CS Atlas GATE 2027
Computer Networks course
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Computer Networks/Phase 6 · Integration & Final Test

Final test 6.4 · Item 90 of 90

Final end-to-end test

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Introduction

30 questions, GATE pattern: 15 one-mark and 15 two-mark; MCQ, MSQ and NAT; 90 minutes. Negative marking only on MCQ (−1/3 for 1-mark, −2/3 for 2-mark). Answers stay locked until you submit; then you get the score, a topic-wise accuracy table, full explanations and the lessons to re-read.

Final test questions

  1. MCQ · 1 mark

    Which layer is responsible for delivering data from a process on one host to a process on another?

    1. Data link
    2. Network
    3. Transport
    4. Session
  2. MSQ · 1 mark

    Which are true of virtual-circuit switching?

    1. A setup phase precedes data transfer.
    2. Link bandwidth is reserved for the circuit.
    3. All packets of a connection follow the same path.
    4. Packets carry a short circuit identifier.
  3. NAT · 2 marks

    6 packets of 1500 bytes cross 4 links of 10 Mbps (3 routers) by store-and-forward; Tp = 2 ms per link; no other delays. Time until the last bit of the last packet arrives (ms)?

    Write your answer before opening the key.

  4. MSQ · 1 mark

    Which socket API calls normally block?

    1. accept() on a server
    2. connect() on a client until the handshake completes
    3. bind()
    4. recv() when no data has arrived
  5. MCQ · 2 marks

    Data 11010011101100, generator 1011. The CRC is:

    1. 100
    2. 011
    3. 110
    4. 001
  6. NAT · 1 mark

    Internet checksum of the 16-bit words 0xB4F2 and 0x6A1C, as a decimal number?

    Write your answer before opening the key.

  7. MCQ · 1 mark

    A code has minimum Hamming distance 4. It can:

    1. detect 4, correct 2
    2. detect 3, correct 1
    3. detect 3, correct 2
    4. detect 2, correct 1
  8. NAT · 2 marks

    Slotted ALOHA with 10 stations, each transmitting in a slot with probability 0.05. Probability that a slot carries exactly one frame (3 decimals)?

    Write your answer before opening the key.

  9. NAT · 2 marks

    CSMA/CD at 1 Gbps on a 400 m cable, v = 2 × 10⁸ m/s. Minimum frame size in bytes?

    Write your answer before opening the key.

  10. NAT · 1 mark

    Stations A and B are both at their 2nd collision. Probability that A transmits first (3 decimals)?

    Write your answer before opening the key.

  11. MSQ · 1 mark

    Which statements about Ethernet are correct?

    1. An Ethernet frame carrying 30 data bytes is 64 bytes long.
    2. The MTU is 1500 bytes.
    3. 01:00:5E:00:00:FB is a unicast address.
    4. Switched full-duplex Ethernet has no collisions.
  12. NAT · 2 marks

    Shared Ethernet at 100 Mbps, 1000-byte frames, longest path 2 km (v = 2 × 10⁸ m/s). Efficiency using 1/(1 + 5a) (3 decimals)?

    Write your answer before opening the key.

  13. MCQ · 1 mark

    The network address of 200.10.20.70/26 is:

    1. 200.10.20.0
    2. 200.10.20.64
    3. 200.10.20.70
    4. 200.10.20.127
  14. MCQ · 2 marks

    The directed broadcast address of the block containing 172.16.45.200/21 is:

    1. 172.16.47.255
    2. 172.16.45.255
    3. 172.16.40.255
    4. 172.16.63.255
  15. MCQ · 2 marks

    A router's table: 10.0.0.0/8 → e0, 10.20.0.0/16 → e1, 10.20.30.0/24 → e2, 0.0.0.0/0 → e3. Packets to 10.20.30.40 and 10.21.30.40 leave on:

    1. e2 and e0
    2. e1 and e0
    3. e2 and e1
    4. e0 and e3
  16. MCQ · 1 mark

    The smallest single block covering 192.168.0.0/24 through 192.168.3.0/24 is:

    1. 192.168.0.0/22
    2. 192.168.0.0/23
    3. 192.168.0.0/21
    4. 192.168.0.0/16
  17. NAT · 1 mark

    A /22 block is divided into 8 equal subnets. Usable hosts per subnet?

    Write your answer before opening the key.

  18. NAT · 2 marks

    A 4000-byte datagram (20-byte header, DF = 0) meets an MTU of 1500. Fragment offset (8-byte units) of the last fragment?

    Write your answer before opening the key.

  19. NAT · 2 marks

    The same datagram's fragments next cross a link with MTU 620. How many fragments reach the destination?

    Write your answer before opening the key.

  20. MCQ · 1 mark

    An IPv4 header has HLEN = 7. How many bytes of options does it carry?

    1. 7
    2. 8
    3. 28
    4. 12
  21. MSQ · 1 mark

    Which are true of NAT (NAPT)?

    1. 192.168.10.5 is a private address.
    2. NAPT rewrites the source port as well as the source IP of outgoing packets.
    3. NAT lets outside hosts open connections to inside hosts without any configuration.
    4. 172.32.0.1 is a private address.
  22. NAT · 2 marks

    Links: A–B 2, A–C 5, B–C 1, B–D 4, C–E 3, D–E 1, D–F 5, E–F 2. Shortest-path cost from A to F (Dijkstra)?

    Write your answer before opening the key.

  23. MCQ · 1 mark

    Which DNS server holds the final A record for www.example.com?

    1. A root server
    2. The .com TLD server
    3. The authoritative server for example.com
    4. The browser cache only
  24. NAT · 2 marks

    Tt = 1 ms, Tp = 12.5 ms. Minimum sequence-number bits for Selective Repeat to reach 100 % utilisation?

    Write your answer before opening the key.

  25. NAT · 2 marks

    Go-Back-N, window 3, 10 frames to send; every 4th transmitted frame is lost (counting retransmissions). Total transmissions?

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  26. NAT · 1 mark

    TCP sequence-number wrap-around time on a 1 Gbps link, in seconds (2 decimals)?

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  27. NAT · 2 marks

    TCP starts with cwnd = 1 MSS and ssthresh = 16 MSS (slow start, then +1 MSS per RTT). A timeout occurs during RTT 8. What is cwnd (in MSS) during RTT 12?

    Write your answer before opening the key.

  28. NAT · 2 marks

    SRTT = 100 ms, RTTVAR = 10 ms, new SampleRTT = 120 ms, α = 1/8, β = 1/4 (RFC 6298: update RTTVAR with the old SRTT first). New RTO in ms?

    Write your answer before opening the key.

  29. NAT · 1 mark

    A client uses ISN 1000 in its SYN and, after the handshake, sends 500 bytes of data in one segment. What acknowledgement number does the server return for it?

    Write your answer before opening the key.

  30. NAT · 2 marks

    DNS lookup 30 ms, RTT to the web server 20 ms, page = base file + 10 small objects, non-persistent HTTP with up to 4 parallel connections. Total time (ms)?

    Write your answer before opening the key.

Answer key

Try the questions first. Open when you are ready to check.
  1. C — Process-to-process = transport.
    • A: Hop to hop.
    • B: Host to host.
    • D: Dialogue control.

    Review lesson 1.1.2

  2. ACD — Setup, fixed path, VCI labels; no bandwidth reservation.
    • B: Only circuit switching reserves bandwidth.

    Review lesson 1.2.3

  3. 18.8 — Tt = 1.2 ms. (6 + 4 − 1) × 1.2 + 4 × 2 = 10.8 + 8 = 18.8 ms.

    Review lesson 1.3.2

  4. ABD — accept, connect and recv wait for the network.
    • C: bind() just assigns an address.

    Review lesson 4.4

  5. A — Append 3 zeros and divide by 1011 modulo 2 → remainder 100.
    • B: Arithmetic slip.
    • C: Arithmetic slip.
    • D: Arithmetic slip.

    Review lesson 2.2.4

  6. 57584 — 0xB4F2 + 0x6A1C = 0x11F0E → 0x1F0F; checksum 0xE0F0 = 57584.

    Review lesson 2.2.3

  7. B — Detect d − 1 = 3; correct ⌊3/2⌋ = 1.
    • A: Needs d = 5.
    • C: Correcting 2 needs d = 5.
    • D: Under-states detection.

    Review lesson 2.2.2

  8. 0.314..0.316 — 10 × 0.05 × 0.95⁹ = 0.315.

    Review lesson 2.3.2

  9. 500 — Tp = 2 µs; 2 × 2 µs × 10⁹ = 4000 bits = 500 bytes.

    Review lesson 2.3.3

  10. 0.375 — 6 favourable pairs of 16.

    Review lesson 2.3.4

  11. ABD — Padding to 46; MTU 1500; first byte 0x01 is odd → multicast; switches remove collisions.
    • C: 0x01 is odd → multicast.

    Review lesson 2.4.1

  12. 0.615..0.616 — Tt = 80 µs, Tp = 10 µs, a = 0.125; 1/1.625 = 0.615.

    Review lesson 2.4.2

  13. B — Block 64; 70 lies in 64–127.
    • A: That is the /24 network.
    • C: A host address.
    • D: That is the broadcast address.

    Review lesson 3.1.3

  14. A — Third-octet block 8: 40–47 → network 172.16.40.0, broadcast 172.16.47.255.
    • B: That is the /24 broadcast.
    • C: Wrong: broadcast fills all host bits including the third octet.
    • D: That is a /18 broadcast.

    Review lesson 3.1.4

  15. A — 10.20.30.40 matches /8, /16, /24 → e2. 10.21.30.40 matches only /8 (and default) → e0.
    • B: First address has a longer /24 match.
    • C: Second address does not match 10.20.0.0/16.
    • D: Default is used only when nothing else matches.

    Review lesson 3.1.5

  16. A — Four contiguous aligned /24s = one /22.
    • B: Covers only two /24s.
    • C: Twice as large as needed.
    • D: Far too large.

    Review lesson 3.1.4

  17. 126 — /22 + 3 bits = /25 → 2⁷ − 2 = 126.

    Review lesson 3.1.3

  18. 370 — Data 3980 → 1480, 1480, 1020; offsets 0, 185, 370.

    Review lesson 3.2.2

  19. 8 — Max data 600: 1480 → 3, 1480 → 3, 1020 → 2; total 8.

    Review lesson 3.2.2

  20. B — Header = 7 × 4 = 28 bytes; options = 28 − 20 = 8.
    • A: HLEN counts 4-byte words.
    • C: That is the whole header.
    • D: Arithmetic slip.

    Review lesson 3.2.1

  21. AB — 192.168/16 is private; NAPT maps (IP, port) pairs.
    • C: Inbound connections need port forwarding.
    • D: Private range is 172.16–172.31.

    Review lesson 3.3

  22. 8 — A–B–C–E–F = 2 + 1 + 3 + 2 = 8 (A–B–D–E–F costs 9).

    Review lesson 3.4.4

  23. C — Authoritative server.
    • A: Refers to TLD.
    • B: Refers to authoritative.
    • D: Caches copies only.

    Review lesson 5.2

  24. 6 — N = ⌈1 + 25⌉ = 26; SR needs 2ᵏ⁻¹ ≥ 26 → k = 6.

    Review lesson 4.2.4

  25. 27 — Simulating: each loss also wastes the rest of the window; total 27 transmissions.

    Review lesson 4.2.3

  26. 34.3..34.4 — 2³² × 8 / 10⁹ = 34.36 s.

    Review lesson 4.3.3

  27. 8 — RTT 1–8: 1, 2, 4, 8, 16, 17, 18, 19. Timeout: ssthresh = 9, cwnd = 1. RTT 9–12: 1, 2, 4, 8.

    Review lesson 4.3.5

  28. 152.5 — RTTVAR = 0.75·10 + 0.25·20 = 12.5; SRTT = 102.5; RTO = 102.5 + 4·12.5 = 152.5.

    Review lesson 4.3.4

  29. 1501 — SYN uses 1000; data bytes 1001–1500; ACK = next expected = 1501.

    Review lesson 4.3.1

  30. 190 — 30 + 2 × 20 + ⌈10/4⌉ × 2 × 20 = 30 + 40 + 120 = 190.

    Review lesson 5.3

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