2 votes 2 votes Consider three processes (process id $0,1,2$ respectively) with compute time bursts $2,4$ and $8$ time units. All processes arrive at time zero. Consider the Longest Remaining Time First (LRTF) scheduling algorithm. In LRTF ties are broken by giving priority to the process with the lowest process id. The average turn around time is $13$ units $14$ units $15$ units $16$ units Operating System nielit2017july-scientistb-it operating-system process-scheduling + – admin asked Mar 30, 2020 • retagged Aug 23, 2020 by Lakshman Bhaiya admin 1.7k views answer comment Share Follow See all 2 Comments See all 2 2 Comments reply smsubham commented Mar 30, 2020 reply Follow Share Previous year question https://gateoverflow.in/1842/gate2006-64 0 votes 0 votes anjli commented Jan 13, 2021 reply Follow Share ANS : A (13 UNITS ) 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes 13 units gantt chart sequence p2,p1,p2,p1,p2,p0,p1,p2,p0,p1,p2 gateankit answered May 24, 2021 gateankit comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes Answer is 13 units as the scheduling is preemptive type and Gantt chart for it clarifies the solution. Ritik Sahu answered Sep 28, 2021 Ritik Sahu comment Share Follow See all 0 reply Please log in or register to add a comment.