GATE CSE · subject-wise

GATE Computer Networks previous year questions

66 questions from GATE papers, newest first, each with the official answer and a worked explanation.

Practise Computer Networks with instant checking
GATE 2026 S1 · Q18MCQ1 mark · -1/3· Easy

With respect to a TCP connection between a client and a server, which one of the following statements is true?

  1. (A)The client and server use a two-way handshake mechanism before the start of data transmission
  2. (B)The server cannot initiate closing of the connection before the client initiates closing of the connection
  3. (C)The TCP connection is half-duplex
  4. (D)The client and server can initiate closing of the connection at the same time
Answer: D - show solution

TCP uses a three-way handshake, not two-way (A false). Either side may send FIN first (B false). TCP is full-duplex (C false). Both sides can send FIN at the same moment; TCP defines simultaneous close (both go FIN_WAIT_1 -> CLOSING -> TIME_WAIT). Answer: (D).

Concept: TCP connection management: three-way open, full-duplex data, and a close that either side or both at once may start. Simultaneous close is a defined state transition.

GATE 2026 S1 · Q19MSQ1 mark · no negative· Easy

Which of the following statements is/are true with respect to the interaction of a web browser with a web server using HTTP 1.1?

  1. (A)HTTP 1.1 facilitates downloading multiple objects of the same webpage over the same TCP connection, if the objects are stored in the same server
  2. (B)HTTP 1.1 facilitates downloading multiple objects of the same webpage over the same TCP connection, even if they are stored in different servers
  3. (C)HTTP 1.1 facilitates sending a request for downloading one object without waiting for a previously requested object to be downloaded completely
  4. (D)HTTP 1.1 facilitates downloading multiple webpages on the same server to be downloaded over a single TCP connection
Answer: A, C, D - show solution

HTTP 1.1 uses persistent connections: one TCP connection to a server carries many requests, so multiple objects (A) and multiple pages (D) from the same server can share it. A TCP connection is between two hosts, so objects on a different server need another connection (B false). HTTP 1.1 supports pipelining: send the next request before the previous response completes (C true). Answer: (A), (C), (D).

Concept: Persistent connections and pipelining are the HTTP 1.1 features. The trap is (B): a TCP connection cannot span servers.

GATE 2026 S1 · Q44MCQ2 marks · -2/3· Moderate

A TCP sender successfully establishes a connection with a TCP receiver and starts the transmission of segments. The TCP congestion control mechanism's slow-start threshold is set to 10000 segments. Assume that the round-trip time is fixed at 1 millisecond. Assume that the sender always has data to send, the segments are numbered from 1, and no segment is lost. Let t denote the time (in milliseconds) at which the transmission of segment number 2000 starts. Which one of the following options is correct?

  1. (A)9 ≤ t < 10
  2. (B)10 ≤ t < 11
  3. (C)11 ≤ t < 12
  4. (D)12 ≤ t < 13
Answer: B - show solution

In slow start the window starts at 1 segment and doubles each RTT (ssthresh 10000 is never reached here). Segments sent in RTT round k (starting at time k ms) are numbered 2ᵏ to 2ᵏ⁺¹ - 1: round 0 sends 1, round 1 sends 2-3, round 2 sends 4-7, and so on. Segment 2000 lies in 1024..2047, which is round 10, so its transmission starts at some time in [10, 11) ms. Answer: (B) 10 ≤ t < 11.

Concept: Slow start doubles cwnd per RTT, so the cumulative segments after k rounds are 2ᵏ - 1. Off-by-one errors in numbering the rounds give (A) or (C).

GATE 2026 S1 · Q45MCQ2 marks · -2/3· Easy

Consider the implementation of sliding window protocol over a lossless link, with a window size of W frames, where each frame is of size 1000 bits (including header). The bandwidth of the link is 100 kbps (1k = 10³) and the one-way propagation delay is 100 milliseconds. Assume that processing times at the sender and receiver are zero and the transmission time of acknowledgements is also zero. Which one of the following options gives the minimum size of W (in number of frames) required to achieve 100% link utilization?

  1. (A)10
  2. (B)21
  3. (C)20
  4. (D)11
Answer: B - show solution

Transmission time per frame Tt = 1000 bits / 100 kbps = 10 ms. Propagation delay Tp = 100 ms. For 100% utilisation the sender must keep transmitting until the first ACK arrives, which is after Tt + 2Tp = 210 ms. Window W ≥ 210/10 = 21 frames. Answer: (B) 21.

Concept: Sliding window efficiency = W·Tt / (Tt + 2Tp). Forgetting the first frame's own Tt gives 20; using one-way delay gives 11.

GATE 2026 S1 · Q56MSQ2 marks · no negative· Moderate

An ISP having an address block 202.16.0.0/15 assigns a block of 6000 IP addresses to a client, using the classless internet domain routing (CIDR) super-netting approach. Which of the following address blocks can be assigned by the ISP?

  1. (A)202.16.0.0/19
  2. (B)202.17.64.0/19
  3. (C)202.16.32.0/19
  4. (D)202.17.24.0/19
Answer: A, B, C - show solution

6000 addresses need the next power of two, 8192 = 2¹³, so a /19 block. The ISP owns 202.16.0.0/15 = 202.16.0.0 to 202.17.255.255. A /19 block must start where the third octet is a multiple of 32 (2¹³ = 32 × 256). 202.16.0.0 (0) valid; 202.17.64.0 (64) valid; 202.16.32.0 (32) valid; 202.17.24.0 (24) is not a multiple of 32, so not a valid /19 boundary. Answer: (A), (B), (C).

Concept: A CIDR block must be aligned to its own size. Check both containment in the parent block and alignment; (D) fails on alignment.

GATE 2026 S2 · Q21MCQ1 mark · -1/3· Easy

Consider a file of size 4 million bytes being transferred between two hosts connected via a path consisting of three consecutive links of bandwidth 2 Mbps, 500 kbps, and 1 Mbps, respectively. All processing delays and propagation delays are negligible. Assume that there is no other background traffic over the path and no other additional overhead to transfer the file. Which one of the following is the total time (in seconds) to transfer the file? Note: 1M = 10⁶, 1k = 10³

  1. (A)731
  2. (B)64
  3. (C)8
  4. (D)16
Answer: B - show solution

File size = 4 x 10⁶ bytes = 32 x 10⁶ bits. With negligible delays and the data streamed along the path, end-to-end throughput equals the bottleneck link rate, min(2 Mbps, 500 kbps, 1 Mbps) = 500 kbps. Time = 32 x 10⁶ / 5 x 10⁵ = 64 s. Answer: (B) 64.

Concept: End-to-end throughput of a path is limited by its slowest link. Traps: forgetting bytes-to-bits (giving 8 or 16), or summing per-link transfer times as if the whole file were stored and forwarded at each hop.

GATE 2026 S2 · Q22MCQ1 mark · -1/3· Easy

Which one of the following protocols may need to broadcast some of its messages?

  1. (A)SMTP
  2. (B)FTP
  3. (C)DHCP
  4. (D)HTTP
Answer: C - show solution

A DHCP client has no IP address and does not know the server's address, so DHCPDISCOVER (and typically DHCPREQUEST) is sent to the broadcast address 255.255.255.255. SMTP, FTP and HTTP run over TCP connections to a known server and are unicast. Answer: (C) DHCP.

Concept: Broadcast is needed when the destination is unknown, as in address configuration. TCP-based application protocols are always unicast.

GATE 2026 S2 · Q33NAT1 mark · no negative· Easy

If an IP network uses a subnet mask of 255.255.240.0, the maximum number of IP addresses that can be assigned to network interfaces is __________. (answer in integer)

Answer: 4094 - show solution

255.255.240.0 = 11111111.11111111.11110000.00000000, a /20 mask, leaving 32 - 20 = 12 host bits. That gives 2¹² = 4096 addresses; excluding the network address (all host bits 0) and the broadcast address (all 1) leaves 4094 assignable. Answer: 4094.

Concept: Usable hosts = 2^(host bits) - 2. The trap is answering 4096 by forgetting the network and broadcast addresses, or miscounting the bits in 240 (four 1s).

GATE 2026 S2 · Q43MCQ2 marks · -2/3· Easy

Consider the transmission of data bits 110001011 over a link that uses Cyclic Redundancy Check (CRC) code for error detection. If the generator bit pattern is given to be 1001, which one of the following options shows the remainder bit pattern appended to the data bits before transmission?

  1. (A)011
  2. (B)101
  3. (C)000
  4. (D)100
Answer: D - show solution

Generator 1001 has degree 3, so append three 0s: 110001011000. Modulo-2 division by 1001: 1100 xor 1001 = 0101; continue bringing down bits and XORing whenever the leading bit is 1. Working through the 12 bits leaves remainder 100. The transmitted frame is 110001011100. Answer: (D) 100.

Concept: CRC: append (degree of generator) zeros, divide with XOR (no borrows), the remainder is the checksum. Traps: appending 4 zeros instead of 3, or doing ordinary subtraction.

GATE 2026 S2 · Q58NAT2 marks · no negative· Moderate

Consider a new TCP connection between a sender and a receiver. The receiver advertised window is constant at 48 KB, the maximum segment size (MSS) is 2 KB, and the slow start threshold for TCP congestion control is 16 KB. Assume that there are no timeouts or duplicate acknowledgements. The number of rounds of transmission required for the congestion control algorithm of the TCP connection to reach the congestion avoidance phase is ___________. (answer in integer) Note: 1K = 2¹⁰

Answer: 4 - show solution

Slow start begins at cwnd = 1 MSS = 2 KB and doubles each round: round 1: 2 KB, round 2: 4 KB, round 3: 8 KB, round 4: 16 KB. In round 4 cwnd reaches ssthresh = 16 KB (8 MSS), so the connection is now in congestion avoidance and grows linearly afterwards. The 48 KB receiver window never limits this. Answer: 4.

Concept: Slow start doubles cwnd per RTT until it reaches ssthresh, then switches to additive increase. The trap is starting cwnd at 0 or counting the number of doublings (3) rather than transmission rounds.

GATE 2026 S2 · Q65NAT2 marks · no negative· Moderate

It is necessary to design a link-layer protocol between two hosts that are directly connected over a lossless link of length 3000 kilometers. Assume that the link bandwidth is 10⁸ bits per second and that the propagation delay in the link is 5 nanoseconds per meter. Every transmitted data byte is assigned a unique sequence number. Let n be the minimum number of bits needed for the sequence number field in the protocol header such that i. the sequence numbers do not wrap around before 60 seconds, and ii. the maximum utilization of the link is achieved. The value of n is ______. (answer in integer)

Answer: 30 - show solution

(i) In 60 s the link carries 10⁸ x 60 = 6 x 10⁹ bits = 7.5 x 10⁸ bytes, each needing a distinct number: 2ⁿ ≥ 7.5 x 10⁸ gives n = 30 (2²⁹ ≈ 5.4 x 10⁸ is too small, 2³⁰ ≈ 1.07 x 10⁹). (ii) One-way delay = 3 x 10⁶ m x 5 ns = 15 ms, RTT = 30 ms; bytes in flight for full utilisation = 10⁸ x 0.03 / 8 = 375,000, needing about 19-20 bits. The binding requirement is (i). n = 30. Answer: 30.

Concept: Sequence space must cover both the wrap-around time and the bandwidth-delay product; take the larger requirement. The trap is computing only the window for utilisation and answering about 19 or 20.

GATE 2025 S1 · Q16MCQ1 mark · -1/3· Easy

Identify the ONE CORRECT matching between the OSI layers and their corresponding functionalities as shown. OSI Layers: (a) Network layer, (b) Transport layer, (c) Datalink layer Functionalities: (I) Packet routing, (II) Framing and error handling, (III) Host to host communication

  1. (A)(a)-(I), (b)-(II), (c)-(III)
  2. (B)(a)-(I), (b)-(III), (c)-(II)
  3. (C)(a)-(II), (b)-(I), (c)-(III)
  4. (D)(a)-(III), (b)-(II), (c)-(I)
Answer: B - show solution

Network layer routes packets across networks: (a)-(I). Transport layer provides end-to-end (host to host / process to process) communication: (b)-(III). Data link layer does framing and error detection on a link: (c)-(II). Answer: (B).

Concept: Layer responsibilities. "Host to host" is the transport layer's end-to-end role; error handling also exists at transport, but framing pins (II) to the data link layer.

GATE 2025 S1 · Q22MSQ1 mark · no negative· Easy

Consider the 3-way handshaking protocol for TCP connection establishment. Let the three packets exchanged during the connection establishment be denoted as P1, P2, and P3, in order. Which of the following option(s) is/are TRUE with respect to TCP header flags that are set in the packets?

  1. (A)P3: SYN = 1, ACK = 1
  2. (B)P2: SYN = 1, ACK = 1
  3. (C)P2: SYN = 0, ACK = 1
  4. (D)P1: SYN = 1
Answer: B, D - show solution

P1 (client to server): SYN = 1, ACK = 0. P2 (server to client): SYN = 1, ACK = 1 (SYN-ACK). P3 (client to server): SYN = 0, ACK = 1. So (B) and (D) are true; (A) wrongly sets SYN in P3 and (C) wrongly clears SYN in P2. Answer: (B), (D).

Concept: Handshake flags: SYN, SYN+ACK, ACK. SYN consumes a sequence number and appears only in the first two segments.

GATE 2025 S1 · Q40MCQ2 marks · -2/3· Easy

A packet with the destination IP address 145.36.109.70 arrives at a router whose routing table is shown. Which interface will the packet be forwarded to? Subnet Address Subnet Mask (CIDR) Interface 145.36.0.0 /16 E1 145.36.128.0 /17 E2 145.36.64.0 /18 E3 145.36.255.0 /24 E4 Default -- E5

  1. (A)E3
  2. (B)E1
  3. (C)E2
  4. (D)E5
Answer: A - show solution

Third octet 109 = 01101101. /16 145.36.0.0: first 16 bits match, so E1 is a candidate. /17 145.36.128.0 needs third-octet top bit 1; 109 has 0: no. /18 145.36.64.0 needs top two bits 01; 109 starts 01: match. /24 145.36.255.0 needs third octet 255: no. Longest prefix match is /18. Answer: E3 (A).

Concept: Longest prefix match. The trap is stopping at the first matching entry (/16, E1) instead of choosing the most specific match.

GATE 2025 S1 · Q57NAT2 marks · no negative· Moderate

Suppose a message of size 15000 bytes is transmitted from a source to a destination using IPv4 protocol via two routers as shown in the figure. Each router has a defined maximum transmission unit (MTU) as shown in the figure, including IP header. The number of fragments that will be delivered to the destination is ________. (Answer in integer) Figure: Source → Router-1 (MTU = 5000 bytes) → Router-2 (MTU = 3000 bytes) → Destination

Answer: 7 - show solution

Assume a 20-byte IP header; fragment data (except the last) must be a multiple of 8. At MTU 5000: max data 4980, rounded down to 4976. 15000 = 3 x 4976 + 72, giving 4 fragments: 4976, 4976, 4976, 72. At MTU 3000: max data 2980, rounded down to 2976. Each 4976 fragment splits into 2976 + 2000 (2 fragments); the 72-byte one passes unchanged. Total = 3 x 2 + 1 = 7. Answer: 7.

Concept: IPv4 fragmentation: payload per fragment is (MTU - header) rounded down to a multiple of 8, and fragments are never reassembled at routers. Trap: dividing 15000 directly by the smallest MTU (6).

GATE 2025 S2 · Q16MCQ1 mark · -1/3· Easy

Consider the following statements: (i) Address Resolution Protocol (ARP) provides a mapping from an IP address to the corresponding hardware (link-layer) address. (ii) A single TCP segment from a sender S to a receiver R cannot carry both data from S to R and acknowledgement for a segment from R to S. Which ONE of the following is CORRECT?

  1. (A)Both (i) and (ii) are TRUE
  2. (B)(i) is TRUE and (ii) is FALSE
  3. (C)(i) is FALSE and (ii) is TRUE
  4. (D)Both (i) and (ii) are FALSE
Answer: B - show solution

(i) ARP maps an IPv4 address to the MAC (link-layer) address on the local network: true. (ii) TCP supports piggybacking: a data segment from S to R carries the ACK flag and acknowledgement number for data received from R, so (ii) is false. Answer: (B) (i) is TRUE and (ii) is FALSE.

Concept: Tests ARP's direction (IP to MAC, not the reverse, which was RARP) and TCP piggybacking of ACKs on data segments.

GATE 2025 S2 · Q17MCQ1 mark · -1/3· Easy

Consider the routing protocols given in List I and the names given in List II: List I List II (i) Distance vector routing (a) Bellman-Ford (ii) Link state routing (b) Dijkstra For matching of items in List I with those in List II, which ONE of the following options is CORRECT?

  1. (A)(i) - (a) and (ii) - (b)
  2. (B)(i) - (a) and (ii) - (a)
  3. (C)(i) - (b) and (ii) - (a)
  4. (D)(i) - (b) and (ii) - (b)
Answer: A - show solution

Distance vector routing (for example RIP) computes routes with the distributed Bellman-Ford algorithm. Link state routing (for example OSPF) floods the topology and each router runs Dijkstra's algorithm. Answer: (A) (i) - (a) and (ii) - (b).

Concept: Distance vector uses neighbour-only information iteratively (Bellman-Ford); link state needs the full map and a single-source shortest path computation (Dijkstra).

GATE 2025 S2 · Q18MCQ1 mark · -1/3· Moderate

A machine receives an IPv4 datagram. The protocol field of the IPv4 header has the protocol number of a protocol X. Which ONE of the following is NOT a possible candidate for X?

  1. (A)Internet Control Message Protocol (ICMP)
  2. (B)Internet Group Management Protocol (IGMP)
  3. (C)Open Shortest Path First (OSPF)
  4. (D)Routing Information Protocol (RIP)
Answer: D - show solution

The IPv4 protocol field names the protocol carried directly inside the IP payload. ICMP (1), IGMP (2) and OSPF (89) are carried directly in IP. RIP runs over UDP (port 520), so the protocol field would say UDP (17), never RIP. Answer: (D) RIP.

Concept: Know which protocols sit directly on IP. The trap is assuming all routing protocols are alike: OSPF is directly over IP, RIP over UDP, BGP over TCP.

GATE 2025 S2 · Q23MCQ1 mark · -1/3· Moderate

Consider a network that uses Ethernet and IPv4. Assume that IPv4 headers do not use any options field. Each Ethernet frame can carry a maximum of 1500 bytes in its data field. A UDP segment is transmitted. The payload (data) in the UDP segment is 7488 bytes. Which ONE of the following choices has the CORRECT total number of fragments transmitted and the size of the last fragment including IPv4 header?

  1. (A)5 fragments, 1488 bytes
  2. (B)6 fragments, 88 bytes
  3. (C)6 fragments, 108 bytes
  4. (D)6 fragments, 116 bytes
Answer: D - show solution

IP payload = UDP header 8 + data 7488 = 7496 bytes. With a 20-byte IP header and MTU 1500, each fragment carries at most 1480 bytes of payload (a multiple of 8, so valid). 7496 = 5 x 1480 + 96, so there are 5 full fragments and a 6th carrying 96 bytes. Last fragment size = 96 + 20 = 116 bytes. Answer: (D) 6 fragments, 116 bytes.

Concept: Fragmentation splits the whole IP payload, which includes the UDP header. Forgetting the 8-byte UDP header gives 88 bytes of payload and the trap answers 88 or 108.

GATE 2025 S2 · Q36MCQ2 marks · -2/3· Easy

Suppose we are transmitting frames between two nodes using Stop-and-Wait protocol. The frame size is 3000 bits. The transmission rate of the channel is 2000 bps (bits/second) and the propagation delay between the two nodes is 100 milliseconds. Assume that the processing times at the source and destination are negligible. Also, assume that the size of the acknowledgement packet is negligible. Which ONE of the following most accurately gives the channel utilization for the above scenario in percentage?

  1. (A)88.23
  2. (B)93.75
  3. (C)85.44
  4. (D)66.67
Answer: A - show solution

Transmission time = 3000 / 2000 = 1.5 s. Propagation delay = 0.1 s, so round trip = 0.2 s (ACK transmission negligible). Utilization = 1.5 / (1.5 + 0.2) = 1.5 / 1.7 = 0.88235, i.e. 88.23%. Answer: (A) 88.23.

Concept: Stop-and-wait efficiency = Tt / (Tt + 2Tp). The trap is using one-way delay, which gives 93.75%.

GATE 2024 S1 · Q16MCQ1 mark · -1/3· Easy

A user starts browsing a webpage hosted at a remote server. The browser opens a single TCP connection to fetch the entire webpage from the server. The webpage consists of a top-level index page with multiple embedded image objects. Assume that all caches (e.g., DNS cache, browser cache) are all initially empty. The following packets leave the user's computer in some order. (i) HTTP GET request for the index page (ii) DNS request to resolve the web server's name to its IP address (iii) HTTP GET request for an image object (iv) TCP SYN to open a connection to the web server Which one of the following is the CORRECT chronological order (earliest in time to latest) of the packets leaving the computer ?

  1. (A)(iv), (ii), (iii), (i)
  2. (B)(ii), (iv), (iii), (i)
  3. (C)(ii), (iv), (i), (iii)
  4. (D)(iv), (ii), (i), (iii)
Answer: C - show solution

The browser first needs the server's IP, so the DNS request (ii) leaves first. Then TCP opens the connection with SYN (iv). Over that connection it requests the index page (i), and only after parsing it does it learn of and request the embedded image (iii). Order: (ii), (iv), (i), (iii). Answer: (C).

Concept: Name resolution precedes connection setup, which precedes HTTP requests; embedded objects are fetched after the base page. The trap is putting SYN before DNS.

GATE 2024 S1 · Q29MSQ1 mark · no negative· Easy

TCP client P successfully establishes a connection to TCP server Q. Let N_P denote the sequence number in the SYN sent from P to Q. Let N_Q denote the acknowledgement number in the SYN ACK from Q to P. Which of the following statements is/are CORRECT?

  1. (A)The sequence number N_P is chosen randomly by P
  2. (B)The sequence number N_P is always 0 for a new connection
  3. (C)The acknowledgement number N_Q is equal to N_P
  4. (D)The acknowledgement number N_Q is equal to N_P + 1
Answer: A, D - show solution

The initial sequence number is chosen (pseudo)randomly to avoid confusion with old segments and spoofing, so (A) true and (B) false. A SYN consumes one sequence number, so Q acknowledges with the next expected number N_P + 1: (D) true, (C) false. Answer: (A), (D).

Concept: SYN and FIN each consume one sequence number even without data. The trap is thinking the ISN starts at 0 (tools like Wireshark show relative numbers).

GATE 2024 S1 · Q31MSQ1 mark · no negative· Easy

Which of the following fields is/are modified in the IP header of a packet going out of a network address translation (NAT) device from an internal network to an external network?

  1. (A)Source IP
  2. (B)Destination IP
  3. (C)Header Checksum
  4. (D)Total Length
Answer: A, C - show solution

Outbound through NAT, the private source IP is replaced by the NAT's public IP (and the transport port, but that is not in the IP header). Changing any header field forces recomputation of the IP header checksum. The destination IP and total length are unchanged. Answer: (A), (C).

Concept: NAT rewrites the source address on the way out and the destination on the way in, and every rewrite changes the checksum. The trap is forgetting the checksum.

GATE 2024 S1 · Q36MCQ2 marks · -2/3· Moderate

Consider a network path P−Q−R between nodes P and R via router Q. Node P sends a file of size 10⁶ bytes to R via this path by splitting the file into chunks of 10³ bytes each. Node P sends these chunks one after the other without any wait time between the successive chunk transmissions. Assume that the size of extra headers added to these chunks is negligible, and that the chunk size is less than the MTU. Each of the links P−Q and Q−R has a bandwidth of 10⁶ bits/sec, and negligible propagation latency. Router Q immediately transmits every packet it receives from P to R, with negligible processing and queueing delays. Router Q can simultaneously receive on link P−Q and transmit on link Q−R. Assume P starts transmitting the chunks at time t = 0. Which one of the following options gives the time (in seconds, rounded off to 3 decimal places) at which R receives all the chunks of the file?

  1. (A)8.000
  2. (B)8.008
  3. (C)15.992
  4. (D)16.000
Answer: B - show solution

There are 10⁶/10³ = 1000 chunks of 8000 bits; each takes 8000/10⁶ = 8 ms per link. P finishes sending the last chunk at 1000 × 8 ms = 8.000 s. Q is store-and-forward and pipelined, so the last chunk then needs one more 8 ms transmission on Q−R. R has everything at 8.000 + 0.008 = 8.008 s. Answer: (B) 8.008.

Concept: Pipelined store-and-forward: total = (number of packets + hops − 1) × per-packet time. The trap is doubling the whole file time (16 s) as if Q waited for the entire file.

GATE 2024 S1 · Q58NAT2 marks · no negative· Moderate

Consider the entries shown below in the forwarding table of an IP router. Each entry consists of an IP prefix and the corresponding next hop router for packets whose destination IP address matches the prefix. The notation "/N" in a prefix indicates a subnet mask with the most significant N bits set to 1. Prefix Next hop router 10.1.1.0/24 R1 10.1.1.128/25 R2 10.1.1.64/26 R3 10.1.1.192/26 R4 This router forwards 20 packets each to 5 hosts. The IP addresses of the hosts are 10.1.1.16, 10.1.1.72, 10.1.1.132, 10.1.1.191, and 10.1.1.205. The number of packets forwarded via the next hop router R2 is _______

Answer: 40 - show solution

Longest prefix match on the last octet. /25 R2 covers 128-255; /26 R3 covers 64-127; /26 R4 covers 192-255. 16: only /24, R1. 72: R3. 132: R2 (not in 192-255). 191: R2. 205: R4 (longer than /25). R2 gets 2 hosts × 20 = 40 packets. Answer: 40.

Concept: Longest prefix match picks the most specific entry. The trap is sending .205 to R2 or .191 to R4.

GATE 2024 S1 · Q65NAT2 marks · no negative· Moderate

Consider sending an IP datagram of size 1420 bytes (including 20 bytes of IP header) from a sender to a receiver over a path of two links with a router between them. The first link (sender to router) has an MTU (Maximum Transmission Unit) size of 542 bytes, while the second link (router to receiver) has an MTU size of 360 bytes. The number of fragments that would be delivered at the receiver is ________

Answer: 6 - show solution

Payload = 1400 B. Link 1 (MTU 542): max payload 522, rounded down to a multiple of 8 = 520, giving 520, 520, 360. Link 2 (MTU 360): max payload 340, rounded to 336. Router refragments each: 520 → 336 + 184; 520 → 336 + 184; 360 → 336 + 24. Fragments are not reassembled at the router, so the receiver gets 6. Answer: 6.

Concept: Fragment payloads (except the last) must be multiples of 8, and only the destination reassembles. The trap is not rounding to 8, or refragmenting the original datagram at the router (giving 5).

GATE 2024 S2 · Q23MSQ1 mark · no negative· Easy

Node X has a TCP connection open to node Y. The packets from X to Y go through an intermediate IP router R. Ethernet switch S is the first switch on the network path between X and R. Consider a packet sent from X to Y over this connection. Which of the following statements is/are TRUE about the destination IP and MAC addresses on this packet at the time it leaves X?

  1. (A)The destination IP address is the IP address of R
  2. (B)The destination IP address is the IP address of Y
  3. (C)The destination MAC address is the MAC address of S
  4. (D)The destination MAC address is the MAC address of Y
Answer: B - show solution

The IP destination is end-to-end, so it is Y's IP (B true, A false). The MAC destination is the next layer-3 hop: Y is on another network, so X sends the frame to its gateway R's MAC. A switch is transparent at layer 2 and is never addressed, so neither S's MAC (C) nor Y's MAC (D) is the destination. Answer: (B).

Concept: IP addresses are end-to-end, MAC addresses are hop-by-hop to the next router. Switches never appear as frame destinations.

GATE 2024 S2 · Q28MSQ1 mark · no negative· Easy

Which of the following statements about IPv4 fragmentation is/are TRUE?

  1. (A)The fragmentation of an IP datagram is performed only at the source of the datagram
  2. (B)The fragmentation of an IP datagram is performed at any IP router which finds that the size of the datagram to be transmitted exceeds the MTU
  3. (C)The reassembly of fragments is performed only at the destination of the datagram
  4. (D)The reassembly of fragments is performed at all intermediate routers along the path from the source to the destination
Answer: B, C - show solution

In IPv4 any router whose outgoing link MTU is smaller than the datagram fragments it (unless DF is set), so A is false and B is true. Fragments may travel different paths, so they are reassembled only at the final destination: C true, D false. Answer: (B), (C).

Concept: IPv4 allows fragmentation inside the network but reassembly only at the destination. Source-only fragmentation is IPv6 behaviour, a common mix-up.

GATE 2024 S2 · Q32MSQ1 mark · no negative· Easy

Which of the following fields of an IP header is/are always modified by any router before it forwards the IP packet?

  1. (A)Source IP Address
  2. (B)Protocol
  3. (C)Time to Live (TTL)
  4. (D)Header Checksum
Answer: C, D - show solution

Every router decrements TTL. Since TTL is covered by the header checksum, the checksum must be recomputed as well. The source address changes only under NAT (not at any router), and the protocol field is never changed in forwarding. Answer: (C), (D).

Concept: The TTL decrement forces a checksum update at each hop. NAT rewriting addresses is neither 'always' nor done by 'any router'.

GATE 2024 S2 · Q38MCQ2 marks · -2/3· Easy

Which one of the following CIDR prefixes exactly represents the range of IP addresses 10.12.2.0 to 10.12.3.255?

  1. (A)10.12.2.0/23
  2. (B)10.12.2.0/24
  3. (C)10.12.0.0/22
  4. (D)10.12.2.0/22
Answer: A - show solution

The range covers 2 × 256 = 512 addresses = 2⁹, so the prefix length is 32 − 9 = 23. The block must start on a multiple of 512: third octet 2 = 00000010, and 2 is a multiple of 2, so 10.12.2.0/23 covers 10.12.2.0 - 10.12.3.255 exactly. /24 is too small, /22 covers 10.12.0.0 - 10.12.3.255 (too large). Answer: (A) 10.12.2.0/23.

Concept: CIDR block size must be a power of two and aligned to its size. 10.12.2.0/22 is not even a valid block start.

GATE 2024 S2 · Q54NAT2 marks · no negative· Easy

Consider a TCP connection operating at a point of time with the congestion window of size 12 MSS (Maximum Segment Size), when a timeout occurs due to packet loss. Assuming that all the segments transmitted in the next two RTTs (Round Trip Time) are acknowledged correctly, the congestion window size (in MSS) during the third RTT will be _________

Answer: 4 - show solution

On timeout, ssthresh = 12/2 = 6 MSS and cwnd resets to 1 MSS, entering slow start. RTT 1: cwnd = 1; all acknowledged, cwnd = 2. RTT 2: cwnd = 2; all acknowledged, cwnd = 4. Still below ssthresh 6, so during the third RTT cwnd = 4 MSS. Answer: 4.

Concept: Timeout means cwnd = 1 and slow start (doubling per RTT) up to ssthresh = half the old window. Fast-recovery halving applies to triple duplicate ACKs, not timeouts.

GATE 2024 S2 · Q55NAT2 marks · no negative· Easy

Consider an Ethernet segment with a transmission speed of 10⁸ bits/sec and a maximum segment length of 500 meters. If the speed of propagation of the signal in the medium is 2 × 10⁸ meters/sec, then the minimum frame size (in bits) required for collision detection is _________

Answer: 500 - show solution

One-way propagation delay = 500 / (2 × 10⁸) = 2.5 µs. For CSMA/CD the transmission time must be at least the round-trip time 2 × 2.5 = 5 µs. Minimum frame = 10⁸ × 5 × 10⁻⁶ = 500 bits. Answer: 500.

Concept: CSMA/CD requires frame transmission time ≥ 2 × propagation delay so the sender is still transmitting when a collision returns. Using one-way delay gives 250, the trap.

GATE 2023 · Q17MCQ1 mark · -1/3· Easy

Suppose two hosts are connected by a point-to-point link and they are configured to use Stop-and-Wait protocol for reliable data transfer. Identify in which one of the following scenarios, the utilization of the link is the lowest.

  1. (A)Longer link length and lower transmission rate
  2. (B)Longer link length and higher transmission rate
  3. (C)Shorter link length and lower transmission rate
  4. (D)Shorter link length and higher transmission rate
Answer: B - show solution

For Stop-and-Wait, utilization U = Tt / (Tt + 2Tp) = 1 / (1 + 2a), where a = Tp/Tt, Tt = L/R is transmission time and Tp is propagation delay (grows with link length). U is lowest when a is largest: Tp large (longer link) and Tt small (higher transmission rate). Answer: (B) Longer link length and higher transmission rate.

Concept: Stop-and-Wait wastes the round trip; the waste dominates when the frame is sent quickly but takes long to propagate. The trap is thinking a faster link always improves utilization.

GATE 2023 · Q25MSQ1 mark · no negative· Easy

Which of the following statements is/are INCORRECT about the OSPF (Open Shortest Path First) routing protocol used in the Internet?

  1. (A)OSPF implements Bellman-Ford algorithm to find shortest paths.
  2. (B)OSPF uses Dijkstra's shortest path algorithm to implement least-cost path routing.
  3. (C)OSPF is used as an inter-domain routing protocol.
  4. (D)OSPF implements hierarchical routing.
Answer: A, C - show solution

OSPF is a link-state protocol: each router floods link-state advertisements and computes least-cost paths with Dijkstra's algorithm, so (B) is correct and (A), which describes distance-vector protocols like RIP, is incorrect. OSPF is an interior gateway protocol used within one autonomous system (intra-domain); inter-domain routing uses BGP, so (C) is incorrect. OSPF supports hierarchical routing through areas and a backbone area, so (D) is correct. The INCORRECT statements are (A) and (C).

Concept: Link-state = Dijkstra (OSPF); distance-vector = Bellman-Ford (RIP); inter-domain = BGP. Note the question asks for INCORRECT statements.

GATE 2023 · Q50MSQ2 marks · no negative· Moderate

Suppose you are asked to design a new reliable byte-stream transport protocol like TCP. This protocol, named myTCP, runs over a 100 Mbps network with Round Trip Time of 150 milliseconds and the maximum segment lifetime of 2 minutes. Which of the following is/are valid lengths of the Sequence Number field in the myTCP header?

  1. (A)30 bits
  2. (B)32 bits
  3. (C)34 bits
  4. (D)36 bits
Answer: B, C, D - show solution

Sequence numbers count bytes and must not wrap around within the maximum segment lifetime. Bytes sent in 2 minutes at 100 Mbps: 100 × 10⁶ / 8 × 120 = 1.5 × 10⁹ bytes. Need 2^b ≥ 1.5 × 10⁹. 2³⁰ ≈ 1.07 × 10⁹ is too small; 2³¹ ≈ 2.15 × 10⁹ suffices. So any length ≥ 31 bits is valid: 32, 34, 36 bits. (The window needed for the 150 ms RTT, about 1.9 MB, is much smaller.) Answer: (B), (C), (D).

Concept: Sequence space must exceed the bytes that can be in flight during one MSL so old segments cannot be confused with new ones. The trap is sizing by RTT instead of MSL.

GATE 2023 · Q52MSQ2 marks · no negative· Moderate

Suppose in a web browser, you click on the www.gate-2023.in URL. The browser cache is empty. The IP address for this URL is not cached in your local host, so a DNS lookup is triggered (by the local DNS server deployed on your local host) over the 3-tier DNS hierarchy in an iterative mode. No resource records are cached anywhere across all DNS servers. Let RTT denote the round trip time between your local host and DNS servers in the DNS hierarchy. The round trip time between the local host and the web server hosting www.gate-2023.in is also equal to RTT. The HTML file associated with the URL is small enough to have negligible transmission time and negligible rendering time by your web browser, which references 10 equally small objects on the same web server. Which of the following statements is/are CORRECT about the minimum elapsed time between clicking on the URL and your browser fully rendering it?

  1. (A)7 RTTs, in case of non-persistent HTTP with 5 parallel TCP connections.
  2. (B)5 RTTs, in case of persistent HTTP with pipelining.
  3. (C)9 RTTs, in case of non-persistent HTTP with 5 parallel TCP connections.
  4. (D)6 RTTs, in case of persistent HTTP with pipelining.
Answer: C, D - show solution

DNS: iterative lookup through root, TLD and authoritative servers costs 3 RTT. Non-persistent HTTP, 5 parallel connections: base HTML needs TCP setup + request = 2 RTT; the 10 objects go in two rounds of 5, each round 2 RTT (new TCP connection + request) = 4 RTT. Total 3 + 2 + 4 = 9 RTT. Persistent HTTP with pipelining: TCP setup 1 RTT, HTML 1 RTT, all 10 objects pipelined in 1 RTT, total 3 + 3 = 6 RTT. Answer: (C) and (D).

Concept: Count RTTs per phase: DNS levels, TCP handshake, each request round. Non-persistent pays a handshake per object; pipelining collapses all object requests into one RTT.

GATE 2023 · Q65NAT2 marks · no negative· Easy

The forwarding table of a router is shown below. Subnet Number Subnet Mask Interface ID 200.150.0.0 255.255.0.0 1 200.150.64.0 255.255.224.0 2 200.150.68.0 255.255.255.0 3 200.150.68.64 255.255.255.224 4 Default 0 A packet addressed to a destination address 200.150.68.118 arrives at the router. It will be forwarded to the interface with ID ________.

Answer: 3 - show solution

AND the destination with each mask. /16: 200.150.0.0, matches (interface 1). /19 (255.255.224.0): 68 AND 224 = 64, giving 200.150.64.0, matches (interface 2). /24: 200.150.68.0, matches (interface 3). /27 (255.255.255.224): 118 AND 224 = 96, giving 200.150.68.96 ≠ 200.150.68.64, no match. Longest prefix match is the /24 entry. Answer: 3.

Concept: Longest prefix matching: among all matching entries choose the longest mask. The trap is assuming 200.150.68.64/27 covers .118; that subnet is .64 to .95.

GATE 2022 · Q22MCQ1 mark · -1/3· Moderate

Consider an enterprise network with two Ethernet segments, a web server and a firewall, connected via three routers as shown below. Topology in the figure: the firewall connects to the Internet and has two further links - one to Router 1 and one to Router 2. Router 1 connects to Ethernet segment 1 (a LAN of hosts). Router 2 connects to the web server and to Router 3. Router 3 connects to Ethernet segment 2 (a LAN of hosts). What is the number of subnets inside the enterprise network?

  1. (A)3
  2. (B)12
  3. (C)6
  4. (D)8
Answer: C - show solution

Every router or firewall interface bounds a subnet, so count the distinct links inside the enterprise: firewall-Router 1, firewall-Router 2, Router 2-web server, Router 2-Router 3, Router 1-Ethernet segment 1, Router 3-Ethernet segment 2. That is 6 subnets (the Internet side of the firewall is outside the enterprise). Answer: (C) 6.

Concept: Routers separate subnets: each point-to-point router link and each LAN behind a router is its own subnet. The trap is counting only the two Ethernet LANs or including the Internet link.

GATE 2022 · Q35NAT1 mark · no negative· Moderate

Consider the resolution of the domain name www.gate.org.in by a DNS resolver. Assume that no resource records are cached anywhere across the DNS servers and that iterative query mechanism is used in the resolution. The number of DNS query-response pairs involved in completely resolving the domain name is_____________. Q.36 � Q.45 Multiple Choice Questions (MCQ), carry TWO marks each.

Answer: 4 - show solution

With nothing cached and iterative resolution, the resolver walks the hierarchy one level at a time: (1) root server, referred to the .in TLD server; (2) .in server, referred to the org.in server; (3) org.in server, referred to the gate.org.in server; (4) gate.org.in authoritative server returns the address of www.gate.org.in. That is 4 query-response pairs. Answer: 4.

Concept: In iterative resolution the resolver contacts each level of the domain hierarchy separately. The trap is counting only root, TLD and authoritative (3) and forgetting the second-level org.in zone.

GATE 2022 · Q55MSQ2 marks · no negative· Moderate

Consider routing table of an organization's router shown below: Subnet Number Subnet Mask Next Hop 12.20.164.0 255.255.252.0 R1 12.20.170.0 255.255.254.0 R2 12.20.168.0 255.255.254.0 Interface 0 12.20.166.0 255.255.254.0 Interface 1 default R3 Which of the following prefixes in CIDR notation can be collectively used to correctly aggregate all of the subnets in the routing table?

  1. (A)12.20.164.0/20
  2. (B)12.20.164.0/22
  3. (C)12.20.164.0/21
  4. (D)12.20.168.0/22
Answer: B, D - show solution

Third-octet ranges: 12.20.164.0/22 covers 164-167; 12.20.166.0/23 covers 166-167 (inside it); 12.20.168.0/23 covers 168-169; 12.20.170.0/23 covers 170-171. Together: 164-171. 12.20.164.0/22 (164-167) plus 12.20.168.0/22 (168-171) cover exactly this range. A /21 must start at a multiple of 8 (160), so 12.20.164.0/21 is not a valid aligned prefix and would mean 160-167. A /20 (160-175) also grabs addresses that should go to the default route R3. Answer: (B) and (D).

Concept: Aggregation must be exact and aligned: a /k block starts at a multiple of its size. The trap is picking a single larger prefix that over-covers or is misaligned.

GATE 2022 · Q57NAT2 marks · no negative· Moderate

Consider a network with three routers P, Q, R connected in a line: P - Q - R. All the links have cost of unity. The routers exchange distance vector routing information and have converged on the routing tables, after which the link Q-R fails. Assume that P and Q send out routing updates at random times, each at the same average rate. The probability of a routing loop formation (rounded off to one decimal place) between P and Q, leading to count-to-infinity problem, is___________.

Answer: 0.5 - show solution

Before the failure, P's distance to R is 2 (via Q). When Q-R fails, Q sets its distance to R to ∞. If Q sends its update first, P learns R is unreachable and no loop forms. If P sends first, it still advertises distance 2 to R, so Q adopts a route via P with cost 3 while P routes via Q: a loop that counts to infinity. With equal rates, each is first with probability 1/2. Answer: 0.5.

Concept: Count-to-infinity arises when a stale route is advertised back to the router that lost the link. Without split horizon, the outcome depends only on who updates first.

GATE 2022 · Q59NAT2 marks · no negative· Easy

Consider a 100 Mbps link between an earth station (sender) and a satellite (receiver) at an altitude of 2100 km. The signal propagates at a speed of 3×10⁸ m/s. The time taken (in milliseconds, rounded off to two decimal places) for the receiver to completely receive a packet of 1000 bytes transmitted by the sender is_________.

Answer: 7.07 to 7.09 - show solution

Transmission time = 1000 × 8 bits / 10⁸ bps = 8 × 10⁻⁵ s = 0.08 ms. Propagation delay = 2.1 × 10⁶ m / (3 × 10⁸ m/s) = 7 × 10⁻³ s = 7 ms. The last bit arrives after transmission + propagation = 7.08 ms. Answer: 7.08 (accepted range 7.07 to 7.09).

Concept: Time to fully receive a packet = transmission delay + propagation delay. The trap is counting only the propagation delay (7.00) or doubling it as for round-trip.

GATE 2022 · Q60NAT2 marks · no negative· Moderate

Consider the data transfer using TCP over a 1 Gbps link. Assuming that the maximum segment lifetime (MSL) is set to 60 seconds, the minimum number of bits required for the sequence number field of the TCP header, to prevent the sequence number space from wrapping around during the MSL is____________.

Answer: 33 - show solution

TCP numbers bytes. In one MSL the link can send 10⁹ bits/s × 60 s = 6 × 10¹⁰ bits = 7.5 × 10⁹ bytes. The sequence space must hold at least that many distinct numbers: 2ⁿ ≥ 7.5 × 10⁹. 2³² ≈ 4.29 × 10⁹ is too small and 2³³ ≈ 8.59 × 10⁹ suffices, so n = 33 bits. Answer: 33.

Concept: Wrap-around time = 2ⁿ / byte rate must exceed MSL. The trap is using bits instead of bytes, which gives 36.

GATE 2018 · Q23MCQ1 mark · -1/3· Easy

Match the following: Field Length in bits P. UDP Header's Port Number I. 48 Q. Ethernet MAC Address II. 8 R. IPv6 Next Header III. 32 S. TCP Header's Sequence Number IV. 16

  1. (A)P-III, Q-IV, R-II, S-I
  2. (B)P-II, Q-I, R-IV, S-III
  3. (C)P-IV, Q-I, R-II, S-III
  4. (D)P-IV, Q-I, R-III, S-II
Answer: C - show solution

UDP port numbers are 16 bits (P-IV). An Ethernet MAC address is 48 bits (Q-I). The IPv6 Next Header field is 8 bits (R-II). The TCP sequence number is 32 bits (S-III). Answer: (C) P-IV, Q-I, R-II, S-III.

Concept: Recall of standard header field sizes. Ports are 16 bits (0 to 65535), MAC 48, TCP sequence 32, IPv6 Next Header 8 (it plays the role of IPv4's Protocol field). The trap is confusing the 8-bit Next Header with the 32-bit sequence number.

GATE 2018 · Q24MCQ1 mark · -1/3· Easy

Consider the following statements regarding the slow start phase of the TCP congestion control algorithm. Note that cwnd stands for the TCP congestion window and MSS denotes the Maximum Segment Size. (i) The cwnd increases by 2 MSS on every successful acknowledgment. (ii) The cwnd approximately doubles on every successful acknowledgement. (iii) The cwnd increases by 1 MSS every round trip time. (iv) The cwnd approximately doubles every round trip time. Which one of the following is correct?

  1. (A)Only (ii) and (iii) are true
  2. (B)Only (i) and (iii) are true
  3. (C)Only (iv) is true
  4. (D)Only (i) and (iv) are true
Answer: C - show solution

In slow start, cwnd grows by 1 MSS for every ACK received, so (i) and (ii) are false. If cwnd is W segments, one RTT brings about W ACKs and adds W MSS, so cwnd roughly doubles each RTT: (iv) true. Growth of 1 MSS per RTT is congestion avoidance (additive increase), not slow start, so (iii) is false. Answer: (C) Only (iv) is true.

Concept: Slow start is exponential per RTT (+1 MSS per ACK); congestion avoidance is linear (+1 MSS per RTT). The trap is mixing the per-ACK and per-RTT descriptions of the two phases.

GATE 2018 · Q35NAT1 mark · no negative· Moderate

Consider a long-lived TCP session with an end-to-end bandwidth of 1 Gbps (= 10⁹ bits-per-second). The session starts with a sequence number of 1234. The minimum time (in seconds, rounded to the closest integer) before this sequence number can be used again is _______.

Answer: 34 to 35 - show solution

TCP numbers bytes with a 32-bit sequence number, so 2³² bytes are sent before a number repeats. At 10⁹ bits/s = 1.25 × 10⁸ bytes/s, the wrap-around time = 2³² / 1.25 × 10⁸ = 4294967296 / 125000000 ≈ 34.36 s. Answer: 34 (accepted range 34 to 35).

Concept: Sequence-number wrap-around time = 2³² bytes / bandwidth in bytes per second. The trap is forgetting that sequence numbers count bytes, not bits (which gives about 4.3 s).

GATE 2018 · Q64NAT2 marks · no negative· Easy

Consider an IP packet with a length of 4,500 bytes that includes a 20-byte IPv4 header and a 40-byte TCP header. The packet is forwarded to an IPv4 router that supports a Maximum Transmission Unit (MTU) of 600 bytes. Assume that the length of the IP header in all the outgoing fragments of this packet is 20 bytes. Assume that the fragmentation offset value stored in the first fragment is 0. The fragmentation offset value stored in the third fragment is _______.

Answer: 144 - show solution

IP payload = 4500 - 20 = 4480 bytes (the TCP header is part of the IP payload). Each fragment may carry at most 600 - 20 = 580 bytes, rounded down to a multiple of 8: 576 bytes. Offsets are stored in 8-byte units: fragment 1 starts at byte 0 (offset 0), fragment 2 at byte 576 (offset 72), fragment 3 at byte 1152 (offset 144). Answer: 144.

Concept: Fragment data sizes must be multiples of 8 (except the last), and the offset field counts 8-byte units. Traps: subtracting the TCP header from every fragment, or reporting the byte offset 1152 instead of 144.

GATE 2018 · Q65NAT2 marks · no negative· Moderate

Consider a simple communication system where multiple nodes are connected by a shared broadcast medium (like Ethernet or wireless). The nodes in the system use the following carrier-sense based medium access protocol. A node that receives a packet to transmit will carrier-sense the medium for 5 units of time. If the node does not detect any other transmission in this duration, it starts transmitting its packet in the next time unit. If the node detects another transmission, it waits until this other transmission finishes, and then begins to carrier-sense for 5 time units again. Once they start to transmit, nodes do not perform any collision detection and continue transmission even if a collision occurs. All transmissions last for 20 units of time. Assume that the transmission signal travels at the speed of 10 meters per unit time in the medium. Assume that the system has two nodes P and Q, located at a distance d meters from each other. P starts transmitting a packet at time t=0 after successfully completing its carrier-sense phase. Node Q has a packet to transmit at time t=0 and begins to carrier-sense the medium. The maximum distance d (in meters, rounded to the closest integer) that allows Q to successfully avoid a collision between its proposed transmission and P's ongoing transmission is _____.

Answer: 50 - show solution

Q senses the medium from t = 0 to t = 5. To avoid colliding it must detect P's signal within that window, after which it defers until P finishes. P's signal reaches Q at t = d/10. The condition d/10 ≤ 5 gives d ≤ 50 m. For any larger d, Q hears nothing during sensing, starts transmitting and collides with P's 20-unit transmission. Answer: 50.

Concept: Carrier sensing only works if the other station's signal arrives within the sensing interval: propagation delay ≤ sensing time. The trap is involving the 20-unit transmission time, which matters for collision detection but not for this deferral condition.

GATE 2016 Set1 · Q34MCQ1 mark · -1/3· Easy

Which one of the following protocols is NOT used to resolve one form of address to another one?

  1. (A)DNS
  2. (B)ARP
  3. (C)DHCP
  4. (D)RARP
Answer: C - show solution

DNS maps host names to IP addresses; ARP maps IP addresses to MAC addresses; RARP maps MAC addresses to IP addresses. DHCP dynamically assigns IP addresses and configuration to hosts; it does not translate one existing address form into another. Answer: (C) DHCP.

Concept: Distinguishing address resolution from address allocation. The trap is thinking of DHCP as a MAC-to-IP lookup like RARP; it is an allocation protocol.

GATE 2016 Set1 · Q35MCQ1 mark · -1/3· Moderate

Which of the following is/are example(s) of stateful application layer protocols? (i) HTTP (ii) FTP (iii) TCP (iv) POP3

  1. (A)(i) and (ii) only
  2. (B)(ii) and (iii) only
  3. (C)(ii) and (iv) only
  4. (D)(iv) only
Answer: C - show solution

HTTP is stateless: each request is independent (state is added via cookies, outside the protocol). TCP is stateful but is a transport-layer protocol, not application layer. FTP keeps session state (login, current directory, transfer mode) across commands, and POP3 keeps session state (authorization, transaction, update states). Stateful application-layer protocols: FTP and POP3. Answer: (C).

Concept: Know which protocols maintain server-side session state and at which layer each operates. The traps are HTTP (stateless) and TCP (stateful but transport layer).

GATE 2016 Set1 · Q62MCQ2 marks · -2/3· Easy

Consider that B wants to send a message m that is digitally signed to A. Let the pair of private and public keys for A and B be denoted by Kₓ⁻ and Kₓ⁺ for x = A, B, respectively. Let Kₓ(m) represent the operation of encrypting m with a key Kₓ and H(m) represent the message digest. Which one of the following indicates the CORRECT way of sending the message m along with the digital signature to A?

  1. (A){m, K_B⁺(H(m))}
  2. (B){m, K_B⁻(H(m))}
  3. (C){m, K_A⁻(H(m))}
  4. (D){m, K_A⁺(m)}
Answer: B - show solution

A digital signature is created by the sender encrypting the message digest with the sender's own private key, so anyone (here A) can verify it with the sender's public key K_B⁺. B is the sender, so the signature is K_B⁻(H(m)), sent along with m. B does not have A's private key, and encrypting with a public key gives confidentiality, not authentication. Answer: (B).

Concept: Signing uses the signer's private key; verification uses the signer's public key. The trap is confusing signing with encryption for confidentiality (receiver's public key).

GATE 2016 Set1 · Q63NAT2 marks · no negative· Easy

An IP datagram of size 1000 bytes arrives at a router. The router has to forward this packet on a link whose MTU (maximum transmission unit) is 100 bytes. Assume that the size of the IP header is 20 bytes. The number of fragments that the IP datagram will be divided into for transmission is ________.

Answer: 13 - show solution

Data in the datagram = 1000 - 20 = 980 bytes. Each fragment carries at most 100 - 20 = 80 data bytes, and 80 is a multiple of 8, as fragment offsets require. Fragments = ⌈980 / 80⌉ = ⌈12.25⌉ = 13 (12 of 80 bytes and one of 20 bytes). Answer: 13.

Concept: Fragment by data payload, with each fragment's data a multiple of 8 except the last. The trap is dividing 1000 by 100 (giving 10) and forgetting the header repeats in every fragment.

GATE 2016 Set1 · Q64NAT2 marks · no negative· Moderate

For a host machine that uses the token bucket algorithm for congestion control, the token bucket has a capacity of 1 megabyte and the maximum output rate is 20 megabytes per second. Tokens arrive at a rate to sustain output at a rate of 10 megabytes per second. The token bucket is currently full and the machine needs to send 12 megabytes of data. The minimum time required to transmit the data is ________ seconds.

Answer: 1.1 - show solution

Burst phase: sending at 20 MB/s while tokens arrive at 10 MB/s drains the full 1 MB bucket at a net 10 MB/s, so the burst lasts S = C / (M - ρ) = 1 / (20 - 10) = 0.1 s, sending 20 × 0.1 = 2 MB. The remaining 10 MB go at the token rate, 10 MB/s: 1 s. Total = 0.1 + 1 = 1.1 s. Answer: 1.1.

Concept: Token bucket burst length S = C / (M - ρ). The trap is computing 12 / 20 = 0.6 s, ignoring that the bucket empties and the rate then falls to the token rate.

GATE 2016 Set1 · Q65NAT2 marks · no negative· Moderate

A sender uses the Stop-and-Wait ARQ protocol for reliable transmission of frames. Frames are of size 1000 bytes and the transmission rate at the sender is 80 Kbps (1 Kbps = 1000 bits/second). Size of an acknowledgement is 100 bytes and the transmission rate at the receiver is 8 Kbps. The one-way propagation delay is 100 milliseconds. Assuming no frame is lost, the sender throughput is ________ bytes/second.

Answer: 2500 - show solution

Frame transmission time = 8000 bits / 80000 bps = 0.1 s. ACK transmission time = 800 bits / 8000 bps = 0.1 s. Propagation: 0.1 s each way, 0.2 s total. Cycle per frame = 0.1 + 0.1 + 0.1 + 0.1 = 0.4 s. Throughput = 1000 bytes / 0.4 s = 2500 bytes/second. Answer: 2500.

Concept: Stop-and-wait cycle time includes the ACK's transmission time when it is not negligible. The trap is ignoring the ACK time and getting 1000 / 0.3 ≈ 3333.

GATE 2016 Set2 · Q33MCQ1 mark · -1/3· Easy

Anarkali digitally signs a message and sends it to Salim. Verification of the signature by Salim requires

  1. (A)Anarkali's public key.
  2. (B)Salim's public key.
  3. (C)Salim's private key.
  4. (D)Anarkali's private key.
Answer: A - show solution

Anarkali signs with her own private key. Anyone can verify the signature with the matching public key, so Salim uses Anarkali's public key. Salim's keys play no role in a signature. Answer: (A).

Concept: Digital signatures: sign with the sender's private key, verify with the sender's public key. Encryption for confidentiality is the reverse (receiver's public key), which is the usual mix-up.

GATE 2016 Set2 · Q34MCQ1 mark · -1/3· Easy

In an Ethernet local area network, which one of the following statements is TRUE?

  1. (A)A station stops to sense the channel once it starts transmitting a frame.
  2. (B)The purpose of the jamming signal is to pad the frames that are smaller than the minimum frame size.
  3. (C)A station continues to transmit the packet even after the collision is detected.
  4. (D)The exponential backoff mechanism reduces the probability of collision on retransmissions.
Answer: D - show solution

(A) is false: CSMA/CD keeps sensing while transmitting in order to detect collisions. (B) is false: the jam signal tells all stations a collision occurred; padding is a separate field. (C) is false: the station aborts transmission on detecting a collision. (D) is true: binary exponential backoff widens the random wait range after each collision, reducing the chance of colliding again. Answer: (D).

Concept: CSMA/CD mechanics: listen while talking, abort and jam on collision, then back off exponentially. Confusing the jam signal with padding is the trap.

GATE 2016 Set2 · Q35MCQ1 mark · -1/3· Easy

Identify the correct sequence in which the following packets are transmitted on the network by a host when a browser requests a webpage from a remote server, assuming that the host has just been restarted.

  1. (A)HTTP GET request, DNS query, TCP SYN
  2. (B)DNS query, HTTP GET request, TCP SYN
  3. (C)DNS query, TCP SYN, HTTP GET request
  4. (D)TCP SYN, DNS query, HTTP GET request
Answer: C - show solution

After a restart the DNS cache is empty, so the host first sends a DNS query to resolve the server name. With the IP address it opens a TCP connection, starting with SYN. Only after the handshake can the HTTP GET be sent over that connection. Answer: (C) DNS query, TCP SYN, HTTP GET request.

Concept: Order of protocols when fetching a page: resolve the name, connect, then request. "Just been restarted" rules out a cached DNS answer.

GATE 2016 Set2 · Q63NAT2 marks · no negative· Easy

A network has a data transmission bandwidth of 20 × 10⁶ bits per second. It uses CSMA/CD in the MAC layer. The maximum signal propagation time from one node to another node is 40 microseconds. The minimum size of a frame in the network is ______ bytes.

Answer: 200 - show solution

For collision detection the transmission time must be at least the round-trip propagation time: L/B ≥ 2 × Tp. L = 2 × 40 × 10⁻⁶ × 20 × 10⁶ = 1600 bits = 200 bytes. Answer: 200.

Concept: CSMA/CD minimum frame: L = 2 × Tp × B. Using one-way propagation gives 100 bytes, the trap.

GATE 2016 Set2 · Q64MCQ2 marks · -2/3· Moderate

For the IEEE 802.11 MAC protocol for wireless communication, which of the following statements is/are TRUE? I. At least three non-overlapping channels are available for transmissions. II. The RTS-CTS mechanism is used for collision detection. III. Unicast frames are ACKed.

  1. (A)All I, II, and III
  2. (B)I and III only
  3. (C)II and III only
  4. (D)II only
Answer: B - show solution

I is true: in the 2.4 GHz band channels 1, 6 and 11 do not overlap. II is false: RTS-CTS is used for collision avoidance (reserving the medium, solving hidden terminals), since wireless stations cannot detect collisions while transmitting. III is true: 802.11 acknowledges every unicast data frame because collisions are not detected. Answer: (B) I and III only.

Concept: 802.11 uses CSMA/CA, not CSMA/CD. RTS-CTS avoids collisions; ACKs replace detection. Calling RTS-CTS collision detection is the trap.

GATE 2016 Set2 · Q65NAT2 marks · no negative· Moderate

Consider a 128 × 10³ bits/second satellite communication link with one way propagation delay of 150 milliseconds. Selective retransmission (repeat) protocol is used on this link to send data with a frame size of 1 kilobyte. Neglect the transmission time of acknowledgement. The minimum number of bits required for the sequence number field to achieve 100% utilization is ______.

Answer: 4 - show solution

Frame transmission time Tt = 8192 bits / 128000 bps = 64 ms. For 100% utilization the sender window must cover Tt + 2Tp = 64 + 300 = 364 ms, i.e. 364/64 = 5.69, so 6 frames. Selective repeat needs a sequence space of at least twice the window: 12, so ceil(log₂ 12) = 4 bits. (Taking 1 KB = 1000 bytes gives 362.5/62.5 = 5.8, again 6 frames and 4 bits.) Answer: 4.

Concept: Window size = 1 + 2a with a = Tp/Tt, rounded up; for selective repeat the sequence number space must be at least 2 × window. Forgetting the doubling gives 3 bits.

GATE 2014 Set1 · Q33MCQ1 mark · -1/3· Moderate

Consider the following three statements about link state and distance vector routing protocols, for a large network with 500 network nodes and 4000 links. [S1] The computational overhead in link state protocols is higher than in distance vector protocols. [S2] A distance vector protocol (with split horizon) avoids persistent routing loops, but not a link state protocol. [S3] After a topology change, a link state protocol will converge faster than a distance vector protocol. Which one of the following is correct about S1, S2, and S3?

  1. (A)S1, S2, and S3 are all true.
  2. (B)S1, S2, and S3 are all false.
  3. (C)S1 and S2 are true, but S3 is false.
  4. (D)S1 and S3 are true, but S2 is false.
Answer: D - show solution

S1: each link state router floods LSAs and runs Dijkstra over the whole 500-node, 4000-link topology, far more computation than a distance vector router's Bellman-Ford update from neighbours. True. S2: split horizon only prevents two-node loops; distance vector can still suffer count-to-infinity loops with three or more nodes. Link state, with a full topology at every node, does not form persistent loops. The statement has it backwards. False. S3: link state floods the change immediately and every router recomputes, so it converges faster than distance vector's iterative exchange. True. Answer: (D) S1 and S3 are true, but S2 is false.

Concept: Trade-offs between LS and DV routing: LS costs more computation but converges fast and is loop-resistant. The trap is believing split horizon fully cures distance vector's loop problem.

GATE 2014 Set1 · Q34MCQ1 mark · -1/3· Easy

Which of the following are used to generate a message digest by the network security protocols? (P) RSA (Q) SHA-1 (R) DES (S) MD5

  1. (A)P and R only
  2. (B)Q and R only
  3. (C)Q and S only
  4. (D)R and S only
Answer: C - show solution

A message digest is produced by a cryptographic hash function. SHA-1 and MD5 are hash functions. RSA is a public-key encryption / signature algorithm and DES is a symmetric block cipher; neither produces a digest. Answer: (C) Q and S only.

Concept: Distinguishing hash functions (digests) from encryption algorithms. The trap is picking RSA because it is used in digital signatures - the signature signs a digest computed by a hash such as SHA-1 or MD5.

GATE 2014 Set3 · Q33MCQ1 mark · -1/3· Easy

In the following pairs of OSI protocol layer/sub-layer and its functionality, the INCORRECT pair is

  1. (A)Network layer and Routing
  2. (B)Data Link Layer and Bit synchronization
  3. (C)Transport layer and End-to-end process communication
  4. (D)Medium Access Control sub-layer and Channel sharing
Answer: B - show solution

Routing is a network layer function, end-to-end process communication is the transport layer's job, and channel sharing (multiple access) is the MAC sub-layer's job. Bit synchronization - agreeing on bit timing on the medium - belongs to the physical layer; the data link layer handles framing and frame-level error control. Answer: (B).

Concept: Mapping functions to OSI layers. The trap is confusing bit synchronization (physical layer) with frame synchronization (data link layer).

GATE 2014 Set3 · Q35MCQ1 mark · -1/3· Moderate

Host A (on TCP/IP v4 network A) sends an IP datagram D to host B (also on TCP/IP v4 network B). Assume that no error occurred during the transmission of D. When D reaches B, which of the following IP header field(s) may be different from that of the original datagram D? (i) TTL (ii) Checksum (iii) Fragment Offset

  1. (A)(i) only
  2. (B)(i) and (ii) only
  3. (C)(ii) and (iii) only
  4. (D)(i), (ii) and (iii)
Answer: D - show solution

Every router decrements TTL, so TTL changes. Because the header changed, each router recomputes the header checksum, so the checksum changes. If a link on the path has a smaller MTU, a router fragments D, and the fragments that reach B carry non-zero fragment offsets. All three may differ. Answer: (D) (i), (ii) and (iii).

Concept: Which IPv4 header fields routers rewrite. The trap is forgetting fragmentation along the path, which changes the fragment offset (the question says 'may').

GATE 2014 Set3 · Q37NAT2 marks · no negative· Moderate

Every host in an IPv4 network has a 1-second resolution real-time clock with battery backup. Each host needs to generate up to 1000 unique identifiers per second. Assume that each host has a globally unique IPv4 address. Design a 50-bit globally unique ID for this purpose. After what period (in seconds) will the identifiers generated by a host wrap around?

Answer: 256 - show solution

Use the 32-bit IPv4 address to make IDs unique across hosts. To distinguish up to 1000 IDs within one second needs ⌈log₂ 1000⌉ = 10 bits. That leaves 50 - 32 - 10 = 8 bits for the clock (seconds counter). An 8-bit counter covers 2⁸ = 256 distinct seconds before repeating, so the IDs wrap around after 256 seconds. Answer: 256.

Concept: Bit budgeting for a unique ID: host part + per-second sequence + time part. The trap is using all 18 non-address bits for time and forgetting the per-second counter.

GATE 2005 · QundefinedMCQ1 mark · -1/3· Easy

An organization has a class B network and wishes to form subnets for 64 departments. The subnet mask would be:

  1. (A)255.255.0.0
  2. (B)255.255.64.0
  3. (C)255.255.128.0
  4. (D)255.255.252.0
Answer: D - show solution

A class B network's default mask is /16 (255.255.0.0). To form 64 subnets, need at least ceil(log2(64)) = 6 extra bits borrowed from the host portion. New prefix length = 16 + 6 = /22. /22 in dotted decimal: the third octet's top 6 bits are network bits (11111100 = 252), so the mask is 255.255.252.0.

Concept: The question gives the department COUNT (64), not a prefix length directly - the first step is always converting a required subnet count into the number of extra bits needed (2^bits >= count), THEN adding those bits to the class's default prefix length.