28 votes 28 votes The process state transition diagram of an operating system is as given below. Which of the following must be FALSE about the above operating system? It is a multiprogrammed operating system It uses preemptive scheduling It uses non-preemptive scheduling It is a multi-user operating system Operating System gateit-2006 operating-system normal process + – Ishrat Jahan asked Oct 31, 2014 • edited Jul 9, 2019 by Lakshman Bhaiya Ishrat Jahan 11.5k views answer comment Share Follow See all 2 Comments See all 2 2 Comments reply Chhotu commented Dec 2, 2017 reply Follow Share Similar -> https://gateoverflow.in/1318/gate2009-32 1 votes 1 votes PreyumKr commented Dec 12, 2023 reply Follow Share Someone just draw the diagram of preemptive and non preemptive transition state. It will surely clear a few doubts here. 0 votes 0 votes Please log in or register to add a comment.
Best answer 52 votes 52 votes Answer (B). Explanation: It is a multiprogrammed operating system. Correct, it has ready state. We can have multiple processes in ready state here so this is Multiprogrammed OS. It uses preemptive scheduling False : There is no arrow transition from running to ready state. So, this is non preemptive. It uses non-preemptive scheduling True. It is a multi-user operating system. We can have multiple user processes in ready state. So, this is also correct. Akash Kanase answered Nov 7, 2015 • edited Jun 29, 2018 by kenzou Akash Kanase comment Share Follow See all 17 Comments See all 17 17 Comments reply Show 14 previous comments John_Smith commented Nov 4, 2023 reply Follow Share A few minor corrections:Just because ready state is present does not mean that the given OS is a multi-programmed OS. The presence of both the ready state (multiple processes may reside in the main memory at the same time), and the fact that a process is blocked when waiting for the release of the required resources, or when waiting for I/O completion (in which case some other process may be fetched now and executed), together implies that the given OS may be a multi-programmed OS. As the question ask for the option that must be FALSE, we pass over option \((A)\).Option \((D)\) is not correct, it’s just that it’s possible, i.e., multiple processes from multiple users may or may not be in ready state. As the question ask for the option that must be FALSE, we pass over option \((D)\). 1 votes 1 votes PreyumKr commented Dec 12, 2023 reply Follow Share Can't you guys see the transition from running to ready state. Its clearly there, it goes to i/o then to ready state. While it's doing that CPU should be given to some other process or else there was no need to go back to ready state again. I don't know why my guide also says the answer is B but I believe the answer is C. It uses non-preemtive scheduling is a false statement as there is a clear transition from running to ready state, so its a preemptive scheduling. 0 votes 0 votes John_Smith commented Dec 13, 2023 reply Follow Share @PreyumKr Consider processes that are entirely CPU bound, requiring no I/O or competing for resources (except CPU). That’s possible, right? Just because there is an edge from running state to blocked state doesn’t mean that every process has to make that transition. In that case, would you still have a preemptive OS? A preemptive OS should make provisions for all sort (and sequence) of processes, right? 1 votes 1 votes Please log in or register to add a comment.
13 votes 13 votes B is False. If it were a preemptive scheduling then there would have been a transition from Running state to Ready state. Gate Keeda answered Jan 15, 2015 Gate Keeda comment Share Follow See all 0 reply Please log in or register to add a comment.