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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
in Computer Networks by Active (2.5k points) | 3.4k views

4 Answers

+24 votes
Best answer

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

by Active (2.4k points)
selected by
0
is ALOHA in 2017 GATE syllabus ?
0
yes it is in GATE syllabus
+3 votes
(b) option

(70*50)/1000
=3.5
by Boss (20.1k points)
+3 votes

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)

by Boss (13.7k points)
+2 votes
70 request/sec

time is slotted in units of 50 msec

channel load = 50 msec * 70 request / sec

                    = 3.5

b option
by Active (4k points)
Answer:

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