Short on time? This page is the recap. For explanations and worked examples, use the full study guide.
Recap
- Computing device: a physical artifact that can run a program. Computing system: devices and programs working together. Computer network: interconnected devices that send or receive data.
- Path: a sequence of directly connected devices from sender to receiver. Routing: finding a path.
- Bandwidth: maximum data per fixed time, usually in bits per second.
- The Internet: interconnected networks using standardised, open protocols. Protocol: an agreed-upon set of rules.
- Routing is usually dynamic; the Internet is scalable.
- Data travel as a stream of packets: a data chunk plus metadata for routing and reassembly. Packets may arrive in order, out of order or not at all.
- IP, TCP and UDP are common protocols.
- The World Wide Web is a system of linked pages, programs and files; it uses the Internet and HTTP.
Key relationships
| Idea | How to work it out |
|---|---|
| Number of paths | List sequences of linked devices from sender to receiver with no repeats |
| Transfer time | File size in bits ÷ bandwidth in bits per second |
| File size to bits | bytes × 8 |
| Reassembly | Use the packet numbers in the metadata |
| Dynamic routing | A different route can be chosen if one is busy or fails |
| Internet versus Web | Internet is the network; the Web is a service on it (HTTP) |
Assumptions and conventions
- Bandwidth problems assume the link is the only limit and the data flow continuously.
- 1 byte = 8 bits. Say which unit you use at each step.
- In diagrams, count links (connections), not devices, when giving the length of a path.
Mistakes to avoid
- Forgetting to multiply bytes by 8 before dividing by bandwidth in bits per second.
- Calling the Web “the Internet”.
- Counting devices instead of links in a path.
- Assuming packets arrive in order.
- Saying a network has one path when the diagram shows alternatives.
- Describing protocols as private to one company.
Two quick worked examples
1. A 12 MB photo crosses a 24 Mbps link. 12 × 8 = 96 megabits; 96 ÷ 24 = 4 seconds.
2. A message is sent as packets 1 to 5 and arrives in the order 2, 1, 5, 3. The receiver can reorder using the packet numbers to get 1, 2, 3, 5 and sees that packet 4 is missing. With a protocol that re-requests missing packets, packet 4 is asked for again; with one that does not, there is a gap.
Quick self-check
- A 10 MB file crosses a 40 Mbps link. How long does it take? (10 × 8 = 80 megabits; 80 ÷ 40 = 2 seconds.)
- Why can a receiver put out-of-order packets back in order? (Each packet’s metadata includes its place in the message.)
- Is email part of the Web? (No; it uses the Internet but is not a linked page system over HTTP.)
Next: practice questions.