exam fehandbook so tay on tap sat hach fe

8/6/2019 Exam FEHandbook So Tay on Tap Sat Hach FE

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FE HANDBOOK

S TAY N LUYN FE

HIRATSUKA RyozoH Ni 7-2005

JETRO VITEC

T CHC XC TIN THNG MI

NHT BN

TRUNG TM ST HCH CNG NGH

THNG TIN V H TRO TO

http://www.ngohaianh.info

http://www.ngohaianh.com

http://www.ngohaianh.net


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MC LC

Li ni uPhn bui sng

Phn loi ni dung .. 4Cu hi v d .. 6p n v gii thch 56

Phn bui chiuPhn loi ni dung . 109Cu hi v d .. 110

p n v gii thch 119





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Ni dung St hch K sCng ngh Thng tin Cbn ( cu hi bui sng )

Lnh vc Mc lc

1-1-1 Chuyn i gi tr v biu din d liu1. L thuyt thng tin cbn

1-1-2 Thng tin v lgic

1-2-1 Cu trc d liu

1. CsKhoa hc

My tnh

2. Cu trc d liu v gii thut

1-2-2 Gii thut

2-1-1 Thit b thng tin

2-1-2 Kin trc b x l

2-1-3 Kin trc b nh

2-1-4 B nhph (ngoi)

2-1-5 Kin trc v thit b vo/ra

2-1-6 Kiu v tnh cht ca my tnh

1.Phn cng

2-1-7 H thng nhng

2-2-1 Cc hiu hnh2.Phn mm cbn

2-2-2 Qun l tp

2-3-1 Cng ngh cu hnh h thng

2-3-2 Hiu nng ca h thng

3. cu hnh h thng v lgic

kin thit

2-3-3 tin cy ca h thng v hiu qu chi ph

2. H thng my

tnh

4. Cc ng dng h thng 2-4-1 H thng a phng tin

3-1-1 Ngn ng

2-1-2 Gi phn mm

3-1-3 Mi trng pht trin

3-1-4 K thut pht trin

3-1-5 Cc kthut phn tch yu cu v thit k

3-1-6 Phng php lp trnh, kim th v xt duyt

3-1-7 Qun l pht trin

1.Pht trin h thng

3-1-8 S dng mi trng bn ngoi

3-2-1 Vn hnh h thng

3. Pht trin v vnhnh h thng

2.Vn hnh v bo tr h thng

3-2-2 Bo tr h thng

4-1-1 Cc giao thc v kim sot truyn tin

4-1-2 M ho v truyn tin

4-1-3 Mng (LAN v WAN)

4-1-4 Thit b truyn thng

4. Cng ngh

mng

1.Cng ngh mng

4-1-5 Phn mm mng

5-1-1 Cc m hnh CSDL

5-1-2 Ngn ng CSDL

5. Cng ngh

CSDL

1.Cng ngh CSDL

5-1-3 Qun tr CSDL





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CU HI





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Cu 1: 1-1-1 Chuyn i gi tr v biu din dliu

Khi cho s nh phn 11001011 dch chuyn s hc sang phi 2 bit, sau tip tc dch

chuyn lgic sang tri 2 bit, th kt qu ca php tnh l p n no di y?

A. 11001000 B. 11001111 C. 00111011 D. 00111100 E. 00111000


Kt qu ca php tnh 100 ca h thp phn (cs 10) tri 2C ca h thp lc phn (16)

v sau tr tip 70 ca h bt phn (cs 8) s l bao nhiu?

A: nh phn 10000000

B: t phn 21

C: bt phn 0

D: thp lc phn A

E: thp lc phn 11


Gi tr tng ng vi kt qu php tr di y l bao nhiu?

(1234)16 - (10101010101)2

A (1100 1101 0000)2

B (6006)8

C (2005)10

D (CDF)16

E (SOS)16


Gi tr biu din s 110 h thp phn trong h bt phn c 4 ch s l gi tr no di

y?

A 0156

B 0701

C 1010D 1100

E 1101


C 2 s biu th s m bng phn b 2 l 01011000 v 10010100. S thp phn th hin

tng ca 2 s ny l p n no? Tuy nhin, bit u s l bit du (0 l dng, 1 l m).

A -11

B +11

C -20

D +20





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E +21


Trong cc php tnh nh phn di y, php tnh no khng pht sinh nh(carry)

A 10001 + 110

B 1111 + 1000

C 1101 + 11

D 1010 + 110


Khi hon chuyn gi tr biu th bng s du phy ng 32-bit (theo xut ca IEEE)

di y sang dng biu th vi lu tha cs 2, th s thu c p n no?.

0 10001010 10100000000000000000000

A -(1.625)10 x 211

B +(1.625)10 x 211C -(1.101)2 x 2

12

D +(1.101)2 x 212

E +(1.101)10 x 212


biu din mt con s no di dng s th p phn khoanh vng (zone decimal

notation), th cn 6 byte. Vy biu din con s di dng s thp phn ng gi

(packed decimal notation), cn bao nhiu byte?A . 1

B . 2

C . 3

D . 4

E . 5

Cu 9: 1-1-2 Thng tin v lgic

Biu Ven cho bit mch NOR tng ng vi hnh v no ?





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Cu 10: 1-1-2 Thng tin v lgic

Trong s biu th s thay i trng thi nhim v (Task State Transition), vic gn

(assignment) quyn s dng CPU cho nhim v, s tng ng l cng on no t A

n E di y?

Cu 11: 1-2-1 Cu trc dliu

p n no di y l cu trc d liu vo trc ra trc (Fist In First Out).

A. Binary Tree (cy nh phn)B. Multiway Tree (cy a hng)C. Array (mng)D. Queue (hng i)E. Stack (ngn xp)

A B A B

A B

A

B

A B

A B

DC

E

Kt thc nhim

Pht sinh nhim

Trng thi chy

Trng thi sn sng Trng thi

A

BC

D

Ngt SVC

E





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Phn t b xo khi danh sch nh phn t c trong s di y gi l g?

Thay i v tr con tr

A. Fragmentation (phn mnh)B. Defragmentation (kh phn mnh)C. Synonym (ng ngha)D. Dead lock (Kha cht)E. Gabbage (thng rc)


Cho 5 s 5,6,7,8,9 vo ngn xp theo th t nh vy, ch sc ly ra th 4 l s no?

A. 5B. 6C. 7D. 8E. 9Cu 14: 1-2-1 Cu trc dliu

Gn cc ch s t 1~9 vo cy nh phn di y. Ch s nhp vo nt C l s bao nhiu?

Vi iu kin phn t nhnh bn tri < phn t trong nt < phn t nhnh bn phi.

A. 5 B. 6 C. 7 D. 8 E. 9

a b X eXc d





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Nu b sung phn t 4 vo v tr nh du * trong ng (heap), th phn tv tr A l

bao nhiu

A. 2 B. 4 C. 6 D. 9 E.12


Biu thc s hc thu c khi v cy nh phn hon chnh (perfect binary tree) theo trnh

t trung gian l p n no?

A. A+ B x C D E FB. (A + B x C) (D - (E - F))

C. (A + B) x C (D - E) F

D. (A + B) x C D - (E - F)

E. A + (B x C) (D - E) F

Cu 17: 1-2-2 Gii thut

Tin hnh sp xp trong my tnh bng thut ton sp xp ni bt (bubble sort), th 1000

d liu mt 1 giy. Nu sp xp 5000 d liu tng t th phi mt bao nhiu lu?

A. 1.5 giyB. 5 giy

2

6 9

13 30

12 19 1714





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C. 10 giyD. 25 giyE. 125 giyCu 18: 1-2-2 Gii thut

Cn sp x p cc d liu di y theo th t tng dn. Trong phng php Basic

Exchange Method (bubble sort), vic thc hin chuyn i v tr ca d liu pht sinh bao

nhiu ln?

Ch s di 1 2 3 4 5

D liu 15 5 11 12 7

A. 5 lnB. 6 lnC. 7 lnD. 8 lnE. 9 lnCu 19: 1-2-2 Gii thut

Khi x l bng my tnh, c mt s thao tc x l rt hay c thc hin. Trong cc thao

tc x l ny, s to ra nhiu thut ton (algorithm), vy p n no di y l thut ton

tm ra ng dn ngn nht bng dng biu ?

A. Quick Sort (sp xp nhanh)B. Binary Search (tm kim nh phn)C. Phng php Boyer - MooreD. Phng php thut ton GreedyE. Phng php tm kim DijkstraCu 20: 1-2-2 Gii thut

Khi tm kim bng cha 5000 d liu bng php tm kim nh phn (binary search

method), s ln so snh bnh qun s l bao nhiu?

Log102 = 0.3010A. 8B. 9C. 10D. 11E. 12Cu 21: 1-2-2 Gii thut

Trong php bm (Hash method), vic thu c gi tr bm ging nhau t cc kha khc

nhau c gi l hin tng xung t (synonym). Phng php trnh hin tng trn

l g?





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XY Z

A. Linear Programming (lp trnh tuyn tnh)B. Chain Method (phng php chui)C. Linear Search Method (phng php tm kim tuyn tnh)D. Invariant assertion method (Phng php xc nhn bt bin )E. Dynamic programming (Phng php lp trnh ng)Cu 22: 1-2-2 Gii thut

Hm bm tng ng ca s c 3 ch s ( a1 a2 a3 ) l mod (a1 + a2 + a3 , 5 ). Vi cc t

hp s di y, gi tr bm b xung t l trng hp no?

A. 881 v 509B. 913 v 426C. 731 v 429D. 102 v 297E. 677 v 388Cu 23: 1-2-2 Gii thut

nh ngha hm quy (recursive function) F(n) nh sau:

Nu n > 0, th F(n) = n x F(n - 1)

Nu n = 0, th F(n) = 1

Khi , F(5) s bng bao nhiu?

A. 120B. 15C. 1D. 5E. 24Cu 24: 2-1-1 Thit b thng tin

K hiu mch s hc AND (AND arithmetic circuit) tng ng vi hnh v no di y?

A. B.

C.

D.

E

XY

ZCHA

X

YZ

X

YZ

S

ZFA





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Cu 25: 2-1-1 Cc hiu hnh

Con chip MPU (CPU) u tin trn th gii do Shima Masatoshi v Ted Hoff lm vic ti

mt doanh nghip M pht minh ra vo nm 1971. Con chip u tin c s dng

tnh ton trong Accumulator c dung lng 4 bit. Sau , doanh nghip tung ra con chip

MPU c dung lng 32 bit u tin vo nm 1988 l doanh nghip no?

A. Texas InstrumentB. IntelC. MotorolaD. AMDE. IBMCu 26: 2-1-2 Kin trc b xl

nh dng lnh sau y ln COMET, my tnh ton o calculation virtual Ni dungti cc thanh ghi trc khi thc hin cc dng lnh s nh sau:

Lnh: LD GRO, GR1

Ni dung ca GRO l 2; ni dung ca GR1 l 5

Ni dung ca GR2 l 8

Con s ghi ti GRO sau khi thc hin dng lnh l p n no?

A: 2 B: 5 C: 7 D: 8 E: 9

Cu 27: 2-1-2 Kin trc b xlThanh ghi l vng lu tm thi nhng d liu trong b x l. Trong cc thanh ghi di

y, ci no s duy tr tnh trng kt qu ca php ton?

A. Thanh ghi a nng (General-purpose Register)B. Thanh ghi yu cu ngt (Interruption Request Register)C. Thanh ghi s hc (Arithmetic Register)D. Thanh ghi ch mc (Index Register)E. Thanh ghi c(Flag Register)Cu 28: 2-1-2 Kin trc b xl

K thut tng tc chy ca my tnh th hin s di y tng ng vi p n

no?

Tm np

(fetch)

Gii m

(decode)

Thc hin

(operation)

Tm n p Gii m Thc hin







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Thc hin lnh th 2

Thc hin lnh th 1

A. ng ng lnhB. MIPSC. Tng nhp ng h (clock-up)D. CISCE. CPU a nngCu 29: 2-1-2 Kin trc b xl

Thi gian x l trong trng hp chy mt chng trnh gm 10 triu dng lnh trn my

tnh 30 MIPS l bao nhiu lu? Vi t l s dng b x l l 85%, khng tnh n ph ti

(overhead) ca OS. S thp phn s lm trn ch s th 3.

A. 0.04B. 0.06C. 0.39D. 0.63E. 0.93Cu 30: 2-1-2 Kin trc b xl

Khi lp thm nhng linh kin di y tng tnh nng x l khi s dng chng trnh

bng tnh (spreadsheet program), linh kin no khng c lin quan n vic tng tnh

nng?A. SIMMB. DIMMC. Over drive processorD. Cc hnh (video card)E. Cc m thanh (sound card)Cu 31: 2-1-2 Kin trc b xl

Ngt c ngha l vic CPU tm dng mt chng trnh ang chy, chy mt chngtrnh khc. Trong s cc lnh ngt sau y, lnh no pht sinh do thao tc ca ngi qun

l h thng?

A.Ngt chng trnhB.Ngt ng hC.Ngt vo-raD.Ngt SVCE. Ngt bn giao tip gia ngi v my (console)Cu 32: 2-1-2 Kin trc b xl

Trong s cc k thut tng tc my tnh di y, ci no cn s dng chip RISC. Hy





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chn hai p n.

A. Cache memory (B nhm n)B. VLIWC. ng ng lnhD. Dch ti uE. Super scalar architecture (Kin trc siu v hng)Cu 33: 2-1-3 Kin trc b nh

B (byte) l n v th hin ln v d liu x l trong my tnh, 1B = 8bit. Vy, 0.5GB

bng khong bao nhiu bit?

A 3,200,000 bit

B 50,000,000 bit

C 3,200,000,000 bit

D 4,000,000,000 bitE 5,000,000,000 bit

Cu 34: 2-1-3 Kin trc b nh

Vi nhng iu kin di y, thi gian bnh qun truy cp vo b nhca CPU l bao

nhiu lu?

T l truy cp b nhm n (cache memory) = 0.7

Thi gian truy cp vo b nhm n Tc = 10ns

Thi gian truy cp vo b nhchnh Tm = 50nsA. 22nsB. 34nsC. 40nsD. 46nsE. 70nsCu 35: 2-1-3 Kin trc b nh

B nht nm gia b x l v b nhchnh, c tc dng gim bt s chnh lch v tc gia hai thit b trn l thit b no di y?

A. Thanh ghiB. B nhm nC. B nhngoiD. Bm aE. Knh ni b (internal bus)Cu 36: 2-1-3 Kin trc b nh

Cc b nhbn dn, ty theo mc ch s dng v c tnh, c chia lm mt s loi

khc nhau. Loi thit b m ngay c khi b ngt in ngun, d liu vn khng b mt v





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c th ghi chng ln bao nhiu ln cng c, l thit b no di y.

A. Mask ROMB. SRAMC. DRAMD. Flash memoryE. SIMMCu 37: 2-1-3 Kin trc b nh

SRAM (Static RAM) l mt loi RAM, c tc truy cp rt nhanh, nu khng b ngt

in ngun, d liu s khng bao gib mt. SRAM c to ra bng cch s dng loi

mch no di y?

A. Mch ANDB. Mch lt (flip-flop)C. B cng y (full adder)D. B cng bn phn (half adder)E. Mch NOTCu 38: 2-1-3 Kin trc b nh

Thut ng th hin thao tc di y trong b nhch l g?

Chng trnh iu khin Chng trnh iu khin

Chng trnh A Chng trnh AVng khng s dng Chng trnh B

Chng trnh B

Vng khng s dng

Vng khng s dng Vng khng s dng

A. Lm gn b nh(memory compaction)B. Kh phn mnh (defragmentation)C. Gch cho (Thrashing)D. Tro i (swapping)E. Phn mnh (Fragmentation)Cu 39: 2-1-3 Kin trc b nh

Trong h thng lu tro, p n no di y l thut ton nh s trang hiu qu nht?

A. Phng php FIFOB. Phng php LIFOC. Phng php LRUD. Giao dch a chng (Dynamic Address Transaction)





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E. Thay trang (Page Replacement)

Cu 40: 2-1-4 B nhph (ngoi)

Trong a t c c im di y, mt rnh c th lu c bao nhiu bn ghi?

c im mt n va t (Magnetic Disk Unit Specification):

Dung lng lu tr tng ng vi mt rnh ghi 28000 bytes

Khong trng gia cc khi (Inter-Block Gap (IBG)) 500 bytes

iu kin ca d liu lu tr

di bn ghi 400 bytes

H s khi (block factor) 10

A. 40 B. 60 C. 80 D. 100 E. 120

Cu 41: 2-1-4 B nhph (ngoi)Trong a t c c im nh di y, thi gian truyn d liu (data transfer time) ca

mt khi l bao nhiu lu?

Dung lng lu tr trn mt rnh ghi (bytes) 20000

Tc quay (vng/pht) 3200

di bn ghi (bytes) 1000

H s khi (block factor) 8

A. 18,75 mili giy B. 15 mili giy C. 12,5 mili giy

D. 10 mili giy E.7,5 mili giy


Thi gian tm kim trung bnh (average search time) ca a t c c im di y l

bao nhiu lu?

Tc quay Thi gian trung bnh nh vu t vo

rnh (average seek time)

3000 vng/1 pht 8 mili giy

A. 5,0 mili giy B. 6,3 mili giy C. 10,0 mili giyD. 13,3 mili giy E. 20,0 mili giy

Cu 43: 2-1-4 B nhph ngoi

Dung lng lu gi ca a mm c c im di y l p n no?

- S mt lu gi 2

- S rnh trn mt mt 80

- S sector trn mt rnh 18

- Dung lng lu gi ca mt sector (byte) 1024A. 640kB B. 1,2MB C.1,4MB D. 1,8MB E. 2,88MB





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Trong s biu din tc bng t di y, phn biu th vic tng gim tc tng

ng l phn no trong a t?

A. Bn ghi vt l B. EOT C. Bn ghi lgc D. BOT E. IBG


lu tp di y vo mt a t vi mt lu tr l 4800BPI, di IBG l 0,25

inch, th a t cn c di l bao nhiu?

- S khi 1000 khi

- di bn ghi vt l 500 byte- H s khi 12

A. 1250 inch B. 1500 inch C. 1750 inch D. 2250 inch E. 2500 inch


tng thng lng (throughput), phng php truyn d liu gia b nhch vi thit

b xut nhp tc thp thng qua b nhph l p n no?

A. POLLingB.

BLOKingC. SWAPPing

D. SPOOLingE. QUEUingCu 47: 2-1-5 Kin trc v thit b vo / ra

Trong s cc thit b cu to nn my tnh di y, thit b kt hp gia biu khin

(controller) v b s hc (arithmetic unit) l thit b no di y?

A. B nh B. Thit b ra C. B x l D. Thit b vo E. Thit b dchCu 48: 2-1-5 Kin trc v thit b vo / ra

Dng

Gim

tc

Gim

tc

Tng

tc

Tc

Tng

tc

T c khngi

T c khngi

T c khngi

Thi gian





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Lu di y cho bit mi lin quan gia cc thit b cu to nn my tnh. Thit b

hp A tng ng vi thit b no di y?

A. B nhB. B x lC. Thit b raD. B s hcE. Biu khinCu 49: 2-1-5 Kin trc v thit b vo / ra

Thao tc x l nh p v xut trong my tnh a nng, c thc hin thng qua knh

(channel). Knh l b x l chuyn nhp v xut, hot ng song song vi CPU. pn no di y th hin vng lu tr nhng ch th do b x l a ra i vi knh?

A. CSWB. CAWC. CAID. CCWE. CAM

Cu 50: 2-1-5 Kin trc v thit b vo / raThit b vo s dng c thng tin c nh du ghi trong bi thi hay phiu iu tra,

l thit b no di y.

A. OMRB. OCRC. Bng (tablet)D. My qut nh (image scanner)E. My nh s (digital camera)Cu 51: 2-1-5 Kin trc v thit b vo / ra

Trong cc thit b vo s dng nhp d liu t bn ngoi vo my tnh, thit b dng

lung d liu lung iu khin

A





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nhp thng tin bng cch chm tay ln mn hnh hin th, mt kiu rt hay c s dng

ti cc my rt tin tng (ATM) ca cc ngn hng l loi no di y?

A. Tivi mn hnh phngB. Qu cu nh du (track ball)C. B s ha (digitizer)D. Mn hnh (Xc gic) (touch screen)E. Bt quang (light pen)Cu 52: 2-1-5 Kin trc v thit b vo / ra

My tnh v cc thit b ngoi c kt ni vi nhau thng qua giao din vo-ra (IO

interface). Trong s cc giao din vo-ra di y, tiu chun dng cho my o c c th

kt ni chui cnh hoa (daisy-chain) l p n no di y?

A. IDEB. CentronicsC. IOBASE-TD. SCSIE. GPIBCu 53: 2-1-5 Kin trc v thit b vo / ra

Thit b mn hnh khng thch hp lm mi trng phn cng s dng GUI (Graphical

User Interface) l thit b no di y?

A. Mn hnh CRTB. Mn hnh tinh th lng LCDC. Mn hnh PlasmaD. Mn hnh k t (character display)E. Mn hnh Multi-scanCu 54: 2-1-5 Kin trc v thit b vo / ra

Lng thng tin trn mt mn hnh hin th di dng High Color (16 bit) trn mn hnh

VGA 640 x 480 l bao nhiu (1 KB = 1024 byte)A. Khong 600 KBB. Khong 900 KBC. Khong 1500 KBD. Khong 2100 KBE. Khong 3200 KB





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Cu 59: 2-2-1 Cc hiu hnh

Mt trong nhng mc ch chnh ca OS l nhm tng khi lng cng vic c th x l

trong mt n v thi gian. Do , p n no di y th hin khi lng cng vic?

A. Thi gian hot ng ca Hiu hnh trong qu trnh x l (Overhead)B. Thi gian p ng (response time)C. Thi gian quay vng (turnaround time)D. Thi gian ri ca my tnh (idling time)E. Thng lng (throughput)Cu 60: 2-2-1 Cc hiu hnh

Khi khi ng my, hiu hnh s lu rt nhiu loi thng tin nh thi gian bt u

cng vic, thi gian kt thc cng vic, thng bo a ra Nhng thng tin ny c gi

l g? Hy chn 2 p nA. Supervisor callB. S nht k (journal)C. Bn ghi (log)D. PSWE. Cun d liu (spool)Cu 61: 2-2-1 Cc hiu hnh

Trong s cc hiu hnh (OS) l phn mm qun l ti nguyn ca my tnh, lm cho nhot ng hiu qu hn, th loi no di y khng c s dng cho my tnh c nhn?

A. Windows XPB. LinuxC. MVSD. MS-DOSE. OS/2

Cu 62: 2-2-1 Cc hiu hnhNgi ta gi cng vic x l mt ln do ngi dng thc hin l job. Ngi dng s s

dng ci g nhp job?

A. Job ManagementB. Job SchedulingC. Job StepD. Job Control LanguageE. JISCu 63: 2-2-1 Cc hiu hnh

Trong hiu hnh (OS), i vi h thng qun l tp tin bng th mc c cu trc phn





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tng, p n no di y cho bit v tr s dng hin ti?

A. Knh (bus)B. Th mc hin thiC. Th mc conD. Th mc gcE. Cy th mcCu 64: 2-2-1 Cc hiu hnh

p n no di y l chng trnh la chn nhim v cn thc hin theo th tu tin,

trong tnh trng c nhiu nhim v cng mt lc?

A. Task control block - Khi kim sot tc vB. Dispatcher - B gi phtC. Round Robin - Quay vngD. Time Slice - Lt thi gianE. Task State - Trng thi tc vCu 65: 2-2-2 Qun l tp

C nhiu cch phn loi tp nh da vo cch dng, cch t chc hay tnh cht Vy tp

no di y l tp c xy dng nhm lu mt cch tm thi cc kt qu trung gian?

A. Tp tm thi (temporary file)B. Tp nn (compressed file)C. Tp chia s (sharing file)D. Tp thng xuyn (permanent file)E. Tp h thng (system file)Cu 66: 2-2-2 Qun l tp

Trong my tnh a nng, ngi ta gi n v lgc to tp tin l g? Trong n v ny c

4 loi l di cnh, di bin i, di khng xc nh, bn ghi mrng (spanned

record).

A.

Trng d liuB. BaC. Bn ghi vt lD. Bn ghi lgcE. KhiCu 67: 2-2-2 Qun l tp

Trong cc hnh thc lu gi tp dng nh di y, nu lu thnh nh tnh, th dung lng

tp ln nht s l p n no?

A. JPEGB. BMP





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C. PNGD. GIFE. MPEGCu 68: 2-2-2 Qun l tp

C nhiu phng php t chc tp (file organization method), mi phng php c mt

c trng ring. Trong s cc phng php t chc tp di y, phng php no c th

s dng cho c trng hp bng t?

A. Phng php truy nhp tun t (sequential access)B. Phng php t chc phn hoch (partional organization)C. Phng php truy nhp trc tip (direct access)D. Phng php t chc theo ch mc (indexed organization)E. Phng php truy nhp lu gio (virtual storage access)Cu 69: 2-2-2 Qun l tp

Tp c cu thnh t mt s vng. p n no di y khng phi l vng cu thnh

nn mt tp t chc theo ch mc? Hy chn hai p n.

A. Th mc (directory)B. Vng d liu chnh (prime data area)C. Vng ch mc (index area)D. Vng trn (overflow area)E. Vng thnh vin (member area)Cu 70: 2-2-2 Qun l tp

T p VSAM (Virtual storage access method - Phng php truy nhp lu gi o) l

phng php t chc tp bao gm cc hnh thc nh truy nhp tun t, truy nhp trc tip

hay t chc theo ch mc p n no di y trnh by phng php t chc tp ny?

A. Cu thnh t vng thnh vin (member area) v th mc (directory)B. Cu thnh t vng d liu chnh, vng ch mc v vng trnC.

Truy nhp trc tip vo bn ghi thng qua a ch ca bn ghiD. C 4 loi l ESDS, RRDS, KSDS, LDS.

E. Truy nhp theo th tc ghi vo theo kiu vt lCu 71: 2-2-2 Qun l tp

Nu nhiu tin trnh cng truy nhp vo mt ngun, th kt qu thu c thng b sai

lch. Chc nng trnh tnh trng ny gi l chc nng kim sot loi tr (exclusive

control). N cn c gi l Semaphore (bin bo hiu). Tuy nhin, do c chc nng ny

m s dn n trng hp c t 2 tin trnh tr ln s cng i i tc tr ngun, khin

khng th tip tc thc hin php tnh c. Hin tng ny gi l g?

A. Partition (Phn hoch)





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B. Swapping (Tro i)C. Deadlock (Cht tc)D. Overlay (Trn)E. Paging (Phn trang)Cu 72: 2-3-1 Cng ngh cu hnh h thng

V phng php x l theo l bng cch tch lu d liu ri x l mt ln, hnh thc x l

minh ho trong s di y gi l g?

A. Phng php cafeteriaB. theo l t xa (remote batch)C. l m(open batch)D. l trung tm (center batch)E. l ng (closed batch)

Cu 73: 2-3-1 Cng ngh cu hnh h thng

H thng x l giao dch trc tuyn (OLTP) c s dng trong nghip v no? Hy chn

p n thch hp.

A. H thng rt tin gi ti ngn hngB. Qun l vn hnh tuC. X l ng thi cc qu trnh t nhiu ngi dng n CPUD. Tch lu d liu ri x l mt lnE. Tch lu d liu ca ngi dng cui, gi n my tnh qua mng, ri x l mt lnCu 74: 2-3-1 Cng ngh cu hnh h thng

Ngi ta gi k thut lp nhiu CPU nng cao tnh nng, l k thut a b x l (multi

processor). Ba x l s dng trong siu my tnh tnh ton khoa hc l p n no?

A. Tandem Connection ProcessorB. Array ComputerC. Tightly Coupled MultiprocessorD. Loosely Coupled MultiprocessorE. Duplex Processor System

Yu cu thao

tcThao tc my

Ng.dng

Thao tc vin My tnh





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Cu 75: 2-3-2 Hiu nng ca h thng

Trong phng php x l theo l, thi gian k t lc yu cu my tnh thc hin mt lot

cng vic, cho ti khi nhn c kt qu x l gi l g?

A. Thi gian p ng (response time)B. Thng lng (Throughput)C. Thi gian quay vng (Turn around time)D. Overhead (Ph ti)E. Overhead time (Thi gian ph ti)Cu 76: 2-3-2 Hiu nng ca h thng:

ng cong biu th s bin ng ca t l s c gi l g?

A. ng cong Gomperz B. Biu hnh ch Z C. S backmanD. ng cong logistic E. ng cong bn tm (bathtub)Cu 77: 2-3-2 Hiu nng ca h thng

Trong l thuyt hng i, phng php k hiu s hc dng m t m hnh hng i l

p n no?

A. Phng php k hiu Kendall (Kendalls Notation)B. Phng php k hiu Ba Lan (Polish Notation)C. Phng php k hiu Ba Lan o ngc (Reverse Polish Notation)D.

Phng php k hiu nh phn (Binary Notation)E. Dng thc Backus Naur (Backus Naur Form)

Cu 78: 2-3-3 tin cy ca h thng v hiu qu chi ph

Trong cc biu di y, biu no biu th ng cong gia tng tin cy

(reliability growth curve).

Thi gian

S c do xung cp

S c ln uS c xy ra ngu nhin

T l s c





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Khi h thng b trc trc, vic dng h thng li m bo an ton, trnh tnh trng b

tn tht ln hn nu tip tc x l, c gi l g?

A. File Protection (Bo v tp)B. Fault Tolerance (Kh nng chu ng sai st)C. Fault Avoidance (Trnh hng)D. Fail Soft (Hng mm)E. Fail Safe (Hng an ton)Cu 80: 2-3-3 tin cy ca h thng v hiu qu chi ph

T sut vn hnh (operating ratio) l n vnh gi kh nng vn hnh. Nu MTBF (thi

gian trung bnh gia cc ln trc trc) ca mt h thng l 190 gi, v MTTR (thi gian

trung bnh sa cha) l 10 gi, th t sut vn hnh ca h thng ny l bao nhiu?

A. 5% B. 9,5% C. 19%

D. 20% E. 95%


Bn h thng con (sub system) W,X,Y,Z c ni vi nhau nh s di y. Nu t

sut vn hnh ca cc h thng ln lt l: W=0,7 X=0,8 Y=0,8 X=0,7 th t sut vn

hnh ca ton h thng l bao nhiu? Lm trn s thp phn, ly n 3 ch s.

A. 0,41B. 0,68C. 0,75D. 0,95E. 0,96

A

C

E

D

B

W

Z

YX





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H thng N trong M c ngha l mt h thng mc song song, trong nu trong s M h

thng con, c N h thng hot ng tt th c h thng s hot ng c. Xc xut hot

ng ca h thng 3 trong 4 l p n no? Lm trn n 2 ch s.

A. 13%B. 25%C. 31%D. 38%E. 44%Cu 83: 2-4-1 H thng a phng tin

Thao tc son tho cc k t, hnh nh, m thnh thnh ni dung a phng tin (multi

media content) l p n no?

A. Trnh by (presentation)B. ho (graphic)C. Hot nh (animation)D. Tc gi (authoring)E. Duyt (browsing)Cu 84: 2-4-1 H thng a phng tin

Lng d liu nh m thanh, hnh nh, nh ng s dng trong h thng a phng tin

l rt ln. Do vy, k thut nn l rt quan trng. Phng php nn nh tnh mu a nngdo ISO tiu chun ho l p n no di y?

A. GIFB. JPEGC. MMRD. MPEGE. Run Length

Cu 85: 3-1-1 Ngn ngNgn ng bc cao c tnh c lp so vi phn cng rt cao, c th m t mt cch n

gin bng mt th ngn ng gn ging vi ngn ng t nhin ca con ngi. Trong s

cc ngn ng sau y, ngn ng no khng phi l ngn ng bc cao? Hy chn 2 p n.

A. COBOLB. Assembly LanguageC. FortranD. PrologE. Machine LanguageCu 86: 3-1-1 Ngn ng





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Ty vo phng php lp trnh, c th chia ngn ng phi th tc thnh nhiu loi nh.

Trong s cc ngn ng di y, loi no bao gm LISP?

A. Object Oriented LanguageB. Functional Program LanguageC. Machine Oriented LanguageD. Tabular Form LanguageE. Logical Program LanguageCu 87: 3-1-1 Ngn ng

Chng trnh dng chuyn chng trnh ngun c vit bng ngn ng bc cao thnh

ngn ng my hay ngn ng trung gian l p n no di y?

A. Chng trnh dch hp ng (Assembler)B. Chng trnh bin dch (Compiler)C. Chng trnh thng dch (Interpreter)D. GeneraterE. TranslaterCu 88: 3-1-1 Ngn ng

Tiu chun k t in t lm c s cho HTML cho CERN ca Thu S pht minh ra,

chuyn dng m t trang web internet l g?

A. UNCB. URLC. HDLCD. ARCE. SGMLCu 89: 3-1-1 Ngn ng

Trong cc th (tag) dng m t HTML di y, ci no biu th s xung dng?

A. B.

C.

D. E.

Cu 90: 3-1-1 Ngn ng

Trong cc loi cu to chng trnh, cu to trong mc d cho chy mt chng trnh

ng thi vi nhiu quy trnh khc nhau, nhng mi quy trnh u cho ra mt kt qu

ng, l cu to no?

A. ng dng (reentrant)B. quy (recursive)





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C. Ti nh v (relocatable)D. Ti s dng (reusable)E. Ti s dng ni tip (serial reusable)Cu 91: 3-1-2 Gi phn mm

Trong chng trnh bng tnh, c th thit lp dng thc cho cc trong tng trang tnh

(sheet). Trong cc mc di y, ci no khng c coi l dng thc?

A. Cng thc tnh tonB. Phng ch (font)C. Kiu hin thD. ng vinE. Bo v kho (lock protection)

Cu 92: 3-1-2 Gi phn mmTrong bng c to bng chng trnh bng tnh, cng thc tnh =SUM ($A1:A3) ghi

ti A4 v hin th nh di y. Khi copy A4 sang C4 th C4 hin th thnh 45. Lc

ny, nhp vo B4 cng thc =AVERAGE ($A1:A3). Nu copy B4 sang C4 th

C4 s hin th s bao nhiu.

A B C

1 1 4 7

2 2 5 8

3 3 6 94 6

A. 3.5 B. 5 C. 6.5 D. 21 E. 45Cu 93: 3-1-2 Gi phn mm

Trong bng c to bng chng trnh bng tnh, C1 l cng thc =SUM(A1:$B$2).

Khi copy C1 C2, th C2 hin th kt qu 7. Nu copy C1 sang C3, th C3 hin th

kt qu l bao nhiu?

A B C

1 1 2 10

2 3 4

3 5 6

4 7 8

A. 11 B. 18 C. 26 D. 33 E. 36





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Cu 94: 3-1-3 Mi trng pht trin

Cng c CASE Computer Aided Software Engineering l cng c h trvic pht trin

phn mm v bo tr phn mm. Trong cc cng c CASE di y, cng c no h tr

vic vn hnh sau pht trin h thng.

A. Cng c CASE trnB. Cng c CASE diC. Cng c CASE tch hpD. Cng c CASE bo trE. Cng c CASE chungCu 95: 3-1-3 Mi trng pht trin

Trong giai on kim th chng trnh, p n no di y khng phi l cng c h tr

kim th nh h tr g ri (debug), chn on chng trnh (cng c g ri, h trg

ri)?A. InspectorB. Dump routineC. Theo di (tracer)D. Trnh to d liu kim th (test data generator)E. Trnh dch v (service program)Cu 96: 3-1-4 K thut pht trin

Trong lnh vc pht trin h thng, c mt s m hnh ty theo cch tin hnh thao tcpht trin. p n no di y l m hnh trong pht trin bng cch lp i lp li cc

quy trnh pht trin?

A. Wire Frame ModelB. Waterfall ModelC. Prototype ModelD. Spiral ModelE. Client Server ModelCu 97: 3-1-4 K thut pht trin

Trnh t thao tc theo m hnh Waterfall l p n no?

A. K hoch cs thit k bn ngoi thit k chng trnh thit k bn tronglp trnh kim th

B. K hoch cs thit k bn ngoi thit k bn trong thit k chng trnhlp trnh kim th

C. Bo gi hp ng sn xut giao hng x l khiu ni qun l khchhng

D. Bo gi x l khiu ni sn xut hp ng giao hng qun l khchhng





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E. K hoch pht trin vn hnh v bo dng sa cha tnh ton chi phhy b v thay th

Cu 98: 3-1-4 K thut pht trin

p n no di y khng phi l thao tc thit k thuc thit k bn ngoi?

A. Thit k m hnh d liu lgcB. Thit k hnh thc giao dinC. Thit k dng thc u ra (ouput format)D. Thit k d liu vt lE. nh ngha h thng conCu 99: 3-1-5 Cc k thut phn tch yu cu v thit k

Trong s lung d liu di y, ci no biu th X1 v X2?

A) Lung d liu (data flow) B) Lu tr d liu (data store)

C) Thc th d liu (data entity) D). Ngun d liu (data source)

E) X l (process)

Cu 100: 3-1-5 Cc k thut yu cu v thit k

Trong cc kiu ghi ca bn ghi ng k vo cc phng tin lu tr, ty theo hnh thc

lu ca bn ghi lgc , chia lm 4 loi di y. Trong , p n no di y c th

kt ni ti nhiu khi ca phng tin lu tr lu tr bn ghi lgc.A. Bn ghi c di cnh (fixed length record)B. Bn ghi bao trm (spanned record)C. Bn ghi c di bin i (variable length record)D. Bn ghi c di khng xc nh (undefined length record)E. Khng phi l cc p n trn.Cu 101: 3-1-5 Cc k thut yu cu v thit k

Phn hoch chng trnh thnh mt s m un. Trong s cc bn (strength) ca

chng trnh ti phng php phn hoch ny, khi chng trnh c hnh thnh nhrt

nhiu chc nng tun t, bn no c tc dng kt ni tng tnh nng vi nhau thng

W X1 X2

Z

Y1 Y2 Y3





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qua d liu?

A. bn thi gian (time strength)B. bn th tc (procedural strength)C. bn trao i (communicative strength)D. bn thng tin (informational strength)E. bn chc nng (functional strength)Cu 102: 3-1-5 Cc k thut yu cu v thit k

Trong s cc b phn nhp trn mn hnh s dng ti GUI ca MS-Windows, loi no

di y c s dng khi phi la chn duy nht mt trong nhiu phng n la chn c

nh?

A. Hp kt hp (combo box)B.Nt bm radio (radio button)C. Hp nh du (check box)D. Hp k t (text box)E. Menu ko xung (pulldown menu)Cu 103: 3-1-5 Cc k thut yu cu v thit k

Tu theo quan im phn hoch (partioning), c rt nhiu phng php phn hoch m

un khi tin hnh thit k cu trc ca chng trnh. Trong s cc phng php di y,

ci no l phng php phn hoch gn vi cu trc ca d liu? Hy chon 2 p nA. Phng php phn hoch STSB. Phng php WarnierC. Phng php phn hoch theo hm dng chungD. Phng php JacksonE. Phng php phn hoch theo giao tcCu 104: 3-1-5 Cc k thut yu cu v thit k

Trong s cc phng php phn hoch m un khi tin hnh thit k cu trc ca chngtrnh, phng php phn hoch STS l phng php trong s phn hoch chng trnh

thnh 3 phn. Cch kt hp cc phn phn phn hoch no l ng?

A. La chn (select), bin i ( transform), b cha (sink)B. La chn (select), dch (translation), b cha (sink)C.Ngun (source), bin i( transform), b cha (sink)D.Ngun (source), kim tra (test), b cha (sink)E. Ngun (source) ,dch (translation), lin tc( serial)Cu 105: 3-1-5 Cc k thut yu cu v thit k

Trong nh l cu trc, c nh ngha 3 cu trc iu khin cbn, v s dng 2 cu trc





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b sung. Kiu cu trc lgc trong s di y tng ng l p n no?

A. Kiu cu trc a nhnhB. (multi-branch type)C. Kiu lp li (repeat until type)D. Kiu vng l p

type)

E. Kiu la chn (selectivetype)

F. Kiu tun t (sequential type)Cu 106: 3-1-5 Cc k thut phn tch yu cu v thit k

Kiu s trong cho bit mi quan h gia cc d liu nh di y gi l g?

A. DFD B. HIPO C. ERDD. Biu tin trnh (lu - flowchart) E. Biu NS (NS chart)

Cu 107: 3-1-5 Cc k thut phn tch yu cu v thit k

Giao din m phn cng cung c p cho ngi dng c gi l giao din con ngi

(human interface). Trong s cc thut ng di y, t no khng c lin quan n giao

din con ngi ?A. WYSIWYGB. MMIC. GUID. CUIE. APICu 108: 3-1-6 Phng php lp trnh, kim thv xt duyt

Tu theo phm vi i tng kim tra v mc ch kim tra, quy trnh kim tra c chia

thnh 5 quy trnh nh na. Trong cc quy trnh kim tra di y, ci no s do ngi

dng tin hnh. Hy chn 2 p n

Khng

X l

X l c thchin X ln?

C

0.0 Tr lng

1.0 X l vic 2.0 Tnh thng

1.1 DATA 1.2 DATA 2.1 Tnh 2.2 Tnh





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A. Kim tra tch hp (integration test)B. Kim tra h thng (system test)C. Kim tra vn hnh (operation test)D. Kim tra ph duyt (agreement test)E. Kim tra n thCu 109: 3-1-5 Cc k thut yu cu v thit k

Trong kim tra t trn xung (top-down test), khi m un kim tra gi m un th cp, th

ci no s m phng tnh nng ca m un c gi.

A. StubB. Trnh to d liu kim tra (test data generator)C. Phn sn (middleware)D. Trnh iu khinE. M phng (simulator)Cu 110: 3-1-6 Phng php lp trnh, kim thv xt duyt

Khi tin hnh chnh sa, thay i, b sung thm tnh nng cho h thng hin c, php

kim th no snh gi cc tnh nng hin c xem chng c c thc hin ng n

hay khng?

A. Kim th tnh chu ng (endurance test)B. Kim th hi quy (regression test)C. Kim th hiu nng (performance test)D. Kim th t di ln (bottom up test)E. Kim th ti (load test)Cu 111: 3-1-6 Phng php lp trnh, kim thv xt duyt

C nhiu cch to d liu cho thao tc kim th hp en. Cch no di y s ly gi

tr bin thuc nhm ng tr lm d liu kim th.

A. Bao trm tt ciu kinB.

Phng php ln k hoch th nghimC. Biu nhn qu

D. Phn tch gi tr gii hnE. Phn tch tng ngCu 112: 3-1-7 Qun l pht trin

Phng php kim tra xem gi tr d liu nhp c nm trong phm vi quy nh hay khng

l p n no?

A. Kim tra trng (overlap check)B. Kim tra gii hn (limit check)C. Kim tra s (numeric check)





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D. Kim tra dng thc (format check)E. Kim tra i chiu

Cu 113: 3-1-7 Qun l pht trin

Cn p k s kim tra cho d liu s 1914. Hng s gia trng l 1234, cs l 10. Tnh

tch ca cc ch s d liu vi cc ch s ca hng s gia trng v tr ch s tng ng,

ri tnh bnh qun. Ly s d ca php chia kt qu bnh qun cho c s lm k s

kim tra (check digit) v gn vo pha ng sau d liu. Kt qu l p n no di y?

A. 19140B. 19141C. 19145D. 19146E. 19148Cu 114: 3-1-7 Qun l pht trin

Trong mt t chc gm 21 thnh vin, nu khong kim sot (span of control) l 4, th

tng qun l sc chia thnh my tng?

A. 1B. 2C. 3D. 4E. 5Cu 115: 3-2-1 Vn hnh h thng

RASIS gm cc ch ci u tin ca cc tnh nng, m phi c cc tnh nng ny mi

c th yn tm s dng h thng my tnh. Tnh cht no di y khng c trong

RASIS?

A. Availability (Tnh c sn)B.

Security (Tnh an ninh)C. Redundancy (Tnh d tha)

D. Reliability (Tn tin cy)E. Serviceability (Tnh phc vc)

Cu 116: 3-2-1 Vn hnh h thng

Trong x l giao tc csd liu, t bn ch ci u ca 4 thao tc bt buc phi thc

hin, ngi ta gi tt thnh c tnh ACID. A l Atomicity, I l Isolation, D l Durability.

Vy C l g?

A. Complementary





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B. CompactC. ConstrutiveD. ClearlyE. ConsistencyCu 117: 3-2-2 Bo tr h thng

Trong quy trnh bo tr, thao tc c ra c t ca h thng, t phn mm c sn (ti liu

v danh sch chng trnh program list document) tng ng l p n no?

A. M hnh quy trnh (process model)B. Gri chng trnh (program debug)C. K ngh ngc (reverse engineering)D. Lm mu (prototyping)E. K ngh tin (forward engineering)Cu 118: 3-2-2 Bo tr h thng

Trong s cc nhn vin thc hin nghip v bo tr, nhn vin ng vai tr l u mi h

trngi dng l ai?

A.Ngi chu trch nhim h thngB.Ngi qun l thay iC.Ngi bo trD.Ngi qun l cu hnhE. Ngi qun l bo trCu 119: 3-3-2 Hiu nng ca h thng

C mt a t c c im sau:

Tc quay 5000 vng/pht

Thi quan quyt nh v tr bnh qun 20 mili giy

Dung lng lu tr trong mt rnh 15,000 bytes

Dung lng lu tr trong mt cung t 500 bytes

c c d liu trong mt cung t, thi gian truy cp bnh qun cn l bao nhiu miligiy?

A. 8.4 msec

B. 12.4 msec

C. 26.4 msec

D. 33.4 msec

E. 35.4 msec

Cu 120: 4-1-1 Cc giao thc v kim sot truyn tin

Trong cc th tuc iu khin thao tc truyn d liu p dng cho truyn k t, th tc no

di y s dng 10 k tiu khin truyn v tin hnh gi nhn di dng tng tin





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Tng ng dng (application layer)

X

Y

Tng giao din mng (network interface layer)

A. X: Tng vn chuyn (transport layer) Y: Tng mc ni d liuB. X: Tng internet Y: Tng mc ni d liuC. X: Tng internet Y: Tng vn chuynD. X: Tng vn chuyn Y: Tng internetE. X: Tng internet Y: Tng MACCu 125: 4-1-1 Cc giao thc v kim sot truyn tin

C mt s phng php truy nhp mng LAN. Trong cc phng php di y, phngphp no cn c gi l phng php Ethernet?

A. Phng php TDMAB. Phng php Token RingC. Phng php Token BusD. Phng php CSMA/CDE. Phng php CSMA/CA

Cu 126: 4-1-1 Cc giao thc v kim sot truyn tinCp Ethernet c chia thnh mt s loi tu theo cht lng cp hay tc truyn. p

n no di y s dng loi cp xon i nhm 3?

A. 10 BASE 2B. 10 BASE 5C. 10 BASE-TD. 100 BASE-FXE. 100 BASE-TXCu 127: 4-1-1 Cc giao thc v kim sot truyn tin

Trong cc giao thc internet di y, giao thc dng cho thin t l g?

A. SMTPB.NNTPC. FTPD. HTTPE. SNMPCu 128: 4-1-2 M ha v truyn tin

Trong trng hp truyn tn hiu k thut s bng ng truyn tng t (analog line),





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cn thc hin thao tc bin i tn hiu s thc (digital signal) thnh tn hiu tng t

(analog signal). Phng php bin i thc hin nh s di y l p n no?

A. PM B. PCM C. FM D. AM E. ADPCM

Cu 129: 4-1-2 M ha v truyn tin

ng b ho nhng li c ngha l khng ng b l p n no di y?

A. ng b ho khung (frame sychronization)

B. ng b ho bt u - kt thc ng b (start stop sychronization)

C. ng b ho k t (character sychronization)

D. ng b ha SYN (SYN sychronization)

E. ng b ho c(flag sychronization)


p n no di y l thao tc kim tra khng ch tin hnh tm kim li, m cn chnh

sa c li?

A. Phng php CRC

B. Kim tra bit chn l l(odd parity check)

C. Kim tra bit chn l theo chiu dc (longitudinal parity check)

D. M Hamming

E. Kim tra bit chn l theo chiu ngang (lateral parity check)


Cn bao nhiu thi gian (start stop) c th truyn d liu vi 600 k t bng phng

php bt u - kt thc, vi mch c tc truyn l 1200 bit / giy? Tuy nhin, vi iu

kin d liu l 8 bit, khng c bit chn l. Bit bt u v bit kt thc cng l bit1.

A. 0,4 giy B. 0,5 giy C. 1,5 giy D. 4 giy E. 5 giy


Trong mch di y, c mt pht li pht sinh mt bit li. T l li bit l bao nhiu? Ly

kt qu l s c hai ch s vi ch s hng th 3 c lm trn.

Tn hiu s thc

Tn hiu iu bin

0 1 0 0 1 0 1 0





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Cu 136: 4-1-3 Mng ( LAN v WAN)

Hnh thc mng my tnh kt ni cc mng cch xa nhau l p n no di y?

A. WANB. MANC. LAND. JANE. VANCu 137: 4-1-3 Mng ( LAN v WAN)

V mng LAN khng dy, U ban IEEE802 lin tc tiu chun ho v to ra mt s

tiu chun. Trong , tiu chun no di y s dng sng truyn thng 5GHz v c tc

truyn ti a l 54Mb/giy?

A. IEEE 802.11B. IEEE 802.11aC.

IEEE 802.11bD. IEEE 802.11g

E. IMT2000Cu 138: 4-1-4 Thit b truyn thng

Trong cc thit b s dng truyn d liu di y, thit b no c lp t gia

modem v ng truyn, thc hin thao tc gi tin v nhn tin, v gn y thng c

lp bn trong modem?

A. DCE B. DTE C. DSU D. NCU E. CCU

A. Hnh sao B. Kiu li C. Kiu vng

D. Kiu Bus E. Hn hp





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Cu 139: 4-1-4 Thit b truyn thng

p no di y sp xp cc rle thao tc truyn mng theo trnh t sau:

- Rle tng vt l ca OSI

- Rle tng lin kt d liu

- Rle cho ti tng mng

A. Cu (brigde), b lp (repeater), b chn ng (router)

B. B lp, cu, b chn ng

C. Cu , b chn ng, b lp

D. Cu, b chuyn mch (switch), b chn ng

E. B lp, b chn ng, kho chuyn mch

Cu 140: 4-1-5 phn mm mng

Khi xem trang ch trn internet, ngi ta s dng mt phn mm l trnh duyt

(Browser). p n no di y khng phi l tn gi ca cng c trnh duyt?A. MosaicB.Netscape CommunicationC. Open OfficeD. OperaE. MozillaCu 141: 5-1-1 Cc m hnh CSDL

M hnh d liu quan h (relational data model) do E.F.Cold ca IBM pht minh ra nm1970. y l m hnh trong ngi ta c th t do thoi mi kt hp cc d liu vi

nhau v c thng dng trn phm vi rng. Trong m hnh ny, s kt hp d liu vi tn

gi no di y?

A. Gc v Nhnh v Nt (Root & Branch & Node)

B. Bn ghi v Mng v Tp (Record & Network & Set)

C. Hng v Ct (Row & Column)

D. Thc th v Quan h v Thuc tnh (Entity & Relation & Attributes)

E. Quan h v B v Thuc tnh (Relation & Tuple & Attribute)

Cu 142: 5-1-1 Cc m hnh CSDL

D liu di y c x l n giai on no ca qu trnh chun ho?

M nhn vin Tn nhn vin M chng ch Tn chng

ch

Ngy t chng

ch

A. Khng chun ho

B. Chun ha th 1

C. Chun ha th 2

D. Chun ha th 3

E. Chun ha th 4





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Cu 143: 5-1-2 Ngn ngCSDL

Khi m t SQL, ni thc hin cc thao tc x l csd liu, ngi ta ch yu l m t

thao tc tnh ton hm v tnh ton tp hp. p n no di y c xp vo thao tc

tnh ton tp hp?

A. La chn (selection)

B. Chia (division)

C. Kt ni (joint)

D. Tch (product)

E. Chiu (Projection)


Trong cc thao tc c s d liu lin quan, m t thch h p nht cho thao tc kt ni

(joint) l p n no di y?

B. Ch ly nhng dng tho mn iu kinC. Ch ly nhng ct c chnhD. To bng mi lin kt cc dng vi cc mc chung c cng mt gi tr, t 2 bng tr

ln

E. Loi b nhng phn trng lp trn hai bng v ly ra tt c cc dng.F. Kt hp cc b ca 2 bng v to thnh mt bng.Cu 145: 5-1-2 Ngn ngCSDL

SQL c cu thnh t 3 yu t l DDL (nh ngha c s d liu c lin quan), DCL(iu khin cc d liu c lin quan) v DML (thao tc csd liu c lin quan). p

n no di y c xp vo nhm DDL?

A. CREATE B. INSERT C. UPDATE D. SELECT E. GRANT


Lnh SQL thc hin thao tc bng di y l p n no?

No. Name Gender Age No. Name Gender Age1401 Nam Male 30 1401 Nam Male 30

1402 Trang Female 25 1402 Trang Female 25

1403 Viet Male 55

===>

1403 Viet Male 45

A. INSERT B. CREATE C. UPDATE D. SELECT E. DELETE


D liu c ly ra theo cu lnh SQL di y l p n no?

Name Gender Age





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Van F 25

Tien M 29

Ha F 31

Loc M 30

Long M 32

SELECT Name FROM Student table WHERE Gender = MAND Age > 29.

A. Loc 30 B. Loc C. Loc M 30 D. Tien 29 E. Tien

Long 32 Long Long M 32 Loc 30 Loc

Long 32 Long


D liu c ly ra theo cu lnh SQL di y l p n no?

SELEC AVG (Age) FROM Employee Table

No Name Age

101 Loan 22

102 Giang 24

103 Anh 26

104 Chi 28

A. 22 B. 25 C. 28 D. 34 E 100

Cu 149: 6-1-1 An ninh

Virt my tnh l chng trnh c lm hng cc chng trnh hay d liu. Chng trnh

pht hin v dit virt my tnh l g?

A. B ci t (installer)B. Thut s(wizard)C. D vt (tracer)D. Chng trnh theo di (watching program)E. Tng la (firewall)Cu 150: 6-1-1 An ninh

Trong phng php mt m, c h thng m kho cng khai (public key crypto system)

v h thng m kho chung (common key cryprto system). Phng php m kho cng

khai s dng 2 loi kho, mt l kho c m ho bng phng php mt m phi i

xng v kho gii m ring bit. Hai loi kho ny gi l g?A. Encode key v Decode key





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C. 200505D.NG80LGE. NLCu 155: 6-2-4 Cc t chc ha

HTML l do mt nh nghin cu ngi Anh tn l Timothy John Berners-Lee lm vic

cho CERN ti Gi-ne-v ca Thu S, pht trin da trn Trang Web ton cu (world

wide web) c cng b vo nm 1989. Cc phin bn ca HTML u do cc t chc

quc t khuyn co s dng. Cc phin bn HTML 3.2 tri do t chc no khuyn co

s dng?

A. IETFB. RFCC. ITU-TD. ISOE. WWW ConsortiumCu 156: 6-2-4 Cc t chc ha

Ethernet l do nh nghin cu mang tn Robert M.Metcalfe lm vic ti Vin nghin cu

Palo Alto ca Xerox, pht minh ra vo khong nm 1972. Cc phin bn sau ca

Ethernet u do mt t chc quc t cng b v tiu chun ho. l t chc no?

A. ITU-TB. RFCC. IEEED. ISOE. 3ComCu 157: 7-1-1 Qun tr kinh doanh

C nhiu phng php thu thp d liu hay thng tin. Phng php thng s dng trong

hi ngh, do Alex F.Osborne ca M pht minh ra l g?

A.

Phng php KJB. Phng php DelphiC. Phng php Least quareD. Brain Storming (ng no)E. S xng c (fish born chart)Cu 158: 7-2-1 K ton ti chnh

V nguyn tc k ton doanh nghip, c hai nguyn tc l nguyn tc cn i ti sn v

nguyn tc tnh ton l li. Trong cc nguyn tc di y, nguyn tc no nm trong c

hai nguyn tc l nguyn tc cn i ti sn v nguyn tc tnh ton l li?

A.Nguyn tc ph hp gia chi ph v thu nhp (matching costs with revenue principles)





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B.Nguyn tc ch ngha tng kim ngch ( Gross amount principles)C.Nguyn tc phn loi v chnh hp ( Section and orrangement principles)D.Nguyn tc tnh ton vn khi cn i ti sn ( B/S Integrity principles)E. Nguyn tc ch ngha pht sinh ( Accrual principles)Cu 159: 7-2-2 K ton qun l

Hy tnh im ho vn theo bng k l li di y?

Hng mc S tin (nghn la)

Doanh thu 5600

Chi ph bin i 2800

Chi ph cnh 2000

Li nhun 800

A. 3850B. 4000C. 4333D. 4667E. 4800Cu 160: 7-2-2 K ton qun l

T l quan h c ngha l vic phn tch t l gia cc hng mc trong bo co ti chnh.

Ci no di y c dng xc nh kh nng thanh ton tim nng di hn, khi phn

tch mc an ton ca mt doanh nghip?A. T l vn ch shu (Owners equity)B. T l tin mt trong ti khon (Acid test)C. T l vn lu ng ( Currenr)D. T l vn cnh (Fixed)E. T l cng n(Dept)Cu 161: 7-3-1 H thng IE v OR (nghin cu hot ng)

Trong d n minh ho trong s PERT di y, ngy bt u tin hnh thao tc L smnht l ngy bao nhiu? Ngy bt u trin khai d n l 0 ngy.





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A. 15 ngy B. 16 ngy C. 17 ngy D. 18 ngy. E. 19 ngy

Cu 162: 7-3-1 H thng IE v OR (nghin cu hot ng)

Trong nghip v qun l cht lng ti doanh nghip sn xut, phng php no l phhp vi thao tc di y. Tng hp v sp xp nhng vn phc tp v nhng kin

hay tng cha c tng hp.

A. Biu nhn qu ( Cause and Effect Diagram Graph)B. Biu phn b (Scatter Plot Graph)C. Biu ParetoD. Bng kim (check sheet)E. th (graph)Cu 163: 7-3-1 H thng IE v OR (nghin cu hot ng)

Trong trng hp ng thi rt ra hai l t b bi 52 l bi khng cha con phng teo, xc

xut c hai l bi u l qun u ngi (tc l J,Q,K) l bao nhiu?

A. 1/17B. 1/26C. 16/119D.

9/169E. 11/221


Bin thin (variance) v lch chun (standard deviation) l gi tr cho bit mc

phn tn ca cc gi tr trong mt nhm d liu. phn tn ca nhm d liu di y

l bao nhiu?

D liu 5 6 9 11 14

A. 5,2B. 8,0C. 10,8

1

3

2

5

6

4

8

9

7C

D

E

G

F J

K

L

H

B I

A

9 da s

1 days3 days

3 days

5 days

6 days

4 days

2 days

4 da s

7 days

4 da s

7 days





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D. 13,9E. 16,9


Tng s nhn vin ca mt cng ty l 500 ngi, gi s s lng ngi ngh vic hng

ngy theo bin thin nh thc vi t l ngh vic P = 0,01 th gi tr k vng v gi tr bin

thin ca s ngi ngh vic hng ngy l bao nhiu?

A. 4,95 v 10B. 4,95 v 5C. 5 v 4,95D. 495 v 100E. 500 v 495Cu 166: 7-3-1 H thng IE v OR ( nghin cu hot ng)

Trong cc s thuc 7 cng c qun l cht lng (hay 7 cng c qun l cht lng

mi) theo phng php thng k, kiu s no trong cc p n di y biu din cc

ct d liu sp xp theo th t gim dn v c ng nt t biu th tng tch lu ca d

liu, thun tin cho vic phn tch ABC?

A. S phn bB. S mi tnC. Biu D. S ParetoE. PDPCCu 167: 7-3-1 H thng IE v OR ( nghin cu hot ng)

Trong trng hp doanh thu ca tng mt hng l nh sau, ta tin hnh phn tch ABC do

nhm A chim 70%, th s mt hng ca nhm A l bao nhiu?

Mt hng 1 2 3 4 5 6 7

Kim ngch 150 100 80 70 50 30 20n v: 1000 yn

A. 2 B. 3 C. 4 D. 5 E. 6

Cu 168: 7-3-1 H thng IE v OR ( nghin cu hot ng)

S di y c s dng khi iu tra hnh nh gi l g?

Kh Hi Trung bnh Hi Kh

Sng Ti

m thanh tt m thanh ti





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52

Din vin tt Din vin ti

A. S JB. S SDC. S qun l (control chart)D. Fan chartE S Gantt

Cu 169: 7-3-1 H thng IE v OR ( nghin cu hot ng)

Tn gi ca s di y l g?

A. S giai on (step chart)B. S tng hnh (pictograph)C. Snh lp th (stereograph)D. S hnh qut (Fan chart)E.

S Rader

Cu 170: 7-4-1 Sdng h thng thng tin

Trng hp thc hin h thng FAC (Factory Automation) cn c vo h thng thng tin

kinh doanh (MIS), ton b h thng thng tin c gi l g?

A. CADEMB. CIMC. H thng OAD.NCE. MRP

120%

110%

100%

90%

80%

70%

2000 year 2005 year

Bus

Auto

Truck





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53

Cu 171: 7-4-2 H thng k ngh

Trong h thng FA dng qun l tng hp ton b h thng sn xut k t khi ln k

hoch sn xut cho ti khi xut hng ra khi kho mt cch hp l, c rt nhiu h thng

ph. H thng ph dng m phng qu trnh phn tch cng , phn tch ccu, v

tin hnh thit k khi qut l g?

A. CAPPB. MISC. CAED. CAPE. CAM

Cu 172: 7-4-3 H thng kinh doanh

H thng kinh doanh l h thng thng tin trong tin hnh lu tr, tch lu v phn tchqu trnh hot ng ca doanh nghip. Trong cc p n di y, ci no khng nm

trong h thng kinh doanh?

A. EOSB. EDIC. OAD. CADE. POSCu 173: 7-4-3 H thng kinh doanh

C rt nhiu hnh thc thng mi in t trn mng internet. Hnh thc u gi trc

tuyn c tin hnh trn trang web, c xp vo hnh thc no di y?

A. B to BB. B to CC. Giao dch trc tuyn (online trade)D. C to CE. E- market placeCu 174: 7-4-3 H thng kinh doanh

Trong h thng POS (Point of Sale = im bn hng), s phi nm bt c s lng mua

vo v s lng sn phm bn ra mt cch thng xuyn. Thit b dng tnh tng v

i chiu tn, gi c v s lng hng mua vo, vi d liu v tn, gi c, s lng c

nhp vo ch dn m vch (barcode leader) hay thit bu cui my tnh POS, ng thi

lin lc vi my tnh ch, l thit b no di y?

A. My scanB. Thit bu cui cm tay (handly terminal)C. Thit b qun l lu tr (store controller)





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54

D. OCRE. BCRCu 175: 7-5-2 Quyn shu tr tu

T chc quc t hin ang hot ng da vo Cng c Pari v quyn s hu cng

nghip vo nm 1883 v Cng c Berne v quyn tc gi nm 1886 l t chc no?

A. WIPOB. UNESCOC. ITUD. BIRPIE. ISOCu 176: 7-5-2 Quyn shu tr tu

Trong quyn shu tr tu c mt s loi quyn. p n no di y khng thuc nhmcc quyn v shu tr tu?

A. Quyn gii php hu chB. Quyn chn dungC. Quyn tc giD. Quyn nhn hiuE. Quyn kiu dng

Cu 177: 7-5-2 Quyn shu tr tuHin nay, tm quan trng ca vic bo v nhng quyn shu tr tu khng phi l i

tng c ph bin trn cc phng tin thng tin i chng, ngy cng tr nn quan

trng. Trong s cc quyn shu tr tu di y, quyn vi i tng bo h l cc k t

hay hnh v dng phn bit hng ho, dch v, v bo v quyn li ca ngi s dng

n l g?

A. Quyn tc giB. Quyn kiu dngC.

Quyn nhn hiuD. Quyn sng ch

E. Quyn gii php hu chCu 178: 7-5-6 Cc lut khc v vn o c ngnh ngh

x l k t, my tnh x l da vo mt loi m, gi l m k t. Trong cc m k t

di y, m no khng s dng c ch Katakana (ch cng).

A. M ISOB. M JISC. M Shift-JISD. M UNICODE





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E. M EUCCu 179: 7-5-6 Cc lut khc v vn o c ngnh ngh

Cng ty t bn c chia thnh hai nhm l cng ty t nhn v cng ty nh nc. Trong

, cng ty t nhn li c chia thnh nhiu nhm nh na. Trong cc cng ty t nhn

di y, cng ty no no c ch shu doanh nghip hon ton tch ri vi ngi iu

hnh doanh nghip?

A. Cng ty hp danhB. Cng ty hp vnC. Cng ty c phnD. Cng ty i bn cng c li (mutual company)E. Cng ty trch nhim hu hn

Cu 180: 7-5-6 Cc lut khc v vn o c ngnh nghChc nng, t chc, b phn trong doanh nghip hay cskinh doanh c chia thnh

hai nhm: ng li (line) v nhn vin (staff). Trong cc b phn di y, ci no c

xp vo nhm nhn vin?

A. B phn mua hngB. B phn nghin cu pht trinC. B phn ch toD. B phn qun l cht lngE. B phn kinh doanh





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56

P N





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p n cu 1: A

Chuyn s hc Chuyn l gic

Gii thch: C th tnh ton nh sau:

11001011

dch chuyn s hc sang phi 1 bit

11100101

dch chuyn s hc sangphi 1 bit 11110010 dch chuyn lgic sang tri 1 bit 1100100 dch chuyn

lgic sang tri 1 bit 11001000

p n cu 2: C

Phn tch: hon chuyn t h n phn sang h thp phn c thc hin nh sau: mi ch

s nhn vi cs h, c m l hng ch s tri 1, sau cng tng chng li.

Gii thch:

2C ca h thp lc phn (16)

= 2 x (16 (2-1)) + C x (16 (1-1)) = 2 x 16 + 12 x 1 = 44 ca h thp phn

70 ca h bt phn

Bt du khng i Thm 0 vo ba tri Thm 0 vo ba triMi bt chuyn

thm ngay vo ba tri ging nh bt Thm 0 vo ba tri

1 1 0 0 0 1 0 0

1 0 1 0 1 0 1 0

1 1 0 1 0 1 0 1

Dch sang tri

Trc khi dch

Dch sang tri

Dch phi

Trc khi dch

Dch phi





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58

= 7 x (8 (2-1)) + 0 x (8 (1-1)) = 7 x 8 + 0= 56

Kt qu php tnh h thp phn l:

100 44 56 = 0

p n cu 3: D

Gii thch: quy tt c cc h cs v h thp phn (cs 10), ri tnh ton v so snh.

(1234)16 = (4660)10 (10101010101)2 = (1365)10

Do : (4660)10 - (1365)10 = (3295)10

Cc kt qu chuyn v dng h thp phn s l:

(1100 1101 0000)2 = (3280)10 (6006)8 = (3078)10

(CDF)16 = (3295)10 (SOS)16: Khng c s no nh vy dng h thp lc phn.

p n cu 4: A

Gii thch: thp phn 110 = 1 x 64 + 5 x 8 + 6

1 x 82 + 5 x 81 + 6 x 80

bt phn

Cch gii khc: quy tt c v h nh phn, ri ct 3 ch s li thnh mt nhm.

001 101 1101 5 6

p n cu 5: C

Cch suy lun: gi tr m th hin phn b 2 c tnh bng cch: ly 100000000 h nh

phn (c s 1 hng u), tri s nh phn thnh gi tr tuyt i ca s.

Gii thch:

(01011000)2 = (88)10(10010100)2 l m, nn

(100000000)2 - (10010100)2 = (01101100)2 = (108)10

Do , 88 108 = -20

p n cu 6: A

im chnh: Carry ngha l vic nhs, trong , kt qu tnh ton mt hng vt qu

cs, khin phi cng thm 1 vo hng ch s pha trc.

Gii thch: ti phn nh du hnh tam gic, l trng hp pht sinh nh(carry).

A 10001 B 1111 C 1101 D 1010 E 101

0 51 6





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59

+ 110 +1000 + 11 + 110 + 110

--------- ---------- ---------- ---------- ---------

10111 10111 10000 10000 1011





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p n cu 7: B

Cng thc: biu th s du phy ng 32 bit do IEEE xng:

1 bit 8bit 23 bit

v tr ca du phy thp phn

Du ca phn nh tr: S (0 hoc 1) = 0

Gi tr phn m: E = (10001010)2 = (138)10

Gi tr phn nh tr: M = 10100000000000000000000

Gi tr s du phy ng: = (-1)s x 2(E-127) x (1 + M)

Gii thch: = (-1)0 x 2(138-127) x (1 + 0.101)2

= +(1.101)2 x 211

= +(1.625)10 x 211

p n cu 8: D

Gii thch: i vi s thp phn khoanh vng, mi 1 byte cho s c mt ch s, v 1 byte

v tr sau cng c phn du (sign part). Cn i vi s thp phn ng gi, mi 1 byte

cho s c hai ch s, v 4 bit v tr sau cng l phn du (sign part).

V di vi s1234:

: Trong trng hp s thp phn khoanh vng th l 4 byte

1111 0001 1111 0010 1111 0011 1100 0100

Zone bit 1 Zone bit 2 Zone bit 3 Sign bit 4

: Trong trng hp s thp phn ng gi th l 3 byte

0000 0001 0010 0011 0100 1100

1 2 3 4 Sign bit

nu khng trn byte th in thm 0 vo.

- dng s thp phn khoanh vng, cn 6 byte. Nu khng c phn du, th dng s

thp phn ng gi, ch cn 3 byte tc l mt na. Cho thm phn du th l 4 byte.

p n cu 9: Dim chnh

A. Php tng lgic loi tr XOR B. Php tch lgic AND C. Php ton i snh

D. Php phnh tng lgic NOR E. Php phnh tch lgic NAND

Gii thch: Php phnh tng lgic (NOR) sai lgic nu t nht mt trong A, B ng.

A + B = A B

S E M





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p n cu 10: B

Chy (run): tnh trng nhim vang c thc hin vi th tu tin cao.

Sn sng (ready): tnh trng lc no cng c th thc hin c.

i (wait): tnh trng ang i kt thc thao tc vo-ra c yu cu.

A: ng k nhim v

B: chuyn i

C: ht thi gian (time out)

D: pht sinh lnh vo-ra

E: hon thnh thao tc vo-ra

Gii thch: trn s, dispatcher l chng trnh nm gia B v C, c chc nng gn

quyn s dng CPU cho nhim v c th tu tin cao nht trong s tt c cc nhim v

c th thc hin c.

p n cu 11: DBinary tree (cy nh phn): cu trc hnh cy ch c t 2 nhnh trxung.

Multiway tree: cu trc hnh cy nhng s nhnh khng gii hn 2 m nhiu hn.

Mng: sp xp cc phn t hu hn thnh bng. C th tham chiu cc phn t theo tn

mng v ch s di.

Ngn xp: cu trc d liu LIFO (vo sau ra trc). Lu tr cc d liu chng ln nhau.

Gii thch: Xp hng l cu trc FIFO (vo trc ra trc), s dng trong danh sch i

ca cng vic (job) v nhim v (task).

p n cu 12: E

Phn mnh:vng cha c s dng nm ri rc trong vng lu tr chnh hay b nhph.

Kh phn mnh: thao tc nhm xo b tnh trng phn mnh trong cng.

ng ngha: l d liu ng sau, trong trng hp thu c gi tr bm ging nhau bng

phng php bm t cc d liu khc nhau.

nh tr hon ton: tnh trng 2 quy trnh c chung ngun cng chi quy trnh kia gii

phng ngun trc.

Gii thch: Thng rc l phn t khng cn thit sau khi xo nhng vn lu li c thkhi phc.

p n cu 13: B

im chnh: ngn xp l cu trc vo sau ra trc (LIFO), nn s ly ra theo th t ngc

vi trnh t lu tr.





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62

Gii thch: s ly ra theo trnh t 9,8,7,6,5 do ch s ly ra ln th 4 l s 6.

p n cu 14: D

Gii thch: Cy tm kim nh phn c c trng l, trong tt c cc , phn t trong bao

gicng ln hn phn t bn tri, v nh hn phn t bn phi.

p n cu 15: C

im chnh: Heap l c tnh i ng cy nh phn ton phn theo mng.

Quan h cha-con bt k nt no cng lm hnh thnh kch cca d liu.

p n cu 16: B

C,B,AA,B,C

Push Pop

C

B

A

Ngn xpnh

y

2

4 9

13 3019

12 1714

6

4

12 19

Thay i

6

19

12 4

Thay i

A





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63

Gii thch: Quy tc khi v cy nh nhp theo trnh t trung gian s nh sau:

- Bt u v t nhnh bn tri, v ly gi tr trong nt xa hn.- Ly gi tr trong nt phn nhnh nu gi tr vt qu gi tr trong nt t tri sang

phi.

p n cu 17: DGii thch: S ln so snh ca Bubble Sort l O(n2), do vy thi gian s l 52=25 ln.

25 ln ca 1 giy l 25 giy.

p n cu 18: B

im chnh: Bubble Sort s tin hnh so snh d liu lin k, nu quan h cha-con

(ln-b) b ngc th s cho hon chuyn v tr v sp xp li.

Gii thch:

Ch s di 1 2 3 4 5Bt u 15 5 11 12 7

So snh hon chuyn v tr ln 1 5 15 11 12 7

Hon chuyn v tr ln 2 5 11 15 12 7





Kt thc

B t u K t thc

2

3 5

6

8

9

10

11





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p n cu 19: E

Quick Sort: l phng php sp xp nhanh trong s dng mt gi tr lm tiu chun

(trc).

Binary Search: l phng php phn chia phm vi nh phn cc d liu c sp xp

sn, tm ra d liu cn.

Phng php Boyer - Moore: cn c gi l phng php BM. y l thut ton tm

kim chui k t vi hiu qu tm kim cao, do R.S.Boyer v J.S.Moore pht minh ra.

Phng php thut ton Greedy: cn c gi l phng php tng thm. N l thut ton

x l vn gn.

Gii thch: Phng php tm kim Dijkstra l thut ton x l vn ng dn ngn

nht do Dijkstra pht minh ra.

p n cu 20: EGii thch: Vi s d liu N, th s ln bnh qun ca php tm kim nh phn bng

(log2N) (du [ ] c ngha l b ch shng thp phn.

log25000 = log210000/2 = log2104 - log22 = 4 x log210 1 = (4/log102) -1 = 4/0.3010 1 =

13.289 -1 = 12.289 = 12 ln.

p n cu 21: B

Linear programming (lp trnh tuyn tnh): l phng php ti u ha trong tin hnh

ti a ha (hoc ti thiu ha) hm i tng, trn cscc iu kin s dng bt ngthc tuyn tnh (th cp).

Phng php tm kim tuyn tnh: phng php tm chui k tc bit t trong mng.

Tm kim tu mng.

Phng php xc nhn bt bin: mt loi phng php kim chng tnh ng n ca

thut ton.

Phng php lp trnh ng: phng php x l gii quyt vn vng dn ngn

nht hay vn gn bin.

Gii thch: Phng php chui cn gi l php bm m(open hash), n ni d liu chacc gi tr bm ng nht thnh danh sch bng cc nt.





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p n cu 22: D

im chnh: mod (a1 + a2 + a3, 5) c ngha l s d ca php chia tng ca a1, a2, a3 cho 5.

Cn xung t c ngha l c cng mt gi tr bm.

Gii thch: Cc t hp gi tr bm s nh sau:

A. Tng cc ch s ln lt l 17 v 14. Gi tr bm l 2 v 4B. Tng cc ch s ln lt l 13 v 12. Gi tr bm l 3 v 2C. Tng cc ch s ln lt l 11 v 15. Gi tr bm l 1 v 0D. Tng cc ch s ln lt l 3 v 18. Gi tr bm l 3 v 3Tng cc ch s ln lt l 20 v 19. Gi tr bm l 0 v 4

p n cu 23: A

Phn tch k hm sc u bi a ra, ta c:

F(n) = n! = n x (n 1) x .. 3 x 2 x 1

Do : F(5) = 5 x 4 x 3 x 2 x 1 = 120

p n cu 24: C

Gii thch: K hiu mch s tng ng nh sau: A l B cng bn phn (half adder), B l

mch OR, C l mch ADN, D l mch phnh, E l biu th B cng y

Mch s hc AND l mch dng tnh ton AND ( logical product operation) , bng

chn l s nh sau:

A B A AND B

0 0 0

0 1 0

1 0 0

1 1 1





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p n cu 25: B

Sn phm ca Intel:

Nm 1971 i4004 4bitMPU


Nm 1978 8086 16bitMPU

Nm 1982 80286 16bitMPU



Nm 1993 Pentium 32bitMPU

Nm 1999 Pentium III 32bitMPU

Nm 2001 Itanium 64bitMPU

p n cu 26: B

LD, GRO, GR1 c ngha l chuyn ni dung Thanh ghi a nng 1 (General Purpose

Register) sang Thanh ghi a nng 0. Tc l:

Np GR1 vo GRO

Do vy, ni dung "s 5" ca GR1 sc nhp vo GRO.

p n cu 27: E

Thanh ghi a nng: l thanh ghi c s dng rng ri tnh ton hay cp nht a ch.

Thanh ghi yu cu ngt: l thanh ghi trong bit s trthnh ON.

Thanh ghi s hc: nm trong khi s hc ALU, n c chc nng ghi li cc con s trcv sau khi thc hin php tnh.

Thanh ghi ch mc: l thanh ghi c s dng cp nht a ch mc lc hay bm.

Gii thch: Thanh ghi cs cung cp cc thng tin nh c pht sinh vic nhs, mn s

hay khng, hoc c b trn b nhhay khng.

p n cu 28: A

MIPS: l n v th hin tnh nng x l ca CPU, n cho bit c th thc hin c bao

nhiu triu lnh trong 1 giy.Tng nhp ng h: l k thut tng tc nhm tng tc nhp ng h ca CPU s

dng trong my tnh c nhn.

CISC: my tnh vi CPU c kh nng x l tp lnh phc tp.

CPU a nng: lp nhiu CPU mt lc, tng tc x l v tng tin cy.

Gii thch: Cho ng ng lnh chng ln nhau tng tc x l lnh. C th p dng

cho my tnh RISC vi tp lnh hn ch, c thi gian thc hin cc lnh tng ng

nhau.





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p n cu 29: C

Cng thc: MIPS l n v th hin tnh nng x l ca CPU, n cho bit c th thc hin

bao nhiu triu lnh trong 1 giy.

Gii thch: S lnh m my tnh thc hin trong 1 giy

= 30 x 1 triu x 0.85 = 25.5 triu lnh

Thi gian x l: 10 triu 25.5 triu = 0.392 giy = 0.39 giy.

p n cu 30: E

SIMM v DIMM: nu b nhca nhng b phn ny tng ln, th c th s dng chng

trnh bng tnh thao tc, t lm tng tnh nng x l.

Over drive processor: Nhvic tng cng tnh nng ca CPU, tnh nng x l cng s

c nng cp mt cch ton din.

Cc hnh: Do tnh nng hin th hnh nh tng ln, nn tnh nng khi v s, biu cng c ci thin tt hn.

Gii thch: Cc m thanh l thit b ch cn thit khi mun s dng tnh nng v mt m

thanh v n khng h tc ng hoc nh hng n cc tnh nng x l khc.

p n cu 31: E

Ngt chng trnh: xut hin khi b trn b nh CPU hoc pht sinh li trong khi thc

hin chng trnh.

Ngt ng h: xut hin khi ng h ra ch th cho CPU.Ngt vo-ra: xut hin thng bo kt thc thao tc vo-ra t bn ngoi knh, do s c khi

x l thao tc vo-ra.

Ngt SVC: xut hin khi chng trnh x l ti CPU pht sinh supervisor call (SVC), do

chng trnh ang chy nhn c yu cu t hiu hnh.

Gii thch: ngt bn giao tip gia ngi v my l hin tng pht sinh khi c s can

thip ca iu hnh vin (operator) t system console (thit b cui) ca my tnh a

dng.

p n cu 32: C, E

Cache memory (b nhm n ): vng lp y khong cch v tc gia CPU v b

nh.

VLIW (verylong instruction word): thay cho ng ng , chp nhn lnh di, cho thc

hin nhiu php tnh ton tng tc .

ng ng lnh: b phn tin hnh x l lnh song song bng cch quy chuyn cc tp

lnh v nhng lnh c cng di v cng thi gian x l, vi tnh nng thp c tn sut

s dng cao. Cn c con chip RISC.

Dch ti u ho: c th pht huy tnh nng ca b x l, bng cch to m chng trnh

ph hp vi thao tc x l trong b x l.





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Super scalar architecture: kin trc siu v hng thit b c th thc hin nhiu lnh

trong mt chu k ti ng ng.

p n cu 33: D

Cng thc: 1KB 1,000B

1MB 1,000,000B

1GB 1,000,000,000B

1TB 1,000,000,000,000B

Gii thch: 0.5GB 1,000,000,000B x 0.5

500,000,000B

500,000,000B x 8bit

4,000,000,000bit

p n cu 34: ACng thc: Thi gian truy cp CPU bnh qun Ta = Tc + (1-) Tm

Gii thch: Cn c vo cng thc trn, ta c thi gian truy cp CPU bnh qun

= 0.7 x 10 + (1 - 0.7) x 50

= 22 nano giy

p n cu 35: B

Thanh ghi: l thit b nh c tc ghi nhanh nht trong phn cp b nh (memory

hierarchy) c ci t trong b x l. Dung lng ca thanh ghi nh.B nhngoi: chnh l b nhh tr. N bao gm a cng, MO v.v.

B m a: l thit b nm gia b nh chnh v b nh h tr, c tc dng lp y

khong cch v tc gia hai thit b trn.

Knh ni b: l ng truyn tn hiu kt ni cc thanh ghi, cng vo/ra nm xung quanh

CPU.

Gii thch: trong b nhm n, ngi ta thng dng SRAM, mt thit b tc cao c

s dng thit b lng cc (bipolar device).

p n cu 36: D

Mask ROM: c s dng trong cc b nhlu tr IPL, hay cc b nhca bng tr chi

trong my tr chi.

SRAM: l thit b nh khng lm mt thng tin lu tr, nu nh khng tt my tnh.

Khng cn phi lu tr d liu (refresh) u n ni dung lu gi khng b mt.

DRAM: thit b nhbn dn lu tr thng tin bng cch np in vo tin. Cn phi

lm ti.

SIMM: Single In-line Memory Module. Mun cha nhiu DRAM.

Gii thch: Flash memory: l ROM c th xa v ghi chng nhiu ln. C kh nng xa

v ghi chng da vo cc n v khi.





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70/130

70

p n cu 37: B

Mch AND: mch thc hin php tnh V

B cng y : mch cng c kh nng tnh c sn v nhgia ch s hng di

v ch s hng trn sau 3 ln nhp.

B cng bn phn: mch cng ch tnh c sn v nhln ch s hng trn sau 2 ln

nhp.

Mch NOT: mch thc hin php tnh lgic phnh.

Gii thch: Mch nhhai trng thi (mch lt) c cu to rt phc tp, do c t cc (cell)

nhp vo IC, nn dung lng lu tr nh hn DRAM, v gi thnh ca n rt cao

p n cu 38: A

Kh phn mnh (defragmentation): thc hin ti nh ph. Xa b nhng th lt vt,

im gin on ca tp tin.

Gch cho (slashing): tnh trng chuyn vo (roll in), chuyn ra (roll out) di dng honi lin tc.

F. Tro i: (swapping) vic hon i ni dung t vng nhchnh sang b lu tr ph.Phn mnh (Fragmentation) : tnh trng cc vng khng s dng vng lu tr chnh b

nm ri rc (phn on).

Gii thch: Lm gn b nh(Memory Compaction) l vic b tr li cc chng trnh b

chia on trong vng lu tr chnh, nhm duy tr vng trng cha s dng.

p n cu 39: CPhng php FIFO - First In First Out (vo trc ra trc): cho ra page-out trang u tin

c lu trong b nhchnh. Cng c trng hp cho page-out ngay trang mi xem gn

nht.

Phng php LIFO Last In First Out (vo sau ra trc): cho page-out ngay trang mi

c page-in gn nht, do vy khng thng qua h thng lu tro.

Dynamic Address Transaction: s dng bng nh s trang (page table) thay i a

ch truy cp khi thc hin lnh. i a cho thnh a ch thc trn vng lu tr chnh.

Page Replacement: vic nh s trang trong vng lu tro. Khng phi l thut ton.Gii thch: Phng php LRU - Least Recently Used: cho page-out trang c thi gian

tham chiu c nht. Thng qua vng lu tro.





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p n cu 40: B

Cng thc:

Dung lng mt khi = di bn ghi x h s khi + khong cch gia cc khi

S bn ghi lu vo mt rnh = (dung lng lu gi ca mt rnh : dung lng ca mt

khi) x h s khi

Gii thch: Dung lng mt khi = 400 x 10 + 500 = 4500

S khi trong mt rnh = 28000 : 4500 = 6,222 = 6 (b s thp phn lm trn)

S bn ghi lu vo mt rnh = 6 x 10 = 60

Minh ho bng hnh v:

Dung lng 1 khi (block)

di mt khi

rnh IBG Bn ghi 1 Bn ghi 2 Bn ghi 10 IBG

Trong mt khi chc sbn ghi tngng vi h skhi.

p n cu 41: E

Cng thc:

Thi gian quay mt vng (mili giy/vng) = 60 x 1000 tc quay (vng/pht)

di khi = h s khi x di bn ghi

Thi gian truyn d liu ca khi = ( di khi dung lng rnh) x thi gian quay mtvng

Gii thch: Theo cng thc trn, ta c:

Thi gian quay mt vng = 18.75 mili giy

di khi = 8000 byte

Thi gian truyn d liu ca khi = (8000 20.000) x 18,75 = 7,5 mili giy.

p n cu 42: C

Cng thc:Thi gian quay mt vng (mili giy/vng) = 60 x 1000 tc quay (vng/pht)

Thi gian tm kim ti a (mili giy) = thi gian quay mt vng

Thi gian tm kim trung bnh (mili giy) = thi gian quay mt vng : 2

Gii thch: Thi gian quay mt vng = 60 x 1000 3000 = 20 (mili giy/vng)

Thi gian tm kim ti a = 20 (mili giy)

Thi gian tm kim trung bnh = 20 2 = 10,0 (mili giy)

Nu ta cng gi tr trn vi thi gian nh vu t vo rnh v thi gian truyn d liu, ta

c thi gian truy nhp.





72/130

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p n cu 43: E

Cng thc:

Dung lng lu gi = s mt lu gi x s rnh trn mt mt x s sector trn mt rnh x

dung lng lu gi ca mt sector

Gii thch: Dung lng lu gi = 2 x 80 x 18 x 1024 = 2 949 120 byte 2,88 MB

p n cu 44: E

Bn ghi vt l: phn ghi d liu. D liu c ghi vi tc khng i.

Bn ghi lgc: l tp hp ca nhiu d liu. Trong mt khi c rt nhiu bn ghi lgc.

BOT beginning of tape: im bt u ca bng t.

EOT end of tape: im kt thc ca bng t

Gii thch: IBG l phn khng lu d liu gia cc khi (bn ghi vt l). Khi ghi d liu,

ti y s xut hin vic tng gim tc , nn n lm nh hng khng ch ti hiu qu

lu gi m cn nh hng ti hiu sut thi gian.

p n cu 45: B

Cng thc:

di bng t lu mt tp = s khi x ( di khi + di IBG)

di khi (inch) = h s khi x di bn ghi lgcmt lu gi

Gii thch: Thay s vo cng thc trn, ta c:

di ca mt khi = 12 x 500 4800 = 1,25 inch. Do s khi l 1000, nn di cn

thit ca bng t = 1000 (1,25 + 0,25) = 1500BPI c ngha l s byte trn mt inch (byte/inch).

p n cu 46: D

POLLing: vic ln lt hi im danh

BLOCKing: vic tng hp nhiu bn ghi l gc vo trong mt khi (bn ghi vt l) ng

khi

SWAPPing: thay ni dung trong b nhch v b nhph tng ng vi mc u tin

ca chng trnh, ri x l. Tro iSPOOLing: SPOOL (simultaneous peripheral operations online): mt phng php

tng thng lng.

Queueing: d phng d liu trong phng php FIFO, xp hng

p n cu 47: C

My tnh c 5 chc nng v ln lt gi tng ng l:

(1) Thit b vo

(2) B nh

(3) B s hc

(4) Biu khin





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(5) Thit b ra

Gii thch: Biu khin + b s hc = B x l (processor), b x l c chc nng iu

khin vic tnh ton v x l

p n cu 48: E

im chnh

Cc mi tn cho bit chu trnh iu khin u xut pht t hp A. Thit b chnh l b

iu khin

p n cu 49: B

CSW (Channel State Word): L ngn ng knh thng bo cho b x l bit tnh hnh

thc hin ca knh (tnh hnh cho bit vic nhp, xut kt thc hay cha).

CAI: h thng dy v hc bng my tnh.

CCW (Channel Command Word): l ngn ng to thnh chng trnh trn knh (channel

program), chnh l ch thi vi knh.

CAM: Computer Aided Manufacturing): sn xut vi s h trca my tnh

Gii thch: CAW (Channel Address Word): cho bit a ch ca chng trnh trn knh.

Sau khi thng tin ny c ghi vo, b nhs ra ch th bt u cho knh v cho tin hnh

x l nhp xut.

p n cu 50: AOCR (Optical Character Reader): thit bc k t quang hc. C thc ch vit tay

hoc ch in.

Bng: thit b vo ta dng phng, c s dng trong CAD

My qut nh: l thit b nhp hnh v trn giy, nh, tranh vo my tnh di dng d

liu nh.

My nh s: l camera lu tr hnh nh trong b nhbn dn.

Gii thch: OMR (Optical Mark Reader: thit bc im quang hc) c kh nng c

nhng phn c nh du bng bt ch trong khu vc chnh.

B s hc

B nh Thit b raThit b vo

lung dliu lungiu khin

B iu khin





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p n cu 51: D

Tivi mn hnh phng: khng phi l thit b vo ca my tnh.

Qu cu nh du: thit b nhp ta da vo vic di chuyn con tr (cursor) trn mn

hnh bng cch xoay qu bng c lp trong thit b.

B s ha: thit b vo ta bng cch, cho con tr (pointer) hay bt chuyn dng di

chuyn trn mn phng ngang. C ngha gn ging nh bng.

Bt quang: bt chuyn dng nhp ta trn mn hnh CRT.

Gii thch: Mn hnh cm gic s vo ta khi ta chm trc tip vo mn hnh. c s

dng ti h thng cho nhng ngi khng quen vi thao tc my tnh.

p n cu 52: E

IDE: c s dng vi mc ch tng t SCSI, nhng ch c th kt ni ti a 2 thit b

(dng mrng c th ln ti 4 thit b).

Centronics: s dng kt ni my in, ch khng kt ni c chui cnh hoa(daisy-chain).

IOBASE-T: h thng cp xon i ca Ethernet. Kt ni kiu dng sao t HUB.

SCSI: c tc dng kt ni chui cnh hoa (daisy-chain) v kt ni cc b nhh tr.

C th kt ni ti a 8 thit b.

Gii thch: GPIB l giao din dng kt ni my o c do cng ty Hewlett-Packard

pht minh ra. C th kt ni chui cnh hoa (daisy-chain) vi s lng ti a l 15 bng

cch truyn song song 8 bit.

p n cu 53: D

Mn hnh CRT: mn hnh s dng ng tia in t.

Mn hnh tinh th lng LCD: l mn hnh s dng tinh th lng, c kiu TFT hay DSTN

v.v.

Mn hnh Plasma: Mn hnh s dng hiu ng pht quang ca plasma. C th lm mn

hnh mng.

Mn hnh Multi-scan: mn hnh c th s dng nhiu phn gii khc nhau.

Gii thch: i vi GUI, bt buc phi trnh by ha di dng ca s (window) hocbiu tng (icon). Ngoi k t ra, mn hnh k t rt kh hin th cc i tng khc, do

vy khng thch hp.

p n cu 54: A

Cng thc: sim nh (dot) mn hnh ca VGA = 640 x 480 = 307200 dot

S mu ca High Color = 216 mu = 65.536 mu

Lng thng tin trn mt dot ca High Color = 16 bit = 2 byte

Gii thch: lng thng tin trn mt mn hnh = 307200 (dot) x 2 byte = 614400 byte =

600 KB





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p n cu 55: E

My in kim: to hnh k t bng cch, cc u kim (pin) trn u my in (printer head)

nh ra chc vo bng mc hnh thnh du mc trn giy.

My in phun: to hnh k t bng cch phun mc t vi (nozzle) trn u my in (printer

head) ln giy.

My in nhit: u my in (printer head) c un nng, lm nng chy mc ca bng

mc dnh ln trn giy.

My in dng: in thnh dng vi nhiu k t cng mt lc.

Gii thch: my in laze dng nhit v p sut in mc dnh trn trng bng tnh in ln

giy.

p n cu 56: A

Bng mch chnh: l bng mch ca my tnh. c lp vo CPU hay b nh

10BASE 2: l mt chun ca Ethernet. S dng cp ng trc v b kt ni BNC.Ethernet: l mt loi h thng cp mng do Gio s Metcalfe ca Xerox pht minh ra

Serial Cable: l loi cp trong s dng cc dy cp ring bit cho gi tin v nhn tin ,

ri gi v nhn d liu mt cch lin tc.

Gii thch: Nhc bus, m tnh c lp ca cc thit b tng ln v chi ph pht trin

cc thit b cng r hn. C bus a ch, bus d liu, bus iu khin

p n cu 57: B v C

cpi (character per inch) cho bit cht lng in. S k t tng ng vi mt inch.cps (character per second) l n v th hin tc in. S k t in c trong mt giy.

ppm (page per minute) l n v th hin tc in ca my in. N cho bit c th in c

bao nhiu trang trong vng 1 pht.

dpi (dot per inch): n v th hin phn gii ca my in. Cho bit s dot in tng ng

vi mt inch.

bps (bit per second) l tc truyn thng tin. S bit gi trong mt giy.

p n cu 58: B.Siu my tnh: l loi my tnh c thit k trong coi trng tc tnh ton.

My trm : l loi my tnh bn c tnh nng cao v cc k s thng s dng loi my

ny.

My tnh a nng: l loi my tnh c s dng rng ri trong nhiu lnh vc, n cn

c gi bng nhng ci tn khc nh my tnh ln (mainframe) hay my ch (host

computer).

My tnh n-ron: l loi my tnh m phng h thng thn kinh ca sinh vt.

Gii thch: nh vt l hc John Von Neumann pht minh ra H thng chng trnh lu

tr v cng b vo nm 1945.





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p n cu 59: E

Overhead: l thi gian m chng trnh iu khin OS s dng. Khng lin quan ti nng

sut cng vic ca ngi dng.

Thi gian p ng: thi gian tnh t khi gi xong d liu t thit b cui my tnh cho ti

khi bt u gi tr li kt qu.

Thi gian quay vng: thi gian k t khi bt u nhp cng vic (d liu) vo my tnh

cho ti khi c c kt quu ra mt cch hon chnh.

Thi gian ri ca my tnh: khi thc hin cng lc nhiu nhim v (task) khc nhau, s

xut hin tnh trng thi gian vo-ra (I/O) b rt ngn. N l thi gian CPU khng hot

ng. Thi gian CPU ngh.

Thng lng: Thi gian cn thit x l mt lng cng vic, mt lng d liu hay

mt cng vic nht nh trong mt n v thi gian.

p n cu 60: B v C

Supervisor call: l vic chng trnh ang chy s ra lnh ngt i vi chng trnh gim

st.

Bn ghi: s lu li tnh hnh hot ng ca h thng. N cn c gi vi mt ci tn

khc l S nht k. Ngi ph trch qun l vn hnh s khng th thiu bn ghi khi

mun phng trnh li, tm hiu nguyn nhn v khi phc.

PSW (program status word): thng tin cho bit tnh trng ca chng trnh bng my tnh

a chc nng.Cun d liu: to b nhph trgi l b nhm (buffer), vic x l vo-ra c thc

hin mt cch tch ri vi b x l.

p n cu 61: C

Windows XP hin nay l hiu hnh ca cng ty Microsoft c s dng nhiu nht

trong cc my tnh.

Linux: l hiu hnh my tnh c pht trin trn csca Unix.

MS-DOS: l hiu hnh my tnh tiu chun trc khi Windows ra i. Thc hin thaotc ch yu bng lnh (command).

OS/2: hiu hnh my tnh ca cng ty IBM.

Gii thch: MVS l hiu hnh my tnh a dng cln do IBM pht minh, n khng

chy trn my tnh c nhn.

p n cu 62: D

Job Management: l vic tin hnh cc thao tc nh x l trc, x l tun t hay x l

sau cn thit thc hin mt cng

exam fehandbook so tay on tap sat hach fe

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