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GATE CSE 2027 · SECTION 10 · COMPUTER NETWORKS

GATE CSE 2027 Computer Networks: Syllabus, Weightage, Subnetting, TCP, Routing & Solved Problems

GATE CSE Computer Networks 2027 guide: Layering, switching, error detection, MAC and Ethernet, routing, IPv4 with CIDR, NAT and fragmentation, TCP flow and congestion control, sockets, DNS and HTTP: the full Computer Networks section for GATE CSE 2027 with worked numericals.

GATE CSE Computer Networks 2027 syllabus, subnetting, TCP and routing
8 to 9
Marks (approx., recent papers)
Mostly NAT
Question style
IPv4 + TCP
Highest-yield topics
₹499
GATE CSE Test Series (50% off)

By Piyush Wairale (M.Tech, IIT Madras) · Former Instructor, IIT Madras BS Degree Programme · Mentor to 20,000+ GATE aspirants · Updated November 2026

Quick Answer

• Computer Networks (CN) is Section 10 of GATE CSE 2027 and typically carries 8 to 9 marks.

• Syllabus: layering; circuit, packet and virtual circuit switching and performance metrics; error detection, MAC, Ethernet; distance vector and link state routing; IPv4 fragmentation, CIDR, NAT; TCP flow and congestion control, socket API; DNS and HTTP.

• The most repeated numericals: subnetting/CIDR, IP fragmentation, sliding window efficiency, TCP congestion window and CSMA/CD minimum frame size.

• CN numericals reward formula fluency. Drill it with the GATE CSE 2027 Test Series by Piyush Wairale.

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GATE CSE Computer Networks Syllabus 2027 (Official, IIT Madras)

Section 10 of the CS paper: Principles of layering; basics of switching (circuit, packet and virtual circuit) and performance metrics; data link layer: error detection, Medium Access Control, Ethernet; distance vector and link state routing; IPv4: fragmentation, CIDR notation, Network Address Translation; TCP: flow control and congestion control, socket API; DNS and HTTP.

Full paper structure: GATE CS 2027 syllabus and exam pattern.

Topic-wise Weightage and Priority (GATE CSE Computer Networks)

Topic Typical question Priority
IPv4, CIDR, fragmentation, NAT Subnet masks, host counts, route aggregation, fragment offsets Very high
TCP flow & congestion control cwnd evolution, slow start/AIMD, sequence numbers, window size Very high
Data link: error detection, MAC CRC remainder, sliding window efficiency, CSMA/CD frame size High
Routing Distance vector table updates, count to infinity, link state (Dijkstra) Medium
Switching & performance Total delay for packet vs circuit switching Medium
DNS, HTTP, sockets Persistent vs non-persistent HTTP RTTs, socket call order Medium

Layering, Switching and Performance Metrics

Each layer provides a service to the one above and adds its own header (encapsulation). Transmission delay = L/B; propagation delay = d/v; with store-and-forward packet switching across k links, end-to-end delay for one packet is about k × (L/B) + total propagation delay, and pipelining several packets reduces total time. Circuit switching adds setup time but no per-hop store-and-forward delay. Throughput, bandwidth-delay product (B × RTT) and utilisation are the core metrics.

Error detection: parity, checksum and CRC. For CRC with a generator of degree r, append r zeros, divide (XOR) and send the remainder; CRC detects all single-bit errors and all burst errors of length ≤ r. Hamming distance d detects d − 1 errors and corrects ⌊(d − 1)/2⌋.

Flow control at the link: with a = Tp/Tt, stop-and-wait efficiency = 1/(1 + 2a); a sliding window of size W gives efficiency min(1, W/(1 + 2a)). Go-Back-N with k sequence bits allows W ≤ 2k − 1; Selective Repeat allows W ≤ 2k−1.

MAC: pure ALOHA max throughput 18.4%, slotted ALOHA 36.8%. For CSMA/CD, minimum frame size = 2 × Tp × bandwidth. Ethernet uses 48-bit MAC addresses and binary exponential backoff.

Distance Vector and Link State Routing

Distance vector (Bellman-Ford style): each router shares its vector with neighbours; Dx(y) = min over neighbours v of (c(x,v) + Dv(y)). Slow to converge and prone to count-to-infinity (mitigated by split horizon and poison reverse). Link state: each router floods its link costs and runs Dijkstra on the full topology; converges faster at the cost of more computation and flooding.

IPv4: Fragmentation, CIDR and NAT

CIDR: a /n block has 232−n addresses and 232−n − 2 usable hosts. Routers use longest prefix match. Route aggregation merges contiguous, aligned blocks into one shorter prefix.

Fragmentation: data per fragment must be a multiple of 8 bytes and fit in MTU − header. The fragment offset field counts in 8-byte units. Only the last fragment has MF = 0. Reassembly happens only at the destination.

NAT maps many private addresses (10.0.0.0/8, 172.16.0.0/12, 192.168.0.0/16) to one or a few public addresses using port numbers, conserving IPv4 space but breaking end-to-end addressing.

TCP: Flow Control, Congestion Control and the Socket API

TCP is byte-oriented: sequence numbers count bytes, and SYN and FIN each consume one sequence number. Flow control uses the receiver’s advertised window (rwnd). Congestion control: slow start doubles cwnd every RTT until ssthresh, then congestion avoidance adds 1 MSS per RTT. On timeout, ssthresh = cwnd/2 and cwnd = 1 MSS; on 3 duplicate ACKs (fast retransmit/fast recovery in Reno), ssthresh = cwnd/2 and cwnd = ssthresh. The effective send window = min(cwnd, rwnd). Socket API: server calls socket, bind, listen, accept; client calls socket, connect; both then read/write and close.

DNS and HTTP

DNS maps names to IP addresses using a hierarchy (root, TLD, authoritative) via recursive or iterative queries, mostly over UDP port 53. HTTP runs over TCP (port 80). Non-persistent HTTP needs 2 RTTs per object (TCP handshake + request); persistent HTTP reuses the connection, and with pipelining several objects can be fetched in about one RTT after the first.

Five GATE-Style Numericals, Solved (GATE CSE Computer Networks)

Q1 (NAT). How many usable host addresses are in a /26 block?
Solution: 26 − 2 = 62.

Q2 (NAT). A 4000-byte IP datagram (20-byte header) crosses a link with MTU 1500. How many fragments, and what is the offset field of the last one?
Solution: Data = 3980 B. Max data per fragment = 1480 (a multiple of 8). Fragments carry 1480, 1480, 1020: 3 fragments. Last offset = 2960/8 = 370.

Q3 (NAT). Transmission time 1 ms, propagation delay 2 ms. Stop-and-wait efficiency, and the window size for full utilisation?
Solution: a = 2, efficiency = 1/(1 + 4) = 20%; window = 1 + 2a = 5.

Q4 (NAT). CSMA/CD on a 10 Mbps link with a round-trip propagation time of 51.2 µs. Minimum frame size?
Solution: 107 × 51.2 × 10−6 = 512 bits = 64 bytes.

Q5 (NAT). TCP cwnd is 20 MSS when a timeout occurs. What are the new ssthresh and cwnd, and what is cwnd after 5 more successful RTTs?
Solution: ssthresh = 10, cwnd = 1. Slow start: 2, 4, 8, then capped at 10 (reaching ssthresh), then congestion avoidance gives 11. cwnd after 5 RTTs = 11 MSS.

CN numericals are free marks if your formulas are automatic.

The GATE CSE 2027 Test Series has Computer Networks subject-wise tests on subnetting, fragmentation, sliding windows and TCP congestion control, with detailed step-by-step solutions.

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Common Mistakes to Avoid (GATE CSE Computer Networks)

Forgetting to subtract 2 (network and broadcast) for usable hosts.

Writing the fragment offset in bytes instead of 8-byte units.

Using RTT instead of one-way propagation (or the reverse) in the CSMA/CD and efficiency formulas: read what the question gives.

Letting slow start overshoot ssthresh.

Forgetting SYN and FIN consume sequence numbers in TCP sequence questions.

4-Week Preparation Strategy (GATE CSE Computer Networks)

Week 1: Layering, switching, delays and data link (error detection, sliding window, MAC). Week 2: IPv4, CIDR, subnetting, fragmentation and NAT (50+ numericals). Week 3: Routing, TCP flow and congestion control, sockets, DNS and HTTP. Week 4: 10 years of GATE CS CN questions and the CN tests in the GATE CSE 2027 Test Series, then full-length mocks across all ten subjects.

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More GATE courses and notes: piyushwairale.com

FAQs: GATE CSE Computer Networks 2027

How many marks does Computer Networks carry in GATE CSE?

About 8 to 9 marks in recent GATE CS papers, mostly numerical answer type.

Is IPv6 in the GATE CSE 2027 syllabus?

No. The syllabus lists IPv4 (fragmentation, CIDR, NAT) only.

Are UDP, email and network security asked?

The 2027 syllabus names TCP, DNS and HTTP explicitly; UDP appears mainly as DNS’s transport. Email protocols and network security are not listed.

Which is the best test series for GATE CSE 2027?

The GATE CSE 2027 Test Series by Piyush Wairale (IIT Madras) on piyushwairale.com offers subject-wise and full-length mocks with MCQ, MSQ and NAT questions, detailed solutions and performance analysis for ₹499. See the test series.

Which CN topic should I master first?

IPv4 addressing with CIDR and subnetting. It appears every year and is needed to understand routing and NAT questions.

GATE CSE 2027: All Subject-wise Guides

1. Engineering Mathematics · 2. Digital Logic · 3. Computer Organization & Architecture · 4. Programming & Data Structures · 5. Algorithms · 6. Theory of Computation · 7. Compiler Design · 8. Operating System · 9. Databases (DBMS) · 10. Computer Networks

That completes this GATE CSE Computer Networks guide. Full paper overview: GATE CS 2027 Syllabus, Exam Pattern & Cutoff. Practise all ten subjects in one place with the GATE CSE 2027 Test Series.

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