GATE CSE 2021 Set 1 | Question: 25

Question

Three processes arrive at time zero with $\text{CPU}$ bursts of $16,\;20$ and $10$ milliseconds. If the scheduler has prior knowledge about the length of the $\text{CPU}$ bursts, the minimum achievable average waiting time for these three processes in a non-preemptive scheduler (rounded to nearest integer) is _____________ milliseconds.

Comments

How to know which alorithm need here

Answer 1

ANS IS 12.

HERE IN QUESTION, IT IS MENTIONED THAT “SCHEDULER HAS PRIOR KNOWLEDGE ABOUT THE LENGTH OF THE CPU BURSTS” SO IT MEANS WE HAVE TO USE SJF.

 

PS: BY MISTAKE I WROTE AVG TAT IT IS AVG WT ONLY. :)

Answer 2

For minimizing average waiting time/ maximize throughput,  SJF(shortest job first) works better for  non-preemptive.(SRTF for preemptive)

so schedule processes in increasing order of their burst time i.e. C(first),A,B(last)

now waiting will be 0ms,10ms,26ms for C,A,B respectively.

avg wait time= 36/3=12ms