DCN Midsem — Mock Exam Papers

Three full-length practice papers, each mixing all six syllabus areas the way a real exam would — not grouped by topic. Attempt a paper closed-book first, then open each answer.

Scope note: every question below sits inside the confirmed syllabus (Forouzan Ch1 1.1–1.6 · Kurose Ch1 1.1–1.5 · Forouzan Ch2 2.1–2.6 except 2.4 · Forouzan Ch3 3.1 / 3.2.1 / 3.3.1–ALOHA-only / 3.3.2). No CSMA(/CD/CA), no CRC/Hamming/Checksum, no HDLC/PPP/ARQ, no ARP, no analog modulation (ASK/FSK/PSK/QAM). Marks split (Part A short, Part B long-with-choice) is a realistic Amrita-style structure for practice — it is not a leaked copy of the real paper, since the actual weightage was never confirmed. Tags mark whether a question rewards byheart recall, understanding, or is a numerical.

Practice Paper 1

Part A: 10 × 2 marks = 20 · Part B: 5 × 8 marks (choice of two per question) = 40 · Total 60

Part A — Short Answer (2 marks each)

  1. List the five components of a data communication system. [2] byheart
    Show answer
    Message, Sender, Receiver, Transmission Medium, Protocol.
  2. State the three elements of a protocol. [2] byheart
    Show answer
    Syntax (structure/format of data), Semantics (meaning of each section), Timing (when to send and how fast).
  3. Differentiate LAN and WAN in one line each. [2] understand
    Show answer
    LAN: privately owned, connects devices in a single office/building/campus, small geographic area. WAN: spans a large geographic area (city/country/globe), often uses third-party carriers to interconnect LANs.
  4. What is meant by traffic intensity in the context of network delay? [2] understand
    Show answer
    Traffic intensity I = (a·L)/R, where a = average packet arrival rate, L = packet length (bits), R = link bandwidth (bps). It must stay below 1 — as I approaches 1, queuing delay grows without bound.
  5. Write Nyquist's formula for the capacity of a noiseless channel. [2] byheart
    Show answer
    C = 2·B·log₂(L), where B = bandwidth in Hz, L = number of signal levels.
  6. At what value of G does Slotted ALOHA reach its maximum throughput, and what is that maximum? [2] byheart
    Show answer
    Maximum throughput = 36.8% (1/e), reached at G = 1.
  7. Name any two causes of signal impairment. [2] byheart
    Show answer
    Any two of: Attenuation (loss of signal energy), Distortion (different frequency components travel at different speeds), Noise (thermal/crosstalk/impulse).
  8. What is byte stuffing and why is it needed? [2] understand
    Show answer
    Inserting an extra ESC byte before a FLAG or ESC byte that occurs naturally inside the data, so the receiver can distinguish real data from the frame-boundary flag pattern.
  9. List the five layers of the TCP/IP protocol suite. [2] byheart
    Show answer
    Application, Transport, Network, Data Link, Physical.
  10. Differentiate FDM and TDM in one line each. [2] understand
    Show answer
    FDM: divides the channel into separate frequency bands; all sources transmit simultaneously, each on its own band. TDM: divides time into slots; sources take turns transmitting in their assigned slot on the full bandwidth.

Part B — Long Answer (8 marks each, answer any one of the two)

  1. Explain the OSI reference model with the function of each of its seven layers. Support with a real-life analogy. [8] understand OR Explain the TCP/IP protocol suite and describe the encapsulation process as data moves from the Application layer down to the Physical layer.
    Show model answer
    OSI (top→bottom): Application (network-facing services, e.g. HTTP/DNS) · Presentation (encryption, compression, format translation) · Session (dialog control, synchronization) · Transport (process-to-process delivery, ports, TCP/UDP) · Network (logical/IP addressing, routing) · Data Link (framing, MAC addressing, media access) · Physical (raw bit transmission over the medium).

    Analogy: Alice writes a letter to Bob. She composes the message (Application), addresses the envelope with Bob's full address (Network layer — stays the same end to end), the postal service routes it through several sorting offices, each relabeling the local delivery instructions at every hop (Data Link — the MAC address changes hop to hop while the IP/postal address does not), and the truck itself just physically carries the paper (Physical).

    Encapsulation (OR answer): Data → Segment (Transport layer adds source/destination port numbers) → Packet (Network layer adds source/destination IP) → Frame (Data Link layer adds source/destination MAC + trailer) → Bits (Physical layer converts to signal). Each layer treats the layer above's output as pure payload and wraps its own header around it without inspecting the contents.
  2. A channel has a bandwidth of 3000 Hz and an SNR of 30 dB. (a) Find the channel capacity using Shannon's theorem. (b) If 8 signal levels are used on a noiseless version of the same channel, find the capacity using Nyquist's formula. [8] numerical OR Compare circuit switching and packet switching, with one real-life example of each.
    Show model answer
    (a) Shannon: SNR(dB)=30 ⇒ SNR(linear) = 1030/10 = 1000.
    C = B·log2(1+SNR) = 3000 × log2(1001) ≈ 3000 × 9.967 ≈ 29,901 bps ≈ 29.9 kbps
    (b) Nyquist: L = 8 ⇒ log2(8) = 3.
    C = 2·B·log2(L) = 2 × 3000 × 3 = 18,000 bps = 18 kbps
    Note the two answers differ because they answer different questions — Shannon gives the noise-imposed ceiling, Nyquist gives what your chosen number of levels actually achieves on a clean channel; the real usable rate is whichever is lower.

    OR answer: Circuit switching reserves a dedicated end-to-end path before data flows (setup phase), guarantees fixed bandwidth, but wastes capacity when idle — e.g. a traditional landline phone call (64 kbps reserved for the whole call). Packet switching sends each packet independently with no reservation, sharing link capacity efficiently for bursty traffic, at the cost of variable delay — e.g. web browsing / WhatsApp messages routed as IP packets.
  3. State and explain the throughput formulas for Pure ALOHA and Slotted ALOHA. Explain why Slotted ALOHA achieves double the maximum throughput of Pure ALOHA. [8] understand OR Explain Reservation, Polling, and Token Passing as controlled-access methods, with one real-life example of each.
    Show model answer
    Pure ALOHA: S = G·e−2G vulnerable time = 2×Tfr max S = 18.4% at G = 0.5 Slotted ALOHA: S = G·e−G vulnerable time = Tfr max S = 36.8% at G = 1.0
    In Pure ALOHA a station transmits the instant it has data, so a collision occurs if any other station starts anywhere within one frame-time before or after — a window of 2×Tfr. Slotted ALOHA forces all stations to start only at synchronized slot boundaries, so a collision can only happen if two stations pick the exact same slot — the vulnerable window shrinks to just Tfr. Halving the vulnerable window doubles the probability of a clean transmission, which is why slotted throughput is exactly double pure throughput at every G.

    OR answer: Reservation — stations reserve a minislot before sending data in the following frame; e.g. VSAT satellite links used by bank branches. Polling — a primary station asks each secondary "Poll" (do you have data?) / "Select" (I have data for you); e.g. old IBM mainframe terminal networks, industrial SCADA systems. Token Passing — a special token circulates among stations and only the current token-holder may transmit; e.g. IBM Token Ring, FDDI.
  4. Explain Mesh, Star, Bus, and Ring topologies with their link-count and port-count formulas. For n = 8 devices, calculate the number of physical links required in a mesh topology. [8] numerical OR Explain transmission delay and propagation delay with a numerical example.
    Show model answer
    Mesh: links = n(n-1)/2, ports = n-1 | Star: links = n, hub ports = n Bus: 1 backbone + n taps + 2 terminators | Ring: n links, 2 ports/device
    Mesh gives maximum fault tolerance (a dedicated line per pair) at the cost of the most cabling; Star is simple to manage but the hub is a single point of failure; Bus is cheap to cable but one break anywhere kills the whole segment; Ring passes data/tokens in one direction and typical implementations use a second counter-rotating ring for self-healing.
    n = 8 ⇒ links = 8×7/2 = 28, ports per device = 7
    OR answer: Transmission delay = L/R (time to push all bits of a packet onto the link — depends on packet size and link rate). Propagation delay = d/s (time for a bit already on the wire to travel the physical distance, at roughly 2×108 m/s). Example: distance = 2500 km, propagation speed = 2×108 m/s, packet = 5000 bits, link rate = 2 Mbps.
    d_prop = 2,500,000 m / 2×10^8 m/s = 12.5 ms d_trans = 5000 bits / 2,000,000 bps = 2.5 ms Total nodal (ignoring proc/queue) = 15 ms
  5. What is bit stuffing? Given the data 0 1111111 0 (a 0, seven consecutive 1s, then a 0), show the stuffed output bit by bit. [8] numerical OR Explain any three line coding schemes among NRZ-L, NRZ-I, Manchester, Differential Manchester, and AMI.
    Show model answer
    Bit stuffing: whenever the sender's data contains five consecutive 1s, it inserts an extra 0 immediately after them, purely so a genuine run of six 1s (the flag pattern 01111110) can never appear by accident inside real data. The receiver reverses this: after any five consecutive 1s it sees, it removes the next 0 automatically.
    Original: 0 1111111 0 (9 bits: one run of seven 1s) Scan the run: 1 1 1 1 1 ← 5th one, insert stuffed 0 here → 1 1 (remaining 2 ones pass through) Stuffed: 0 11111 0 11 0 = 0111110110 (10 bits)
    OR answer (pick any three): NRZ-L: 0 = high voltage, 1 = low voltage (level itself encodes the bit). NRZ-I: 1 = a transition at the start of the bit interval, 0 = no transition. Manchester: a transition always occurs at the middle of every bit (self-clocking); 0 = high-to-low, 1 = low-to-high. Differential Manchester: always has a mid-bit transition for clocking, but the bit value is shown by whether there's an extra transition at the start of the interval (0 = extra transition, 1 = none). AMI: 0 = zero voltage, 1 = alternating positive/negative voltage.

Practice Paper 2

Part A: 10 × 2 marks = 20 · Part B: 5 × 8 marks (choice of two per question) = 40 · Total 60

Part A — Short Answer (2 marks each)

  1. State the four criteria used to judge the effectiveness of a network. [2] byheart
    Show answer
    Performance, Reliability, Security, and (for data communication effectiveness specifically) Delivery, Accuracy, Timeliness, Jitter — the standard set taught is Delivery, Accuracy, Timeliness, Jitter.
  2. Give the formula for propagation delay and name each term. [2] byheart
    Show answer
    d_prop = d/s, where d = distance the signal must travel, s = propagation speed of the medium (≈2×108 m/s in copper/fiber).
  3. What is the vulnerable time for Pure ALOHA, and why is it that value? [2] understand
    Show answer
    2×Tfr — a collision occurs if any other station starts transmitting anywhere from one full frame-time before to one full frame-time after your own transmission.
  4. Write Shannon's capacity formula for a noisy channel. [2] byheart
    Show answer
    C = B·log₂(1+SNR), where B = bandwidth in Hz, SNR = signal-to-noise ratio (linear, not dB).
  5. Define attenuation. [2] byheart
    Show answer
    Loss of signal energy/strength as it travels through a medium over distance; compensated for using amplifiers/repeaters.
  6. What rule does bit stuffing follow, and what pattern is being avoided? [2] byheart
    Show answer
    Insert a 0 after any five consecutive 1s in the data, so the six-1s flag pattern (01111110) marking a frame boundary can never appear accidentally inside real data.
  7. Differentiate Star and Bus topology in terms of fault tolerance. [2] understand
    Show answer
    Star: only the failing device's own link is affected — other devices keep working (unless the central hub itself fails, which takes everything down). Bus: any break in the single shared backbone cable takes the entire segment down.
  8. Give the mnemonic used to remember the OSI layers top to bottom. [2] byheart
    Show answer
    "All People Seem To Need Data Processing" — Application, Presentation, Session, Transport, Network, Data Link, Physical.
  9. What is meant by "packet switching," in one line? [2] understand
    Show answer
    Each packet is routed independently through the network with no dedicated path reserved in advance, sharing link capacity efficiently across many flows.
  10. Name the two multiplexing techniques covered and one example application of each. [2] understand
    Show answer
    FDM — cable TV (each channel on its own frequency band). TDM — T1 phone trunks (24 calls cycling through fixed repeating time slots).

Part B — Long Answer (8 marks each, answer any one of the two)

  1. A noiseless channel has a bandwidth of 2 MHz and uses 64 signal levels. Find the maximum data rate using Nyquist's formula. [8] numerical OR Explain the concept of protocol layering and why networks are built as a stack of layers rather than one monolithic system.
    Show model answer
    L = 64 ⇒ log2(64) = 6 C = 2·B·log2(L) = 2 × 2,000,000 × 6 = 24,000,000 bps = 24 Mbps
    OR answer: Layering separates concerns — each layer solves exactly one problem (e.g. Physical only pushes bits, Data Link only gets a frame across one link, Network only handles end-to-end addressing/routing) and offers a clean, fixed interface to the layer above/below it. This lets each layer's technology evolve independently (you can swap Wi-Fi for Ethernet without touching the Transport or Application layer), makes the system easier to design, test, and troubleshoot, and lets protocols at different layers be mixed and matched (e.g. the same IP layer running over many different Data Link technologies).
  2. Given the data 11111 0 111111 0 (5 ones, a 0, 6 ones, a 0 — 13 bits total), apply bit stuffing and show the final output. [8] numerical OR Explain the difference between attenuation, distortion, and noise, each with a real-life cause.
    Show model answer
    Original (13 bits): 11111 0 111111 0 Run 1 (5 ones): 11111 → insert stuffed 0 after the 5th one → 111110, then original 0 follows → 1111100 Run 2 (6 ones): after the 5th one, insert stuffed 0, remaining 1 one passes through, then original 0 follows 11111 [stuff 0] 1 0 → 11111010 Stuffed output (15 bits): 1111100 11111010 = 111110011111010
    Two stuffed 0s were inserted (one per run that reached five consecutive 1s), so the output grew from 13 to 15 bits.

    OR answer: Attenuation — signal weakens with distance, e.g. a long unamplified cable run losing strength. Distortion — different frequency components of a composite signal travel at slightly different speeds and arrive out of phase, e.g. signal smearing over a long analog line. Noise — unwanted energy from an external source corrupts the signal, e.g. crosstalk between adjacent cables or thermal noise in the receiver electronics.
  3. Compare Pure ALOHA and Slotted ALOHA on: mechanism, vulnerable time, and maximum throughput. If G = 1 in Pure ALOHA, calculate the actual throughput and compare it to the maximum achievable throughput of Pure ALOHA. [8] numerical OR Explain circuit switching with a real-life example, including what happens during its setup phase.
    Show model answer
    Pure ALOHA: stations transmit whenever ready, vulnerable time = 2Tfr, max S = 18.4% at G=0.5 Slotted ALOHA: stations transmit only at slot boundaries, vulnerable time = Tfr, max S = 36.8% at G=1.0 At G=1 (Pure ALOHA): S = G·e^(-2G) = 1 × e^-2 = 1 × 0.1353 = 0.1353 = 13.5%
    13.5% is lower than Pure ALOHA's own maximum of 18.4% — this shows the maximum only occurs at the optimal load G=0.5; pushing more traffic (G=1) past that point increases collisions faster than it increases successful sends, so throughput falls.

    OR answer: Circuit switching first runs a setup phase that reserves a dedicated path with fixed bandwidth end-to-end before any data is sent (e.g. dialing a traditional landline call — the exchange reserves a 64 kbps path for the full duration of the call, whether or not anyone is speaking); after setup, data flows with guaranteed bandwidth and no further routing decisions; a teardown phase releases the reserved resources when the call ends.
  4. For a star topology with n = 12 devices, state the number of links and the number of ports required at the hub and at each device. Then explain why Star is easier to troubleshoot than Bus. [8] numerical OR Explain the five components of a data communication system with a real-life example tying all five together.
    Show model answer
    n = 12 ⇒ links = n = 12, hub ports = n = 12, each device needs 1 port
    Star is easier to troubleshoot because every device has its own dedicated link back to the hub — a fault shows up as exactly one dead link, and testing/isolating it doesn't disturb any other device. In Bus, all devices share one physical cable, so a fault anywhere on it (or a bad connector/tap) can silently affect the entire segment, and locating the exact break usually means physically testing cable sections.

    OR answer: Message = the information (e.g. an email's text), Sender = the device that creates and sends it (a laptop), Receiver = the device that gets it (a phone), Transmission Medium = the physical path it travels (Wi-Fi radio waves / fiber backbone), Protocol = the agreed rules both ends follow (e.g. SMTP/TCP/IP) so the message is interpreted correctly at the other end.
  5. Explain the traffic intensity formula and compute it for a link with average packet arrival rate a = 1000 packets/sec, packet length L = 1500 bytes, and link capacity R = 150 Mbps. Comment on the result. [8] numerical OR Explain Reservation and Polling as controlled-access methods, including what happens during a "Select" and a "Poll" in polling.
    Show model answer
    L = 1500 bytes = 12,000 bits I = (a·L)/R = (1000 × 12,000) / 150,000,000 = 12,000,000 / 150,000,000 = 0.08
    I = 0.08 is well below 1, meaning the link is lightly loaded — packets arrive far slower than the link can drain them, so queuing delay stays small. As I approaches 1, the queue (and delay) would grow without bound.

    OR answer: Reservation: time is divided into intervals; each interval begins with a reservation frame made of minislots (one per station) where a station can flag that it wants to send, and the primary then grants data slots to only the stations that reserved. Polling: works with one primary and several secondaries; the primary sends a Select frame when it has data for a secondary (to alert/wake it up before sending), and sends a Poll frame to ask a secondary whether it has data to send.

Practice Paper 3

Part A: 10 × 2 marks = 20 · Part B: 5 × 8 marks (choice of two per question) = 40 · Total 60 — slightly harder, integrative mix

Part A — Short Answer (2 marks each)

  1. A noiseless channel has bandwidth 1000 Hz and capacity 6000 bps. How many signal levels are being used? [2] numerical
    Show answer
    C=2B·log2(L) ⇒ 6000=2×1000×log2(L) ⇒ log2(L)=3 ⇒ L=2³=8 levels.
  2. What does "jitter" mean as a network effectiveness criterion? [2] byheart
    Show answer
    Variation in the delay of arrival of packets belonging to the same flow — even if average delay is fine, uneven arrival spacing (jitter) degrades real-time traffic like voice/video.
  3. In Slotted ALOHA, what does it mean for a station to transmit at the "beginning of a time slot"? [2] understand
    Show answer
    All stations are synchronized to a common clock dividing time into fixed-length slots equal to one frame's transmission time; a station with data must wait until the next slot boundary to begin sending, rather than sending the instant it becomes ready.
  4. For a mesh network with n=6 devices, how many I/O ports does each device need? [2] numerical
    Show answer
    ports = n-1 = 5.
  5. Name the three types of network by geographic scope discussed. [2] byheart
    Show answer
    LAN (Local Area Network), WAN (Wide Area Network), and the Internet (a network of networks).
  6. Differentiate NRZ-L and NRZ-I in one line. [2] understand
    Show answer
    NRZ-L encodes the bit value directly as a voltage level (0=high, 1=low); NRZ-I encodes it as the presence/absence of a transition at the start of the interval (1=transition, 0=no transition), independent of absolute voltage level.
  7. What is the primary drawback of TDM compared to FDM when a source has no data to send? [2] understand
    Show answer
    In (synchronous) TDM, a source's reserved time slot goes unused (wasted) if it has nothing to send at that moment, since slots are fixed per source rather than shared dynamically.
  8. Give the token-passing rule for medium access. [2] byheart
    Show answer
    A special control frame (the token) circulates among stations in a defined order; only the station currently holding the token is permitted to transmit data.
  9. What is meant by "nodal delay," and list its four components. [2] byheart
    Show answer
    The total delay experienced at one node (router/switch) before a packet leaves toward the next hop: processing delay + queuing delay + transmission delay + propagation delay.
  10. Why can real data never legitimately contain six consecutive 1s after bit stuffing has been applied? [2] understand
    Show answer
    Because the stuffing rule forces a 0 to be inserted the instant five consecutive 1s occur, capping every run in the transmitted data at five — so a genuine run of six 1s can only ever be the flag pattern (01111110) marking a frame boundary.

Part B — Long Answer (8 marks each, answer any one of the two)

  1. A link has bandwidth 4000 Hz and SNR(dB) = 24 dB. (a) Find capacity using Shannon. (b) If this capacity is to be achieved using Nyquist with the smallest possible integer number of signal levels, estimate that L. [8] numerical OR Explain packet switching and datagram delivery, contrasting it with the reserved-path nature of circuit switching.
    Show model answer
    SNR(linear) = 10^(24/10) = 10^2.4 ≈ 251 C = B·log2(1+SNR) = 4000 × log2(252) ≈ 4000 × 7.98 ≈ 31,900 bps Nyquist: C = 2B·log2(L) ⇒ log2(L) = C/(2B) = 31,900/8000 ≈ 3.99 ≈ 4 ⇒ L = 2^4 = 16 levels (rounding up to the nearest achievable integer level count)
    OR answer: In packet switching, each packet carries its own destination address and is routed independently hop by hop as a self-contained datagram — no path is reserved beforehand, so packets belonging to the same message may take different routes and arrive out of order. This is efficient for bursty traffic since idle sources don't waste reserved bandwidth, unlike circuit switching where a fixed path (and its bandwidth) is locked for the entire session regardless of whether data is actively flowing.
  2. Explain why Slotted ALOHA's throughput formula is S = G·e-G while Pure ALOHA's is S = G·e-2G, connecting the difference directly to their vulnerable-time windows. At G = 2, compute the throughput for both. [8] numerical OR Explain the OSI and TCP/IP models' relationship — which OSI layers does TCP/IP's Application layer absorb, and why?
    Show model answer
    The e-kG term models the probability that no other station transmits during the vulnerable window; since Pure ALOHA's window (2Tfr) is twice as wide as Slotted ALOHA's (Tfr), the exponent carries a factor of 2 for Pure ALOHA and 1 for Slotted ALOHA, directly reflecting "twice the exposure to collision, per unit of offered load G."
    At G=2: Pure: S = 2 × e^-4 = 2 × 0.0183 = 0.0367 = 3.67% Slotted: S = 2 × e^-2 = 2 × 0.1353 = 0.2707 = 27.07%
    Both are already well past their individual optimum G (0.5 and 1 respectively), so both are declining from their peaks, but Pure ALOHA collapses far more sharply because its wider vulnerable window punishes higher load much more severely.

    OR answer: TCP/IP's single Application layer absorbs OSI's Presentation and Session layers because, in practice, formatting/encryption (Presentation) and dialog/session management (Session) ended up being handled inside application software itself (e.g. TLS and session cookies are implemented by the application/library, not by a separate universal network layer) — TCP/IP was designed pragmatically from working protocols rather than as a theoretical seven-layer model, so it only keeps the layer boundaries that mapped to real, separately-implemented pieces of the stack.
  3. Explain bus and ring topology, including what happens to the network when one device fails in each case. Which requires termination and why? [8] understand OR Explain the concept of encapsulation and decapsulation with an example of a web request travelling from a browser to a server.
    Show model answer
    Bus: all devices tap into one shared backbone cable; if a device fails, the rest of the network is unaffected since each tap is a passive connection — but if the cable itself breaks, the whole segment goes down. The two ends of the backbone need terminators to absorb signal and prevent reflections bouncing back down the wire and corrupting data.
    Ring: devices are connected in a closed loop, each repeating the signal to the next; if one device (or its ring interface) fails and there's no redundancy, the entire ring is broken because the signal path is a single closed chain — this is why practical ring networks (e.g. FDDI) often use a second counter-rotating ring for self-healing. Ring needs no separate terminators since the loop is inherently closed.

    OR answer: A browser request: Application layer creates the HTTP GET request; Transport layer wraps it in a TCP segment with source/destination ports (encapsulation); Network layer wraps that in an IP packet with source/destination IP addresses; Data Link layer wraps that in a frame with MAC addresses for the next hop; Physical layer sends raw bits. At the server, each layer strips (decapsulates) exactly the header its peer layer added, until the original HTTP request is recovered — each layer only ever reads the header meant for it, never the payload above.
  4. A mesh network needs to support n = 10 devices. Find the number of links and ports required. If the design were changed to a star topology instead, restate both values and explain the main practical trade-off between the two choices. [8] numerical OR Explain any two line coding schemes not covered in Practice Paper 1's Part B, and state one real device/standard that uses each.
    Show model answer
    Mesh, n=10: links = 10×9/2 = 45, ports/device = 9 Star, n=10: links = 10, hub ports = 10, device ports = 1 each
    Trade-off: Mesh needs far more cabling and ports (45 links vs 10) but survives any single link failure without affecting other pairs — no single point of failure. Star needs drastically less cabling and simpler device hardware (1 port each), but the entire network depends on the central hub; if the hub fails, every device loses connectivity.

    OR answer (example — Differential Manchester & AMI): Differential Manchester always has a transition at mid-bit for self-clocking, and encodes the bit value using whether there's an additional transition at the start of the bit interval (0=extra transition, 1=none) — used in classic Token Ring (IEEE 802.5) networks. AMI (Alternate Mark Inversion) represents 0 as zero voltage and 1 as alternating positive/negative voltage pulses — used in T1 telephone trunk lines.
  5. Explain the two forms of framing (bit stuffing and byte stuffing) and state which is used with bit-oriented protocols versus byte-oriented protocols. [8] understand OR A packet of 2000 bits is sent over a 500 kbps link across a distance of 1000 km at propagation speed 2×108 m/s. Find the total delay (transmission + propagation only).
    Show model answer
    Bit stuffing operates at the level of individual bits — it's used with bit-oriented protocols where the frame boundary flag is a specific bit pattern (01111110); a 0 is stuffed after every five consecutive 1s in the data. Byte stuffing operates at the level of whole bytes — used with byte-oriented protocols where the flag is a specific reserved byte value; an ESC byte is inserted before any byte in the data that accidentally matches the FLAG or ESC byte value. Both exist for the same reason at different granularities: guaranteeing the receiver can never mistake real data for a frame-boundary marker.

    OR answer:
    d_trans = L/R = 2000 / 500,000 = 0.004 s = 4 ms d_prop = d/s = 1,000,000 / 2×10^8 = 0.005 s = 5 ms Total = 4 + 5 = 9 ms