operating systems 11) When the job scheduler starts to schedule a proess, which
ID: 3737897 • Letter: O
Question
operating systems
11) When the job scheduler starts to schedule a proess, which has not run et, how will the status of that process change? 11.1 from hold to ready 11.2 from ready to running 11.3 from running to ready 11.4 from running to waiting 11.5 from running to finished 12) When the process scheduler assigns the CPU to a proces, how will the status of thnt twe change? 12.1 from hold to ready 12.2 from ready to running 12.3 from running to finished 12.4 from running to waiting 12.5 from waiting to ready Assume that three jobs (Job A, Job B, and Job C) are submitted for execution as follows Job id: Arrival time(in milliseconds): CPU cycle time needed: ABC 0 12 734 13) Assume the operating system is using the first-come first-served scheduling algorithm (FCFS) to assign use of the CPU. The average turnaround time is 13.1 10.33 13.2 4.67 13.3 9.33 13.4 5.67 13.5 31 14) Assume the operating system is using the round robin scheduling algorithm (RR) to assign use of the CPU. Assume the time slice being used is 3 milliseconds. The average turnaround time is: 14.1 10.0 14.2 10.67 14.3 19.0 14.4 14.0 14.5 4.7 15) Four conditions must be present simultaneously in a system with a deadlock 15.1 Mutual exclusion. Starvation. Resource holding and waiting, Circular wait 15.2 Resource holding and waiting. Spooling. Mutual exclusion. Circular wait 15.3 Resource holding and waiting. Blocking. Busy waiting. Critical section. 15.4 Buffering. Blocking. Paging. No preemption. 15.5 Resource holding and waiting. Circular wait. No preemption. Mutal exclusionExplanation / Answer
11
11.1 from hold to ready - When a process first arrives which is not yet ran it will be in hold status untill scheduler takes the process and makes ready for processing
15
15.5 - at least one resource must be held and because of one resource is held another process may wait for the resource which can not be preempted and because of this there may be a circular waiting
12
12.2 from ready to running - when a scheduler gets a processor it will go to run mode from ready
13
13.2 4.67. in FCFC process runs when it gets processor. So in this case A starts at 7 B starts when A completed i.e. 7 and finishes at 10 then C starts at 10 finishes at 14. So 14/3 = 4.67
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14.1 10 in round robin scheduling each process given a processor for brief amount of time until its gets completed.
A0 TO 3
B 3 TO 6 ARRIVES AT 1 SO TOTAL 5
C 6 TO 9
A 9 TO 12
B Completed already
C 12 TO 13 ARRIVES AT 2 SO TOTAL 11
A 13 TO 14 ARRIVES AT 0 SO TOTAL 14
5 + 11 + 14 = 30
FOR THREE PROCESS 30/3 ITS 10
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