0 votes 0 votes If there are ‘n’ processes in a system, with the time-quanta less than the CPU burst times of all these processes in a given round, then the number of preemptions is at least: (A) 2n (B) n (C) 2n–1 (D) n–1 Operating System operating-system + – amit166 asked Jan 9, 2022 amit166 333 views answer comment Share Follow See 1 comment See all 1 1 comment reply palashbehra5 commented Jan 9, 2022 reply Follow Share Asked “Least preemptions” Constraints “Time quanta is less than CPU BT of all processes” Assumption “There was no process using CPU hence initial preemption is ignored” Best Case, We will need only 2 passes of all the processes as time quanta is less than all of them. 1st Pass : P1 || CS || P2 || CS ||…. || Pn ie n-1 Context Switches 2nd Pass : || CS || P1 || CS || P2 || CS ||….. || Pn ie n Context Switches n+n-1 = 2n-1. This is the rough idea, IDK how to do formal proofs. 1 votes 1 votes Please log in or register to add a comment.