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Module 5: CPU Scheduling.• Basic Concepts.• Scheduling Criteria.• Scheduling Algorithms.• Multiple-Processor Scheduling.• Real-Time Scheduling.• Algorithm Evaluation. 5.1 Silberschatz and Galvin 1999 Basic Concepts.• Maximum CPU utilization obtained with multiprogramming.• CPU–I/O Burst Cycle – Process execution consists of a cycle of. CPU execution and I/O wait• CPU burst distribution. 5.2 Silberschatz and Galvin 1999 Alternating Sequence of CPU And I/O Bursts. 5.3 Silberschatz and Galvin 1999 Histogram of CPU-burst Times. 5.4 Silberschatz and Galvin 1999 CPU Scheduler.• 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 under 1 and 4 is nonpreemptive• All other scheduling is preemptive 5.5 Silberschatz and Galvin 1999 Dispatcher.• 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 5.6 Silberschatz and Galvin 1999 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.• Response time – amount of time it takes from when a request. was submitted until the first response is produced, not output. (for time-sharing environment). 5.7 Silberschatz and Galvin 1999 Optimization Criteria.• Max CPU utilization.• Max throughput.• Min turnaround time.• Min waiting time.• Min response time. 5.8 Silberschatz and Galvin 1999 First-Come, First-Served (FCFS) Scheduling. • Example: Process Burst Time. P1 24. P2 3. P3 3 • Suppose that the processes arrive in the order: P1 , P2 , P3. The Gantt Chart for the schedule is: P1 P2 P3. 0 24 27 30 • Waiting time for P1 = 0; P2 = 24; P3 = 27 • Average waiting time: (0 + 24 + 27)/3 = 17. 5.9 Silberschatz and Galvin 1999 FCFS Scheduling (Cont.).Suppose that the processes arrive in the order. P2 , P3 , P1 .• The Gantt chart for the schedule is: P2 P3 P1.0 3 6 30• 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. 5.10 Silberschatz and Galvin 1999 Shortest-Job-First (SJR) Scheduling.• Associate with each process the length of its next CPU burst Use thes
