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A certain population of ALOHA users manages to generate $70$ request/sec. If the time is slotted in units of $50$ msec, then channel load would be

  1. $4.25$
  2. $3.5$
  3. $450$
  4. $350$

5 Answers

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34 34 votes

Answer: (b)
Explanation: In slotted ALOHA we divide the time into slots and force the station to send only at the beginning of the time slot. Here slot time is 50 ms, so number of slots in 1 second = 1/(50 X 10^-3) => 20 slots/sec

  • Requests per second = 70
  • Time slots per second = 20
  • Channel load = No. of Requests / No. of Slots = 70 / 20 = 3.5

Reference: http://www.cs.wichita.edu/~chang/lecture/cs742/homework/hwk3-sol.txt

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Answer is (B) part.

70 request per second so number of request in one time slot(50 ms) will be (70*50)/1000 = 3.5 (which is channel load means request per time slot)

1 1 vote
in 1 sec it is generating 70 request means to generate 1 request it will take 14.3 msec

so in 1 slot time(50 msec) it will have (1/14.3)*50 =3.5

answers should be (B)
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