mechanism. IP alone cannot put bits on a wire; frames need a MAC (or equivalent) address to be accepted by a NIC, so the last hop of delivery would be impossible.
- Every device must store a global routing table. With a flat, non-hierarchical scheme each node would need an entry for all other nodes — state per device for hosts, infeasible at internet scale.
- Broadcast storm / scanning problem. Locating a host would require broadcasting to every device on every network, wasting bandwidth and destroying scalability.
- Loss of technology independence. LAN hardware would have to understand IP directly, tightly coupling L2 and L3 and making new link technologies impossible to add transparently.
- No bootstrapping. At power-on a host knows its own MAC but not yet its IP; protocols like ARP map IP → MAC precisely because both addresses must coexist.
Conclusion: the IP address answers "which network is the destination on, and how do I route there?", while the MAC address answers "which interface on this link should receive the frame?". Both are indispensable for scalable, layered, technology-independent communication.
Q4a. What do you mean by Random Access Protocols? Explain the limitations of using ALOHA as random access protocol.201914m
Module 2: Data Link Layer & MAC Sublayer
View this question on its own page →What do you mean by Random Access Protocols? Explain the limitations of using ALOHA as random access protocol.
Worked SolutionRandom Access Protocols
Random access (also called contention-based) protocols are MAC sublayer protocols in which no station has a scheduled turn or fixed priority to transmit. Any station with data ready may attempt to send at any time, without needing explicit permission from a central controller or a pre-allocated time slot. Because multiple stations can transmit simultaneously, collisions are possible, and the protocol defines how stations detect/avoid them and recover (typically through random backoff before retrying). Access is purely a function of chance and channel conditions — hence "random access." Examples: ALOHA, CSMA, CSMA/CD, CSMA/CA.
Limitations of ALOHA as a random access protocol
Low maximum throughput — pure ALOHA achieves only ~18.4% channel utilization at best (slotted ALOHA improves this to ~36.8%, but both remain far below what controlled-access or channelization protocols can achieve).
No carrier sensing — ALOHA stations transmit without first checking whether the channel is already busy, unlike CSMA-based protocols. This causes many avoidable collisions that a simple "listen before talk" check could have prevented.
Large vulnerable time — in pure ALOHA especially, the vulnerable window (2 x frame transmission time) is large, making collisions more likely as the number of active stations or offered load increases.
Poor scalability — as the number of stations (and hence offered load G) increases beyond the optimal point, throughput actually decreases rather than saturating gracefully, since collision probability grows faster than useful transmissions.
No collision detection during transmission — a station doesn't know a collision happened until it fails to receive an ACK (often after a timeout), wasting the full frame transmission time even when the frame was doomed to collide, unlike CSMA/CD which can abort mid-transmission upon detecting a collision.
Requires synchronization for the slotted variant — while slotted ALOHA improves throughput, it requires all stations to share a precise, common time reference, adding system complexity.
These limitations motivated the development of Carrier Sense Multiple Access (CSMA) variants, which listen before transmitting to substantially reduce (though not eliminate) collision probability.
Q5a. In the context of Transmission Control Protocol (TCP), explain the statement, "TCP help us to establish a reliable connection over an unreliable network".20197m
Module 4: Transport Layer & Congestion Control
View this question on its own page →In the context of Transmission Control Protocol (TCP), explain the statement, "TCP help us to establish a reliable connection over an unreliable network".
Worked SolutionTCP: Reliable Connection over an Unreliable Network
The underlying IP service is best-effort: it can lose, duplicate, reorder or corrupt datagrams and offers no timing guarantees. TCP sits above IP and adds connection-oriented, end-to-end reliability so that the application sees a dependable, in-order byte stream despite that hostile network.
How TCP builds reliability
- Connection establishment & termination (3-way handshake). SYN → SYN-ACK → ACK fixes initial sequence numbers on both sides (avoiding old duplicate segments), establishes negotiated parameters (MSS, window scale), and symmetrically closes the connection with FIN/ACK so no data is lost.
- Sequence numbers and acknowledgements. Every byte carries a sequence number; the receiver ACKs the next expected byte number. This detects lost segments and eliminates duplicates.
- Retransmission on timeout/RTO. A retransmission timer (with RTT estimation and Karn's algorithm / exponential back-off) resends unacknowledged data. Fast retransmit on 3 duplicate ACKs avoids waiting for the full timeout.
- Flow control (sliding window). The advertised receive window prevents the sender from overrunning the receiver's buffer — reliability at the endpoints.
- Checksum. A 16-bit header+data checksum detects corruption; the segment is discarded and retransmitted.
- Ordered delivery. The receiver re-sequences segments that arrive out of order using the sequence numbers before handing data to the application.
- Congestion control (slow start, congestion avoidance, fast recovery). Even though the network is unreliable and may drop packets under load, TCP adapts its sending rate so that congestion (the main cause of loss) is minimised — the sender cooperates with the network instead of overloading it.
Why the network can stay unreliable
Reliability is placed end-to-end at the transport layer rather than in every router. Routers simply forward best-effort datagrams; TCP endpoints detect and repair loss. This is the end-to-end argument: a simpler, faster core plus intelligence at the edges yields a reliable service over an unreliable network.
In short: TCP converts IP's unreliable, connectionless datagram service into an error-free, ordered, flow- and congestion-controlled byte stream using connections, sequencing, ACKs, timers and retransmission.
Q5b. Outline the difference between TCP and UDP.20197m
Module 4: Transport Layer & Congestion Control
View this question on its own page →Outline the difference between TCP and UDP.
Worked SolutionDifference between TCP and UDP
Parameter TCP (Transmission Control Protocol) UDP (User Datagram Protocol) Connection Connection-oriented (3-way handshake first) Connectionless — no setup/teardown Reliability Reliable: ACKs, retransmission, sequencing Unreliable: best-effort delivery Ordering Segments delivered in order using sequence numbers No ordering guarantee Error/flow control Yes — checksum + retransmission, sliding-window flow control Only a checksum; no retransmission, no flow control Congestion control Yes — slow start, congestion avoidance, fast recovery None Header overhead 20 bytes minimum (up to 60 B with options) Fixed 8 bytes Data unit Segment (byte stream, no message boundaries) Datagram (message-oriented, preserves boundaries) Duplication Duplicates detected and discarded Possible duplicates passed to the app Application HTTP, SMTP, FTP control, file transfer, web, e-mail — where accuracy matters DNS, DHCP, SNMP, streaming voice/video, VoIP, online games — where speed matters Speed / latency Slower due to handshakes and ACK overhead Faster, low overhead Broadcast/multicast Not supported Supported Termination Graceful connection release (FIN) No termination concept In brief: TCP trades speed for guaranteed, ordered, error-free delivery over a virtual connection; UDP is lightweight and fast, delivering datagrams with no guarantees — chosen when timeliness matters more than perfect accuracy (or when the application implements its own reliability).
Q6a. Why there is a need of doing error control at transport layer despite of the fact that we are doing the same thing at the data link layer? Why we are doing this error correction multiple times?201914m
Module 2: Data Link Layer & MAC Sublayer
View this question on its own page →Why there is a need of doing error control at transport layer despite of the fact that we are doing the same thing at the data link layer? Why we are doing this error correction multiple times?
Worked SolutionIt might seem redundant to perform error control at both the Data Link layer and the Transport layer, but each operates at a fundamentally different scope, and neither alone can guarantee true end-to-end reliability.
Data Link layer error control is only per-hop (link-local)
- The Data Link layer's error checking (e.g. CRC) only verifies that a frame arrived correctly across a single physical link — from one node to its immediate neighbor (e.g. host to router, or router to router).
- Once a router receives a frame, verifies it, and strips the Data Link header, it re-encapsulates the data into a brand-new frame (with a new CRC) for the next link. This new frame's correctness only re-verifies that this specific hop was error-free.
- Crucially, the router's own internal processing (buffer memory corruption, software bugs, hardware faults during forwarding) is not covered by any Data Link layer check — an error introduced inside a router between de-framing and re-framing would go completely undetected by Data Link layer checks alone.
Transport layer error control provides genuine end-to-end assurance
- The Transport layer's error checking (e.g. TCP checksum) is computed once at the original source and verified only at the final destination — it isn't recomputed and reset at every intermediate hop.
- This means it can catch errors introduced anywhere along the entire path — not just on the physical wire, but also from bugs, memory corruption, or misconfigured devices inside any of the intermediate routers themselves.
- It is the only layer that gives the actual communicating applications (not just adjacent link partners) a guarantee that the data they exchange is correct.
Why not rely on just one layer
- Relying only on the Data Link layer leaves the "inside the router" gap uncovered, as explained above.
- Relying only on the Transport layer, without any Data Link layer error control, would mean that a badly corrupted frame due to noise on a single hop wouldn't be caught (and potentially retransmitted) until the entire end-to-end path's data arrives and fails the final Transport-layer check — far less efficient than catching and locally retransmitting the error immediately at the hop where it occurred.
Conclusion: Data Link layer error control provides fast, local, per-hop reliability (catching and fixing noise-induced errors quickly, hop by hop), while Transport layer error control provides the only genuine end-to-end guarantee, covering failures that occur anywhere in the path — including inside intermediate devices. The two are complementary, not redundant.
Q7a. Explain type of delays in computer networks. Suppose a node A wants to send data of 200 mb to node B. The bandwidth of the channel is 1 Gbps. There are three routers in between node A to node B. Assume other delays equal to 0. Calculate the total delay involved.201914m
Module 1: Physical Layer & Data Communication
View this question on its own page →Explain type of delays in computer networks. Suppose a node A wants to send data of 200 mb to node B. The bandwidth of the channel is 1 Gbps. There are three routers in between node A to node B. Assume other delays equal to 0. Calculate the total delay involved.
Worked SolutionTypes of delay in computer networks
When a packet travels from source to destination through a network, it experiences four types of delay at each hop (router/link):
Processing delay — time a router takes to examine the packet header, check for bit errors, and determine the outgoing link (routing table lookup).
Queuing delay — time a packet waits in the router's output buffer before it can be transmitted, depending on current congestion at that router.
Transmission delay — time to push all of the packet's bits onto the link, equal to
Packet size / Bandwidth.Propagation delay — time for a bit to physically travel from one end of the link to the other, equal to
Distance / Propagation speed.
Total delay per hop = Processing + Queuing + Transmission + Propagation delay.
Numerical problem
Data to send = 200 Mb = 200 x 10^6 bits Bandwidth = 1 Gbps = 1 x 10^9 bits/sec Routers in between = 3 -> number of links (hops) = 3 + 1 = 4 (A -> R1 -> R2 -> R3 -> B) Other delays (processing, queuing, propagation) = 0Since only transmission delay is non-zero, and (assuming store-and-forward switching at each router, the standard assumption) each router must fully receive the data before forwarding it, the data must be transmitted in full once per link:
Transmission delay per link = Data size / Bandwidth = (200 x 10^6) / (1 x 10^9) = 0.2 sec (200 ms) Total delay = Transmission delay per link x Number of links = 0.2 x 4 = 0.8 sec (800 ms)Total delay involved = 0.8 seconds (800 ms)
Q8a. With the help of neat figures, explain sliding window mechanism.201914m
Module 2: Data Link Layer & MAC Sublayer
View this question on its own page →With the help of neat figures, explain sliding window mechanism.
Worked SolutionThe sliding window mechanism allows a sender to transmit multiple frames before receiving an acknowledgement for the first one, pipelining transmission to make efficient use of the link instead of idling after every single frame (as in Stop-and-Wait).
Sender and Receiver Windows
Sequence numbers: 0 1 2 3 4 5 6 7 (wrap around, mod 2^m) Sender's window (size N=4), showing state at some instant: [ACKed] [ACKed] [sent,unACKed] [sent,unACKed] [sent,unACKed] [can send] [not yet] 0 1 2 3 4 5 6 <----------------- window (N=4) ------------------>The sender may transmit any frame within its current window without waiting for an ACK. Each time the oldest unacknowledged frame (left edge of window) gets ACKed, the window slides forward by one position (or more, for cumulative ACKs), opening room for new frames on the right edge.
Receiver's window defines which sequence numbers it is currently willing to accept:
- Go-Back-N: receiver window size = 1 (accepts only the next expected frame in strict order; discards anything else).
- Selective Repeat: receiver window size = N (matches sender), allowing it to buffer out-of-order frames.
Step-by-step operation (Go-Back-N example, window=3)
Time --> Sender: send0 send1 send2 [window full, wait] Receiver: recv0(ACK1) recv1(ACK2) recv2(ACK3) As ACK1 arrives, window slides -> sender sends frame3 As ACK2 arrives, window slides -> sender sends frame4 ...and so on, continuously, as long as ACKs keep arriving in time.Before ACK1 arrives: After ACK1 arrives (window slides by 1): [0][1][2][ ][ ] [0][1][2][3][ ] ^ base ^ base (now at 1) window: {0,1,2} window: {1,2,3}Handling a lost frame
If frame 1 is lost:
Receiver gets: 0 (accept, ACK1) -> 2 (discard, out of order, resend ACK1) -> 3 (discard, out of order, resend ACK1) Sender's timer for frame 1 expires -> resends frames 1, 2, 3 (go back to 1)Why it improves throughput
By keeping up to N frames "in flight" simultaneously rather than one at a time, the sliding window mechanism keeps the transmission link continuously busy across the round-trip time, rather than leaving it idle while waiting for each individual ACK — this is especially critical on links with a large bandwidth-delay product (high speed and/or long distance), where Stop-and-Wait would waste the vast majority of available capacity.
Q9a. Compare the following terms: (a) ARP vs RARP201914m
Module 3: Network Layer & Routing
View this question on its own page →Compare the following terms: (a) ARP vs RARP
Worked SolutionARP vs RARP (Address Resolution Protocol vs Reverse Address Resolution Protocol)
Both are network-layer support protocols that map between a device's IP (logical) address and its MAC (physical) address, but they perform exactly opposite mappings.
1. Basic definitions
- ARP resolves a known IP address → unknown MAC address. When a host knows the destination's IP (e.g. from DNS) but must build an Ethernet frame, it broadcasts an ARP request on the LAN; the owner of that IP replies with its MAC, which is then cached in the ARP table.
- RARP resolves a known MAC address → unknown IP address. A diskless station that knows only its own MAC (read from the NIC) broadcasts a RARP request asking "what is my IP?"; a RARP server on the LAN replies with the assignment.
2. Comparison table
Basis ARP RARP Direction of mapping IP address → MAC address MAC address → IP address Who uses it Any host/router that needs to frame a packet Diskless/thin client that boots without local IP configuration Protocol layer Network layer, just above the data link layer (operates over Ethernet) Network layer, just above the data link layer Message types ARP request (broadcast) / ARP reply (unicast), also reverse ARP & gratuitous ARP RARP request (broadcast) / RARP reply (unicast) Server needed No server required — any host answers for itself Dedicated RARP server required on every LAN Configuration stored In each host's ARP cache (dynamic + static entries) Centralised on the RARP server (file/database) Message format Same 28-byte packet for both request and reply (hardware type, protocol type, hlen/plen, opcode, SHA/SPA/THA/TPA) Same format as ARP with opcode 3/4 Broadcast used Only when the MAC for a target IP is not in the cache Always, since the client has no IP and cannot use IP-based services Limitations Cannot cross routers (broadcast domain); prone to ARP spoofing/poisoning LAN-limited, servers must be on the same segment; largely obsolete Replaced/superseded by Still fundamental to IPv4 (proxy ARP, gratuitous ARP) BOOTP and DHCP, which also supply subnet mask, default gateway, DNS, etc. Direction of query "Who has 192.168.1.5? Tell 192.168.1.10" "I am 00:AA:BB:CC:DD:EE; what is my IP?" 3. Operational flow
ARP (IP → MAC):
- Host A wants to send to 192.168.1.5 and checks its ARP cache — entry missing.
- A broadcasts an ARP request: "Who has 192.168.1.5?".
- Every host on the LAN receives it; only the owner of 192.168.1.5 replies unicast: "192.168.1.5 is at 00:11:22:33:44:55".
- A stores the pair in its ARP cache (typically 15–45 min TTL) and now frames the actual data.
RARP (MAC → IP):
- Diskless machine B reads its MAC 00:AA:BB:CC:DD:EE at boot.
- B broadcasts a RARP request: "My MAC is 00:AA:BB:CC:DD:EE, what is my IP?".
- The RARP server on that LAN looks up the MAC in its table and unicasts back the assigned IP (and the server's own IP).
- B configures its network interface and continues booting.
4. Key differences in one line
ARP lets an already-configured host reach other hosts on the link by discovering hardware addresses for known IP addresses; RARP lets an unconfigured host discover its own IP from its hardware address — a capability now provided by BOOTP/DHCP.
Q9b. Compare the following terms: (b) Unicast vs Multicast201914m
Module 5: Application Layer & Security
View this question on its own page →Compare the following terms: (b) Unicast vs Multicast
Worked SolutionUnicast vs Multicast
Both are IP delivery modes describing how many receivers a single stream of data reaches. Unicast is one-to-one communication between exactly one sender and one receiver; multicast is one-to-many, where one sender transmits a single copy of the data that is replicated by routers only along the branches leading to subscribed receivers.
1. Comparison table
Basis Unicast Multicast Communication type One-to-one (single sender, single receiver) One-to-many (single sender, group of receivers) Address used Class A/B/C unicast address (host-specific) Class D address, range 224.0.0.0 – 239.255.255.255 Address identifies One specific host interface A group, not a host — no fixed location Receiver membership Implicit — the address names the receiver Explicit join/leave via IGMP (Internet Group Management Protocol) Packet replication One copy sent to that host only Router replicates packets only on links with group members Bandwidth usage receivers ⇒ copies of the traffic on the source link path One copy leaves the source; total network load ≪ copies Router behaviour Standard forwarding along the unicast route Needs multicast routing (e.g. PIM, DVMRP) and a shared/distribution tree Routing table Simple per-destination unicast route Separate multicast forwarding table built from group state Use of receiver address Destination address is the receiver's unicast IP Destination address is the group address; receiver IP never appears in the header Example protocols/applications HTTP, FTP, SMTP, SSH IPTV, video conferencing, live streaming, OSPF/MOSPF updates, news feeds Scalability of delivery Scales with number of receivers (source sends repeatedly) Scales well — sender effort is constant regardless of group size Error/ack behaviour Each receiver can be acknowledged individually No per-receiver ACK (would defeat the purpose); unreliable, best-effort Address classes (IPv4) A, B, C D only Requires Nothing beyond routing IGMP at host/router, multicast-capable routers and protocols 2. Why multicast is efficient
If a source must send a 1 Mbps video stream to 1000 users:
- Unicast: the source (or its access link) transmits Mbps.
- Multicast: the source transmits 1 Mbps once; the network duplicates packets only at routers where members exist, so link carrying depends on topology, not on group size.
3. Diagram (conceptual)
Unicast (3 copies) Multicast (1 copy, replicated) S ──► R1 ──► H1 S ──► R1 ──┬──► H1 (member) │ (one copy) ├──► H2 (member) ├──► R2 ──► H2 │ │ R2 ──► H3 (member) └──► R3 ──► H3 (non-member: no copy)4. Summary
Unicast addresses a device and is used for ordinary request–response traffic where reliability and individual delivery matter. Multicast addresses a group and is used for bandwidth-hungry one-to-many distribution, where sending a single copy and letting routers replicate it saves enormous network capacity — at the cost of needing IGMP plus multicast routing.
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Q9c. Compare the following terms: (c) Flooding vs Broadcasting201914m
Module 2: Data Link Layer & MAC Sublayer
View this question on its own page →Compare the following terms: (c) Flooding vs Broadcasting
Worked SolutionAspect Flooding Broadcasting Definition A routing technique where a node sends an incoming packet out on every outgoing link except the one it arrived on; each neighbor does the same, so the packet spreads through the network via all possible paths Sending a packet to all nodes/hosts in a network, typically via a single dedicated broadcast address, without hop-by-hop re-forwarding by every intermediate node's own logic Mechanism Achieved implicitly through repeated forwarding at every node (no destination address needed — relies on the network topology itself) Achieved explicitly using a reserved broadcast address (e.g. the IP broadcast address, or a MAC broadcast address FF:FF:FF:FF:FF:FF) that all hosts on a segment recognize and accept Path used Multiple/all paths through the network simultaneously (packet duplication at every hop) Typically a single logical delivery, or hop-by-hop forwarding controlled by the network layer's broadcast/multicast rules Redundancy Very high — the same packet can arrive at a destination multiple times via different paths, requiring duplicate suppression (e.g. sequence numbers, hop counters) Lower — delivery is a single well-defined operation per broadcast domain Guaranteed delivery Yes (given a connected network and enough hop-limit) — since every path is tried, at least one copy is guaranteed to reach the destination, useful for highly robust/military applications Depends on network configuration — broadcast is typically confined to a local network/broadcast domain (routers usually don't forward broadcasts across subnets) Typical use Routing table-free dissemination (e.g. some ad-hoc/military networks), or as a fallback when no routing info is available Address Resolution Protocol (ARP) requests, DHCP discovery, and other LAN-local announcements In short: flooding is a routing/forwarding strategy that relies on redundant path exploration at every node, while broadcasting is a delivery mode aimed at reaching every host on a network, usually confined to a local broadcast domain and controlled by a designated address rather than blind re-forwarding.
Q9d. Compare the following terms: (d) Amplifiers vs Repeaters201914m
Module 1: Physical Layer & Data Communication
View this question on its own page →Compare the following terms: (d) Amplifiers vs Repeaters
Worked SolutionAspect Amplifier Repeater Signal type Boosts analog signals Regenerates digital signals Method Increases the amplitude/strength of the entire incoming waveform, including any noise picked up along the way Detects the original bit pattern and reconstructs a clean, fresh digital signal, discarding accumulated noise Noise handling Noise is amplified along with the signal — degrades over multiple hops Noise does not accumulate — each hop starts fresh Layer Physical layer, used in analog transmission (e.g. old telephone lines) Physical layer, used in digital transmission (e.g. Ethernet, digital telecom) Typical use Long-distance analog communication links LANs and digital transmission lines In short: an amplifier blindly strengthens whatever signal (and noise) arrives, while a repeater reads the digital data and retransmits a clean copy — making repeaters more reliable for digital networks over long distances.
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