(SEM VI) THEORY EXAMINATION 2024-25 COMPUTER NETWORKS

BCS603 – COMPUTER NETWORKS

Time: 3 Hours | Max Marks: 70

SECTION A – Short Answer Questions

(2 × 7 = 14 marks | Attempt ALL)

Write formula / definition + 1–2 clear points

a. Links in Fully Connected Mesh Topology                                     For n nodes, number of links:

Links=n(n−1)2\text{Links} = \frac{n(n-1)}{2}Links=2n(n−1)​             For n = 10 → 45 links

b. Received Power after Attenuation

Given:                                                                                               Initial power = 2 W, Attenuation = −3 dB

−3 dB means power becomes half                                                 Pr=1 WP_r = 1 \text{ W}Pr​=1 W 

c. When Contention-Based MAC Protocols Are Suitable                Light traffic

Burst data transmission                                                                   When few nodes transmit simultaneously
Example: CSMA/CD, CSMA/CA

d. Piggybacking

Piggybacking is the technique of attaching acknowledgment (ACK) with outgoing data frames to improve channel efficiency.

e. Functions of OSI Layers                                                         Physical – transmission of bits

Data Link – framing, error detection                                         Network – routing

Transport – end-to-end delivery                                               Session – session control

Presentation – encryption, compression                                   Application – user interface

(Draw neat OSI diagram in exam)

f. Congestion Control Policies                                                  Admission control

Traffic shaping                                                                          Packet scheduling

Load shedding

g. Role of Stub in RPC

Stub acts as a communication proxy that packs and unpacks parameters for remote procedure calls.

SECTION B – Medium Answer Questions

(7 × 3 = 21 marks | Attempt ANY THREE)
Write steps + explanation + numerical where needed

a. CRC Error Detection (Numerical – VERY IMPORTANT)

Append CRC bits using generator polynomial                         Divide message by generator

Transmit codeword                                                                   Receiver repeats division

Non-zero remainder ⇒ error detected                                     (Show bit inversion detection clearly)

b. Subnetting of 192.168.10.0/24 into 4 Subnets

Borrow 2 bits → /26

c. Leaky Bucket vs Token Bucket                                             Leaky Bucket

Constant output rate                                                              No burst allowed

Token Bucket                                                                         Allows bursts

Better traffic shaping

More flexible

d. Public Key Cryptography & RSA                                         Uses public and private keys.

RSA Steps:                                                                             Choose two primes p, q

Compute n = pq                                                                     Choose public key e

Compute private key d

Advantages: Security, authentication                                   Disadvantages: Slow, complex computation

e. SNMP, DNS, Data Compression                                         SNMP: Network management

DNS: Domain name to IP resolution

Data Compression: Reduces data size for efficient transmission

SECTION C – Attempt ANY ONE

(7 marks)

a. Slotted ALOHA (Numerical)                                                Given: 10% slots idle

Idle probability = e⁻ᴳ = 0.1                                                     G=2.3G = 2.3G=2.3

Throughput:

S=Ge−G=2.3×0.1=0.23S = G e^{-G} = 2.3 \times 0.1 = 0.23S=Ge−G=2.3×0.1=0.23 

b. Total Delay Calculation (Numerical)

Include:                                                                                    Transmission delay

Propagation delay                                                                    Processing delay

Queuing delay

Total delay=∑(All delays)\text{Total delay} = \sum (\text{All delays})Total delay=∑(All delays)