CPU Scheduling
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The document discusses CPU scheduling, which involves allocating CPU resources to processes in a way that maximizes efficiency, speed, and fairness. It categorizes CPU scheduling into non-preemptive and preemptive types, and details various scheduling algorithms such as First Come First Serve (FCFS), Shortest Job First (SJF), and priority scheduling. Additionally, it outlines scheduling criteria including CPU utilization, throughput, turnaround time, and waiting time, providing examples to demonstrate average turnaround and wait times.
CPU Scheduling
- 1. CPU SCHEDULING T.Y.B.Sc. (CS) By: Ms. Farhat Rahat Ali Shaikh
- 2. CONTENTS Introduction CPU Scheduling: Dispatcher Types of CPU Scheduling Non – Preemptive Scheduling Preemptive Scheduling CPU Scheduling: Scheduling Criteria Scheduling Algorithms First Come First Serve(FCFS) Scheduling Shortest-Job-First(SJF) Scheduling Shortest Remaining Time First (SRTF) Scheduling Priority Non - Preemptive Scheduling Priority Preemptive Scheduling Round Robin(RR) Scheduling 2
- 3. Introduction CPU scheduling is a process which allows one process to use the CPU while the execution of another process is on hold(in waiting state) due to unavailability of any resource like I/O etc., thereby making full use of CPU. The aim of CPU scheduling is to make the system efficient, fast and fair. Whenever the CPU becomes idle, the operating system must select one of the processes in the ready queue to be executed. The selection process is carried out by the short-term scheduler (or CPU scheduler). The scheduler selects from among the processes in memory that are ready to execute, and allocates the CPU to one of them. 3
- 4. CPU Scheduling: Dispatcher The dispatcher is the module that gives control of the CPU to the process selected by the short-term scheduler. This function involves: Switching context Switching to user mode Jumping to the proper location in the user program to restart that program from where it left last time. The dispatcher should be as fast as possible, given that it is invoked during every process switch. The time taken by the dispatcher to stop one process and start another process is known as the Dispatch Latency. 4
- 5. Types of CPU Scheduling CPU scheduling decisions may take place under the following four circumstances: When a process switches from the running state to the waiting state (for I/O request or invocation of wait for the termination of one of the child processes). When a process switches from the running state to the ready state (for example, when an interrupt occurs). When a process switches from the waiting state to the ready state (for example, completion of I/O). When a process terminates. In circumstances 1 and 4, there is no choice in terms of scheduling. A new process(if one exists in the ready queue) must be selected for execution. There is a choice, however in circumstances 2 and 3. When Scheduling takes place only under circumstances 1 and 4, we say the scheduling scheme is Non-preemptive; otherwise the scheduling scheme is Preemptive. 5
- 6. Types of CPU Scheduling Non-Preemptive Scheduling: In this type of scheduling, once the CPU has been allocated to a process, the process keeps the CPU until it releases the CPU either by terminating or by switching to the waiting state. It is the only method that can be used on certain hardware platforms, because It does not require the special hardware(for example: a timer) needed for preemptive scheduling. Preemptive Scheduling: In this type of Scheduling, the tasks are usually assigned with priorities. At times it is necessary to run a certain task that has a higher priority before another task although it is running. Therefore, the running task is interrupted for some time and resumed later when the priority task has finished its execution. 6
- 7. CPU Scheduling: Scheduling Criteria There are many different criteria's to check when considering the "best" scheduling algorithm, they are: CPU Utilization: To make out the best use of CPU and not to waste any CPU cycle, CPU would be working most of the time(Ideally 100% of the time). Considering a real system, CPU usage should range from 40% (lightly loaded) to 90% (heavily loaded.) Throughput: It is the total number of processes completed per unit time or rather say total amount of work done in a unit of time. This may range from 10/second to 1/hour depending on the specific processes. Turnaround Time: It is the amount of time taken to execute a particular process, i.e. The interval from time of submission of the process to the time of completion of the process(Wall clock time). Waiting Time: It is the total time spent by the process in the ready state waiting to get control on the CPU. 7
- 8. CPU Scheduling: Scheduling Criteria Load Average: It is the average number of processes residing in the ready queue waiting for their turn to get into the CPU. Response Time: Amount of time it takes from when a request was submitted until the first response is produced. Remember, it is the time till the first response and not the completion of process execution(final response). In general CPU utilization and Throughput are maximized and other factors are reduced for proper optimization. 8
- 9. Scheduling Algorithms To decide which process to execute first and which process to execute last to achieve maximum CPU utilization, we have various scheduling algorithms as follows : First Come First Serve(FCFS) Scheduling Shortest Job First(SJF) Scheduling Priority Scheduling Round Robin(RR) Scheduling 9
- 10. First Come First Serve(FCFS) Scheduling In the "First come first serve" scheduling algorithm, the process which arrives first, gets executed first, or we can say that the process which requests the CPU first, gets the CPU allocated first. It is a Non-preemptive Scheduling algorithm. Easy to understand and implement. Its implementation is based on FIFO queue. Poor in performance as average wait time is high. This is used in Batch Systems. 10
- 11. First Come First Serve(FCFS) Scheduling Example 1: Find out Average Turn – around time (T.A.T.) and Average Wait time (W.T.) for the given jobs: 11 Jobs Arrival Time (A.T.) Burst Time (B.T.) J1 1 5 J2 0 3 J3 2 2 J4 3 4 J5 2 8
- 12. First Come First Serve(FCFS) Scheduling Solution: Gantt Chart 12 Jobs A.T. B.T. T.A.T. (F.T. – A.T.) W.T. (S.T. – A.T.) J1 1 5 J2 0 3 J3 2 2 J4 3 4 J5 2 8 J2 0 3 J1 8 J3 10 J5 18 J4 22 S.T. F.T. 8 – 1 = 7 3 – 0 = 3 10 – 2 = 8 22 – 3 = 19 18 – 2 = 16 S.T. F.T. S.T. F.T. S.T. F.T. S.T. F.T. 3 – 1 = 2 0 – 0 = 0 8 – 2 = 6 18 – 3 = 15 10 – 2 = 8 Avg. T.A.T. = 53 5 =10.6 ms Avg. W.T. = 31 5 =6.2 ms Total 53 ms. 31 ms. S.T. = Start/Service Time F.T. = Finish Time
- 13. First Come First Serve(FCFS) Scheduling Example 2: Find out Average Turn – around time (T.A.T.) and Average Wait time (W.T.) for the given process: 13 Process Arrival Time (A.T.) Burst Time (B.T.) P1 1 5 P2 0 6 P3 1 2 P4 0 4 Avg. T.A.T. = 11.5 ms Avg. W.T. = 7.2 ms
- 14. Shortest Job First(SJF) Scheduling Shortest Job First scheduling works on the process with the shortest burst time or duration first. This is the best approach to minimize waiting time. Easy to implement in Batch systems where required CPU time is known in advance. Impossible to implement in interactive systems where required CPU time is not known. Processer should know in advance how much time process will take. It is of two types: Non Pre-emptive (SJF) Pre-emptive (SRTF) 14
- 15. Shortest Job First (SJF) Non - Preemptive In this type of scheduling CPU allocated to the process(min burst time) cannot be preempted until completes its CPU burst. Example 1: Find out Average Turn – around time (T.A.T.) and Average Wait time (W.T.) for the given jobs: 15 Jobs Arrival Time (A.T.) Burst Time (B.T.) J1 0 4 J2 1 1 J3 2 2 J4 3 1
- 16. Shortest Job First (SJF) Non - Preemptive Solution: Gantt Chart 16 Jobs A.T. B.T. T.A.T. (F.T. – A.T.) W.T. (S.T. – A.T.) J1 0 4 J2 1 1 J3 2 2 J4 3 1 J1 0 4 J2 5 J4 6 J3 8 S.T. F.T. 4 – 0 = 4 5 – 1 = 4 8 – 2 = 6 6 – 3 = 3 S.T.F.T. S.T. F.T. S.T. F.T. 0 – 0 = 0 4 – 1 = 3 6 – 2 = 4 5 – 3 = 2 Avg. T.A.T. = 17 4 =4.2 ms Avg. W.T. = 9 4 =2.2 ms Total 17 ms. 9 ms. S.T. = Start/Service Time F.T. = Finish Time
- 17. Shortest Job First (SJF) Non - Preemptive Example 2: Find out Average Turn – around time (T.A.T.) and Average Wait time (W.T.) for the given process: 17 Jobs Arrival Time (A.T.) Burst Time (B.T.) P1 3 5 P2 0 2 P3 4 2 P4 5 3
- 18. Shortest Job First (SJF) Non - Preemptive Solution: Gantt Chart 18 Jobs A.T. B.T. T.A.T. (F.T. – A.T.) W.T. (S.T. – A.T.) P1 3 5 P2 0 2 P3 4 2 P4 5 3 P2 0 2 P1 14 P4 9 P3 6 S.T. F.T. 14 – 3 = 11 2 – 0 = 2 6 – 4 = 2 9 – 5 = 4 S.T.F.T. S.T. F.T.S.T. F.T. 9 – 3 = 6 0 – 0 = 0 4 – 4 = 0 6 – 5 = 1 Avg. T.A.T. = 19 4 =4.75 ms Avg. W.T. = 7 4 =1.75 ms Total 19 ms. 7 ms. S.T. = Start/Service Time F.T. = Finish Time 4 Idle state
- 19. Shortest Job First (SJF) Non - Preemptive Example 3: Find out Average Turn – around time (T.A.T.) and Average Wait time (W.T.) for the given process: 19 Process Arrival Time (A.T.) Burst Time (B.T.) P1 1.5 5 P2 0 1 P3 2 2 P4 3 4 Avg. T.A.T. = 4.875 ms Avg. W.T. = 1.875 ms
- 20. Shortest Remaining Time First (SRTF) Preemptive In Preemptive Shortest Job First Scheduling, jobs are put into ready queue as they arrive, but as a process with short burst time arrives, the existing process is preempted or removed from execution, and the shorter job is executed first. Example 1: Find out Average Turn – around time (T.A.T.) and Average Wait time (W.T.) for the given jobs: 20 Jobs Arrival Time (A.T.) Burst Time (B.T.) J1 0 4 J2 1 1 J3 2 2 J4 3 1
- 21. Shortest Remaining Time First (SRTF) Preemptive Solution: Gantt Chart 21 Jobs A.T. B.T. T.A.T. (F.T. – A.T.) W.T. (T.A.T. – B.T.) J1 0 4 J2 1 1 J3 2 2 J4 3 1 J1 0 1 J2 2 J4 5 J3 4 8 – 0 = 8 2 – 1 = 1 4 – 2 = 2 5 – 3 = 2 F.T. F.T. F.T. 8 – 4 = 4 1 – 1 = 0 2 – 2 = 0 2 – 1 = 1 Avg. T.A.T. = 13 4 =3.25 ms Avg. W.T. = 5 4 =1.25 ms Total 13 ms. 5 ms. F.T. = Finish Time J3 3 J1 8 F.T. CPU J1 = J2 = J3 = J4 = 3 => 1 => 2 => 1 => 0 1 => 0 0 0
- 22. Shortest Remaining Time First (SRTF) Preemptive Example 2: Find out Average Turn – around time (T.A.T.) and Average Wait time (W.T.) for the given process: 22 Process Arrival Time (A.T.) Burst Time (B.T.) P1 1.5 5 P2 0 1 P3 2 2 P4 3 4
- 23. Shortest Remaining Time First (SRTF) Preemptive Solution: Gantt Chart 23 Jobs A.T. B.T. T.A.T. (F.T. – A.T.) W.T. (T.A.T. – B.T.) P1 1.5 5 P2 0 1 P3 2 2 P4 3 4 P2 0 1 1.5 P3 4 P3 3 12.5 – 1.5 = 11 1 – 0 = 1 4 – 2 = 2 8 – 3 = 5 F.T. F.T. F.T. 11 – 5 = 6 1 – 1 = 0 2 – 2 = 0 5 – 4 = 1 Avg. T.A.T. = 19 4 =4.75 ms Avg. W.T. = 7 4 =1.75 ms Total 19 ms. 7 ms. F.T. = Finish Time P1 2 P4 8 F.T. CPU P2 = P1 = P3 = P4 = 4.5 => 0 2 => 4 => 0 1 => 0 0 Idle state P1 12.5
- 24. Shortest Remaining Time First (SRTF) Preemptive Example 3: Find out Average Turn – around time (T.A.T.) and Average Wait time (W.T.) for the given process: 24 Jobs Arrival Time (A.T.) Burst Time (B.T.) P1 3 5 P2 0 2 P3 4 2 P4 5 3 Avg. T.A.T. = 4.5 ms Avg. W.T. = 1.5 ms
- 25. Priority Scheduling Priority scheduling is one of the most common scheduling algorithms in batch systems. Each process is assigned a priority. Process with highest priority is to be executed first and so on. Processes with same priority are executed on first come first serve basis. Priority can be decided based on memory requirements, time requirements or any other resource requirement. It is of two types: Non – Preemptive Scheduling Preemptive Scheduling 25
- 26. Non –Preemptive Priority Scheduling In case of non – preemptive priority scheduling algorithm if a new process arrives with a higher priority than the current running process, the incoming process is put at the head of the ready queue, which means after the execution of the current process it will be processed. Example 1: Find out Average Turn – around time (T.A.T.) and Average Wait time (W.T.) for the given jobs: 26 Jobs Arrival Time (A.T.) Burst Time (B.T.) Priority J1 0 10 3 J2 4 5 0 (High) J3 3 2 1 J4 5 16 2 J5 2 8 4 (Low)
- 27. Non –Preemptive Priority Scheduling Solution: Gantt Chart 27 Jobs A.T. B.T. Priority T.A.T. (F.T. – A.T.) W.T. (T.A.T. – B.T.) J1 0 10 3 J2 4 5 0 (High) J3 3 2 1 J4 5 16 2 J5 2 8 4 (Low) J1 0 10 J2 15 J5 41 J4 33 10 – 0 = 10 15 – 4 = 11 17 – 3 = 14 33 – 5 = 28 F.T. F.T. F.T. 10 – 10 = 0 11 – 5 = 6 14 – 2 = 12 28 – 16 = 12 Avg. T.A.T. = 102 5 =20.4 ms Avg. W.T. = 61 5 =12.2 ms Total 102 ms. 61 ms. F.T. = Finish Time J3 17 F.T. F.T. 41 – 2 = 39 39 – 8 = 31
- 28. Non –Preemptive Priority Scheduling Example 2: Find out Average Turn – around time (T.A.T.) and Average Wait time (W.T.) for the given process: 28 Jobs Arrival Time (A.T.) Burst Time (B.T.) Priority P1 1 5 1 (High) P2 0 6 2 P3 1 2 1 P4 0 4 3 (Low) P5 2 3 2
- 29. Non –Preemptive Priority Scheduling Solution: Gantt Chart 29 Jobs A.T. B.T. Priority T.A.T. (F.T. – A.T.) W.T. (T.A.T. – B.T.) P1 1 5 1 (High) P2 0 6 2 P3 1 2 1 P4 0 4 3 (Low) P5 2 3 2 P2 0 6 P1 11 P4 20 P5 16 11 – 1 = 10 6 – 0 = 6 13 – 1 = 12 20 – 0 = 20 F.T. F.T. F.T. 10 – 5 = 5 6 – 6 = 0 12 – 2 = 10 20 – 4 = 16 Avg. T.A.T. = 62 5 =12.4 ms Avg. W.T. = 42 5 =8.4 ms Total 62 ms. 42 ms. F.T. = Finish Time P3 13 F.T. F.T. 16 – 2 = 14 14 – 3 = 11
- 30. Non –Preemptive Priority Scheduling Example 3: Find out Average Turn – around time (T.A.T.) and Average Wait time (W.T.) for the given process: 30 Jobs Arrival Time (A.T.) Burst Time (B.T.) Priority P1 0 4 3 P2 2 3 1(Low) P3 1 7 4 (High) P4 3 15 2 Avg. T.A.T. = 16 ms Avg. W.T. = 8.75 ms
- 31. Preemptive Priority Scheduling If the new process arrived at the ready queue has a higher priority than the currently running process, the CPU is preempted, which means the processing of the current process is stopped and the incoming new process with higher priority gets the CPU for its execution. Example 1: Find out Average Turn – around time (T.A.T.) and Average Wait time (W.T.) for the given jobs: 31 Jobs Arrival Time (A.T.) Burst Time (B.T.) Priority J1 0 10 3 J2 4 5 0 (High) J3 3 2 1 J4 5 16 2 J5 2 8 4 (Low)
- 32. Preemptive Priority Scheduling Solution: Gantt Chart 32 Jobs A.T. B.T. Priority T.A.T. (F.T. – A.T.) W.T. (T.A.T. – B.T.) J1 0 10 3 J2 4 5 0 (High) J3 3 2 1 J4 5 16 2 J5 2 8 4 (Low) J1 0 3 J1 33 J4 26 33 – 0 = 33 9 – 4 = 5 10 – 3 = 7 26 – 5 = 21 F.T. F.T. F.T. 33 – 10 = 23 5 – 5 = 0 7 – 2 = 5 21 – 16 = 5 Avg. T.A.T. = 105 5 =21 ms Avg. W.T. = 61 5 =12.2 ms Total 105 ms. 61 ms. F.T. = Finish Time J2 9 F.T. F.T. 41 – 2 = 39 39 – 8 = 31 J3 10 J5 41 J3 4
- 33. Preemptive Priority Scheduling Example 2: Find out Average Turn – around time (T.A.T.) and Average Wait time (W.T.) for the given process: 33 Jobs Arrival Time (A.T.) Burst Time (B.T.) Priority P1 1 5 1 (High) P2 0 6 2 P3 1 2 1 P4 0 4 3 (Low) P5 2 3 2
- 34. Preemptive Priority Scheduling Solution: Gantt Chart 34 Jobs A.T. B.T. Priority T.A.T. (F.T. – A.T.) W.T. (T.A.T. – B.T.) P1 1 5 1 (High) P2 0 6 2 P3 1 2 1 P4 0 4 3 (Low) P5 2 3 2 P2 0 1 P1 6 P4 20 P5 16 11 – 1 = 10 6 – 0 = 6 13 – 1 = 12 20 – 0 = 20 F.T.F.T. F.T. 10 – 5 = 5 6 – 6 = 0 12 – 2 = 10 20 – 4 = 16 Avg. T.A.T. = 62 5 =12.4 ms Avg. W.T. = 42 5 =8.4 ms Total 62 ms. 42 ms. F.T. = Finish Time P3 8 F.T. F.T. 16 – 2 = 14 14 – 3 = 11 P2 13
- 35. Preemptive Priority Scheduling Example 3: Find out Average Turn – around time (T.A.T.) and Average Wait time (W.T.) for the given process: 35 Jobs Arrival Time (A.T.) Burst Time (B.T.) Priority P1 0 4 3 P2 2 3 1(Low) P3 1 7 4 (High) P4 3 15 2 Avg. T.A.T. = 17 ms Avg. W.T. = 9.75 ms
- 36. Round Robin Scheduling Round Robin is the preemptive process scheduling algorithm. Each process is provided a fix time to execute called quantum. Once a process is executed for given time period. Process is preempted and other process executes for given time period. Context switching is used to save states of preempted processes. 36
- 37. Round Robin Scheduling Example 1: Find out Average Turn – around time (T.A.T.) and Average Wait time (W.T.) for the given process with Time Slice (Quantum) = 3ms. 37 Jobs Arrival Time (A.T.) Burst Time (B.T.) P1 1 5 P2 0 7 P3 3 3 P4 2 10
- 38. Round Robin Scheduling Solution: Gantt Chart Time Slice (Quantum) = 3ms. 38 Jobs A.T. B.T. T.A.T. (F.T. – A.T.) W.T. (T.A.T. – B.T.) P1 1 5 P2 0 7 P3 3 3 P4 2 10 P2 0 3 P1 6 P1 17 P2 15 17 – 1 = 16 21 – 0 = 21 12 – 3 = 9 25 – 2 = 23 F.T.F.T. F.T. 16 – 5 = 11 21 – 7 = 14 9 – 3 = 6 23 – 10 = 13 Avg. T.A.T. = 69 4 =17.25 ms Avg. W.T. = 44 4 =11 ms Total 69 ms. 44 ms. F.T. = Finish Time P4 9 F.T. P3 12 P4 20 P2 21 P4 24 P4 25
- 39. Round Robin Scheduling Example 2: Find out Average Turn – around time (T.A.T.) and Average Wait time (W.T.) for the given process with Time Slice (Quantum) = 2ms. 39 Jobs Arrival Time (A.T.) Burst Time (B.T.) P1 1 5 P2 0 3 P3 2 2 P4 3 4 P5 13 2
- 40. Round Robin Scheduling 40 Jobs A.T. B.T. T.A.T. (F.T. – A.T.) W.T. (T.A.T. – B.T.) P1 1 5 P2 0 3 P3 2 2 P4 3 4 P5 13 2 P2 0 2 P1 4 P1 11 P2 9 16 – 1 = 15 9 – 0 = 9 6 – 2 = 4 13 – 3 = 10 F.T.F.T.F.T. 15 – 5 = 10 9 – 3 = 6 4 – 2 = 2 10 – 4 = 6 Avg. T.A.T. = 40 5 =8 ms Avg. W.T. = 24 5 =4.8 ms Total 40 ms. 24 ms. F.T. = Finish Time P3 6 F.T. P4 8 P4 13 P5 15 P1 16 Solution: Gantt Chart Time Slice (Quantum) = 2ms. F.T. 15 – 13 = 2 2 – 2 = 0
- 41. Round Robin Scheduling Example 2: Five jobs arrive at 0 arrival time the order is given below. Find out Average Turn – around time (T.A.T.) and Average Wait time (W.T.) for the given process with Time Slice (Quantum) = 10ms. 41 Jobs Burst Time (B.T.) J1 10 J2 29 J3 3 J4 7 J5 12 Avg. T.A.T. = 35.2 ms Avg. W.T. = 23 ms
- 42. THANK YOU 42