TNH TON PHN TNHc vin: Nguyn Th Mai Hng Lp : M11CQCT01

T N NHKhi nim t n nh l, bt k tnh trng ban u, h thng c m bo hi t n mt trng thi thch hp trong gii hn thi gian ca chnh n m khng cn bt k s can thip bn ngoi no

T N NH- Dijkstra cng a ra mt v d v cc khi nim t n nh bng cch s dng mt h thng t n nh token ring - gii thch cc khi nim v t n nh ta s dng mt nhm tr em v yu cu chng ng trong mt vng trn.

T N NHCc nguyn tc t n nh p dng i vi bt k h thng c xy dng trn mt s lng ln ca cc thnh phn c pht trin c lp vi nhau hoc nhng thnh phn c hp tc hoc cnh tranh t c mc tiu chung

2. M HNH H THNGMt m hnh h thng phn phi bao gm mt tp n cc my trng thi c gi l b vi x l giao tip vi nhau B vi x l th i trong h thng c gi l Pi. Ln cn ca mt b x l l b vi x l trc tip kt ni vi n. Mt b x l c th giao tip trc tip vi cc ln cn ca n

2. M HNH H THNGCc giao tip gia cc b vi x l ln cn c th c thc hin hoc qua thng ip hoc b nh chia s - B nh chia s thng thch hp vi cc h thng c cc b x l gn nhau. V d nh my tnh a x l - M hnh phn tn thng qua thng ip ph hp vi cc b x l t gn nhau cng nh c phn phi rng ri qua mng

M HNH THNG QUA THNG IP- Hng i First-in first-out (FIFO) c s dng m hnh chuyn giao thng ip khng ng b - Lin kt 1 chiu Pi n Pj s dng hng i firstin first-out (FIFO) Qij cha cc thng ip c gi t Pi ti cc ln cn Pj - Lin kt 2 chiu t Pi ti Pj s dng 2 hng i FIFO. Mt hng i cha cc thng ip t Pi

M HNH THNG QUA THNG IPH thng phn tn qua thng ip c k hiu l c:

c = (s1, s2, sn, q1,2, q1,3.....qi,j,.qn,n-1) Trong : si, 1 i n l trng thi ca Pi qi,j (ij) trong hng i Qi,j l thng ip gi t Pi ti Pj

SHARED MEMORY MODEL- B x l giao tip s dng thanh ghi truyn thng chia s. - Cu hnh mt h thng vi n b x l v m thanh ghi truyn thng c k hiu l c c = (s1, s2, s3...sn, r1, r2rm)si : 1 i n, l trng thi ca Pi ri:1 j m,l ni dung ca thanh ghi truyn thng

3. NH NGHA T N NHMt h thng A c gi l t n nh vi thuc tnh P nu tha mn hai c tnh sau: Khp kn: P b hp trong nhng thc thi ca S. l, mt khi P c thnh lp trong S, n khng th sai lch. Hi t: Bt u t mt trng thi ty , S c m bo t c mt trng thi P p ng

3. NH NGHA T N NHS p ng QP ( c l Q n nh P) nu tha mn: Khp kn: P b hp trong nhng thc thi ca S. l, mt khi P c thnh lp trong S, n khng th sai lch. Hi t: Nu S bt u t bt k trng thi ton cu p ng Q, S c m bo t c mt trng thi P p ng trong mt s hu hn cc trng thi hu hn

T N NH NGU NHIN V T N NH XC SUT T n nh ngu nhin: h thng c cho l t n nh ngu nhin, khi v ch khi n t n nh v s lm trn cn t ti trng thi chnh xc c gii hn bi mt s k khng i.

T N NH NGU NHIN V T N NH XC SUT T n nh xc sut: h thng S c cho l t n nh xc sut i vi thuc tnh P, nu c hai c tnh sau y: - Tnh khp kn: P c ng di s thc hin ca S. l, mt khi P c thnh lp vo S, n khng th c gi mo. - Hi t: tn ti mt hm f t tp s t nhin [0,1] tha mn lim k f (k) = 0, nh vy m xc sut t trng thi P, t trng thi ton cc ty chn trong qu trnh chuyn i trng thi k l 1-f (k).

T N NH NGU NHIN V T N NH XC SUTMt h thng gi n nh l mt h thng, nu bt u trong mt trng thi ty chn c m bo t c mt trng thi m sau n khng b lch ra khi v tr c nh sn

4. TRNG THI TRONG CC THNH PHN N LDijkstra a ra 3 gii php cho vng mt chiu vi n my, 0, 1, ........., n-1, c K trng thi, (i)K n, (ii)K = 3, (iii)K = 4 Ghosh ch ra rng mt vng t n nh yu cu ti thiu 3 trng thi.

(i) K ni vi bt k my no, chng ta s dng k hiu: S: Trng thi ca my L: Trng thi ca my bn tri R: Trng thi ca my bn phi

(i) K nCc my tnh ngoi l: If L = S then S := (S+1) mod K End If; Cc my khc: If L := S then S := L End If;

(ii) K 3Gii php ny ch s dng 3 my trng thi, trng thi trong mi my l {0,1,2} My 0: gi l my cn di My n-1: c gi l cn trn.

(ii) K 3My cn di, my 0:

If (S+1) mod 3 = R then S := (S1) mod 3My cn trn, my n1:

If L = R and (L+1) mod 3 S then S := (L+1) mod 3 S:= R

(ii) K 3The other machines:

If (S+1) mod 3 = L then S := L If (S+1) mod 3 = R then S:= R

(ii) K 3If (S+1) mod 3 = R then S := (S1) mod 3 Trng thi ca my di cng ph thuc vo trng thi hin ti ca n v my ln cn bn phi ca n. Vi s = 0, 1, 2 th (s + 1) mod 3 = 1, 2, 0 T nhng kt qu ny a ra 3 kh nng sau: 1. Nu s = 0 v r = 1, sau trng thi ca s -> 2.

(ii) K 3If L = R and (L+1) mod 3 S then S := (L+1) mod 3 S:= R Trng thi my cn trn ph thuc vo ln cn tri v ln cn phi ca n Nh vy trng thi ca my cn trn l: - 1, khi ln cn bn tri ca n l 0. - 2, khi ln cn bn tri ca n l 1. - 0, khi ln cn bn tri ca n l 2.

(ii) K 3 If (S+1) mod 3 = L then S := L If (S+1) mod 3 = R then S:= R 1. Nu s = 0 v L = 1, sau s = 0. 2. Nu s = 1 v L = 2, sau s = 2. 3. Nu s = 2 v L = 0, sau s = 1.

(ii) K 3

5. IU KIN CA T N NHGouda v Evangelist a ra hai khi nim lin quan n iu kin t n nh: Khong hi t: s lng ti a ca qu trnh chuyn tip c th c thc hin trong mt h thng, bt u t mt trng thi ty , trc khi n t n mt trng thi an ton.

5. IU KIN CA T N NHKhong p ng: p ng m rng s lng ti a ca qu trnh chuyn tip c th c thc hin trong mt h thng t c trng thi mc tiu, bt u t mt s trng thi ban u. S la chn trng thi ban u v trng thi mc tiu ph thuc vo ng dng Mc ch ca thit k ca mt thut ton t n nh l gim khong hi t v tng khong p ng.

TNH TON PHN TNHc vin: Nguyn Th Yn Lp : M11CQCT01

Cc phng php thit k h thng t n nh T n nh c c trng trong iu khon ca mt k th c hi m mc tiu l ph v hot ng bnh thng ca h thng. iu ny i th c th tiu dit mt s phn ca h thng, hoc lm gin on hot ng ca mt hoc nhiu phn. Hn na, n c th lm cho mt h thng khng pht hin c rng n b tn cng, ngay sau khi cuc tn cng xut hin. c gi l t n nh, mt h thng phi c kh nng khi phc li hot ng bnh thng khi tip xc vi cc cuc tn cng nh vy.

Cc phng php thit k h thng t n nh Nu h thng hoc cc b phn ca n b ph hy hon ton, n khng cn c th cho h thng hot ng, sau khng c h thng t n nh c th lm vic. K nghch th thnh cng trong vic t c mc tiu ca mnh. Tuy nhin, nu c cc thnh phn c li m bo cho h thng hot ng, sau mt h thng t n nh t t s tip tc hot ng bnh thng sau v tn cng. l cc nh thit k quyt nh theo nhng iu kin h thng c th c gi l "ph hy hon ton" hoc "vn c kh nng vn hnh."

Phng php phn lp v m-un ha tng c bn l phn chia h thng thnh phn nh hn, lm cho mi thnh phn t n nh c lp, v sau tch hp chng to nn h thng. n nh tun theo lp, mi quan h t n nh l bc cu. V vy, cc lp khc nhau ca cc chng trnh t n nh (ca chnh n t n nh) c th c bao gm. Bc u tin l xy dng t n nh "nn tng" v bt k chng trnh c vit trn nn tng s t ng tr nn t n nh. tng c bn ng sau mt nn tng t n nh l cung cp nguyn thy c th c s dng vit cc chng trnh khc.

T n nh phn phi cy bao trm Trong cc h thng phn tn, mt cy bao trm l c s cho nhiu giao thc phn tn phc tp. xc nh mt cy bao trm, mng c m phng nh mt th G=(V,E) trong V l tp hp ca cc nt mng (nh) v E l tp hp cc lin kt truyn thng (cnh) gia cc nt mng. Mt cy bao trm T=(V,E) ca G l mt th bao gm cng mt tp hp cc nt V, E l mt tp hp con ca E. Mt cy bao trm ca th l ni chung khng phi l duy nht (ngay c khi nt gc l c nh). Mt cy bao trm trong mt mng thng l mt iu kin tin quyt cho cc giao thc mng tham gia nhiu hn nh nh tuyn hoc lu hnh th. N thng lm tng hiu qu ca cc giao thc mng.

Thut ton t n nh cho cy bao trmDolev, Israeli, and Moran algorithm Ni dung thut ton: Hai nc lng ging Pi v Pj giao tip vi nhau bng cch c v vit ti hai s ng k chia s, rij v rji. giao tip, Pi vit vo rij v c t rji. Pj vit vo rji v c t rij. Nt gc vit gi tr 0 trong s ng k ca tt c cc nc lng ging. Tt c cc b x l khc lin tc thc hin cc bc sau y:

Ni dung thut ton: - Trong mi ln lp, b x l c trong s ng k ca tt c cc nc lng ging v tnh ton gi tr cho bin dist. - Tnh gi tr cho bin dist bng cch chn khong cch ti thiu ca cc nc lng ging ca n, thit lp bin dist ca n n mc ti thiu khong cch cng thm 1. - Cp nht s ng k ca cc nc lng ging. Bin cc b tng ng ng k rij c k hiu l lrij. N lu tr cc gi tr cui cng ca r c c bi Pi

Thut ton t n nh cho cy bao trm

Dolev, Israeli, and Moran algorithm

Spanning-tree, is Self-stabilizing Thut ton n nh bt u t qu trnh gc. Sau khi nt gc vit gi tr 0 trong s ng k ca tt c cc nc lng ging gi tr ny s khng thay i na. Cc nc lng ging trc tip ca gc sau khi kim tra cc bin u vo ca h s thy rng nt gc c khong cch ti thiu ca tt c cc nt khc (cc nt khc c khong cch t nht 1) . Do , tt c cc nc lng ging trc tip ca gc s chn th mc gc l ph huynh ca h v cp nht khong cch ca h mt cch chnh xc n 1. Sau khi tt c cc nt c khong cch t gc l d tnh ton khong cch t gc mt cch chnh xc v vit n trong s ng k ca n, gi tr ny khng cn thay i v cc nt cch gc d+1 sn sng tnh ton khong cch t gc. Sau khi chu k cp nht hon tt th ton b cy c n nh.

TNH TON PHN TNHc vin: Th Nhn Lp : M11CQCT01

1. A Probabilitic Self-Stabilization Leader Election AlgorithmBi ton: H thng phn tn bao gm n v tr v chng cn bu chn ra v tr i u truyn thng ip gia chng bng vic s dng Probabilitic SelfStabilization Leader Election Algorithm Trong sut mt nv thi gian, cc v tr c th xy ra ba kh nng: Im lng: khng c v tr no c gng chuyn mt thng ip. Thnh cng: ch c duy nht mt v tr s dng kinh truyn truyn thng ip. Va chm: c t nht 2 v tr ang c gng truyn thng ip

1. A Probabilitic Self-Stabilization Leader Election Algorithm - Thut ton c m phng li nh sau:

1. A Probabilitic Self-Stabilization Leader Election Algorithm V d m phng thut ton

2. T n nh xem nh l mt gii php khng li Kh nng chu li: Kh nng chu li c nh ngha l kh nng chu ng s tht bi tm thi, trong trng thi ca mt thnh phn thay i mt cch t nhin, nhng cc thnh phn cn li vn chnh xc. H thng t n nh a ra mc khng li m h thng vn c th tip tc vn hnh khi c li xy ra. T n nh l mt cch khc xem xt kh nng khng li ca h thng phn tn, n cung cp mt ch Xy dng trong s bo v chng li nhng s tht bi tm thi nhng tht bi c th gy ra li d liu trong h thng phn tn, t n nh lm cho cc h thng c th t phc hi t nhng tht bi mt cch t ng m khng c bt c s can thip ca bn ngoi.

2. T n nh xem nh l mt gii php khng li T n nh a ra mt phng php hp nht chng li nhng li nht thi bng vic hp nht nhng li vo trong m hnh thit k. Nhng li sau y c th c x l bi mt h thng t n nh Khi to khng nht qun: cc tin trnh khc nhau trong mt chng trnh c th c khi to cc trng thi cc b m khng nht qun vi cc tin trnh khc. Ch thay i: C cc ch khc nhau ca s thc thi mt h thng. Trong vic thay i ch ca s vn hnh, tt c cc tin trnh khng th em li s thay i cng mt thi im. Mt chng trnh nht nh t c mt trng thi ton cc trong mt vi tin trnh thay i trong khi mt s tin trnh khc th khng. Li truyn: Nhng li ny bao gm mt, tht bi hoc t li th t ca cc thng ip v c th dn n s khng nht quan ca bn gi v bn nhn. Li tin trnh v phc hi: Nu mt tin trnh b sp v phc hi sau , trng thi c b ca n c th khng nht qun vi nhng phn cn li ca h thng/chng trnh. S c b nh: S c b nh c th l nguyn nhn dn n mt trng thi cc b, to ra s khng nht qun vi cc phn cn li ca h thng/chng trnh.

3. Nhng nhn t ngn chn s t n nh

a.

Tnh i xng:S t n nh yu cu tt c cc tin trnh khng nn l ng nht/i xng bi v gii php t n nh nhn chung da vo mt tin trnh c bit. Mt h thng c th l khng i xng bi v trng thi hoc tnh khng i xng bi v nh danh. Mt h thng khng i xng bi v trng thi khi tt c cc tin trnh l ng nht; tuy nhin, chng khi to t cc trng thi cc b ban u khc nhau. Mt h thng khng i xng bi nh danh khi khng c tin trnh no l ng nht. Nhn chung, mt th thng khng i xng bi trng thi th khng t n nh, trong khi mt h thng khng i xng bi nh danh th c th t n nh

a. b. c.

S kt thc Tnh c lp Cu hnh m phng

3. Nhng nhn t ngn chn s t n nh

a. b.

Tnh i xng: S kt thcS t n nh nhn chung l khng hp vi s kt thc. Nu bt k trng thi ton cc khng an ton no l trng thi kt thc th h thng s khng c th l n nh. Nh c s t n nh khng hp vi s kt thc nhn chung nn c mt trng hp ngoi l trong t n nh c th c thc hin khi c s xut hin ca s kt thc. Trong cc chng trnh tun t trng thi xc nh (khi s lng trng thi l xc nh ) th trnh bin dch c th b i tt c nhng trng thi khng an ton.

a. b.

Tnh c lp Cu hnh m phng

3. Nhng nhn t ngn chn s t n nh

a. b. c.

Tnh i xng: S kt thc Tnh c lpTnh c lp xy ra trong mt h thng khi trng thi cc b v s tnh ton ca mi tin trnh l ph hp vi mt s s tnh ton v trng thi ton cc an ton; tuy nhin, s tnh ton v trng thi ton cc kt qu li khng an ton. Trong trng hp ny, h thng khng th n nh do giao tip v phi hp khng ph hp gia cc tin trnh ca n.

d.

Cu hnh m phngVi nhng cu hnh m phng kt qu l khi s tnh ton ging nhau (chui cc hnh ng) vi hai trng thi khc nhau th khng c cch no phn bit gia chng. Nu mt trong hai trng thi l khng an ton th h thng khng m bo hi t t trng thi khng an ton.

4. Hn ch ca t n nh S cn thit ca mt my tnh c bit: Hu nh tt c cc thut ton t n nh da trn thc t rng c t nht mt my c bit trong h thng. iu ny c th kh khn t c trong mt s h thng, nhng n khng phi l mt nhc im ln trong hu ht cc h thn phn tn. Cn bng gia hi t v phn hi n nh gi i khi rt tn km thit k cc h thng t n nh. Gim bt cc yu cu ca h thng c th gim bt chi ph. Mt h thng c cho l n nh khi v ch khi mi tnh ton u c mt vi trng thi trong bt c tnh ton no bt u t trng thi ny s thuc mt tp cc tnh ton hp php. Mt khc, mt h thng gi n nh, mi tnh ton ch cn c mt vi trng thi cc hu t ca s tnh ton bt u trng thi ny thuc tp hp cc tnh ton hp php. c tnh ca gi n nh r rng l yu hn so vi yu cu n nh mc d n t tn km hn ci t. Xc minh tnh ng n ca h thng t n nh

Trn trng cm n!