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1、.,Chapter 5: CPU Scheduling,Chapter 5: CPU Scheduling,Basic Concepts Scheduling Criteria Scheduling Algorithms Multiple-Processor Scheduling Real-Time Scheduling Thread Scheduling Operating Systems Examples Java Thread Scheduling Algorithm Evaluation,Basic Concepts,Maximum CPU utilization obtained w
2、ith multiprogramming CPUI/O Burst Cycle Process execution consists of a cycle of CPU execution and I/O wait CPU burst distribution,Alternating Sequence of CPU And I/O Bursts,Histogram of CPU-burst Times,CPU Scheduler,Selects from among the processes in memory that are ready to execute, and allocates
3、 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 under 1 and 4 is nonpreemptive All other scheduling is preemptive,Dispatcher,Dispat
4、cher 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
5、 another running,Scheduling Criteria,CPU utilization keep the CPU as busy as possible Throughput # 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 Respons
6、e time amount of time it takes from when a request was submitted until the first response is produced, not output (for time-sharing environment),Optimization Criteria,Max CPU utilization Max throughput Min turnaround time Min waiting time Min response time,First-Come, First-Served (FCFS) Scheduling,
7、ProcessBurst Time P124 P2 3 P3 3 Suppose that the processes arrive in the order: P1 , P2 , P3 The Gantt Chart for the schedule is: Waiting time for P1 = 0; P2 = 24; P3 = 27 Average waiting time: (0 + 24 + 27)/3 = 17,FCFS Scheduling (Cont.),Suppose that the processes arrive in the order P2 , P3 , P1
8、The Gantt chart for the schedule is: Waiting time for P1 = 6; P2 = 0; P3 = 3 Average waiting time: (6 + 0 + 3)/3 = 3 Much better than previous case Convoy effect short process behind long process,Shortest-Job-First (SJF) Scheduling,Associate with each process the length of its next CPU burst. Use th
9、ese 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 sche
10、me is know as the Shortest-Remaining-Time-First (SRTF) SJF is optimal gives minimum average waiting time for a given set of processes,ProcessArrival TimeBurst Time P10.07 P22.04 P34.01 P45.04 SJF (non-preemptive) Average waiting time = (0 + 6 + 3 + 7)/4 = 4,Example of Non-Preemptive SJF,Example of P
11、reemptive SJF,ProcessArrival TimeBurst Time P10.07 P22.04 P34.01 P45.04 SRTF (preemptive) Average waiting time = (9 + 1 + 0 +2)/4 = 3 Can only estimate the length,Determining Length of Next CPU Burst,Can only estimate the length Can be done by using the length of previous CPU bursts, using exponenti
12、al averaging,Prediction of the Length of the Next CPU Burst,Examples of Exponential Averaging, =0 n+1 = n Recent history does not count =1 n+1 = tn Only the actual last CPU burst counts If we expand the formula, we get: n+1 = tn+(1 - ) tn -1 + +(1 - )j tn -j + +(1 - )n +1 0 Since both and (1 - ) are
13、 less than or equal to 1, each successive term has less weight than its predecessor,Priority Scheduling,A priority number (integer) is associated with each process The CPU is allocated to the process with the highest priority (smallest integer highest priority) Preemptive nonpreemptive SJF is a prio
14、rity scheduling where priority is the predicted next CPU burst time Problem Starvation (indefinite blocking) low priority processes may never execute Solution Aging as time progresses increase the priority of the process,Round Robin (RR),Each process gets a small unit of CPU time (time quantum), usu
15、ally 10-100 milliseconds. After this time has elapsed, the process is preempted and added to the end of the ready queue. If there are n processes in the ready queue and the time quantum is q, then each process gets 1/n of the CPU time in chunks of at most q time units at once. No process waits more
16、than (n-1)q time units. Performance q large FIFO q small q must be large with respect to context switch, otherwise overhead is too high,Example of RR with Time Quantum = 20,ProcessBurst Time P153 P2 17 P368 P4 24 The Gantt chart is: Typically, higher average turnaround than SJF, but better response,Time Quantum and Context Switch Time,Turnaround Time Varies With The Time Quantum,Multilevel Queue,Ready queue is partitioned into separate queues:foreground (interactive)background (batch) Ea
