Giao Trinh Co So Mang Thong Tin

8/14/2019 Giao Trinh Co So Mang Thong Tin

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C s mng thng tin

Gio trnh dnh cho sinh vin i hc ngnh in t - Vin thng

Khoa in t Vin ThngTrng i hc Bch khoa H ni


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.........

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.........Cc t vit tt

FAS Frame Alignment SignalIEEE Institute of Electronics and

Electrical EngineeringITU International Telecommunication

UnionMFAS Multi-Frame Alignment SignalPDF Probability Density Function

pdf probability distribution functionTDMA Time Division Multiple Access

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.........Bng i chiu thut ng Anh - Vit

Ting Vit Ting Anh

Bng tn thng di Band PassBng tn c s BasebandTrm gc Base StationKnh ChannelVa p CollisionCuc ni Connection

M ho iu khin li Error Control CodingMt ph nng lng Energy Spectral DensityKhung Frame p ng tn s Frequency ResponseGiao thoa gia cc k t Intersymbol Interference a khung Multi-frame a truy nhp Multiple AccessB ghp knh, b hp knh Multiplexer Hiu ng xa - gn Near Far EffectKt ni, lin kt Link u thu, phn thu Sender u thu, phn thu, ch SinkM ho ngun Source CodingGhp knh phn chia theo thi gian Time Division MultiplexingB pht, khi pht Transmitter

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.........Mc lcCc t vit tt.............................................................................................................................................3Bng i chiu thut ng Anh - Vit ........................................................................................................Mc lc.....................................................................................................................................................5Mc lc hnh v.........................................................................................................................................7Mc lc bng biu.....................................................................................................................................8Chng 1 Gii thiu ...................................................................................................................................1

1.1. Mc ch ca vic m hnh ha v nh gi c tnh hot ng ca h thng ............................................11.2. Cc khi nim c bn trong h thng thong tin ...........................................................................................11.3. Cc bc v phng php nh gi mt mng thng tin ............................................................................1

1.3.1. o c, thu tp k qu thng k ..................................................................................................................................11.3.2. M hnh ha ton hc ...................................................................................................................................................11.3.3. M phng ....................................................................................................................................................................1

1.4. Cc cng c phc v cho vic nh gi cht lng hot ng ca mng .....................................................1Chng 2 Hng i Cc h thng thi gian lin tc ...............................................................................2

2.1. Gii thiu l thuyt hng i ........................................................................................................................22.1.1. Hng i v c im ..................................................................................................................................................22.1.2. Cc tham s hiu nng trung bnh .................................................................................................................................5

2.2. Nhc li cc khi nim thng k c bn .....................................................................................................102.2.1. Tin trnh im ..........................................................................................................................................................102.2.2. Tin trnh Poisson ......................................................................................................................................................12

2.3. nh lut Little ...........................................................................................................................................142.3.1. Cng thc Little .........................................................................................................................................................142.3.2. Chng minh cng thc Little ......................................................................................................................................15

2.4. Cc m hnh hng i ................................................................................................................................162.4.1. K hiu Kendall .........................................................................................................................................................162.4.2. Qu trnh Sinh-T (Birth-Death) .................................................................................................................................172.4.3. Hng i M/M/1 ........................................................................................................................................................172.4.4. Hng i M/M/1/K ....................................................................................................................................................202.4.5. Hng i M/M/C .......................................................................................................................................................20

2.5. L thuyt lu lng ....................................................................................................................................212.5.1. Khi nim v lu lng v n v Erlang ...................................................................................................................212.5.2. H thng tn tht (Loss System) v cng thc Erlang B ..............................................................................................232.5.3. H thng tr (Delay) v cng thc Erlang C ................................................................................................................26

2.6. H thng hng i c u tin ......................................................................................................................282.6.1. Qui tc v t chc hng i ........................................................................................................................................292.6.2. u tin ca khch hng trong hng i u tin ......................................................................................................322.6.3. Duy tr qui tc hng i, lut Kleinrock ......................................................................................................................322.6.4. Mt s hng i n server ........................................................................................................................................332.6.5. Kt lun ....................................................................................................................................................................33

2.7. Bi tp (Pending) ........................................................................................................................................34Chng 3 Mng hng i ........................................................................................................................ 35

3.1. Mng ni tip ..............................................................................................................................................35

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.........Chng 4 nh tuyn trong mng thng tin ............................................................................................ 36

4.1. Yu cu v nh tuyn trong mng thng tin .............................................................................................364.1.1. Vai tr ca nh tuyn trong mng thng tin ...............................................................................................................364.1.2. Cc khi nim trong l thuyt graph ...........................................................................................................................36

4.2. Cc m hnh nh tuyn qung b (broadcast routing) ..............................................................................384.2.1. Lan trn gi (flooding) ...............................................................................................................................................384.2.2. nh tuyn bc ngu nhin (random walk) ...............................................................................................................394.2.3. nh tuyn khoai ty nng (hot potato) ......................................................................................................................394.2.4. nh tuyn ngun (source routing) v m hnh cy (spanning tree) ..............................................................................404.2.5. Duyt cy .................................................................................................................................................................40

4.3. Cc m hnh nh tuyn thng dng ..........................................................................................................614.3.1. nh tuyn ngn nht (Shortest path Routing) .............................................................................................................61

4.4. Bi tp (Pending) ........................................................................................................................................84Chng 5 iu khin lung v chng tc nghn .....................................................................................85

5.1. Tng quan ..................................................................................................................................................855.1.1. M u ......................................................................................................................................................................855.1.2. Khi nim iu khin lung ........................................................................................................................................88

5.1.3. Khi nim chng tc nghn ........................................................................................................................................895.1.4. Nhim v ch yu ca iu khin lung v chng tc nghn ...................................................................................895.1.5. Phn loi iu khin lung v trnh tc nghn ............................................................................................................90

5.2. Tnh cng bng ...........................................................................................................................................915.2.1. nh ngha .................................................................................................................................................................915.2.2. Tnh cng bng v mt bng truyn ............................................................................................................................915.2.3. Tnh cng bng v mt b m ...................................................................................................................................915.2.4. C ch pht li ARQ ..................................................................................................................................................935.2.5. Stop-and-Wait ARQ ..................................................................................................................................................945.2.6. Go-back-N ARQ ......................................................................................................................................................1005.2.7. Selective repeat ARQ ...............................................................................................................................................107

5.3. iu khin lung v trnh tc nghn theo phng php ca s ...............................................................1095.3.1. iu khin lung theo ca s (Window Flow Control) ..............................................................................................1095.3.2. iu khin tc nghn s dng ca s thch ng (adaptive window) ............................................................................115

5.4. iu khin lung v chng tc nghn da trn bng thng (rate-based flow control) ..............................1205.4.1. Khi nim ................................................................................................................................................................1205.4.2. iu khin bng thng theo thut ton go r (leaky bucket) .....................................................................................1215.4.3. Thut ton GPS (pending) ........................................................................................................................................125

5.5. Bi tp (Pending) ......................................................................................................................................125Chng 6 K thut m phng ................................................................................................................ 126

6.1. Gii thiu ..................................................................................................................................................1266.2. M phng da trn cc s kin ri rc v cc cng c ............................................................................126

6.2.1. Phng php m phng da trn s kin ri rc ........................................................................................................1266.2.2. Cc cng c m phng thng dng da trn s kin ri rc ......................................................................................129

6.3. Cng c m phng mng NS2 ..................................................................................................................1306.3.1. Cu trc ..................................................................................................................................................................1306.3.2. Cc tin ch trong NS h tr cho m phng mng [Pending] .....................................................................................1326.3.3. Th d (Pending) ......................................................................................................................................................132

6.4. Kt lun (Pending) ....................................................................................................................................1326.5. Bi tp (Pending) ......................................................................................................................................133

Ti liu tham kho.................................................................................................................................134

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.........Mc lc hnh v Hnh 2-1 M hnh chung ca h thng hng i.....................................................................................2

Hnh 2-2: V d v mng hng i m......................................................................................................3 Hnh 2-3 V d v mng hng i ng....................................................................................................3 Hnh 2-4 H thng hng i n gin.......................................................................................................5 Hnh 2- 5. Cc s kin n trong thi gian t............................................................................................6 Hnh 2-6: Cc s kin i trong thi gian t...............................................................................................6 Hnh 2-7 Khong thi gian s dng nh ngha tin trnh.................................................................13 Hnh 2-8..................................................................................................................................................15

Hnh 2-9. Chui Markov ca mt qu trnh sinh-t................................................................................17 Hnh 2-10 Chui Markov ca hng i M/M/1.......................................................................................17 Hnh 2-11.................................................................................................................................................20 Hnh 2-12.................................................................................................................................................20 Hnh 2-13 Lu lng mang (mt )( bng s thit b bn) l mt hm thi gian (ng cong C). Lulng trung bnh trong khong thi gian T (ng cong D)...................................................................22 Hnh 2-14 Thut ton xp hng theo mc u tin...................................................................................31Hnh 2-15 Xp hng cn bng trng s..................................................................................................31

Hnh 2-16 Mt s loi hng i n server thng gp...........................................................................33 Hnh 4-17. Hng ch bn trong router....................................................................................................39 Hnh 4-18. Duyt cy...............................................................................................................................41Hnh 4-19. Cc thnh phn....................................................................................................................45

Hnh 4-20. Php tnh Minimum Spanning Tree ( MST)..........................................................................53

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.........Mc lc bng biu

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Chng 1 Gii thiu1.1. Mc ch ca vic m hnh ha v nh gi c tnh hot ng ca h thng

1.2. Cc khi nim c bn trong h thng thong tin

1.3. Cc bc v phng php nh gi mt mng thng tin

1.3.1. o c, thu tp k qu thng k

1.3.2. M hnh ha ton hc

1.3.3. M phng

1.4. Cc cng c phc v cho vic nh gi cht lng hot ng ca mng


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Chng 2 Hng i Cc hthng thi gian lin tc2.1. Gii thiu l thuyt hng i

2.1.1. Hng i v c im

Trong bt c mt h thng no th khch hng i n cc im cungcp dch v v ri khi h thng khi dch v c cung cp.V d:Cc h thng in thoi: khi s lng ln khch hng quay s ktni n mt trong nhng ng ra hu hn ca tng i.Trong mng my tnh: khi m gi tin c chuyn t ngun ti ch vi qua mt s lng cc nt trung gian. H thng hng i xut hinti mi nt qu trnh lu tm thng tin ti b m.H thng my tnh: khi cc cng vic tnh ton v tuyn lm vic cah thng yu cu dch v t b x l trung tm v t cc ngun khc.Nhng tnh hung ny c din t bng hnh v sau:

S kin n Server

Hng i

S kin i

Hnh 2-1 M hnh chung ca h thng hng i

Ngi ta m t tin trnh n v tin trnh phc v nh th no?H thng c bao nhiu server?

C bao nhiu v tr i trong hng i?C bt k quy tc ni b c bit no khng (yu cu dch v, mc u tin, h thng cn ri khng)?

c im ca h thng hng i

Miu t ca tin trnh n (phn b khong thi gian n)Miu t ca tin trnh phc v (phn b thi gian phc v)S lng server S lng cc v tr iCc quy tc hng i c bit:

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Quy tc phc v (FCFS, LCFS, RANDOM)Thi gian ri (phn b thi gian ri, khi m thi gian ri bt u )Mc u tinNhng lut khcVi mt mng c th ca hng i gm c cc thng tin sau:S kt hp gia cc hng iChin lc nh tuyn:

Xc nh (Deterministic)Da vo mt lpThng k

X l nghn mng (khi b m ti ch b y)S lng khch hng b suy gim

Hng i gc b nghnTi nh tuyn

Chng ta s xem xt v d v cc mng hng i n gin khc

S

S

S

S M

Hnh 2-2: V d v mng hng i m

S

S

S

S

Hnh 2-3 V d v mng hng i ng

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Phn tch h thng hng i hoc mng hng i bao gm:

Phn tch gii tchQu trnh m phng

C hai phng php trnKt qu gii tch t c:

Yu cu t tnh ton a ra kt qu chnh xc (khng xy ra li xc sut)

Nhng kt qu thu c (cc thng s dch v) c chia thnh hai nhm ln:

Dnh cho ngi s dngDnh cho cc nh cung cp phc vThng s quan trng cho ngi s dng:Tr hng iTng tr (bao gm tr hng i v tr phc v )S lng khch hng trong hng iS lng khch hng trong h thng (gm khch hng ch v khch

hng ang c phc v )Xc sut nghn mng (khi kch thc b m hu hn)

Xc sut ch phc vThng s quan trng cho cc nh cung cp dch v:Kh nng s dng server Kh nng s dng b mLi ch thu c (thng s dch v v cc xem xt v kinh t)Li ch b mt (thng s dch v v cc xem xt v kinh t) p ng nhu cu ca ngi s dngCht lng dch v (QoS):Tn tht (PDF, mean)Tr (PDF, mean)Jitter (PDF, mean)

a ra cc thng s trn thu c:

Hm phn b xc sutCc gi tr trung bnh o c cc thi im cc i, cc tiu

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Cc hm phn b xc sut cha ng y cc thng tin lin quann cc thng s quan tm. Tuy nhin, vic thit lp c cc hmny l kh thc hin.Phn tch h thng hng i c chia thnh:

Phn tch thi gian ngn (da trn mt thi im nht nh)Phn tch trong mt khong thi gian (trng thi n nh) (da trn

tham s v hn)Cu trc logic ca phn tch h thng hng i o c nhiu thng s thng k: mean-mean, moments, transform,

pdf Phn tch thi gian ngn s dng cho cc trong hp n gin- s

dng cc phng php m phng hay xp x Vic phn tch chnh xc khng th cho p dng cho qu trnh n nh-

s dng cc phng php xp x, nu khng th dng cc phngphp m phng.

Tip theo chng ta s c cc kt lun sau:

Kt lun chung : cc gi thit lin quan n c tnh v cu trcca h thng hng i t c kt qu chnh xc t nht l chocc thng s hiu nng trung bnh vi iu kin n nh.

2.1.2. Cc tham s hiu nng trung bnh

V d v h thng hng i n gin

Mt Server

S v tr trong hng il v hn

S n vi tc trung bnh

S i

Hnh 2-4 H thng hng i n gin

- tc n trung bnh , thi gian n trung bnh -1/ - tc phc v trung bnh, thi gian phc v trung bnh 1/Vi kch thc ca b m l v hn, quy tc phc v l FCFS(n trc phc v trc )Xt khong thi gian t, v xt nhng s kin n trong khong thigian ny:

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S kin ACc s n

S kin B

S kin C

t

t

t

Hnh 2-5. Cc s kin n trong thi gian t

S kin A: C 1 s kin n trong t S kinB: khng c s kin n trong t S kinC : C nhiu hn 1 s kin n trong t Gi s rng t 0 . Nh vy ta s c:- Pr{A}= t - Pr{B}= 1- t - Gi thitP{C}= 0 ,vi1/ l khong thi gian n trung bnh (thc t c phn b theohm m ca tin trnh n Poisson).Xt khong thi gian t v xt nhng s kin i trong khong thi gianny

S kin ACc s i

S kin B

S kin C

t

t

t

Hnh 2-6: Cc s kin i trong thi gian t

S kin A: C 1 s kin i trong t S kinB: khng c s kin i no trong t S kinC : C nhiu hn 1 s kin i trong t Gi s rng t 0. Nh vy ta s c:

Pr{A}= t

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Pr{B}= 1- t Gi thitPr{C}= 0 , vi 1/ l thi gian phc v trung bnh (thc tc phn b theo hm m.D l s kin ca 1 hoc nhiu s n AND vi s kin ca 1 hoc

nhiu s i trong khong t Gi sPr{D}=0 , (2-1)Thc ra, n ch ra rng khi t nh, s kin nhn (va i va n) lkhng xy ra.Ngoi cc gi thit trn v c tnh ca tin trnh n v tin trnh phcv, cn c thm cc gi thit sau:Tin trnh n l tin trnh Poisson vi tham s Khong thi gian n phn b theo hm m vi tham s1/

Thi gian phc v phn b theo hm m vi tham s1/Tin trnh n l c lp vi tin trnh phc v v ngc li phn tch h thng hng i cn hiu khi nim Trng thi hthng. C th nh ngha thng qua bin thch hp m t S phttrin theo thi gian ca h thng hng i. thun tin cho h thnghng i bin c chn s l s khch hng trong h thng ti thiim t.

Trng thi h thng tit = N(t)=S lng khch hng ti thiimt (2-2)

Tc l : pN (t)=Pr{N(t)=N} (2-3)

vi

pN (t) l k hiu ca trng thi thN ca h thng ti thi imt .

Pr{N(t)=N}l xc sut c N khch hng trong h thng ti thiimt .

C ngha l c N khch hng trong h thng ti thi imt .S dng trng thi u tin tit=0 , nu ta c th tm pN(t) th c thm t h thng c quan h v mt thi gian nh th no?Tip theo, cho thi gian t 0 .Xt cc trng thi c th ca h thng{0,1,}(bng ng s lngkhch hng trong h thng) ti thi im t ta c th tm trng thi cah thng ti thi im t+t nh sau:

p0 (t+t )= p0 (t)(1-t)+p1(t)t, N=0.

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pN (t+t )= pN (t)(1- t-t)+pN-1(t)t+ pN+1(t)t,N>0 (2-4)

ta lun c iu kin phn b chun:0,1)( =

t t pi

i (2-5)

Tc l chun ha cc pi(t), t0,thnh cc tnh cht phn b ri rctheo thi gian.Ta c th tnh gii hn khi t 0 v c h phng trnh vi phn:

0),()()()()(

0),()()(

11

100

>+++=

=+=

+ N t pt pt pdt

t dp

N t pt pdt

t dp

N N N N

(2-

6)

gii ta pho cho iu kin ban u.Gi s rng h thng hng i bt u ti thi im t=0 vi N khchhng trong h thng, iu kin ban u c vit nh sau:

pi (0)=0, vi iN

pN (0)=1, vi i=N (2-7)S dng iu kin ban u ph hp h thng c th c gii c gii php thi gian ngn (transient solution), mt gii php phc

tp thm ch cho cc h n gin nht.By gi ta xt gii php trng thi n nh (equilibrium solution), t.Khi ta c:

0,0)(

0,0)(0

>=

==

N dt

t dp

N dt

t dp

N

(2-8)

V vy, p0 (t)=p0 , vi N=0 pN (t)=pN , vi N>0 (2-9)

nh ngha = / vi ng rng h thng hng i n nh vi 0 (2-10)

Ga s tun theo iu kin phn b chun, ta c: pi = i (1- ), i=0,1, (2-11)

vi gii php trng thi n nh cho phn b trng thi vi


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gii php trng thi n nh khng ph thuc iu kin phn b banu. Tuy nhin, n cn iu kin rng tc n nh hn tc phcv.

Cc tham s hiu nng trung bnh

S lng trung bnh ca khch hng trong h thngNhc li rng phn b ca trng thi n nh cho s lng khch hngtrong h thng khi t. V vy, c th suy ra s khch hng trung bnhtrong h thng t phn b trng thi n nh ca h thng nh sau:

===

=

= 1)1(][

00 i

i

ii iip N E (2-

12)Kt qu trn khng p dng cho s trung bnh khch hng trong hthng ti mt khong thi gian ngnt (arbitrary timet ).

S lng trung bnh ca khch hng trong hng iCh rng s lng khch hng trong hng i th bng vi s lngkhch hng trong h thng tr i 1. S dng cng cc gi thit ta c:

=

=

===

=

=

= 11)1(

1)1(][

2

0111

p pip pi N E i

ii

ii

iQ

(2-13)Ch rng tng bt u ti=1, do s kin khch hng i ch ngkhi c nhiu hn 0 khch hng trong h thng.Ch rng(i-1)! , do ang tm s lng khch hng trung bnh trong

hng i.Thi gian trung bnh trong h thngThi gian ny c th c phn chia thnh hai thnh phn :Thi gian iThi gian phc vTnh ton cc tham s hiu nng ny i hi nhng gi thit thm datrn c tnh ca h thng hng i :Quy tc phc v khch hng : Gi s quy tc first-come, first served

l khch hng c phc v theo th t nh khi n h thngPhn b trng thi n nh pk, k=0,1,, cng ging nh phn b xc

sut ca s lng khch hng trong h thng.Thi gian phc v d trung bnh ca khch hng s dng phc v

khi tin trnh n xy ra vi tc 1/, cng ging nh vy. V vyc gi l c tnh khng nh.

S dng cc gi thit cho thi gian trung bnh trong h thng cakhch hng :

[ ])1(

111

000

=+=+=

=

=

=k

k

k

k

k

k

pk

p pk

W E (2-

14)

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Thi gian trung bnh trong hng i (thi gian i c phc v)Vi cc gi thit trn ta c:

[ ])1(0

==

=k

k Q p

k W E (2-

15)Ch rng thi gian trung bnh trong hng i bng vi thi gian trungbnh h thng tr i thi gian phc v:

[ ] [ ])1(

1)1(

11

=

== W E W E Q (2-16)

C th c kh nng rng khch hng phi ch c phc vS dng phn b trng thi n nh pk, k=0,1,ta ch rng lngkhch hng n lun phi i c phc v nu s lng khchhng ln hn 0 trong h thng.

V vy,P wait =1-p0 = (2-17)

S dng server ngha vt l ca tham s hiu nng l n a ra khong thi gian khiserver bn. v vy,

P busy =1-p0 = (2-18)Cc cch tip cn trnh by c s dng phn tch bt k mth thng hng i u phi c cc gi thit sau:

Tin trnh n l tin trnh poisson, c ngha l khong thi gian nc phn b theo hm m.Tin trnh n vi tc n thay i.H thng c mt hoc nhiu server Thi gian phc v c dng phn b hm mTin trnh n l c lp vi cc tin trnh phc v v ngc liC v hn cc v tr i hu hn trong h thngTt c cc gi thit to thnh lp n gin nht ca h thng hng i.

2.2. Nhc li cc khi nim thng k c bn

2.2.1. Tin trnh im

Cc tin trnh n l mt tin trnh im ngu nhin, vi tin trnh nychng ta c kh nng phn bit hai s kin vi nhau. Cc thng tin vs n ring l (nh thi gian phc v, s khch hng n) khng cnbit, do vy thng tin ch c th dng quyt nh xem mt s n cthuc qu trnh hay khng.

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M t tin trnh

Chng ta xem xt qui lut ca tin trnh im thng thng, ngha lloi tr cc tnh hung n kp. Xt s ln cuc gi n vi cuc gith i ti thi imTi :

0 = T0 < T1 < T2 < < ..< Ti < Ti+1< (2-19)Ln quan st th nht tiT0 = 0 .S cc cuc gi trong na khong thi gian m [0, t ] l Nt, y Nt lmt bin ngu nhin vi cc tham s thi gian lin tc v thi gian rirc, khi t tng th Nt khng bao gi gim.Khong thi gian gia hai ln n l:

X i = T i - T i-1 (2-20)Khong thi gian ny gi l khong thi gian gia hai ln n. S

phn b ca tin trnh ny gi l s phn b khong n.Tng ng vi hai bin ngu nhinNt v Xi , hai tin trnh ny c thc m t theo hai cch:Cch biu din sN t : khong thi giant gi khng i, v ta xt bin

ngu nhinN t cho s cuc gi trong khong thi giant .Cch biu din khongt i : s cc cuc gi n l hng s (n), v ta xt

bin ngu nhint i l khong thi gian din ran cuc gi.Mi quan h cn bn gia hai cch biu din th hin n gin nhsau:

Nt < nkhi v ch khi =

>= n

iin t X T

1

iu ny c biu din bng ng thc Feller - Jensen :{ } { }t T pn N p nt >=< vi n = 1, 2,.. (2- 21)

Phn tch tin trnh im c th da trn c hai cch ny, v nguyntc chng tng ng vi nhau. Cch biu din khong thi giantng ng vi vic phn tch chui thi gian thng thng.Cch biu din s khng song song vi phn tch chui thi gian. Sliu thng k c tnh ton trn mi n v thi gian v ta c cc mctrung bnh thi gian.

c tnh ca tin trnh im

Phn ny chng xem xt c tnh ca n thng qua cch biu din s.Tnh dng (tnh ng nht thi gian)(Stationarity-time homogeneity) :Tnh cht ny c th m t l cho d v tr no trn trc thi giancng vy, phn b xc sut tin trnh im l c lp vi thi imquan st. nh ngha sau y c s dng trong thc t:

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nh ngha : Cho tu t2 > 0 v vi mi 0k > . Xc sut m kcuc gi n trong khong thi gian [t1, t1+t2] l c lp vi t1,ngha l vi mi t, k ta c:

{ } { }k N N pk N N p t t t t t t t t === ++++ )()( 121121 (2-22)

y l mt trong nhiu nh ngha v tnh dng ca tin trnh im cccuc gi n.Tnh c lp (Independence)Tnh cht ny th hin l: tng lai ca tin trnh ch ph thuc votrng thi hin ti.

nh ngha : xc sut c k s kin (vi k nguyn v ln hn

hoc bng 0) trong khong [t1, t1+t2] l c lp vi cc s kintrc thi im t1 :

{ } { }k N N pn N N k N N p t t t t t t ==== )(|)( 120112 (2-23)

Nu iu ny ng vi mi t th tin trnh ny l tin trnh Markov: trngthi tip theo ch ph thuc vo trng thi hin ti, nhng c lp vivic n c c nh th no. y chnh l tnh cht khng nh.Nu tnh cht ny ch xy ra ti cc thi im no (v d thi imn), th nhng im ny c gi l cc im cn bng hay cc imti to. Khi tin trnh c nh gii hn, v ta cn lu li im ti togn nht.Tnh u n (Regularity)Nh ni ta loi tr cc tin trnh ca nhiu cuc gi vo mt thiim, vy ta c nh ngha sau:

nh ngha : mt tin trnh im c gi l u n nu xcsut xy ra vi nhiu hn mt s kin cng mt thi im bngkhng:

{ } 0)(,0:),(2)( =+ t ot khit o N N p t t t (2-24)

2.2.2. Tin trnh Poisson

Tin trnh Poisson ltin trnh imquan trng nht bi v vai tr can cng quan trng nh vai tr ca phn b chun trong phn bthng k. Tt c nhng tin trnh im ng dng khc u l dngtng qut ho hay dng sa i ca tin trnh Poisson. Tin trnhPoisson m t rt nhiu tin trnh trong i sng thc t, do n c tnhngu nhin nht.

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c tnh ca tin trnh Poisson :

Nhng c tnh c bn ca tin trnh Poisson l:Tnh dng

Tnh c lp ti mi thi imTnh u nHai tnh cht sau l tnh cht c bn, t tin trnh Poisson c cng ph thuc thi gian.T cc tnh cht trn ngi ta c th a ra cctnh cht khc biu din tin trnh Poisson, l:Biu din s: l s cc s kin n trong mt khong thi gian vi

di c nh c phn b theo tin trnh Poisson.Biu din khong thi gian: l cc khong thi gian Xi gia cc s

kin lin tip nhau c phn b theo hm m.

Tin trnh n Poisson s dng trong lu lng vin thng ca mngchuyn mch gi v mng my tnh. Thm vo tin trnh Poisson c s dng m t cc tin trnh nhiu v nghin cu hintng cc h in t xut hin trong cht bn dn, v trong cc ngdng khc Ba vn c bn c s dng nh ngha tin trnh n Poisson.Xt mt khong thi gian nh t (vi 0t ), nhHnh 2-7.

Hnh 2-7 Khong thi gian s dng nh ngha tin trnh l:Xc sut ca mt tin trnh n trong khong thi giant c nh

ngha l )t(ot + , vi 1t


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Phng sai : )()( 222 k E k E k = hay:

T k E k == )(2 (2-27)

Tham s l hng s t l, c xem l tham s tc :T

k E )(=

Phng trnh (2-25) m t tc n trung bnh ca tin trnhPoisson. Bnh thng gi tr trung bnhE(k) tin ti khng tngng vi T ln: T k E k ./1)(/ = vi ngha l T ln, phn bc quan h cht ch vi gi tr trung bnh T . Do nu mt thng s(ngu nhin) s cc tin trnh n n trong khong thi gianT ln (lntheo ngha T >>1, hocT >> 1/ ), n/T c th nh gi . Cngch l Te)0( p = . Khi T tng vi phn b nhE (k) = T , xcsut khng c tin trnh n no trong khong thi gianT tin nkhng vi e mT .

2.3. nh lut Little

Xem xt mt h thng hng i, khch hng n l mt tin trnh ngunhin. Cc khch hng n h thng cc thi im ngu nhin vch c phc v th khch hng s ri khi h thng.

2.3.1. Cng thc Little

Chng ta c k hiu nh sau:)( t N = S cuc gi n h thng ti thi imt .

t = S cuc gi i n h thng trong khong thi gian t (0,t ).t = S cuc gi ri khi h thng trong khong thi gian t (0,t ).iT = Thi gian ca cuc gi th i trong h thng (thi gian phc

v).Nh vy:

t N - S lng cuc gi trung bnh n h thng trong (0,t ) l :

=t

t t dt N t N

0

1

t - Mt cuc gi trong khong (0,t) l :t

t t

=

t T - Thi gian trung bnh ca cui gi trong h thng l :

== t i it

t T T

11

Gi s cc gii hn sau y tn ti :

t t t t t t T T N N

=== lim;lim;lim

C cng thc sau:T N = (2-28)

Cng thc trn c tn gi l nh l Little

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S cuc gi trung bnh trong h thng bng tch mt cuc givi thi gian chim knh trung bnh.

2.3.2. Chng minh cng thc LittleChng minh cng thc Little bng phng php hnh hc theo nhminh ha di y.

(t )

T 1

N (t )

T 2

T ii

(t )

Hnh 2-8

Xt trong khong (0,t ) :

Din tch phn gch cho: [ ] ==t

t t t t dt N S 0

)()(

Mt khc din tch ny cng bng : S= 1.=

t

i

Ti

1

Nh vy t

dt t N 0

)( = =

t

i

Ti

1

t

iit

o

t t

t

T

t dt N

t

== 11

tc l : t t t T N = (*)Nu gii hn sau y tn ti :

t t t t t t T T N N

=== lim;lim;lim (**)

T (*) v (**) T N = Cng thc c chng minh

15


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2.4. Cc m hnh hng i

2.4.1. K hiu Kendall

Bt k h thng xp hng no cng c m t bi :Tin trnh n

Nu cc khch hng n vo cc thi imt1, t2 tj th cc bin sngu nhin Pj=tj-tj-1c gi l cc thi im gia cc ln n. Ccthi im ny thng c gi thit l cc bin s ngu nhin c lpv c phn b ng nht IID (Independent and Identycallydistributed). Cc tin trnh n thng dng nht l :M : Tin trnh m (l tin trnh Markov hay tin trnh khng nh)Er : Tin trnh Erlang bcr Hr : Tin trnh siu s m bcr D: Tin trnh tt nh (deterministic)G: Tin trnh chung

Tin trnh phc v

Thi gian m mi cng vic tiu tn cn thit ti server gi l thi gianphc v. Cc thi gian phc v thng gi thit l cc bin s ngunhin IID. Cc tin trnh phc v thng dng nht cng ging nh thigian n.S lng cc b server: S lng cc server phc v cho hng i

Dung lng h thng Kch thc b nh m cc i

Qui m mt

S lng cc cng vic n ti hng i. Qui m mt lun l huhn trong cc h thng thc. Tuy nhin phn tch h thng vi qui mmt ln s d dng hn nu gi thit rng qui m mt l v hn.

Qui tc phc v

Th t m theo cc cng vic trong hng xp c phc v. Ccqui tc ph bin nht l n trc phc v trc FCFS (First ComeFirst Served), n sau phc v trc LCFS (Last Come First Served),theo vng trn RR (Round Robin), thi gian x l ngn nht phc vtrc SPT (Shortest Procesing Time First) v thi gian x l ngn nhtc c SRPT (Shortest Remaining Processing Time First)

K hiu Kendall

A/S/m/B/K/SDc s dng rng ri m t h thng xp hng A: Phn b thi gian gia cc ln n

S: Phn b thi gian phc vm: S lng server

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B:Kch thc b mK : Quy m mt SD: Quy tc phc vV d hng iM/D/1: M c ngha tin trnh n l tin trnh Markovkhng nh (vi thi gian gia cc ln n theo hm m);D thi gianphc v lun nh nhau (tt nh); 1 c mt server duy nht phc v.Phn B/K/SDca k hiu b loi tr cho thy rng dung lng cah thng v qui m mt l v hn v qui tc phc v l FCFS.

2.4.2. Qu trnh Sinh-T (Birth-Death)

Trng thi ca h thng c biu din bng s cc khch hng ntrong mt h thng. Khi c mt khch hng mi n th trng thi cah thng s thay i sang n+1, khi c mt khch hng ra i th trngthi h thng s thay i sang n-1, ta c lc chuyn tip trng thil qu trnh sinh t.

0 ii-1 i+121

2 0 1

1 2 3

...

i-2

i-1

i-1 i

i i+1

Hnh 2-9. Chui Markov ca mt qu trnh sinh-t

n : Tc ca ln nn

n : Tc ca ln iP n: Xc sut n nh trng thi n ca qu trnh sinh t ti trng thin

P n =n

n

...

...

21

110 .P 0 (2-29)

P0 - xc sut trng thi 0,Pn - xc sut trng thin

2.4.3. Hng i M/M/1

Lc trng thi

0 ii-1 i+121

...

Hnh 2-10 Chui Markov ca hng i M/M/1

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Tt c cc tc n u l ,

: Tc ca ln n : Tc ca ln i

P n=( ) n P 0 = n P 0

(2-30)Pn: Xc sut n nh trng thinP0 : Xc sut n nh trng thi 0

: Mt lu lng =

Trong trng hp ny s knh phc v bng 1, ch c 1 server Cc cng thc tnh ton:Xc sut c n khch hng trong h thng

P n= (1- ) n ; n=1,2,... (2-31)P 0 = (1- )

S lng trung bnh cc khch hng trong h thng

L=E(n)= 1 (2-33)

Phng sai: 2n = 2)1(

(2-34)

Tham s thi gian

Thi gian trung bnh ca 1 khch hng trong h thng: W

W = L = )1(

=

1 (2-35)

Thi gian phc v trung bnh cho mt khch hng : WS

W S =

1 =

(2-36)

Thi gian trung bnh ca khch hng trong hng i

W q = W- W S = )1( -

=)1(

2

(2-37)Chiu di hng i

S lng trung bnh cc khch hng trong h thng

L= 1 (2-38)

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S lng trung bnh cc job trong server:L S L S = 1P(n>=1) =1- P(n=0) =1-(1- ) = (2-39)

S lng trung bnh ca cc cng vic trong hng i Lq

L q = L- LS =

1 = 1

2

(2-40)V d: Cho Switch nhn cc bn tin n tc 240bn tin/pht. dibn tin c phn b hm m vi chiu di trung bnh l 100 k t. Tc truyn bn tin i khi h thng l 500 k t/giy. Tnh cc tham ssau :Thi gian trung bnh ca bn mt tin trong h thngS bn tin trung bnh trong h thngTnh chiu di hng i v thi gian i trung bnh

Bi gii : Xt h thngM/M/1:Tc n 4

60240 == bn tin/giy

Tc phc v 5100500 ==

Mt lu lng 8.054 ===

S bn tin trong h thng

L=E(n)= 48.01

8.01

=

= bn tin

Thigian trung bnh ca bn tin trong h thng

W = 144

== L

(s)

Chiu di hng i Lq

L q = 2,38,018,0.8,0

1

2

=

=

bn tin

Thi gian i trung bnh Wq

W q = 8,042,3

)1(

2

===

q L (s)

19


28/143


29/143

Po= [ )1(!

)(!

1)(

1

0

+

= cc

nc

cc

n

n ] 1 (2-48)

Xc sut xut hin hng i

Pq =)1(!

)(

cc Po c (cng thc Erlang) (2-49)

di hng i:

Lq = Pq. 1 (2-50)

Thi gian i:

Wq =

Lq (2-51)

2.5. L thuyt lu lng

2.5.1. Khi nim v lu lng v n v Erlang

nh ngha

Trong l thuyt lu lng vin thng chng ta thng s dng thutng lu lng biu th cng lu lng, tc l lu lng trongmt n v thi gian. Thut ng v lu lng c ngun gc t ting

v c ngha l bn rn.Theo (ITU-T,1993) nh ngha nh sau:

Cng lu lng : Mt lu lng tc thi trong mtnhm ti nguyn dng chung l s ti nguyn bn ti thi im.

Nhm ti nguyn dng chung c th l mt nhm phc v nh ngtrung k. Tin hnh thng k mt lu lng hin ti c th tnh toncho mt chu k T, ta c cng lu lng trung bnh l:

=T

dt t nT

T Y 0

)(1

)( (2-52)

Vin(t) l s thit b s dng ti thi im tLu lng mang

Ac = Y = A c gi l lu lng c thc hin bi mt nhmphc v trong khong thi gianT (hnh 3.1).Trong thc t, thut ng cng lu lng thng c ngha l

cng lu lng trung bnh.

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Hnh 2-13 Lu lng mang (mt )( bng s thit b bn) l mt hmthi gian (ng cong C). Lu lng trung bnh trong khong thi gian

T (ng cong D)

n v ca cng lu lng lErlang (k hiu lErl), y l nv khng c th nguyn. (Ra i 1946 ghi nh cng n ca nhton hc ngi an mch A.K Erlang (1878-1929), ngi tm ra lthuyt lu lng in thoi).

Khi lng lu lng : l tng lu lng mang trong chu kT vc o bng n v Erlang - gi (Eh) (theo nh tiu chun ISOnhng n v tiu chun c th l Erlang giy, nhng thng thngn v Erlang gi thng s dng nhiu hn).Lu lng mang khng th vt qu s lng ca ng dy. Mtng dy ch c th mang nhiu nht mt Erlang. Doanh thu ca ccnh khai thc t l vi lu lng mang ca mng vin thng. i vi in thoi c nh thng th c Ac =0,01 0,04 Erl i vi c quan : 0,04 0,06 Erl

Tng i c quan: 0,6 Erl in thoi tr tin : 0,7 Erl

Lu lng pht sinh A

Lu lng pht sinh l lu lng c mang nu khng c cuc gino b t chi do thiu ti nguyn, v d nh vi s knh khng b giihn.Lu lng pht sinh l mt gi tr l thuyt khng o lng c ch

c th c lng thng qua lu lng mang.

22


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Ta gi mt cuc gi l , l s cuc gi trung bnh n trong mtn v thi gian v gi s l thi gian phc v trung bnh. Khi lulng pht sinh l:

s A . = (2-53)

T phng trnh ny ta thy rng n v lu lng khng c thnguyn. nh ngha ny ph hp vi nh ngha trn vi iu kinknh phc v khng b gii hn. Nu s dng cho mt h thng vinng lc gii hn ta c s xc nh ph thuc vo h thng.Ngoi ra c th c tnh: A = / ( : tc phc v)

Lu lng tn tht Ar

Lu lng tn tht l chnh lch gia lu lng pht sinh v lulng mang. Gi tr ny ca h thng gim khi nng lc ca h thngtng.

Ar = A Ac (2-54)Lu lng pht sinh l mt tham s s dng trong tnh ton l thuytnh c. Tuy nhin, ch c lu lng mang thng ph thuc vo hthng thc mi l tham s o lng c trong thc t.Trong h thng truyn dn s ta khng ni v thi gian phc v m ch ni v cc tc truyn dn. Mt cuc giao dch c th l qu trnhtruyn s n v (nh bits hay bytes).Nng lc h thng l , ngha l tc bo hiu s liu, c tnhbng n v trn giy (v d bt/s). Nh vy thi gian phc v cho mt

giao dch nh th tc l thi gian truyn s l s/ n v thi gian (vd nh giy-s); ngha l ph thuc vo .Nu trung bnh c cuc giao dch n trong mt n v thi gian, th s dng h thng s l:

s.= (2-55)

Vi: 01 .

2.5.2. H thng tn tht (Loss System) v cng thc Erlang BCng thc Erlang B

Cng thc Erlang c m t bng ba thnh phn: cu trc, chinlc v lu lng:Cu trc : Ta xem xt mt h thng c n knh ng nht hot ngsong song v c gi l nhm ng nht (cc server, knh trung k,khe slot).Chin lc: Mt cuc gi ti h thng c chp nhn nu cn tnht mt knh ri (mi cuc gi ch cn mt knh ri). Nu tt c ccknh u bn th cuc gi s b hu b v n s b loi b m khng

23


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gy mt nh hng no sau (cuc gi b loi b c th c chpnhn trn mt tuyn khc). Chin lc ny c gi l m hnhLoss(tn tht) Erlang hay m hnhLCC(Lost Calls Cleared).Lu lng : Gi s rng trong khong thi gian dch v c phn btheo hm m (s m ), v tin trnh s dng l tin trnh Poisson vitc . Loi lu lng ny c gi l PCT -I (Pure Chance TrafficType I). Tin trnh lu lng ny s tr thnh tin trnh Mackov ngin x l bng ton hc.Cng thc Erlang B biu th mi quan h gia lu lng xut hin,lng thit b, v xc sut tn hao nh mt hm s c s dngrng ri nh l l thuyt tiu chun cho vic lp k hoch trong hthng vin thng, v vy cng thc Erlang B cha ng nhng tiuchun sau:Cc cuc gi xut hin mt cch ngu nhin:

Xc sut xy ra s c cuc gi l lun c nh bt chp thi gian (xcsut c nh xy ra s c ca cuc gi).Xc sut xy ra s c ca cuc gi khng b nh hng bi cc cuc

gi trc (khng cn st li nhng c im ca cuc gi trc).Trong thi gian rt ngn, khng c cuc gi no xut hin hoc ch c

mt cuc gi xut hin (cc cuc gi ri rc).Dng tn hao trong khi vn hnh khi tt c cc mch u bn:Trong dng tn hao vn hnh ny, cuc gi khng th lin lc c

khi tt c cc mch u bn. Trong trng hp tn hiu c

gi ra ngoi v d ng ra tr nn thng sut sau khi tn hiu bnc gi ra th cuc gi vn khng c kt ni.Nhm mch ra l nhm trung k c kh nng s dng ht.Thi gian chim dng ca cc cuc gi gn ng vi phn b hmm.Cc mch vo th v hn, cn cc mch ra th hu hn.Xc sut tn hao cuc gi trong cng thc Erlang B c trnh bytrong cng thc sau:

En(A)= E n,1 (A) = P(n) =

!...

!21

!2

n A A

A

n A

n

n

++++=

=

n

i

i

n

i A

n A

0 !

!

(2-56)Vi A -Lu lng pht sinh (A= .s)n - S knhVic tnh ton cng thc trn khng ph hp c khi c An v n! tngqu nhanh, khi my tnh s b trn s do vy ngi ta thng p

dng mt s kt qu tnh ton v a ra cng thc sau:

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)(.)(.

)(1

1

A E A x A E A

A E x

x x

+= vi E0 (A) = 1

(2-57)T quan im ton ng dng, hm tuyn tnh c n nh cao nhtta c:

)(1)( 1 A I A x A I x x += viI0 (A) = 1 (2-58)

y In (A) = 1/ En (A) (2-59)Cng thc ny hon ton chnh xc, thm ch vi cc gi tr(n.A) lnvn khng xut hin li. y l cng thc c bn cho rt nhiu bngs ca cng thc Erlang BV d : Cho tc gi n bng mt cuc gi trn 1 pht, thi giantrung bnh ca 1 cuc gi l 3 pht, s knh phc v bng 4. Tnh xcsut tn tht P theo 2 cng thc trn.Cch 1:Lu lng pht sinh A= Erl t 33.1. ==

P(n)= 206,0

!43

!33

23

31

!43

432

4

=++++

ngha : c 1/5 cc cuc gi ti s thu bao b tn tht (b bn)Cch 2:

E )(.4 )(.)( 334 A E A A E A A +=

E 1)(0 = A

E 43

313

)(.1)(.

)(0

01 =+

=+

= A E A

A E A A

E 179

43

.32

43

.3

)(.2)(.

)(1

12 =

+=

+=

A E A A E A

A

E 7827

179

.33

179

.3

)(.3)(.

)(2

23 =

+=

+=

A E A A E A

A

E 2061.039381

179

.34

179.3

)(.4)(.

)(3

34 =

+=

+=

A E A

A E A A

Cc c tnh lu lng ca cng thc Erlang B

Bit c xc sut trng thi ta c th bit c cc s o hiu nng.25


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nghn theo thi gian:l xc sut m tt c cc trung k b chim timt thi im bt k bng vi phn thi gian tt c cc trung k bchim trn tng thi gian (3.13) nghn theo cuc gi:xc sut m mt cuc gi bt k b mt bngt l s cuc gi b chn trn tng cc cuc gi.

nghn lu lng : )( A E A

Y AC n=

=

Ta c E = B = C, bi v cng cuc gi c lp vi trng thi, ychnh l tnh cht PASTA (Poisson Arrival See Time Average), n phhp vi tt c cc h thng tun theo tin trnh Poisson. Trong tt ccc trng hp khc, t nht c ba tham s o tc nghn l khc nhau.V d : Cho thi gian xem xtT l 1h ,lu lng pht sinh A l 1 Erl,s knh l n=3, thi gian phc v trung bnh cho mt cuc gi l 3pht. Tnh s lng cuc gi b nghn trong khong thi gianT , tnhlu lng tn tht, lu lng mang?Bi gii :S cuc gi tn tht :

N loss = B.N=P(n).N

N= 2060.31

T.SA

T. === cuc gi

31==

S A

cuc gi/pht

B=P(n)=161

!31

!21

11

!31

!iA!n

A

3

3

n

0i

i

n

2=

+++=

=

N loss = 25.120.161 cuc gi

ngha : Trong 20 cuc gi dn c 1.25 cuc gi b nghn khngc phc v.Lu lng tn tht :

Ar= A.C = 1.161

161

= (Erl)

Lu lng mang

Ac = Y= A(1-P(n)) = 1.(1-161 )= 15/16 (Erl)

2.5.3. H thng tr (Delay) v cng thc Erlang C

Xt lu lng vi tin trnh poisson (Khng gi hn v ti nguyn).Phn b thi gian phc v l PCT-1. H thng hng i ny c tn l

26


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h thng tr Erlang.Trong h thng ny th lu lng mang s bnglu lng pht sinh v khng c khch hng no b nghn.

Cng thc Erlang C

Giw l bin ngu nhin ca thi gian i th ta c xc xut binw>0 l:

E n,2 (A) = P(w>0) =

Ann

n A

n A A

A

Ann

n A

nn

n

+

++++

.!)!1(

...!2

1

.!

12

(A


36/143

Xc sut cuc gi c phc v:

Sn = 1- E 95

94

1)(,2 == An

Cch 2:

)(1

)(11

2,13,13,2 A E A E E =

E 1)(0,1 = A

E 31

212

)(.21

)(.2)(

0,1

0,11,1 =+

=+

= A E

A E A

E 52

104

3

2.22

32

.2

)(.22

)(.2)(

1,1

1,12,1 ==

+=

+=

A E

A E A

E 194

52

.23

52

.2

)(.3)(.

)(2,1

2,13,1 =

+=

+=

A E A

A E A A

49

25

.4

19)(

1)(

1)(

1

2,13,13,2

=== A E A E A E

E 94

)(3,2 = A

2.6. H thng hng i c u tin

Cc khch hng sau khi n h thng c th phi ng vo hng i,do cn c cc qui tc nht nh m bo khch hng c phcv mt cch nhanh nht. Tuy nhin kch thc ca hng i khngphi l mt gi tr v hn, chnh nguyn nhn ny l ngun gc cacc thng s khc lin quan n hng i v t chc hng i.Hng i l mt quan im ton hc v tnh hung trong th gii thc,n a ra cc phn tch c kh nng nh gi hiu sut lu lng cakhch hng (nh cc cuc gi, cc t bo ATM, hay cc mng LAN)khi i qua hng i.C t nht 7 tham s thng s dng trong h thng l:Kt cu cc mc u tin (cc lp) ca khch hng n, nu c hn

mt mc u tin trong hng i (v d trong ca hng th nam giiv ph n l hai lp) do thi gian phc v trong cc mc utin l khc nhau.

Vi mi mc u tin khch hng c phn b tin trnh n ring.Vi mi mc u tin, kch thc hay s khch hng to ra lu lng.

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Phn b thi gian phc v ca Server hng i (hnh ng caServer). Trong nhiu mng truyn thng thng gi l phn bchiu di.

Cc qui tc ca hng i.

Chiu di ti a ca hng i (ph thuc vo kch thc ca Buffer).Phn ng ca khch hng khi b tr, tc nghn,

2.6.1. Qui tc v t chc hng i

Mt cch cc phn t mng x l cc dng lu lng n l sdng cc thut ton xp hng sp xp cc loi lu lng.Khch hng ang i trong hng i c phc v c th cla chn theo nhiu cch, u tin chng ta quan tm n 3 loi quitc sau:

FCFS (First Come First Served ) n thng c gi l hng i cngbng hay hng i gi v qui tc ny thng xut hin trong cucsng hng ngy ca chng ta. N c xem nh l FIFO, ch lFIFO ch s dng trong hng i khng s dng cho ton hthng.

LCFS ( Last Come First sever) l chu trnh ngn xp, nh vic xphng trn gi ca ca hng.v.v qui tc ny cng xem nh LIFO( Last In First Out)

SIRO (Sevice In Random Order) tt c cc khch hng ang i tronghng i c xc sut c chn phc v nh nhau. N cnc gi l RANDOM hay RS (Random Selection).

Hai qui tc u tin ch s dng trong ln n m c xt, trong khiqui tc th 3 khng c xem nh tiu chun v khng yu cu nh.(Ngc vi hai qui tc u).Nh ba trng hp cp trn tng thi gian i cho tt c cckhch hng l nh nhau. Qui tc ca hng i ch quyt nh lm sao xc nh tng thi gian i ca khch hng. Trong chng trnhiu khin h thng hng i c th c nhiu qui tc phc tp. Trongl thuyt hng i chng ta gi thit l tng lu lng pht sinh l clp vi qui tc ca hng i.Vi h thng my tnh chng ta thng c gng gim tng thi giani, n c th thc hin khi s dng thi gian phc v nh l tiuchun:SJF (Shortest Job First): Vic u tin ngn nht.SJN (Shortest Job Next): Vic tip theo ngn nht.SPF (Shortest Processing Time First): Thi gian x l u tin ngnnht.Qui tc ny c gi thit nh l chng ta bit thi gian phc v trong

s pht trin, qui tc hng i ny tiu hnh ho tng thi gian i chott c cc khch hng.

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Nh ni trn qui tc nh hng ti thi gian n hoc thi gian phcv. Mt s tho hip gia cc qui nh c c bi:RR (Round Robin): mt khch hng c phc v cho trong mt

khong thi gian c nh (Time slice). Nu dch v khng honthnh trong khong thi gian ny, th khch hng tr li hng i lFCFS.

PS (Processor Sharing): tt c khch hng chia s dung lng dch vbng nhau.

FB (Foreground-Background): qui tc ny c gng thc hin SJF mkhng bit n thi gian phc v sau ny. Server s cung cp dchv khch hng c thi gian phc v t nht. Khi tt c cc khchhng c c thi gian phc v ging nhau, FB c xc nhnh l PS.

Qui tc cui cng l qui tc ng do qui tc hng i ph thuc volng thi gian s dng trong hng i.T cc qui tc trn nhng thut ton xp hng hay dng l:Xp hng vo trc ra trc (FIFO Queuing).Xp hng theo mc u tin (PQ - Priority Queuing).Xp hng tu bin (CQ - Custom Queuing).Xp hng theo cng bng trng s (WFQ - Weighted Fair Queuing).

Xp hng vo trc ra trc (FIFO Queuing)

Trong dng n gin nht, thut ton vo trc ra trc lin quan nvic lu tr gi thng tin khi mng b tc nghn v ri chuyn tip ccgi i theo th t m chng n khi mng khng cn b tc na. FIFOtrong mt vi trng hp l thut ton mc nh v tnh n gin vkhng cn phi c s thit t cu hnh nhng n c mt vi thiu st.Thiu st quan trng nht l FIFO khng a ra s quyt nh no vtnh u tin ca cc gi cng nh l khng c s bo v mng nochng li nhng ng dng (ngun pht gi) c li. Mt ngun pht gili pht qu ra mt lu lng ln t ngt c th l tng tr ca cclu lng ca cc ng dng thi gian thc vn nhy cm v thi gian.FIFO l thut ton cn thit cho vic iu khin lu lng mng trong

giai on ban u nhng vi nhng mng thng minh hin nay i hiphi c nhng thut ton phc tp hn, p ng c nhng yu cukht khe hn.

Xp hng theo mc u tin (PQ - Priority Queuing)

Thut ton PQ m bo rng nhng lu lng quan trng s c cs x l nhanh hn. Thut ton c thit k a ra tnh u tinnghim ngt i vi nhng dng lu lng quan trng. PQ c th thchin u tin cn c vo giao thc, giao din truyn ti, kch thc gi,a ch ngun hoc i ch ch ...Trong thut ton, cc gi c tvo 1 trong cc hng i c mc u tin khc nhau da trn cc mc u tin c gn (V d nh bn mc u tin l High, Medium,Normal, v Low) v cc gi trong hng i c mc u tin cao s

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2.6.2. u tin ca khch hng trong hng i u tin

Khch hng c chia thnh p lp u tin. Khch hng lp u tin

k c u tin cao hn so vi khch hng lp u tin k+1. Hngi u tin li oc chia thnh cc nhm sau:Khng u tin phc v trc (Non-preemptive hay l HOL - Headof the Line): Khch hng n vi mc u tin cao hn so vi khchhng ang c phc v th vn phi ch cho n khi server phc vxong khc hng ny (v phc v xong tt c cc khch hng khc cmc u tin cao hn n).u tin phc v trc (preemptive): Vic phc v khch hng cquyn u tin thp s b ngng li khi c mt khch hng m quynu tin ca n cao hn n h thng. u tin phc v trc li c thchia thnh cc nhm nh sau:Phc hi u tin (preemptive resume), khi m s phc v c tip

tc t thi im m n b ngt qung trc .u tin khng ly mu li (preemptive without resampling), khi m s

phc v bt u li t u vi khong thi gian phc v khng i.u tin ly mu li (preemptive with resampling), khi m s phc v

bt u li vi khong thi gian phc v mi.

2.6.3. Duy tr qui tc hng i, lut Kleinrock

Gi thit thi gian phc v ca khch hng l c lp vi qui tc cahng i. Do dung lng ca Server l hn ch v c lp (chiu dihng i) v sau mt thi gian Server t n ngng v tc phcv b gim.Chng ta gii thiu hai hm thng p dng rng ri trong l thuythng i:

Hm ti U (t)

L hm ph thuc thi gian, n phc v khch hng n ti thiim t , hm U(t) c lp vi qui tc ca hng i. Gi tr trung bnh

ca hm ti lU(t)= E{U(t)}.Thi gian i o W (t)

L thi gian i ca khch hng khi anh ta n ti thi im t, thigian i o ph thuc vo qui tc hng i, gi tr trung bnh lW=E{W(t)}. Nu qui tc hng i l FCFS thU(t)=W(t), trong tin trnhPoisson th thi gian i o s bng thi gian i thc t.

nh l: Lut Kleinrock:

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= AV A

W A ii 1.

. =const

(V l thi gian phc v trung bnh thi im bt k)

Thi gian i trung bnh cho tt c cc loi khch hng (lp) btc ng bi lu lng ti ca lp ang xt l c lp vi qui tcca hng i.

2.6.4. Mt s hng i n server

Hnh 2-16 Mt s loi hng i n server thng gp

2.6.5. Kt lun

L thuyt hng i c nghin cu ngay t trong mng chuynmch knh, tuy nhin vic p dng trong mng chuyn mch knh cnhn ch, sau c nghin cu su rng trong mng chuynmch gi vi vic ng gi d liu. Cc tn hiu thoi truyn thngc s ho, ng gi v chuyn ti trong mng gi nh l mt phnc s ca mng d liu.Tin trnh im l tin trnh quan trng n cho php phn bit cckhch hng n (cc s kin) v n l mt tin trnh ngu nhin vicc tnh cht nh: tnh c lp, tnh u n ti mi thi im v tnhdng. Tin trnh Poisson l mt tin trnh im v l tin trnh quantrng nht. Cc tin trnh khc ch l rt gn hay pht trin ca tintrnh Poisson. Tin trnh Poisson l tin trnh m t nhiu tin trnh

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trong i sng thc t nn n l tin trnh ngu nhin nht do vy nng vai tr nh l mt tin trnh chun trong phn b thng k.Cc khch hng n (gi hay cuc gi) mt Server n c th cphc v ngay hoc phi mt mt khong thi gian ch no cho nkhi Server ri v thc hin tip nhn x l. Cc qui tc phc v cckhch hng i c phc v c thit lp cho cc Server qua cc khch hng ln lt c phc v theo mc u tin ca mnh dovy cc khch hng c u tin khc nhau th c thi gian ch khcnhau. Cc thng s ny c quyt nh bi thut ton xp hng cahng i v cng t nh hng ti QoS ca cc loi dch v cungcp trn mng.Cc thng s ca hng i c xc nh thng qua l thuyt xc sutthng k, nh l Little, qui tc duy tr hng i Kleinrock v quan trnghn c l cc tin trnh i - n ca khch hng l cc tin trnhPoisson vi phn b hm m cng vi thut ton xp hng ca n.Xc nh cc thng s hng i nh: chiu di hng i cc thiim bt k hoc ngay c khi c khch hng, qua a ra ccphng n iu khin lu lng trn mng cho ph hp nhm gimthiu cc s c trn mng nh gi c hiu sut s dng tinguyn ng thi xc nh c cp QoS m c th cung cp trnmng, l c s cho vic thit k cc mng h thng vin thng sauny.

2.7. Bi tp (Pending)

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Chng 3 Mng hng i3.1. Mng ni tip

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Chng 4 nh tuyn trong

mng thng tin4.1. Yu cu v nh tuyn trong mng thng tin4.1.1. Vai tr ca nh tuyn trong mng thng tin

4.1.2. Cc khi nim trong l thuyt graph

Phn ny gii thiu cc thut ng v cc khi nim c bn nhm mt cc mng, graph, v cc thuc tnh ca n. L thuyt graph l mtmn hc xut hin t lu, nhng l thuyt ny c mt s thut ngc chp nhn khc nhau dng cho cc khi nim c bn. V th cth s dng mt s thut ng khc nhau lp m hnh graph chomng. Cc thut ng c trnh by di y ny l cc thut ng c cng nhn v c s dng thng xuyn chng ny.Mtgraph G, c nh nghi bi tp hp cc nhV v tp hp cccnh E . Cc nh thng c gi l ccnt v chng biu din v tr(v d mt im cha lu lng hoc mt khu vc cha thit b truynthng). Cc cnh c gi l cc lin kt v chng biu dinphng tin truyn thng. Graph c th c biu din nh sau:

G=(V, E)Hnh 4.1 l mt v d ca mt graph.

Hnh 4.1. Mt graph n gin

Mc d theo l thuyt,V c th l tp hp rng hoc khng xc nh,nhng thng thngV l tp hp xc nh khc rng, ngha l c thbiu din

V={v i | i=1,2,......N}Trong N l s lng nt. Tng tE c biu din:

E={ei | i=1,2,......M}

Mt lin kt,ej , tng ng mt kt ni gia mt cp nt. C th biudin mt lin ktej gia nt i v k bi

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e j =(v i ,v k )hoc bi

e j =(i,k)Mt lin kt gi l i ti mt nt nu nt l mt trong hai imcui ca lin kt. Nti v k gi lk nhau nu tn ti mt lin kt (i,k ) gia chng. Nhng nt nh vy c xem l cc ntlng ging .Bc ca nt l s lng lin kt i ti nt hay l s lng nt lngging. Hai khi nim trn l tng ng nhau trong cc graph thngthng. Tuy nhin vi cc graph c nhiu hn mt lin kt gia cngmt cp nt, th hai khi nim trn l khng tng ng. Trongtrng hp , bc ca mt nt c nh ngha l s lng lin kti ti nt .Mt lin kt c th c hai hng. Khi th t ca cc nt l khng c nghi. Ngc li th t cc nt c ngha. Trong trng hp th tcc nt c ngha, mt lin kt c th c xem nh l mtcung vc nh ngha

a j =[v i ,v k ] hoc n gin hn

a j =[i,k] k c gi lcn k hng ra i vii nu mt cung [i,k ] tn ti vbc hng raca i l s lng cc cung nh vy. Khi nimcn khng vo v bc cn k hng vo cng c nh nghatng t.

Mt graph gi l mtmng nu cc lin kt v cc nt c mt tronglin kt c cc thuc tnh (chng hn nh di, dung lng, loi...).Cc mng c s dng m hnh cc vn cn quan tm trongtruyn thng, cc thuc tnh ring bit ca nt v lin kt th lin quann cc vn c th trong truyn thng.S khc nhau gia cc lin kt v cc cung l rt quan trng c vvic lp m hnh cho mng ln qu trnh hot ng bn trong ca ccthut ton, v vy s khc nhau cn phi lun c phn bit r rng.V mt hnh hc cc lin kt l cc ng thng kt ni cc cp ntcn cc cung l cc ng thng c mi tn mt u, biu dinchiu ca cung.Mt graph c cc lin kt gi lgraph v hng , tuy nhin mtgraph c cc cung gi l graph hu hng . Mt graph hu hngc th c c cc lin kt v hng. Thng thng , cc graph cgi s l v hng, hoc s phn bit l khng c ngha.C th c kh nng xy ra hin tng xut hin nhiu hn mt lin ktgia cng mt cp nt (iu ny tng ng vi vic c nhiu knhthng tin gia hai chuyn mch). Nhng lin kt nh vy c gi lcc lin kt song song . Mt graph c lin kt song song gi l mtmultigraph .

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Cng c kh nng xut hin cc lin kt gia mt nt no v chnhnt . Nhng lin kt c gi l ccself loop . Chng t khi xuthin v thng xut hin do vic xem hai nt nh l mt nt trong qutrnh lp m hnh graph cho mt mng hoc pht sinh trong qu trnhthc hin mt thut ton c vic hp nht cc nt. Hnh 4.2 minh homt graph c cc lin kt song song v cc self loop. Mt graph khngc cc lin kt song song hoc cc self loop gi l mtgraph ngin . Vic biu din v vn dng cc graph n gin l tng i ddng, v vy gi thit rng cc graph c xem xt l cc graph ngin. Nu c s khc bit vi gi thit ny, chng s c ch ra.

4.2. Cc m hnh nh tuyn qung b (broadcast routing)

4.2.1. Lan trn gi (flooding)

Mt dng mnh hn ca nh tuyn ring bit l lan trn gi. Trong

phng thc ny, mi gi i n router s c gi i trn tt c ccng ra tr ng m n i n. Phng thc lan trn gi ny hinnhin l to ra rt nhiu gi sao chp (duplicate). Trn thc t, s giny l khng xc nh tr khi thc hin mt s bin php hn chqu trnh ny.Mt trong nhng bin php l s dng b m bc nhy trongphn tiu ca mi gi. Gi tr ny s b gim i mt ti mi bcnhy. Gi s b loi b khi b m t gi tr khng. V mt l tng,b m bc nhy s c gi tr ban u tng ng vi di tngun n ch. Nu nh ngi gi khng bit di ca ng i,

n c th t gi tr ban u ca b m cho trng hp xu nht. Khi gi tr ban u s c t bng ng knh ca mng con.Mt k thut khc ngn s lan trn gi l thm s th t vo tiu cc gi. Mi router s cn c mt danh sach theo nt ngun ch ranhng s th t t ngun c xem xt. trnh danh schpht trin khng gii hn, mi danh sch s tng ln bi s mk ch ra rng tt c cc s th t nk c xem. Khi mt gi i ti,rt d dng c th kim tra c gi l bn sao hay khng. Nu nggi l bn sao th gi ny s b loi b.Lan trn gi c u im l lan trn gi lun lun chn ng ngn

nht. C c u im ny l do v phng din l thuyt n chn ttc cc ng c th do n s chn c ng ngn nht. Tuynhin nhc im ca n l s lng gi gi trong mng qu nhiu.S dng lan trn gi trong hu ht cc ng dng l khng thc t.Tuy vy lan trn gi c th s dng trong nhng ng dng sau.Trong ng dng qun s, mng s dng phng thc lan trn gi

gi cho mng lun lun hot ng tt khi i mt vi qun ch.Trong nhng ng dng c s d liu phn b, i khi cn thit phi

cp nht tt c c s d liu. Trong trng hp s dng lantrn gi l cn thit. V d s dng lan trn gi gi cp nht bnnh tuyn bi v cp nht khng da trn chnh xc ca bngnh tuyn.

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Phng php lan trn gi c th c dng nh l n v so snhphng thc nh tuyn khc. Lan trn gi lun lun chn ngngn nht. iu dn n khng c gii thut no c th tmc tr ngn hn.

Mt bin i ca phng php lan trn gi l lan trn gi c chn lc.Trong gii thut ny, router ch gi gi i ra trn cc ng m i theohng ch. iu c ngha l khng gi gi n nhng ng mr rang nm trn hng sai.

4.2.2. nh tuyn bc ngu nhin (random walk)

Trong phng php nh tuyn ny, router s chuyn gi i n trnmt ng u ra c chn mt cch ngu nhin. Mc tiu caphng php ny l cc gi lang thang trong mng cui cng cngn ch. Vi phng php ny gip cho qu trnh cn bng ti giacc ng. Cng ging nh phng php nh tuyn lan trn gi,phng php ny lun m bo l gi cui cng s n ch. So viphng php trc th s nhn rng gi trong mng s t hn. Nhcim ca phng php ny l ng t ngun n ch c th dihn ng ngn nht. Do tr ng truyn s di hn s tr ngnnht thc s tn ti trong mng.

4.2.3. nh tuyn khoai ty nng (hot potato)

nh tuyn ring bit l loi nh tuyn m router quyt nh tuyn ich da vo thng tin bn thn n lm lt c.

y l mt thut ton tng thch ring bit (isolated adaptivealgorithm). Khi mt gi n mt nt, router s c gng chuyn gi icng nhanh cng tt bng cch cho n vo hng ch u ra ngnnht. Ni cch khc, khi c gi i n router s tnh ton s gi cnm ch truyn tren mi ng u ra. Sau n s gn gi mivo cui hng ch ngn nht m khng quan tm n ng s iu. Hnh 4-17biu din cc hng ch u ra bn trong mt router timt thi im no . C ba hng ch u ra tng ng vi 03ng ra. Cc gi ang xp hng trn mi ng ch c truyni. Trong v d y, hng ch nF l hng ch ngn nht vi ch c mt gi nm trn hng ch ny. Giu thut khoai ty nng do st gi mi n vo hng ch ny.

Hnh 4-17. Hng ch bn trong router

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C th bin i tng ny mt cht bng cch kt hp nh tuyntnh vi gii thut khoai ty nng. Khi gi i n, router s tnh n cnhng trng s tnh ca ng dy v di hng ch. Mt kh nngl s dng la chn tnh tt nht tr khi di hng ch ln hn mtngng no . Mt kh nng khc l s dng di hng ch ngnnht tr trng s tnh ca n l qu thp. Cn mt cch khc l spxp cc ng theo trng s tnh ca n v sau li sp xp theo di hng ch ca n. Sau s chn ng c tng v tr sp xp lnh nht. D gii thut no c chn i chng na cng c c tnhl khi t ti th ng c trng s cao nht s c chn, nhng slm cho hng ch cho ng ny tng ln. Sau mt s lu lngs c chuyn sang ng t ti hn.

4.2.4. nh tuyn ngun (source routing) v m hnh cy (spanning tree)

Chng ta s xt mt s thut ton c bn dng cho vic tm kim cccy c s dng thit k v phn tch mng. Mt cy l mt graphkhng c cc vng; bt k mt cp nt no cng ch c duy nht mtng i. y ch yu xem xt cc graph v hng, nhng graph c cc lin kt c s dng c hai chiu trong qu trnh to ra ccng i.V mt s l do, cc cy rt hu dng v c s dng nh l graphc bn cho cc thut ton v cc k thut phn tch v thit k mng.Th nht, cc cy l mng ti thiu; cung cp mt s kt ni mkhng mt lin kt no l khng cn thit. Th hai, do vic ch cungcp duy nht mt ng i gia mt cp nt bt k, cc cy gii quyt

cc vn v nh tuyn (ngha l quyt nh vic chuyn lu lnggia hai nt). iu lm n gin mng v dng ca n. Tuy nhin,v cc cy lin thng ti thiu nn cng n gin v c tin cy tithiu. l nguyn nhn ti sao cc mng thc t thng c tnh linthng cao hn. Chnh v vy, vic thit k mt mng thng bt ubng mt cy.

4.2.5. Duyt cy

Cho trc mt cy no , chng ta c th i ti mi nt ca n. Qutrnh gi l mt qu trnh duyt cy. Trong qu trnh thc hin, cc

cnh trong cy c duyt hai ln, mi ln theo mt hng khcnhau. C nhiu cch duyt khc nhau. u tin, ch ra mt nt ca cylm nt gc. Vic duyt c thc hin xoay quanh nt . C mt siu kin la chn nt gc ny (chng hn nt gc l mt khu vcmy tnh trung tm). Ngoi ra, nt gc c th c chn mt cchngu nhin.Gi s nt A trong hnh 4.1 l nt gc ca cy. T A chng ta c thln lt i ti cc nt k cn ca n nh lB, C hoc D. Sau , li itheo cc nt k cn ca chng (B, C v D) l E , F , G v H . Tip tc iti ln lt cc nt k cn khc bn cnh cc nt ny. Khi , vicduyt ny s kt thc khi ti cc ntI , J , K v L. Qu trnh ny cgi l tm kim theo chiu rng. Trong qu trnh tm kim theo chiurng mt c im cn ch l nhng nt gn nt gc nht s c

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ti trc. Vic tm kim s thc hin theo mi hng cng lc. iu i khi c ch v c thc hin d dng.Mt thut ton nhm i ti mi nt ca cy th c gi l thut tonduyt cy. Thut ton sau y, Bfstree , thc hin mt qu trnh tmkim theo chiu rng. (Chng ta quy c rng, cc tn hm c k tu tin l k t hoa phn bit chng vi cc tn bin).Bfstrees s dng mt danh sch k cnn_adj_list , danh sch ny lit ktt c cc nt k cn ca mi nt thuc cy. n gin hn, gi srng cy ny l mt cy hu hng hng ra nhn t gc v do n_adj_list s ch bao gm cc nt k cn vi mt nt no mcc nt k cn xa gc hn so vi nt ang xt.

Hnh 4-18. Duyt cy

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Visit l mt th tc trong thc hin mt s qu trnh no i vimi nt (chng hn nh in ln mn hnh cc thng tin ca mi nt.v.v).Thut ton ny c thc hin cng mt hng i. Hng i l mtFIFO; trong cc phn t c thm vo t pha sau hng i vchuyn ra t pha trc. Cc th tcInitializeQueue , Enqueue ,Dequeue , NotEmpty lm vic trn cc hng i.InitializeQueuethit lp mt hng i rng.Enqueue , Dequeue l cc th tc thm mt phn t vo cui hng i v chuyn mt phn t ra tu hng i. Hm NotEmpty tr v TRUE hoc FALSE tu thucvo hng i c rng hay khng.n_adj_list l mt chui m mi phn t ca chui l mt danhsch. n_adj_list[n] l mt danh sch cc nt k cn ntn. Nh ni chng trc, for_each(element, list) , l mt cu

trc iu khin thc hin vng lp i vi tt c cc phn t calistv thc hin cc m bn trong vng lp, trong vng lp cc phnt ca list ln lt c s dng. Th tc trn hot ng vi gi thitl n_adj_list c thit lp trc khi th tcBfsTree cgi.Tng t, ta c th nh ngha mt qu trnh tm kim theo chiu su.Qu trnh ny cng bt u t nt gc. Qu trnh duyt tip tc thchin nt lng ging cha c duyt ca nt va mi c duyt. Tacng gi s rng cy bao gm cc lin kt c hng i ra xa nt gc.V d 4.1 :

Tr li vi graph trong hnh 4.1, ta c th ti ntB t nt A. Sau , tati ntE , k cn vi ntB-nt c duyt gn thi im hin ti nht.Nt E ny khng c nt k cn cha duyt no, do vy ta phi quaytr li ntB i sang nt F . Ta tip tc i ti cc ntI, J, K (cng vivic quay li ntI ), v nt L. Sau ta quay tr v nt A, tip tc ticc nt cn li lC, D, G v H . Do vy, ton b qu trnh duyt l:

A, B, E, F, I, J, K, L, C, D, G, H Nh rng th t ca qu trnh duyt l khng duy nht. Trong qutrnh duyt trn ta chn cc nt k cn xm nhp theo th t t triqua phi. Nu chn theo th t khc, qu trnh duyt l:A, B, F, I, J, K, L, E, D, H, G, CTrt t thc t ca qu trnh duyt ph thuc vo tng thut ton cth. iu ny cng ng vi mt qu trnh tm kim theo chiu rng.Kim tra thut tonBfsTree , trt t ny l mt hm ca trt t ccnt cn k trongn_adj_list .Thut ton DfsTree sau s thc hin mt qu trnh tm kim theochiu su.

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Visit(root)for each(neighbor, n_adj_list[node])

DfsTree(n, neighbor, n_adj-list)

Qu trnh tm kim ny s c thc hin vi s tr gip ca mtngn xp theo kiu LIFO, ngha l phn t c thm vo v chuynra t nh ngn xp. Trong trng hp ny, chng ta thng gi quy DfsTree , thc t chng ta s dng ngn xp h thng, nghal s dng loi ngn xp m h thng s dng lu gi cc li gihm v i s.C hai loi duyt trnh by trn u l qu trnh duyt thun (ngha lcc qu trnh ny duyt mt nt ri sau duyt ti nt tip theo cant ). Qu trnh duyt ngc i khi cng rt cn thit, trong qutrnh duyt ngc mt nt c duyt sau khi duyt nt tip cant . D nhin, cng c th thnh lp mt danh sch thun v sau o ngc danh sch . Cng c th thay th trt t tm kim mtcch trc tip nh th tc sau:

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Visited


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duyt trc khi duyt nt to ra mt cy, nu graph l khng linthng th to ra mt rng.

Hnh 4-19. Cc thnh phn

Hnh 4-19 biu din mt graph c 4 thnh phn. Gi s vng trn tpcc nt i theo tun t alphabet, cc thnh phn c nh s theotrt t cc nt c ch ci "thp nht" v ch s thnh phn c biudin bn cnh nt.Vi mi thnh phn, thut ton trn s giDfs kim tra thnh phn. Trong , thut ton cng kim tra cc cnh, mi cnh mt ln. Vth, phc tp ca n c bc bng bc ca tng s cc nt cng vis cc cnh trong tt c cc thnh phn (ngha l phc tp cathut ton bngO(N+E )).

Cy bc cu ti thiu (Minimum Spanning Tree)

C th s dngDfs tm mt cy bc cu nu c mt cy bc cutn ti. Cy tm c thng l cy v hng. Vic tm cy "tt nht"thng rt quan trng . Chnh v vy, chng ta c th gn mt " di"cho mi cnh trong graph v t ra yu cu tm mt cy c di tithiu. Thc t, " di" c th l khong cch, gi, hoc l mt ilng nh gi tr hoc tin cy. Mt cy c tng gi l ti thiu

c gi l cy bc cu ti thiu.Ni chung, nu graph l mt graph khng lin thng, chng ta c thtm c mt rng bc cu ti thiu. Mt rng bc cu ti thiu l mttp hp cc cnh ni n graph mt cch ti a c tng di l tithiu. Bi ton ny c th c xem nh l vic la chn mt graphcon ca graph gc cha tt c cc nt ca graph gc v cc cnhc la chn. u tin, to mt graph cn nt, n thnh phn vkhng c cnh no c. Mi ln, chng ta chn mt cnh thm vograph ny hai thnh phn lin thng trc cha c kt ni clin kt li vi nhau to ra mt thnh phn lin thng mi (ch khng

chn cc cnh thm vo mt thnh phn lin thng trc v to ramt vng). V vy, ti bt k giai on no ca thut ton, quan h:

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n=c+elun c duy tr, y n l s lng nt trong graph,e l s cnhc la chn tnh cho ti thi im xt v c l s lng thnh phntrong graph tnh cho ti thi im xt. cui thut ton,e bng n tri s thnh phn trong graph gc; nu graph gc l lin thng, chngta s tm c mt cy c(n-1) cnh. Nh gii thch trn,Dfs stm ra mt rng bc cu. Tuy nhin, chng ta thng khng tm ccy bc cu c tng di ti thiu.Thut ton "hu n"Mt cch tip cn kh d tm mt cy c tng di ti thiu l, mi giai on ca thut ton, la chn cnh ngn nht c th. Thutton gi l thut ton "hu n". Thut ton ny c tnh cht "thincn" ngha l khng lng trc c cc kt qu cui cng do ccquyt nh m chng a ra mi bc gy ra. Thay vo , chng

ch a ra cch chn tt nht cho mi qu trnh la chn. Ni chung,thut ton "hu n" khng tm c li gii ti u cho mt bi ton.Thc t thut ton thm ch cn khng tm c mt li gii kh thingay c khi li gii tn ti. Tuy nhin chng hiu qu v d thchin. Chnh v vy chng c s dng rng ri. Cc thut ton nycng thng to c s cho cc thut ton c tnh hiu qu v phctp hn.V th, cu hi u tin t ra khi xem xt vic ng dng mt thutton gii quyt mt bi ton l liu bi ton y c hay khng cutrc no m bo cho thut ton hot ng tt. Hy vng rng thutton t ra cng m bo c mt li gii kh thi nu li gii tn ti.Khi , n s m bo tnh ti u v m bo yu cu no v thigian thc hin. Bi ton tm cc cy bc cu ti thiu thc s c mtcu trc mnh cho php thut ton "hu n" m bo c tnh ti ucng nh m bo phc tp tnh ton mc va phi.Dng chung ca thut ton "hu n" l:

Bt u bng mt li gii rng s.

Trong khi vn cn c cc phn t cn xt,

Tm e, phn t "tt nht" vn cha xtNu vic thm e vo s l kh thi th e c thm vo s, nu vic

thm khng kh thi th loi b e.

Cc yu cu cc kh nng sau:So snh gi tr ca cc phn t xc nh phn t no l "tt nht"Kim tra tnh kh thi ca mt tp cc phn tKhi nim "tt nht" lin quan n mc ch ca bi ton. Nu mcch l ti thiu, "tt nht" ngha l b nht. Ngc li, "tt nht" nghal ln nht.

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Thng thng, mi gi tr gn lin vi mt phn t, v gi tr gn linvi mt tp n gin ch l tng cc gi tr i cng ca cc phn ttrong tp . l trng hp cho bi ton cy bc cu ti thiu cxt trong phn ny. Tuy nhin, khng phi l trng hp chung.Chng hn, thay cho vic ti thiu tng di ca tt c cc cnhtrong mt cy, mc ch ca bi ton l ti thiu ho di cc cnhdi nht trong cy. Trong trng hp , gi tr ca mt cnh l dica cnh v gi tr ca mt tp s l di ca cnh di nht nmtrong tp.Mun tm c cnh "tt nht" b sung, hy nh gi cc cnhtheo nh hng v gi tr ca n ti gi tr ca tp. Gi sV(S) lgi tr ca tpS v v(e,S) l gi tr ca mt phn te th v(e,S) cquan h vi tpS bi cng thc

v(e,S)= V(S e) - V(S)

Trong trng hp ti thiu di ca cnh di nht trong mt cy.v(e,S) bng 0 i vi bt k cnh no khng di hn cnh di nht c chn. Ngc li, n s bng hiu di gia cnh vi cnh dinht c chn, khi hiu ln hn 0.Trong trng hp chung, gi tr ca tp c th thay i mt cch ngunhin khi cc phn t c b sung vo n. Chng ta c th gn gitr 1 cho cc tp c s lng phn t l chn v 2 cho cc tp c slng phn t l l. iu lm cho cc gi tr ca cc phn t ch lmt trong hai gi tr +1 v -1. Trong trng hp ny, thut ton "hun" khng c s dng. By gi gi s rng "trng lng" ca mttp bin i theo mt cch hp l hn th khi , s c mt c s hpl hn cho vic ch ra phn t "tt nht". Mt iu quan trng cn ch l, khi tp ln ln, gi tr ca phn t m trc khng c xemxt c th thay i do cc phn t thm vo tp . Khi iu ny xyra, thut ton "hu n" c th mc li trong cc la chn ca n v snh hng ti cht lng ca li gii m chng ta nhn c.Tng t, trong hu ht cc trng hp, tnh kh thi c th b nhhng mt cch ngu nhin do s b sung phn t. Chnh v vy,trong cc bi ton m nhng tp c s lng phn t chn c thc xem l kh thi v nhng tp c s phn t l l c th c xeml khng kh thi th thut ton "hu n" hoc bt k thut ton no c

b sung cc phn t, mi ln mt phn t, s khng hot ng. V vychng ta s gi thit cc tnh cht sau, nhng tnh cht ny lun cduy tr trong mi trng hp xem xt:Tnh cht 1:Bt k mt tp con no ca mt tp kh thi th cng kh thi, c bittp rng cng l mt tp kh thi.Ngoi ra gi thit rng phc tp ca thut ton tnh ton gi trca mt tp v kim tra s kh thi ca chng l va phi, c bit, khi phc tp ny l mt a thc ca s nt v cnh trong graph.

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cc ng c. C rt nhiu cch thc hin cc qu trnh ny. properties c th l mt dy v cc hm Append , Delete v

ElementsOf c th hot ng vi cc danh sch ch s (danh schm cc phn t l cc ch s mng). Trong thc t cch thc hinc chn l cch lm sao cho vic thc hin cc hmTest vSelectBestElement l tt nht. on gi m trn gi thit rng thut ton "hu n" s dng li khikhng cn phn t no xem xt. Trong thc t, c nhiu nguynnhn thut ton dng li. Mt trong nhng nguyn nhn l khi ktqu xu i khi cc phn t c tip tc thm vo. iu nay xy ra khitt c cc phn t cn li u mang gi tr m trong khi chng ta angc tm cho mt gi tr ti a. Mt nguyn nhn khc l khi bit rngkhng cn phn t no trong tp ng c c kh nng kt hp vicc phn t va c chn to ra mt li gii kh thi. iu ny xy rakhi mt cy bc cu ton b cc nt c tm thy.Gi s rng thut ton dng li khi iu l hp l, cn nu khng,cc phn t khng lin quan s b loi ra khi li gii.Gi thit rng, cc li gii cho mt bi ton tho mn tnh cht 1 v gitr ca tp n gin ch l tng cc gi tr ca cc phn t trong tp.Ngoi ra, gi thit thm rng tnh cht sau c tho mn:Tnh cht 2:Nu hai tpSp v Sp+1 ln lt c p v p+1 phn t l cc li gii vtn ti mt phn t e thuc tp S p+1 nhng khng thuc tp S p thS p { e }l mt li gii.

Chng ta thy rng, cc cnh ca cc rng tho mn tnh cht 2,ngha l nu c hai rng, mt c p cnh v rng kia c p+1 th luntm c mt cnh thuc tp ln hn m vic thm cnh vo tpnh hn khng to ra mt chu trnh.Mt tp cc li gii tho mn cc tnh cht trn gi l mt matroid. nh l sau y l rt quan trng (chng ta ch tha nhn ch khngchng minh).

nh l 4.1

Thut ton hu n m bo m mt li gii ti u cho mt biton khi v ch khi cc li gii to ra mt matroid.

C th thy rng, tnh cht 1 v tnh cht 2 l iu kin cn v m bo tnh ti u ca thut ton hu n . Nu c mt li gii chomt bi ton no m n tho mn hai tnh cht 1 v 2 th cch ngin nht l dng thut ton hu n gii quyt n. iu ngvi mt cy bc cu.Sau y l mt nh l khng km phn quan trng.

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nh l 4.2

Nu cc li gii kh thi cho mt bi ton no to ra mtmatroid th tt c cc tp kh thi ti a c s lng phn t nh

nhau.

Trong , mt tp kh thi ti a l mt tp m khi thm cc phn tvo th tnh kh thi ca n khng c bo ton; N khng nht thitphi c s lng phn t ti a cng nh khng nht thit phi ctrng lng ln nht. nh l o ca nh l trn cng c th ng nghi l nu tnh cht 1c tho mn v mi tp kh thi ti a c cng s lng phn t, thtnh cht 2 c tho mn.

nh l 4.2 cho php chng ta chuyn i mt bi ton ti thiuP thnh mt bi ton ti aP' bng cch thay i cc gi tr ca ccphn t. Gi thit rng tt cv(xj)trong P c gi tr m. Li gii ti ucho bi ton P c s lng phn t ti a lm th chng ta c th tora mt bi ton ti aP' t P bng cch thit lp cc gi tr ca ccphn t trongP' thnh -v(xj). Tt c cc phn t u c gi tr dngv P' c mt li gii ti u cham phn t. Thc ra, th t ca cc ligii ti a phi c o li: li gii c gi tr ti a trongP' cng l ligii c gi tr ti thiu trongP .Gi s lc nay ta cn tm mt li gii c gi tr ti thiu, tun theo iukin l c s lng ti a cc phn t. S tnh c cc phn t c gi trdng. C th gii quyt bi tonP nh l mt bi ton ti aP' bngcch thit lp cc gi tr ca cc phn t thnhB-v(xj)viB c gi trln hn gi tr ln nht ca xj . Khi cc gi tr trongP' u dng vP' l mt li gii ti u cm phn t. Th t ca tt c cc tp kh thiti a b o ngc: mt tp c gi tr lV trong P th c gi tr lmB-V trong li giiP' . Mt gi tr ti a trongP' th c gi tr ti thiutrong P. Quy tc ny cng ng vi cc cy bc cu tho mn tnhcht 1 v tnh cht 2 v c th tm mt cy bc cu ti thiu bng cchs dng mt thut ton hu n.Thut ton KruskalThut ton Kruskal l mt thut ton hu n c s dng tmmt cy bc cu ti thiu. Tnh ng n ca thut ton da trn ccnh l sau:

nh l 4.3

Cc rng th tho mn tnh cht 1 v 2.

Nh chng ta bit, mt rng l mt tp hp cc cnh m tp hp

khng cha cc chu trnh. R rng l bt k mt tp con cc cnh

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no ca mt rng (thm ch c tp rng) cng l mt rng, v vy tnhcht 1 c tho mn. thy rng tnh cht 2 cng tho mn, xt mt graph c biu dintrong hnh 4.4.

Hnh 4.3.

Gi s c mt rngF1 c p cnh. Rng{2,4}l mt v d vi p=2 , vn c biu din bng nt t trong hnh 4.4. Khi xt mt rngkhc F2 c p+1 cnh. C hai trng hp c xt.Trng hp 1:F2 i ti mt ntn, nhng F1 khng i ti nt . Mtv d ca trng hp ny l rng{1, 4, 6}, rng ny i tiE cn F1 thkhng. Trong trng hp ny, c th to ra rng{2, 4, 6}bng cch

thm cnh6 vo rng {2,4}.Trng hp 2: F2 ch i ti cc nt mF1 i ti. Mt v d ca trnghp ny l rng {1. 4. 5}. Xt S, mt tp cc nt mF1 i ti. Chorng c k nt trong tpS. V F1 l mt rng nn mi cnh trongF1gim s lng thnh phn trongS i mt, do tng s lng thnhphn l k-p. Tng t, F2 to ra k-(p+1) thnh phn tS (s lngthnh phn va ni b hn vi s lng thnh phn caF1). V vy,mt cnh tn ti trongF2 m cc im cui ca n nm cc thnhphn khc nhau trong F1 th c th thm cnh voF1 m khngto ra mt chu trnh. Cnh 3 l mt cnh c tnh cht trong v dny (cnh 1 v 5 cng l nhng cnh nh vy).V th, chng ta thy rng nu tnh cht 1 v 2 c tho mn th mtthut ton hu n c th tm c mt li gii ti u cho c bi toncy bc cu ti thiu ln bi ton cy bc cu ti a. Ch rng mtcy bc cu l mt rng c s cnh ti aN-1 cnh viN l s nttrong mng. Sau y chng ta s xt bi ton ti thiu.Thut ton Kruskal thc hin vic sp xp cc cnh vi cnh u tinl cnh ngn nht v tip theo chn tt c cc cnh m nhng cnhny khng cng vi cc cnh c la chn trc to ra cc chutrnh. Chnh v th, vic thc hin thut ton n gin l:

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p=next[node]q=next[p]while ( p!=q )

next[node]= q node = q p=next[node]q=next[p]

return (p)

FindComponent tr v nt gc ca thnh phn chanode . Hm nycng iu chnhnext , nt hng v nt gc cha nt . c bit,hm ny iu chnh next hng ti im tng cao hn. Tarjan ch rarng, bng cch , th lm gn ng i ti nt gc mt cc honton cn hn l khng lm gn mt cht no c v ton b kt qu

trong vic tm kim v cp nht next ch ln hn so viO(n+m) mtcht vin l s lng nt vm l s lng cnh c kim tra.V d 4.3:

Hnh 4-20. Php tnh Minimum Spanning Tree ( MST)

Xt mt mng c biu din trong hnh 4.4. cc du * trong hnhc gii thch di y. u tin, sp xp cc cnh v sau lnlt xem xt tng cnh, bt u t cnh nh nht. V th, chng taxem ( A, C ) l cnh u tin. GiFindComponent cho nt A ta thyc p lnq u l A nn FindComponent tr v A nh l nt gc cathnh phn cha nt A. Tng t,FindComponent tr vC nh lnt gc ca thnh phn cha ntC . V th, chng ta mang A v C vocy v thit lpnext[A] bng C . Sau , xt (B, D). Hm cng thchin tng t vB, D c thm vo cy, next[B] bng D. Chngta xt (C, E), chp nhn n v thit lpnext[C] bngE .By gi, xt(A, E). Trong FindComponent , p l C cn q l E . V thchng ta chy vo vng lpwhile , thit lpnext[A] bng E v rtngn ng i t A tiE viE l nt gc ca thnh phn cha chng.Node , p v q c thit lp thnhE v FindComponent tr vE

nh l nt gc ca thnh phn cha nt A. FindComponent cng tr

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vE nh l nt gc ca thnh phn chaE . V th, c hai im cuica (A, E)l cng mt thnh phn nn(A, E)b loi b.Tip n, xt(A, B). Trong qu trnh giFindComponent i vi nt A, chng ta thy rng p=q=E v next khng thay i. Tng t, qu

trnh giFindComponent i vi ntB ta c p=q=D. V th, chngta thit lpnext[E] bng D. Ch rng, chng ta khng thit lpnext[A] bngB, m li thit lpnext i vi nt gc ca thnh phnca A bng vi nt gc ca thnh phn caB.Cui cng,(C, D)c kim tra v b loi b.Trong hnh 4.4 nhng cnh trong cy bc cu c phn bit bi mtdu * ngay bn cnh cc cnh . Ni dung ccnext c ch rabng cc cung (cc cnh hu hng) c mi tn. Chng hn,next[B] bng D c ch ra bng mt mi tn tB ti D. Ch rng, cc cung c nh ngha binext to ra mt cy, nhng nichung cy khng phi l mt cy bc cu ti thiu. Thc vy, vitrng hp c mt cung(E, D), ngay c khi cc cung khng cnthit phi l mt phn graph. V vy, bn thnnext ch nh ngha cutrc thnh phn khi tin hnh thc hin thut ton. Chng ta to mtdanh sch hin cc cnh c chn dnh cho vic bao gp trong cy.Gi ca cy c nh ngha binext tng i bng phng, nghil cc ng i ti cc nt gc ca cc thnh phn l ngn khinFindComponent hot ng hiu qu.Hin nhin, s phc tp ca thut ton Kruskal c quyt nh bivic sp xp cc cnh, s sp xp c phc tp lO(m log m).Nu c th tm c cy bc cu trc khi phi kim tra tt c cccnh th chng ta c th ci tin qu trnh bng cch thc hin spxp phn on. C th, chng ta c th lu gi cc cnh trong mtkhi (heap) v sau ly ra, kim tra mi cnh cho n khi mt cyc to ra. Chng ta d dng bit c qu trnh dng vo lcno; ch n gin l theo di s lng cnh c xt v dng likhi c n-1 cnh c chp nhn.Chng ta gi s rng, cc qu trnh qun l khi (heap) nh thit lp,b xung v ly d liu ra l n gin. iu quan trng cn ch yl phc tp ca vic thit lp mt khi (heap) cm phn t lO(m),

phc tp ca vic tm phn t b nht lO(1) v phc tp cavic khi phc mt khi (heap) sau khi b xung, xo, hoc thay i mtgi tr lO(logm). Chnh v vy, nu chng ta xtk cnh tm cy bccu, phc tp trong vic duy tr mt khi (heap) bngO(m+klogm), phc tp ny b hnO(mlogm)nu k c bc b hn bc ca m. kti thiu bngO(n) nn nu graph l kh mng th vic s dng khi(heap) s khng c li. Nu graph l dy c th vic lu tr c thc xem xt. y l phin bn cui cng ca thut ton Kruskal,thut ton ny tn dng cc hiu ng ni trn.

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a nt vo cy v ghi li cnh ni nt vi cy

Thut ton Prim da trn nhng nh l sau y:

nh l 4.4 .

Mt cy l mt MST nu v ch nu cy cha cnh ngn nhttrong mi cutset chia cc nt thnh hai thnh phn.

thc hin thut ton Prim, cn phi theo di khong cch t mint khng thuc cy ti cy v cp nht khong cch mi khi cmt nt c thm vo cy. Vic c thc hin d dng; ngin ch l duy tr mt dy d_tree c cc thng tin v khong cch ni trn. Qu trnh tun theo:

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FindMin tr v mt nt khng thuc cy v gn cy nht.Scan cpnht khong cch ti cy i vi cc nt khng thuc cy.C th thy rng phc tp ca thut ton ny lO(n2); c hai hmFindMin v Scan c phc tp lO(n) v mi hm c thc hin

n ln. So snh vi thut ton Kruskal ta thy rng phc tp cathut ton Prim tng nhanh hn so vi phc tp ca thut tonKruskal num, s lng cc cnh, bngO(n2),cn nu m c cngbc vin th phc tp ca thut ton Kruskal tng nhanh hn.C th tng tc thut ton Prim trong trng hp graph l mt graphmng bng cch ch quan tm n cc nt lng ging ca nti vac thm vo cy. Nu sn c cc thng tin k lin, vng lpfor trong Scan c th tr thnh.

for each (j , n_adj_list[i] ) phc tp caScan tr thnh O(d) vid l bc ca nti . Chnh vth phc tp tng cng caScan gim tO(n2) xungO(m).Thit lp mt tp k lin cho ton b mt graph l mt php ton c phc tp bngO(m):

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Hnh 4.2. Graph c lin kt song song v self loop

Bng 4.1Nt init. A C E B DA 0 0(-) 0(-) 0(-) 0(-) 0(-)B 100 10(A) 10(A) 10(A) 10(A) 10(A)C 100 2(A) 2(A) 2(A) 2(A) 2(A)D 100 100(-) 11(A) 11(A) 5(B) 5(B)E 100 7(A) 6(C) 6(C) 6(C) 6(C)

V d 4.4 :

Tr li hnh 4.4, gi s rng cc cnh khng c biu din c dibng 100. Thut ton Kruskal s chn(A, C), (B, D), (C, E), v loi(A,E) bi v n to ra mt chu trnh vi cc cnh c chn l(A, C)v (C, E), chn (A, B)sau dng li v mt cy bc cu hon ton c tm thy.Thut ton Prim bt u t nt A, nt A s c thm vo cy. Tiptheo l cc nt C, E, B v D. Bng 4.1 tng kt cc qu trnh t

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