Computer Network Fundamentals and Architecture
The fundamentals domain is the frame everything else hangs on. This chapter develops whether you can place a function, a PDU, a device or a standard at the correct layer, and whether you can compute channel capacity and delay - not whether you can merely recite layer names.
Layered models and PDUs
- OSI: physical, data link, network, transport, session, presentation, application. TCP/IP: network interface, internet, transport, application. The upper three OSI layers collapse into the TCP/IP application layer.
- PDU names by layer: bit, frame, packet, segment (TCP) or datagram (UDP). Encapsulation introduces protocol control information such as headers or trailers at the relevant layers; decapsulation reverses it at the receiver.
- Presentation handles encoding, compression and encryption; session handles dialogue control and resynchronisation - a frequent discrimination question.
- Devices by layer: repeater/hub (1), bridge/switch/NIC (2), router (3); gateway is a context-dependent term, including an IP default gateway.
Topologies and standardisation
- Star: easy management, central-node single point of failure. Bus: shared medium, collisions. Ring: deterministic token access. Mesh: potentially greater path redundancy and greater link cost, used in WAN backbones and data-centre fabrics.
- Bodies: ISO (OSI, also ISO/IEC joint standards), ITU-T (telecom recommendations, X series), IEEE 802 (LAN/MAN), IETF (RFCs), TIA/EIA (cabling), China's GB/T and YD series.
- Number anchors: 802.3 Ethernet, 802.11 WLAN, 802.1 Q VLAN tagging, 802.1 D STP, 802.1 X port access control.
Data-communication arithmetic
- Bit rate R = baud rate B × log 2 N, where N is the number of signal states per symbol.
- Nyquist (noiseless low-pass): C = 2 W·log 2 N.
- Shannon: C = W·log 2(1 + S/N), with S/N as a power ratio, not in dB.
- Total delay = transmission delay + propagation delay + processing + queueing. Transmission delay = frame length / bandwidth; propagation delay = distance / signal velocity.
Worked example
A 3 kHz line with a 30 dB signal-to-noise ratio. Convert: S/N = 10^(30/10) = 1000. Then C = 3000 × log 2(1001) ≈ 3000 × 9.97 ≈ 29.9 kb/s. Substituting 30 directly into the logarithm gives ≈14.9 kb/s - the classic distractor answer.
Traps that cost marks
- Treating dB as a ratio in Shannon's formula.
- Confusing transmission delay (depends on frame size and bandwidth) with propagation delay (depends on distance and medium).
- Assuming every switch is one broadcast domain: VLAN partitioning and Layer 3 functionality change the answer.
- Assuming a perfect one-to-one mapping between OSI's three upper layers and TCP/IP's single application layer.