do an giao thuc x.25

K Thut Truyn S Liu

K thut chuyn mch gi dng giao thc X.25

NHN XT CA GIO VIN HNG DN ..

H Ni, ngy thng nm 2011 Gio vin hng dn

Khoa in T

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K Thut Truyn S Liu


LI NI UNgy nay vic ng dng k thut truyn s liu rt quan trng vi cuc sng con ngi, cuc sng ngy nay cng ngy cng pht trin i hi cc nh truyn thng phi c mt ng truyn ln m bo d liu khng b tc nghn hay b mt v mt l do no . i su vo tm hiu r hn v k thut truyn s liu nhm sinh vin chng em chn ti Tm Hiu k thut chuyn mch gi dng giao thc X.25. K thut ny cho php chng ta c th truyn gi tin d liu theo mc ch s dng ca chng ta Mc d ht sc c gng, xong do kh nng c hn, nn bi n ca em chc chn cn nhiu thiu xt, em rt mong c s ch bo thm ca cc thy c. Em xin chn thnh cm n c Phm Th Qunh Trang cng nh cc thy c trong khoa in t tn tnh gip trong sut thi gian chng em thc hin n. Sinh vin thc hin:

V Quc Ton Nguyn Hu Tnh Nguyn Tin Tng

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Trang

PHN I: TNG QUAN V K THUT CHUYN MCH GI X.251.1. Gii thiu v k thut mng X.25. X.25 nh ngha chun giao din gia cc thit b u cui s liu ngi s dng DTE vi thit b cui knh d liu DCE. X.25 c chc nng va iu khin giao din DTE/DCE va thc hin chc nng truyn d liu gia DTE vi node ca mng chuyn mch gi. Cc mng X.25 cung cp cc la chn cho chuyn mch o hoc c nh. X.25 cung cp dch v tin cy cng nh iu khin lung d liu t node ti node(End to End). Cc mng X.25 c tc ti a 64Kbps. Tc ny thch hp vi cc tin trnh truyn thng chuyn giao tp v cc thit b u cui c lng lu thng ln. Tuy nhin vi tc nh vy khng thch hp vi vic cung cp cc dch v ng dng LAN trong mi trng WAN. Giao thc X.25 c ng dng trong cc mng chuyn mch gi cng cng. Nhim v ca mng l chuyn cc gi tin n ch ng th t v ng a ch. m bo khng li trong gi nhn c bn ch, X.25 tin hnh pht hin v chnh sa li

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K Thut Truyn S Liu


Hnh 1: S mng X.25 n gin

1.2. c im k thut mng X.25. - Ph hp trong mi trng truyn dn cht lng km -Bng thng hn ch, tc chun ca X.25 l 64kbps, tuy nhin ngy nay c mt s mng X.25 c bng thng ln n 2Mbps. Kiu truyn Dng dch v Bng thng ti a Knh lgic Bo hiu UNI Bo hiu NNI Kh nng di ng Internet Gi D liu 2Mbps VC,PVC X.25 X.75 X.25 trong mng di ng C h tr

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PHN II: T CHC PHN LP CA X.252.1. T chc phn lp cho X.25. X.25 l k thut chuyn mch gi hot ng trn 3 tng thp nht ca m hnh OSI: tng vt l, tng kin kt d liu v tng cp mng.

Hnh 2: Phn lp cho X.25

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-Bn tin c thit b u cui phn thnh cc gi c chiu di v thng tin a ch. Sau cc gi c ng li thnh cc khung vi cc thng tin h tr cho vic truyn dn hkoong c li. Tip cc khung c truyn trn mi trng truyn dn.

Hnh 3: knh logic trong X.25 2.2. X.25 lp 1-lp vt l Lp vt l xc nh cc vn v in, th tc kiu cc b chuyn c s dng. Bao gm cc chun ca CCITT X26/27 v EIA( USA Electronic Institue Association ), RS: X.21, X.21 Bis, V.32

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Hnh 4: Mi quan h gia X.25 vi m hnh OSI Lp vt l giao tip gia trm v tuyn ni vi node (lin quan n ng truyn gia DTE v DCE). N nh ngha cc vn nh bo hiu in, cc kiu, chun ca cc b u chuyn. 2.3. X.25 lp 2- lp lin kt d liu. Cung cp mt ng thng tin iu khin dng, khng c li gia hai u cui ca mt tuyn lin lc. N to iu kin cho cc cp cao hn lm vic m khng qun ngi v vic s liu b sai lc v cho cp di iu khin lung. Giao thc cp tuyn s dng mt s khi nim t giao thc HDLC (giao thc iu khin tuyn s liu cp cao). C hai kiu giao thc X.25 lp 2: LAP v LAPB.LAP c ngha l: th thc xm nhp tuyn (Link access procedure). Cn LAPB c ngha l th thc xm nhp tuyn c cn bng (Link access procedure balanced). LAPB hon thin hn LAP mt t v l kiu m hu ht mi ngi s dng. 2.3.1. Th thc khung ca LAPB n v tin giao thc LAPB l "khung". Hnh 6.1 trnh by cu trc ca cc khung LAPB. Trng F cha 1 byte c. Khi cc khung cha c pht i, cc byte c lin tc c chuyn i (byte mu nh phn 01111110).Khoa in T in T 2-K3

K Thut Truyn S Liu


Trng "A" cha a ch gi tin. Vng ny c th cha hoc 00000011 (a ch A) hoc 00000001 (a ch B). Vic s dng a ch A v B s c m t sau ny. Cc trng C l trng iu khin khung. N c s dng xc nh khung cha nhng g. Ch rng hnh 6.1.a v 6.1.b trng iu khin lun di 8 bits, trong khi hnh 6.1.c v 6.1.d, trng iu khin ny c th di n 8 n 16 bits. l do c s thay i thm ca giao thc m hin cha c nhc ti. Kiu LAPB chun ny cho php kch thc ca s ti a (xem chng 2 dnh gii thch cc ca s giao thc) ca 7 s lin tip t 0 ti 7. Mt vng 3 bit cn cho cng vic ny, n ghp khp trong trng iu khin. C th xy ra trng hp kch thc ca s ln hn s hay hn. c iu kiu LAPB m rng c xc nh, n c th tr gip cc kch thc ca s ti 127. Khi cn phi c trng 7 bits. Khi trng iu khin c di thay i th nhiu iu khon ca X.25 khng tr gip c cho phng thc lm vic m rng ny. trng hp hnh 6.1.a v 6.1.b ch c mt trng "I" c dng chuyn tin ca giao thc cp cao hn cc gi X.25 cp 3.Trng FCS cha dy kim tra khung. N c s dng b thu khung kim tra m bo n thu m khng c li. Thit b pht khung a thm vo FCS, tr s ca n c tnh ton theo ni dung khung.Cui cng c mt trng "F" khc. C ny xc nh im cui ca khung. Hon ton c kh nng mt khung khc tip theo ngay sau c ny, v vy ch c mt c gia cc khung. C mt vn ny sinh t cu trc khung ny. Gi s ni dung ca khung gia cc trng c c kiu bit 01111110, l kiu bt c. V c nh du im cui ca khung, v vy c th khung khng thu c chnh xc. khc phc vn ny, s liu c pht i theo cch ring. Nu ni dung ca khung cha 5 hoc hn 5 bits 1 mt dy th my pht s b sung vo mt bit 0 sau 5 bit 1. iu ny m bo khng bao gi xy ra 6 bit 1 lin tip gia ca mt khung. My thu nhn bit c iu my pht thc hin, nu n thy bit 0 theo sau 5 bit 1 th n bit rng bit 0 ny cn b loi b i v n c my pht a thm vo. K thut ny c coi nh k thut chn bit. Th t bit phi 12345678 C F 01111110 12345678 12345678 a ch A 8 bits iu khin C 16 bits 16 ti 1 FCS FCS 16 bit s F 01111110 12345678 C

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Th t bit phi

12345678 C F 0111111 0

12345678 a ch A 8 bits

12345678 iu khin C 16 bits Thng tin INFO N bits

16 ti 1 FCS FCS 16 bit s

12345678 C F 01111110

Th t bit phi

12345678 C F 01111110

12345678 a ch A 8 bits

1 ti 16 ti *) 1 iu FCS khin C FCS *) bit s 16 bit s

12345678 C F 01111110

Th t bit phi

12345678 C F 01111110

12345678 a ch A 8 bits

1 n 16 ti *) 1 iu FCS khin C FCS *) bit s 16 bit s

12345678 C F 01111110

*) 16 i vi th thc khung cha a ch dy lin tip, 8 cho th thc khung khng cha a ch dy lin tip. Hnh 6.1. Cc th thc khung 2.3.2. Cc kiu khung LAPBKhoa in T in T 2-K3

K Thut Truyn S Liu


Giao thc LAPB xc nh mt kiu khung chnh thng c dng chuyn tin theo giao thc LAPB v chuyn tin theo giao thc cp cao hn.Kiu khung ny c xc nh trng iu khin. Bng 6.1 trnh by cc loi trng iu khin hp thc LAPB. Tu theo phng thc LAPB a ra c hai dng khc nhau ca cc kiu khung. Cc chc nng khung vn gi nguyn, ch cc chc nng c a ra mi c m t bng ny. Th thc Chuyn tin Lnh I (Tin) p ng 0 M ho N(S) P P/F P/F P/F N(R) N/R N/R N/R

RR (sn RR (sn sng thu) sng thu) RNR (cha RNR (cha Gim st sn sng thu) sn sng thu) REJ (khng REJ (khng chp nhn) chp nhn) SABM (thit lp phng thc cn ... bng khng ng b) DISC Ct tuyn ni ... (gii to) Khng DM (phng nh s thc khng u ni) UA (xc nhn khng nh s FRMR (khng chp nhn khung

1.......0.......0.......0 1.......0.......1.......0 1.......0.......0.......1

1.......1.......1.......1

P

1....0....0

1.......1.......0.......0

P

0....1....0

1.......1.......1.......1 1.......1.......0.......0 1.......1.......1.......0

F F F

0....0....0 1....1....0 0....0....1

Bng 6.1: Th thc trng iu khin Ch yu c hai kiu khung: Khung lnh v khung p ng. Khung p ng c pht xc nhn cng vic thu mt lnh. V d nh cc khung I l cc khung lnh . Sau khi thu c mt khung I hay nhiu khung I, mt p ng cn c chuyn i xc nhn rng, khung hoc cc khung thu c chnh xc. Ch

Khoa in T

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K Thut Truyn S Liu


rng, cc khung S c th l cc khung lnh hoc khung p ng (tr li). Chng c s dng lm vai tr g tu theo iu kin c th. Cc lnh v cc p ng c phn bit nh gi tr trng A ca khung. Lu l trng ny c th cha a ch ca A hoc a ch B. p ng c pht cho mt lnh thu c lun c cng trng A v n l ca lnh ny. Nu DCE pht lnh th dng a ch A. Nu DTE pht lnh th dng a ch B.Thc ra cp tuyn s liu th y l s khc nhau ch yu gia DTE v DCE.By gi n lt m t cc kiu khung khc nhau. Khung "I" l "khung tin". N c dng chuyn tin cho giao thc cp cao hn. Cc khung S gi l cc khung gim st. C 3 kiu khung S: RR (my thu sn sng lm vic), RNS (my thu cha sn sng lm vic) v REJ (khung pht li). Cc khung ny lin quan ti cng vic iu khin lung cho khung I v khc phc li tuyn thng tin do hng khung. Cc khung "U" gi l cc khung khng c nh s. Chng c gi nh vy v chng khng cha a ch dy. Cc khung ny c dng khi xng, chn tuyn (SABM, SABME, DISC, DM, v UA) v bo co nhng s vi phm giao thc (FRMR). Lnh SABM (Set Asynchronous Balanced Mode) thit lp phng thc cn bng khng ng b v SABME (Set Asynchronous Balanced Mode Extended) thit lp phng thc cn bng khng ng b m rng) dng thit lp tuyn vo trng thi chuyn tin (tc l ti trng thi cao). S khc nhau duy nht gia hai lnh ny l : SABM i hi phng thc lm vic thng thng (kch c ca s ti a l 7) cn SABME i hi phng thc lm vic m rng (kch c ca s ti a 127). Khung lnh DISC (gii to) dng a tuyn v trng thi thp (di) v nh vy chng mc no n ngc vi cc lnh SABM v SABME. p ng DM (phng thc gii to) dng tr li cho SABM hoc SABME thu c nu my pht DM khng mun a tuyn vo trng thi chuyn tin. p ng UA (xc nhn khng nh s) dng khng nh lnh DISC hoc SABM thu c. p ng FRMR (khng chp nhn khung) dng ch th lnh sau cng hoc p ng sau cng khng hp l v mt no . FRMR mang thng tin m t l do. 2.3.3 Cc trng (vng) N (R) v N (S)

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K Thut Truyn S Liu


Cm N(R) do b pht khung s liu s dng bo cho my thu s th t ca khng tin tip theo m my thu ang i. Cc khung RR v RNR dng cm ny khng nh cng vic thu cc khung tin c th t ti N(R). Khung REJ dng yu cu pht li cc khung tin c s th t bt u t N(R). Cm N(S) dng ch th s th t ca mt khung tin. 2.3.4 Bit P Bt P (hoc bit u/cui) c s dng chung ch th mt khung c pht li. Khi s dng trong mt lnh th bt ny gi l bit u, cn khi s dng trong mt p ng th n gi l bit cui. Khi mt p ng c to ra cho mt lnh th bit cui phi bng bit u ca lnh. Tng qut, lc u pht mt lnh, bit u l khng (xo). Khi lnh c pht i, cn c mt p ng. Nu khng thu c p ng trong mt khong thi gian xc nh th lnh s c pht li, ln ny bit u l lp. Khong thi gian quy nh, trong phi thu c mt p ng gi l T1. l mt trong cc tham s cu hnh cc tuyn c bit. Mc cc tham s h thng sau ny s cp nhiu hn v vn ny. Cc bit ca khi tin - Trng iu khin khung khng c chp nhn l cm m iu khin ca khung thu, gy ra s t chi khung. - V(S) l bin s trng thi pht hin thi DCE hoc DTE bo co trng thi t chi (bit 10=bit th t thp). - CIR thit lp mt ch th khung b t chi l mt p ng, cn RIS thit lp 0 ch th khung b t chi l mt lnh. - V(R) l bin s trng thi thu hin thi DCE hoc DTE bo co trng thi t chi (bit 14=bt th t thp). - W trng thi 1 ch th trng iu khin thu c v quay v cc bit t 1 ti 8 khng c xc nh hoc khng c thc hin. - X trng thi 1 ch th trng iu khin thu c v quay v cc bit t 1 ti 8 b coi l khng hp l do khung cha trng tin khng cho php khung ny hoc khung ny l mt khung gim st hay mt khung khng c nh s c di khng chun xc . Bit W cn trng thi 1 phi hp vi bit ny.Khoa in T in T 2-K3

K Thut Truyn S Liu


- Y trng thi 1 ch th trng tin thu c vt qu dung lng thit lp cc i. - Z trng thi 1 ch th trng iu khin thu c v quay v cc bt t 1 ti 8 cha N(R) khng hp l. 12345678 9 Trng iu khin khung 0 khng chp thun 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

V(S)

C/R V(R)

W

X

y

Z

0

0

0

0

Hnh 6.2 Trng tin ca FRMR 2.3.5. Thao tc cp tuyn s liu C hai cung on thao tc chnh: cung on lp tuyn v cung on chuyn tin. Theo c im hnh thi ca giao thc th hai cung on ny c chia nh thnh mt s ln cc trng thi. V vy giao thc ny c xc nh theo bng trng thi. iu ny c ngha l nu bin c ny xy ra theo trng thi ny th lm nh vy v chuyn sang trng thi mi . Thc ra cc bng trng thi ch cn cho ngi thc hin giao thc, v vy chng ta khng quan tm ti cc bng trng thi y. Hai cung on ca tuyn s liu s c m t di y. Thao tc i vi DTE cng ging nh i vi DCE. V vy thut ng DXE c dng cho c DTE v DCE. Cung on lp tuyn Khi mt DXE mi c khi ng, l cung on lp tuyn. trng thi ny ph bin l pht DISC theo chu k. iu ny ch yu ni "ti ang vo cuc". Nu khng c tr li trong khong T1 th DISC c pht ln na nhng c thit lp bit P. N c vit l DISC (P). Hnh 6.3 m t trng thi ny. Nu mt DXE thu mt DISC hoc DISC(P) v mun khi ng tuyn, n tr li bng mt UA hoc UA(F) (tc l mt UA c lp mt bt cui). DXE thu UA hoc UA(F) ny s ch mt khong thi gian l T3. Nu mt SABM hocSABME thu c trong khong thi gian ny th p ng UA c pht i v tuyn s liu chuyn sang cung on chuyn tin. Nu mt SABM(P) hoc SABME(P) thu c th mt UA(F) c pht i v tuyn chuyn sang cung on chuyn tin. Lu

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K Thut Truyn S Liu


rng nu s chm tr hn xy ra th iu ny c ngha l mt SABM hoc SABME b mt v s thit lp bit u ch th rng khung c pht i. Cung on chuyn tin

Hnh 3-10 m t qu trnh trnh thit lp mt tuyn a tuyn vo cung on chuyn tin, tip theo l a tuyn quay v cung on lp tuyn. cung on chuyn tin I, cc khung RR, RNR v REJ c dng iu khin cng vic chuyn giao s liu giao thc cp cao hn qua tuyn. Nu thu c mt khung I chun xc v DXE c th tip nhn na th n tr li cho khung I ny bng mt khung p ng RR. Nu DXE khng th tip nhn na th n tr li bng mt p ng RNR, bo cho DXE kia rng hin n bn v khng th tip nhn thm s liu thi im ny. p ng REJ dng yu cu pht li mt hoc nhiu khung I b DXE nghi l mt (c th b loi b do mt li FCS sinh ra trong khi thu). Cc khung RR, RNR v REJ c dng tr li khung I l cc p ng. Dng lnh ca cc khung RR, RNR v REJ dng hi DXE kia v trng thi hin ti ca n hoc bo cho n nu trng tha ca DXE ny thay i. Khi c s dng l lnh th cc khung RR, RNR, v REJ lun c s thit lp bt u. V vy cc p ng to ra bn thu lun c gn bt cui. xem xt n lm vic ra sao, gi thit rng mt DXE tr li cho mt khung tin bng mt p ng RNR do n khng th tip nhn s liu na. Khi li c th thu s liu, n c th pht mt lnh RR(P) cho DXE kia, thng bo cho n v trng thi mi. Sau DXE thu c th tr li bng mt p ng RR(F), RNR(F) hoc REJ(F), (tu thuc vo trng thi ca n) v li tip tc pht cc khung I,. C DTE ln DCE c th chuyn tuyn sang trng thi thit lp nh pht i mt lnh DISC vo bt c lc no. Nu mt DXE i hi phc hi tuyn th n pht i lnhKhoa in T in T 2-K3

K Thut Truyn S Liu


SABM hoc SABME. Cng th, iu ny c th xy ra bt c lc no.Pha thu pht mt UA tr li v tuyn li tr v cung on chuyn tin. Trng thi t chi khung Trng thi t chi khung c a vo khi thu mt khung khng hp l. iu c ngha l mt khung khng c thu nhn cng vi trng a ch A hoc B trng A v khng c li FCS, nhng ni dung ca khung vn khng chun xc hoc khng tng ng i vi trng thi ca pha my thu. Hin nhin y l trng thi tng i trm trng, c th biu hin s vi phm giao thc v cn phi ti lp tuyn. Mc d tuyn c th c ti lp ngay nh pht i lnh SABM hoc SABME, nhng cng khng th bo cho DXE kia v sao tuyn li phi ti khi ng. V vy khi mt DXE thu mt khung khng hp l th n pht mt p ng FRMF bo cho DXE kia bit ci g b sai. Ch yu y l mt s lun ti: "bn pht cho ti mt khung b sai v v sao vy". p ng FRMF l mt bt c bit bi v n l p ng duy nht c th pht i tr li mt p ng - tt, c phi khng? Ngay trng thi t chi khung, tuyn c th c ti khi ng bng mt lnh SABM hoc SABME. 2.3.6. Cc tham s h thng Cc tham s h thng l cc tham s cu hnh, n xc nh cc kha cnh no ca s thao tc cp tuyn s liu. i lng T1 l khong thi gian m my pht khung lnh ch mt p ng trc khi li pht i mt lnh c gn bt u. i khi gi y l khong ti th. T1 phi ln hn thi gian dng pht mt khung c di cc i v nhn mt p ng cho khung ny, n c th l mt khung cc i. N tu thuc vo tc pht cc bt theo tuyn thng tin v khong tr x l my thu. Cn c khong nh thi na, l T2, n c xc nh nh l thi gian cc i cn dng trc khi my thu thu mt khung v pht i mt khung xc nhn vic thu khung ny. N lun ngn hn T1. iu ny thc t thch hp pht i mt khung xc nhn vic thu mt khung cng sm cng tt. Khong nh thi gian T3 xc nh mt DXE phi ch bao lu i vi lnh thit lp tuyn trc khi bt u pht i cc DISC cung on lp tuyn. Gi tr ny l T1xN2. N2 l s ln cc i mt khung lnh c pht li trc khi tuyn c ti khi ng.

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K Thut Truyn S Liu


Thc cht nu T1 ht N2 ln th my pht t b v ti khi ng tuyn bi SABM hoc SABME. PSS dng gi tr 20 i vi N2. Cc mng khc nhau c th quy nh cc gi tr N2 khc nhau nhng chng hon ton ging PSS. N1 l s bt cc i c th c trong mt khung I. N bao gm cc cm IA, C, I, v FCS. V d nu kch c cm I cc i cho mt tuyn l 128 bytes th N1 s l 1064. Tham s h thng k l s lng cc i ca cc khung I c nh s tun t m mt DXE c th pht i nhng khng c xc nhn bt c ln no (tc l kch c ca s). PSS xc nh gi tr k l 7 v n khng b tr cho phng thc lm vic m rng. 2.4. X.25 lp 3-lp mng. X.25 lp 2 to ra phng thc chuyn tin giao thc cp cao hn (trong cc khung tin ) gia hai u cui ca mt tuyn thng tin m bo chun xc, iu khin lu lng chuyn s liu. X.25 lp 3 to cho s liu c pht i trong cc khung tin. n v s liu cp mng l gi. Giao thc cp mng trn c bn xc nh thao tc gi o qua giao thc cp tuyn. Mi cuc gi o c lp mng to ra cho cc giao thc cp cao hn l mt tuyn c iu khin theo lung gia DXE ni ht v mt DXE xa qua mng. X.25 lp 3 thc t c nh ngha l mt giao thc gia mt DTE v mt DCE u ni trc tip qua mt tuyn thng tin. DTE c th nh l mt PAD cn DCE c th l mt thit b chuyn mch gi X.25. C th c nhiu kiu mng khc nhau dng c s dng cung cp tuyn ni gia hai DXE. Th nhng iu quan trng l giao tip cp mng gia DTE v DCE phi gi ging nhau d cho cc mng bao gm c cc tuyn truyn gia DXE ni ht v DXE xa. 2.4.1. Khun mu gi cp mng Mi mt gi cp mng c cng khun mu u 3 bytes m t hnh 6.8. Cm nhn dng khun mu chung (GFI) l khi 4 bt c dng ch th khun mu chung cho phn cn li ca u . Cng vic m ho ca cm GFI s c m t trong khi m t cc kiu gi. GFI+LCGN LCN PTI Phn cn li ca gi

Hnh 6.8. Khun mu gi cp mng Cm th hai ca bytes u ny ca gi l a ch nhm knh lgic (LCGN). N ko sang c bytes th hai to thnh a ch knh lgic (LCN) 12 bit, n dng

Khoa in T

in T 2-K3

K Thut Truyn S Liu


nhn dng cho tng cuc gi o ring bit. Byte th ba l cm nhn dng kiu gi (PTI), n nh ra chc nng ca gi. 2.4.2. Cc kiu gi cp mng Cc gi thit lp v xo cuc gi Gi gi vo v yu cu gi dng yu cu thit lp mt cuc gi o gia DXE pht gi ny v DXE thu gi ny. Gi ch cuc gi c u ni hay cuc gi c tip nhn c dng tr li cho gi yu cu gi hoc gi ch cuc gi vo ch th rng cuc th ni c tip nhn v by gi cuc gi c tin hnh. Gi yu cu gii to v biu th gii to c dng hoc kt thc mt tuyn ni ang lm vic hoc t chi mt yu cu thit lp gi (tc l tr li cho gi yu cu gi hoc gi vo). Gi xc nhn gii to dng xc nhn rng thu c gi ch th gii to trc hoc gi yu cu gii to. Cc gi s liu v ngt Gi s liu c dng chuyn s liu cho giao thc cp cao hn gia hai DXE u ni vi nhau bi cuc gi o. Gi ngt c dng chuyn mt phn nh s liu (ti a 32 bytes) gia hai DXE vi u tin rt cao. Gi ngt c kh nng nhy qua cc gi s liu v khng ph thuc vo s iu khin lu lng cp mng. Gi xc nhn ngt c dng xc nh vic thu mt gi ngt. Ch c th c mt gi ngt khng c xc nhn bt k ln no. Cc gi iu khin lung v ti lp Cc gi RR v RNR c dng xc nhn vic thu cc gi s liu. S dng gi RR khi my thu c th thu thm cc gi s liu. Gi RNR c s dng khi my thu tm thi b bn v khng th thu thm s liu. Gi REJ c th c DTE s dng yu cu chuyn cc gi s liu. Dch v REJ khng cn thit b tr tt c cc DCE v thc t n khng cn cho thao tc chun xc ca nghi thc. S dng gi REJ c ng l mt gi s liu thu c chun xc bi cp tuyn s liu b DTE lm mt v l do no , c th do n b y ra khi vng nh m dnh cho gi tin thu c. Gi ch th ti lp/yu cu ti lp dng chuyn cuc gi o v trng thi trc ca n khi cuc gi c thit lp lc ban u. Ton b cc vic cha gii quytKhoa in T in T 2-K3

K Thut Truyn S Liu


xong ca s liu b vt b, cc a ch dy c lp khng v cc trng thi iu khin lung b xo. Gi ny thng c s dng khi li giao thc c pht hin hoc iu g xo s liu b "mc kt" mt cuc gi m khng cn phi xo cuc gi hin thi. Gi xc nhn ti lp c dng xc nhn vic thu ca gi ch th ti lp/yu cu ti lp v nh vy th thc ti lp c thc hin. Cc gi ti khi ng Gi ch th ti khi ng/yu cu ti khi ng c dng xo i tt c cc cuc gi o ang xc tin v chuyn ti ton b cp mng v trng thi khi u ca n. Gi ny l gi u tin c cp mng pht i khi cp tuyn s liu chuyn sang cung on chuyn tin. Gi xc nhn ti khi ng c dng xc nhn cng vic thu mt gi ch th ti khi ng/yu cu ti khi ng v ch th rng cp mng hin ang hot ng Kiu gi T DCE ti DTE T DTE ti DCE Byte 3 Cc bit 8.....7.....6.....5.....4.....3......2......1

Thit b v gii to cuc gi Yu cu gi Gi vo 0.....0.....0.....0.....1.....0......1......1 Tip nhn cuc gi u ni cuc gi 0.....0.....0.....0.....1.....1......1......1 Yu cu gii to Ch th gii to 0.....0.....0.....1.....0.....0......1......1 Xc nhn gii ta ca Xc nhn gii to DCE 0.....0.....0.....1.....0.....1......1......1 DTE S liu v ngt S liu DTE S liu DCE x.....x.....x.....x.....x.....x......x......0 Ngt ca DTE Ngt ca DCE 0.....0.....0.....0.....0.....0......1......1 Xc nhn ngt ca Xc nhn ngt ca DCE 0.....0.....1.....0.....0.....1......1......1 DTE iu khin lung v ti lp DCE RR (module 8) DTE RR (module 8) x.....x.....x.....0.....0.....0......0......1 DCE RR (module 128) DTE RR (module 128) 0.....0.....0.....0.....0.....0......0......1 DCE RNR (module 8) DTE RNR (module 8) x.....x.....x.....0.....0.....1......0......1 DCE RNR (module 128) DTE RNR (module 0.....0.....0.....0.....0.....1......0......1 128) x.....x.....x.....0.....1.....0......0......1 DTE REJ (module 8) 0.....0.....0.....0.....1.....0......0......1 Ch th ti lp DTE REJ (module 0.....0.....0.....1.....1.....0......1......1Khoa in T in T 2-K3

K Thut Truyn S Liu


Xc nhn ti lp DCE

Ch th ti khi ng Xc nhn ti khi ng DCE Phn on

Xc nhn ng k

128) Yu cu ti lp 0.....0.....0.....1.....1.....1......1......1 Xc nhn ti lp DTE Ti khi ng Yu cu ti khi ng 1.....1.....1.....1.....1.....0......1......1 Xc nhn ti khi 1.....1.....1.....1.....1.....1......1......1 ng DTE Phn on 1.....1.....1.....1.....0.....0......0......1 ng k 1.....1.....1.....1.....0.....0......1......1 Yu cu ng k 1.....1.....1.....1.....0.....1......1......1

Bng 6.2. Cc tr s ca cm m PTI Cc gi phn on li v ng k dch v Gi phn on li do DCE pht cho DTE khi DCE thu mt gi tin b li trm trng. V d : Khi thu c mt gi c trng GFI khng chun xc, DCE c th pht mt gi phn on li cho DTE, gi ny cha m phn on, li thch hp: Khng phi ton b cc DCE u to ra cc gi phn on li. Gi yu cu ng k dch v c th c DTE pht cho DCE yu cu c s dng hay khng s dng mt s dch v no trong khong thi gian no . Cc dch v lin quan s c m t sau ny. Gi xc nhn ng k do DCE pht cho DTE tr li cho mt gi yu cu ng k dch v t DTE. 2.4.3. Cc a ch dy cp mng Cng nh cp tuyn s liu, cc kiu gi xc nh u mang theo chng cc a ch dy. Cc a ch dy ny (ch s th t) c dng m bo cho cc gi s liu c chuyn i khng b mt v theo mt th t chun xc. C hai a ch dy c ti i, l a ch dy P(S) v a ch dy P(R). a ch dy P(S) ch c mang theo cc gc s liu v dng nhn dng t gi s liu ring.

Khoa in T

in T 2-K3

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a ch dy P(R) c mang theo gi s liu, gi RR, RNR, v REJ. Vng m P(R) cc gi ny chuyn a ch dy ca gi s liu tip theo m my pht s chuyn cho my thu. Ging nh cp tuyn s liu, c h thng nh s dy thng dng, n s dng cm 3 bt v c a ch dy t 0 ti 7, v mt h thng nh s dy m rng n s dng cm m 7 bt v c a ch dy t 0 ti 127. 2.4.4. Trng m nhn dng khun mu. Chng ta i qua cc kiu gi cp mng, by gi chng ta chuyn sang cng vic m ho cm m GFI. Bng 6.3 trnh by cc gi tr m cm GFI c th nhn Bit "Q" ch xut hin cc gi s liu v c dng phn bit gi s liu theo hai loi khc nhau: Cc gi s liu thng thng v cc gi s liu "nh phm cht". Cc gi s liu nh phm cht thng c s dng cho php chuyn thng tin iu khin giao thc cp cao hn m khng nh hng ti s liu giao thc cp cao hn m chng c pht i cc gi s liu thng thng. Bt D l bt xc nh chuyn giao. Bit ny c th xut hin cc gi thit lp gi nhng thc t chc nng ca n ch lin quan n vic chuyn giao cc gi s liu. Bit 5 v 6 ca cm m GFI c s dng ch th h thng nh s dy no c s dng. H thng nh s dy m rng l mt trong cc kiu t chn, gi l kiu t trc. Tc l h thng nh s c dng cn phi c quyt nh khi tuyn X.25 c thit lp. Ton b cc cuc gi o trn tuyn cn phi s dng h thng nh s ny t trc n. Nu dch v ng k c hiu lc th n c th chuyn i h thng nh s hin thi theo nhng iu kin nht nh. Phn ln cc trng hp s dng h thng nh s thng thng v ch cn rt t iu b sung b tr cho h thng nh s m rng. Byte 1 Cc bit 7 6 0 0 0 1 0 0 1

H thng a ch dy module 8 H thng a ch dy module 128 Gc gii to, iu khin H thng a ch dy module 8 lung, ngt, ti lp, ti khi ng ng k v phn H thng a ch dy module 128 on H thng a ch dy module 8 Cc gi s liu H thng a ch dy module 128 M rng nhn dng khun mu thng thng Cc gi thit lp gi

8 0 0 0

5 1 0 1 0

0 0 0 0 0 0 0 0

0 1 1 0 1 1

Khoa in T

in T 2-K3

K Thut Truyn S Liu


Dng cho cc ng dng khc * Khng xc nh Bng 6.3. Tr s ca cm m GFI 2.4.5. Cung on ti khi ng

*

*

0

0

Ngay khi cp tuyn s liu chuyn sang trng thi chuyn tin, cp mng cn c ti khi ng. Mt yu cu ti khi ng/ch th ti khi ng DCE hoc DTE pht i, sau n c xc nhn bi gi xc nhn ti khi ng. V n l chung cho gi ch th ti khi ng/gi yu cu ti khi ng m cn pht i ngay khi cp tuyn lm vic, nn thng xy ra "va vp". Nu mt DXE ang ch s xc nhn ti khi ng m n thu c ch th ti khi ng / yu cu ti khi ng th "va vp" ti khi ng b xy ra v gi thu c coi nh l s xc nhn ti khi ng. Hnh 6.9 m t th thc ca cm m ch th ti khi ng / yu cu ti khi ng v cc gi xc nhn ti khi ng. Lu l cm m LCGN v LCN c m ho ton b s 0. a ch y ca knh lgic 000000000000 c th khng dng c bi mt cuc gi o tt c cc iu khin b sung v cn loi ra bt c ni no c th c.

Bng 6.4 m t gi tr c th c ca cm m nguyn nhn ti khi ng trong gi ch th ti khi ng. Cc bit 8......7......6......5......4......3......2......1 0......0......0......0......0......0......0......0 0......0......0......0......0......0......1......1 0......0......0......0......0......1......1......1 0......1......1......1......1......1......1......1

Li th tc ti ch mng Mng lm vic Xc nhn ng k/ xo

Bng 6.4. M ho cho cm m ch nguyn nhn ti khi ng. ca cc gi ch th ti khi ng Cc byte

Khoa in T

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K Thut Truyn S Liu


1 2 3 4 5

Cc bit 8......7......6......5......4......3......2......1 Nhn dng khun mu ......0......0......0......0 chung 0......0......0......0......0......0......0......0 Nhn dng kiu gi 1......1......1......1......1......0......1......1 Nguyn nhn ti khi ng M phn on li

a) Gi ch th ti khi ng / gi yu cu khi ng Cc bit 8......7......6......5......4......3......2......1 Cc byte 1 2 3 Nhn dng khun mu ......0......0......0......0 chung 0......0......0......0......0......0......0......0 Nhn dng kiu gi 1......1......1......1......1......1......1......1 b) Gi xc nhn ti khi ng Hnh 6.9 Khi c dng gi yu cu ti khi ng, bt 8 cn c thit lp gi tr 1 thay cho gi tr 0. V vy y l s hon thin ng k so vi X.25, n ni ln rng cm m ny gi yu cu ti khi ng cn phi c gi tr m ton b khng. iu ny thng c ngha l khi s dng cc tuyn ghp X.25 th l do ti khi ng mt i. Cm m phn on li c th c dng cung cp thm thng tin v l do khi ng li. Ch c gi khc c lin quan ti cung on ny l gi phn on li. Hnh 6.10 m t th thc ca gi ny. Cm gii ngha phn on bao gm 3 byte u ca cm m u trong gi ch th nguyn nhn phi pht gi phn on li. Nu ch c di 3 byte tc l ch c nhng g thu c. Cc byte Cc bit 8......7......6......5......4......3......2......1

Khoa in T

in T 2-K3

K Thut Truyn S Liu


1 2 3 4 5

Nhn dng th thc ......0......0......0......0 chung 0......0......0......0......0......0......0......0 Nhn dng kiu gi 1......1......1......1......0......0......0......1 M phn on li Gii ngha phn on li Hnh 6.10

2.4.6. Thit lp cc cuc gi thc Gi gi vo/gi yu cu gi yu cu mt cuc gi thc c chuyn mch cn c thit lp gia hai DXE c ni vo mng. Hnh 6.11 m t th thc ca gi ny. Lu l bt 7 ca GFI c nhn l "D". Nu bt ny c thit lp tc l thu bao ch gi yu cu s dng dch v bit D (xem mc cung on chuyn giao tin chng ny v vn ). Nu n b xo tc l ch gi khng c nh s dng n. DXE pht gi gi vo/yu cu gi phi chn mt a ch nhm knh lgc (LCGN) v a ch knh logic (LCN) nhn dng duy nht mt cuc gi ring. Hin nhin n khng th l a ch dng bi cuc gi ang tin hnh. Dch v PSS chia s lng tng th knh thnh mt nhm. Cc nhm khc nhau c nhn dng bi trng m LCGN ny. Cc nhm ny l: LCGN, Kiu, Tn, S dng N dng DCE nh v LCNs t a ch knh thp nht tr ln, cn DTE li nh v LCNs t knh cao nht xung. Thng thng th c DCE v DTE u s dng SVC c hai cnh. V d nu ch cn 16 knh lgic thi th LCGN 4 c th c s dng cng vi cc LCN t 0 ti 15. DCE s bt u phn b cc knh t 0 tr ln, cn DTE s bt u phn b cc a ch knh t 15 tr xung. Cc bit 8......7......6......5......4......3......2......1 Nhn dng th thc a ch nhm knh logic chung a ch knh logicKhoa in T in T 2-K3

K Thut Truyn S Liu


Nhn dng kiu gi 0......0......0......0......1......0......1......1 (Cc) a ch DTE 0......0......0......0............. Chiu di dch v Cc dch v S liu thu bao gi Hnh 6.11. Gi gi vo/gi yu cu gi C hai cm m gi l a ch DTE b gi v a ch DTE ch gi. Mi a ch c th ti 15 ch s, mi ch s c th t 0 ti 9. Thng ch ti a 14 ch s c s dng 14 ch s ny c chia thnh mt a ch 12 ch s cng vi 2 ch s cui cng l a ch di cp. Mng X.25 thng to tuyn trn c s a ch 12 ch s u. Cn 2 ch s sau dng cho a ch mt thc th ring trong DXE c a ch . Nh vy a ch ny c th l a ch DXE ch gi v a ch DXE b gi. L do v sao DTE c s dng l c im ca DTE cng nh cc thit b c u ni vo mng c DCE l giao tip cho mng. V vy bn thn ca cc DCE khng bao gi l b gi. Tuy nhin trong thc t th c cc DTE ln cc DCE vn c th l b gi. Cc a ch ny c thit lp dng gi s dng th thc s thp phn c m ho nh phn (BCD). iu ny c ngha l 2 ch s trong mi byte: 4 bit cao (8-5) cho ch s th nht, 4 bit thp (4-1) cho ch s th hai. a ch u tin l a ch DTE b gi. N l a ch ca thit b m thu bao ch gi mun thit lp cuc gi o ti . a ch th 2 l a ch thu bao ch gi (DTE). N l a ch ca thit b bt ngun ca gi tin. c im ny cho php DTE khng cha a ch DTE ch gi mc d DCE lun lun cn phi bao gm c n. Nh th cng vic thc hin X.25 c th l DCE hoc DTE, a ch DTE ch gi lun bao gm hu ht. Mt vn nh xy ra khi tng s cc ch s trong cm a ch DTE ch gi v b gi l l v ch mt na byte cui c dng. Quy nh ny chn s 0 a thm vo lp vo na byte cui cng. Tip theo cc a ch DTE l cm m dch v m u bi byte ch chiu di dch v. Chiu di ny l tng s byte ca cm m dch v c php (chiu di ny c th thay i c). Chiu di ti a ca trngdch v l 100 byte. Trng ny dng m ho cc thng tin c bit v cuc gi v cc yu cu dch v trong khong thi gian tin hnh cuc gi. Cc dch v ny s c m t sau ny.

Khoa in T

in T 2-K3

K Thut Truyn S Liu


Trng cui cng ca gi l trng s liu thu bao ca cuc gi (CUD). Trng ny cha cc s liu tu chn, n c truyn dn khng bin i gia DCX ch gi v b gi. cc gi gi vo thng thng/cc gi yu cu gi trng CUD di 16 byte l ti a. Nu dch v "chn nhanh" c s dng th trng ny c th di ti 128 bytes. Dch v chn nhanh l mt trong cc dch v c th c trong trng dch v ca gi. Trng CUD c th c dng cho nhiu mc ch khc nhau tu theo ng dng. Mt ng dng chung l cung cp kh nng to lp a ch m rng khi mt phn ca tuyn gi bao gm cc mng khng dng X.25. Nu cuc gi c pha thu gi gi vo/gi yu cu gi chp nhn th mt gi ch th cuc gi c u ni/cuc gi c tip nhn c pht tr li. Hnh 6.12 m t th thc ca gi ch th cuc gi c u ni/cuc gi c tip nhn. Lu l c hai dng gi u ni gi/tip nhn gi. Th thc thng thng c th cha cc i l 3 bytes v khng c trng s liu thu bao b gi. Th thc m rng c t nht 5 bytes (nu khng c a ch v cc dch v th chiu di dch v v a ch vn phi c) v c th cha trng s liu thu bao b gi nu gi gi vo/gi yu cu gi quy nh s dng dch v chn nhanh. Cc bit 8......7......6......5......4......3......2......1 Nhn dng th thc a ch nhm knh chung logic a ch knh logic Nhn dng kiu gi 0......0......0......0......1......0......1......1 di a ch DTE di a ch DTE b ch gi gi (Cc) a ch DTE 0......0......0......0......... Chiu di dch v Cc dch v S liu thu bao b gi Hnh 6.12 Gi ch cuc gi c u ni/c tip nhn

1 2 3 Cc byte 4 5

Khoa in T

in T 2-K3

K Thut Truyn S Liu


Nhc li l trng GFI c bit 7 c gn nhn D. Nu ch gi c yu cu dch v ny v DXE b gi c kh nng cung cp th cn phi thit lp bit ny trong gi. Nu khng th bit ny phi xo i. p ng nhc nh cho gi gi vo/gi yu cu gi l gi ch th xo/gi yu cu xo. Hnh 6.13 m t th thc ca gi ny. Cng c hai kiu bin th nh vi gi u ni cuc gi v gi tip nhn cuc gi. Cc quy tc s dng a ch, dch v v xo s liu thu bao dn ra t cc quy tc ca gi u ni cuc gi v gi tip nhn cuc gi. Cc bit 8......7......6......5......4......3......2......1 Nhn dng th thc a ch nhm knh chung logic a ch knh logic Nhn dng kiu gi 0......0......0......1......0......0......1......0 Nguyn nhn gi to M phn on li Chiu di a ch Chiu di a ch DTE ch gi DTE b gi (Cc) a ch DTE 0......0......0......0.......... Chiu di dch v Cc dch v S liu thu bao xo Hnh 6.13. Gi ch th xo/gi yu cu xo Trng l do gii to (xo) c dng ch th l do v sao cuc gi b xo. Bng 6.5 m t cc tr s c th dng cho trng nguyn nhn xo (mong mun ca cc tc gi l khng b phm li). Lu DTE c th to lp trng ny nu bit 8 c thit lp. Cng nh i vi trng l do ti khi ng, y l mt trong cc tnh nng mi, quan trng ca X.25. Gi xc nhn xo dng xc nhn vic thu gi ch th xo/gi yu cu xo. Th thc ca n m t hnh 6.14. y li c hai bin th ca gi ny. Lc ny th thc gi thng thng lun c s dng. Gi th thc m rng ch c th do mtKhoa in T in T 2-K3

1 2 3 Cc byte 4 5 6

K Thut Truyn S Liu


DCE pht cho mt DTE v n ch c dng lin quan ti dch v. "thng tin tnh cc". khc cc dch v tu chn. Cc bit 8......7......6......5......4......3......2......1 0......0......0......0......0......0......0......0 x......x......x......x......x......x......x......x 0......0......0......0......0......0......0......0 0......0......0......0......1......0......0......1 0......0......0......1......0......0......0......1 0......0......0......1......1......0......0......1 0......0......1......0......0......0......0......1 0......0......1......1......1......0......0......1 0......0......0......0......0......0......1......1 0......0......0......0......1......0......1......1 0......0......0......1......0......0......1......1 0......0......0......0......0......1......0......1 0......0......0......0......1......1......0......1 0......0......0......1......0......1......0......1

DTE khi xng gi DTE khi xng gi a ch bn Khng c khai thc Li th tc hng ra B gi khng chp nhn tr cc ch gi khng tng thch Khng chp nhn chn nhanh Khng li Yu cu dch v khng hp l Chn tip cn Li th tc ti ch ti mng Khng c kh nng tip cn RPOA

Bng 6.5 Bng gi tr ca trng nguyn nhn xo Cc bit 8......7......6......5......4......3......2......1 Nhn dng th thc a ch nhm knh 1 chung logic 2 a ch knh logic Nhn dng kiu gi Cc byte 3 0......0......0......1......0......1......1......1 Chiu di a ch Chiu di a ch 4 DTE ch gi DTE b gi 5 (Cc) a ch DTE Chiu di dch v Cc dch v Hnh 6.14. Gi xc nhn xo

Khoa in T

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K Thut Truyn S Liu


Cc byte

1 2 3

Cc bit 8......7......6......5......4......3......2......1 Nhn dng th thc chung a ch nhm knh logic Q......D......1......1 P(R) M (P)S S liu thu bao 0

Cc byte

1 2 3 4

Cc bit 8......7......6......5......4......3......2......1 Nhn dng th thc chung a ch nhm knh logic Q......D......1......0..... a ch knh logic P(S) 0 P(R) M S liu thu bao (Khi m rng module ti 128) D: Bt xc nhn phn pht; Q: Bt nh tiu chun M: Bt tng thm s liu Hnh 6.16. Gi s liu

2.4.7 Cung on chuyn giao tin Mt cuc gi o s mt nhiu thi gian nht cung on chuyn s liu. Cc gi s liu c trao i gia cc DXE mi u ca tuyn gi. Hnh 6.16 m t th thc ca gi s liu. Lu y li c hai th thc c th xy ra. N c to ra do ta nh s dy m rng v thng thng. Bit M l bit ch th thm s liu. N dng to ra kh nng chuyn cc bn tin s liu i. Cc bn tin ny di hn mt gi cp mng n. Bng tin ny c chia

Khoa in T

in T 2-K3

K Thut Truyn S Liu


thnh cc gi cp mng v gi cui cng s to lp bit M. Gi cui m khng c bit M th hin y l cui ca dy. Khi mt DXE thu mt gi s liu n phi pht tr li cho DXE u kia tin xc nhn cuc gi. C th c hai gi, gi RR hoc RNR. Hnh 6.17 m t th thc ca cc gi ny. Gi RNR c pht i khi DXE khng c kh nng thu thm cc gi tin na, thng do n ht tim nng. Gi RR c pht i khi DXE c kh nng thu tip tc cc gi s liu. Nh trc y nhc ti, mt DTE c th pht i mt gi REJ cho DCE yu cu pht li cc gi s liu. Th thc ca n hnh 6.18. S hin din ca bt D c dng m bo rng DXE u kia thc s thu c gi s liu. V sao iu ny li cn thit ? Vn l ch mc d DXE pht gi s liu ny c th nhn c gi tr li xc nhn l RR hay RNR, n khng bit chnh xc gi ny t u ti. Nu bit D c thit lp th yu cu s dng dch v "xc nhn phn pht", khi lp mng khng pht i s xc nhn cho ti khi DXE i phng thu c gi s liu v xc nhn iu . Cc bit 8.....7.....6.....5.....4.....3......2......1 Nhn dng th thc i ch nhm knh chung logic 0.....0.....0.....1 i ch knh logic Nhn dng kiu gi P(R) 0.....0.....0.....0......1 (Module 8) Cc bit 8......7.....6.....5.....4.....3......2......1 Nhn dng th thc i ch nhm knh chung logic 0.....0.....1.....0.... i ch knh logic Nhn dng kiu gi 0.....0.....0.....0.....0.....0......0......1 P(R) D

Cc byte

1 2 3

1 Cc byte 2 3 4

Khoa in T

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a) Gi RR Cc bit 8.....7.....6.....5.....4.....3......2......1 Nhn dng th thc i ch nhm knh chung logic 0.....0.....0.....1 i ch knh logic Nhn dng kiu gi P(R) 0.....0.....1.....0......1

Cc byte

1 2 3

1 Cc byte 2 3 4

Cc bit 8......7.....6.....5.....4.....3......2......1 Nhn dng th thc i ch nhm knh chung logic 0.....0.....1.....0.... i ch knh logic Nhn dng kiu gi 0.....0.....0.....0.....0.....1......0......1 P(R) D

b) Gi RNR (Khi module m rng 128) Hnh 6.17. Gi RR(a) v gi RNR(b) Cc bit 8.....7.....6.....5.....4.....3......2......1 Nhn dng th thc i ch nhm knh chung logic 0.....0.....0.....1 i ch knh logic Nhn dng kiu gi P(R) 0......1......0......0......1in T 2-K3

1 Cc byte 2 3

Khoa in T

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a) Module 8 Cc bit 8......7.....6.....5.....4.....3......2......1 Nhn dng th i ch nhm knh thc chung logic 0.....0.....0.....1... i ch knh logic Nhn dng kiu gi 0.....0.....0.....0.....1.....0......0......1 P(R) D b) Module m rng 128 Hnh 6.18. Gi REJ

1 Cc byte 2 3

Cc gi cui lin quan ti cung on chuyn tin l cc gi ch th ti lp/yu cu ti lp v xc nh ti lp. Hnh 6.20 m t th thc ca gi tin ny. Bng 6.6 l bng m ho ca trng l do ti lp. DTE c th thit lp trng ny khi bt 8 c thit lp. M phn on li s cung cp thm thng tin v l do ti lp. Th thc ca cc gi yu cu ng k v xc nhn ng k m t hnh 6.21.

1 Cc byte 2 3 4

Cc bit 8......7.....6.....5.....4.....3......2......1 Nhn dng th thc i ch nhm knh chung logic i ch knh logic Nhn dng kiu gi 0.....0.....1.....0.....0.....0......1......1 S liu thu ngt a) Gi ngt

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1 Cc byte 2 3

Cc bit 8......7.....6.....5.....4.....3......2......1 Nhn dng th thc i ch nhm knh chung logic i ch knh logic Nhn dng kiu gi 0.....0.....1.....0.....0.....1......1......1 b) Gi xc nhn ngt

Hnh 6.19. M t th thc cc gi mgt v xc nhn ngt

DTE khi xng DTE khi xng Khng lm vic tt Li th thc pha xa Li th thc ti ch mng Li thao tc DTE pha xa Li thao tc mng ch khng tng thch Mng khng lm vic

Cc bt 8......7.....6.....5.....4.....3......2......1 0.....0.....0.....0.....0.....0......0......0 1.....x.....x.....x.....x.....x......x......x 0.....0.....0.....0.....0.....0......0......1 0.....0.....0.....0.....0.....0......1......1 0.....0.....0.....0.....0.....1......0......1 0.....0.....0.....0.....0.....1......1......1 0.....0.....0.....0.....1.....0......0......1 0.....0.....0.....0.....1.....1......1......1 0.....0.....0.....1.....0.....0......0......1 0.....0.....0.....1.....1.....1......0......1

Bng 6.6. Cc tr s m ca trng l do ti lp Cc byte 1 2 3 4 0.....0.....0.....1.....1.....0......1......1 L do ti lp Cc bit 8......7.....6.....5.....4.....3......2......1 Nhn dng th thc i ch nhm knh chung logic i ch knh logic Nhn dng kiu gi

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5

M on li

a) Gi ch th ti lp 1 yu cu ti lp Cc bit 8......7.....6.....5.....4.....3......2......1 Nhn dng th thc i ch nhm knh chung logic i ch knh logic Nhn dng kiu gi 0.....0.....0.....1.....1.....1......1......1 b) Gi xc nhn ti lp Hnh 6.20

1 Cc byte 2 3

2.4.8. Trng m dch v Mt s gi cp mng c trng dch v trong chng. Trng m dch v ny cho php DXE ch gi yu cu mt s dch v c th dng c trong lc thc hin cuc gi. Trng dch v ny to thnh t mt hay nhiu "phn t dch - v". C 4 dng phn t dch v c bn.(Hnh 6.22) Bit 7 v 8 ca byte u tin ca phn t dch v m ho kiu dch v. Trong thc t c mt s lng kh ln m phn t dch v. Cc kiu nu ra y l kiu ng ch nht b tr dch v mng OSI. Byte u, byte nhn dng kiu dch v, ri tip ti l mt hay nhiu byte ca trng tham s dch v.

Cc byte 1

Cc bit 8......7.....6.....5.....4.....3......2......1 Nhn dng th thc 0.....0......0......0 chungin T 2-K3

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2 3 4

i ch knh logic Nhn dng kiu gi 1.....1.....1.....1.....0.....0......1......1 Chiu di a ch DTE Chiu di a ch DCE a ch DCE v DTE 0.....0......0......0 Chiu di ng k ng k

0

a) Gi yu cu ng k Cc bit 8......7.....6.....5.....4.....3......2......1 Nhn dng th thc 0.....0......0......0 chung 0......0.....0.....0.....0.....0......0......0 Nhn dng kiu gi 1.....1.....1.....1.....0.....1......1......1 Nguyn nhn Phn on Chiu di Chiu di a ch a ch DCE DTE a ch DCE v DTE 0.....0......0......0 Chiu di ng k ng k

1 2 3 Cc byte 4 5

0

b) Gi xc nhn ng k Hnh 6.21. Gi yu cu v xc nhn ng k

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Cc byte

0 1

Cc bit 8......7.....6.....5.....4.....3......2......1 0......0.....x.....x.....x.....x......x......x Trng tham s dch v/ ng k Loi A Bit 8......7.....6.....5.....4.....3......2......1 0......1.....x.....x.....x.....x......x......x Dch v/ ng k Trng tham s Loi B Bit 8......7.....6.....5.....4.....3......2......1 1......0.....x.....x.....x.....x......x......x Trng tham s dch v/ ng k

Cc byte

0 1 2

Cc byte

0 1 2 3

Loi C Bit 8......7.....6.....5.....4.....3......2......1 1......1.....x.....x.....x.....x......x......x Chiu di trng tham s dch v ng k Trng tham s dch v ng k Loi C Hnh 6.22. Cc kiu phn t dch v M rng a ch b gi Dch v ny c a thm vo chuyn b tr cho a ch OSINSAP (im tip cn dch v mng), a ch ny cn nhiu v tr hn trng a ch DTE b gi.Khoa in T in T 2-K3

Cc byte

0 1 2

K Thut Truyn S Liu


Byte u tin sau trng m dch v l chiu di ca trng tham s (ch s byte). N nhiu hn s lng byte cn gi phn m rng ca a ch. Byte u tin ca trng tham s c chia thnh hai vng. Cc bit t 1 ti 6 ch th s lng ch s phn m rng a ch (mi ch s i hi 4 bit). Bit 7 v 8 m ho kiu s dng phn m rng a ch nh sau: Bit 8 Bit 7 S dng m rng a ch b gi 0 0 chuyn i ton b a ch OSINSAP b gi 0 1 chuyn i mt phn a ch OSINSAP b gi 1 0 Ch th khng chuyn i phn m rng a ch OSI b gi 1 1 D tr S lng cc i cc ch s cho php l gia 32 v 40 tu thuc vo iu lin quan ti nh chiu di hp l ti a ca a ch OSINSAP (CCITT v ISO bt ng). Ch s a ch m rng c ghp vo trng tham s ging nh cc a ch DTE.

M rng a ch ch gi iu ny ging ht cu trc dch v trc, loi tr ton b cc t "b gi" c chuyn thnh "ch gi". u tnh cc v chn nhanh Dch v ny c mt trng tham s 1 byte. Cc bit trng tham s i hi tu theo vic s dng dch v. Chng c xc nh nh sau: Th t bit Dch v yu cu 1 o tnh cc. Nu mng ang dng tnh cc cho ch gi (tc PSS), khi o tnh cc cc c tnh cho DXE b gi thay v cho DXE ch gi. 7 Hn ch tr li. Ch tr li khi bit ny c thit lp xo cuc gi. 8 Chn nhanh. N xc nh trng s liu m rng ca thu bao ch gi v trng s liu ca thu bao xa c hiu lc s dng. Tt c cc bit khc trong byte ny cn lp "O" v chng c th c phn phi cho cc dch v mi trong tng lai ca nghi thc ny. Thu xp kch c gi Byte u ca trng tham s ny l kch c ti a ca cc gi s liu t DXE b gi, cn byte th 2 l kch c ti a ca cc gi s liu t DXE ch gi. Cc tr s c th c ca trng ny l:Khoa in T in T 2-K3

K Thut Truyn S Liu


00000100 = 16 bytes 00000101 = 32 bytes 00000110 = 64 bytes 00000111 = 128 bytes 00001000 = 256 bytes 00001001 = 512 bytes 00001010 = 1024 bytes 00001011 = 2048 bytes 00001100 = 4096 bytes Nu mt cuc u ni c tham s ny trng dch v ca n th DXE ch gi yu cu s dng kch c gi ti a c ch ra. DXE b gi c th xem xt cc gi tr ny nh chuyn v cc phn t dch v theo gi u ni/tip nhn cuc gi (nu cuc gi c th c tip nhn). Nu n chp nhn cc gi tr yu cu th n s chuyn v cc gi tr ging th. Nu khng th n c th thu xp mt gi tr gim ti mt kch c m n c th qun l c. Lu l n khng th tng gi tr ny loi tr trng hp c bit. Tt c cc khon b sung cn c kh nng tr gip cho mt kch c gi 128 bytes, ch yu ch c th tr gip cho kch c gi ny. Nu yu cu to lp kch c gi cc i t hn 128 bytes th DXE b gi c php tr li theo gi tr dnh cho 128 bytes. Nu gi u ni/tip nhn gi khng bao hm mt p ng cho cuc th gi v s tho thun kch c gi, th iu ny hm l DXE b gi c th chp nhn kch c gi theo yu cu. Mc d dch v ny cho php cc kch c gi khc nhau c th thu xp theo mi hng nhng thng thng th hai kch c gi ny l nh nhau. Kch c ca s Tham s ny rt ging tham s trc. Byte u ca trng tham s ny l kch c ca s cho gi s liu t DXE b gi. Byte th hai l kch c ca s cc gi s liu t DXE ch gi. Khi s dng h thng a ch dy thng thng th kch c ca s c th t 1 ti 7. Cn i vi h thng a ch dy m rng th kch c ca s c th trong phm vi t 1 n 127. H thng tho thun i vi tham s ny cng nhiu nh i vi kch c gi. Ngoi ra DXE b gi ch c th chp thun gi tr nh th hoc t hn loi tr trng hp khi kch c ca s l 1 yu cu. Tt c cc iu b sung cn chp thun cho kch c ca s l 2 nhng khng cn thit cho 1. Giao thc ny cho php DXE b gi chuyn v gi tr 2 tr li cho s yu cu c gi tr 1 nu n khng chp nhn kch c ca s .

Khoa in T

in T 2-K3

K Thut Truyn S Liu


Tho thun loi tip thng ti thiu iu ny cho php DXE ch gi yu cu mt mc tip thng no cho cuc gi. Trng hp tham s l mt byte n, n m ho loi tip thng (tim nng khc nhau) cho mi hng. Tr chuyn tip u cui Khon ny cho php cc DXE thit lp tr cho gi tin t u ti cui cuc gi. Byte u ca trng tham s ny ch th s byte nh sau: n c th l 2; 4 hoc 6. Mt gi yu cu gi vo/yu cu gi c th c tt c 6 byte, cn gi u ni/tip nhn gi c th ch c 2 byte. Hai byte u l tr tch gp gia hai DXE. Hai byte tip theo cha tr chuyn tip theo yu cu. Hai byte cui cng l tr u - cui ti a c th chp nhn c. Ton b cc gi tr ny c biu th theo ms. Tho thun xc tin s liu Dch v ny c th c s dng quyt nh dch v xc tin s liu c hiu lc khng. S liu cn xc tin c ti i cc gi ngt, v vy n quyt nh s liu cc gi ngt c th c dng hay khng. Trng tham s ny c th phc tp thm. Bit 1 l 0 nu dch v khng c s dng, hoc bit 1 l 1 nu s dng dch v ny. Cc bit khc c th phn phi cho cc dch v khc trong tng lai theo c tnh k thut nu ra. Hin ti chng cn c m ho ton s 0. 2.5. HOT NG CA GIAO THC X.25 X.25 hot ng da trn c s knh c nh PVC v knh o chuyn mch SVC. PVC thay th chc nng cho knh lin kt im-im c nh gia cc thit b u cui. S dng loi knh ny, giao din c hiu qu hn nh s lin kt c m bo v khng b tr cuc gi. SVC s dng ti a s mm do linh hot ca chuyn mch gi trong thc t. Hot ng ca X.25 theo cc giai on: giai on thit lp knh o, giai on trao i thng tin v giai on phng knh o. Ngay sau khi thit lp knh o, mt s thng bo tm tt ca cu trc gi tin s c node ngun gi n node ch. Nu chp nhn c, node ch s hin th v thng bo li cho node ngun. ng truyn song hng c thit lp. Giai on trao i d liu: node ngun gi khung thng tin, node ch s tin hnh kim tra tnh nng hp l ca khung thng qua cc bit FCS. Nu khng hp l th loi b khung v gi thng bo li cho

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in T 2-K3

K Thut Truyn S Liu


node ngun bit v yu cu truyn li. Nu khung hp l th node ny tip tc cc th tc truyn gi khung ti node tip theo trong mng, ng thi thng bo li cho cc node ngun bit l nhn c thng tin. Node ngun sau khi nhn c thng bo m t node ch, tip tc gi gi tin tip theo Sau khi kt thc, knh o s c gii phng

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do an giao thuc x.25

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