Introduction and Layering
What a network is, the OSI and TCP/IP models, encapsulation, which device works at which layer, and the four kinds of delay.
A computer network is a set of devices (hosts, switches, routers) connected by links that carry data between them. The Internet is a network of networks, all speaking the same protocols.
Why layers
Sending a web page involves many separate problems: turning bits into signals, getting a frame to the next machine, finding a route across the world, recovering lost data, and understanding HTTP. Layering gives each problem to one layer. Each layer:
- offers a service to the layer above it,
- uses the service of the layer below it, and
- talks to the same layer on the other machine using a protocol.
A layer can change its implementation (for example, Wi-Fi instead of Ethernet) without the others noticing.
The OSI and TCP/IP models
| # | OSI layer | Job | Unit of data (PDU) | TCP/IP model |
|---|---|---|---|---|
| 7 | Application | Services for applications (HTTP, DNS, SMTP) | Message | Application |
| 6 | Presentation | Data format, encryption, compression | Message | Application |
| 5 | Session | Dialogue control, checkpoints | Message | Application |
| 4 | Transport | Process-to-process delivery, reliability, flow and congestion control | Segment (TCP) / datagram (UDP) | Transport |
| 3 | Network | Host-to-host delivery across networks: addressing and routing | Packet (datagram) | Internet |
| 2 | Data link | Node-to-node delivery on one link: framing, MAC addresses, error detection | Frame | Link |
| 1 | Physical | Bits as signals on the medium | Bit | Link |
The TCP/IP model merges OSI layers 5–7 into one application layer, and layers 1–2 into the link layer (some books count physical separately and call it a five-layer model).
A mnemonic for OSI from the bottom: Please Do Not Throw Sausage Pizza Away.
Encapsulation
On the way down, each layer adds its own header (and the data link layer also a trailer) to what it gets from above. The receiver strips them in reverse order.
Addresses at each layer
| Layer | Address | Example | Identifies |
|---|---|---|---|
| Transport | Port number (16 bits) | 443 | A process on a host |
| Network | IP address (32 bits in IPv4) | 192.168.1.10 | A host interface, globally |
| Data link | MAC address (48 bits) | 3c:22:fb:1a:09:7e | A network card, on the local link |
Devices and layers
| Device | Highest layer it understands | What it does |
|---|---|---|
| Repeater, hub | Physical (1) | Regenerates bits; a hub repeats them to every port |
| Bridge, switch | Data link (2) | Forwards frames using MAC addresses |
| Router | Network (3) | Forwards packets between networks using IP addresses |
| Gateway, firewall, proxy | Up to application (7) | Translates or filters at higher layers |
A hub forms one collision domain and one broadcast domain. A switch splits collision domains (one per port) but keeps one broadcast domain. A router splits broadcast domains too.
Delays
The time for one packet to cross one link is the sum of four delays:
| Delay | Formula | Depends on |
|---|---|---|
| Transmission delay | Tt = L / R (packet length / bandwidth) | Packet size and link speed |
| Propagation delay | Tp = d / v (distance / signal speed) | Link length (v ≈ 2 × 10⁸ m/s in copper or fibre) |
| Queuing delay | Waiting in the router's buffer | Traffic load |
| Processing delay | Checking headers, looking up routes | Router speed |
Transmission delay is the time to push all bits onto the link; propagation delay is the time for one bit to travel to the other end. A wide motorway (high bandwidth) gets the cars on faster, but does not make the journey shorter.
The bandwidth–delay product R × Tp is the number of bits "in flight" on the link at once: the link's capacity as a pipe.
L = 1000 * 8 # packet: 1000 bytes, in bits
R = 1_000_000 # bandwidth: 1 Mbps
d = 2_000_000 # distance: 2000 km, in metres
v = 2e8 # signal speed in m/s
Tt = L / R
Tp = d / v
print(f"transmission delay = {Tt * 1000:.1f} ms")
print(f"propagation delay = {Tp * 1000:.1f} ms")
print(f"total (no queuing) = {(Tt + Tp) * 1000:.1f} ms")
print(f"bandwidth-delay product = {R * Tp:.0f} bits")transmission delay = 8.0 ms propagation delay = 10.0 ms total (no queuing) = 18.0 ms bandwidth-delay product = 10000 bits
Practice questions
A 1500-byte packet is sent over a 10 Mbps link that is 3000 km long, with a signal speed of 2 × 10⁸ m/s. Ignoring queuing and processing, how many milliseconds does it take from the first bit leaving the sender to the last bit arriving?
Show answer and explanation
Answer: 16.2Transmission delay = 12,000 bits / 10⁷ bps = 1.2 ms. Propagation delay = 3 × 10⁶ m / 2 × 10⁸ m/s = 15 ms. The last bit leaves after 1.2 ms and arrives 15 ms later: 16.2 ms.
A 100 Mbps link has a round-trip time of 40 ms. How many bytes can the sender transmit during one round-trip time?
Show answer and explanation
Answer: 500000Bits in one RTT = 10⁸ × 0.04 = 4 × 10⁶ bits = 500,000 bytes. This is why TCP needs large windows on fast long-distance links.
Which layer is responsible for process-to-process delivery?
Show answer and explanation
Answer: BThe network layer delivers packets from host to host; the transport layer uses port numbers to deliver data to the right process on that host.
Which statements are true? (Select all that apply.)
Show answer and explanation
Answer: A, B, DEach switch port is its own collision domain, but broadcasts still reach every port. A router does not forward link-layer broadcasts. A hub repeats every bit to every port, so all its ports share one collision domain.
Doubling the bandwidth of a link changes which delay?
Show answer and explanation
Answer: ATransmission delay is L / R, so doubling R halves it. Propagation delay d / v depends only on distance and the medium.