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This final lesson is a synthesis and exam-technique session. Rather than introducing new material, it takes the whole networking topic — communication methods, the TCP/IP stack, IP addressing, protocols, network models, web technologies, encryption, security, and the Internet versus the Web — and works it through the way the examiner sees it: as multi-mark questions that demand precise vocabulary, applied reasoning and, in the longer items, genuine evaluation.
We work four full exam-style questions below, each spanning several areas of the specification, and each with a complete set of banded model answers (Mid-band, Stronger, Top-band), an AO breakdown, and examiner-style commentary explaining why each answer sits where it does. The recurring lesson is that A-Level marks are earned by explaining mechanisms and justifying choices, not by listing labels — so study not just what the top-band answers say but how they reason.
This lesson synthesises the entire Communication and networking area of the AQA A-Level Computer Science (7517) specification. It draws together communication methods (serial/parallel, duplex, synchronous/asynchronous), the four-layer TCP/IP stack and encapsulation, IP addressing and subnetting, network protocols (TCP/UDP, HTTP/HTTPS, DNS, DHCP, SMTP/IMAP, FTP/SFTP, SSH), client-server and peer-to-peer architectures, web technologies, encryption and digital signatures, network security, and the Internet/Web distinction with packet switching. Its purpose is exam technique: applying this combined knowledge to multi-part, multi-AO questions.
Every networking question is marked against three assessment objectives, and recognising which one a part is testing tells you how much depth to give:
| AO | What it rewards | Signalled by command words |
|---|---|---|
| AO1 | Knowledge and understanding of concepts | State, define, describe, give |
| AO2 | Applying knowledge to a specific scenario | Explain (in this context), calculate, show how |
| AO3 | Analysis and evaluation — weighing options, justifying a judgement | Evaluate, justify, discuss, compare and recommend |
The single biggest cause of dropped marks is answering an AO2/AO3 question at AO1 level — stating a fact when the question demanded that you apply it to the scenario or justify a choice. Watch the command word, and always tie your reasoning back to the specifics given in the question.
Networking exams almost always include a short calculation, and these three are the staple types. Knowing the method cold turns them into guaranteed, fast marks.
1. Transmission time — time = data size ÷ bit rate. The trap is units: convert bytes to bits (×8) and keep bits-per-second straight.
// How long to send a 10 MB file over a 100 Mbit/s link?
10 MB = 10 x 8 = 80 Mbit // bytes -> bits
time = 80 Mbit / 100 Mbit/s = 0.8 seconds
2. Usable hosts in a subnet — usable = 2^(host bits) − 2 (subtract the network and broadcast addresses).
// A /26 network — how many usable hosts?
host bits = 32 - 26 = 6
usable = 2^6 - 2 = 64 - 2 = 62 hosts
3. Subnet / host sizing — to make at least N subnets, find the smallest s with 2s≥N and add s to the prefix; to fit at least N hosts, find the smallest h with 2h−2≥N and the prefix is 32−h.
// Need at least 6 subnets from a /24?
2^2 = 4 (too few), 2^3 = 8 >= 6 -> borrow 3 bits -> /27
// Need at least 50 hosts per subnet?
2^5 - 2 = 30 (too few), 2^6 - 2 = 62 >= 50 -> 6 host bits -> /26
A method shown clearly earns the working marks even if a final number slips, so always write the steps, not just the answer.
Before writing any answer, identify the command word and pitch the depth to match it:
| Command word | What the examiner wants |
|---|---|
| State / give / name | A fact or term — one line, no explanation needed |
| Describe | An account of how something works, in steps |
| Explain | Reasons — why something is so, with a causal chain |
| Compare | Point-by-point similarities and differences |
| Justify / evaluate / discuss | A weighed judgement — pros and cons leading to a reasoned conclusion |
Misreading the command word is the most avoidable error in the whole paper: writing a one-line state answer to an explain question, or merely describing when asked to evaluate, leaves marks on the table no matter how accurate the content.
A weather station on a remote hillside sends sensor readings every few seconds over a long cable to a base unit, which then uploads the collected data to a cloud server over the Internet using a reliable connection.
(a) The cable from the sensor uses serial, asynchronous transmission. Explain why serial transmission, rather than parallel, is appropriate for this long cable. (3 marks) (b) Explain the purpose of the start and stop bits used in the asynchronous transmission. (2 marks) (c) When the base unit uploads data to the cloud server, the data passes down the four layers of the TCP/IP stack. Name the four layers in order and state the data-unit name produced at the transport and network layers. (4 marks)
AO breakdown: Part (a) is AO1/AO2 — knowledge of data skew applied to the long-cable scenario. Part (b) is AO1 — the role of framing bits. Part (c) is AO1 — recall of the layers and the data-unit names (segment, packet).
(a) Serial is better over a long cable because parallel can get muddled up over distance, so serial is more reliable. (b) The start bit shows the byte is starting and the stop bit shows it has ended. (c) The layers are application, transport, internet and link. The transport layer makes segments and the internet layer makes packets.
(a) Over a long cable, parallel transmission suffers from data skew — the separate wires differ slightly so bits arrive at slightly different times and cannot be reliably reassembled. Serial sends one bit at a time down a single channel, so there is no skew between bits, making it reliable over distance; it also needs fewer wires, so the cable is cheaper. (b) The start bit signals the receiver that a character is beginning and triggers its sampling; the stop bit marks the end of the character and returns the line to an idle state before the next start bit. (c) In order: application, transport, internet (network), link. The transport layer produces a segment (TCP); the network/internet layer produces a packet.
(a) Parallel transmission sends several bits at once over separate wires sampled on a shared clock; over a long cable, differences in wire length, resistance and capacitance cause data skew, so bits drift out of alignment with the clock and a bit from one cycle can fall into the next sampling window, corrupting the data (crosstalk between the closely packed wires adds further noise). Serial transmission sends one bit at a time over a single channel, so there is no inter-bit skew, making it far more reliable over this distance; it also uses fewer wires, so the cable is cheaper and thinner — and modern serial links run at high clock rates, so throughput is not sacrificed. For a long hillside cable, reliability and cost make serial the correct choice. (b) Because asynchronous transmission has no continuously shared clock, each character must be delimited individually: the start bit is a transition announcing that a character begins, allowing the receiver to lock its sampling clock to that character; the stop bit(s) mark the end of the character and guarantee a known idle state before the next start bit, so the next character's start is unambiguous. (c) Descending order: application → transport → network/internet → link. At the transport layer a TCP header (with port numbers and a sequence number) is added, producing a segment; at the network/internet layer an IP header (with source and destination IP addresses) is added, producing a packet. (The link layer would then frame this as a frame with MAC addresses and a CRC.)
Examiner-style commentary: The mid-band answer uses the right key terms ("muddled up", "start/stop", the layer names) but does not explain the mechanism of skew or why the framing bits are needed, so it caps at AO1 level. The stronger answer earns full marks on (b) and (c) and most of (a) by explaining skew and the framing-bit roles precisely. The top-band answer adds the underlying cause of skew (wire differences, sampling windows), notes crosstalk and the high-clock-rate counterpoint, and gives the data-unit names with the headers that create them — the precision and completeness expected at the top band.
A college is allocated the private network
192.168.20.0/24. The network manager needs to divide it into subnets so that six separate teaching departments each have their own subnet to contain their broadcast traffic.(a) State how many bits must be borrowed from the host portion to provide at least six subnets, and give the resulting CIDR prefix and subnet mask. (3 marks) (b) For the third subnet, give the network address, the usable host range and the broadcast address. (3 marks) (c) Explain why each subnet has two fewer usable host addresses than the total number of addresses it contains. (2 marks)
AO breakdown: Part (a) is AO2 — applying the bits-to-subnets relationship. Part (b) is AO2/AO3 — a multi-step calculation requiring correct method. Part (c) is AO1/AO2 — knowledge of reserved addresses explained.
(a) You need 3 bits because 3 bits give 8 subnets which is more than 6. The prefix is /27. (b) The third subnet starts at 192.168.20.64. The broadcast is 192.168.20.95. (c) Two addresses can't be used — the network address and the broadcast address.
(a) Six subnets need 3 borrowed host bits, because 22=4<6 but 23=8≥6. Borrowing 3 bits from /24 gives a /27 prefix, mask 255.255.255.224. (b) A /27 has 32−27=5 host bits, so 25=32 addresses per subnet (block size 32). Subnet 1 is .0, subnet 2 is .32, so the third subnet is network 192.168.20.64, usable range 192.168.20.65–192.168.20.94, broadcast 192.168.20.95. (c) The all-zeros host address is the network address and the all-ones host address is the broadcast address; neither can be assigned to a device, so there are two fewer usable hosts.
(a) The number of subnets created by borrowing b host bits is 2b. Here 22=4 is too few but 23=8≥6, so 3 bits must be borrowed, extending the prefix from /24 to /27 (mask 255.255.255.224, last octet 11100000). This yields 8 subnets for 6 departments — a slight, unavoidable waste, since subnet counts must be powers of two, leaving two spare subnets for future growth. (b) A /27 leaves 32−27=5 host bits, giving 25=32 addresses per subnet, so the block size in the last octet is 32. The subnets begin at .0, .32, .64, .96, ... The third subnet therefore has network address 192.168.20.64 (host bits all 0), usable host range 192.168.20.65 to 192.168.20.94, and broadcast address 192.168.20.95 (host bits all 1 — one below the next subnet's network address, .96). (c) Within every subnet two addresses are reserved and cannot be assigned to a host: the address with all host bits set to 0 identifies the network itself (the network address), and the address with all host bits set to 1 is the broadcast address, used to reach every host on that subnet at once. Hence a subnet of 2h total addresses provides only 2h−2 usable host addresses — here 25−2=30 per department.
Examiner-style commentary: The mid-band answer reaches the right numbers but skips the working — it does not derive the block size or show why /27, and (c) names the two addresses without explaining their function — so method marks are at risk. The stronger answer shows the 2b≥6 reasoning, derives the block size from the host bits, and correctly identifies the reserved addresses by role. The top-band answer is fully rigorous: it states the subnet-count formula, notes the power-of-two waste as an evaluative aside, derives every figure from the host-bit count, and explains the purpose of the network and broadcast addresses, concluding with the 2h−2 result applied to give 30 usable hosts.
A company lets staff work from home. They log in to an internal web application over HTTPS, transfer files using SFTP rather than plain FTP, and connect through a VPN. The web application's database was recently probed by an SQL injection attempt.
(a) Explain how HTTPS uses both asymmetric and symmetric encryption to protect the login. (4 marks) (b) Explain why the company uses SFTP instead of plain FTP. (2 marks) (c) Describe how an SQL injection attack works and state the most effective way to prevent it. (4 marks)
AO breakdown: Part (a) is AO1/AO2 — hybrid encryption applied to the login. Part (b) is AO2/AO3 — applying the plaintext-vulnerability idea and evaluating the security gain. Part (c) is AO1/AO2 — the attack mechanism and its named fix.
(a) HTTPS uses asymmetric encryption to share a key and then symmetric encryption for the data because symmetric is faster. (b) SFTP is used because it encrypts the files, but plain FTP sends them in plaintext which is not secure. (c) SQL injection is when an attacker types SQL into an input box to attack the database. You can prevent it with parameterised queries.
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