操作系統(tǒng)全英文期末考試題(帶答案)

1、一選擇題 (20 分，每題 1 分）1.enerally speakng, which one snothe ajoconcrnfor a perat ng syst m i he o lo ing four o ions?( d ) amanag the computer bmangeth syste resurces c.design nd apeinterfacebtwnusrpogram and c mpter rdwae y em .ig evlproramming langage mplier 2. the mai diadvantg of bchsytem s （c ) a.cp

2、u utliztio i lb. not cocrent c.lkf itain d.lw degre f utmtion 3. a roestrsfrms from wait ngsateo redy sate i aus d b e (b）a.inter eent b.roces chedulinc.craea pr ssfrrrad.witi g orme event 4. the ncrt proces is referso( c）a.te pres anb r i rlll b.th press can be rn norder c.the rocss can berun inthe

3、 sa e tie d.the proces can n tbe interrupte5. in ult-roce syste, n ode tonsure the integity o public vriabls，the proess shoulbeutualyecl veccestrtical aeas.the so-ale ital area s ( ）a.abufer b.a date aea c.synchrniztionmecanism d.a pogam 6. teoderl use f reso cs ocation sraean destroy he onii( d ) o

4、 avi dedloc. a.mu ua exclusive b.hod and wait c.no prem od.circular wait 7. use applications use thesyse resourc tocomplet its operaton y thesuppond se vices o ( c ) a.clic igthe oue b.keybard comad c.ssm call d.graphuse iterface 8. tear fou jbs arrie at t sae ieand t xecuin ime ofeachbis 2h nowt un

5、 on onprocessort sngle hanel,then v ge turaroundimis (b ) .h b.5h c.5h dh 9. ong th jo scedling algrts, ( )is r tedt the josetmae runninime. a.fcs sched lg algorithm b.ort-jbfrtshulng algorithm c.hih respon e ati a rthm d.balaned s heduli10.in memormagmet, he prpsfusingthe overay nd swapig i （c) a.s

6、harig n mmrb.expndingmain memorypysically c.svng ain meryspce d.mproving cpu utlition 11.in th pgerplaceme t agorithm,whihnecn cue thbeadhenomnon?( a ) a.ifo b.lru c.lokin d.opt 12. th fllowing descrpion the system in afe sta e,whc oneis co t？( b) a.ust cau e ddl ifthe system s in isecure te b. may

7、cause deadck if the systemi in insecur state c.it may caus dadlk if the sstm s in scre tate d.al ar wrng 13.gene aly，wn we talk boutmoy rotecion, the sc mea ig is ( c ）a.pr nt adwar emofrm damagig b.pve prgram om oig in emry c.prvent e crs-borercalbeeen prog ams d.preent the program f obeing peeped

8、14.the a tualcpaity irtual emory iequal to ( ) a.th capacit of externa moy(ik) b.h sm ofthe ca aci eenlemoy andmain mory c.hspae tha he cpulical ddres givs d.thmaller n beween th opton b ad c 15.phyicl files rganizatio is dtemn y(d) a.a icatns b.main memor cp ity c.exernal me rycapacity d.opering sy

9、stem 16. copute syste conigued with wo plotrsnd ree printes, ore prorl drvehes evie,syste shouldpro e ( c ) deicrive pogam. a.5 b.c.2 d.17.wh thee are en er of chanels insyse ,t aycasebttlneck.to olve t problem,wch of hfollo otins o theffetvewy?（）a.mpvng thspeed of pb.using thvirtua evice tchnolgy c

10、.adig som hardwar ffenthedvices d.incr in the ath be eendevices an chnnels 18.whei/ deics and main memory ar xchanndat，it ca baceed withtcp frequently int entin,hiswa of exhaing tais caled ( c）a.polinb.intrrps c.dect emoracess d.one f he19. the folowing descrii o e managem nt,whicoe noorrec？（b ）a.al

11、 externa devisare ma ed b hesysem in uiform b.hnnel is asoftware of ontrlng input n otut c.thi int tevnt fomthe i/o chanelare manedydvice maagmnt d.one of he esp nsiilityfthe opera ig syte is to use te rdwr fctvely 20.h opertingsysm used （a ） , t realized a mech nism t w can use more pacetosavemoe t

12、imea.soolingb.vir ual stoagec.canel d.overlay 二填空題 (20 分，每空 1 分) 1.softwar maytrggeran inerupt b exe ting specil operatin clled sytm a (p) 2.if th e i onlyon eneral upos u,thenhe systemis a single-p ssor s stem(2) 3. a prcsca b thougt o sa progra execuin. (79) 4.asa poess exe ue，itchanges stat ech p

13、rocess ma be i one of te fol ing tes:w,runni,waiting,eady or trinated .(p83) 5.long-term(jb) shduing is thslctionof processes thatill elloed to ontend frecu and horterm(pu) shedlin is the slcionof one prcess from the eady qee. (116) 6.the proes exeuing in th oeraingsstmm e ether indepednt ocesses or

14、cooperatigpoceses. ooperatig poce erquire an interpros comnication mechanismc mun e with ech othr.pncpally,ommuniction s achive throgh t schemes: haememry d mesg assing. (p16) 7.in odernoperatngys ,rsrce alocati uit is proce,pocesor cheung nit ishrad (27) 8. st o rn pe tng ystmsrovdek nesupport for

15、teads ；mog tese are w ndow,asweas solais and linx .(p146) 9.cpu chedulng is thebsis ofmultpgrmmed operating syste s(p13) .he cfs alor m is onpreempve;he rralorit see pive. somtimes,a waitng proess is nev r ganabl tocne stte，becase te eors it hasqust red y oter waitng pocesses this sitaton is ledeadl

16、ck (p24）12.temain purpose o a oputesytem is txeuteprogam.these programs,togeth r ith the daa they acss,ust e in ai memo y(a le parti lly) uringexecution.(p274) 13. variousemory-management goriths f in masecs.i cmprig dferent memory-mangeent straegie,we ue t folowcoderations:hrdwaport,performnc,frage

17、natin,relocation, swping,sharing and p otetion. (p310）14.aprocess is thrsigi i is pendg moe imepaing tan executi g. 15.virtumemory s te iq tat enablesuso map a lareogical addresssaeo aler physiaemory.(p36) 16.when we solve the or problemsofpe repace e nd framellocton,the proeesgn of a agng system rq

18、uirs tht wecnside page ize,i/,ocking,rocess reaio,progrm sructue,nd other isue.(p36) 17.teoperatng ystm abstract fo tepysicl proeries f it storge devics to dfi a logicl torage u t,thefile.（p373) 18.sice fles r theain informa n-soag mchaim insompter systm,ileprotectin is neded(p4) 19.h eetime isthe t

19、mefo the diskarm to oe the heds to thecylidr contani t desred secto .(45) 20.the rdwar ea m tatabls vce tontiy the u is alled nintrrupt （p499) 三簡答題（分，每題6 分）1.wt the opertin system? hat role oes he erating systm play acmputer? 開放題，解釋操作系統(tǒng)概念, 操作系統(tǒng)可以實現(xiàn)哪些基本功能？關鍵詞：a. 管理系統(tǒng)資源，控制程序運行，改善人機界面, 為其他應用軟件提供支持。 b.

20、基本功能 : 進程與處理機管理、作業(yè)管理、存儲管理、設備管理、文件管理。2.herear fivejob， c,d,eto be running， each o he estimated runni gime ofa,b,e is ,6 ,3 ，5 ,x. write dow eir runig order which s the shrtest aveage waititime? 由于短作業(yè)優(yōu)先調(diào)度算法可以使作業(yè)的平均周轉(zhuǎn)時間最短,同樣使平均等待時間最短 ,所以采用段作業(yè)優(yōu)先算法。下面對的取值進行討論: 當 0 x3 時，作業(yè)的運行順序應該為e,b,；當 35 時,作業(yè)的運行順序應該為,e,

21、d,b,a；當 5=x=6 時，作業(yè)的運行順序應該為c， ,b,a; 當 6x9 時,作業(yè)的運行順序應該為 ,d,b,e 3.whts e criteaof se ting a pr per oces sedulng agorithm？由于各種調(diào)度算法都有自己的特性, 因此, 很難評價哪種算法是最好的。一般說來，選擇算法時可以考慮如下一些原則： c利用率 (cpu utilizaton)； 吞吐量 (tr ugpt ）; 等待時間 ( ng ti e） ；? 響應時間（esons t me)。 周轉(zhuǎn)時間 (u ar d ti e)?在選擇調(diào)度算法前 , 應考慮好采用的準則，當確定準則后 , 通

22、過對各種算法的評估， 從中選擇出最合適的算法。4. explai thefolowingtrm：1)ytem alls )vrtual mmoy 3)ddess rel caion 1)系統(tǒng)調(diào)用提供了應用程序和操作系統(tǒng)的接口。系統(tǒng)調(diào)用把應用程序的請求傳給內(nèi)核，調(diào)用相應的的內(nèi)核函數(shù)完成所需的處理，將處理結(jié)果返回給應用程序。2)虛擬存儲器是指一種實際上并不存在的虛假存儲器,它是系統(tǒng)為了滿足應用對存儲器容量的巨大需求而構(gòu)造的一個非常大的地址空間,從而使用戶在編程時無須擔心存儲器的不足,就好像有一個無限大的存儲器供其使用一樣。3）指當程序裝入到與其邏輯地址不同的主存空間時,將程序地址空間的邏輯地址變換

23、為主存空間的物理地址的變換過程。5 a funciond yu thinthe file syste can relize?pease u e your own wds. (1) 提供方便的文件系統(tǒng)應用接口; ()將邏輯文件映射為物理文件; (3) 管理文件存儲空間的使用; （4）保證文件存儲的安全性和可靠性。四應用題（ 30 分，每題 0 分) 1.theris a tict halln acmmodate10 peopl.i it is ls ta 100pepl n the hall, buyrs cano nto ad bu kes，then leave he hall. if ter

24、e hs100pople, th buershou wait otde he all.now p eae aner the followigquston: (1)i itsynchro zato or muulexclsionbetwe thebuyer？(2)descrthe process of buyer wit（) ad signa(）operaon: (1)購票者直接是互斥關系。(2)一個售票廳可容納10人購票，說明最多允許10人共享售票廳 ;可引入一個信號量emy,初值為 00.由于購票者必須是互斥地進行購票,故應再設一個 u x,初值為 1。用 wait（) 和igna() 表達

25、如下：mp， mtex：ema hoe; emp： =00;utex：1； n: wi(mpty)；a（ utex); 進入購票 ,購票后退出ignl(py） ；sinl(muex）; nd 2.ia compute stem,herei on cu an twoexern devices o1 ando2,notree jos j，j2 and j are aleady n hema mr o b ru,te ririy ofthe is jj2j3.t order nd timtey sthecp and decesare sflowng: j1:io2(0ms）,pu(0s)，i1（0s

26、),cpu(0s). j2：i1(2ms),cp(20ms）,io(40ms). 3:cpu（30s),1(20ms）. owhey run at th ae time and he se f esus i preempiv,plase answer the follo in q tions（y eed not to onsider oter oprt g tie)：(1)calulte the time1,j2, cost frostart to fishprael(2)what is cu utizatio hn ee b are finisi? (3)wht is he os tilaion en threobs e nsing? a: (1)j1占用完 io2 30ms之后,搶占的 cp， 運行 10m,再占用 i 30s,然后占用 pu 1ms，完成。所用時間為 +103010=m。j2占用 io 20ms，搶占了 j的 cp,運行 10ms后，被 j1 搶占 cp，等待 10m，之后再次得到 pu，運行 0ms,之后占用 io2 40ms,所用時間為 0+0+1+10+0=0ms。j3占用 pu 2