CN GATE Practice Set
Mixed GATE-style questions on delays, error detection, sliding windows, MAC, subnetting, fragmentation, routing, TCP and application protocols.
These questions mix every lesson of the Computer Networks track, in the three GATE formats. Most are numericals: write down the formula, put in the units carefully (bits versus bytes, ms versus µs), and only then calculate. Every numerical answer on this page is recomputed by a program when the site is built.
Links and the data link layer
[Delay] A 1000-byte packet crosses 2 links (one store-and-forward router in between). Each link is 8 Mbps with a propagation delay of 1 ms. Ignoring queuing and processing, what is the end-to-end delay in ms?
Show answer and explanation
Answer: 4Per link: transmission 8000 bits / 8 × 10⁶ bps = 1 ms, plus 1 ms propagation, so 2 ms. The router must receive the whole packet before forwarding, so the two links add up: 4 ms.
[Sliding window] In Go-Back-N, the window is 7 frames and the propagation delay is 5 times the frame transmission time. What is the link utilisation? Give the answer to two decimal places.
Show answer and explanation
Answer: 0.64a = 5, so one cycle lasts 1 + 2a = 11 frame times, of which 7 are used: η = 7/11 ≈ 0.64.
[CRC] The message 1010001101 is sent with the generator 110101. What is the CRC appended to it?
def crc(message, generator):
bits = list(message + "0" * (len(generator) - 1))
for i in range(len(message)):
if bits[i] == "1":
for j, g in enumerate(generator):
bits[i + j] = "0" if bits[i + j] == g else "1"
return "".join(bits[-(len(generator) - 1):])
print(crc("1010001101", "110101"))Show answer and explanation
Answer: BThe generator has 6 bits (degree 5), so 5 zeros are appended and the remainder of the modulo-2 division is 5 bits long: 01110. The frame sent is 101000110101110.
[CSMA/CD] A 100 Mbps CSMA/CD network uses a 2 km cable with a signal speed of 2 × 10⁸ m/s. What is the minimum frame size in bytes?
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Answer: 250Tp = 2000 / (2 × 10⁸) = 10 µs. Lmin = 2 Tp R = 2 × 10⁻⁵ × 10⁸ = 2000 bits = 250 bytes.
[ALOHA] A slotted ALOHA channel runs at 1 Mbps with 1000-bit frames. What is the maximum number of frames per second it can carry successfully? (Round down to a whole number.)
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Answer: 367There are 10⁶ / 1000 = 1000 slots per second, and at best a fraction 1/e ≈ 0.368 of them succeed: about 367.9, so at most 367 complete frames per second.
Network layer
[Subnetting] What is the network address of the host 172.16.45.130/20?
import ipaddress
print(ipaddress.ip_interface("172.16.45.130/20").network.network_address)Show answer and explanation
Answer: C/20 leaves 4 network bits in the third byte, so the mask there is 240. 45 = 00101101, and 00101101 AND 11110000 = 00100000 = 32. The network is 172.16.32.0 (broadcast 172.16.47.255).
[Subnetting] How many usable host addresses does a /23 network have?
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Answer: 5109 host bits give 2⁹ = 512 addresses, minus the network and broadcast addresses: 510.
[Fragmentation] A datagram with total length 3000 bytes and a 20-byte header must cross a link with MTU 1000. What is the fragment offset of the last fragment?
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Answer: 366Each fragment carries at most ⌊980 / 8⌋ × 8 = 976 data bytes. The 2980 bytes split as 976 + 976 + 976 + 52, starting at bytes 0, 976, 1952 and 2928. The last offset is 2928 / 8 = 366.
[Routing] Which statement about distance vector routing is true?
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Answer: BDistance vector routers exchange vectors with neighbours and apply Bellman-Ford. Flooding link costs and running Dijkstra is link state routing. Count-to-infinity is a known problem of distance vector.
Transport and application layers
[TCP] A TCP segment has sequence number 5000 and carries 1460 bytes. The receiver had already received everything before it. What acknowledgement number does the receiver send?
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Answer: 6460The segment carries bytes 5000 to 6459, so the receiver asks for byte 6460 next.
[Congestion control] TCP Reno starts with cwnd = 1 MSS and ssthresh = 8 MSS. Three duplicate ACKs arrive in round 6. Using the rules of the congestion control lesson (slow start stops at ssthresh), what is cwnd in round 9?
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Answer: 7Rounds 1–6: 1, 2, 4, 8, 9, 10. In round 6 the duplicate ACKs halve the window: ssthresh = cwnd = 5. Then congestion avoidance adds 1 per round: 5, 6, 7 in rounds 7, 8 and 9.
[HTTP] A page consists of a base file and 6 embedded objects, fetched over one persistent HTTP connection without pipelining. Ignoring transmission times and DNS, how many RTTs are needed?
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Answer: 81 RTT for the TCP handshake, 1 for the base file, and 1 for each of the 6 objects requested one after another: 8 RTTs.
[Protocols] Which pairs of protocol and transport are correct? (Select all that apply.)
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Answer: A, B, DDNS queries use UDP 53, SMTP uses TCP 25 and HTTP uses TCP 80. DHCP uses UDP (server 67, client 68), since the client has no address to set up a TCP connection with.