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Bi n t p ph n cu h i l thuy t1. Process l g? Process khc chng
trnh im g? Cho bit cc cng vic chnh ca b phn Process Management. 2.
Cho bit cc cng vic chnh ca b phn Memory Management, file
management, secondary storage management? 3. Cho bit cc cng vic
chnh ca b phn I/O system management v ca H thng bo v? 4. System
call l g? hy a ra mt s v d v system call. System program l g v n
khc application program nh th no? lit k cc system program c bn. 5.
Kernel mode l g ? User mode l g ? Cho bit s khc nhau gia chng ?Cc
trng thi ca mt qu trnh? v s v gii thch s chuyn i gia cc trng
thi?Khoa KTMT V c Lung 1
cu h i l thuy t6. Ti sao cc h iu hnh hin i h tr mi trng a nhim ?
Phn bit multitask, multiprogramming v multiprocessing. 7. Khi nim
tin trnh c xy dng nhm mc ch g ? S khc bit, mi quan h gia tin trnh v
tiu trnh ? 8. Thng tin lu tr trong PCB v TCB ? T chc iu phi tin
trnh ? Phn tch u, khuyt ca cc chin lc iu phi 9. Cc c ch trao i thng
tin : tnh hung s dng, u, khuyt ?Cc yu cu ng b ho ? 10. Phn bit nhm
gii php busy waiting v Sleep&Wakeup. Phn bit cch s dng
semaphore, monitor.
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cu h i l thuy t11. Gii thch s khc bit gia a ch logic v a ch
physic?Gii thch s khc bit gia hin tng phn mnh ni vi v ngoi vi? 12.
Khi no th xy ra li trang ? M t x l ca h iu hnh khi c li trang. Gi s
c mt chui truy xut b nh c chiu di p vi n s hiu trang khc nhau xut
hin trong chui. Gi s h thng s dng m khung trang ( khi ng trng). Vi
mt thut ton thay th trang bt k : Cho bit s lng ti thiu cc li trang
xy ra ? Cho bit s lng ti a cc li trang xy ra ?
Cc d ng bi t pBi tp nh thi CPU vi cc thut tonTnh Response time,
Turnaround time, Waiting time.
Bi tp ng b ha cc tin trnh Chng minh c quyn truy xut, cc tnh cht
yu cu, ng b ha hot ng cc tin trnh s dng semaphore,..
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FCFS SchedulingV d : Th t vo
Non-Preemptive SJF SchedulingV d :Process Arrival TimeBurst Time
P1 0 7 P2 2 4 P3 4 1 P4 5 4Gantt Chart for Schedule P1 P3 7 8 P2 12
P4 16
Process P1 P2 P3
Burst Time 24 3 3
P1, P2, P3
Thi gian ch P1 = 0; P2 = 24; P3 = 27;
Thi gian ch trung bnh (0+24+27)/3 = 17
Gantt Chart for Schedule P1 0 P2 24 P3 27 30 0
Average waiting time = (0+6+3+7)/4 = 4
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Preemptive SJF Scheduling(SRTF)V d :
Round RobinTime Quantum = 20
Process Arrival TimeBurst Time P1 0 7 P2 2 4 P3 4 1 P4 5 4
Process Burst Time P1 53 P2 17 P3 68 P4 24Gantt Chart for
Schedule P1 P2 37 P3 57 P4 P1 77 P3 P4 P1 P3 P3 154 162
Gantt Chart for Schedule P1 0 2 P2 4 P3 P2 5 7 P4 11 P1 16 0
20
97 117 121 134
Average waiting time = (9+1+0+2)/4 = 3
turnaround time trung bnh ln hn SJF, nhng ap ng tot hnV c Lung 7
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Khoa KTMT
Bi t p 1: i u ph i nhi u c pMt hh iu phi tin trnh bng chin lc
"iu phi nhiu cp u tin xoay vng", dng 3 cp u tin: Cp 1 u tin cao
nht, s dng gii thut Round Robin vi quantum=2ms Cp 2 s dng gii thut
SJF khng c quyn Cp 3 s dng gii thut FIFO Mt tin trnh nu cp cao 4ms
s c chuyn xung cp tht hn, nu tin trnh ang cp cui cng th sau khong
thi gian ny s c chuyn ln cp u tin. Xt tp cc tin trnh sau (vi thi
gian yu cu CPU v u tin km theo): Tin trnh P1 P2 P3 P4 Thi im vo
RL(ms) 0 1 2 4 Thi gian CPU(ms) 5 4 3 2 u tin 3 1 1 3
Bi 2.Gi s b nh chnh c phn thnh cc phn vng c kch thc l 600K,
500K, 200K, 300K ( theo th t ), cho bit cc tin trnh c kch thc 212K,
417K, 112K v 426K ( theo th t ) s c cp pht b nh nh th no, nu s dng
: a) Thut ton First fit b) Thut ton Best fit c) Thut ton Worst fit
Thut ton no cho php s dng b nh hiu qa nht trong trng hp trn ?
Cho bit kt qu iu phi hot ng ca cc tin trnh trn v thi gian lu li
trong h thng (tnh t lc vo RL) v thi gian ch trong RL ca tng tin
trnhKhoa KTMT V c Lung 9 Khoa KTMT V c Lung 10
Bi t p qu n l b nhBi 03:Xt mt khng gian a ch c 8 trang, mi trang
c kch thc 1K. nh x vo b nh vt l c 32 khung trang. a) a ch logic gm
bao nhiu bit ? b) a ch physic gm bao nhiu bit ?
Bi t p qu n l b nhBi 05: Xt bng phn on sau y : Cho bit a ch vt l
tng ng vi cc a ch logique sau y : a. 0,430 b. b. 1,10 c. c. 2,500
d. d. 3,400 e. e. 4,112Segment 0 1 2 3 4 Base 219 2300 90 1327 1952
Length 600 14 100 580 96
Bi 04:Xt mt h thng s dng k thut phn trang, vi bng trang c lu tr
trong b nh chnh. a) Nu thi gian cho mt ln truy xut b nh bnh thng l
200nanoseconds, th mt bao nhiu thi gian cho mt thao tc truy xut b
nh trong h thng ny ? b) Nu s dng TLBs vi hit-ratio ( t l tm thy) l
75%, thi gian tm trong TLBs xem nh bng 0, tnh thi gian truy xut b
nh trong h thng ( effective memory reference time)Khoa KTMT V c
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Virtual memoryBi 06: Xt chui truy xut b nh sau: 1, 2 , 3 , 4 , 2
, 1 , 5 , 6 , 2 , 1 , 2 , 3 , 7 , 6 , 3 , 2 , 1 , 2 , 3 , 6 C bao
nhiu li trang xy ra khi s dng cc thut ton thay th sau y, gi s c 1,
2, 3, 4, 5, 6, 7 khung trang ? a) LRU b) FIFO c) Chin lc ti u
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