Gyanrush
Start freeOpen Gyanrush
Start freeOpen Gyanrush
  1. LPU MCA Semester 1
  2. CAP7001
  3. Unit 3

CAP7001 · Unit 3: Data link layer

Multiple access: random access, controlled access and channelization

19 practice questions · in the MTE and ETE syllabus

Practise this topicPractise this topic

Key points

  • Random: ALOHA, CSMA family; controlled: reservation, polling, token; channelization: FDMA, TDMA, CDMA
  • Pure ALOHA max 0.184; slotted max 0.368
  • CSMA/CD: Tfr >= 2Tp
  • CDMA codes are orthogonal

The data link layer shares a medium using multiple access protocols.

  • Random access: ALOHA, CSMA, CSMA/CD (wired Ethernet), CSMA/CA (wireless).
  • Controlled access: reservation, polling, token passing.
  • Channelization: FDMA, TDMA, CDMA.

Pure ALOHA: vulnerable time 2 Tfr, S = G e^(-2G), maximum 0.184 at G = 0.5. Slotted ALOHA: vulnerable time Tfr, S = G e^(-G), maximum 0.368 at G = 1.

CSMA/CD needs frame time Tfr >= 2 Tp, so minimum frame = bit rate x 2Tp (Ethernet: 512 bits = 64 bytes). CDMA uses orthogonal chip sequences (inner product 0 with another code, N with its own).

Worked example: at 1000 frames/s with 1 ms frames (G = 1), pure ALOHA delivers 1000 x e^-2, about 135 frames/s; slotted ALOHA gives about 368.

Exam pitfall: the 18.4% maximum is at G = 0.5, not G = 1. At G = 1 pure ALOHA gives only about 0.135 (13.5%).

Try two questions

Question 1 · Tier 2, Understand

Maximum throughput of pure ALOHA and slotted ALOHA are respectively about:

  1. 18.4% and 36.8%
  2. 10% and 20%
  3. 36.8% and 18.4%
  4. 50% and 100%
Show the answer

A. Pure: 1/(2e) ~ 0.184 at G=0.5; slotted: 1/e ~ 0.368 at G=1.

Question 2 · Tier 3, Apply

In pure ALOHA the throughput is S = G e^(-2G). At G = 0.5 the throughput is approximately:

  1. 0.5
  2. 0.25
  3. 0.368
  4. 0.184
Show the answer

D. S = 0.5 x e^(-1) = 0.184. The exponent is -2G = -1, so S = 0.5 x 0.368 = 0.184, not 0.368.