1. H

VI N

NG NGH

U H NH VI N TH NG

KHOA: ng ngh thng tin

n hc phn:..H iu hnh.. M hc phn:............ gnh o to : . ng ngh thng tin.. rnh o to: .i hc. 1. g n hng c u h i thi u h i loi 1 im hng 1 : u hi 1.1: hng trnh ng dng gi dch v ca h iu hnh bng cch no? Hy ly mt v d v giao din lp trnh cho mt h iu hnh thng dng. u hi 1.2: Trnh by k thut x l theo m (l) v u im ca k thut ny. H thng x l theo m c cn h iu hnh khng ? u hi 1.3: a chng trnh l g ? L do s dng a chng trnh trong my tnh ? Yu cu i vi phn cng khi s dng a chng trnh? hng 2 : u hi 1.4 : Trnh by khi nim tin trnh v ch r im khc nhau gia tin trnh vi chng trnh. Nu tn t nht bn thao tc lin quan ti qun l tin trnh (ch cn nu tn, khng cn trnh by chi tit). u hi 1.5 : Trnh by v thao tc to mi tin trnh. Tin trnh c th b kt thc trong nhng trng hp no ? u hi 1.6 : Trnh by v thao tc v qu trnh chuyn i gia cc tin trnh. u hi 1.7: Th no l dng (thread) mc ngi dng v mc nhn. Nu u nhc im ca mi loi. u hi 1.8 : Trnh by v iu quay vng. ho v d minh ha v tnh thi gian ch i trung bnh khi iu theo kiu ny. u hi 1.9 : Th no l b tc ? iu kin xy ra b tc l g ? hng 3 : u hi 1.10 : Th no l a ch l gic v a ch vt l ? u hi 1.11 : Trnh by k thut phn chng c nh b nh. u hi 1.12 : Trnh by c ch nh x a ch khi s dng k thut phn chng b nh. u hi 1.13 : Trnh by phng php kt hp phn trang vi phn on. V s v gii thch c ch nh x a ch. . hng 4 u hi 1.14 : Vic nh ngha v s dng khi nim file em li nhng u im g ? Khi t tn cho file cn quan tm ti nhng quy nh g ? u hi 1.15 : Trnh by khi nim th mc ? Thng tin trong cc khon mc c nht thit phi lu tr gn nhau khng ? u h i loi 2 im hng 1 : u hi 2.1: Trnh by ngn gn v cc thnh phn c bn ca h iu hnh. u hi 2.2 : Trnh by v nhn ca h iu hnh ? Th no l ch nhn v ch ngi dng ? u hi 2.3 : Trnh by v cu trc nguyn khi v cu trc phn lp ca h iu hnh. Phn tch so snh u nhc im hai kiu cu trc ny. u hi 2.4: Trnh by v cu trc vi nhn ca h iu hnh. Phn tch so snh cu trc ny vi cu trc nguyn khi v cu trc phn lp. hng 2 : u hi 2.5 : Trnh by v nm trng thi ca tin trnh. V s v gii thch v vic chuyn i gia nm trng thi ny

u hi 2.6 : iu tin trnh l g ? iu dng c khc iu tin trnh khng ? Trnh by v iu c phn phi li v khng phn phi li. u hi 2.7 : Trnh by hai bin php ngn nga b tc (chn 2 bin php bt k trong s cc bin php c th). u hi 2.8 : Trnh by mt gii php gip khng xy ra b tc khi s dng c hiu cho bi ton trit gia n cm. hng 3 : u hi 2.9 : Trnh by k thut gip tng tc truy cp bng trang v bng trang nhiu mc. u hi 2.10: Trnh by l do phi i trang, v cc bc tin hnh khi i trang. u hi 2.11: Trnh by k thut i trang ti u v i trang vo trc ra trc. u hi 2.12 : Trnh by cc phng php xc nh s lng khung trang ti a cp cho mi tin trnh v xc nh phm vi cp pht hng 4 : u hi 2.13 : Trnh by cc cu trc d liu dng cho t chc bn trong ca th mc. u hi 2.14 : Trnh by cch kim sot truy cp file s dng mt khu v s dng danh sch qun l truy cp u hi 2.15 : Trnh by cc thao tc c bn vi file. Phn tch r mt h thng file c nht thit phi c thao tc m file hay khng. u h i loi 3 im Chng 1 : u hi 3.1: Trnh by khi nim h iu hnh. Phn tch r hai chc nng c bn ca h iu hnh. u hi 3.2 : Dch v ca h iu hnh l g ? Trnh by nhng dch v in hnh m h iu hnh cung cp. Lm r v qu trnh ti v chy h iu hnh khi mi khi ng. hng 2 : u hi 3.3 : Trnh by khi nim dng (thread) v m hnh a dng. Vn s hu ti nguyn ca tin trnh v dng. Phn tch u im ca m hnh a dng. u hi 3.4 : Phn tch cc vn cn quan tm trong s dng v qun l tin trnh ng thi (concurrent processes) i vi ba dng tin trnh : tin trnh c lp c cnh tranh ti nguyn, tin trnh hp tc nh chia s ti nguyn, v tin trnh hp tc nh trao i thng ip. u hi 3.5 : Trnh by gii thut Peterson cho on nguy him. Phn tch u nhc im ca phng php ny. u hi 3.6 : Trnh by phng php pht hin v x l b tc bng cch s dng th. Phn tch u im ca phng php ny so vi phng php ngn nga b tc. hng 3 : u hi 3.7 : Trnh by k thut ti trong qu trnh thc hin. Trnh by k thut lin kt ng v th vin dng chung. Phn tch r u im m tng phng php em li. u hi 3.8 : Trnh by k thut phn chng ng b nh. Phn tch u nhc im ca phng php ny so vi phn chng c nh. Khi di chuyn chng sang v tr khc cn thay i thng tin g trong khi nh x a ch. u hi 3.9 : Trnh by k thut phn chng s dng phng php k cn (buddy). Phn tch r cc im ging/khc nhau v u nhc im ca phng php ny so vi phn chng c nh v phn chng ng (lu khng cn trnh by li hai phng php sau). u hi 3.10 : Trnh by k thut phn on b nh bao gm c cu trc v cch nh x a ch. Phn tch so snh u nhc im ca phn on vi phn trang. u hi 3.11 : Trnh by k thut np trang theo nhu cu dng cho b nh o. Phn tch r cng mt lnh c th xy ra nhiu s kin li trang khng. hng 4 : u hi 3.12 : Trnh by cu trc th mc dng cy v dng th khng c chu trnh. u trc th mc dng khng chu trnh c u im g so vi dng cy ? Th no l ng dn tuyt i v ng dn tng i. u hi 3.13 : Trnh by phng php cp pht khng gian cho file s dng danh sch kt ni v s dng khi ch s (I-node). Hai phng php ny c im g ging v khc nhau. u hi 3.14 : Trnh by v yu cu phi m bo tnh ton vn ca h thng file v cc phng php

m bo tnh ton vn. u h i loi 4 im hng 2 : u hi 4.1: a) Trnh by cc tiu ch nh gi thut ton iu . b) Trnh by thut ton iu n trc phc v trc v iu c mc u tin. c) ho cc tin trnh vi thi gian ( di) chu k PU tip theo v s u tin nh trong bng sau (s u tin nh ng vi u tin cao). it rng cc tin trnh cng xut hin vo thi im 0 theo th t P1, P2, P3, P4. Tin trnh Thi gian ( di) S u tin P1 9 3 P2 1 1 P3 2 2 P4 3 1

V biu th hin th t v thi gian cp pht PU cho cc tin trnh khi s dng thut ton : 1) iu quay vng vi di lng t = 1 ; 2) iu theo mc u tin khng c phn phi li. Tnh thi gian ch i trung bnh cho tng trng hp. u hi 4.2 : a) Trnh by thut ton iu u tin tin trnh ngn nht, thi gian cn li ngn nht. b) iu theo mc u tin c phn phi li v khng phn phi li khc nhau th no ? c) ho cc tin trnh vi di v thi im xut hin nh trong bng sau Tin trnh Thi im xut hin di P1 8 P2 2 4 P3 4 2

V biu th hin th t v thi gian cp pht PU cho cc tin trnh khi s dng thut ton : 1) iu u tin tin trnh ngn nht ; 2) iu u tin thi gian cn li ngn nht. Tnh thi gian ch i trung bnh cho tng trng hp. u hi 4.3: a) Trnh by v cc gii php phn cng (cm ngt, s dng lnh my c bit) cho vn loi tr tng h v on nguy him. b) S dng Test_and_Set thc hin loi tr tng h cho bi ton cc trit gia n cm. c) Phn tch r gii php s dng Test_and_Set s dng trn c th gy b tc hoc i khng. u hi 4.4 : a) Trnh by phng php s dng c hiu (semaphore) cho vn loi tr tng h v on nguy him. b) S dng c hiu thc hin ng b ha cho bi ton Ngi sn xut, ngi tiu dng vi b m hn ch. u hi 4.5 : a) Trnh by gii php s dng monitor cho vn loi tr tng h v on nguy him b) S dng monitor thc hin loi tr tng h cho bi ton Ngi sn xut, ngi tiu dng vi b m hn ch. u hi 4.6 : a) Phng trnh b tc l g ? Phn tch u im ca vic phng trnh b tc so vi ngn nga b tc. b) Trnh by phng php phng trnh b tc s dng thut ton ngi cho vay (bankers algorithm), cho v d minh ha c th. hng 3 : u hi 4.7 : a) Trnh by khi nim phn trang b nh. b) Trnh by v nh x a ch khi phn trang b nh.

c) Gi s khng gian nh l gic gm 4 trang, mi trang kch thc 1024 , b nh vt l gm 32 khung. ng trang c cho di y : 3 1 2 3 5

biu din a ch l gic trong trng hp ny cn bao nhiu bit ? Tnh a ch vt l cho nhng a ch l gic sau : 1052, 2500, 4000. u hi 4.8 : a) Trnh by chin lc i trang t s dng trong thi gian cui. b) Trnh by chin lc i trang s dng thut ton ng h. c) Gi s tin trnh c cp 4 khung nh vt l, cc trang ca tin trnh c truy cp theo th t sau : 1,2,3,4,5,3,4,1,6,7,8,7,8,9,7,8,9,5. Hy xc nh th t np v i trang nu s dng hai thut ton ni trn. hng 4 : u hi 4.9 : a) Trnh by phng php cp pht khng gian cho file s dng cc khi lin tip. Khi no nn s dng phng php ny cho h thng file ? b) Trnh by cc bc cn thit (trnh by bng li, khng cn vit m) c bng FAT t th nh US vo b nh. c) Gi s bng FAT c c vo b nh ti a ch void *fat , vit on chng trnh trn / ++ lit k tt c cc cluster trng trong s N cluster u tin. Gi s mt file bt u ti cluster n, vit on chng trnh lit k cc cluster thuc v file . u hi 4.10 : a) Trnh by phng php s dng danh sch kt ni trn bng ch s khi cp pht khng gian cho file. b) Trnh by cc bc cn thit (trnh by bng li, khng cn vit m) c th mc gc t th nh US vo b nh trong trng hp h thng file ca th nh l FAT 16. c) Gi s th mc gc ca h thng file FAT 16 s dng tn file di ti a 8 k t c c vo b nh ti a ch void *root . Vit on chng trnh trn / ++ thc hin hai vic : in tn v di cc file trong th mc gc, tm mt file c tn cho trc trong th mc gc v s th t cluster u tin ca file .

hi ch : K hiu (m) cu hi c quy nh X.Y Trong : + X tng ng s im cu hi (X chy t 1 n 5). + Y l cu hi th Y (Y chy t 1 tr i)

2. u t c c ph ng n t h p c u h i thi thnh c c thi (Nu thy cn thit) :................... T hp 1 cu 1 im, 1 cu 2 im, 1 cu 3 im, 1 cu 4 im. 3. ng d n cn thi t h c: .................................................. .................................................. ........ Nn t hp cu hi trong mt thi phn b u theo c 4 chng

Ngn hng cu hi thi ny c thng qua b mn v nhm cn b ging dy hc phn. H Ni, ngy . . . thng . . . . nm 20 . .r ng hoa r ng b mn i ng vin ch tr bin son

Cn y l bi gii thng bn gi cho mnh. 1.1: h ng trnh ng dng gi dch v c a h i u hnh bng c ch no? y l y mt v d v giao din lp trnh cho mt h i u hnh thng dng.+ c chng trnh ng dng gi dch v ca HDH thng qua li gi h thng c cha trong giao din lp trnh m HDH cung cp+ li gi h thng l dng lnh c bit, c thc hin qua th vin hm gi v cc hm h thng khc+V d mt

hm ca Win32 API cho php yu cu dch v ghi ra file ca Windows: OOL WINAPI WriteFile(__in HANDLE hFile,__in LPCVOID lpBuffer,__in DWORD nNumberOfBytesToWrite,__out-opt LPDWORD lpNumberOfBytesWritten,__inout-opt LPOVERLAPPED lpOverlapped 1.2: Trnh by thut l theo m (l) v u im c a thut ny. thng l theo m c cn h i u hnh hng ? Sau khi nhn dc chng trnh k thut vin s phn ct thnh cc m,mi m gm nhng ct c yu cu ging nhau.Ton b m sau c np vo bng t v ti vo my thc hin ln lt.Mi khi ct ca m kt thc ct gim st(mt chng trnh nh c gi thng xuyn trong b nh) t ng np ct tip theo ca m vo my v cho php ct ny chy.Sau khi ton b m c thc hin xong k thut vin ly bng t cha m v np tip m mi vo thc hin.+u in ca kt : gim ng k thi gian chuyn i gia 2 ct // cng mt m do momitor c th t ng np vo ct nhanh hn k thut vin.Hiu sut s dng PU do c ci thin ng k+h thng x l theo m cn HDH gim st vic np ct v chy ct. 1.3: a ch ng trnh l g ? do s dng a ch ng trnh trong m y tnh ? Yu cu i v i phn cng hi s dng a ch ng trnh? a chng trnh hay cn gi l a nhim c s dng.H thng cha ng thi nhiu chng trnh trong b nh.+l do : mc d vic x l theo m cho php gim thi gian chuyn i gia cc ct ng dng xong hiu sut s dng PU vn tng i thp.Mi khi c yu cu vo ra PU phi dng vic x l DL ch qu trnh vo ra kt thc .Do tc vo ra lun thp hn tc PU rt nhiu nn PU thng xuyn phi ch i trong nhng khong thi gian di hn ch tnh trang ni trn ta s dng k thut a chng trnh.+ vic thc hin a chng trnh i hi nhng s h tr nht nh t phn cng c bit l kh nng vo ra bng ngt v c ch DMA. Nu ko c c ch ny PU s phi trc tip iu khin qu trnh vo ra thng tin v DL.Hiu qu ca a ct do s bng 0. 1.5 : rnh by v thao t c to m i ti n trnh. i n trnh c th b t th c trong nhng tr ng h p no ? To mi tin trnh: Gn s nh danh cho tin trnh c to mi v to mt trong bng tin trnh To khng gian nh cho tin trnh v P .Khi to P . Lin kt P ca tin trnh vo cc danh sch qun l Kt thc tin trnh: Kt thc bnh thng: yu cu HDH kt thc mnh bng cch gi li gi h thng exit() kt thc: tin trnh cha kt thc .Do cc li .Yu cu nhiu b nh hn so vi s lng h thng c th cung cpThc hin lu hn thi gian gii hn Do qun tr h thng hoc h iu hnh kt thc. 1.6 : rnh by v thao t c v qu trnh chuyn i gia c c ti n trnh.Thng tin v tin trnh hin thi (cha trong P ) c gi l ng cnh (context) ca tin trnh Vic chuyn gia tin trnh cn c gi l chuyn i ng cnh Xy ra khi: ngt Tin trnh gi li gi h thng Trc khi chuyn sang thc hin tin trnh khc, ng cnh c lu vo P Khi c cp pht PU thc hin tr li, ng cnh c khi phc t P vo cc thanh ghi v bng tng ng Thng tin no phi c cp nht v lu vo P khi chuyn tin trnh? => Ty tng trng hp: H thng chuyn sang thc hin ngt vo/ra ri quay li thc hin tip tin trnh: Ng cnh gm thng tin c th b hm x l ngt thay i => ni dung thanh ghi, trng thi PU Sau khi thc hin ngt, h thng thc hin tin trnh khc Thay i trng thi tin trnh p nht thng tin thng k trong P huyn lin kt P ca tin trnh vo danh sch ng vi trng thi mi p nht P ca tin trnh mi c chn p nht ni dung thanh ghi v trng thi PU => huyn i tin trnh i hi thi gian. 1.7: h no l dng (thread) mc ng i dng v mc nh n. u u nh c im c a mi loi.dng mc ngi dng do trnh ng dng t to ra v qun l HDH ko bit v s tn ti ca nhng dng nh vy. to ra dng mc ngi dng trnh ng dng s dng th vin ngn ng lp trnh cung cp VD nh trn java.+u im :-do dng to ra v qun l trong khng gian ngi dng nn vic chuyn i dng ko i hi phi chuyn sang ch nhn v do vy tit kim (t) hn-trnh ng dng c th iu dng theo c im ring ca mnh ko ph thuc vo phng thc iu ca HDH-c th s dng dng mc ngi dng c trn nhng HDH ko h tr a dng bng cch b sung th vin dng m cc ng dng c th dng chung.+ nhc : khng tn dng u v tnh p ng ca m hnh a lung, k tn dng kin truc n` PU Dng mc nhn c h iu hnh to ra v qun l.HDH s cung cp giao din lp trnh bao gm mt s li gi h thng m trnh ng dng c th gi yu cu to/xa dng thay i tham s lin quan ti qun l dng .HDH windows v linux l hai VD trong vic h tr dng mc nhn. +u im : khc phc c cc nhc im ca dng mc ngi dng. th l vic s dng dng mc nhn cho php tng tnh p ng v kh nng thc hin ng thi ca cc dng trong cng 1 tin trnh

.Trn h thng vi nhiu PU dng mc nhn c th c cp PU khc nhau thc hin //+ nhc : tc chm 1.8 : rnh by v i u quay vng. ho v d minh ha v tnh thi gian ch i trung bnh hi i u theo iu ny. Sa i F FS dng cho cc h chia s thi gian thm c ch phn phi li bng cch s dng ngt ca ng h H thng xc nh nhng khong thi gian nh gi l lng t/ lt ct thi gian t Khi PU c gii phng, HDH t thi gian ca ng h bng di lng t, ly tin trnh u hng i v cp PU cho n Tin trnh kt thc trc khi ht thi gian t: tr quyn iu khin cho HDH Ht lng t thi gian m tin trnh cha kt thc: ng h sinh ngt Tin trnh ang thc hin b dng li Quyn iu khin chuyn cho hm x l ngt ca HDH HDH chuyn tin trnh v cui hng i, ly tin trnh u v tip tc i thin thi gian p ng so vi FCFS Thi gian ch i trung bnh vn di La chn di lng t thi gian? VD : 2468101214 22222222

P1P2P P1P2P1P1P1 Thi gian ch i ca P1,P2,P3 ln lt l 6,6,4Thi gian ch i Tb=(6+6+4)/3=5,33 1.9 : h no l b tc ? i u in y ra b tc l g ?nh ngha : Tnh trng mt nhm tin trnh c cnh tranh v ti nguyn hay c hp tc phi dng v hn Do tin trnh phi ch i mt s kin ch c th sinh ra bi tin trnh khc cng ang trong trng thi ch iiu kin xy ra b tc :ng thi xy ra 4 iu kin: 1. Loi tr tng h: c ti nguyn nguy him, ti 1 thi im duy nht 1 tin trnh s dng 2. Gi v ch: tin trnh gi ti nguyn trong khi ch i 3. Khng c phn phi li (no preemption): ti nguyn do tin trnh gi khng th phn phi li cho tin trnh khc tr khi tin trnh ang gi t nguyn gii phng ti nguyn 4. h i vng trn: tn ti nhm tin trnh P1, P2, , Pn sao cho P1 ch i ti nguyn do P2 ang gi, P2 ch ti nguyn do P3 ang gi, , Pn ch ti nguyn do P1 ang gi 1.10 : h no l a ch l gic v a ch vt l ? a ch logic: Gn cho cc lnh v d liu khng ph thuc vo v tr c th tin trnh trong b nh hng trnh ng dng ch nhn thy v lm vic vi a ch logic ny L a ch tng i tc l mi phn t ca chng trnh c gn mt a ch tng i i vi mt v tr no a ch vt l: L a ch chnh xc trong b nh my tnh c mch nh s dng truy nhp ti chng trnh v d liu a ch logic c chuyn thnh a ch vt l nh khi nh x a ch 1.11 : rnh by thut ph n ch ng c nh b nh . hia b nh thnh cc chng vi s lng nht nh, khng thay i, mi chng cha ng 1 tin trnh. th chia thnh cc chng c kch thc bng nhau hoc khng bng nhauKch th c c c ch ng bng nhau: n gin Kch thc chng trnh > kch thc chng => khng th cp pht Gy phn mnh trong Kch th c c c ch ng h c nhau: - hn chng c kch thc nh nht: cn c hng i lnh cho mi chng:Gim phn mnh trong, ti u cho tng chng H thng khng ti u - Kch thc cc chng khc nhau: Dng hng i chung cho mi chng: hng sn c nh nht s c cp pht Khi 1 chng c gii phng: chn tin trnh gn u hng nht v c kch thc ph hp nhtu im: n gin, t x l h c im: S lng chng xc nh ti thi im to h thng hn ch s lng tin trnh hot ng Kch thc chng thit lp trc: khng hiu qu 1.12 : rnh by c ch nh a ch hi s dng thut ph n ch ng b nh S dng 2 thanh ghi ca Processor : thanh ghi gii hn v thanh ghi c s+ Thanh ghi gii hn cha di chng cha tin trnh+ Thanh ghi c s cha v tr bt u ca tin trnhKhi 1 a ch logic c np vo, PU s kim tra xem n c nh hn thanh ghi gii hn khng. Nu c th n s c cng vi thanh ghi c s tn c a ch vt l 1.14 : Vic nh ngha v s dng h i nim file em li nhng u im g ? Khi t tn cho file cn quan t m t i nhng quy nh g ? u im: c thao tc vi file s d dng hn i vi ngi lp trnh ng dng HDH c th kim sot c cc thao tc vi file t tn cho file: Cho php xc nh file L thng tin ngi dng thng s dng nht khi lm vic vi file Quy tc t tn cho file ca mt s HDH: H iu hnh, di ti a ,Phn bit ch hoa ch thng ho php s dng du cch , c k t cm 1.15 : rnh by h i nim th mc ? hng tin trong c c ho n mc c nh t thi t ph i l u tr gn nhau hng ?+Th mc = cc khon mc ~ files|Khon mc cha cc thng tin v file: tn, kch thc, v tr, kiu file,hoc con tr ti ni lu tr thng tin ny. oi th mc nh 1 bng, mi dng l khon mc ng vi 1 file.+ Thng tin trong cc khon mc phi nht thit lu tr gn nhau v :

Ton b thuc tnh ca File dc lu trong th mc, file ch cha data=> kch thc khon mc, th mc ln.-Th mc ch lu thng tin ti thiu cn thit cho vic tm kim v tr file trn a=> kch thc gim. 2.1: rnh by ngn gn v c c thnh phn c b n c a h i u hnh. c thnh phn c bn ca h iu hnh:+ Qun l tin trnh : Tin trnh l chng trnh ang thc hinQun l tin trnh bao gm mt s thao tc : to v xa tin trnh, treo v khi phc tin trnh, ng b ha lin lcgiua cc tin trnh, gii quyt b tcqun l b nh : HDH c phn phi v cp pht b nh cho cc tin trnh, to ra va nh x b nh o vo b nh thc, cung cp v gii phng b nh theo yu cu ca tin trnh, qun l b nh cp v cn trng+ Qun l vo ra : nhm n gin ha v tng hiu qu qu trnh trao i thng tin vi thit b vo ra + Qun l tp v th mc : gip ngi dung s dng thng tin m khng cn quan tm n cc c im cu b nh ngoi bn vic s ng cc khi nim tp v th mc. c chc nng ch yu l to, xa tp hoc th mc ; c ghi tp ; nh x tp vo th mc ra b nh ngoi+ H tr mng v qun l phn tn : gip kt ni my tnh vi cc mng my tnh+ Giao din vi ngi dng : l giao din gia HDH vi ngi dng+ c chng trnh tin ch v ng dng : l cc thnh phn ko sn c trong HDH, khng bt buc phi c 2.2 : rnh by v nh n c a h i u hnh ? h no l ch nh n v ch ng i dng ? Nhn (kernel) l phn ct li, thc hin cc chc nng c bn nht, quan trng nht ca HDH v thng xuyn c gi trong b nh h nhn l ch m chng trnh thc hin trong c y quyn truy cp v iu khin phn cng h ngi dng l ch m chng trnh b han ch rt nhiu v quyn truy cp va iu khin phn cngVic chia lm e ch nhm trnh cho chng trnh v tnh hoc c thc hin thao tc lm nh hng h thng 2.6 : i u ti n trnh l g ? i u dng c h c i u ti n trnh hng ? rnh by v i u c ph n phi li v hng ph n phi li.iu tin trnh l vic HDH quyt nh xem tin trnh no c s dng ti nguyn phn cng v trong bao luiu tin trnh quyt nh th t v thi gian s dng PUiu tin trnh c th coi l iu ngiu phn phi l iu m HDH c th can thip vo qu trnh thchin ca tin trnh thu hi c PU, iu cung cp phn phi li lm cho qun l tin trnh phc tp iu khng phn phi li l dang iu m HDH phi ch i vic tgii phng PU ca tin trnh do 1 s l do nh : tin trnh kt thc hay chuyn sang trng thi i. th xy ra trng hp PU b chim hu v h 2.7 : rnh by hai bin ph p ngn nga b tc (chn 2 bin ph p b t trong s c c bin ph p c th). oi tr t ng h: hng th ngn nga Gi v ch: ch 1: Yu cu tin trnh phi nhn ton b ti nguyn cn thit trc khi thc hin tip Nu khng nhn , tin trnh b phong ta ch cho n khi c th nhn ti nguyn ch 2: Tin trnh ch c yu cu ti nguyn nu khng gi ti nguyn khcTrc khi yu cu thm ti nguyn, tin trnh phi gii phng ti nguyn c cp v yu cu li (nu cn) cng vi ti nguyn mi Khng c ph n phi li: ch 1: Khi mt tin trnh yu cu ti nguyn nhng khng c do b cp pht, HDH s thu hi li ton b ti nguyn n ang gi Tin trnh ch c th thc hin tip sau khi ly c ti nguyn c cng vi ti nguyn mi yu cu ch 2: Khi tin trnh yu cu ti nguyn, nu cn trng, s c cp pht ngay Nu ti nguyn do tin trnh khc gi m tin trnh ny ang ch cp thm ti nguyn th thu hi li cp cho tin trnh yu cu Nu hai iu kin trn u khng tha th tin trnh yu cu ti nguyn phi ch 2.8 : Trnh by mt gi i ph p gi p hng y ra b tc hi s dng c hiu cho bi to n tri t gia n c m. trnh b tc cho bi ton trit gia n c ta ch th dng bin php ti mi thi im lun ch c 4 ngi ngi vo bn cn 1 ngi khng ngi . Khi c 5 chic a v 4 ngi -> s c 1 ngi cm c c 2 chi a n cm. Khi ngi ny n xong th s phi ng dy ri khi bn ngi kia c ngi vo bn. thc hin c phng php ny chng ta ch cn thm 1 semaphore cho table kim sot s ngi vo bn trc khi ly a u h i 2.9 : rnh by thut gi p tng tc truy cp b ng trang v b ng trang nhi u mc.Do mi thao tc truy cp b nh u i hi truy cp bn phn trang=> n c c ch t chc bng phn trang lm tng tc truy cp+ Ta c th s dng thanh ghi lm bng phn trang v cc thanh ghi c tc truy cp cao nhng s lng li c hn+ Ta cng c th lu bng trang b nh chnh v s dng 1 con tr tr ti bng phn trang. V tc b nh chm nn ta c th sdng b nh cache tng tc truy cp-Do h thng my tnh hin i c khng gian a ch lgic rt ln nn vic truy cp b nh t bn trang s rt tn thi gian. gii quyt vn ny ta c th chia nh mi khon mc ca bng trn l ch ti bng di m hng ch ti trang 2.10: rnh by l do ph i i trang, v c c b c ti n hnh hi i trang.-V mi b nh c cp pht 1 s khung trang nht nh cha trang, Khi tin trnh c thc hin th tng trang ca trang tin trnh uc np vo b nh. Vic np trang vo b nh c th xy ra thng hp thiu trang v HDH phi tm khung trng thm trang thiu vo, nhng khi b nh h khung th HDH phi y 1

trang khi b nh theo 1 chin lc no np trang mi vo. Do dn ti vic i trang.Qu trnh i trang: 1: Xc nh trang cn np vo trn a 2: Nu c khung trng th chuyn sang 4 3: La chn 1 khung gii phng, theo 1 thut ton no Ghi ni dung khung b i ra a (nu cn), cp nht bng trang v bng khung 4: c trang cn np vo khung va gii phng; cp nht bng trang v bng khung 5: Thc hin tip tin trnh t im b dng trc khi i trang 2.11: rnh by thut i trang ti u v i trang vo tr c ra tr c. i trang ti u (OP ): hn trang s khng c dng ti trong khong thi gian lu nht i ho php gim ti thiu s kin thiu trang v do l ti u theo tiu chun ny HDH khng on trc c nhu cu s dng cc trang trong tng lai => khng p dng trong thc t m ch so snh vi cc chin lc khcVo tr c ra tr c (F FO): Trang no c np vo trc th b i ra trc n gin nht Trang b trao i l trang nm lu nht trong b nh -> khng cn ti na 2.12 : rnh by c c ph ng ph p c nh s l ng hung trang ti a c p cho mi ti n trnh v c nh phm vi c p ph t p ph t slg hung c nh : p cho tin trnh mt s lng c nh khung cha cc trang nh S lng c xc nh vo thi im to mi tin trnh v khng thay i trong qu trnh tin trnh tn ti p pht bng nhau: - c tin trnh c cp s khung ti a bng nhau -S lng c xc nh da vo kch thc MEM v mc a chng trnh mong mun p pht khng bng nhau: - c tin trnh c cp s khung ti a khc nhau - p s khung t l thun vi kch thc tin trnh - mc u tin p ph t slg hung thay i :S lng khung ti a cp cho mi tin trnh c th thay i trong qu trnh thc hin Vic thay i ph thuc vo tnh hnh thc hin ca tin trnh ho php s dng b nh hiu qu hn phng php c nh => n theo di v x l thng tin v tnh hnh s dng b nh ca tin trnh _ Phm vi c p ph t: p pht ton th: ho php tin trnh i trang mi vo bt c khung no (khng b kha), k c khung c cp pht cho tin trnh khc p pht cc bTrang ch c i vo khung ang c cp cho chnh tin trnh Phm vi cp pht c quan h mt thit vi s lng khung 2.13 : rnh by c c c u tr c d liu dng cho t chc bn trong c a th mc.+Danh sch: T chc th mc di dng danh sch cc khon mc Tm kim khon mc c thc hin bng cch duyt ln lt danh sch Thm file mi vo th mc: Duyt c th mc kim tra xem khon mcvi tn file nh vy c cha Khon mc mi c thm vo cui danh sch hoc 1 trong bng M file, xa file Tm kim trong danh sch chm ache th mc trong MEM+ y nh phn: Tng tc tm kim nh TDL c h tr sp xp H thng file NTFS ca WinNT ng bm (hash table): Dng hm bm tnh v tr ca khon mc trong th mc theo tn file Thi gian tm kim nhanh Hm bm ph thuc vo kch thc ca bng bm => kch thc bng c nh+T chc th mc ca DOS: Mi a logic c cy th mc ring, bt u t th mc gc ROOT Th mc gc c t phn u ca a, ngay sau sector khi ng OOT v bng FAT Th mc gc cha files v cc th mc con Th mc con c th cha files v cc th mc cp di na c t chc di dng bng: mi khon mc chim 1 dng trong bng v c kch thc c nh 32 bytes+T chc th mc ca Linux: Th mc h thng file Ext2 ca Linux c cch t chc n gin Khon mc cha tn file v a ch I-node Thng tin cn li v cc thuc tnh file v v tr cc khi d liu c lu trn I-node ch khng phi th mc Kch thc khon mc ph thuc vo di tn file Phn u ca khon mc c trng cho bit kch thc khon mc 2.15 : rnh by c c thao t c c b n v i file. Ph n tch r mt h thng file c nh t thi t ph i c thao t c m file hay hng.+To file: To file trng cha c data; c dnh 1 ch trong th mc +Xa file: Gii phng khng gian m d liu ca file chim Gii phng ch ca file trong th mc +M file: Thc hin trc khi ghi v c file c cc thuc tnh ca file vo MEM tng tc +ng file: Xa cc thng tin v file ra khi bng trong MEm+Ghi vo file +c file1 h thng k nht thit phi c thao tc m file v khi c, ghi file nu file cha c m th HDH s t ng tm kim file trong th mc, c cc thuc tnh ca file v xd v tr ca file 3.1 rnh by h i nim h i u hnh. Ph n tch r hai chc nng c b n c a h i u hnh.H iu hnh l h thng phn mm ng vai tr trung gian gia ngi s dng v phn cng my tnh nhm to mi trng gip thc hin chng trnh thun tin v hiu quHDH gm 2 chc nng chnh l qun l ti nguyn v qun l vic chy chng trnhQun l ti nguyn l m bo cho ti nguyn h thng s dng c ch v hiu qu.Ti nguyn phn cng bao gm PU,b nh chnh,b nh th cp,thit b vo ra.Qun l ti nguyn trc ht l phn phi ti nguyn ti cc ng dng hiu qu.yu cu c hdh thu nhn v p ng bng cch cp pht ti nguyn cho chng trnh tng ng, cp pht c ti nguyn hdh cn phi lu gi tnh trng ti nguyn bit c hin ti ti nguyn no cn trng,ti nguynno ang c sd.Qun l ti nguyn cn m bo cho chng trnh khng xm phm n ti nguyn cp cho chng trnh khcNhim v quan trng ca my tnh l chy chng trnh,1 chng trnh ang chy gi l tin trnh.HDH gip vic chy chng trnh d dng hn. mi trng thun li

cho chng trnh hdh to ra cc my o.My o l my logic vi nhng ti nguyn o c tnh cht v kh nng khc ti nguyn thc:d sd hn,d lp trnh hn,s lng nhiu hn,kh nng c th vt qua kh nng ti nguyn thc.ti nguyn o l bn m phng ti nguyn thc c thc hin bng phn mm.ti nguyn o cung cp cc dch v ging ti nguyn thc nhng n sd hn v s lng nhiu hn. s dng ti nguyn phn cng d dng ngi ta tru tng ha chng bng cch loi b nhng chi tit khng quan trng,ch gi li nhng kha cnh m ngi sd quan tm,sau khi tru tng ha ti nguyn phn cng vn cung cp chc nng nh ban u nhng d sd hn v nhng chi tit c th c du i. 3.2 Dch v c a h i u hnh l g ? rnh by nhng dch v in hnh m h i u hnh cung c p. m r v qu trnh t i v chy h i u hnh hi m i hi ngDch v hdh l nhng cng vic m hdh thc hin gip ngi dng hoc chng trnh ng dng.Nhng dv chnh m hdh cc:+Ti v chy chng trnh+Giao din vi ngi dng:gd dng lnh,gd ha+Thc hin vo ra d liu+Lm vic vi file+Pht hin v x l li:pht hin v s l kp thi nhng li xy ra vi phn cng cng nh phn mm m bo ch h thng hot ng n nh an ton+Truyn thng:cc dch v cho php thit lp lin lc v truyn thng tin+ p pht ti nguyn+Dch v an ninh v bo mt thc hin 1 chng trnh,chng trnh phi c ti t a vo b nh sau c trao quyn thc hin lnh,sau khi kt thc ctrinh phi gii phng b nh v ti nguyn m ctrinh chim gi,hdh s thc hin cc cng vic ny.Do hdh l ctrinh u tin c thc hin khi h thng khi ng nn hdh phi t ti chnh mnh vo b nh trong.vc ti din ra nh sau:hdh c 1 chng trnh nh gi l ctrinh ti hay ctrinh mi.ctrinhf ny nm 1 v tr xc nh trn a hoc b nh ngoi.sau khi h thng khi ng,1 chng trnh nm sn trong b nh ROM c kch hot v c chng trnh mi 1 v tr quy c trn a vo b nh trong.sau ctrinh mi s chu trch nhim ti nhng phn cn li ca hdh vo b nh v trao quyn k h thng cho hdh 3.3 : rnh by h i nim dng (thread) v m hnh a dng. V n s hu ti nguyn c a ti n trnh v dng. Ph n tch u im c a m hnh a dng.Dng : mi v thc hin ca tin trnh tc l 1 chui lnh c cp PU v thc hin c lp gi l dng thc hin.a dng l nhiu chui lnh c thc hin cng 1 lc trong phm vi 1 tin trnh.M hnh a dng :mi dng cn c kh nng qun l con tr lnh v ni dung thanh ghi.mi dong cng c trng thi ring c cha trong khi qun l lnh.mi dng c ngn xp ring.tt c cc cc dng chia s chung khng gian nh v ti nguyn ca tin trnh.cc dng c cng khng gian c c th truy cp d liu ca tin trnhTi nguyn ca tin trnh v dng :Trong h thng cho php a dng,tin trnh vn l n v c HDH phn phi ti nguyn,cc tin trnh s hu chung 1 s ti nguyn :khng gian nh ca tin trnh,v 1 s ti nguyn khc nh file ang m v thit b vo ra. u im ca m hnh a dng :tng hiu nng v tit kim thi gian,tng tnh p ng,d dng chia s ti nguyn v thng tin,tn dng c kin trc x l vi nhiu PU,thun li cho vc t chc ctrinh 3.4 : Ph n tch c c v n cn quan t m trong s dng v qu n l ti n trnh ng thi (concurrent processes) i v i ba dng ti n trnh : ti n trnh c lp c cnh tranh ti nguyn, ti n trnh h p t c nh chia s ti nguyn, v ti n trnh h p t c nh trao i thng ip c tin trnh cnh tranh ti nguyn ln nhau :+Loi tr tng h :khi cc tin trnh ng thi truy cp ti nguyn m kh nng cse ca ti nguyn hn chm bo khi tin trnh truy cp ti nguyn th tin trnh khc k c php truy cp.cc ti nguyn nh vy gi l ti nguyn nguy him,nhng don ctrinh yc sd ti nguyn nguy him gi l on nguy him+Khng xy ra b tc :b tc l tnh trng 2 hay nhu t.trnh k th tip tc thc hin do phi i ch ln nhau+Khng i ti nguyn :l tnh trng t.trinh phi ch i qu lu m cha n lt cp ti nguynTin trnh hp tc vs nhau thng qua ti nguyn chung : c t.trnh hp tc vs nhau trao i t.tin vs nhau thng qua cse vng nh dng chung.vc cc tin trnh truy cp ng thi d.liu dng chung i hi m bo tnh nht qun d.liu.vc cp nhp dlieu chung cung yc b loi tr tng h, ngoi ra c th xy ra tnh trng b tc v i.k chy ua l tnh hung 1 dng or tin trnh c v ghi dl v kt qu ph thuc vo th t cc thao tc vo ra.vn xy ra khi c dik chy ua c th gq bng cch t ton b thao tc vo on nguy him v sd loi tr tng h thao tc ny k b tin trnh khc xen ngangTin trnh c lin lc nh gi thng ip : c tin trnh hp tc vs nhau c th ll trc tip vi nhau bng cch gi thng ip.vc gi thng ip cc t.trnh k c.s hay cnh tranh ln nhau nn k yc loi tr tng h,nhng c th xy ra b tc v i 3.5 : rnh by gi i thut Peterson cho on nguy him. Ph n tch u nh c im c a ph ng ph p ny.Thut ton peterson ool flag[2] ;Int turn ;Void P0(){// tien trinh P0For( {//lap vo hanFlag[0]=true ;turn =1 ;while(flag[1]&&turn==1) ;//lap den khi dk khng tha mnFlag[0]=false ;}}Void P1(){// tien trinh P1For( {//lap vo hanFlag[1]=true ;turn =0 ;while(flag[0]&&turn==0) ;//lap den khi dk k tha mnFlag[1]=false ;}}Void main(){Flag[0]=flag[1]=0 ;Turn=0 ;//tt tin trnh chnh,chy ng thi P1 v P2Startprocess(P0) ;Startprocess(P1) ;}+ u im :Tha mn cc y/c :k loi tr tng h k tin trin :+P0 ch c th b P1 ngn cn vo on nguy him nu flag[1]=true

v turn =1 lun ng+ 2 kh nng vi P1 ngoi on nguy him :-P1 cha sn sng vo ong nguy him=>flag[1]=false, P0 c th vo ngay on nguy him- P1 t flag[1]=true v ang trong vng lp while => turn =1 hoc 0:Turn = 0 0 vo on nguy him ngayTurn =1: P1 vo on nguy him, sau t flag[1]=false=>quay li kn 1- h i gii hn+ Nhc im-sd trn thc t tng i phc tp-i hi tin trnh ang yu cu vo on nguy him phi nm trong trng thi ch i tch cc. h i tch cc; tin trnh vn phi sd PU ktra xem c th vo on nguy him?=>lng ph PU 3.6 : rnh by ph ng ph p ph t hin v l b tc bng c ch s dng th. Ph n tch u im c a ph ng ph p ny so v i ph ng ph p ngn nga b tc. Tr li :+ Pht hin b tc :TH mi dng ti nguyn ch c mt ti nguyn duy nht=> sd th bd quan h ch i ln nhau gia tin trnh-Xy dng th cp pht ti nguyn :+ c nt l tin trnh v ti nguyn +Ti nguyn dc ni vi tin trnh bng cung c hng nu ti nguyn c cp cho tin trnh .+Tin trnh dc ni vi ti nguyn trnh bng cung c hng nu tin trnh ang c cp cho ti nguyn .+ th ch i :-c xd t th cp pht ti nguyn = cch b i cc nt t/ vi ti nguyn v nhp cc cung i i qua nt b b- ho php pht hin tnh trng tin trnh ch i l iu kin sinh ra b tc-SD thut ton pht hin chu trnh trn th c hng pht hin b tc trn th ch iV s :+ Thi im pht hin b tc ; - tc ch c th xh sau khi 1 tin trnh no yu cu ti nguyn v ko c tha mn.-=> hy thut ton pht hin b tc mi khi c y/c cp pht ti nguyn ko c tm=> cho php pht hin b tc ngay khi va xy ra.- hy thng xuyn lm gim h/nang ht-=>Gim tn sut chy thut ton pht hin b tc :+Sau tng chu k t vi chc pht ti vi gi+Khi c mt s u hiu nh hiu sut sd PU gim xung di 1 ngng no + X l khi bi b tc :-Kt thc all tin trnh ang b b tc-Kt ln lt tng tin trnh ang b b tc n khi ht b tc+HDH phi chy li thut ton pht hin b tc sau khi kt 1 tin trnh+HDH c th chn th t kt tin trnh da trn tiu tr no .-Khi phc tin trnh v thi im trc khi b b tc sau cho cc tin trnh thc hin li t im ny :+i hi HDH lu tr trng thi c th thc hin quay lui v khi phc v cc im ktra trc .+Khi chy li, cc tin trnh c th li ri vo b tc tip.Ln lt thu hi li ti nguyn t cc tin trnh b tc cho ti khi ht b tc. 3.7 : rnh by thut t i trong qu trnh thc hin. rnh by thut lin t ng v th vin dng chung. Ph n tch r u im m tng ph ng ph p em li. Tr li :+ Ti trong qu trnh thc hin :-Hm cha b gi th cha ti vo b nh - hng trnh chnh dc load vo b nh v chy-Khi c li gi hm :+ hng trnh s ktra hm dc ti vo cha+Nu cha, chng trnh s tin hnh ti sau nh x a ch hm vo ko gian chung ca chng trnh v thay i bng a ch ghi li cc nh x . -Lp trnh vin m nhim, HDH cung cp cc hm th vin cho ti ng+ Lin kt ng v th vin dng chung-Lin kt tnh : cc hm v th vin c lk lun v chng trnh-Lng ph ko gian c trn a v MEM trong-Trong gd lk, ko kt ni cc ham th vin vo cha trnh m ch chn cc thng tin v hm th vin .- c modul th vin c lk trong qu trnh thc hin:+Ko gi bn sao cc modul th vin m tin trnh gi stub cha thng tin v modul th vin.+Khi stub dc gi, n ktra modul t/ c trong b nh cha. Nu cha, th ti phn cn li v v dng.+Ln tip theo cn sd, modul th vin s c chy trc tip +Mi modul th vin ch c 1 bn sao duy nht cha trong MEM- n h tr t HDH 3.8 : rnh by thut ph n ch ng ng b nh . Ph n tch u nh c im c a ph ng ph p ny so v i ph n ch ng c nh. Khi di chuyn ch ng sang v tr h c cn thay i thng tin g trong hi nh a ch.Tr Li :+ K thut :-Kch thc, s lng v v tr chng u c th thay i-Khi c y/c, HDH cp cho tin trnh 1 chng c kch thc ng bng tin trnh .-Khi tin trnh kt s to vng trng trong MEM- c vng trng nm cnh nhau c nhp li thnh vng ln hn-Sd cc chin lc cp chng :+ hn vunf thch hp u tin+Vng thich hp nht+Vng ko thch hp nht+ u im : Trnh phn mnh trong+ Nhc im : Gy phn mng ngoi : dn nhng vng trng nh thnh ln(nn)+ Nu chng b di chuyn th ni dung ca thanh ghi c s b thay i cha a ch v tr m 3.9 : rnh by thut ph n ch ng s dng ph ng ph p cn (buddy). Ph n tch r c c im ging/ h c nhau v u nh c im c a ph ng ph p ny so v i ph n ch ng c nh v ph n ch ng ng (l u hng cn trnh by li hai ph ng ph p sau).Tr li :+ Trnh by k thut :- c chng v khi trng c kch thc l ly tha ca 2^k(L trit gia cha kp ly a th trit gia khc ly mt => i ti nguyn u h i 4.8 : a) rnh by chi n l c i trang t s dng trong thi gian cui. b) rnh by chi n l c i trang s dng thut to n ng h. c) i s ti n trnh c c p 4 hung nh vt l, c c trang c a ti n trnh c truy cp theo th t sau : 1,2,3,4,5,3,4,1,6,7,8,7,8,9,7,8,9,5. y c nh th t np v i trang n u s dng hai thut to n ni trn. Gii : A, chin lc i trang t sd trong thi gian cui : - Trang b i l trang m tgian t ln truy cp cui cng n thi hin ti l lu nht. theo nguyn tc cc b y l trang t c kh nng s dng nht trong tg lai - ch xd trang c s ln truy cp cui : + sd bin m : Mi khon mc ca bng phn trang thm 1 trg cha tgian truy cp cui PU cng thm b m thi gian logic ny h s b m T khi c truy cp. Mi trang nh c truy cp th ch s ca b m s c ghi vo trg` tgian truy cp trong khon mc Trg` tgian lun cha thi gian truy cp trong ln cui Trang b i l trang c gi tr tgian nh nht + sd ngn xp 1 ngn xp c sd cha s trang Mi khi c 1 trang nh c truy cp, s trang s chuyn ln nh ngn xp y ngn xp l trang cn i, nh ngn xp l trang truy cp gn nht b, chin lc i trang dng thut ton ng h : - ci tin t thut ton FIFO - mi trang c gn thm bit s dng U. Khi trang c truy cp th U=1=> trang mi np vo U=1 - cc khung c th i c lin kt vo 1 ds vng - khi c nh cu i trong HDH ktra trang ang b tr ti Nu U=0 : trang b i ngay Nu U=1 : t U=0 v tr sang trang tip theo - nu tt c U=1 th con tr quay 1 vng i ht U=0 ri i trang hin thi ang b tr