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8/4/2019 Vu Minh Ngoc

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Mc lc

PHN 1: TNG QUAN C S D LIU PHN TN ......................................................................... 3

1.1. nh ngha CSDL phn tn ................................................................................................ 3

1.2. H qun trc s d liu .................................................................................................. 4

1.3. H qun trCSDL phn tn ............................................................................................... 5

1.4. Kin trc c bn ca mt CSDL PT ................................................................................... 5

1.5. Tnh trong sut ca CSDL PT............................................................................................ 6

PHN 2. CC VN VPHN MNH ........................................................................................... 7

1. CC CHIN LC THIT K CSDL PT ................................................................................. 7

2. PHN MNH D LIU.......................................................................................................... 7

2.1. Mnh quan hl g? ...................................................................................................... 7

2.2. Cc l do phn mnh ..................................................................................................... 7

2.3. Cc kiu phn mnh...................................................................................................... 7

2.4. Cc yu cu ca vic phn mnh .................................................................................. 8

PHN 3. PHN MNH NGANG......................................................................................................... 9

1. THNG TIN CN THIT CA PHN MNH NGANG ............................................................ 9

2. PHN MNH NGANG NGUYN THY ................................................................................. 113. TNH Y V TNH CC TIU ...................................................................................... 11

4. THUT TON XC NH TNH Y V CC TIU ...................................................... 12

5. THUT TON PHN MNH NGANG NGUYN THY - PHORIZONTAL .............................. 13

6. THUT TON PHN MNH NGANG DN XUT.................................................................15

7. KIM TRA TNH NG N CA PHN MNH NGANG ................................................... 18


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PHN 1: TNG QUAN C S D LIU PHN TN

1.1. nh ngha CSDL phn tn

Mt CSDL phn tn l mt tp hp nhiu CSDL c lin quan v mt logic vc phn btrn mt mng my tnh:

- Tng quan v logic- D liu c quan h vi nhau, bao gm:o Gia cc .. vi nhau nh 1 1, 1 n, n no Gia cc thuc tnh ca . bao gm:

Cc rng buc Gi trton hc Ph thuc hm

-Tnh cht phn tn:

o L tp hp nhiu mi quan hlogic c phn btrn mng my tnho Tnh phn tn th hin :

Phn tn phn cng Xl phn tn

Trm 1

Trm 2

Trm 3Trm 4

Trm 5

Mng my tnh

Hnh 1.1 Mi hnh hCSDL phn tn


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1.2. H qun trc s d liuKin trc ca mt h qun trCSDL nh sau:

Trong :

Mc trong (Internal Level) hoc mc vt l (Physical Level): xc nhcch thc lu tr d liu v cc phng php truy cp vo ;

Mc quan nim(Conceptual Level): cn gi l m hnh quan nim cad liu (MQD) hoc m hnh logic ca d liu (MLD). N xc nh cch

sp xp thng tin bn trong CSDL; Mc ngoi (External Level) hoc mc quan st (View Level): xc nh

cc giao din nh nhng ng dng, tng tc v hin thcho ngi sdng.

Gia ba mc ny c hai nh x (mapping):

nh x gia mc ngoi v mc quan nim:nh xngoi nh x gia mc trong v mc quan nim:nh x trong


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Tnh c lp logictnh trong sut: mc quan nim c thc hiu chnhm khng ph thuc vo mc vt l, ngha l ngi qun tr CSDL c th phttrin, thay i n m khng lm phin g n ngi s dng;

1.3. H qun trCSDL phn tnL h thng phn mm qun l CSDL PT. N lm cho sphn tn trn l

trong sut vi ngi s dng, tc l thao tc trn CSDL PT nh trn CSDL tptrung.

1.4. Kin trc c bn ca mt CSDL PT

a. Lc tng th: Xc nh tt ccc d liu sc lu trtrong c s d liu phn

tn. S tng th c nh ngha theo cch nh trong CSDL tp

trung.

N m tbi ton tng th, khng c lin quan ti phn mnh.b. Lc phn on:

Mi quan h tng th c th chia thnh mt vi phn khng giaonhau gi l phn on (fragment).

S phn on m tcc nh x gia cc quan h tng thv ccon c nh ngha trong s phn on (fragmentationSchema),

Lc tng th

Lc phn on

Lc nh v

Lc nh xa phng 2Lc nh xa phng1

DBMS ca vtr 1

CSDL a phng ti vtr 1

Cc vtr khc

DBMS ca vtr 2

CSDL a phng ti vtr 2

Hnh1.2 Kin trc c bn ca CSDL phn tn


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Cc kiu phn mnh d liu bao gm :i. Phn mnh ngang

ii. Phn mnh dcc. Lc nh v:

Cc on l cc phn logic ca mt quan h tng thc nh vvt l trn mt hay nhiu trm. S nh vxc nh on d liu no c nh v ti trm no

trn mng.

d. Lc nh xa phng: Thc hin nh x vt l vocc on c phn trna phng.

1.5. Tnh trong sut ca CSDL PTTnh trong sut ca mt hphn tn c hiu nh l vic che khut i cc

thnh phn phn tn ca hi vi ngi s dng v nhng ngi lp trnh ngdng.

Khi d liu c phn on th vic truy cp vo CSDL c thc hinbnh thng nh l cha bphn tn v khng nh hng ti ngi sdng.


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PHN 2. CC VN V PHN MNH

1. CC CHIN LC THIT KCSDL PT1.1.Thit kphn tn d liu theo m hnh ttrn xung (TOP DOWN)

Thit k t tng thn ring bit Phn r mt h thng ln thnh cc h thng con Phn tch cc yu cu nhm nh ngha mi trng h thng Thu thp cc yu cu v d liu v nhu cu xl ca cc trm c s dng

CSDL

1.2.Thit kphn tn d liu theo m hnh tdi ln (BOTTOM UP)Nhn xt

Phng php thit ktrn xung thc sc hiu qukhi xy dng mt hthng mi.

Trong thc t, mt s CSDL tn ti trc, c t chc trong mitrng tp trung v CSDL phn tn c pht trin bng cch lin ktchng li thnh mt CSDL mi thng nht (Cc DBMS a phng khcnhau c s dng)

Cch thit k

Chn mt m hnh d liu chung m tlc tng th Chuyn mi lc a phng theo m hnh d liu chung chn Tch hp cc lc a phng vo lc tng th

2. PHN MNH D LIU2.1.Mnh quan hl g?

Vic chia mt quan h thnh nhiu quan h nh hn c gi l phnmnh quan h.

2.2.Cc l do phn mnh Khung nhn hoc n v truy xut ca cc ng dng khng phi l ton b

quan hm thng l mt mnh. Vic phn r mt quan hthnh nhiu mnh, mi mnh c xl nh

mt n v, scho php thc hin nhiu giao dch ng thi. Vic phn mnh cc quan h scho php thc hin song song mt cu vn

tin bng cch chia n ra thnh mt tp cc cu vn tin con hot tc trncc mnh.

2.3.Cc kiu phn mnh Phn mnh ngang chnh l php chn trn quan h Phn mnh dc - chnh l php chiu trn quan h


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2.4.Cc yu cu ca vic phn mnh Tnh y : quy tc ny m bo cho cc mc d liu trong quan h R phn

rhon ton vo cc mnh. Tc l Nu mt quan hR c phn r thnh cc mnh R1, R2, ..., Rk

th mi mc d liu c trong R phi c trong t nht mt mnh Ri no. Tnh phc hi : Th hin khnng ti to cc mi quan h tcc mnh ca

n. Tnh tch bit : Cc mnh khng c cc phn t ging nhau hay khng giao

nhau


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PHN 3. PHN MNH NGANG

1. THNG TIN CN THIT CA PHN MNH NGANGMun phn mnh CSDL, th trc ht phi bit cc thng tin cn thit. Cc thngtin bao gm: Thng tin vCSDL, Thng tin vng dng.

o Thng tin v CSDL phc v cho mc ch thit kphn tno Thng tin vng dng phc v cho vic phn mnh cho tng quan

h

a) Thng tin v CSDLo L cc thng tin vlc khi nim ton cc ca CSDLo N biu din cc quan h bng cc cung c hng

o Quan h ti in cui ca ng ni : l quan h cho Quan h ti im u : quan hthnh vin

b) Thng tin vng dngi. Thng tin nh tnh hng dn cho cc hot ng phn mnh

V tc bn C dng : Pj: Ai Value Trong thuc {=, , } Value c chn t min gi tr ca Ai Pri: biu th tp tt ccc v tn gin c nh ngha

trn quan h Ri V d:

Trong quan hPROJ cc v tsau y l cc v tn gin:PNAME = Maintenance

BUDGET 200000 V t hi s cp l hi tng cp ca cc v tc bn


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Cho tp Pri = {pi1, pi2, ...., pim}

Tp cc v t hi s cp Mi = {mi1, mi2, ..., miz} c nhngha nh sau

Mi ={ mij | mij = p*ik} , 1 k m ; 1 j z . Trong = p hoc = V d :

Xt mt s v tn gin c thnh ngha c trn quanh PAY :

p1: TITLE = Elect.Engp2: TITLE = Syst. Analp3: TITLE = Mech. Engp4: TITLE = Programmerp5: SAL 30000p6: SAL > 30000

Hi v ts cp da trn v tn gin:

m1: TITLE = Elect.Eng ^ SAL 30000m2: TITLE = Elect.Eng ^ SAL > 30000m3: (TITLE = Elect.Eng) ^ SAL 30000m4: (TITLE = Elect.Eng) ^ SAL > 30000m5: TITLE = Programmer ^ SAL 30000m6: TITLE = Programmer ^ SAL > 30000

ii. Thng tin nh lng ch yu s dng trong m hnh cp pht tuyn hi s cp

N: slng cc b tha mn iu kin mt hi v ts cp cho

K kin sel(mi) V d:

m1: TITLE = Elect.Eng ^ SAL 30000m2: TITLE = Elect.Eng ^ SAL > 30000

sel(m1)=0, sel(m2)=1


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ii. Nu thm vo BUDGET 200000 votrong Pr

Nhn xt: Hiu nm na th tnh y l s phkn min gi tr cacc trng ca quan h R nu n c xut hin trong Pr

b) Tnh cc tiu Mt v tc tnh lin ti, l mt v tc khnng xc nh mt phn

mnh. Nu tt ccc v ttrong Pr u c tnh lin ti th Pr c gi l tp

v tn gin c tnh cc tiu. V d:

Pr = { LOC = Montreal, LOC = New York,LOC = Paris , BUDGET 200000

};

l y v cc tiu

Nu thm vo PNAME = Intrumentation vo Pr th Pr skhng y

4. THUT TON XC NH TNH Y V CC TIUa) Pht biu thut ton theo hiu

Bc 1:Tm mt v tc lin ti v phn hoch quan h cho trongPr

Bc 2:Cho thm mt v t vo Pr v loi bn khi Pr Bc 3: Kim tra xem trong Pr c tn ti v tno tng ng vi

v t mi khng ? Nu c th loi b v t mi khi Pr

Lp li bc 1 cho n khi Pr = rng -> m bo cho Pr lun cctiu ti mi bc.


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Cui vng lp, tp Pr y v cc tiub) Nhn xt

Loi bcc v t hi s cp tha, bng cch xc nh nhng v tmuthun vi cc php ko theo

c)

V dGi sPr = { P1, P2 }P1 : A = value_1

P2 : A = value_2

Min gi tr ca A l { value_1, value_2 }C 2 php ko theo lun ng sau:

(A = value_1) => (A = value_2)(A = value_1) => (A = value_2)

C 4 v t hi s cp sau c nh ngha theo Pr: m1: (A = value_1) (A = value_2)

m2: (A = value_1) (A = value_2)



Cc v t hi s cp m1 v m4 b loi bv mu thun vi php ko theo

5. THUT TON PHN MNH NGANG NGUYN THY - PHORIZONTALa) Nhn xt

Thut ton phn mnh ngang nguyn thy, thc cht l tm tp cc hiv ty v ti thiu, loi bcc hi v tmu thun vi php ko theo

b) Cc bc: Bc 1: S dng thut ton COM_MIN tm tp cc hi v ty

v ti thiu Bc 2:Xc nh cc php ko theo Bc 3: Kim tra xem trong Pr c tn ti v tno mu thun vi

cc php ko theo khng ? Nu c th loi b v t mi khi Pr

c) V d:Phn mnh d liu vi tp cc v tn gin sau Pr = P1 : LOC = Montreal P2 : LOC = New York P3 : LOC = Paris P4 : BUDGET 200000

S dng thut ton COM_MIN ta cPr = { P1, P2, P3, P4, P5}

Tp v t hi s cp M nh sau: m1 : LOC = Montreal BUDGET


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m2 : LOC = New York BUDGET 200000 m6 : LOC = Paris P5 : BUDGET > 200000

Cc php ko theo : i1 : P1 P2 P3 i2 : P2 P1 P3 i3 : P3 P2 P1 i4 : P4 P5 i5 : P5 P4

i6 : P4 P5 i7 : P5 P4

NX: cc hi v tu tha mn php ko theo

Vy ta c hi v ts cp sau:M = { m1, m2, m3, m4, m5, m6 }

Kt quphn mnh ngang nguyn thy s to ra 6 mnh tng ng :FPROJ = { PROJ1, PROJ2, PROJ3, PROJ4, PROJ5, PROJ6}

Trong PROJ2v PROJ5 rng


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6. THUT TON PHN MNH NGANG DN XUTa. nh ngha

i. L php phn mnh mt quan hton cc ra nhiu mnh quan hngang m khng da vo tnh cht ca cc thut tnh ca n.Nhng c suy din t kt quphn mnh ngang ca mt quanhkhc.

ii. N da trn cc quan hthnh vin ca mt ng ni theo phpton chn ca mt quan h ch

iii. S dng : Equijoin Semijoin

b. nh ngha theo cng thc ton:i. Cho mt ng ni L trong

S l quan h ch R l quan hthnh vin

ii. Phn mnh ngang dn xut R c nh ngha :Ri= R Si 1 i k

Si = Fi(S) : Fi l biu thc nh ngha phn mnh ngang

c. V d 1:

Xt quan h chPAY v quan hthnh vin EMP, nhm ccnhn vin thnh 2 nhm :

C lng SAL 30000Ta c 2 bng:

PAY1 = (PAY)


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PAY2 = (PAY) Hai mnh ca EMP1 v EMP2 :

EMP1 = EMP PAY1 EMP2 = EMP PAY2

Nhn xt : Mi mnh chc mt ng n hoc i => thn gin

u im : Kt ni gia cc mnh n gin C th thc hin song song mt cu truy vn

d. V d 2:Trng hp c nhiu hn 2 kt ni n/i

VD : quan h ASGNh vy sc nhiu cch phn mnh quan hny


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C 2 tiu ch sau : PM c c tnh kt ni tt hn PM c s dng cho nhiu ng dng hn

V d ng dng 1 : phn mnh ASG theo cc mnh PROJ1,PROJ3, PROJ4, PROJ6

Cc mnh:

Cc Phn mnh dn xut theo {PROJ1, PROJ3, PROJ4, PROJ6 }:

Phn mnh dn xut ca ASG


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7. KIM TRA TNH NG N CA PHN MNH NGANGa. Tnh y

Tnh y c tp cc v tn gin Pr trong thut ton phnmnh ngang nguyn thy m bo cho sphn mnh cng c thamn tnh y .

Quy tc ton vn tham chiu m bo cho tt ccc btrong ccquan h mnh ca quan hthnh vin u c mt trong quan h ch

b. Tnh phc hiQuan h btch c thkhi phc bng cch thc hin php ton

hp cc quan h mnh kh hp trong trng hp phn mnh ngang

nguyn thy v phn mnh ngang dn xut.c. Tnh tch bit

Mt quan hR c phn r ngang thnh cc mnh theo phngphp nguyn thy hay dn xut, th mt b bt k nm trong mt mnh Rith n skhng nm trong mt mnh khc.

Quy tc ny m bo cc mnh phn r vi nhau.

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