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Original, AI-authored English practice with worked guidance. Selected topics only; independent teacher review is pending. This is not a complete syllabus, official past-paper collection or predicted exam score. Check your own exam-board specification: these concepts are not certified against a particular board or exam year.
151 original questions across 21 selected topic areas. No account or payment required to practise.
Cambridge International AS & A Level Computer Science (9618), examination year 2027: AI-assisted conceptual alignment to official sources, not teacher or exam-board certification. Syllabus mapping and remaining coverage.
5 questions in this topic
Exact English revisions checked in an AI-assisted academic review; independent teacher approval is still pending.
All topics
Original English exercises targeting selected 9618 2027-29 requirements. Conceptual alignment only; independent teacher approval, practical competence and complete teaching coverage are not established.
Subtopics in this topic (3)
IPv4 subnet boundaries
Client/server and cloud trade-offs
TCP/IP service roles
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Question 1A host has IPv4 address192.168.10.130/26. Assume ordinary subnet network/broadcast reservations, no special routes, and one directly attached subnet. Find the dotted mask, network address, broadcast address, usable host range and usable count. Is192.168.10.200 directly on the same subnet? Explain routing and why a private address alone is not a complete security policy.6 indicative marks
Construct, trace and explain6 indicative marksNo calculator
A host has IPv4 address192.168.10.130/26. Assume ordinary subnet network/broadcast reservations, no special routes, and one directly attached subnet. Find the dotted mask, network address, broadcast address, usable host range and usable count. Is192.168.10.200 directly on the same subnet? Explain routing and why a private address alone is not a complete security policy.
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Worked guidance
/26 means24 network bits plus two in the last octet. Its mask there is11000000=192; blocks start0,64,128,192. ANDing130 with192 gives128. The six host bits run0..63, so this block ends191 and reserves its first/last address.
There are2^6-2=62 ordinary usable hosts. Octet200 AND192 gives192, different from130's network128, hence the router is needed under the assumptions. A private address can reduce direct public routability but cannot enforce which users/services are authorised.
Self-marking points
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Mask is255.255.255.192, leaving six host bits.
Network address192.168.10.128 and broadcast192.168.10.191.
Destination200 is in the192..255 block, not the128..191 block.
Send via an appropriate router/default gateway rather than treating200 as directly local.
Private addressing is not authentication/access control; local attackers or exposed services still need safeguards.
Examiner tip
Indicative distinct points, not an official Cambridge mark allocation. State bounds, assumptions and intermediate states; practical teacher review remains required.
Question 2A school needs centrally managed file permissions, twenty classroom clients and remote staff access. Propose client-server versus peer-to-peer and a wired star topology, with a failure trade-off. Compare a public cloud service with a private school-hosted service using availability and governance considerations. Name all four TCP/IP layers, place HTTP/SMTP/IMAP/FTP in the appropriate layer and state each service purpose. Explain one packet-switching benefit without claiming every packet arrives in order.6 indicative marks
Construct, trace and explain6 indicative marksNo calculator
A school needs centrally managed file permissions, twenty classroom clients and remote staff access. Propose client-server versus peer-to-peer and a wired star topology, with a failure trade-off. Compare a public cloud service with a private school-hosted service using availability and governance considerations. Name all four TCP/IP layers, place HTTP/SMTP/IMAP/FTP in the appropriate layer and state each service purpose. Explain one packet-switching benefit without claiming every packet arrives in order.
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Worked guidance
Choose central client-server accounts for this stated permissions requirement; document backups and server availability rather than implying the server cannot fail. A star switch connects each client on its own link: one disconnected classroom cable does not remove every other link, but the shared switch remains a failure concentration.
A public cloud service delegates infrastructure to a provider and depends on internet access and appropriate data governance. A private service gives the school more direct control but still needs patching, capacity and recovery planning. Neither option automatically guarantees privacy or zero downtime.
The TCP/IP sequence is Application -> Transport -> Internet -> Link when sending, reversed on receipt. HTTP/SMTP/IMAP/FTP are application services, not routing protocols. Packet switching can multiplex traffic and allow different paths; retransmission/reassembly are separate responsibilities when an ordered reliable service is needed.
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Client-server supports central accounts/permissions; peers do not provide that central administration by themselves.
A switch-centred star isolates a single client-cable failure, while central switch failure can disconnect many clients.
Public cloud may reduce local maintenance/provide remote scalability but adds provider/connectivity/governance dependence; private hosting retains control with local cost/resilience responsibility.
TCP/IP layers are Application, Transport, Internet, Link.
Packet switching shares links and can route separate packets flexibly; reliable ordering requires suitable transport mechanisms rather than a blanket IP guarantee.
Examiner tip
Indicative distinct points, not an official Cambridge mark allocation. State bounds, assumptions and intermediate states; practical teacher review remains required.
Question 3A prerecorded stream plays at constant 3 Mb/s from a buffer holding 12 Mb at playback start. During a 10 s interval the network supplies only 2 Mb/s. Use decimal bits, ignore overhead and assume uninterrupted delivery. Calculate buffer contents at10s and time to empty if these rates persist. Explain why lowering encoding bitrate could help but may affect quality.4 indicative marks
Analyse4 indicative marksNo calculator
A prerecorded stream plays at constant 3 Mb/s from a buffer holding 12 Mb at playback start. During a 10 s interval the network supplies only 2 Mb/s. Use decimal bits, ignore overhead and assume uninterrupted delivery. Calculate buffer contents at10s and time to empty if these rates persist. Explain why lowering encoding bitrate could help but may affect quality.
Cambridge 2027: Direct conceptual match
Simulated buffer12-t Mb to zero at12s; streaming bitrate is explicitly in existing9618 section2.1.
Playback consumes 3 Mb each second while delivery restores 2 Mb. The difference is a drain of 1 Mb/s, not a combined drain of 5.
After 10 s, 12-(1*10)=2 Mb remain. Maintaining the same rates exhausts 12 Mb in 12 s from the start, or two seconds after the stated interval.
Encoding at or below the sustained input rate can prevent this drain in the simplified model, but fewer transmitted bits can reduce visual/audio detail. Buffering only postpones a persistent deficit.
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Net buffer drain is1 Mb/s.
After10s buffer holds2 Mb.
It empties after12s from playback start.
Lower playback bitrate reduces drain but may reduce fidelity.
Examiner tip
Convert bytes to bits if necessary; all quantities here are already in bits.
Question 4A supplied reliable receiver buffers packets by sequence number and outputs only consecutive packets starting at Expected=1. After output it increments Expected; duplicates below Expected are ignored. Arrivals are (2,'B'),(1,'A'),(2,'B'),(3,'C'). Trace output after each arrival and final Expected. Explain why packet routing alone cannot implement this guarantee.4 indicative marks
Analyse4 indicative marksNo calculator
A supplied reliable receiver buffers packets by sequence number and outputs only consecutive packets starting at Expected=1. After output it increments Expected; duplicates below Expected are ignored. Arrivals are (2,'B'),(1,'A'),(2,'B'),(3,'C'). Trace output after each arrival and final Expected. Explain why packet routing alone cannot implement this guarantee.
Cambridge 2027: Direct conceptual match
Executed buffer/Expected state machine with out-of-order and duplicate arrivals; limited to selected packet-delivery concepts, not a full TCP implementation.
Packet2 cannot be output while packet1 is missing, so it waits in the buffer. On packet1, output1 then the buffered2, obtaining AB.
Expected is now3. The repeated packet2 is below that value and is ignored; packet3 then extends the output to ABC and advances Expected to4.
The outputs after arrivals are empty, AB, AB, ABC. Routers choose forwarding paths; the stated receiver additionally tracks delivery order and duplicates. This example alone does not specify a complete retransmission protocol.
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First arrival produces no output and buffers2.
Second outputs AB and sets Expected=3.
Duplicate2 changes nothing; packet3 gives ABC and Expected=4.
Ordering/duplicate control belongs to the supplied end-to-end receiver, not routing alone.
Examiner tip
A duplicate arrival must not cause duplicate application output.
Question 5A school's diagnostic program records ResolveOK and FetchOK booleans. It outputs NAME if NOT ResolveOK, otherwise SERVICE if NOT FetchOK, otherwise OK. Trace (FALSE,FALSE),(TRUE,FALSE),(TRUE,TRUE). Explain what successful DNS resolution establishes and why the second case does not by itself identify a failed web server.4 indicative marks
Analyse4 indicative marksNo calculator
A school's diagnostic program records ResolveOK and FetchOK booleans. It outputs NAME if NOT ResolveOK, otherwise SERVICE if NOT FetchOK, otherwise OK. Trace (FALSE,FALSE),(TRUE,FALSE),(TRUE,TRUE). Explain what successful DNS resolution establishes and why the second case does not by itself identify a failed web server.
Cambridge 2027: Direct conceptual match
Checked nested branch coverage and scoped DNS to9618 section2.1 (unlike AP CSP); no current domain records or external service tests used.
IF NOT ResolveOK THEN
OUTPUT "NAME"
ELSE
IF NOT FetchOK THEN
OUTPUT "SERVICE"
ELSE
OUTPUT "OK"
ENDIF
ENDIF
The first input fails the outer test; the second passes resolution but fails retrieval; the third passes both. This produces NAME,SERVICE,OK.
A resolved address does not establish that an application request can reach a working service. The SERVICE label is a diagnostic branch, not evidence distinguishing server failure from a blocked route or request.
Self-marking points
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Traces NAME,SERVICE,OK in order.
DNS supplies an address associated with a name.
Resolution is not successful retrieval of web content.
Fetch failure may involve connectivity, filtering or the service, so cause is not uniquely located.
Examiner tip
A diagnostic label is not proof of a unique underlying cause.
