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
- MCQ · 1 mark
Which layer is responsible for delivering data from a process on one host to a process on another?
- Data link
- Network
- Transport
- Session
- MSQ · 1 mark
Which are true of virtual-circuit switching?
- A setup phase precedes data transfer.
- Link bandwidth is reserved for the circuit.
- All packets of a connection follow the same path.
- Packets carry a short circuit identifier.
- 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.
- MSQ · 1 mark
Which socket API calls normally block?
- accept() on a server
- connect() on a client until the handshake completes
- bind()
- recv() when no data has arrived
- MCQ · 2 marks
Data 11010011101100, generator 1011. The CRC is:
- 100
- 011
- 110
- 001
- NAT · 1 mark
Internet checksum of the 16-bit words 0xB4F2 and 0x6A1C, as a decimal number?
Write your answer before opening the key.
- MCQ · 1 mark
A code has minimum Hamming distance 4. It can:
- detect 4, correct 2
- detect 3, correct 1
- detect 3, correct 2
- detect 2, correct 1
- 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)?
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- NAT · 2 marks
CSMA/CD at 1 Gbps on a 400 m cable, v = 2 × 10⁸ m/s. Minimum frame size in bytes?
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- NAT · 1 mark
Stations A and B are both at their 2nd collision. Probability that A transmits first (3 decimals)?
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- MSQ · 1 mark
Which statements about Ethernet are correct?
- An Ethernet frame carrying 30 data bytes is 64 bytes long.
- The MTU is 1500 bytes.
- 01:00:5E:00:00:FB is a unicast address.
- Switched full-duplex Ethernet has no collisions.
- 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)?
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- MCQ · 1 mark
The network address of 200.10.20.70/26 is:
- 200.10.20.0
- 200.10.20.64
- 200.10.20.70
- 200.10.20.127
- MCQ · 2 marks
The directed broadcast address of the block containing 172.16.45.200/21 is:
- 172.16.47.255
- 172.16.45.255
- 172.16.40.255
- 172.16.63.255
- 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:
- e2 and e0
- e1 and e0
- e2 and e1
- e0 and e3
- MCQ · 1 mark
The smallest single block covering 192.168.0.0/24 through 192.168.3.0/24 is:
- 192.168.0.0/22
- 192.168.0.0/23
- 192.168.0.0/21
- 192.168.0.0/16
- NAT · 1 mark
A /22 block is divided into 8 equal subnets. Usable hosts per subnet?
Write your answer before opening the key.
- 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?
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- NAT · 2 marks
The same datagram's fragments next cross a link with MTU 620. How many fragments reach the destination?
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- MCQ · 1 mark
An IPv4 header has HLEN = 7. How many bytes of options does it carry?
- 7
- 8
- 28
- 12
- MSQ · 1 mark
Which are true of NAT (NAPT)?
- 192.168.10.5 is a private address.
- NAPT rewrites the source port as well as the source IP of outgoing packets.
- NAT lets outside hosts open connections to inside hosts without any configuration.
- 172.32.0.1 is a private address.
- 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)?
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- MCQ · 1 mark
Which DNS server holds the final A record for www.example.com?
- A root server
- The .com TLD server
- The authoritative server for example.com
- The browser cache only
- NAT · 2 marks
Tt = 1 ms, Tp = 12.5 ms. Minimum sequence-number bits for Selective Repeat to reach 100 % utilisation?
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- 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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- NAT · 1 mark
TCP sequence-number wrap-around time on a 1 Gbps link, in seconds (2 decimals)?
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- 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?
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- 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?
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- 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?
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- 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)?
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Answer key
Try the questions first. Open when you are ready to check.
- C — Process-to-process = transport.
- A: Hop to hop.
- B: Host to host.
- D: Dialogue control.
- ACD — Setup, fixed path, VCI labels; no bandwidth reservation.
- B: Only circuit switching reserves bandwidth.
- 18.8 — Tt = 1.2 ms. (6 + 4 − 1) × 1.2 + 4 × 2 = 10.8 + 8 = 18.8 ms.
- ABD — accept, connect and recv wait for the network.
- C: bind() just assigns an address.
- A — Append 3 zeros and divide by 1011 modulo 2 → remainder 100.
- B: Arithmetic slip.
- C: Arithmetic slip.
- D: Arithmetic slip.
- 57584 — 0xB4F2 + 0x6A1C = 0x11F0E → 0x1F0F; checksum 0xE0F0 = 57584.
- B — Detect d − 1 = 3; correct ⌊3/2⌋ = 1.
- A: Needs d = 5.
- C: Correcting 2 needs d = 5.
- D: Under-states detection.
- 0.314..0.316 — 10 × 0.05 × 0.95⁹ = 0.315.
- 500 — Tp = 2 µs; 2 × 2 µs × 10⁹ = 4000 bits = 500 bytes.
- 0.375 — 6 favourable pairs of 16.
- ABD — Padding to 46; MTU 1500; first byte 0x01 is odd → multicast; switches remove collisions.
- C: 0x01 is odd → multicast.
- 0.615..0.616 — Tt = 80 µs, Tp = 10 µs, a = 0.125; 1/1.625 = 0.615.
- B — Block 64; 70 lies in 64–127.
- A: That is the /24 network.
- C: A host address.
- D: That is the broadcast address.
- 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.
- 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.
- A — Four contiguous aligned /24s = one /22.
- B: Covers only two /24s.
- C: Twice as large as needed.
- D: Far too large.
- 126 — /22 + 3 bits = /25 → 2⁷ − 2 = 126.
- 370 — Data 3980 → 1480, 1480, 1020; offsets 0, 185, 370.
- 8 — Max data 600: 1480 → 3, 1480 → 3, 1020 → 2; total 8.
- 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.
- 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.
- 8 — A–B–C–E–F = 2 + 1 + 3 + 2 = 8 (A–B–D–E–F costs 9).
- C — Authoritative server.
- A: Refers to TLD.
- B: Refers to authoritative.
- D: Caches copies only.
- 6 — N = ⌈1 + 25⌉ = 26; SR needs 2ᵏ⁻¹ ≥ 26 → k = 6.
- 27 — Simulating: each loss also wastes the rest of the window; total 27 transmissions.
- 34.3..34.4 — 2³² × 8 / 10⁹ = 34.36 s.
- 8 — RTT 1–8: 1, 2, 4, 8, 16, 17, 18, 19. Timeout: ssthresh = 9, cwnd = 1. RTT 9–12: 1, 2, 4, 8.
- 152.5 — RTTVAR = 0.75·10 + 0.25·20 = 12.5; SRTT = 102.5; RTO = 102.5 + 4·12.5 = 152.5.
- 1501 — SYN uses 1000; data bytes 1001–1500; ACK = next expected = 1501.
- 190 — 30 + 2 × 20 + ⌈10/4⌉ × 2 × 20 = 30 + 40 + 120 = 190.