About this previous year paper
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
| Subnet | Range |
|---|---|
| 1 | 192.168.10.0 – 63 |
| 2 | 192.168.10.64 – 127 |
| 3 | 192.168.10.128 – 191 |
| 4 | 192.168.10.192 – 255 |
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)