CS Atlas GATE 2027
Computer Networks course
0 / 90 complete

Computer Networks/Phase 2 · Data Link Layer

Chapter review L2.3 · Item 36 of 90

Medium Access Control

Read at your own pace. Try the questions before opening the answers.

On this page

Chapter recap

Partitioning: TDMA/FDMA give R/N, no collisions, idle waste. ALOHA: pure S = G·e⁻²ᴳ (vulnerable 2T, max 0.184 at G = 0.5); slotted S = G·e⁻ᴳ (vulnerable T, max 0.368 at G = 1). CSMA: listen first (1-, non-, p-persistent); CSMA/CD: abort + jam, Lmin = 2·Tp·B. Backoff: after n collisions K ∈ {0..2^min(n,10) − 1}, abort at 16. Taking turns: polling; token ring ETR 1/(1 + a/N), DTR 1/(1 + a(N+1)/N).

Merged rule card

From 2.3.1 The multiple access problem; collisions; channel partitioning

  • Collision: overlapping transmissions on a shared link; all involved frames lost.
  • MAC families: partitioning (TDMA/FDMA), random access (send freely, recover from collisions), taking turns (an orderly schedule).
  • TDMA/FDMA: R/N per station, no collisions, idle shares wasted, wait up to (N − 1) slots.

From 2.3.2 Pure and slotted ALOHA: vulnerable time, throughput, maximum efficiency

  • T = frame time; G = attempts per T.
  • Pure: vulnerable 2T, S = G·e⁻²ᴳ, max 1/(2e) ≈ 0.184 at G = 0.5.
  • Slotted: vulnerable T, S = G·e⁻ᴳ, max 1/e ≈ 0.368 at G = 1.
  • Finite N: P(success) = N·p·(1 − p)ᴺ⁻¹.

From 2.3.3 CSMA and its persistence variants; CSMA/CD; minimum frame size = 2·Tp·B

  • CSMA: sense before sending; collisions still possible within T_p.
  • 1-persistent / non-persistent / p-persistent: what to do when busy.
  • CSMA/CD: listen while sending; on collision jam, stop, back off.
  • L_min = 2·T_p·B; d_max = L_min·v/(2B); i.e. T_t ≥ 2T_p.
  • CSMA/CA (Foundation): wireless avoids collisions instead of detecting them.

From 2.3.4 Binary exponential backoff; probability of winning

  • After the n-th collision: K uniform in {0, …, 2^min(n,10) − 1}; wait K × slot; abort after 16.
  • Slot = 512 bit times (51.2 µs at 10 Mbps).
  • P(A wins) = #{K_A < K_B} / (a·b); both at range 2: win 1/4, tie 1/2.
  • Capture: A at range 2 vs B at range 4 → A wins 5/8.

From 2.3.5 Controlled access: polling, token passing and its efficiency

  • Polling: primary asks each station; η = T_t/(T_t + T_poll); single point of failure.
  • Token passing: only the token holder sends (token holding time).
  • Ring latency = d/v + N·b/R.
  • ETR η = 1/(1 + a/N); DTR η = 1/(1 + a(N+1)/N).

Practice questions

  1. NAT · 2 marks

    Slotted ALOHA, frame time 1 ms, stations attempt 2000 frames/s. Throughput in frames/s (nearest whole)?

    Write your answer before opening the key.

  2. NAT · 2 marks

    CSMA/CD at 1 Gbps, longest path 200 m, v = 2 × 10⁸ m/s. Minimum frame size in bytes?

    Write your answer before opening the key.

  3. MCQ · 1 mark

    Pure ALOHA reaches its maximum throughput at G =

    1. 0.25
    2. 0.5
    3. 1
    4. 2
  4. NAT · 2 marks

    A at its 1st collision, B at its 3rd. P(A wins the next round) (3 decimals)?

    Write your answer before opening the key.

  5. MSQ · 1 mark

    Which protocols are collision-free?

    1. TDMA
    2. Token passing
    3. Slotted ALOHA
    4. Polling
  6. NAT · 1 mark

    Token ring, N = 8, a = 0.4, delayed token release. Efficiency (3 decimals)?

    Write your answer before opening the key.

  7. MCQ · 1 mark

    In CSMA/CD the minimum frame size exists so that:

    1. frames fit in one slot of TDMA
    2. a sender is still transmitting when a collision report can return
    3. the CRC fits
    4. tokens are not lost
  8. NAT · 1 mark

    TDMA among 25 stations on a 5 Mbps link. Rate per station in kbps?

    Write your answer before opening the key.

  9. MCQ · 1 mark

    A 1-persistent station finding the channel busy:

    1. waits a random time then senses again
    2. keeps sensing and sends the moment it is idle
    3. sends with probability p
    4. gives up

Answer key

Try the questions first. Open when you are ready to check.
  1. 271 — G = 2 → S = 2e⁻² = 0.2707 per frame time → 0.2707 / 1 ms ≈ 271 frames/s.

    Review lesson 2.3.2

  2. 250 — T_p = 1 µs; 2 × 1 µs × 10⁹ = 2000 bits = 250 bytes.

    Review lesson 2.3.3

  3. B — S = G·e⁻²ᴳ peaks at G = 1/2.
    • A: Below the peak.
    • C: That is slotted ALOHA's peak.
    • D: Far past the peak.

    Review lesson 2.3.2

  4. 0.8125 — K_A ∈ {0,1}, K_B ∈ {0..7}: K_A = 0 → 7 wins, K_A = 1 → 6 wins; 13/16 = 0.8125.

    Review lesson 2.3.4

  5. ABD — Partitioning and taking turns are collision-free.
    • C: Two stations may pick the same slot.

    Review lesson 2.3.1

  6. 0.689..0.69 — 1/(1 + 0.4 × 9/8) = 1/1.45 = 0.6897.

    Review lesson 2.3.5

  7. B — T_t ≥ 2·T_p.
    • A: Unrelated.
    • C: Unrelated.
    • D: Unrelated.

    Review lesson 2.3.3

  8. 200 — 5000 / 25 = 200 kbps.

    Review lesson 2.3.1

  9. B — Definition of 1-persistent.
    • A: Non-persistent.
    • C: p-persistent.
    • D: No.

    Review lesson 2.3.3

Ready for the next idea?

Mark this item complete when you’ve finished reading and working through it.