Computer Networks Quick Revision Notes

These notes follow the order most university syllabi and placement interviews use: models first, then one layer at a time from the bottom up. Each section ends with the points examiners and interviewers ask most. The solved numericals were checked by running code, so you can trust the final answers.

1. Network models

OSI vs TCP/IP

Data units: bits (physical), frames (data link), packets (network), segments for TCP or datagrams for UDP (transport), messages (application).

Devices by layer: hub and repeater (1), switch and bridge (2), router (3). A gateway can work at any layer up to 7.

Encapsulation: each layer adds its header on the way down (and the data link layer also adds a trailer); the receiver strips them on the way up.

Key terms

• Bandwidth: the range of frequencies a channel carries, in Hz. In computing it's often used loosely for data rate in bits per second.
• Throughput: the data rate actually achieved.
• Latency: total delay = transmission + propagation + queuing + processing.
• Transmission delay = frame length ÷ data rate.
• Propagation delay = distance ÷ signal speed (about 2 × 10⁸ m/s in cable and fibre).

2. Physical layer

• Nyquist (noiseless channel): maximum data rate C = 2 × B × log₂(L), where B is bandwidth in Hz and L the number of signal levels.
• Shannon (noisy channel): C = B × log₂(1 + SNR), with SNR as a plain ratio. Convert from decibels: SNR = 10^(dB/10).
• Guided media: twisted pair, coaxial cable, optical fibre (highest bandwidth, immune to electromagnetic interference). Unguided: radio, microwave, infrared.
• Switching: circuit switching reserves a path for the whole call (old telephones). Packet switching sends independent packets that share links (the internet).

3. Data link layer

Framing and error control

• Framing methods: character count, byte stuffing with flag bytes, bit stuffing (insert a 0 after five consecutive 1s), physical-layer coding violations.
• Parity detects any odd number of bit errors, but misses an even number.
• CRC: append r zero bits (r = degree of the generator), divide by the generator using XOR, and append the remainder. The receiver divides again; remainder 0 means no error was detected.
• Hamming code: places parity bits at positions 1, 2, 4, 8 and so on. The syndrome gives the position of a single-bit error, so it can correct it.
• Detecting d errors needs a minimum Hamming distance of d + 1; correcting d errors needs 2d + 1.

Flow control: ARQ protocols

Let a = propagation delay ÷ transmission delay.

Go-Back-N resends the lost frame and every frame after it. Selective Repeat resends only the lost frame, so it needs a buffer at the receiver.

Media access

• Pure ALOHA: transmit any time. Maximum throughput 1 ÷ (2e), about 18.4%.
• Slotted ALOHA: transmit only at slot boundaries. Maximum 1 ÷ e, about 36.8%.
• CSMA/CD (classic Ethernet): listen before sending, detect collisions, back off. The minimum frame must last at least one round trip: L_min = 2 × propagation delay × data rate.
• CSMA/CA (Wi-Fi): avoids collisions with random backoff and optional RTS/CTS, because wireless stations can't reliably detect collisions.

Ethernet, switches and ARP

• A MAC address is 48 bits, written in hexadecimal, and identifies the network interface.
• Switches learn which MAC address sits on which port and forward frames only there. Hubs repeat every frame to every port.
• ARP finds the MAC address for an IPv4 address on the local network by broadcast.

4. Network layer

IPv4 addressing

127.x.x.x is loopback. Private ranges: 10.0.0.0/8, 172.16.0.0/12 and 192.168.0.0/16.

Subnetting: with a /n prefix there are 2^(32 − n) addresses per subnet and 2^(32 − n) − 2 usable hosts, because the first address is the network address and the last is the broadcast address. CIDR drops classes and uses any prefix length.

Routing and related protocols

• Distance vector (RIP): each router shares its table with neighbours. Simple, but slow to converge and prone to count-to-infinity. Fixes include split horizon and poison reverse. RIP's maximum is 15 hops.
• Link state (OSPF): each router floods link information and computes shortest paths with Dijkstra's algorithm. Converges faster.
• BGP: routes between autonomous systems on the internet (path vector).
• ICMP: error and diagnostic messages; ping and traceroute use it.
• NAT: maps private addresses to a public address, saving IPv4 space.
• DHCP: assigns IP addresses automatically through Discover, Offer, Request, Acknowledge (DORA).
• IPv6: 128-bit addresses, a simpler fixed header, no broadcast (multicast and anycast instead), and no fragmentation by routers.

5. Transport layer

TCP vs UDP

• Three-way handshake: SYN, SYN-ACK, ACK. Closing: FIN and ACK in each direction.
• Flow control protects the receiver, using the advertised window.
• Congestion control protects the network: slow start doubles the congestion window each round trip until it reaches the threshold (ssthresh), then congestion avoidance adds 1 per round trip. On a timeout, ssthresh becomes half the current window and the window resets to 1. On three duplicate ACKs, fast retransmit and fast recovery halve the window instead of resetting it.
• Ports: 0–1023 are well-known (80 HTTP, 443 HTTPS, 22 SSH, 25 SMTP, 53 DNS). A socket is an IP address plus a port.

6. Application layer

DNS resolution: your resolver asks a root server, then the top-level domain server (.com), then the authoritative server for the domain. Recursive means the resolver does the work for you; iterative means each server refers you to the next one.

HTTP methods: GET (read), POST (create), PUT (replace), PATCH (update part), DELETE. GET, PUT and DELETE are idempotent: repeating them has the same effect as doing them once. Status codes: 2xx success, 3xx redirect, 4xx client error, 5xx server error.

7. Security basics

• Symmetric encryption (AES) uses one shared key and is fast. Asymmetric (RSA, elliptic-curve) uses a public and private key pair and solves key exchange.
• TLS uses asymmetric cryptography to agree a session key, then symmetric encryption for the data. Certificates prove the server's identity.
• Hashing (SHA-256) gives a fixed-size fingerprint and can't be reversed. Digital signatures are a hash encrypted with the sender's private key.
• Firewalls filter traffic by rules. VPNs tunnel encrypted traffic over a public network.

Solved numericals

Every answer below was checked by running code.

1. Transmission and propagation delay

A 1,500-byte frame is sent at 10 Mbps over a 2,000 km link (signal speed 2 × 10⁸ m/s). Find both delays.

Transmission delay = (1,500 × 8) ÷ (10 × 10⁶) = 1.2 ms. Propagation delay = (2,000 × 10³) ÷ (2 × 10⁸) = 10 ms.

2. Stop-and-wait efficiency

For the link in example 1, find the efficiency of stop-and-wait.

a = 10 ÷ 1.2 = 8.33. Efficiency = 1 ÷ (1 + 2 × 8.33) = 1 ÷ 17.67 = 5.66%.

3. Go-Back-N with window 7

Same link, sender window 7. Efficiency = 7 ÷ 17.67 = 39.6%.

4. Window and sequence bits for 100% efficiency

The window must be at least 1 + 2a = 17.67, so 18 frames. Go-Back-N needs 2ᵏ − 1 ≥ 18, so k = 5 bits. Selective Repeat needs 2ᵏ⁻¹ ≥ 18, so k = 6 bits.

5. Nyquist

A noiseless 3,000 Hz channel uses 4 signal levels. C = 2 × 3,000 × log₂4 = 12,000 bps.

6. Shannon

A 3,000 Hz channel has an SNR of 30 dB. SNR = 10³ = 1,000. C = 3,000 × log₂(1,001) ≈ 29,902 bps, about 30 kbps.

7. CSMA/CD minimum frame size

10 Mbps Ethernet, maximum cable length 2,500 m, signal speed 2 × 10⁸ m/s. Propagation delay = 12.5 µs. L_min = 2 × 12.5 × 10⁻⁶ × 10 × 10⁶ = 250 bits.

8. CRC

Data 1101011011, generator 10011 (x⁴ + x + 1). Append four zeros and divide by XOR: the remainder is 1110, so the transmitted frame is 11010110111110. Dividing the received frame by 10011 leaves 0000, so no error is detected.

9. Hamming (7,4), even parity

Encode data 1011, with bits in the order p1 p2 d1 p4 d2 d3 d4. p1 checks positions 1, 3, 5, 7; p2 checks 2, 3, 6, 7; p4 checks 4, 5, 6, 7. The codeword is 0110011. If bit 6 flips (received 0110001), recomputing the checks gives a syndrome of 6, which points at the error.

10. Subnetting a /24 into four

Split 192.168.10.0/24 into 4 equal subnets. Borrow 2 bits, so each is a /26 with 64 addresses and 62 usable hosts.

11. Network and broadcast address from a host

For 172.16.45.200/20: the mask is 255.255.240.0. The third octet 45 falls in the block 32–47, so the network is 172.16.32.0, the broadcast is 172.16.47.255, the usable hosts run from 172.16.32.1 to 172.16.47.254, and there are 2¹² − 2 = 4,094 of them.

12. TCP congestion window

ssthresh = 16 segments, the window starts at 1, and a timeout happens in the round where the window reaches 20. Window per round: 1, 2, 4, 8, 16 (slow start), then 17, 18, 19, 20 (congestion avoidance). After the timeout, ssthresh = 20 ÷ 2 = 10 and the window restarts: 1, 2, 4, 8, 10, 11.

Bandwidth-delay product: a 100 Mbps link with a 40 ms round trip holds 100 × 10⁶ × 0.04 = 4,000,000 bits (500 KB) in flight. That's the window needed to keep the link full.

Last-minute checklist

• Draw the OSI and TCP/IP models side by side, with one protocol per layer.
• Write the formulas for transmission delay, propagation delay, Nyquist, Shannon and ARQ efficiency from memory.
• Do one CRC division and one Hamming encoding by hand.
• Find the network, broadcast and host range for any IP with a prefix.
• Explain the difference between Go-Back-N and Selective Repeat in two sentences.
• Trace the TCP window through slow start, congestion avoidance and a timeout.
• List the ports for HTTP, HTTPS, DNS, SMTP, SSH and FTP.
• Explain what happens when you type a URL: DNS, TCP, TLS, HTTP request, response.

Revising other subjects too? See the DBMS quick revision notes and the operating systems important questions.