CPU Scheduling: Part I ( 5, SGG) Operating Systems. Autumn CS4023

Transcription

1 Operating Systems Autumn

2 Outline 1 CPU Scheduling: Part I ( 5, SGG)

3 Outline CPU Scheduling: Part I ( 5, SGG) 1 CPU Scheduling: Part I ( 5, SGG)

4 Basic Concepts Typical program behaviour CPU Scheduling: Part I ( 5, SGG) Maximum CPU utilization obtained with multiprogramming CPU-I/O Burst Cycle: Process execution consists of a cycle of CPU execution and I/O wait

5 Histogram of CPU-burst Times

6 CPU Scheduler CPU Scheduling: Part I ( 5, SGG) Selects from among the processes in memory that are ready to execute, and allocates the CPU to one of them CPU scheduling decisions may take place when a process: 1 Switches from running to waiting state 2 Switches from running to ready state 3 Switches from waiting to ready 4 Terminates Scheduling according to conditions (1) and (4) is nonpreemptive, or cooperative All other scheduling is preemptive

7 CPU Scheduler (contd.) Non-preemptive (cooperative) 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 MS Windows 3.x; Mac OS prior version X The only method on some hardware platforms Preemptive scheduling incurs cost associated with: Access to shared data (count variable in prev. lecture) More complicated design of the kernel

8 Dispatcher CPU Scheduling: Part I ( 5, SGG) Dispatcher module gives control of the CPU to the process selected by the short-term scheduler; this involves: switching context switching to user mode jumping to the proper location in the user program to restart that program Dispatch latency: time it takes for the dispatcher to stop one process and start another running

9 Scheduling Criteria CPU utilization: keep the CPU as busy as possible: 40% - 90% Throughput: no. of processes that complete their execution per time unit Turnaround time: amount of time to execute a particular process Waiting time: amount of time a process has been waiting in the ready queue = completion time - arrival time - execution time Response time: amount of time it takes from when a request was submitted until the first response is produced, not output (excludes speed of output device) (for time-sharing environment)

10 Optimization Criteria Max CPU utilization Max throughput Min turnaround time Min waiting time Min response time

11 Outline CPU Scheduling: Part I ( 5, SGG) 1 CPU Scheduling: Part I ( 5, SGG)

12 First-Come, First-Served (FCFS) Scheduling Suppose we have three processes, P 1, P 2, P 3 in the ready state with CPU burst times 24, 3, 3, respectively. Suppose that the processes arrive in the order P 1, P 2, P 3 The Gantt Chart for this schedule is: P 1 P 2 P Waiting time for P 1 = ; w.t. for P 2 = = 24; and for P 3 = = 27 Average waiting time is ( )/3 = 17

13 First-Come, First-Served (FCFS) Scheduling (contd.) Suppose that the processes arrived in the order P 2, P 3, P 1 instead The Gantt Chart for this schedule is: P 2 3 P 3 6 P 1 30 Waiting time for P 1 = = 6; w.t. for P 2 = = 0; and for P 3 = = 3 Average waiting time is (6+0+3)/3 = 3 Much better than previous case FCFS is nonpreemptive leading to......convoy effect: short processes build up behind long process

14 Shortest-Job-First (SJF) Scheduling Associate with each process the length of its next CPU burst. Use these lengths to schedule the process with the shortest time Two schemes: nonpreemptive: once CPU given to the process it cannot be preempted until completes its CPU burst preemptive: if a new process arrives with CPU burst length less than remaining time of current executing process, preempt. This scheme is know as the Shortest-Remaining-Time-First (SRTF) SJF is optimal: gives minimum average waiting time for a given set of processes

15 Example of Non-Preemptive SJF Process Arrival Time Burst Time P P P P SJF (non-preemptive): Process Start Time End Time P P P P Average Waiting time: ( )/4 = 4

16 Example of Preemptive SJF Process Arrival Time Burst Time P P P P SJF (preemptive): Process Start Time Preempt Time P P P P P P Average Waiting time: ( )/4 = 3