BSc CSIT (TU) Science BIT Network and Data Communications (BIT254) – 4th Semester (Model) Question Paper 2079 Nepal
This is the official BSc CSIT (TU) (Science stream) BIT Network and Data Communications (BIT254) – 4th Semester (Model) question paper for 2079, as set in the model model examination. It carries 60 full marks and a time allowance of 180 minutes, across 12 questions. On Kekkei you can attempt this BIT Network and Data Communications (BIT254) – 4th Semester (Model) past paper online with a timer, get instant AI feedback and step-by-step solutions, and track the topics where you lose marks — completely free. Whether you are revising for your BSc CSIT (TU) BIT Network and Data Communications (BIT254) – 4th Semester (Model) exam or solving previous years' question papers, this 2079 paper is a great way to practise under real exam conditions.
Section A
Attempt any two questions. (2 × 10 = 20)
Explain TDMA, FDMA and WDMA with necessary diagrams. Represent bit sequence 100100111 by the following waveform: [6+4]
a. NRZ-L b. NRZ-I c. Manchester d. Differential Manchester
Channelization / multiple-access techniques (6 marks)
Channelization is a multiple-access method in which the available bandwidth of a link is shared in time, frequency or through code among different stations.
1. FDMA (Frequency Division Multiple Access) The total available bandwidth is divided into non-overlapping frequency bands, and each station is permanently allocated one band. All stations can transmit simultaneously but on different frequencies. Guard bands separate adjacent channels to prevent interference.
Frequency
^ | st4 |
| | st3 |
| | st2 | (each station = a fixed frequency band, all at the same time)
| | st1 |
+---------------------> Time
2. TDMA (Time Division Multiple Access) The whole bandwidth is just one channel shared in time. Each station is allocated a repeating time slot during which it has the entire bandwidth. Stations transmit one after another in their slots; synchronization (guard times) is required.
Frequency
^ |st1|st2|st3|st4|st1|st2|...
| +------------------------> Time
(each station = a fixed time slot, full bandwidth)
3. WDMA (Wavelength Division Multiple Access) The optical-fibre equivalent of FDMA. The high optical bandwidth of a fibre is divided into several wavelengths (colours of light), and each station transmits on its own wavelength. A multiplexer (prism/grating) combines the wavelengths onto one fibre and a demultiplexer separates them at the receiver, allowing many simultaneous high-speed optical channels.
Comparison
| Technique | Shares | Domain | Medium |
|---|---|---|---|
| FDMA | Frequency bands | Frequency | Radio/electrical |
| TDMA | Time slots | Time | Any |
| WDMA | Wavelengths | Frequency (optical) | Optical fibre |
Line coding of 100100111 (4 marks)
Conventions: NRZ-L — bit value sets the level (here 1 = high, 0 = low). NRZ-I — invert the level on a 1, no change on a 0. Manchester — mid-bit transition: 0 = high→low, 1 = low→high. Differential Manchester — always a mid-bit transition; additionally a transition at the start of the bit for 0 and no start transition for 1.
| Bit | 1 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 1 |
|---|---|---|---|---|---|---|---|---|---|
| NRZ-L (level) | H | L | L | H | L | L | H | H | H |
| NRZ-I (level, start high) | L | L | L | H | H | H | L | H | L |
| Manchester (mid transition) | L→H | H→L | H→L | L→H | H→L | H→L | L→H | L→H | L→H |
| Diff. Manchester (mid always) | start:no, mid flip | start:yes, mid flip | start:yes, mid flip | start:no, mid flip | start:yes, mid flip | start:yes, mid flip | start:no, mid flip | start:no, mid flip | start:no, mid flip |
For NRZ-I, starting from a low reference: the first 1 inverts the level, each subsequent 1 inverts again, and 0s hold the previous level — giving the level pattern shown. The Manchester and Differential Manchester rows show the within-bit transition for each bit; both are self-clocking codes.
Explain leaky-bucket algorithm. Illustrate it with an example. What are its advantages and disadvantages? [4+3+3]
Leaky-bucket algorithm (4 marks)
The leaky-bucket algorithm is a traffic-shaping technique used for congestion control. The analogy is a bucket with a small hole in the bottom: no matter the rate at which water (data) is poured in, it leaks out at a constant rate, and if the bucket is full any extra water overflows and is lost.
In networking the bucket is a finite FIFO queue (buffer):
- Incoming packets are placed in a queue (the bucket) of fixed capacity.
- The network removes/sends packets from the queue at a fixed (constant) output rate, regardless of how bursty the input is.
- If the queue is full when a packet arrives, the packet is discarded.
Thus a bursty input stream is converted into a smooth, fixed-rate output stream. Implementation uses a counter/queue: for a byte-counting version, a counter is initialized to the number of bytes allowed per tick and packets are sent only while the counter permits.
Example (3 marks)
Suppose a host sends data in bursts but the network can accept only 2 packets per clock tick.
| Tick | Input (arrivals) | Bucket queue after input | Output (leak rate = 2) | Remaining in bucket |
|---|---|---|---|---|
| 1 | 5 packets | 5 | 2 | 3 |
| 2 | 1 packet | 4 | 2 | 2 |
| 3 | 0 packets | 2 | 2 | 0 |
A bursty input of 5,1,0 packets is smoothed into a steady output of 2,2,2 packets per tick. (If the bucket capacity were, say, 4, the 5th packet in tick 1 would overflow and be dropped.)
Mathematically, if the bucket sends data at rate for a maximum burst of over time , the maximum output is , enforcing a constant average rate.
Advantages (1.5 marks)
- Smooths bursty traffic into a constant-rate output, protecting the network from congestion.
- Simple to implement (a FIFO queue with a fixed drain rate).
- Provides a hard upper bound on the average data rate sent into the network.
Disadvantages (1.5 marks)
- The output rate is rigidly fixed, so it cannot take advantage of idle network capacity (no bursts allowed even when the network is free) — this is inflexible compared with the token-bucket algorithm.
- When the bucket is full, packets are simply discarded, which can cause data loss for legitimate bursty applications.
Explain the procedure for pure ALOHA. Highlight on the vulnerable time of pure ALOHA. A pure ALOHA network transmits 200-bit frames on a shared channel of 200 kbps. What is the requirement to make this frame collision-free? [5+3+2]
Procedure for pure ALOHA (5 marks)
Pure ALOHA is the original random-access (MAC) protocol in which any station may transmit whenever it has data, with no coordination.
- Whenever a station has a frame, it transmits it immediately on the shared channel.
- Because stations transmit at will, two or more frames may overlap in time, causing a collision, and all colliding frames are destroyed.
- After sending a frame the station waits for an acknowledgment for a time-out period (about , twice the maximum propagation time).
- If no acknowledgment arrives, the station assumes a collision occurred. It waits a random back-off time and then retransmits.
- To avoid repeated collisions, the back-off uses a random multiple of the maximum propagation time (e.g. binary exponential back-off), and after a maximum number of attempts () the station aborts.
The maximum throughput of pure ALOHA is
Vulnerable time (3 marks)
The vulnerable time is the period during which no other station may transmit if the current frame is to survive without collision. For pure ALOHA a frame can collide with a frame that started just before it or just after it, so the vulnerable time is two frame transmission times:
where is the time to transmit one frame. (For slotted ALOHA this reduces to one .)
Numerical: collision-free requirement (2 marks)
Given: frame size bits, channel rate bps.
Frame transmission time:
Vulnerable time:
Requirement: To keep this frame collision-free, no other station must transmit during the 2 ms vulnerable window — i.e. the channel must remain idle for around the frame's transmission. Equivalently, the frame rate must be low enough that, on average, less than one frame is generated per 2 ms vulnerable period.
Section B
Attempt any eight questions. (8 × 5 = 40)
Explain different layers of OSI. [5]
The OSI (Open Systems Interconnection) reference model, defined by ISO, divides network communication into seven layers, each with a specific function:
| # | Layer | Main function | Example unit / protocol |
|---|---|---|---|
| 7 | Application | Provides network services directly to user applications | HTTP, FTP, SMTP (data) |
| 6 | Presentation | Translation, encryption/decryption and compression of data; ensures common format | JPEG, ASCII, SSL |
| 5 | Session | Establishes, manages, synchronizes and terminates sessions between applications | NetBIOS, RPC |
| 4 | Transport | End-to-end delivery, segmentation, flow control and error control | TCP, UDP (segment) |
| 3 | Network | Logical addressing and routing of packets across networks | IP (packet) |
| 2 | Data Link | Framing, physical (MAC) addressing, error and flow control on a link | Ethernet (frame) |
| 1 | Physical | Transmits raw bits over the medium; defines voltages, cables, connectors | bits, cables, hubs |
- The upper layers (5–7) deal with application-level concerns and are usually implemented in software.
- The lower layers (1–4) handle data transport and are partly implemented in hardware.
- At the sender, data flows down through the layers (each adding a header — encapsulation); at the receiver it flows up (each removing its header — decapsulation).
Differentiate between circuit switching and packet switching. What are their practical implications? [3+2]
Difference between circuit switching and packet switching (3 marks)
| Basis | Circuit switching | Packet switching |
|---|---|---|
| Path | A dedicated physical path is set up for the whole session | No dedicated path; each packet routed independently |
| Phases | Setup, data transfer, teardown | No setup; store-and-forward |
| Resource use | Bandwidth reserved (wasted when idle) | Bandwidth shared on demand (efficient) |
| Data unit | Continuous bit stream | Packets with headers |
| Delay | Constant delay after setup; setup delay first | Variable delay (queuing) per packet |
| Reliability | Path fixed; fails if link breaks | Robust; packets reroute around failures |
| Example | Traditional telephone network (PSTN) | Internet (IP), X.25, frame relay |
Practical implications (2 marks)
- Circuit switching guarantees constant bandwidth and low, predictable delay, so it suits real-time, continuous traffic such as voice calls; but it wastes capacity during silence and scales poorly.
- Packet switching maximizes utilization by statistically multiplexing many flows over shared links, so it suits bursty data traffic (web, email, file transfer); but it introduces variable delay and possible packet loss, requiring higher-layer protocols (TCP) for reliability. This efficiency and fault-tolerance is why the Internet is built on packet switching.
Briefly explain satellite network. Consider a noiseless channel with a bandwidth of 3400 Hz transmitting a signal with two signal levels. Calculate the maximum bit rate. [3+2]
Satellite network (3 marks)
A satellite network uses one or more communication satellites in orbit as relay stations (repeaters in the sky) to provide long-distance and wide-area communication.
- An earth station transmits a signal on an uplink frequency to the satellite.
- The satellite's transponder receives, amplifies and shifts the frequency, then retransmits the signal back to earth on a downlink frequency.
- Satellites are classified by orbit: GEO (geostationary, ~35,786 km, fixed over one point), MEO (medium earth orbit, e.g. GPS), and LEO (low earth orbit, e.g. Iridium, Starlink).
- Uses: television broadcasting, telephony, GPS/navigation, weather monitoring, military communication and Internet access to remote areas.
- Advantages: very wide coverage (a single GEO satellite covers ~1/3 of the earth) and reach to remote/inaccessible regions. Drawbacks: high propagation delay (≈250 ms one-way for GEO) and high cost.
Maximum bit rate — Nyquist (noiseless channel) (2 marks)
For a noiseless channel the Nyquist theorem gives:
where = bandwidth and = number of signal levels.
Given and levels:
Maximum bit rate = 6800 bps (6.8 kbps).
Explain about ARP operation. [5]
ARP (Address Resolution Protocol) maps a known logical (IP) address to its corresponding physical (MAC) address on the same local network. A host needs the destination MAC address to build a data-link frame, but it usually knows only the IP address — ARP resolves this.
Operation:
- A host (sender) wants to send a packet to an IP address on its LAN but does not know the receiver's MAC address.
- It first checks its ARP cache (a table of recently learned IP→MAC pairs). If found, it uses the cached MAC and stops.
- If not found, the sender broadcasts an ARP request frame on the LAN. The request contains the sender's IP and MAC and the target IP address, with the target MAC field empty; the destination MAC is the broadcast address
FF:FF:FF:FF:FF:FF. - Every host on the LAN receives the broadcast and compares the target IP with its own. All hosts except the intended one discard it.
- The host whose IP matches sends back a unicast ARP reply directly to the requester, containing its MAC address.
- The sender stores the resolved IP→MAC mapping in its ARP cache (with a timer for aging) and now transmits the data frame to that MAC address.
Cached entries expire after a timeout so that changes (e.g. a replaced NIC) are eventually relearned. ARP works only within a single broadcast domain; to reach a remote host the sender resolves the MAC of its default gateway instead.
Describe user datagram format. List at least four ports used with UDP. [4+1]
User datagram (UDP) format (4 marks)
UDP is a connectionless, unreliable transport protocol. A user datagram has a fixed 8-byte header followed by the data, with four 16-bit fields:
| Field | Size | Description |
|---|---|---|
| Source port number | 16 bits | Port of the sending application (0–65535) |
| Destination port number | 16 bits | Port of the receiving application |
| Total length | 16 bits | Length of the entire user datagram (header + data) in bytes; min 8, max 65535 |
| Checksum | 16 bits | Optional error detection over a pseudo-header, the UDP header and data |
0 15 16 31
+------------+------------+
| Source Port| Dest. Port |
+------------+------------+
| Length | Checksum |
+------------+------------+
| Data ... |
+-------------------------+
Because the header is small and there is no connection setup, flow control or retransmission, UDP has very low overhead and is fast — suitable for real-time and query/response traffic.
Well-known ports used with UDP (1 mark)
| Port | Service |
|---|---|
| 53 | DNS (Domain Name System) |
| 67 / 68 | DHCP (server / client) |
| 69 | TFTP (Trivial File Transfer Protocol) |
| 161 | SNMP |
| 520 | RIP (Routing Information Protocol) |
(Any four of the above are acceptable, e.g. 53, 69, 161, 520.)
Briefly explain recursive resolution, iterative resolution and caching. [5]
These are the techniques used by the DNS (Domain Name System) to resolve a domain name into an IP address.
1. Recursive resolution The client (resolver) sends the query to its local DNS server and demands a complete answer. The local server then takes full responsibility: it queries other DNS servers on the client's behalf (root → TLD → authoritative) and only returns the final resolved IP address (or an error) to the client. The client makes one request and gets the final result — the burden of chasing the answer falls on the server.
Client ---query---> Local DNS (resolves on client's behalf) ---> returns final IP
2. Iterative resolution The client itself does the work step by step. If a queried server does not know the answer, it does not ask others; instead it returns a referral — the address of the next server to ask (e.g. "ask this TLD server"). The client then queries that next server, and so on, until it reaches the authoritative server that gives the final IP. Each server answers with the best it knows.
Client -> Root (referral to .com) -> TLD (referral to authoritative) -> Authoritative (IP)
3. Caching Resolving a name from scratch is slow, so every DNS server stores (caches) the mappings it learns. When a server receives a query whose answer it has cached, it returns the cached record immediately without contacting other servers. Each record carries a TTL (Time To Live) that limits how long it may be cached; when the TTL expires the entry is removed so stale data is not served. Caching greatly reduces DNS traffic and lookup time.
List one difference between open-loop and closed-loop congestion control. Explain acknowledgment policy and discarding policy. [1+4]
Difference between open-loop and closed-loop congestion control (1 mark)
| Open-loop congestion control | Closed-loop congestion control |
|---|---|
| Prevents congestion before it happens (preventive); does not use feedback from the network. | Removes/alleviates congestion after it has occurred (reactive), using feedback from the network. |
(Open-loop policies are applied at the source/destination in advance; closed-loop policies react to detected congestion.)
Acknowledgment policy (open-loop) (2 marks)
The acknowledgment policy imposed by the receiver can affect congestion. If a receiver acknowledges every packet it receives, the extra ACK traffic itself adds load to the network. To reduce congestion, the receiver may acknowledge only N packets at a time (cumulative/selective acknowledgment) or send an ACK only when it has a packet to send or a timer expires. Fewer acknowledgments mean less traffic injected into the network, helping to prevent congestion. Thus the acknowledgment policy is an open-loop (preventive) mechanism.
Discarding policy (2 marks)
A good discarding policy by routers can prevent congestion while limiting damage to data integrity. When buffers begin to fill, a router selectively discards less-sensitive packets rather than important ones. For example, in audio/video transmission a router may drop low-priority frames (e.g. some less critical packets) while preserving the packets needed to keep the stream intelligible. By dropping the right packets early, the discarding policy reduces the load and prevents congestion from worsening without seriously degrading quality — making it an open-loop preventive policy.
What is MAC-address? The message sequence is 11001000 and generator polynomial . Calculate the transmitted encoded frame. [2+3]
MAC address (2 marks)
A MAC (Media Access Control) address is the unique physical/hardware address burned into a network interface card (NIC) by the manufacturer. It is 48 bits (6 bytes) long, usually written as 12 hexadecimal digits (e.g. 00:1A:2B:3C:4D:5E). The first 3 bytes are the OUI (manufacturer ID) and the last 3 bytes are the device serial number. It operates at the data-link layer and is used to deliver frames to the correct device within a local network.
CRC encoded frame (3 marks)
Generator bit pattern 1011 (degree 3, so ).
Message . Append zeros: dividend .
Perform modulo-2 (XOR) division by 1011:
11001000000 ÷ 1011
11001000000
1011
----
0111100... step-by-step XOR (binary long division)
Working it out:
| Step | Current bits | XOR with | Result |
|---|---|---|---|
| 1 | 1100 | 1011 | 0111 |
| 2 | bring down → 1110 | 1011 | 0101 |
| 3 | bring down → 1011 | 1011 | 0000 |
| 4 | bring down → 0000 | 0000 (no divide, leading 0) | 0000 |
| 5 | bring down → 0000 | 0000 | 0000 |
| 6 | bring down → 0000 | 0000 | 000 |
The remainder (last 3 bits) is 000 (the message is exactly divisible by the generator here).
CRC (remainder) = 000.
Transmitted encoded frame = message + CRC = 11001000 000 = 11001000000.
At the receiver, dividing the received 11001000000 by 1011 again should give a zero remainder if no error occurred.
Write short notes on: [2 × 2.5 = 5]
a. Attenuation b. SNMP
a. Attenuation (2.5 marks)
Attenuation is the loss of signal strength (energy) as a signal travels through a transmission medium. As distance increases, the amplitude of the signal weakens because the medium offers resistance and converts part of the signal energy into heat.
- It is measured in decibels (dB): , where is the input power and the output power. A negative dB value indicates loss.
- Attenuation increases with distance and frequency and limits the maximum length of a cable segment.
- To compensate for it, amplifiers (for analog signals) or repeaters/regenerators (for digital signals) are placed along the line to restore signal strength.
b. SNMP (2.5 marks)
SNMP (Simple Network Management Protocol) is an application-layer protocol used to monitor and manage network devices (routers, switches, servers, printers) from a central station.
- It follows a manager–agent model: a manager (network management station) communicates with agents running on managed devices.
- Each agent maintains a database of variables called the MIB (Management Information Base), whose objects are named using SMI (Structure of Management Information) and OIDs.
- Main operations: GET (read a variable), SET (change a variable), GET-NEXT/GET-BULK (read multiple), and TRAP (an agent's unsolicited alert to the manager about an event).
- It runs over UDP (ports 161 for requests and 162 for traps), making management lightweight and efficient.
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