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ID:13655463
大小:1.08 MB
页数:24页
时间:2018-07-23
《操作系统基础基本理论、基本方法复习61290》由会员上传分享,免费在线阅读,更多相关内容在教育资源-天天文库。
1、Chapter1Introduction1.AbstractViewofSystemComponents2.MultiprogrammedBatchSystemsSeveraljobsarekeptinmainmemoryatthesametime,andtheCPUismultiplexedamongthem.Chapter2:Computer-SystemStructures1.Computer-SystemArchitecture2.InterruptTimeLineForaSingleProcessDoingOutput3.Moving-HeadDiskMech
2、anism4.DirectMemoryAccessStructureSixStepProcesstoPerformDMATransfer5.Storage-DeviceHierarchy6.UseofABaseandLimitRegister7.HardwareAddressProtectionChapter3:Operating-SystemStructures1.MS-DOSExecution2.MS-DOSLayerStructureChapter4:Processes1.DiagramofProcessState2.ProcessControlBlock(PCB
3、)3.CPUSwitchFromProcesstoProcess4.RepresentationofProcessScheduling5.AdditionofMediumTermSchedulingChapter5:Threads1.SingleandMultithreadedProcesses2.Many-to-OneModel3.One-to-oneModel4.Many-to-ManyModelChapter6:CPUScheduling1.AlternatingSequenceofCPUAndI/OBursts2.First-Come,First-Served(
4、FCFS)SchedulingProcessBurstTimeP124P23P33lSupposethattheprocessesarriveintheorder:P1,P2,P3TheGanttChartforthescheduleis:lWaitingtimeforP1=0;P2=24;P3=27lAveragewaitingtime:(0+24+27)/3=173.ExampleofNon-PreemptiveSJFProcessArrivalTimeBurstTimeP10.07P22.04P34.01P45.04lSJF(non-preemptive)lAve
5、ragewaitingtime=(0+6+3+7)/4-44.ExampleofPreemptiveSJFProcessArrivalTimeBurstTimeP10.07P22.04P34.01P45.04lSJF(preemptive)lAveragewaitingtime=(9+1+0+2)/4-35.ExampleofRRwithTimeQuantum=20ProcessBurstTimeP153P217P368P424lTheGanttchartis:lTypically,higheraverageturnaroundthanSJF,butbetterresp
6、onse.6.DispatchLatencyChapter7:ProcessSynchronization1.Implementation•Semaphoreoperationsnowdefinedaswait(S):S.value--;if(S.value<0){addthisprocesstoS.L;block;}signal(S):S.value++;if(S.value<=0){removeaprocessPfromS.L;wakeup(P);}2.Deadlock•–twoormoreprocessesarewaitingindefinitelyforanev
7、entthatcanbecausedbyonlyoneofthewaitingprocesses.•LetSandQbetwosemaphoresinitializedto1P0P1wait(S);wait(Q);wait(Q);wait(S);MMsignal(S);signal(Q);signal(Q)signal(S);3.Bounded-BufferProblem•Shareddatasemaphorefull,empty,mutex;Initially:full=0,empty=n,mutex=1ProblemProducerP
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