9 votes 9 votes 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 $4.25$ $3.5$ $450$ $350$ Computer Networks isro2015 computer-networks slotted-aloha + – ajit asked Oct 12, 2015 edited Dec 4, 2022 by Lakshman Bhaiya ajit 6.9k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 29 votes 29 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 Mangilal Saraswat answered Jun 19, 2016 selected Jul 1, 2016 by Desert_Warrior Mangilal Saraswat comment Share Follow See all 2 Comments See all 2 2 Comments reply pC commented Jul 1, 2016 reply Follow Share is ALOHA in 2017 GATE syllabus ? 1 votes 1 votes shweta1920 commented Apr 16, 2017 reply Follow Share yes it is in GATE syllabus 0 votes 0 votes Please log in or register to add a comment.
5 votes 5 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) Chhotu answered Apr 20, 2017 Chhotu comment Share Follow See all 0 reply Please log in or register to add a comment.
3 votes 3 votes (b) option (70*50)/1000 =3.5 focus _GATE answered Oct 12, 2015 focus _GATE comment Share Follow See all 0 reply Please log in or register to add a comment.
2 votes 2 votes 70 request/sec time is slotted in units of 50 msec channel load = 50 msec * 70 request / sec = 3.5 b option akankshadewangan24 answered Apr 29, 2017 akankshadewangan24 comment Share Follow See all 0 reply Please log in or register to add a comment.