🌐 Computer Networks

MAKAUT B.Tech (ECC602 / PCC-CS602) 2024–25 – Detailed Solutions

✅ Group-A (1 Mark) | Group-B (5 Marks) | Group-C (15 Marks) | Exam-ready answers
📌 Group-A (Very Short Answer – 1 Mark Each)
(i) Define Data Rate in terms of transmission.

Answer: Data rate is the number of bits transmitted per second (bps) over a communication channel.

(ii) Name one intradomain routing algorithm.

Answer: OSPF (Open Shortest Path First) [Also acceptable: RIP, IS-IS]

(iii) Which transport layer protocol is used for SMTP?

Answer: TCP

(iv) Write one advantage of VPN.

Answer: Provides secure communication over a public network through encryption.

(v) If there are n nodes in a network, how many physical links are required in Mesh Topology?

Answer: n(n-1)/2

(vi) What is the maximum efficiency achievable in Pure ALOHA?

Answer: 18.4% or 1/(2e)

(vii) What is the purpose of Class D address?

Answer: Multicasting.

(viii) The technique of temporarily delaying outgoing acknowledgements so that they can be hooked onto the next outgoing data frame is called ________.

Answer: Piggybacking

(ix) What is DNS?

Answer: DNS (Domain Name System) translates domain names into IP addresses.

(x) Which technique is used to detect and correct single-bit error?

Answer: Hamming Code

(xi) Which protocol is used to get the MAC address given an IP address?

Answer: ARP (Address Resolution Protocol)

(xii) Which layer is responsible for Host-to-Host delivery?

Answer: Transport Layer

✍️ Group-B (Short Answer – 5 Marks)
2. Compare and Contrast Circuit Switching and Packet Switching
Circuit SwitchingPacket Switching
Dedicated path establishedNo dedicated path
Resources reservedResources shared
Fixed delayVariable delay
Suitable for voice callsSuitable for data communication
Example: Telephone networkExample: Internet

Circuit Switching advantages: guaranteed bandwidth, predictable delay. Packet Switching advantages: better bandwidth utilization, more efficient.

3. Discuss Mesh Topology with Diagram

Definition: Every node directly connected to every other node. Number of links = n(n-1)/2. Advantages: highly reliable, no congestion, easy fault isolation, high security. Disadvantages: expensive, complex wiring. Applications: military, WAN backbone.

A — B — C
| \ | / |
D — E — F
4. Advantage of Independent ACK and Cumulative ACK. Go-Back-N when Sender Window = Receiver Window = 1

Independent ACK: each frame acknowledged separately → precise error detection. Cumulative ACK: one ACK confirms multiple frames → reduces overhead. When both windows =1, Go-Back-N behaves like Stop-and-Wait ARQ.

5. Explain CRC with Example

CRC (Cyclic Redundancy Check): binary division, remainder appended. Example: Data=1101011011, generator=10011 → append 4 zeros, divide → remainder=1110 → transmitted frame: 1101011011 1110. Detects burst errors, highly reliable.

6. What are the two possible transport services? What is meant by segment? What is congestion?

Transport services: Connection-oriented (TCP: reliable, ordered) and Connectionless (UDP: fast, no guarantee). Segment: PDU of Transport Layer. Congestion: network traffic exceeds resources → delay & packet loss.

📘 Group-C (Long Answer – 15 Marks)
7(a) Explain Bit Stuffing Method with Example

Bit stuffing: after five consecutive 1s, a 0 is inserted. Flag pattern: 01111110. Example: 1111101111110 → after stuffing: 1111100 11111010. Receiver removes stuffed bits. Prevents false flag detection, used in HDLC, PPP.

7(b) Data = 1110011011, Received Code = 1100011011. Correct Using Hamming Code

Comparison: transmitted: 1110011011, received: 1100011011 → bit position 3 differs. Error position: 3rd bit. Corrected code = 1110011011. Single-bit error detected & corrected via Hamming code.

7(c) Functionalities of MAC Sublayer

Physical addressing (MAC), frame delimiting, access control (CSMA/CD, CSMA/CA), error detection (CRC), collision handling, media access management. Ensures efficient shared medium access.

8(a) Explain Distance Vector Routing with Example

Based on Bellman-Ford; routers exchange tables with neighbors. Example: A–1–B–2–C, A–4–C. Path A→B→C cost=3 (best). Advantages: simple, low complexity. Disadvantages: slow convergence, count-to-infinity.

8(b) Count-to-Infinity Problem and Its Solution

Routers continuously increase hop count for failed route (2→3→4...). Solutions: Split Horizon, Poison Reverse, Triggered Updates, Hold-down Timer, Maximum Hop Count.

9(a) Functions of Transport Layer

Process-to-process delivery, segmentation & reassembly, flow control, error control, multiplexing, connection management, congestion control. Ensures reliable end-to-end communication.

9(b) Explain Leaky Bucket Algorithm

Traffic shaping: packets stored in bucket, leaks at constant rate. Excess packets discarded. Smooth traffic, reduces congestion, simple but does not support bursts.

Incoming → Bucket → Constant rate output
9(c) Open Loop Congestion Control Techniques

Preventive: Retransmission policy, Window policy, Acknowledgement policy, Discard policy, Admission policy. Prevents congestion before occurrence, improves QoS.

10. Short Notes: (i) DNS (ii) FTP (iii) Firewall (iv) TELNET

DNS: domain → IP translation, distributed hierarchical database. FTP: file transfer, ports 20 (data) & 21 (control). Firewall: monitors/filters traffic; packet filtering, stateful, proxy. TELNET: remote login (port 23), insecure plain text.

11(a) ISP Address Allocation Problem (8 Marks) – 190.100.0.0/16

Total addresses: 65536. Group1 (64×256): 16384 addresses → 190.100.0.0 – 190.100.63.255. Group2 (128×128): 16384 → 190.100.64.0 – 190.100.127.255. Group3 (128×64): 8192 → 190.100.128.0 – 190.100.159.255. Remaining: 65536 – (16384+16384+8192) = 24576 addresses.

11(b) CIDR Routing Decision – Destination 131.23.151.76

Longest prefix match: 131.23.151.76 falls under 131.22.0.0/15 (covers 131.22.0.0 – 131.23.255.255) → Interface = 1.

11(c) What are Subnetting and Supernetting?

Subnetting: dividing large network into smaller subnetworks → better management, reduced broadcast. Supernetting: combining multiple networks into one → reduces routing table size, efficient aggregation.

📖 Exam-focused Summary

✔ Complete coverage of MAKAUT Computer Networks (ECC602/PCC-CS602) 2024–25 – all groups.
✔ Includes data rate, intradomain routing, VPN, mesh topology, CRC, bit stuffing, Hamming code, distance vector, count-to-infinity, leaky bucket, open loop control, DNS, FTP, firewall, CIDR, subnetting, supernetting, and ISP allocation.
✔ Answers structured for 1, 5, and 15-mark questions with clear headings & exam-style presentation.