Chapter recap
Tt = L/B (size and rate, not distance). Tp = d/v (distance, v = 2 × 10⁸ m/s). Nodal delay = processing + queuing + Tt + Tp. N packets over h equal links: (N + h − 1)·Tt + h·Tp; unequal links: the slowest sets the spacing. RTT = 2·Tp; BDP = B × Tp; a = Tp/Tt; send-and-wait U = 1/(1 + 2a). Circuit total = setup + M/R + h·Tp; packets pipeline but pay headers; optimal p ≈ √(MH/(h − 1)).
Merged rule card
From 1.3.1 Bandwidth, data rate, throughput and transmission delay
- Bandwidth / data rate B = link capacity in bps (decimal prefixes).
- Tt = L / B (L in bits) — independent of distance.
- Throughput = bits delivered / time ≤ min link rate on the path (bottleneck).
From 1.3.2 Propagation, queuing and processing delay; multi-hop pipelining
- Tp = d / v, v = 2 × 10⁸ m/s default; independent of size and rate.
- Nodal delay = d_proc + d_queue + Tt + Tp.
- One packet, h equal links: h·Tt + h·Tp.
- N packets, h equal links: (N + h − 1)·Tt + h·Tp (+ (h − 1)·d_proc).
- Unequal links: first packet ΣTt + ΣTp, then + (N − 1) × max Tt.
- Traffic intensity L·a/R → 1: queues explode.
From 1.3.3 RTT, bandwidth-delay product, utilisation and a = Tp/Tt
- RTT = 2·Tp for tiny signals (add Tt / queuing if stated).
- BDP = B × Tp bits in flight; B × RTT before any reply.
- a = Tp / Tt.
- Send-and-wait: U = Tt/(Tt + 2Tp) = 1/(1 + 2a); throughput = U·B.
- To keep the link busy: about 1 + 2a units in flight.
From 1.3.4 Circuit vs packet switching: delay numericals
- Circuit: setup + M/R + h·Tp.
- Message switching: h·(M + H)/B + h·Tp.
- Packet switching: (⌈M/p⌉ + h − 1)·(p + H)/B + h·Tp.
- Optimal p ≈ √(M·H/(h − 1)); in MCQs evaluate each option.
Practice questions
- NAT · 1 mark
A 1250-byte packet on a 10 Mbps link, 400 km long (v = 2 × 10⁸ m/s). Time until the last bit arrives, in ms?
Write your answer before opening the key.
- NAT · 2 marks
20 packets, each Tt = 0.5 ms, over 5 equal links, Tp = 0.2 ms per link. Total in ms?
Write your answer before opening the key.
- NAT · 2 marks
B = 2 Mbps, Tp = 15 ms, units of 1500 bytes, send-and-wait. Utilisation (3 decimals)?
Write your answer before opening the key.
- MCQ · 1 mark
Which quantity is unchanged when a link is made twice as long?
- Propagation delay
- Transmission delay
- a = Tp/Tt
- Bandwidth-delay product
- NAT · 1 mark
Bandwidth-delay product (in bytes) of a 400 Mbps link with Tp = 2 ms?
Write your answer before opening the key.
- MSQ · 1 mark
Which are true?
- End-to-end throughput ≤ the slowest link rate.
- Circuit total time includes M/R once, not h times.
- Message switching beats packet switching on multi-hop paths.
- RTT for a tiny signal on one link is 2·Tp.
- NAT · 2 marks
Circuit: setup 100 ms, file 5 × 10⁵ bits on a circuit of 250 kbps, 3 links with Tp = 10 ms. Total in seconds?
Write your answer before opening the key.
- NAT · 1 mark
How many units must be in flight to keep a link with a = 4.5 fully busy (whole number)?
Write your answer before opening the key.
- MCQ · 1 mark
Two hosts are separated by 3 routers. How many links does the path have?
- 2
- 3
- 4
- 6
Answer key
Try the questions first. Open when you are ready to check.
- 3 — Tt = 1 ms, Tp = 2 ms, total 3 ms.
- 13 — (24)(0.5) + 5(0.2) = 12 + 1 = 13 ms.
- 0.165..0.168 — Tt = 6 ms; U = 6/(6 + 30) = 0.1667.
- B — Tt = L/B has no distance.
- A: Tp doubles.
- C: a doubles.
- D: B × Tp doubles.
- 100000 — 8 × 10⁵ bits = 10⁵ bytes.
- ABD — Bottleneck rule; no store-and-forward on circuits; RTT definition.
- C: Packets pipeline; whole messages wait at every hop.
- 2.13 — 0.1 + 2 + 0.03 = 2.13 s.
- 10 — 1 + 2a = 10.
- C — Routers + 1 = 4 links.
- A: Too few.
- B: Counting routers, not links.
- D: Double counting.