Notes
The last mixed set before the drill and the final test.
Practice questions
- NAT · 1 mark
One packet, 5 links store-and-forward, Tt = 2 ms and Tp = 1 ms per link. Arrival time (ms)?
Write your answer before opening the key.
- NAT · 1 mark
Bits that can be sent in one RTT of 50 ms on a 1 Gbps link, in millions (B × RTT)?
Write your answer before opening the key.
- MSQ · 1 mark
A router has two interfaces; one leads to a switch with 5 hosts. Which are true?
- The switch side has 5 host collision domains plus its router link.
- All hosts on the switch are one broadcast domain.
- The switch blocks broadcasts.
- The router splits broadcast domains.
- NAT · 1 mark
Ethernet frame size (bytes) carrying 100 data bytes?
Write your answer before opening the key.
- NAT · 1 mark
Usable host addresses in one classful Class B network?
Write your answer before opening the key.
- MCQ · 2 marks
Aggregate 200.1.1.0/26, 200.1.1.64/26, 200.1.1.128/26, 200.1.1.192/26:
- 200.1.1.0/24
- 200.1.1.0/25
- 200.1.0.0/23
- 200.1.1.0/26
- NAT · 1 mark
IPv4 header with HLEN = 6 and total length 1024. Data bytes?
Write your answer before opening the key.
- NAT · 2 marks
NAPT using source ports 1024–65535 on one public IP: maximum simultaneous mappings?
Write your answer before opening the key.
- NAT · 1 mark
TCP sequence wrap-around time at 10 Gbps, in seconds (3 decimals)?
Write your answer before opening the key.
- NAT · 2 marks
SRTT = 50 ms, RTTVAR = 5 ms, new sample 70 ms, α = 1/8, β = 1/4 (RFC 6298 order). New RTO (ms)?
Write your answer before opening the key.
- MCQ · 1 mark
HTTP 304 means:
- OK
- Not Modified
- Not Found
- Moved Permanently
- NAT · 2 marks
Both stations at their 3rd collision. P(station A wins) (4 decimals)?
Write your answer before opening the key.
Answer key
Try the questions first. Open when you are ready to check.
- 15 — 5 × (2 + 1) = 15 ms.
- 50 — 10⁹ × 0.05 = 5 × 10⁷ bits.
- ABD — Switch splits collisions; router splits broadcasts.
- C: Switches flood broadcasts.
- 118 — 14 + 100 + 4.
- 65534 — 2¹⁶ − 2.
- A — Four aligned /26 blocks = one /24.
- B: Covers only two.
- C: Too large.
- D: Only one block.
- 1000 — Header 24 bytes; 1024 − 24 = 1000.
- 64512 — 65535 − 1024 + 1 = 64,512.
- 3.43..3.44 — 2³² × 8 / 10¹⁰ = 3.436 s.
- 87.5 — RTTVAR = 0.75·5 + 0.25·20 = 8.75; SRTT = 52.5; RTO = 52.5 + 35 = 87.5.
- B — Cached copy still valid.
- A: 200.
- C: 404.
- D: 301.
- 0.4375 — Pairs with KA < KB: 28 of 64.