cau hoi on tap_pttkht(1)

8/6/2019 Cau Hoi on Tap_pttkht(1)

1/63

n tp Phn tch Thit k H thng DH07DT

[email protected] Page 1

CU HI N TPPart 1

1. Tnh cht no sau y l khng phi l ca use caseA. Ph thuc vo vic thc thi h thng ( implementation-independent):B. Th hin chc nng ca h thng theo quan im ca ngi dngC. Cho xem h thng mc cao ( high-level view of the system)D. Ch tp trung vo nhu cu mong i ca user, khng quan tm n cc bc phi

thc hin2. Cc s no th hin kha cnh ng ca h thng:

A. use case, class, object, component, deployment

B. class, object, component, deployment

C. sequence, use case, class, statechart, activityD. sequence, collaboration, statechart, activity

3. UML l g?A. L mt ngn ng lp trnh gn ging vi Java B. L mt ngn ng c bit dng to m t ng cho cc project C. L cng c dng kim tra s ng nht gia m hnh v m ngunD. C 3 u sai

4. Vic du thng tin trong lp trnh hng i tng (OOP) c thc hin trong khinim ca:

A. EncapsulationB. Polymorphism

C. Abstraction

D. Inheritance

5. Xem xt m hnh Activity sau.Hy xc nh nhng hat ng no c th xy ra ng thi?


2/63



a. Request product, Receive order, Pay bill.b. Receive order, Bill customerc. Process order, Pull materials, Ship order, Bill customer, Pay bill.d. Pay bill, Close order.e. Tt c cc hot ng trn c th xut hin ng thi.

6. Chn cu m t v tc nhn ng nht trong m hnh use case: a. Tc nhn b gii hn i vi ngi dng tng tc vi h thng ang c thit kb. Tc nhn l bt k thc th no bn ngai h thng ang c thit k m

tng tc vi h thngc. Tc nhn b gii hn i vi cc h thng m tng tc vi h thng ang c thit

k.d. Tc nhn cung cp d liu cho h thng.

7. Dng i hat ng (life line):a. Hnh ch nht hp ng nhn mnh rng mt i tng ch hat ng trong sut

phn kch bn trong m hnh sequenceb. Thng ip


3/63



c. ng thng ng di mt i tng trong m hnh sequence ch r khang thigian tri qua ca i tng

d. Thi gian sng ca i tng.8. Cu no ng (mt hoc nhiu) khini v trng thi v s chuyn trng thi trong UML

a. M hnh th trng thi ch c to cho cc lp c hnh vi ng quan trngb. M hnh th phi c to cho tt c cc lp c xc nh cho vn tinguync. Nhng s kin gy ra s chuyn trng thi t trng thi ny sang trng thi

khc c th hin trong m hnh th trng thi d.Nhng hnh ng do vic thay i trng thi gy ra cng c th hint trong m

hnh th trng thi

9. Trong hnh (i) l

a. M rngb. Bao gmc. Lin ktd. S tng qut ha

10.S khc nhau gia cc actor v cc object cng tn la. Actor bn ngoi cn object bn trong h thngb. Object bn ngoi cn actor bn trong h thngc. Object c hnh vi cn actor th khng c.d. Actor c hnh vi cn object th khng c.

11.Di y l lp Circle c v bng UML cho bit isVisible (hnh vi)) l


4/63



a. protectedb. privatec. publicd. inherit

12.Mt mi quan h has a l m t tt nht cho:a. Polymorphismb. Compositionc. Aggregationd. Inheritance.

13.Vic o sc mnh ca s kt ni gia hai thnh phn h thng c bit nh l:a. Couplingb. Cohesionc. Aggregationd. Bonding.

14.Mi quan h (relationship) gia class 10 v class 11 lNewClass10 NewClass11

a. Dependencyb. Aggregationc. Compositiond. Realization

15.Mi quan h (relationship) gia class 12 v class 13 lNewClass12 NewClass13

a. Dependencyb. Realization


5/63



c. Compositiond. Aggregation

16.S khc nhau trong m hnh ho use case nghip v v m hnh ho use case h thng:

A. Khng c s phn bitB. C th thc hin mt trong haiC. Cc tc nhn (actor) trong hai m hnh c xc nh khc nhauD. Cc tha tc vin (worker) trong hai m hnh c xc nh khc nhau

17.1S no sau y khng phi l s ca UML :A. Component diagram

B. State-chart diagramC. Deployment diagram

D. Relationship diagram

18.Loi no sau y l khng phi l actora. Khch hngb. H thngtn dngc. ng h h thngd. Tt c u sai

19.Mi message trong lc sequence s c nh x thnha. Mt thuc tnh (attribute)b. Mt quan h ph thuc (dependency relationship)c. Mt operationd. Tt c u sai

20.S kt hp gia tn ca mt thao tc (operation) v cc tham s (parameter) ca nc xem l _____________ ca thao tc .a. Argumentsb. Signaturec. Responsibilitiesd. Attributes

21.Ci no khng phi l m hnha. Concordeb. M hnh thu nh ca Concorde dng trong ng hmc. Bn v ct lp phn thn ca Concorded. C 3 u l m hnh


6/63



22.K hiu ny l:Class1 Class2

a. Generalizationb. Aggregationc. Composited. Association

23.Mi quan h gia Class v Interface l mi quan hClass

Interface

a. Unidirectional Associationb. Associationc. Realized. Dependency

24.Trong RationalRose, kim tra li tnh tng thch ca ton b m hnh, dng lnh:a. Tool/Check Modelb. Tool/Check Syntaxc. Tool/Option/Check Modeld. Tool/Customize

25.Cc mu thit k phn thnh my loi mu da vo mc ch s dng a. 2b. 3c. 4d. 5

26.on m sau l m sinh t m hnh no? public class Class1

{public Class2 theClass2;

public Class1()

{


7/63



}}

public class Class2

{

public Class2()

{

}

}

a.

Class1 Class2+theClass2

b.

Class1 Class2

0..*

+theClass2

0..*

c.

Class1 Class2+theClass2

zzzzzzzz

d.

Class1 Class2

11

+theClass2

1

Part 2

1) S no sau y khng phi l s ca UML : a. Component diagram

b. State-chart diagramc. Deployment diagram

d. Relationship diagram

2) Thnh phn no sau y khng l c tnh ca mt i tnga. Identity

b. Behaviour

c. Actiond. State


8/63



3) S ng gi c hiu l : a. S che du thong tinb. S t chc cc thnh phn ca mt s votrong mt gi (package)c. Vic xy dng mt lp cha da trn cc thuc tnh v cc hnh vi chung ca cc lp con d. Vic xy dng giao din gm tp cc hnh vi m ta mun s dng li nhiu ln trn m hnh.

4)

Mt lp c m t l tp cc i tng chia x cng cca. Attributes (thuc tnh), behaviour (hnh vi) and operations (hnh ng) b. Identity(c tnh), behaviour and state (trng thi)c. Attributes, operations and relationships (mi quan h)d. Relationships, operations and multiplicity (bn s)

5) Cc use-cases nghip v (Business use-cases) v cc tc nhn (actors) cng m t : a. Cc phn t tnh (static elements) ca cng vic (work) trong tin trnh (process) b. Cc phn t ng (dynamic elements) ca cng vic (work) trong tin trnh (process)c. Khung nhn lun l (logical view) ca cng vic (work) trong tin trnh (process) d. Cc tin trnh nghip v (business processes) m t chc h tr.

6 Cu pht biu no sau y khng ng : :a. Mi use case nghip v (business use case) trong m hnh nghip v (business model) c

chuyn thnh mt h thng con trong m hnh phn tch (analysis model)b. Mi business worker trong m hnh nghip v (business model) c chuyn thnh mt tc nhn

h thng (system actor) trong m hnh phn tch (analysis model)c. Mi thc th nghip v (business entity) trong m hnh nghip v (business model) c chuyn

thnh 1 lp trong m hnh phn tch. (analysis model)d. Mi m hnh i tng (object model) trong m hnh nghip v (business model) c

chuyn thnh s tng tc (interaction diagram) trong m hnh phn tch (analysismodel)

7 Cu pht biu no sau y ng :

a. Mt lp l s ng gi ca mt i tngb. Mt lo biu din s phn cp ca mt i tngc. Mt lp l mt th hin ca mt i tngd. Mt lp l mt nh ngha tru tng ca mt i tng

8 Tnh a hnh c th c m t nh l : a. Che du nhiu ci t khc nhau da trn cng mt giao din.b. Cc thuc tnh v phng thc khc nhau ca cc lp con c cng lp cha.

c. Cc lp kt hp (association class) vi rng buc {or}c. S tng qut ho (Generalization) cc lp con tha k

9 Cm t tt nht biu din mi quan h tng qut ho l : a. Is a part ofb. Is a kind ofc. Is a replica ofd. Is composed of


9/63



10. Mt lp con tha k t lp cha cc :a. Attributes, links

b. Attributes, operations

c. Attributes, operations, relationshipsd. Relationships, operations, links

11. t chc cc phn t (elements) vo bn trong cc nhm (groups) ta s dng :a. Packageb. Classc. Class v interface

d. Component

12. Cc s no sau y m t cc hnh vi ng (dynamic behaviour) ca h thng phn mm?a. S lp v s i tngb. S Use-case v s lpc. S cng tc v s trin khaid. S tun t v s cng tc

13. Vng i ca 1 lp (Life cycle of a class) c trnh by bi :a. S cng tcb. S trng thic. S lpd. S trin khai

14. Trong s use-case, mt tc nhn c trnh by bi :

a. Mt vai tr l 1ngi, 1 thit b phn cng hoc h thng khc.b. Cng mt ngi dung thc hin nhiu hnh ng khc nhau.

c. Mt ngi dung, khch hng v khng quan tm n vai tr ca h. d. Mt h thng vt l hoc 1 thit b phn cng cng vi cc giao din ca n.

15. Cng vic u tin xc nh cc phn ca h thng v nhng quan h gia chng, t chc cc phnvo trong cc tng vi cc ph thuc xc nh c gi l :

a. Use-case analysis

b. Architectural analysisc. Structural analysis

d. Dependency analysis

16. Stereotype package c th biu din cho :

a. Mt giao dinb. Mt s trng thic. Mt tng kin trcd. Mt s use-case

17. ng sinh tn (lifeline) ca 1 i tng c trnh by trong s : a. S i tng


10/63



b. S trng thic. S tun td. S trin khai

18. Pht biu no sau y khng ng ?a. S m t ca cc use-cases tm v phn tch cc lp cng cc i tng ca n.

b. C t nht mt boundary object cho mi actor hay use-case pairc. C mt lp iu khin (control class) ng vi mi use-cased. Cc i tng thc th c nhn din bi vic xem xt cc danh t v cm danh t trong use -

cases

19. Pht biu no sau y ng ?

a. Khng c s hn ch no trn nhiu mi kt hp (multiple associations) gia cng 2 lp.b. C th c nhiu mi kt hp gia cng 2 lp, nhng chng phi mang cc ngha khc

nhau.c. Khng cho php biu din nhiu mi kt hp trn cng 2 lp. d. Cc mi kt hp gia cng 2 lp phi c tp hp li thnh 1 mi kt hp

20. Nu ta mun t chc cc phn t (elements) vo trong cc nhm c th s dng li c vi tt c ccthng tin c che du, ta c th s dng mt trong cc cu trc no ca UML :

a. Package

b. Class

c. Class hoc Interfaced. Subsystem hoc Componentb. C th c nhiu mi kt hp gia cng 2 lp, nhng chng phi mang cc ngha khc

nhau.c. Khng cho php biu din nhiu mi kt hp trn cng 2 lp.

d. Cc mi kt hp gia cng 2 lp phi c tp hp li thnh 1 mi kt hp

21. Pht biu no sau y khng ng :a. Ch nhngpublic classes mi c th c truy xut t nhng phn bn ngoi ca package.chan.

b. Khng tn ti lp ca cc h thng con (classes of a subsystem)c. Cc gi (Packages) tng (layer) thp hn c th ph thuc vo cc gi tng cao hn. d. Giai on thit k l s tinh ch ca giai on phn tch. N thm vo nhng chi tit c th cnhn thc trong giai on thit k.

22. Nu ta mun t chc cc phn t vo trong cc nhm c s dng li vi s che du thng tin y , ta c th s dng mt trong cc cu trc no ca UML ?

a. Package

b. Class

c. Class v interface

d. Subsystem hoc Component

23. Trong giai on no ca quy trnh pht trin phn mm, ta xc nh chi ph v thi gian ca d n, xcnh cc ri ro v mi trng h thng

a) Khi to (Inception)


11/63



b) Tinh ch (Elaboration)c) Xy dng (construction)d) Chuyn giao (transition)

24. Trong giai on no ca quy trnh pht trin phn mm, ta nh gi ri ro, cc thnh phn s dng,

a) Khi to (Inception)b) Tinh ch (Elaboration)c) Xy dng (construction)d) Chuyn giao (transition)

25. Trong giai on no ca quy trnh pht trin phn mm, ta xy dng h thng qua qu trnh gm nhiuvng lp theo quy trnh xon c, mi vng lp l mt d n nh. Bn s qun l ti nguyn, kim sot vthc hin ti u ho, hon thnh vic pht trin cc sn phm v cc t hnh phn ca sn phm, nh gisn phm ci t t cc tiu chun c tha thun.


26. Trong giai on no ca quy trnh pht trin phn mm, ta thc hin ci t h thng, th nghim snphm trin khai, thu thp cc phn hi t pha ngi dung, bo tr h thng


27. Mt .. l dy cc bc m t s tng tc gia ngi dung v h thnga) Kch bnb) Use case

c) Mc tiud) S kin

28. S no m t cc kiu ca cc i tng v cc mi quan h tnh khc nhau gia chng ? a) S lp

b) S tng tcc) S trng thi

d) S hot ng

29. Pht biu no sau y khng ng v mi kt hp trong s lp a) Mi kt hp biu din cc quan h gia cc th hin ca cc lpb) Mi kt hp l cc quy trnh m mt lp s thc hinc) Mi kt hp c th v hngd) Mi kt hp c th c c hai hng


12/63



30. Trong s tun t ---------> biu din :a) Thng ip (message )b) iu kin (condition)c) Lp (iteration)d) Xa i tng (deletion)

31. Trong s tun t [some_text] biu din :a) Thng ip (message )b) iu kin (condition)c) Lp (iteration)d) Xa i tng (deletion)

32. Trong s tun t du *biu din :a) Thng ip (message )b) iu kin (condition)c) Lp (iteration)d) Xa i tng (deletion)

33. Trong s tun t du Xbiu din :a) Thng ip (message )b) iu kin (condition)c) Lp (iteration)d) Xa i tng (deletion)

34. Mnh no sau y ng v gI thng ip khng ng b trong s tun t ? a) Cc thng ip khng ng b c th to mt lung mi, c th to mt i tng mi, v

c th lin lc vi cc lung khc ang chy. b) Cc thng ip khng ng b c th to mt i tng mi nhng khng th to mt lung

mi v khng th lin lc vi cc lung khc ang chy.c) Cc thng ip khng ng b khng th to mt i tng mi nhng c th to mt lung

mi v c th lin lc vi cc lung khc ang chy.d) Cc thng ip khng ng b c th to mt i tng mi, c th to mt lung mi nhng

khng th lin lc vi cc lung khc ang chy.

35. biu dinrng bt k mt th hin no ca lp cha (lp tru tng) phi l mt th hin ca mttrong nhng lp con ca n, ta s dng ?

a) {complete}b) [complete]c) (complete)d)

36. Vi mi quan h no mt thnh phn thuc v duy nht mt tng th. Cc thnhphn ny lun luntn ti v mt i cng vI tng th ?

a) aggregationb) Compositionc) Classification


13/63



d) Generalization37. Pht biu no sau y ng v mi kt hp nh tnh (Qualified Associations) trn s lp ?

a) Mi kt hp nh tnh cung cp chc nng tng t nh ch mc. b) c tnh thng tin nh danh (qualifier) l mt thuc tnh ca mt lp. Biu tng ca n l

hnh ch nht nh k bn class m thc hin vic tm kim

c) Mi kt hp nh tnh n gin ha s nh hng qua mi kt hp phc tp bng cch cungcp cc kha thu hp vic la chn cc i tng kt hp.d) Cc cu trn u ng

38. Pht biu no sau y ng v mi kt hp nh tnh (Qualified Associations) trn s lp ? a) Mi kt hp nh tnh xc nh s lng cc i tng.c th tham gia trong mi kt hpb) Mi kt hp nh tnh cha cc lut bt buc phi m bo cc rng buc ca mi quan h.c) S dng cc thng tin nh danh (qualifier) gim mt bn s one-to-many thnh bn

s one-to-one.d) Cc cu trn u ng

39) Pht biu no sau y ng v lp kt hp trn s lp ?a) N cho php ngi thit k thm vo cc thuc tnh, hnh vi, cng nh nhng tnh cht khc

ca mi kt hp.b) Mt lp kt hp ng gi thng tin v mi kt hpc) Mt lp kt hp c ni n mi kt hp bng ng t nt. (dashed

line)

d) Tt c cc cu trn u ng 40) Pht biu no sau y ng v lp kt hp trn s lp ?

a) Mt lp kt hp c ni n mi kt hp bng ng lin nt. (solid line) b) Mt lp kt hp c ni n mi kt hp bng ng chm .(dotted line)c) Mt lp kt hp cha cc lut bt buc phi m bo cc rng buc ca mi quan h.d) Khng c pht biu no ng

41) Pht biu no sau y ng v s trng thi ?a) S trng thi m t tt c cc trng thi m mt i tng c th c v s chuyn dch

ca cc trng thi nhl kt qu ca cc s kin. b) S trng thi m t hnh vi ca nhiu i tng trong cng mt Use Case.c) S trng thi m t cc loi i tng trong h thng v cc loi mi quan h khc nhau

gia chng.d) S trng thi th hin nhiu thnh phn trong h thng v cc ph thuc ca chng.

42) Hon chnh cu sau v thng tin (artifact) trong s trng thi :Mt . lin quan n mt s chuyn dch v c xem nh mt tin trnh xut hin nhanh v khngb ngt bi mt s s kin.

a) Hnh ng (Action)b) Hot ng (Activity)c) iu kin che chn (Guard)d) S kin (Event)


14/63



43) Hon chnh cu sau v thng tin (artifact) trong s trng thi :Mt . lin quan vi mt trng thi v c th din ra trong thi gian di. N c th b ngt bi mts s kin.

a) Hnh ng (Action)b) Hot ng (Activity)c) iu kin che chn (Guard)d) S kin (Event)

44) Hon chnh cu sau v thng tin (artifact) trong s trng thi :Mt . l mt iu kin lun l v n s tr v chtrue hay false.

a) Hnh ng (Action)b) Hot ng (Activity)c) iu kin che chn (Guard) d) S kin (Event)

45) Mnh no sau y ng v s trng thi ?

a) S trng thi l cch biu din tt hnh vi ca mt i tng qua nhiu Use Cases.b) S trng thi l cch biu din tt hnh vi ca nhiu i tng qua nhiu Use Cases.c) S trng thi l cch biu din tt hnh vi ca nhiu i tng trong mt Use Cases.d) S trng thi l cch biu din tt dy cc hnh ng cho nhiu i tng v Use Cases.

46) Hon chnh cu sau : l cch biu din tt m t hnh vi ca mt i tng qua nhiu Use Cases. N rt tt

m t hnh vi lin quan n mt si tng hp tc vi nhau.a) S trng thi (State Diagrams)b) S tun t (Sequence Diagrams)c) S hp tc (Collaboration Diagrams)d) S hot ng (Activity Diagrams)

47) Hon chnh cu sau : l cch biu din tt m t hnh vi ca nhiu i tng trong mt Use Casea) S trng thi (State Diagrams)b) S tng tc (Interaction Diagrams)c) S hot ng (Activity Diagrams)d) S lp (Class Diagrams)

47) Hon chnh cu sau : l cch biu din tt th hin dy cc hnh ng cho nhiu i tng v Use Case

a) S trng thi (State Diagrams)b) S tun t (Sequence Diagrams)c) S hp tc (Collaboration Diagrams)d) S hot ng (Activity Diagrams)

48) K hin * trong UML biu din :a) Biu din cc bc lp li m khng c cu trc vng lpb) N ch ra rng cc hot ng c thc hin nhiu ln


15/63



c) Biu din nhiu hot ng cn cng c thc hin trong mt vi trng thi.d) Cu a v b ng

49) S hot ng c s dng trong nhng tnh hung sau :a) Phn tch mt use caseb) M t thut ton tun t phc tp, xy dng lu c) Lin quan n cc ng dng a lungd) Tt c cc cu trnu ng

50) S hot ng c s dng trong nhng tnh hung sau :a) Biu din cc i tng cng tc vi nhau nh th no.b) Biu din cc hnh vi ca i tng qua thi gian sng ca chng.c) Biu din iu kin logic phc tpd) Tt c cc cu trnu sai

51) biu din cc quan h vt l gia phn mm v cc thnh phn phn cng trong mt h thng bns dng s no ca UML ?

a) S trin khai (Deployment Diagram)b) S hot ng (Activity Diagram)c) S lp (Class Diagram)d) S trng thi (State Diagram)

52) Mc ch ca s hot ng l :a) Biu din hnh vi vi cu trc iu khin. S hot ng c th biu din nhiu i

tng trong mt use case.b) Biu din cu trc tnh ca cc khi nim, cc loi v cc lpc) Gip nm c mc ch cbn ca lp, tt cho vic khm ph vic ci t use case nh th

no.

d) Biu din cch b tr cc thnh phn trn cc nt phn cng52) Mc ch ca s lp l :

a) Biu din hnh vi vi cu trc iu khin. S hot ng c th biu din nhiu i tngtrong mt use case.

b) Biu din cu trc tnh ca cc khi nim, cc loi v cc lpc) Gip nm c mc ch cbn ca lp, tt cho vic khm ph vic ci t use case nh th

no.

d) Biu din cch b tr cc thnh phn trn cc nt phn cng53) Mc ch ca s trin khai l :

a) Biu din hnh vi vi cu trc iu khin. S hot ng c th biu din nhiu i tngtrong mt use case.b) Biu din cu trc tnh ca cc khi nim, cc loi v cc lpc) Gip nm c mc ch cbn ca lp, tt cho vic khm ph vic ci t use case nh th

no.

d) Biu din cch b tr cc thnh phn trn cc nt phn cng 54) Mc ch ca k thut thit k bng hp ng "Design by Contract" l :


16/63



a) Cung cp nh ngha cht ch mc ch cc hnh vi v trng thi hp l ca lp. b) Biu din cc i tng cng tc trong mt use case nh th no.c) Biu din nhm cc lp v cc ph thuc gia chngd) Cung cp mt vi k thut hu ch cho phn tch, thit k v vit m.

55) Mc ch ca s tng tc (Interaction Diagram) l :

a) Cung cp nh ngha cht ch mc ch cc hnh vi v trng thi hp l ca lp.b) Biu din cc i tng cng tc trong mt use case nh th no. c) Biu din nhm cc lp v cc ph thuc gia chngd) Cung cp mt vi k thut hu ch cho phn tch, thit k v vit m.

56) Mc ch ca s gi (Package Diagram) l :a) Cung cp nh ngha cht ch mc ch cc hnh vi v trng thi hp l ca lp. b) Biu din cc i tng cng tc trong mt use case nh th no.c) Biu din nhm cc lp v cc ph thuc gia chng d) Cung cp mt vi k thut hu ch cho phn tch, thitk v vit m.

57) Mc ch ca s trng thi (State Diagram) l :a) Biu din mt i tng thay i trng thi qua nhiu use cases. b) Biu din nhiu i tng thay i trng thi qua nhiu use cases. c) Biu din cu trc tnh ca cc khi nim, cc loi v cc lpd) Gip cung cp mc ch chnh yu ca lp

58) Mc ch ca Use Case l :a) Ch ra nhng yu cu y ngha t ngi dng . b) Cung cp mt vi k thut hu ch cho phn tch, thit k v vit m. c) L nn tng cho vic kim tra h thng d) Cu a v c ng

59) Trong giai on xy dng (Construction phase), mi ln lp s bao gm :

a) Analysis, design, coding, testing, integrationb) Design, coding, testing, integrationc) Analysis, design, coding, testingd) Design, coding

60) Hon chnh cc mnh sau v mi quan h gia cc Use CasesS dng khi chng ta mun gim cc bc trng lp gia cc use case, ly nhng bc

chung to nn use case ph.

a) Includeb) Generalizationc) Extendd) Delegation

61) Hon chnh cc mnh sau v mi quan h gia cc Use CasesS dng khi chng ta mun to mt use case mi bng cch thm mt s bc vo mt

use case c sn.


17/63



a) Includeb) Generalizationc) Extendd) Delegation

62) Pht biu no sau y ng v use cases ?

a) Use Cases biu din mt cch nhn bn ngoi (external view) ca h thngb) C s tng quan gia Use Cases v lp bn trong h thng c) Khng c s tng quan gia Use Cases v lp bn trong h thng d) Cu a v c ng

63) Pht biu no sau y ng v S tun t ? a) Mi thng ip c biu din bng mt mi tn gia ng sng ca hai i tng.b) Mi thng ip c biu din bng mt ng thng ng t nt c) Mi thng ip phi c nhn vi tn thng ip i km mt con sd) Cu a v c ng

64) S no biu din cc th hin trong mt h thng ti mt thi im ? a) S i tng (Object Diagram)b) S lp (Class Diagram)c) S thnh phn (Component Diagram)d) S h thng (System Diagram)

65) Pht biu no ng v s hot ng ?a) Mt phn nhnh c mt hot ng n v nhiu hot ng i theo iu kin. b) Mt s hp nhtng b ca cc lung ng thi. c) C 2 cu trn u ngd) C 2 cu trn u sai

65) Pht biu no ng v sthnh phn (Component Diagrams) ?a) Mt thnh phn c th c nhiu hn mt giao din, n trnh by mt n v (module) vt

l ca m lnhb) Mt thnh phn khng th c nhiu hn mt giao din. c) Mt thnh phn trnh by vi loi ca n v phn cngd) Cu a v c ng


18/63



66) Pht biu no ng v Use Cases ?a) Nhng yu cu l nhng ghi nhn trc tin trong Use Cases b) Cng vic ca mt ln lp c xc nh bi vic chn la mt s kch bn Use Cases hay ton

b Use Cases.c) Cng vic ca mt ln lp c xc nh bi vic chn la mt s c trng t danh sch cc

c trng hn l vic chn la mt s kch bn Use Cases hay ton b Use Cases. d) Cu a v b ng

67) Kt qu ca giai on tinh ch (Elaboration phase) l :a) Cc yu cu chc nng (Use Cases)b) Nn tng cho vic nh gi chi ph n cui d n. c) Cc yu cu phi chc nng trong bng chi tit b sung.d) Tt c cc cu trn

68) Cc khi nim v phm vi h thng c tm thy trong : a) Use Casesb) S Use Casesc) S tng tc (Interaction Diagrams)d) S tun t (Sequence Diagrams)

69) S no l thng tin (artifact) hu ch nht trong vic phn tch tm ra cc hnh vi ca phn mmc xem nh l hp en ?

a) S tun t h thng (System Sequence Diagram) b) S lp quan nim (Conceptual Class Diagram)c) S trin khai (Deployment Diagram)d) S thnh phn (Component Diagram)

Part 3

1. S tch bit gia h thng v mi trng ca n l mt v d ca: a. System boundary b. Automation boundary.c. Production boundary.d. Information boundary.

2. Mi quan h ngha l g? a. Ging nh mt th tc con, n l mt use case c s dng bi mt use case khc b. Vai tr c th hin bi mt ngi c th khi ngi tng tc vi h thng. c. Nhn mnh s tun t hoc th t cc thng ip. Mt collaboration diagram lm ni bt tp hp

cc i tng cng tc cng nhau thc hin mt use case.d. Gip nh r phm vi h thng bng cch nhn dng cc vai tr ca tc nhn (actor) tng tc vi

h thng v mt tp hp cc quyn v chc nng c cung cp cho cc tc nhn .


19/63



3. iu no sau y khng phi l chc nng ca giai an cn thit cho xy dng mt h c s d liu? a. nh r mu tin d liu (data item).b. Xc nh cc rng buc v quy tc (rule).c. Xy dng m hnh d liu.d. Tt c cc cu trn l chc nng ca giai an cn thit

4. Khi xy dng mt m hnh d liu, mc ch chnh l to ra mt m hnh tt nht theo ngi dng. a. ngb. Sai

5. Mt use case diagram ls m rng ca:a. Bng s kin b. M hnh ng cnh (context).c. M hnh tun t (sequence).d. M hnh class.

6. M hnh Collaborationa. M hnh th hin s tun t ca cc thng ip gia cc i tng trong sut mt use case b. M hnh th hin cc i tng cng tc cng nhau thc hin mt use case c. M hnh collaboration hoc m hnh sequence th hin s tng tc gia cc i tng d. Mt m hnh th hin cc vai tr khc nhau ca user v cch thc cc vai tr s dng h thng

7. S thnh cng ca mt h thng thng tin ph thuc gn nh han tan vo cht lng ca n khic xem xt bi ____________a. phn mmb. th tcc. thng tind. ngi dng

8. Use Case:a. Mt kch bn.b. Mt vai tr c thc hin bi ngi dng bn ngai h thng. c. Mt cng vic hoc chc nng n l c thi hnh bi h thng cho nhng ngi dng h thng

d. Mt bc trong chui s kin

9. Xem xt m hnh sau:Gi s mi kt hp tham gia (enrolment) gia Student v Subject c nh ngha.

Chn bn s thch hp cho m1,m2,m3,m4.


20/63



a. *,1,*,1b. *,1,1,*c. 1,*,*,1 d. 1,*,1,*

10.Nhng ci no sau y l thnh phn ca UML? a. M hnh Activity.b. M hnh Collaboration.c. M hnh Sequence.d. M hnh Use-case.e. Tt c

11.______________ c dng nhm cc lp li vi nhau d s dng, bo tr v s dng li. a. Trng thi (States)b. Trng hp s dng (Use Cases) c. M hnh Contextd. Gi (Packages)


21/63



12.Xem xt m hnh Activity sau.Hy xc nh nhng hat ng no c th xy ra ng thi?

a. Request product, Receive order, Pay bill.b. Receive order, Bill customer c. Process order, Pull materials, Ship order, Bill customer, Pay bill.d. Pay bill, Close order.e. Tt c cc hot ng trn c th xut hin ng thi.

13.Nhng l do hp l nhm cc class vo trong cng mt package l, cc class : a. h tr cc dch v cng cp cao. b. c quan h vi nhau theo s chuyn mn ha (specialisation)c. c quan h vi nhau theo khi kt hp (aggregation).d. lin quan v mt ng ngha v c khuynh hng cng nhau thay i e. c xy dng bi cng nhm cc nh pht trin

14.S sau cho thy cc tnh trng ca vic ng k ca sinh vin trong h thng HND. Cu no sauy khng c th hin trong s ? a. Ch nhng sinh vin (student) trong trng thi active mi c th kt thc kha hc v tr thnh mt

HND tt nghip (graduate)b. Mt sinh vin ch c php bo lu mt ln trong kha hc


22/63



c. Mt sinh vin vi tnh trng ng k ang hat ng c th thay i tnh trng bng vic bo lu,b hc, hoc han thnh kha hc.

d. Khi mt sinh vin trong tnh trng bo lu quay tr li kha hc, s ng k ca h hat ng trli.

15.Trong vng i ca phn mm, giai an no ni chung l ng gi nht? a. S bo tr b. S thc thic. Cc lut ld. Thit k

16.M hnh Use Case:a. M hnh th hin cc i tng cng tc cng nhau thc hin mt use case. b. M hnh th hin dng i ca i tng trong cc trng thi v chuyn trng thi. c. S giao tip gia cc i tng trong mt use case d. M hnh th hin cc vai tr ca user v cch thc cc vai tr s dng h thng e. M hnh thhin s tun t ca cc thng ip gia cc i tng trong use case.

17.Chn cu m t v tc nhn ng nht trong m hnh use case: a. Tc nhn b gii hn i vi ngi dng tng tc vi h thng ang c thit k b. Tc nhn l bt k thc th no bn ngai h thng ang c thit k m tng tc vi h thng c. Tc nhn b gii hn i vi cc h thng m tng tc vi h thng ang c thit k. d. Tc nhn cung cp d liu cho h thng.

18.Khi chn gi phn mm, phi xem xt:i) Gi phn mm c ph hp vi yu cu ngi dng khng? ii) Mc ng tin ca nh cung cpiii)Nh pht trin h thng c phi l mt t chc n nh, ng tiniv)Gi phn mm c n nh khng hoc n c c sa li (duyt li) ln cui? v) Mc tch hp vi cc phn mm/ ng dng khc


23/63



vi)S h tr sn cvii) i ng nhn vin sn sng.

a. i, iv, vb. ii, iv, v, vic. i, iv, v, viid. i, ii, iiie. Tt c cu trn

19.Thuc tnh l th ta c th nhn thy, chm c hoc nu khng th cm nhn c v l th mngi dng lu tr d liu v nhng hnh vi lin quan. a. ng b. Sai

20.i tng tp hp l g?a. L mt thc th i tng cha cc i tng khc.b. L mt thc th i tng ch c nhng phng thc tnh. c. L mt i tng ch c nhng thuc tnh ch yu v cc phng thc th hin d. L mt thc th i tng ch c nhng thuc tnh ch yu

21.Kch bn tan cnha. Cc kch bn c pht trin c xem xt v u tin theo nhng iu kin nht nh b. Hng suy ngh v cc vn bng cch dng m hnh t chc cc han cnh trong th gii thc c. Xy ra trong sut giai an phn tch ca vng i h thngd. m bo khng to ra nhng gi thuyt v quyt nh khc nhau trong nhng khung nhn khc

nhau trong vic pht trin phn mm 22.Trong m hnh trng thi, hnh vi m i tng th hin ra trong khi ang trong trng thi c th

c gi l mt Activitya. ng b. Sai

23.S khc nhau gia m hnh sequence v m hnh collaboration? a. Gip nh r phm vi h thng bng cch nhn dng cc vai tr tc nhn (actor roles) tng tc

vi h thng v mt tp hp cc quyn v chc nng c cung cp cho cc tc nhn b. Nhn mnh s tun t hoc th t cc thng ip c. Mt m hnh collaboration nhn mnh tp hp cc i tng cng tc cng nhau thc hin mt

use case

d. Vai tr c thc thi bi mt ngi c th khi ngi tng tc vi h thng 24.Mt mi lin h gia hai lp trong m hnh lp

a. Th hin s k tha gia hai lpb. Cho php s tng tc gia hai lpc. Cho php s tng tc gia cc i tng ca hai lp d. Th hin s ph thuc gia hai lp

25.Giai an no sau y nh ngha cc quy tc kinh doanh cho mt h thng mi? a. Giai an phn tch h thng


24/63



b. Giai an phn tch yu cu c. Gii an nghin cu s bd. Giai an phn tchquyt nh

26.Dng i hat ng (life line):a. Hnh ch nht hp ng nhn mnh rng mt i tng ch hat ng trong sut phn kch bn

trong m hnh sequenceb. Thng ipc. ng thng ng di mt i tng trong m hnh sequence ch r khang thi gian tri qua

ca i tng d. Thi gian sng ca i tng.

27.Thng ip s kina. S tun t hp l ca vic chun trng thi v cc trng thi trong th trng thi b. ch n ca s chuyn trng thi c ni vi u mi tn ca biu tng chuyn trng thi c. Mt thnh phn trong th trng thi bo hiu s di chuyn t trng thi ny sang ci k d. S chuyn trng thi trong mt trng thi m khng loi b i tng khi trng thi e. S kch hot s thay i trng thi cha thng ip m c ccthuc tnh ca mt s kin

28.M hnh Collaboration th hin nhng tng tc ca i tng sp xp theo trnh t thi gian a. ngb. Sai

29.Mt thc th c vai tr c xc nh r trong ti nguyn ca ng dng v c trng thi, c hnh vi, vnh dng th xc nh:a. Lpb. Tc nhnc. i tng d. Thuc tnh

30.c cu sau:Mt ngn hng cho php khch hng c nhiu ti khan hoc khng c ti khan no, nhng khng cho

php xy ra ti khon chung, mt ti khan phi thuc v ng mt khch hng. Trong nhng m hnh sau, Lp Khch hng c i din bi A v Lp Ti khan i din bi B. M hnh no m t cho cu trn


25/63



a. ab. bc. c d. d

31.Mt h thng m trong cc thnh phn d liu, x l v giao din (interface) ca h thng thng tinc phn tn ra nhiu khu vc trong mt mng my tnh c xem nh l: a. Mt h thng cnh tranhb. Mt h thng trung tm ha c. Mt h thng giao tipd. Mt h thng c s d lu e. Khng c cu no ng

32.Sara mun lm cho cuc phng vn ci m hn mt cuc phng vn bnh thng v thu thp ucnhiu thng tin. C y nn hi nhng cu hi ________a. ngb. khng thch hpc. md. thm d

33.Cu no ng (mt hoc nhiu) khi ni v trng thi v s chuyn trng thi trong UML a. M hnh th trng thi ch c to cho cc lp c hnh vi ng quan trng b. M hnh th phi c to cho tt c cc lp c xc nh cho vn ti nguyn c. Nhng s kin gy ra s chuyn trng thi t trng thi ny sang trng thi khc c th hintrong m hnh th trng thi d. Nhng hnh ng do vic thay i trng thi gy ra cng c th hint trong m hnh th

trng thie. M hnh th trng thi th hin trng thi ca mt i tng n l

34.cu no sau y l ng nht?a. name l mt thuc tnh ca lp Borrower


26/63



b. membershipNum l phng thc ca lp Borrowerc. name l thuc tnh ca getNamed. setName l thuc tnh ca getName

35.Con s ca s kt hp m xut hin gia 2 vt c th th c gi l mt : a. bn s b. thuc tnhc. mi quan hd. mi quan h nh phn

36.X l lun l (logical process) l cng vic m cn c thc hin nhng khng quan tm n cchthc (hnh thc) thc hin.

a. Saib. ng 37.ng gch nt t di hnh ch nht th gi l ______________


27/63



a. dng i (life line) b. ng ca tc nhn (actor line)c. ng bin (boundary line)d. ng iu khin (controller line)e. ng thc th (entity line)

38.Vic xem xt trn cc form v report c sn l mt phn trong vic xc nh yu cu h thng . a. Saib. ng

39.mi quan h ___________ th c dng ch hnh vi ty chn m ch chy di nhng iu kinnht nh chng hn nh l khch hot mt cnh bo.a. bao gm (include)b. kt tp (aggreation)c. m rng (extend) d. tha k (inheritance)e. kt hp (assocition)

40.M hnh ho d liu l mt k thut cho vic t chc v ti liu ho cc m hnh lun l v vt l camt h thng.a. ng b. Sai

41.Cc giao din (interface) c th tha k trc tip t mt hoc nhiu Interfaces khca. ng


28/63



b. sai42.Nhng cu hi sau cp n biu tun t :

F l mt i tng _______________

a. iu khin b.

bin .c. tc nhn .

d. thc th .e. dng i .

43.Mt cch tip cn tch mt h thng thnh cc h thng con gi la. thit k h thngb. lp trnhc. qun l d liud. phn r chc nng

44.Vai tr quan trng nht trong cng vc ca ngi phn tch vin h thng la. gii quyt vn b. bit r d liu no cn c lu tr v s dngc. c k nng lp trnh ttd. hiu nhng kin thc chuyn mn v h thng thng tin

45.Nhng cu no sau y l ng v biu k tha trong hnh sau?


29/63



a. Lp A l cha ca lp D.b. Lp D l tin bi ca lp F. c. Lp A l tin bi ca lp F v D d. Lp A l cha ca lp B, E v F. e. Lp C l lp con ca lp D .

46.Mt cng c th bao gma. Nhng m hnh dng qun l qu trnh pht tri nb. Phnmn h tr gip to nhng m hnh hoc nhng thnh phn yu cu trong d n c. nhng hng dn thc thi m hnhd. Nhng m hnh ca thnh phn h thng

47.Mt h thng tin tc phi c phn cng v phn mn my tnh mi hp la. Saib. ng

48.Lp Student l mt


30/63



a. Lp char b. Chac. Lp cond. Kiu cone. Con

49.Ch ra cu sau y l yu cu chc nng hay phi chc nng H thng s h tr 500 ngi dng cnglc a. phi chc nng b. chc nng

50.Nhng cu hi sau y cp n biu tun t : G l mt i tng _______________

a. bin (boundary)b. dng i (life line)c. thc th (entity)d. tc nhn (actor)e. iu khin (controller)


31/63



51.Trong hnh (i) l

a. M rngb. Bao gmc. Lin lc d. S tng qut ha

52.Mt kch bn (scenario) l mta. Tp hp cc bc trong phm vi ca mt use case b. Tp hp cc use casec. Mt use case.d. Bao gm cc use case.

53.S khc nhau gia cc actor v cc object cng tn l


32/63


33/63



a. 1, 1.1, 1.1.1, 1.1.2, 1.1.3, 1.1.4, 1.1.4.1, 1.1.5, 1.1.5.1b. 1, 1.1, 1.1.1, 1.1.2, 1.1.3, 1.1.4, 1.1.4.1c. 1, 1.1, 1.1.1, 1.1.2, 1.1.3, 1.1.5, 1.1.5.1 d. 1, 1.1, 1.1.1, 1.1.2, 1.1.3

57.Ci no di y khng phi l mc tiu ca m hnh ha?a. c t (Specifying)b. Ti liu ho (Documenting)c. Trc quan ha (Visualizing)d. Xy dng (Constructing)e. Lp trnh (Coding)

58.Qu trnh sinh m (generating code) t mt m hnh UML c gi l: a. Reverse Engineeringb. Forward Engineeringc. Code generation d. Syntax Checkinge. Visual Coding

59.M t no di y ph hp nht cho mt mi kt hp (association)? a. Mt quan h ng ngha (semantic relationship) gia hai vt, trong mt thay i ca vt ny s

nh hng ng ngha n vt kia. b. Mt quan h c cu trc (structural relationship) m t cc lin kt (link) c. Mt quan h gia mt thnh phn chuyn bit vi mt thnh phn tng qut.


34/63



d. Mt quan h ng ngha gia cc classifiers.60.Mt th hin (instance) ca mt class c gi l mt:

a. Object b. Classifierc. Factory Classd. Stereotype

61.Vic du thng tin trong lp trnh hng i tng (OOP) c thc hin trong khi nim ca: a. Encapsulationb. Polymorphismc. Abstractiond. Inheritance.

62.S tha k (inheritance) cn c bit n nh: a. Generalization b. Polymorphismc. Associationd. Stereotyping

63.Mt mi quan h has a l m t tt nht cho: a. Polymorphismb. Compositionc. Aggregation d. Inheritance.

64.Mt mi quan h is a l m ttt nht cho:a. Inheritance b. Compositionc. Aggregationd. Encapsulation.

65.Vic o sc mnh ca s kt ni gia hai thnh phn h thng c bit nh l: a. Coupling b. Cohesionc. Aggregationd. Bonding.

66.Ci nhn tnh (static view) ca h thng c m hnh ha bng: a. class diagrams b. use case diagramsc. state chart diagramsd. activity diagrams.

67.Trong mt use case diagram, cc thc th ngoi (external entities) h thng c gi l: a. use casesb. boundary objects


35/63



c. actors d. stick men

68.Cc Use case m hnh ha:a. Yu cu chc nng (functional requirements) b. Ci nhn tnh (static view)c. Yu cu ngi dng (user requirements) d. Ti liu ha (Documentation)

69.Sequence diagrams c dng trong : a. static viewb. interaction view c. deployment viewd. model management view.

70.Cuc i (life history) ca mt object c m hnh ha trong mt:a. statechart diagram b. use case diagramc. sequence diagramd. object diagram

71.Ci no sau y khng c xem l mt association? a. Mt u l 1 class, u cn li l 3 class b. Cng mt class c hai uc. Hai class khc nhau hai ud. Hai class mt u; hai class khc u cn li

72.S kt hp gia tn ca mt thao tc (operation) v cc tham s (parameter) ca n c xem l_____________ ca thao tc .

a. Argumentsb. Signature c. Responsibilitiesd. Attributes

73.Ci no di y khng l lng s (multiplicity) hp l? a. 0..4b. 1c. 3..*d. i..j

74.Ci no di y l khai bo thuc tnh ng qui tc? a. # balance: Double b. usernamec. *pointerd. startingPoint = (1,2,3)e. initialBalance: 0.0


36/63



75.Tr phi c c t khc i, cc thuc tnh lun c tnh cht mc nh l:a. Changeable b. addOnlyc. frozend. isQuery

76.Loi relationship no nn dng m hnh ha kt ni gia mt giao din (interface) v mt class?a. Dependency b. Realizationc. Generalizationd. Association

77.Ci no sau y l l do chnh xy d ng cc use case?a. c t cc hnh ng m ngi dng s thc hinb. nh ngha cc t vng ca h thng c. gip cc lp trnh vin v khch hng hiu nhau d. dng xc minh h thng khi n tin trin e. biu din cch thc cc objects lm vic vi nhau

78.Loi diagram no thng dng nm bt cc dng iu khin (flow) kt hp vi mt use case? a. Interactionb. Sequencec. use cased. activity

79.S UML diagram trn l biu din ca mt :a. Class b. Objectc. Circled. Package

80.S UML diagram trn l biu din ca mt :a. Classb. Object c. Actord. Use case

Circle: mycircle

Circle


37/63



NewClass3

NewClass4 NewClass5

NewClass6NewClass7

81.Trong s trn biu din loi relationship no?

a. Generalization b.

Aggregationc. Composition

d. Multiple Inheritance.

NewClass8 NewClass9

82.Mi quan h (relationship) gia class 8 v class 9 l:

a. Aggregation b. Compositionc. Inheritanced. Realization

NewClass10 NewClass11

83.Mi quan h (relationship) gia class 10 v class 11 l a. Dependencyb. Aggregationc. Compositiond. Realization

NewClass12 NewClass13

84.Mi quan h (relationship) gia class 12 v class 13 l

a. Dependency


38/63



b. Realizationc. Compositiond. Aggregation

employee employer

85.Trong mi kt hp trn, t employee v employer c xem l a. Role b. Entityc. Objectd. Association

Part 4

PHN 1:c p n1. Yu cu ch yu nht ca phng php tip cn h thng l

phi xem xt h thng trong tng th vn c ca n cng vi cc mi linh ca cc phn h ni ti cng nh mi lin h vi cc h thng bnngoi.

phi xem xt mt cch ton din cc vn kinh t, k thut v t chc

ca h thng qun l. Trong mt h thng phc tp nhiu phn h nh

h thng kinh t, vic ch xem xt mt s phn h m b qua cc phn

h khc, vic ti u ho mt s b phn m khng tnh n mi lin h

rng buc vi cc b phn khc s khng mang li hiu qu ti u

chung cho ton b h thng.

trc ht phi xem xt t chc nh l mt h thng thng nht v mtkinh t, t chc, k thut, sau mi i vo cc vn c th trong

Person Company


39/63



tng lnh vc. Trong mi lnh vc li phn chia thnh cc vn c thhn na, ngy cng chi tit hn. y chnh l phng hng tip cni t tng qut n c th (top-down)

Tch hp cc yu cu trn2. Phng php phn tch thit k c cu trc bt ngun t:

Cch tip cn h thngCch tip cn k thutCch tip cn x hiCch tip cnc cu trc

3. Mc tiu nghin cu h thng nhm: hiu bit r hn v h thng tc ng ln h thng mt cch hiu qu

hon thin h thng hay thit k h thng mi

Bao gm cc mc tiu trn4 H thng

l tp hp cc phn tl tp hp cc phn t v cc mi quan hGm cc phn t, to thnh th thng nht, sinh ra tnh tri h thng

Gm cc phn t, cc mi quan h, to thnh th thng nht, sinh ra tnhtri h thng

5. H thng c s trao i gia input v output vi mi trng qua giao din l

H thng m

H thng ngH XibecniticH thch nghi6. Tnh cht ca h thng quyt nh c ch vn hnh ca n l:

Tnh c cu trc

Tnh nht thTnh t chc c th bcTnh thng nht

7. Thc cht ca vic xy dng HTTT trong 1 t chc l thit k li t chc :

ng


40/63



Sai

Khng bit

C th ng c th sai8. Cc giai on trung tm trong qu trnh pht trin 1 HTTT l:

Phn tch

Thit k

Phn tch v thit k

M ho v kim th

9. M hnh c a vo yu t phn tch ri ro l m hnh

Vng i c in

Lm bn mu

Xon c

4GT

10. khng c xut l thuc qu thi hn l :quy tc T chc

quy tc K thut

quy tc Qun l

khng thuc cc quy tc k trn11. l thuc sp ht hn phi xut trc l:

quy tc T chc

quy tc K thut

quy tc Qun l


41/63



khng thuc cc quy tc k trn

12. s dng my in lin tc khng qu 1 gi l:

quy tc T chc

quy tc K thut

quy tc Qun l

khng thuc cc quy tc k trn13. ch xut hng vo cc bui sng, nhp hng vo cc bui chiu l:

quy tc T chc

quy tc K thut

quy tc Qun l

khng thuc cc quy tc k trn14. ngh hc phi xin php l:

quy tc T chc

quy tc K thut

quy tc Qun l

khng thuc cc quy tc k trn15. Nhim v ca giai on phn tch l phi tr li nhng cu hi:

-u vo (input) ca h thng l g, u ra (output): kt qu x l cah thng l g?

-Nhng qu trnh cn x l trong h thng, hay h thng phn mm sx l nhng ci g?

-Nhng rng buc trong h thng, ch yu l mi quan h gia uvo v u ra nh th no?

Tr li tt c nhng cu hi k trn


42/63



16. M hnh phn cp chc nng dng

Xc nh phm vi h thng c nghin cu

Xc nh tin trnh x l

Phn nh d liu

M t t chc17. T tng trong k thut phn mc m hnh s dng:

Cch tip cn top-down

Cch tip cn down-top

Cch tip cn c cu trc

Cch tip cn khc18. Tn ca chc nng trong m hnh phn cp chc nng l

ng t

Danh t

Tnh t

Trng t19. Trong s lung d liu (DFD) thnh phn no c th v li nhiu ni:

Kho d liu v tc nhn ngoi

Dng thng tin

tin trnh

tin trnh v tc nhn ngoi20. Cho Lung d liu:

Khc

hvay

1. Cho

va

Tr li nva


43/63



Sai u:

Lung thng tin i vo kho d liu khng c tn

Tn ca tin trnh l ng t, tn tc nhn ngoi v kho l danh t

Khng c dng thng tin ni b trc tip

Tin trnh no thiu lung d liu vo hoc ra

21.Cc cng c ch yu din t d liu bao gm:

M ho d liu

T in d liu

M hnh thc th lin kt, m hnh quan h

Tt c cc cng c trn22. Thuc tnh dng phn bit bn th l

Thuc tnh tn gi

thuc tnh nh danh

thuc tnh m t

Thuc tnh lp23. Cc c trng ca quan h l

Bc

Bn s

S n

2. Thu

n

nva

Nhon


44/63



Loi

c 3 yu t trn24. M hnh khi nim d liu v s E_R khc nhau ch:

Bn cht

K hiu

Cch biu din

phc tp25. Trong phng php xy dng m hnh khi nim d liu ca Blanpre, vic nhn

din mt quan h kiu 1-N gia 2 thc th th hin bng s hin hu ca mt phthuc hm:

Gia 2 phn t ca tp kho

Nu gc ca ph thuc hm bao gm t nht 2 phn t thuc kho

Phn t v tri ph thuc hm khng phi l kho

Phn t v phi ph thuc hm khng phi l kho26. ChoBng Yu cu tuyn dng ca cng ty:. . . . . . .

Ngy

yu cuTn

nghBc hc Loi tt

nghipTn

ngoing

Trnh ngoing

M tYu cutuyndng &cc

thng tin

c linquan

Slngtuyndng

Trn bng thng k, c rng buc: Gi tr ca thuc tnh M t Yu cu tuyndngv S lng tuyn dng ca tng cng ty ph thuc khng nhng vo ngyyu cu m cn ph thuc c vo ngh, bc hc, loi tt nghip, loi ngoi ngv c trnh ngoi ng

Chuyn sang lc quan h :


45/63



yu cu (M cng ty, ngy yu cu, m ngh, mbc hc, m loi ttnghip, m loi ngoi ng, m trnh ngoi ng, tn cng ty, M tYu cu tuyn dng, S lng, tn ngh, bc hc, loi tt nghip, tnngoi ng, trnh ngoi ng )

yu cu (M cng ty, tn cng ty, ,ngy yu cu, m ngh, mbc hc,m loi tt nghip, m loi ngoi ng, m trnh ngoi ng, M tYu cu tuyn dng, S lng, tn ngh, bc hc, loi tt nghip, tnngoi ng, trnh ngoi ng )

yu cu (M cng ty, ngy yu cu, m ngh, m bc hc, m loi ttnghip, m loi ngoi ng, m trnh ngoi ng, M t Yu cutuyn dng, S lng, tn ngh, bc hc, loi tt nghip, tn ngoing, trnh ngoi ng )

yu cu (Tn cng ty, ngy yu cu, tn ngh, bc hc, loi tt nghip,tn ngoi ng, trnh ngoi ng, M t Yu cu tuyn dng, Slng)

27. c t quy tc qun l : Cng ty c nhiu chi nhnh nhng mi chi nhnh ch thucvo 1 cng ty

CHINHANH CONGTY

R1

CHINHANH CONGTYR1

CHINHANH CONGTYR1

CHINHANH CONGTYR1


46/63



28. Cho chng t

Phiu nhp

S phiu . . . Ngy nhpTn khch

i ch M khoi ch kho

S TT M vtt

Tn vtt

Slng

n vtnh

ngi

Thnh

tin

(Gi thit: n gi ch ph thuc vo vt t)chuyn Phiu nhp thnh lc quan h:

PHIEUNHAP (sophieu, ngaynhap, ten_kh, diachi_kh, makho, iachikho,S TT, mavattu*, tenvattu*, soluong*,donvitinh*, ongia*, thnh tin*)

PHIEUNHAP (sophieu, ngaynhap, makhach, ten_kh, diachi_kh,makho, iachikho, S TT, mavattu*, tenvattu*, soluong*,donvitinh*,ongia*, thnh tin*)

PHIEUNHAP (sophieu, ngaynhap, ten_kh, diachi_kh, makho,iachikho, mavattu*, tenvattu*, soluong*,donvitinh*, ongia* )

PHIEUNHAP (sophieu, ngaynhap, makhach, ten_kh, diachi_kh, makho,iachikho, mavattu*, tenvattu*, soluong*,donvitinh*, ongia* )

(Cc thuc tnh c du * l thuc tnh lp)29. Cho s E_R:

M khch

Tn khch a ch khch

Khch

giao

Ngygiao

S phiu giao

Ni giao


47/63



Chuyn mi quan h GIAO sang lc quan h:

GIAO (s phiu,m khch, ngy giao, ni giao,m hng*, n gi*,

s lng giao*)

GIAO (s phiu,m khch, ngy giao, ni giao, m hng*, n gi*, slng giao*)

GIAO (s phiu,m khch, ngy giao, ni giao, m hng, n gi*, slng giao*)

GIAO (s phiu,m khch, ngy giao, ni giao, m hng, n gi*, s

lng giao*)

30. Ba hot ng k thut l

Lp k hoch,phn tch, thit k

phn tch, thit k, lp trnh

thit k , lp trnh, kim th

Lp trnh, kim th, ci t

31. M hnh lung d liu h thng khc m hnh lung d liu nghip v ch:

Ch r chc nng no ngi lm, my lm

Ch r chc nng no my v ngi chung lm

n gigiao

Slnggia

Hng

M hng

Tn hng n v tnh M t hng


48/63



Khng ch r chc nng no ngi lm, Chc nng no my lm

M hnh biu din

32. Thit k hng i tng khc thit k hng chc nng ch

C d liu dng chung

Khng c vng d liu dng chung

Cc i tng l cc thc th ph thuc nhau rt cht

Quan tm nhiu hn n mi lin kt

33. Chn c t m hnh E-R ng, theo th t t trn xung di1 2 3 4

DONG_DON

____________SO_DON

MA_HANG

DON_HAN

G

_________SO_DON

NGUOI_CC

_________MA_NCC

TEN_NCC

MAT_HANG

____________MA_HANG

MO_TA_HANGDV_TINH

DONG_DON

____________SO_DONMA_HANG

DON_HAN

G

_________SO_DON

NGUOI_CCMAT_HANG


49/63



NGUOI_CC

_________MA_NCCTEN_NCC

MAT_HANG

____________MA_HANGMO_TA_HANG

DV_TINH

DONG_DON

____________SO_DONMA_HANG

DON_HAN

G

_________

SO_DON

NGUOI_CC

_________MA_NCC

TEN_NCC

MAT_HANG

____________MA_HANG

MO_TA_HANG

DV_TINH

DONG_DON

____________SO_DON

MA_HANG

DON_HAN

G

_________SO_DON


50/63



34. Trong cc nhn t nh hng n d hiu, nhn t no c tp trung nht trongcng vic v o cht lng thit k:

phc tp

t chc d liukiu cch m t thit k

Nhn t khc

35.Cho m hnh khi nim:

Ghi ch: thuc tnh c du * l thuc tnh lp

Mi quan h AT s c chuyn thnh:

S n, ngy t, m khch, m hng, SL t*

S phiuM kho

/C khoNgy giao

SL giao*

n gi giao*

Khch

t giao

Hng

1,n1,n

1,n1,n

N-N N-N

M khch

Tn khch

a ch khch

M hng

Tn hng

n v tnh

S nNgy tSLt*


51/63



S n, ngy t, m khch, m hng , SL t*

S n, ngy t, m khch, m hng* , SL t*

S n, ngy t, m khch,m hng*, SL t*

36. Trong m hnh E_R, quan h th hin bng ng ni gia 2 thc th c xc nhkhi

Ch cn trong 2 thc th c nhm thuc tnh chung

Nhm thuc tnh chung ti mt thc th l kho chnh

Kho chnh ca cc thc th tham gia vo quan h hon ton khc

nhau

Nhm thuc tnh chung khng cn l kho chnh ca mt thc th noc

37. ng ba ch trong m hnh E_R t 3NF nm

bn thc th c kho ngoi

bn thc th c kho chnh

Khng c bn thc th no c

Bn tt c cc thc th tham gia vo quan h

38. Mi quan h:

khng c thuc tnh

phi c t nht mt thuc tnh

c th c thuc tnh hay khng c thuc tnh

Bao gm tt c cc thuc tnh ca cc thc th tham gia vo quan h

39. Cc thuc tnh c th thm vo l


52/63



thuc tnh nh danh

thuc tnh tn gi

thuc tnhm t

Thuc tnh lp40. Mi quan h bin thnh thc th s c

1 thuc tnh lm kho,

Nhiu thuc tnh lm thnh kho

S thuc tnh lm kho ph thuc vo s kho ca cc thc th tham

gia vo quan h

Khng c kho

41. Gii php ttnht chuyn thit k d liu mc logic sang mc vt l l

S dng mt ngn ng lp trnh no

S dng mt h qun tr CSDL

S dng mt h qun tr CSDL v mt ngn ng lp trnh no

S dng k thut c bit

42. Thit k u ra c cch tip cn:

Cht lng tng th

C cu trc

Bao gm 2 cch tip cn trn

c cch tip cn khc vi cc cch tip cn trn

43. Sn phm trung gian ca h thng l


53/63



u ra ngoi h thng

u ra trong h thng

Bt k u ra noca h thngu vo ngoi h thng


54/63



44. Cho cc chng t sau:

a. n t hngS: Tn ngi t hng:

Ngy: a ch:

S tt M hng Tn hng M thng

n v tnh S lng

b. phiu giao hng

S: M khch:

Ngy: Tn khch:

Ni giao hng : a ch:

Stt

M hng Tnhng

n v tnh n gi S lng Thnhtin

Gi thit rng n gi ch ph thuc vo hng. Ghi ch: khch no t hng mi c giao hng

y, Tn ngi t hng v M khch l 2 loi d liu :

ng danh

ng ngha

Khc nhau hon ton

Khng c lin quan g n nhau

45. Cng c chnh c s dng thit k khun dng u ra l

Cc ti liu c t yu cu ca bo co

Cc trang nh dng in ra

c hai cng c trn

cng c khc vi cc cng c k trn


55/63



46. Trong thit k module, l cc thnh phn ca phn mm, th module:

ch cn c tn ring bit

ch cn nh a ch cva c tn ring bit, va nh a ch c

Khng cn c cc yu t k trn

47. Module c c t v thit k sao cho thng tin c cha trong module ny

C th thm nhpt cc module khc

Khng th thm nhp

Thm nhp khi cn thit

iu khin48. H thng tng tc trn mn hnh cho php :

NSD la chn ng dn ph hp

NSD iu khin hin mn hnh

NSD lin kt vi cc module x l

NSD thc hin thao tc

49. Gi cho n gin l mt trong6 quy tc vng ca

Shneiderman

7

8

9

50. Trong thit kgiao din, phi mu no ca Courier:


56/63



en nn xanh

en nn

en nn trng

en nn vng

51.Phng php tinh chnh tng bc phn nh thit k:

Kiu top-down

kiu down- top

phi hp hai kiu trn

Khng cn kiu no c

52. Cc c trng cha c th hin trong s lung d liu :

chc nng logic

chuyn giao d liu

c trng iu khin

c trng tin trnh

53. Mt lc chng trnh c lp i hi

Ch cn thm cc module vo ra, c bit l thm module iu khin

Ch cn thit lp cc li gi

i hi c 2 yu cu trn

Khng i hi g c

54. Vic chuyn mi chc nng x l thnh mt module chng trnh l


57/63



khng nht thit

bt buc

Lun i hi

Khng th thiu

55. Vic phn tch xc nh kim sot nhm thc hin

Xc nh im h v kiu e do t im h trong h thng

la chn cng nh thit k kim sot cn thit

Xc nh tnh trng e do

Thc hin tt c cc k trn

56.Trong cc cch thit k m bo s an ton ca thng tin, cch xy dng cc thtc phc hi l:

Lun c ch

khng phi bao gi cng c ch thc s

Khng c ch

C ch khng ng k

57. Trong cc chin lc chuyn i h thng thng tin c sang mi, chin lc no mang tnh ri ro cao nht:

+ Chuyn i trc tuyn

+Chuyn i song song

+ Chuyn i b phn

+Chuyn i theo giai on


58/63



58. Chn th t u tin ng cc bin php thc hin an ton d liu:

Thit lp cc nguyn tc qun l,S dng cc thit b lu tr anton,S dng cc trang thit b an ton

Thit lp cc nguyn tc qun l, S dng cc trang thit b an ton,S dng cc thit b lu tr an ton

S dng cc thit b lu tr an ton, Thit lp cc nguyn tc qun l,S dng cc trang thit b an ton

S dng cc thit b lu tr an ton,S dng cc trang thit b anton, Thit lp cc nguyn tc qun l

59. Vic ci t h thng

Ch bao gm cc phn mm ng dng

Ch bao gm cc phn mm h thng

Ch bao gm cc thit b my tnh v cc thit b truyn thng

Bao gm tt c cc ci t k trn

60. Vic t chc mt mng tin hc bao gm c cng tc qun l mng tin hc nh qunl cc khi thit b qun l thng nht hot ng ca mng, pht hin trc trc v c cim nghn thng tin trn mng:

ng

Sai

Khng r

C th ng c th sai

PHN 2:cha c p n

1. Tip cn h thng l 1 phng php khoa hc trong nghin cu v gii quyt : + Cc vn kinh t x hi+ Ring vn k thut


59/63



+ Ring vn qun l2. Phng php phn tch h thng c cu trc bt ngun t:

+ Cch tip cn h thng+ Cch tip cn k thut+ Cch tip cn x hi

3. Mc tiu nghin cu h thng nhm:+ Ch hiu bit r hn v h thng+ Ch tc ng ln HT mt cch hiu qu hoc hon thin HT hay thit k HT

mi

+ Bao gm cc ni dung trn4. H thng l

+ Ch l tp hp cc phn t+ Ch l tp hp cc mi quan h v to thnh th thng nht+ Bao gm c cc ni dung trn

5. H thng c s trao i gia input v output vi mi trng qua giao din l+ H thng m + H thng ng + H Xibecnitic6. Tnh cht ca h thng quyt nh c ch vn hnh ca n l:

+ Tnh c cu trc + Tnh nht th + Tnh t chc c th bc7. Phn mm c hiu l

+ Chng trnh + Chng trnh v CSDL+ Chng trnh v CSDL v cc ti liu PT_TK, hng dn s dng

5. Thc cht ca vic xy dng HTTT trong 1 t chc l thit k li t chc :+ ng + Sai + Khng bit

9. Cc giai on trung tm trong qu trnh pht trin 1 HTTT l:+ Phn tch + Thit k + C hai giai on trn10. M hnh c a vo yu t phn tch ri ro l m hnh

+Vng i c in +Lm bn mu +Xon c11. Quy tc khng c xut l thucqu thi hn l quy tc:

+ T chc + K thut + Qun l12. Quy tc l thuc sp ht hn phi xut trc l quy tc:

+ T chc + K thut + Qun l13. Quy tc Vo khch sn trc ht phi qua l tn l quy tc:

+ T chc + K thut + Qun l

14. Quy tc trc khi xut hng phi nhp kho l quy tc:+ T chc + K thut + Qun l

15. Quy tc ngh hc phi xin php l quy tc:+ T chc + K thut + Qun l

16. M hnh phn cp chc nng l hay cha xc nh phm vi h thng c nghin cu

17. T tng trong k thut phn mc m hnh s dng:


60/63



+ Cch tip cn top-down+ Cch tip cn down-top+ C 2 cch tip cn trn

18. Tn ca chc nng trong m hnh phn cp chc nng l+ ng t + Danh t + Tnh t

19. Trong s DFD thnh phn no c th v li nhiu ni:+Kho d liu +Dng thng tin + tin trnh20. Tn ca lung d liu l:

+Tnh t +Danh t + ng t21.Cc cng c ch yu din t d liu bao gm:

+ M ho d liu, t in d liu+ M hnh thc th lin kt, m hnh quan h+ Tt c cc ni dung trn

22. Thuc tnh dng phn bit bn th l

+ Thuc tnh tn gi + thuc tnh nh danh + thuc tnh m t23. Cc c trng ca quan h l+ Bc + Bn s + C 2 yu t trn

24. M hnh khi nim d liu v s E_R khc nhau ch:

+ Bn cht + K hiu + Cch biu din

25. Trong phng php xy dng m hnh khi nim d liu ca Blanpre, vic nhndin mt quan h kiu 1-N gia 2 thc th th hin bng s hin hu ca mt phthuc hm:

+ Gia 2 phn t ca tp kho+ Nu gc ca ph thuc hm bao gm t nht 2 phn t thuc kho

26. Kho thng + di hn+ ngn hn+ khng bit di hn hay ngn hn so vikho ti tiu

27. Trong quan h c s kho ti tiu:+ ch c mt + c nhiu + c th khng c

28. Thut ton tnh kho ti tiu lun xut pht t+ mt kho + tp thuc tnh khng cn l kho + c khng gian

29. Chun ho l mtqu trnh chuyn mt cu trc d liu phc tp thnh cc cu trc

d liu:+ tt hn + n gin hn + va tt va n gin hn30. Mt bng d liu bt k c th coi l quan h ?

+ ng + sai + khng bit31. Ba hot ng k thut l

+ phn tch, thit k, lp trnh+ thit k , lp trnh kim th


61/63



+ Lp trnh, kim th, ci t32. M hnh lung d liu h thng khc m hnh lung d liu nghip v ch:

+ Chc nng no ngi lm,+ Chc nng no my lm+ Bao gm c 2 ni dung trn

33. Thit k hng i tng khc thit k hng chc nng ch+ C d liu dng chung+ Khng c vng d liu dng chung+ Cc i tng l cc thc th ph thuc

34. Trong cc nhn t nh hng n d hiu, nhn t no c tp trung nht trongcng vic v o cht lng thit k:

+ phc tp + t chc d liu + kiu cch m t thit k35. Phng php chc nng c dng nhiu hn phng php i tng v

+ Vic xc nh thc th kh khn hn+ Tnh che du thng tin trong tng i tng+ Tnh tha k

36. Trong m hnh E_R, quan h c xc nh khi+ Ch cn trong 2 thc th c nhm thuc tnh chung+ Nhm thuc tnh chung ti mt thc th l kho chnh+ Nhm thuc tnh chung khng cn l kho chnh ca mt thc th no c

37. ng ba ch trong m hnh E_R nm + bn c kho ngoi + bn c kho chnh

38. Mi quan h: + khng cn c thuc tnh+ phi c t nht mt thuc tnh

+ c th c thuc tnh hay khng c thuc tnh39. Cc thuc tnh c th thm vo l + thuc tnh nh danh + thuc tnh tn gi + thuc tnh m t

40. Mi quan h bin thnh thc th s c+ 1 thuc tnh lm kho,+ Nhiu thuc tnh lm thnh kho+ S thuc tnh lm kho ph thuc vo s kho ca cc thc th tham gia vo quan

h41. Gii php tt nht chuyn thit k d liu mc logic sang mc vt l l

+ S dng mt ngn ng lp trnh no + S dng mt h qun tr CSDL+ S dng c hai cch trn

42. Thit k u ra c cch tip cn:+ Cht lng tng th v c cu trc+ Ch c 1 trong 2 cch tip cn trn+ Khng c cch tip cn no c


62/63



43. Sn phm trung gian ca h thng l+ u ra ngoi h thng+ u ra trong h thng+ Bt k u ra no ca h thng

44. Phng tin u ra ph bin l

+ Giy + microfilm + vi phim45. Cng c chnh c s dng thit k khun dng u ra l+ Cc ti liu c t yu cu ca bo co+ Cc trang nh dng in ra+ c hai cng c trn

46. Module l cc thnh phn ca phnmm:+ ch cn c tn ring bit+ ch cn nh a ch c+ va c tn ring bit, va nh a ch c

47. Module c c t v thit k sao cho thng tin c cha trong module ny+ C th thm nhp+ Khng th thm nhp+ Thm nhp khi cn thit t cc module khc

48. H thng tng tc trn mn hnh cho php :+ NSD la chn ng dn ph hp+ NSD iu khin hin mn hnh+ NSD lin kt vi cc module x l

49. Gi cho n gin l mt trong +6 + 7 +8quy tc vng ca Shneiderman

50. Trong thit k biu mu, phi mu no ca Courier:+ en nn xanh + en nn + en nn trng51.Phng php tinh chnh tng bc phn nh thit k:

+ Kiu top-down +kiu down- top +phi hp hai kiu trn52. Cc c trng cha c th hin trong s lung d liu :

+ chc nng logic + chuyn giao d liu +c trng iu khin53. Mt lc chng trnh c lp i hi

+ Ch cn thm cc module vo ra, c bit l thm module iu khin + Ch cn thitlp cc li gi+ i hi c 2 yu cu trn

54. Vic chuyn mi chc nng x l thnh mt module chng trnh l+ khng nht thit + bt buc

55. Vic phn tch xc nh kim sot nhm thc hin + Xc nh im h v kiu e do t im h trong h thng+ Xc nh tnh trng e do v la chn cng nh thit k kim sot cn thit + Thc hin tt c cc k trn


63/63


56.Trong cc cch thit k m bo s an ton ca thng tin , cch xy dng cc thtc phc hi l:

+ Lun c ch + khng phi bao gi cng c ch thc s57. Xp th t u tin (nh s t 1 n 4) cc chin lc chuyn i h thng thngtin c sang mi:

+ Chuyn i trc tuyn +Chuyn i song song+ Chuyn i b phn +Chuyn i theo giai on58. Xp th t utin (nh s t 1 n 3) cc bin php thc hin an ton d liu :

+ Thit lp cc nguyn tc qun l+ S dng cc thit b lu tr an ton + S dng cc trang thit b an ton

59. Vic ci t h thng+ Ch bao gm cc phn mm ng dng v cc phn mm h thng+ Ch bao gm cc thit b my tnh v cc thit b truyn thng + Bao gm tt c cc ci t k trn

60. Vic t chc mt mng tin hc bao gm c cng tc qun l mng tin hc nh qunl cc khi thit b qun l thng nht hot ngca mng, pht hin trc trc v ccim nghn thng tin trn mng:

+ ng + Sai + Khng r

cau hoi on tap_pttkht(1)

Documents