操作系統(tǒng)概念課件ch(文件)
【正文】 e programming a model of the puter system. ? Software data structures represent the major ponents of the system. ? The simulator has a variable representing a clock。 (., serve all from foreground then from background). Possibility of starvation. ? Time slice – each queue gets a certain amount of CPU time which it can schedule amongst its processes。 P2 = 24。 this involves: ? switching context (saving context and restoring context) ? switching to user mode (from monitor mode ? user mode) ? jumping to the proper location in the user program to restart that program (PC counter) ? Dispatch latency – the time it takes for the dispatcher to stop one process and start another running. SCHEDULING CRITERIA (標準 ) ? CPU utilization (使用率 ): keep the CPU as busy as possible. ? CPU throughput (吞吐量 ): number of processes that plete their execution per time unit. ? Process turnaround time (周轉時間 ): amount of time to execute a particular process. ? Process waiting time (等時間 ): amount of time a process has been waiting in the ready queue. ? Process response time (響應時間 ): amount of time it takes from when a request was submitted until the first response is produced, not output (for timesharing environment). Scheduling criteria ? To maximize or minimize some average measures: ? Maximize CPU utilization. ? Maximize CPU throughput. ? Minimize process turnaround time. ? Minimize process waiting time. ? Minimize process response time. ? To maximize or minimize more average measures: ? Peak value (峰值 ) ? Expectation (數學期望 ) (Average) ? Variance (方差 ) SCHEDULING ALGORITHMS ? Scheduling algorithms ? First e first served (FCFS) （先到先行調度） ? Shortest job first (SJF) （最短作業(yè)優(yōu)先調度） ? Priority scheduling （優(yōu)先權調度） ? Round robin (RR) （輪轉法調度） ? Multilevel queue algorithm （多級隊列調度） ? Multilevel feedback queue algorithm （多級反饋隊列調度） Scheduling algorithms: FCFS(最先先行 ) Process Burst Time (區(qū)間時間 , ms) P1 24 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。 P3 = 3. ? Average waiting time: (6 + 0 + 3)/3 = 3 (ms) ? Much better than previous case. P1 P3 P2 6 3 30 0 Scheduling algorithms: FCFS ? Convo
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