CS Atlas GATE 2027
Computer Networks course
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Computer Networks/Phase 1 · Layering, Switching & Performance

Chapter review L1.3 · Item 22 of 90

Performance metrics

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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

  1. 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.

  2. 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.

  3. 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.

  4. MCQ · 1 mark

    Which quantity is unchanged when a link is made twice as long?

    1. Propagation delay
    2. Transmission delay
    3. a = Tp/Tt
    4. Bandwidth-delay product
  5. NAT · 1 mark

    Bandwidth-delay product (in bytes) of a 400 Mbps link with Tp = 2 ms?

    Write your answer before opening the key.

  6. MSQ · 1 mark

    Which are true?

    1. End-to-end throughput ≤ the slowest link rate.
    2. Circuit total time includes M/R once, not h times.
    3. Message switching beats packet switching on multi-hop paths.
    4. RTT for a tiny signal on one link is 2·Tp.
  7. 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.

  8. 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.

  9. MCQ · 1 mark

    Two hosts are separated by 3 routers. How many links does the path have?

    1. 2
    2. 3
    3. 4
    4. 6

Answer key

Try the questions first. Open when you are ready to check.
  1. 3 — Tt = 1 ms, Tp = 2 ms, total 3 ms.

    Review lesson 1.3.2

  2. 13 — (24)(0.5) + 5(0.2) = 12 + 1 = 13 ms.

    Review lesson 1.3.2

  3. 0.165..0.168 — Tt = 6 ms; U = 6/(6 + 30) = 0.1667.

    Review lesson 1.3.3

  4. B — Tt = L/B has no distance.
    • A: Tp doubles.
    • C: a doubles.
    • D: B × Tp doubles.

    Review lesson 1.3.1

  5. 100000 — 8 × 10⁵ bits = 10⁵ bytes.

    Review lesson 1.3.3

  6. ABD — Bottleneck rule; no store-and-forward on circuits; RTT definition.
    • C: Packets pipeline; whole messages wait at every hop.

    Review lesson 1.3.4

  7. 2.13 — 0.1 + 2 + 0.03 = 2.13 s.

    Review lesson 1.3.4

  8. 10 — 1 + 2a = 10.

    Review lesson 1.3.3

  9. C — Routers + 1 = 4 links.
    • A: Too few.
    • B: Counting routers, not links.
    • D: Double counting.

    Review lesson 1.3.2

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