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S4 ICT: Computer Networks and Internet Protocols (Paper 3 (challenge))
Tutor Prep Guide · LAN versus WAN, IPv4 and IPv6 address length, the 8-bit range of each octet, subnet masks and usable hosts, HTTP/HTTPS/FTP/SMTP with their ports, TCP versus UDP, and bandwidth to download time
S4 ICT: Computer Networks and Internet Protocols (Paper 3 (challenge))
Paper 3 (challenge): ten questions on networks and internet protocols, with answers and notes.
Direct answers Level: S4 ICT; format: IP addresses, subnet masks, usable hosts, ports, TCP and UDP, bandwidth arithmetic.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 |
| Suggested time | 20 |
| Difficulty mix | Foundation 3 · Intermediate 3 · Challenge 3 · Error redo 1 |
| Chapter | Chapter: S4 ICT: Computer Networks and Internet Protocols |
| Level | S4 |
02 Questions (10 items)
- FoundationWith an MTU of 1 500 bytes and 1 460 bytes of payload per packet, how many packets does a 20 MB (20 000 000 bytes) file need?
- FoundationHow many usable host addresses does a /28 subnet provide?
- FoundationUnder a /24, are 192.168.1.100 and 192.168.1.200 in the same subnet?
- IntermediateA department needs 32 computers. A /27 gives only 30, so what is the shortest prefix that fits?
- IntermediateHow many seconds (whole number) does an 8 GB (take it as 8 000 MB) file need at 120 Mbps?
- IntermediateWhat does NAT mainly do?
- ChallengeSubnets 192.168.10.0/24 and 192.168.11.0/24 are merged with the mask 255.255.254.0. How many hosts fit?
- ChallengeWith a round-trip time of 40 ms, how long does the TCP three-way handshake (about 1.5 RTT) take?
- Challenge100 computers each need 2 Mbps on average. With 4 to 1 oversubscription, how much uplink is needed?
- Error redoA student says IPv6 simply widens IPv4 from 32 bits to 64 bits. What is the correct length?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | 13699 packets | 20 000 000 divided by 1 460 is 13 698.6, and a partial packet still has to be sent, so round up to 13699. |
| 2 | 14 | A /28 leaves 4 host bits: 2 to the power 4 is 16, and minus 2 gives 14. |
| 3 | Yes, the first 24 bits are identical | A /24 compares only the first three octets, and both are 192.168.1, so they share a subnet and can talk through a switch without a router. |
| 4 | /26 (62 hosts) | A /27 offers 30 usable addresses, two short, so step up one level to a /26 with 62. |
| 5 | 533 seconds | 8 000 MB times 8 is 64 000 Mb; dividing by 120 Mbps gives 533.3, so 533 seconds, nearly nine minutes. |
| 6 | It lets many private devices share one public IP address | NAT rewrites the source address at the router so many internal hosts share one public IP, saving addresses and hiding the inside network. |
| 7 | 510 hosts | The merged block is a /23 with 9 host bits: 2 to the power 9 is 512, minus 2 gives 510. |
| 8 | 60 ms | SYN, SYN-ACK and ACK take one and a half round trips, so 40 times 1.5 gives 60 ms. |
| 9 | 50 Mbps | Total demand is 100 times 2, that is 200 Mbps, and dividing by 4 gives 50 Mbps, assuming users rarely peak together. |
| 10 | 128 bits | IPv6 uses 128 bits: four times the bits of IPv4 and 2 to the 96 more addresses, written as eight hexadecimal groups. |
04 What to do after marking
- After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
- Repeat the same item two days later; move to the next paper only when both attempts are correct.
- Parents only need the answer and note columns; there is no need to explain every item.
First ask how many bits the address has and how many host bits remain, then decide between 2 to the power minus two and a plain subtraction; name both the protocol and its port, and convert MB to Mb before dividing by bandwidth.Tutor Registration →
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