《操作系统复习题7》由会员分享，可在线阅读，更多相关《操作系统复习题7（8页珍藏版）》请在金锄头文库上搜索。
1、6pe;eajo si ssaoojd e uaq/v pasn oso 中 iuoj )o#!P Aaq; op moh 乙 poiEan si peajq; e uaq/v pasn aje saojnose；eq/v g g 乙。中。中 ueaAaq adA; auo si saouejsiuno opun 乙 speai中 Qg-|ou6| pue speai 中 |8Aa|-jasn uooMioq saouajajjip om； aje ;eq/v g g uORrqos papeajm-a|6u!S eo aoueiujopad aAOjdiui ;ou p|no/v JE+ b
2、iqpEai + ! niu jo sa|diuexa 6 u!iuiubj6ojcI om; apiAOJd 乙 Q UORrqos popEai 中-oibiqs eo aoueiujopad paAOjdiui 6UIAI6 biqpEai + ! niu jo sa|diuexa 6u!iuiubj 6ojc I OM; apiAOJd |/gsassaoojd uaaMjaq qojiMS jxajuoo o;sqAe e suoipe aq; aquosaQ ? 6ui|npoips iuja;-6uo|pue lujaj-iurqpoiu lujaj-ijoqs 6uoiue s
3、aouajajjip aq; aquosaQ 乙 t6u6isap tuajsAs o; qoeojdde laujaAojoiiu oi|i jo sa6e;ueApe uieiu aq; aje ;eq/v 乙 |/ 6u6isap tuajsAs o; qoeojdde pajaAei 。中 jo a6e;ueApe uieiu o 中 si ;eq/v |/ 6S脂。tuajsAs jo asodjnd o 中 si ;eq/v z e 6ieuja| o 中 tuojj ajejedas A|ensn s Ajqjajdjaju! pueiuiuoo o + jo asodjnd 0
4、 tP si ;eq/v g g 6; uaiua6eueiu o 冉 o; pje6aj ui tuajsAs 6unejado ue jo saijiAipe oeiu aAij。中 aje ;eq/v 6 ; uaiua6eueiu a6ejo;s-/tiepuooas o; pje6aj ui tuajsAs 6u!; ejado ue jo saijiAipe oeiu aajq; 0 中 aje ;eq/v 眈 6;uaiua6eueiu Ajoiuaiu oi pje6aj ui tuajsAs Buijejado ue jo saijiAipe oeiu aajq; o 中 a
5、je ;eq/v 乙 6； uaiua6eueiu ssaoojd o; pje6aj ui tuajsAs 6unejado ue jo saijiAipe oeiu aAij。中 aje ;eq/v |/ AeoiAap 0 中 aAeuiiuiia pue a6je| 中 “ jou Aq/v ( S!p e se a6je| se aqoeo e aouejsuioj aoiAap aq; se a6je| se apeiu eq ueo aqoeo e j| 乙 os neo Aaq;op siuaiqojd ;eq/v 乙 oaqs Ao 中 op siuaiqojd ;eq/v
6、|npsn aje saqoeo Aq/v suoseaj om； oaiq 6N jaMSuej noA uieidxA 乙ot 忆乙 乙 uoqounj qoea joosn o 中 si ;eq/v 乙dnjjajui ue pue dej; e uaaMjaq saouajajjip o 中 aje ;eq/vZlAluejsAs (A;uroes)uoipajojd jo iujo)(阴李聋)Ajejuaiuipru e se uoipunj apoiuosn pue apoiuojiuoiu uaaMjaq uoipuijsip aq; saop moh 乙乙pajnqujsjQ
7、J中 OMQN9aiuR |eayp6uueqs aiuiiaAipejajui q ire 日e:siuajsAs 6unejado jo sadA; 6uiM0|0j o + jo saipadojd leijuassa o 中 auijaa 94 乙 soindiuoo leuosjad pue soindiuoo aiuejjuieiu oj siuajsAs Buijejado uaaMjaq saouajajjip uieiu 0 中 aje ;eq/v y 66u!iuiuej6ojdmniu 40 a6e;ueApe uieiu 0 中 si ;eq/v j (SiuejsAs
8、 6u! ; ejado ue 40 sasodjnd uieiu aajq; 0 中 aje ;eq/v 1/ L6.1 A CPU scheduli ng algorithm determ in es an order for the executi on of its scheduled processes. Give n n processes to be scheduled on one processor, how many possible differe nt schedules are there? Give a formula in terms of n.6.2 Defin
9、e the differe nee betwee n preemptive and non preemptive scheduli ng. State why strict non preemptive scheduli ng is un likely to be used in a computer cen ter.6.3 Consider the following set of processes, with the length of the CPU-burst time given in millisec onds:ProcessBurst Time Priority10P2P3P4
10、1, P2, P3, P4, P5, all at time 0.P5The processes are assumed to have arrived in the order Pa. Draw four Gantt charts illustrating the execution of these processes using FCFS, SJF,a non preemptive priority (a smaller priority nu mber implies a higher priority), and RR (qua ntum = 1) scheduli ng.b. Wh
11、at is the turnaround time of each process for each of the scheduling algorithms in part a?c. What is the waiting time of each process for each of the scheduling algorithms in part a?d. Which of the schedules in part a results in the minimal average waiting time (over all processes)?6.4 Suppose that
12、the follow ing processes arrive for executi on at the times in dicated. Each processwill run the listed amount of time. In an sweri ng the questi ons, use non preemptive scheduli ng and base all decisions on the information you have at the time the decision must be made.ProcessArrival TimeBurst Time
13、PiP20.00.4P3a. What is the average1.0turnaround time1for these processes with the FCFS schedulingalgorithm?b. What is the average tur naround time for these processes with the SJF scheduli ng algorithm?c. The SJFalgorithm is supposed to improve performanee, but notice that we chose to run process P
14、i at time 0 because we did not know that two shorter processes would arrive soon. Compute what the average tur naround time will be if the CPU is left idle for the first 1 unit and the n SJF scheduli ng is used. Remember that processes P 1 and P 2 are wait ing duri ng this idle time, so their wait i
15、ng time may in crease. This algorithm could be known as future-k no wledge scheduli ng.Remember that turnaround time is finishing time minus arrival time, so you have to subtract the arrival times to compute the tur naround times. FCFS is 11 if you forget to subtract arrival time.6.6 What advantage is there in having different time-quantum sizes on different levels of a multilevel queue ing system?6.10 Explain the differences in the degree to which the following scheduling algorithms discrim in ate in favor of short processes:a.FCFSb.RRc.Multilevel feedback queues8.1 List three
