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Analyss of MetaRng: a real-tme protocol for Metropoltan Area etwork Marco Cont Lorenzo Donatello 2 Marco Furn 2 Techncal Report UBLCS-98-5 May 998 Abstract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onsglo azonale delle Rcerche, Isttuto CUCE, Va S Mara 36, 5626 Psa, Italy Department of Computer Scence, Mura Anteo Zambon 7, 4027 Bologna, Italy
Introducton Introducton MetaRng ([OFEK9], [OFEK94]) s a MAC (Meda Access Control) protocol developed at IBM TJ Watson Research Center Desgner s man motvaton was to ncrease throughput of a rng-based local area network MetaRng s able to reach a speed up to Gb/s and t s compatble wth the emergng standard ATM [PRYC9] Snce recent year have seen a great usage of synchronous nformaton, also known as real-tme nformaton (nformaton wth tme constrants, namely nformaton that must be sent, by the communcaton protocol, whtn specfed tme wndows), MetaRng s able to handle two types of traffc: synchronous (or real-tme) and asynchronous (generc traffc wth no tme constrants) To handle these type of traffc, MetaRng uses nnovatng characterstcs, such as: concurrent access, spatal reuse polcy, control sgnals that crculate n opposte way wth respect to the data they control These characterstcs should yeld MetaRng a protocol able to effcently handle real-tme traffc n a metropoltan envronment In ths paper we present a deeply nvestgaton about the performance reached by the protocol, when t handles real-tme traffc We hghlght ad we solve some problems of MetaRng The paper s organzed as follow In secton 2 we present the orgnal verson of MetaRng In secton 3 we show an analyss of MetaRng and we hghlght some problems In secton 4 we present propertes of MetaRng obtaned by our analyss and we modfed the orgnal verson of MetaRng to solve problem showed n secton 3 In secton 5 we present propertes of the modfed verson of MetaRng and we gve the gudelnes for developng a synchronous bandwdth allocaton scheme Conclusons are drawn n secton 5 2 MetaRng Protocol MetaRng ([OFEK9], [OFEK94]) s a MAC (Meda Access Control) protocol for LA and MA, wth a speed up to Ggabt/s The network s based on a dual fber-optc rng topology Specfcally (Fgure 2) the network conssts of two hgh-speed undrectonal rngs carryng nformaton n opposte drectons Transmsson on a rng s undrectonal The access to a rng, let us assume rng A, s managed through control nformaton travellng on the opposte rng (e rng B) MetaRng can operate under two basc access control modes: buffer nserton mode for varable sze packets and slotted mode for packets wth a fxed sze The man motvaton for developng MetaRng s to ntegrate tme-constraned applcatons and classcal data applcatons n a hgh-speed LA envronment, to utlze the network resources at the maxmum extent 5LQJ$ $GDWDDQG%VLJQDOFRQWURO 5LQJ% %GDWDDQG$VLJQDOFRQWURO Fgure 2 Dual rng topology UBLCS-98-5 2
2 MetaRng Protocol The ntegraton of traffcs wth dfferent Qualty of Servce (QoS) constrants s acheved by provdng a connecton-orented and connectonless servce to the synchronous (real-tme) and asynchronous (no realtme) traffc, respectvely Specfcally, resources are reserved (f avalable) for a real-tme applcaton n the connecton set-up phase Furthermore, a set of protocol mechansms has been desgned whch ensure that non-reserved asynchronous traffc wll not prevent the synchronous-traffc access An effcent utlzaton of the network resources s acheved va the spatal reuse and the concurrent access to the network Spatal reuse means that packets are removed from the network at ther destnaton A concurrent access s obtaned by enablng statons to transmt n all the empty slots they observe Each staton has two queues: one for the synchronous and one for the asynchronous traffc Packets from the asynchronous queue are transmtted only f the synchronous queue s empty Whenever a staton observes an empty slot, t always can transmt the synchronous traffc Before transmttng the asynchronous packets a staton must collect an authorzaton Specfcally, asynchronous transmssons on a rng A are authorzed by a control sgnal, called SAT (from SATsfed), travellng on the rng B When a staton receves the SAT sgnal n rng B, t performs dfferent actons dependng on ts status A staton, when t receves the SAT on rng B, can be n the satsfed state or not-satsfed state A staton s sad to be satsfed f between two vsts of the SAT sgnal the staton has transmtted, on the rng A, at least l packets or ts output (asynchronous) buffer s empty A staton that s satsfed, when receves the SAT, t forwards the SAT sgnal up-stream wthout any delay On the other hand, a not-satsfed staton wll hold the SAT untl t s satsfed, and only then t wll forward the SAT sgnal up-stream After a staton forwards a SAT, t can send up to k (k l) addtonal asynchronous packets, before recevng and forwardng agan the SAT sgnal SAT and send packets algorthms are shown n Fgure 2 and n Fgure 23 MetaRng has a mechansm for ntegratng two types of traffc over the full-duplex rng: synchronous or real-tme traffc that requres connecton or reservatons set-up, a gven bandwdth and bounded delay, and asynchronous traffc wth no real-tme constrants that can use the remander of the bandwdth n a far manner The mechansm s based on two control messages SAT and ASYC-E whch crculate n the opposte drecton to the data traffc they regulate The ASYC-E (ASYChronous Eable) s used for enablng the ntegraton of the asynchronous traffc, and the SAT s used for ensurng farness of the asynchronous traffc A complete descrpton of MetaRng can be found n [OFEK9], [OFEK94], [CIDO93] R 6$7RU 7LPHRXW 6L R O SDFHWVVHQW R 2XWSXW%XIIHUHPSW\ 6L )RUZDUG6$7 SDFHWVVHQW Fgure 22 Forward SAT algorthm UBLCS-98-5 3
2 MetaRng Protocol 6L 2XWSXW%XIIHU HPSW\ R R SDFHWVVHQW 6L 6HQGRQHSDFHW LQFSDFHWVVHQW Fgure 23 Send packets algorthmanalyss of MetaRng In ths secton we study the MetaRng performance; more precsely our studes are focused on the performance reached when MetaRng handles real-tme traffc In partcular we analyze f MetaRng s able to satsfy the deadlnes of the traffc t handles It s known that a frst protocol requrement, n a real-tme area, s the presence of a known upper bound to the network access tme (e the transfer delay must have a known upper bound: determnstc or statstc [FERR90]) Snce MetaRng's desgners state that MetaRng provdes ths upper bound, t mples that the protocol can be used, wthout problems, n a Real-Tme envronment Unfortunately ths mplcaton s not true; n fact n ths secton we show, through a performance analyss, that t exsts at least one scenaro n whch the synchronous traffc watng tme exceeds the desgners' upper bound In [OFEK9] desgners state: Theorem For a gven node, let ρ denote the amount of synchronous bandwdth reserved (through node ) by the other network nodes, and let ε the synchronous bandwdth reserved by node, then f ρ+ε<, the watng tme of node synchronous traffc s less or equal to T max, where: (2) T max Thres + ( r + 2) Trng = ρ Trng s the rotaton tme, Thres ndcates a threshold beyond that an enqueued packet must be sent as soon as possble, r s the number of round that the ASYC-E makes n a partcular state A complete descrpton and the proof on ths theorem can be found n [OFEK9] In the followng, we wll show a scenaro n whch, even-though, the condtons of theorem are satsfed, the synchronous-traffc watng tme at node exceeds T max The scenaro we consder s a slotted rng wth S slots and statons Let P the nterarrval tme (e the perod) of the staton real-tme traffc Furthermore we assume that: the deadlne D of the staton packets concded wth the traffc perod (e D =P ); 2 each real-tme message can be transmtted n a slot Hence, accordng to these hypotheses, the synchronous bandwdth reserved by a staton s ρ = /P UBLCS-98-5 4
2 MetaRng Protocol Accordng to theorem, let ε=/p, ρ = = P D = P Tmax (22), ρ + ε = + ε < = P, then f the followng condtons hold staton synchronous-packets should never mss ther deadlnes Ths s not true For example, let us consder a rng wth =00 statons, S=0 slots n the rng, Thres=0 slots, r=0 and suppose that statons generate synchronous messages accordng to the followng laws: (23) P = to + S + 75 ℵ P = to + 200 ℵ, 2 00 In ths confguraton, f we consder the staton (e ), accordng to (2)-(23) we have ρ = 30 = 0 495, ε = 0 03, and the upper bound for ts synchronous traffc s T max = < 59 5 = 0505 2 P P slots T max s less than the staton deadlne, and hence, there should be no problem to handle the staton traffc n ths MetaRng confguraton The example presented n Fgure 2 shows that, n some case, the staton deadlne s mssed In ths example all the network nodes are transmttng to a gateway staton (staton G) whch does not transmt any packet Furthermore let us assume that slots crculate clockwse Therefore statons form to 2 observe the empty slots before staton As shown n Fgure 22, at tme t 0, a synchronous message s queued for transmsson at the staton 2, 3,, When a message arrves at staton, at tme t 0 +0, all slots n the network are used by other statons to transmt ther messages Up to tme t 0 +99 all the slots observed by staton are busy and so ts transmsson cannot occur before ths tme To satsfy ts deadlne, the staton traffc must be transmtted before t 0 +S+75=t 0 +85 slots, and ths means that, n the scenaro presented n Fgure 2 and Fgure 22 the staton traffc deadlne s mssed Ths means that MetaRng can t be used to handle real-tme traffc In next secton we re modfyng the orgnal verson of MetaRng to solve these real-tme problems L L Control Sgnals GDWD * Fgure 2 Scenaro studed: staton G s the gateway UBLCS-98-5 5
3 MetaRng propertes * 0HVVDJHJHQHUDWLRQ 'HDGOLQHPHVVDJH * * * * W W6 W6 W HQGXSVWUDPWUDQVPLVVLRQV Fgure 22 Deadlne of message G s not meet 3 MetaRng propertes In ths secton we present some propertes of MetaRng, obtaned by a deeply nvestgaton of the protocol In partcular we are gong to solve problems hghlghted n the prevous secton The soluton to these problems allows MetaRng to handle real-tme traffc The study of MetaRng has shown that one control sgnal (SAT) has a smlar functon to the one of the FDDI token It s known that FDDI [ASI87] can handle real-tme transmssons because the token provdes an upper bound to the network access tme and each tme a staton receves the token t can always transmt a fxed quota of synchronous traffc (see eg [AGRA94], [MALC94]) MetaRng, lkewse, can handle the real-tme traffc transmsson, only f the control sgnal has a known upper bound to the network access tme Hence the goal s to show the presence of an upper bound to the control sgnal round trp tme That s why we nvestgate f the SAT round trp has a tme upper bound; f so the sgnal can control the real-tme traffc yeldng MetaRng to be able to support dstrbuted real-tme applcatons In ths way we ntroduce a change wth respect to the orgnal defnton of the protocol In fact we are tryng to control the real-tme traffc by the SAT sgnal control In the orgnal verson ths sgnal only controls the asynchronous traffc Furthermore we have to note that n MetaRng the SAT control sgnal crculates n opposte way wth respect to the data t controls, whle n FDDI data and control sgnal (the token) have the same drecton However SAT crculaton doesn t affect protocol archtecture and so t s possble to mplement both solutons To understand whch soluton can be the best one (e has a lower value for the upper bound to the network access tme), we analyze both solutons Intally we have consdered a MetaRng approach and then an FDDI approach A comparson between these approaches shows that wth FDDI approach s possble to reach a lower value of the upper bound to the SAT rotaton tme 6
3 MetaRng propertes 3 SAT goes n opposte data crculaton Ths secton shows results obtaned when the SAT control sgnal crculates n opposte way wth respect to the data t controls Partcularly we present a property of MetaRng that show the presence of the upper bound to the SAT rotaton tme; the proof of ths property, together wth other propertes, can be found n [FURI95] and n appendx A of ths paper In the followng we consder a slotted rng wth statons and S slots (Fgure 3) L L 6$7 GDWD L L Fgure 3 etwork scenaro Before showng the upper bound to the SAT rotaton tme, we need to ntroduce two defntons: Prmary delay: s the delay caused to a staton by a slot that has never caused delay to others statons In ths delay we also nclude the slot transmtted whle a staton holds the SAT Secondary delay: s the delay caused to a staton, by a slot that has already caused delay to some others staton Lemma 3 When a staton receves the SAT sgnal, the maxmum secondary delay s less or equal to S δ, ( ), where prec() denotes the ndex of the last staton who has held the SAT If, n the last round, no staton has held the SAT, we have prec()= The value δ, prec( ) denotes the tme t takes the SAT to move from the staton prec() to the staton In the worst case prec()= and so δ, prec ( ) = S Theorem 3 prec Let SAT_TIME be the tme elapsed between two consecutve vsts of the SAT sgnal at the same staton SAT_TIME has an upper bound and the followng holds: (3) SAT _ TIME S + ( S δ + k + l ) + ( S δ + k + l ) + 2 k, prec( ), prec( ) Τ where T s the set contanng all the ndex statons that hold the SAT The ndex s gven to any staton n accordng to the SAT rotaton Τ UBLCS-98-5 7
3 MetaRng propertes Proposton 3 Equaton (3) can be smplfed f we consder a scenaro n whch all the statons can send, n the rng, the same quota of traffc, such as: f S δ, k l, l =l and k =k for each staton and each staton, then the upper bound s equal to: (32) ( S + k + l) The upper bound obtaned n Theorem 3 shows that s possble to have an upper bound to the network access tme, f the rng traffc s controlled by the SAT control sgnal In ths way t s possble to mplement a synchronous bandwdth allocaton scheme that uses SAT control sgnal to handle real-tme traffc In fact at every SAT round, each staton can send, wth certanty, l packets If we gve a prorty to the synchronous traffc, these l packets can be seen as a synchronous quota reserved to staton Hence there s a change to the orgnal verson of MetaRng; n that verson the SAT control sgnal only controls the asynchronous traffc, whle n the modfed verson the SAT s also used to control the synchronous traffc The modfed verson solves problems hghlghted n secton 3 ote that, n ths new verson, the SAT controls both synchronous and asynchronous traffc, and so there s no need of havng a second sgnal (ASYC-E) lke n the orgnal verson of the protocol 3 Studes of real scenaro In ths secton we present a study of a real-scenaro (see appendx B for more detals) Ths study confrms the presence of the upper bound descrbed by equaton (3) In fact t was proved (n a rng wth statons and S slots) that the follow (T max represents the maxmum synchronous traffc watng tme) holds: S Tmax = S + ( ) l + ( ) ( ) + k 2 Partcularly, f >> we have: T S + ( ) l + ( 2) S + ( ) k T ( S + l + k), max that concde wth equaton (32) Moreover, t s easy to prove that the followng proposton holds Proposton 32 The maxmum tme elapsed between two consecutve vsts of the SAT at the same staton belong to Tmax, SAT_ MAX range Analytcally: SAT_ TIME T, SAT_ MAX where: [ ) [ max ) S Tmax = S + ( ) l + ( ) ( ) + k 2 max UBLCS-98-5 8
3 MetaRng propertes SAT_ MAX = S + ( S δ + k + l ) + ( S δ + k + l ) + 2 k, prec( ), prec( ) Τ Τ The proof can be found n [FURI95] and n the Appendx B of ths paper 32 SAT goes n same data drecton In ths secton we analyze MetaRng wth the SAT control sgnal that crculates n the same way wth respect to the data t controls (lke the FDDI token) Our goal s fndng a new upper bound wth a lower value than the one of equaton (3) In fact, ths equaton depends from the number (S) of the slots present n the rng and, n a heavy manner, from the number () of the rng statons (wth synchronous traffc) In the remanng of ths secton we only present the propertes of MetaRng that show the exstence of an upper bound, called SAT_TIME, to the tme elapsed between two consecutve vsts of the SAT at the same staton The proofs of these propertes, together wth other propertes of MetaRng, can be found n [FURI95] and n the Appendx C of ths paper In the followng we consder a slotted rng wth statons and S slots (Fgure 32) L L 6$7DQGGDWD L L Fgure 32 etwork scenaro Theorem 32 Let SAT_TIME be the tme elapsed between two consecutve vsts of the SAT at the same staton SAT_TIME has an upper bound and the followng holds: (33) SAT _ TIME S + l + l + 2 k k, The ndex s gven to any staton n accordng to the SAT rotaton UBLCS-98-5 9
3 MetaRng propertes Proposton 33 If all the rng statons have the parameters l and k wth the same value (l =l and k =k for each staton and each staton ) then the maxmum tme elapsed between two consecutve vsts of the SAT at the same staton has an upper bound equal to: (34) S + k ( 2 3) + 2 l As n the prevous secton, the upper bound obtaned by Theorem 32 shows that t s possble to have an upper bound to the network access tme f the rng traffc s controlled by the SAT sgnal The reason why ths second analyss was made, s for fndng a new upper bound whose value was lower than the one showed n the prevous secton To do that we changed the SAT rotaton way A comparson between these upper bound wll be made n next secton 33 Comparson between the two SAT drectons In the prevous sectons we defned the equatons of the upper bound to the SAT rotaton tme (equatons (3) and (33)) ow, to compare these equatons, we analyze a rng network wth 50 slots (S) Each staton n the rng have the same parameters (l=50 and k=50) These values let us to use equatons (32) and (34) Intally, case A, the SAT goes n opposte drecton wth respect to the data t controls, and n a second analyss, case B, we change the SAT crculaton way (same way of the data t controls) As we can note n Fgure 33 the upper bound has a lower value when the SAT crculates wth the same drecton of the data t controls, than when the SAT crculates n the opposte drecton of the data t controls Ths mean that an other nnovatng characterstc of MetaRng can be dropped In fact t s recommended that data and SAT crculate n the same drecton (n the orgnal verson, the SAT goes n opposte way wth respect to the data t controls) 6$7URWDWLRQWLPH 6$7B7,0( WKRXVDQGRIVORWV &DVH$ &DVH% 6WDWLRQV Fgure 33 Maxmum tme elapsed between two consecutve SAT arrvals at the same staton UBLCS-98-5 0
4 Synchronous Bandwdth Allocaton Scheme 4 Synchronous Bandwdth Allocaton Scheme The obectve of a bandwdth allocaton scheme s to assgn, at each staton and n a partcular tme wndow, a quota of the avalable bandwdth for the staton traffc transmssons For ths reason the defnton of the allocaton scheme s very mportant for the real-tme protocol [CHE92] In fact, a wrong synchronous allocaton could lead a staton unable to satsfy the deadlne of ts own synchronous traffc In prevous sectons we ve shown that the orgnal verson of MetaRng s not able to handle real-tme traffc However, wth the upper bound (Theorem 3 or Theorem 32) showed n last secton, s possble to defne a synchronous allocaton scheme that allows to the modfed verson of MetaRng to correctly manage the real-tme traffc As we re gong to show, the upper bound s not suffcent; that s why we re showng some new propertes (proved n Appendx D) of the modfed verson of MetaRng Snce results obtaned n the prevous secton, we consder data and control sgnal wth the same crculaton way These propertes, together wth the upper bound (Theorem 32) allows the development of a synchronous bandwdth allocaton scheme Snce the am of ths paper s not the defnton of that scheme, we only present some allocaton scheme present n the lterature, and we show gudelnes to develop one of those, n the modfed verson of MetaRng 4 Allocaton schemes In ths secton we show some allocaton scheme present n the lterature Before showng the schemes, t s mportant to note that n [MALC94] the real-tme traffc present n all the staton can be characterzed as follow: C = length message (n slots); D = deadlne message (n slots); P = nterarrval message (n slots) Moreover we can suppose, wthout loss of generalty, that the deadlne of every message s equal to the nterarrval tme (e D =P ) These notatons allow us to better understand the followng scheme 4 Full Length Wth ths scheme, the synchronous bandwdth allocated to a staton s equal to ts total tme requred for transmttng ts synchronous messages Ths scheme attempts to transmt a synchronous message n a sngle turn rather than spttng t nto chunks and dstrbutng ts transmsson over ts perod P Although the synchronous bandwdth allocated s suffcent, the worst case achevable utlzaton s zero [AGRA94]
4 Synchronous Bandwdth Allocaton Scheme 42 Proportonal Wth ths scheme, the synchronous bandwdth allocated to a staton s proportonal to the rato of C and P Intutvely speakng, ths scheme dvdes the transmsson of ts message nto as many parts as the number of tmes the token s expected to arrve at staton wthn ts perod P The worst case achevable utlzaton of ths scheme can asymptotcally approach 0% [AGRA94] 43 Equal Porton In ths scheme the usable bandwdth s dvded equally among the n statons for allocatng ther synchronous capactes The worst case achevable utlzaton of ths scheme s approxmately 0% [AGRA94] 44 ormalzed Proportonal Wht ths scheme, the synchronous bandwdth s allocated accordng to the normalzed load of the synchronous message on a staton Ths scheme uses both local and global nformaton The worst case achevable utlzaton s equal to 33% of the avalable rng utlzaton [AGRA94] 45 MCA (Mnmum Capacty Allocaton) Ths scheme [CHE92a] always assgn the mnmum capacty requested by statons It works under the assumpton D =P In [CHE92b] t s showed that the one obtaned by the normalzed proportonal scheme 46 EMCA (Enhanced MCA) Ths scheme uses an upper bound between n consecutve arrvals of the token at the same staton In [ZHA94] EMCA s defned as optmum allocaton scheme All these allocaton schemes don t use exclusvely the upper bound to the token rotaton tme, but they also use other parameters that strctly depend by the protocol they are embedded to In fact, all these protocols have to satsfed three constrants, called protocol, deadlne and buffer constrant: Protocol constrant: t assures that the total synchronous bandwdth allocated s not greater that the synchronous bandwdth avalable; Deadlne constrant: t assures that each synchronous message s sent wthn ts deadlne; Buffer constrant: t assures that the buffer dmenson n each staton can contan the maxmum number of synchronous messages that can be queued; n other words t assures that synchronous messages wll not lost by overflow To satsfy these constrants, we need to know, at least, the synchronous bandwdth avalable, the network access tme and the buffer sze UBLCS-98-5 2
4 Synchronous Bandwdth Allocaton Scheme Furthermore, t s to note that an allocaton scheme presented (EMCA) needs one more parameter; n fact t uses an upper bound between n consecutve token arrvals at the same staton In fact, usng n token rotatons, ths upper bound allows a better allocaton resources [CHE92] Wth the followng propertes we gve all the necessary parameters for the synchronous bandwdth allocaton schemes presented above Havng these parameters t wll be easy to mplement one of the prevous allocaton schemes In fact our ntent s to gve gudelnes to mplement t n the modfed verson of MetaRng A frst consderaton lead us to consder only the frst two requrements (protocol and deadlne), because these are fundamental to correctly handle real-tme traffc, whle the last one (buffer) s not consder crtcal (ths problem can be solved wth an adequate szng resource) Protocol constrant tells us that the sum of the synchronous bandwdth allocated to each node can not be greater than SAT_TIME (that s the amount of bandwdth avalable between two consecutve SAT arrvals) To defne deadlne constrant we need to know the tme elapsed between message arrval and message transmsson In fact, whle n the token passng protocols, the maxmum token rotaton tme equals the maxmum tme that a message can spend n the output queue before beng transmtted, n MetaRng ths s not true; the SAT arrval doesn t mean transmsson, but only permsson for transmsson The packets wth permsson wll be transmtted as soon as possble (when the staton sees empty slots) Moreover the orgnal verson of MetaRng doesn t gve any guarantee to the synchronous traffc: the only guarantee t gves to each staton, s the transmsson of l packets (even only asynchronous packets) at every SAT rotaton That s why we must gve prorty to the synchronous traffc; n ths way, the synchronous traffc can be transmtted before the asynchronous one In next secton, we present some new propertes of the modfed verson of MetaRng These propertes allow to establsh how satsfed the constrants to correctly handle real-tme traffc Constrants and propertes descrbe here allow the development of a synchronous bandwdth allocaton scheme, that yeld the modfed verson of MetaRng as a protocol able to correctly handle the real-tme traffc 42 Propertes Followng propertes establsh the upper bound to the tme elapsed between n consecutve vsts of the SAT at the same staton These propertes are fundamental for the mplementaton of the EMCA synchronous bandwdth allocaton scheme The proof of these propertes can be found n Appendx D of ths paper Theorem 4 Let SAT_TIME [n] be the tme elapsed between n consecutve vsts of the SAT at the same staton SAT_TIME [n] has an upper bound and the followng holds: SAT_ TIME [ n] n S + l + l + ( n + ) k UBLCS-98-5 3
4 Synchronous Bandwdth Allocaton Scheme Proposton 4 If all the rng statons have the parameters l and k wth the same value (l =l and k =k for each staton and each staton ) then the maxmum tme elapsed between n consecutve vsts of the SAT at the same staton has an upper bound equal to: ( ) n S + 2 l + n + k As we know, the prevous propertes are necessary, but not suffcent for the respect of the real-tme traffc deadlne For ths reason we present the followng propertes (proved n Appendx D) that establsh the upper bound to the network access tme Ths upper bound s used to verfy f the protocol can meet the real-tme traffc deadlne (deadlne greater or equal to the network access tme) Wth the followng propertes t wll be avalable all the parameters needed to mplement one of the allocaton schemes presented n secton 5 In ths way the modfed verson of MetaRng, can be ntegrated wth one allocaton scheme, and hence t can be effcently used n whchever real-tme envronments Lemma 4 Let us consder the frst synchronous packet n the output queue of staton and let T Wat_per be the tme that ths packet has to wat before recevng the permsson of transmsson The followng holds: Twat _ per = S + l + 2 k k Proposton 42 Let us suppose that each synchronous message of staton s dvded nto l packets Let T Wat_per(n) be the tme elapsed n the synchronous queue by the n-th message before recevng the permsson of transmsson The followng holds: [ ] TWat _ per ( n) SAT_ TIME n Lemma 42 Let us consder staton and let T Wat_tx be the tme elapsed between the permsson and the complete transmsson of the frst group of l synchronous packets n the queue The followng holds: TWat _ tx S + l + l + k + k 2 UBLCS-98-5 4
5 Conclusons Proposton 43 If the frst group of l synchronous packets n the queue of staton have wated for a tme equal to T wat _ per, before havng the permsson of transmsson, the maxmum tme elapsed between the permsson and the complete transmssons s equal to l + 2 k Lemma 43 Let us consder the frst synchronous message n the output queue of staton (dvded nto l packets) and let T Wat be the tme elapsed between ts arrval n the frst queue poston and ts complete transmsson The followng holds: T S + l + l k + 2 k Wat Theorem 42 Let us suppose that each synchronous message of staton s dvded nto l packets Let T Wat(n) be the tme elapsed between ts arrval n the queue and ts complete transmsson The followng holds: T ( n) ( n + 2) S + l + l + ( n + 3) k Wat 5 Conclusons In ths work we ve presented a study of a MAC protocol, named MetaRng, developed for supportng the real-tme traffc transmssons n a dstrbuted envronment MetaRng was developed to overcome the neffcency problems caused by usng the real-tme protocol, desgned for LA (lke FDDI), n a metropoltan area Our studes have shown that, despte the ntentons' desgners, MetaRng s not able to support the transmsson of the real-tme traffc n whchever scenaro In fact we've presented a scenaro n whch MetaRng doesn't meet all the deadlnes of the real-tme traffc present n the network; ndeed the synchronous traffc watng tme exceeds the desgners' upper bound to the network access tme For ths reason we ve deeply analyzed the protocol for fndng a soluton to ths problem Ths soluton doesn t requre new technology, because t uses only MetaRng technology Moreover we preserve the compatblty wth ATM standard, and we use some of the nnovatng characterstc presented by MetaRng, lke concurrent access and spatal reuse polcy Our studes have shown that the traffc can be controlled by only one control sgnal Ths sgnal can crculate both n the same and n opposte way wht respect to the data t controls However, for havng a lower value of the upper bound to the SAT rotaton tme, t s recommended that data and SAT sgnal crculate n the same drecton To handle real-tme traffc, an access control protocol must have a synchronous bandwdth allocaton scheme Snce the am of ths paper was the analyss of MetaRng and not the defnton of an allocaton 5
Appendx A scheme, we haven t deeply nvestgated ths ssue However, we have showed some bandwdth allocaton scheme present n the lterature To mplement one of these some new parameters are needed That s why we ve nvestgated the modfed verson of MetaRng, to fnd out propertes from whch those parameters are deducted These propertes are presented and they are fully proved In ths way t s easy to mplement a synchronous bandwdth allocaton scheme In concluson, propertes descrbed n ths paper allow the modfed verson of MetaRng to be used n any real-tme scenaro wth certanty that all the deadlnes wll be meet Appendx A Lemma A The maxmum number of busy slots that a staton can see, when t holds the SAT control sgnal, s equal to: S + k We consder the scenaro presented n Fgure A and we suppose that the SAT sgnal leaves the staton - and goes towards staton Staton - releases the SAT sgnal when t s satsfed, such as when t has transmtted a fxed quota of packets Moreover, after releasng the SAT, staton - can send up to k - packets before blockng tself All the others network staton ( { }, ), can have up to k packets to transmt The worst case s presented when, whle the SAT s gong to the staton from the staton -, all the slots n the rng are marked busy Ths stuaton can happen f the quota (l) that make the staton satsfed s greater than the number of the slot n the rng (S) When a staton receves the SAT, t sees up to S busy slots, and t sees k - slots whch contan packets sent by staton - Ths mean that we do not consder the quota l -, but only the number S In the worst case all the statons (from -2 to +) wll transmt ther own k packets Hence, staton sees ts frst empty slot after a maxmum number of slots equal to: k S + UBLCS-98-5 6
Appendx A L L 6$7 GDWD L L Fgure A etwork scenaro Proposton A A staton can hold the SAT sgnal for a maxmum tme equal to: S + l + k slots From Lemma A we know that the maxmum number of busy slots that a staton can see, whle t s k holdng the SAT sgnal, s equal to S + Staton leaves the SAT sgnal when t s satsfed, such as when t has transmtted l packets In concluson a staton can hold the SAT sgnal for a maxmum tme equal to S + l + k Proposton A2 If a staton holds the SAT, then the maxmum number of packets that a staton has transmtted n the prevous cycle s equal to l Staton holds the SAT sgnal f t s not satsfed, namely t has sent a number of packets less than l Staton holds the SAT untl the number of packets transmtted reached the quota l, and then t releases slots the SAT sgnal So the maxmum number of packets transmtted n the prevous cycle s equal to l In Fgure A2 we present the stuaton of a generc staton At tme a staton releases the SAT sgnal From tme b to tme c, the staton transmts a part of ts l quota At tme d the SAT come back to the staton The staton s not satsfed, and so t holds the SAT The SAT wll be released when the staton s satsfed (e when t sends a number of packets equal to l ); ths happen at tme e UBLCS-98-5 7
Appendx A XPEHURISDFHWVWREHVDWLVILHG O TXRWDRISDFHWVWREH SDFHWVVHQW VDWLVILHG D E F G H WLPH 5HOHDVHRIWKH6$7 6$7DUULYDO 5HOHDVHRIWKH6$7 Fgure A2 Transmssons when a staton holds the SAT Lemma A2 If a staton holds the SAT for a tme equal to S + l + k If a staton holds the SAT for S + l + k, then the SAT wll come back after S slots slots, t means that all the statons n the rng are satsfed because staton has transmtted l packets and all the others statons have sent ther k packets Hence any staton won t hold the SAT sgnal, and t wll come back to staton after S slots Lemma A3 Durng the tme elapsed between two consecutve arrvals of the SAT sgnal at staton, the number of packets that ths staton can send s not greater than k + l To be satsfed a staton must send l packets In the worst case ths packets are transmtted when the staton holds the SAT When a staton releases the SAT, t can send up to k new packets, and so when the SAT come back to the staton, the number of packets sent s not greater than k + l Lemma A4 Let SAT_TIME be the tme elapsed between two consecutve arrvals of the SAT sgnal n the same staton; ths tme has an upper bound and we have: UBLCS-98-5 8
Appendx A SAT _ TIME < S + l + k = Proposton A says that a staton can hold the SAT sgnal for a maxmum tme equal to S + l + k and so f we consder all the statons n the rng we have: SAT _ TIME S + l + k = k If a staton holds the SAT for a tme equal to S + (Lemma A) then all the statons present n the rng (wth the excepton of staton ) are satsfed (Lemma A2) and so t s mpossble for the SAT to be held, n the same round, n more than one staton, and so: SAT _ TIME < S + l + k = Lemma A5 The maxmum number of busy slots that a staton can observe from the start to the next arrval of SAT has an upper bound equal to: 2 k We consder a staton ; when the SAT leaves staton, t can send up to k packets before blockng tself However these slots don't have to be consdered because, by the spatal reuse, ths slots, for the staton, are empty All the others staton send ther k packets, so they use k At ths pont all the statons are blocked, but when the SAT transts, these statons can send others k packets slots Ths means that the maxmum number of busy slot that a staton can observe s gven by 2 k UBLCS-98-5 9
Appendx A Proposton A3 If all the rng statons have the parameters l and k wth the same value (l =l and k =k for each staton and each staton ) the maxmum number of busy slot that a staton can observe n equal to 2 k ( ) Trval and therefore omtted Lemma 3 When a staton receves the SAT sgnal, the maxmum secondary delay s less or equal to S δ, prec ( ), where prec() denotes the ndex of the last staton who has held the SAT If, n the last round, no staton has held the SAT, we have prec()= The value δ, prec( ) denotes the tme t takes the SAT to move from the staton prec() to the staton In the worst case prec()= and so δ =, prec ( ) S If a staton receves the SAT at tme t o, the busy slots observed by staton at tme t can't have caused prmary delay n the others statons o + S δ, prec ( ) Lemma A6 If a staton doesn't hold the SAT, t doesn't ntroduce addtonal delay to the SAT propagaton tme Trval and therefore omtted Lemma A7 The maxmum delay that a staton can add to the SAT tme propagaton s less or equal to S + l + x, where x s the prmary delay experenced from staton It follows from the prevous lemma Lemma A8 Let t o denote the SAT releasng tme from staton, and let t the next SAT departure tme at the same staton In [t o, t ] the maxmum slots marked busy by statons are: ( k + l ) + ( k + l ) + 2 k, Τ Τ where T s the set whch contan the statons that hold the SAT The frst term belong to the staton : f at tme t the staton s satsfed, t doesn't hold the SAT sgnal and so n [t o, t ] t can send up to k packets, but, f t holds the SAT, ts transmssons can be k +l Lkewse, f a staton n [t o, t ] holds the SAT, t sends l packets before recevng the SAT and k after releasng the SAT UBLCS-98-5 20
Appendx B All the statons that don t hold the SAT gve the last term Theorem 3 Let SAT_TIME be the tme elapsed between two consecutve vsts of the SAT sgnal n the same staton SAT_TIME has an upper bound and the followng holds: SAT _ TIME S + ( S δ + k + l ) + ( S δ + k + l ) + 2 k, prec( ), prec( ) Τ Τ where T s the set whch contan all the ndex statons that hold the SAT The ndex s gven to any staton n accordng to the SAT rotaton The frst term s the SAT propagaton tme The term ( S δ prec ) Τ the prmary delay bound, ( ) s the secondary delay bound, and the term k + l + k + l + 2 k s Τ Τ Proposton 3 Equaton (3) can be smplfed f we consder a scenaro n whch all the statons can send, n the rng, the same quota of traffc, such as: f S δ, k l, l =l and k =k for each staton and each staton, then we have: SAT _ TIME ( S + k + l) Trval and therefore omtted Appendx B Let us consder a partcular scenaro: a slotted rng wth S crculatng slots and 4 statons equally spaced It s know that wth spatal reuse polcy, packets are removed from rng by ther destnatons However, usng a broadcast approach, destnatons matches sources; ths mean that each packet travels once all over the rng Intally rng contans no data and the SAT control sgnal s n staton S (Fgure B) UBLCS-98-5 2
Appendx B 6$7 2 6$7 GDWD 4 3 Fgure B Startup scenaro Furthermore we suppose that each staton behaves lke follow: t has no data to delver; 2 t generates real-tme traffc (n asymptotc manner) at tme t-ε f t s the SAT tme arrval Usng a worst case approach, we suppose that the satsfed quota l flls the rng (e l S); ths mean that staton S releases SAT (sendng t to staton S 2 ) when the rng s full of ts own packets (Fgure B2) SAT control sgnal and statons state are llustrated n Fgure B3 In the followng, we analyze, step-by-step, all the rng stuatons When staton S s satsfed, t sends, to staton S 2, the SAT, and t starts to transmt others k packets towards staton S 4 When SAT arrves at staton S 2, the frst of the k packets sent by S s arrved at staton S 4 ; ths packet S wll arrve at staton S 2 after a tme equal to: ( 2 ), where ndcates the number of statons Ths mean that staton S 2 sees the frst empty slot only after the k packets of staton S Analytcally, after S k + ( 2 ) slots It s to note that, we ve supposed that staton S 2 has generated ts packets ust before recevng the SAT sgnal 6$7 6SDFHWV 2 6$7 GDWD 4 3 Fgure B2 Rng s full of staton S packets UBLCS-98-5 22
Appendx B After seeng the frst empty slot, staton S 2 transmts l packets, then t sends the SAT to the staton S 3 and t starts to transmt hs new k packets towards staton S When the SAT arrves at staton S 3, ths staton starts to generate messages, and when S 3 receves the SAT, the frst of the k packets, sent by staton S 2 towards staton S, s at staton S Ths mean that staton S 3 S wll see ths packet after a tme equal to ( 2 ) tme unts, and the frst empty slot wll be seen after a S tme equal to: k + ( 2 ) slots Staton S 3 holds the SAT to transmt ts l packets, after that t sends the SAT to staton S 4 and t starts to transmt ts k packets toward staton S 2 Staton S 4 starts to generate packets ust a tme before recevng the SAT When the SAT arrves at staton S 4 the frst of the k packets, sent from staton S 3 toward staton S 2, s at staton S 2, and t wll arrve at S staton S 4 after a tme equal to: ( 2 ) Also n ths case, staton S 4 sees ts frst empty slot after a tme S equal to k + ( 2 ) Staton S 4 transmts ts l packets, and then t sends the SAT sgnal to staton S and t starts to transmt ts new k packets towards staton S 3 After a tme equal to the slot-dstance between staton S 4 and staton S the SAT sgnal wll come back to staton S Staton S s now satsfed and so t mmedately releases the SAT Ths sgnal wll be at staton S 2, after a tme equal to S, and so, tme elapsed between two consecutve arrvals of the SAT at the same staton S 2 s equal to: S Tmax = S + ( ) l + ( ) ( ) + k 2 Partcularly, f >> we have: T S + ( ) l + ( 2) S + ( ) k T ( S + l + k) max T max value concde wth the one showed n Proposton 3 Lookng at Fgure B3 t s easy to note that the worst case s realzed when there s no concurrent access to the rng and so each staton s able to transmt l+k packets Ths stuaton can be obtaned wth the supposton that f t s the SAT arrval tme, traffc s generated at tme t-ε, wth ε ( 0, T slot ), where T slot s the temporal slot length Ths reasonng lead us to the followng property Proposton 32 The maxmum tme elapsed between two consecutve vsts of the SAT at the same staton belong to Tmax, SAT_ MAX range Analytcally: SAT_ TIME T, SAT_ MAX where: [ ) [ max ) max UBLCS-98-5 23
Appendx B S Tmax = S + ( ) l + ( ) ( ) + k 2 SAT_ MAX = S + ( S δ + k + l ) + ( S δ + k + l ) + 2 k, prec( ), prec( ) From Theorem 3we know that: SAT _ TIME S + ( S δ + k + l ) + ( S δ + k + l ) + 2 k Τ Τ, prec( ), prec( ) and we know that SAT_ TIME T max, where Τ S Tmax = S + ( ) l + ( ) ( ) + k 2 SAT_ TIME T, SAT_ MAX Ths means that: [ ) max Τ 6$7SRVLWLRQ 6 7PD[ 6 6 6 6 7UDQVPLWWLQJVWDWLRQV 6 6 6 OSDFHWV OSDFHWV 6 OSDFHWV Fgure B3 Rng stuaton 7LPHLQVORW 6VORW UBLCS-98-5 24
Appendx C Appendx C Lemma C The maxmum number of busy slots that a staton can see, when t holds the SAT control sgnal, s equal to: k We consder the scenaro n Fgure C and we suppose that the SAT sgnal leaves the staton - and goes towards staton Staton - releases the SAT when t s satsfed such as when t has sent 0 x k packets Moreover, after releasng the SAT, the staton - can send up to k - packets before blockng tself Staton sees the SAT at the end of the x slots and so we don't consder the quota x All the other statons wth + 2 can have up to k packets to transmt, and so at the end of the k - slots there could be all the others k packets Therefore staton wll see the frst empty slot after a maxmum numbers of slots equal to: k slots L L 6$7DQGGDWD L L Fgure C etwork scenaro Proposton C A staton can hold the SAT sgnal for a maxmum tme equal to: l + k slots UBLCS-98-5 25
Appendx C From Lemma C we know that the maxmum number of busy slots that a staton can see, whle t's holdng the SAT, s equal to staton (e the l packets) k At ths value we have to add the quota of packets able to satsfy the Hence a staton can hold the SAT sgnal for a maxmum tme equal to l Proposton C2 If a staton holds the SAT, then the maxmum number of packets that a staton has transmtted n the prevous cycle s equal to l + k slots s analogous to the one presented n Proposton A2 Lemma C2 If a staton holds the SAT for a tme equal to l + k, then the SAT wll come back after S slots s analogous to the one presented n Lemma A2 Lemma C3 Durng the tme elapsed between two consecutve arrvals of the SAT sgnal at staton, the number of packets that ths staton can send s not greater than k + l s analogous to the one presented n Lemma A3 Lemma C4 Let SAT_TIME be the tme elapsed between two consecutve arrvals of the SAT sgnal at the same staton; ths tme has an upper bound and the followng holds: SAT _ TIME < l + k = UBLCS-98-5 26
Appendx C Proposton C says that a staton can hold the SAT sgnal for a maxmum tme equal to l so f we consder all the statons n the rng we have: SAT _ TIME l + k = If a staton holds the SAT for a tme equal to k + k, and (Lemma C) then all the statons present n the rng (wth the excepton of staton ) are satsfed (Lemma C2) and so t s mpossble for the SAT to be held, n the same round, n more than n one staton, and so: SAT _ TIME < l + k = Lemma C5 The maxmum number of busy slots that a staton can observe between the release to the next arrval of the SAT sgnal, has an upper bound equal to: 2 k k We consder a staton When the SAT leaves staton, t can send up to k packets before blockng tself However these slots don't have to be consdered because, by the spatal reuse, ths slots, for the staton, are empty All the other statons send ther k packets, so they use k At ths pont all the statons are blocked, but when the SAT transts, these statons can send others k packets Staton - sends ts own k packets after releasng the SAT; ths mean that we don t have to consder the term k - slots In concluson the maxmum number of busy slot that a staton can observe s gven by 2 k k Proposton C3 If all the statons present n the rng have the parameters l and k wth the same value (l =l and k =k for each staton and each staton ) then the maxmum number of busy slots that a staton can observe from the start to the next arrval of SAT, has an upper bound equal to k ( 2 3 ) UBLCS-98-5 27
Appendx C Theorem 32 Let SAT_TIME be the tme elapsed between two consecutve vsts of the SAT at the same staton SAT_TIME has an upper bound and the followng holds: SAT _ TIME S + l + l + 2 k k The ndex s gven to any staton n accordng to the SAT rotaton If a staton s not satsfed (e t was not able to transmt ts l packets), when t sees the SAT, t sezes the SAT The slots whch contan the packets of the staton can produce a delay only to the staton k (the staton that holds the SAT) If staton holds the SAT then the maxmum delay t adds to the SAT propagaton tme s gven by l +k So, f all the rng statons hold the SAT, the SAT crculaton has an upper bound and we have: SAT _ TIME S + ( l + k ) If a staton h doesn't hold the SAT, ths staton can send 2 k h packets and we have: SAT _ TIME * S + ( l + k ) + 2 kh h If SAT _ TIME * > SAT _ TIME means that the staton h (who ddn't hold the SAT) has ncreased the SAT propagaton tme; we have ths stuaton when k Let Τ = { } { h k = l } h > l h h h the set of statons that ncrease the SAT propagaton tme when they hold the SAT sgnal The orderng of the set T s gven n respect wth the SAT crculaton ote that the staton s always present n T; ths happens because f the SAT sgnal arrves to ths staton and ths staton s satsfed, the number of slots to be consdered s k ; for ths reason the T set always contan the ndex of the staton After ths consderatons the upper bound can be rewrte as: SAT _ TIME S + ( l + k ) + 2 k Τ Τ ow, we can do an other consderaton: when the SAT moves from the staton, whose ndex s the last of the T set, towards the staton, all the statons vsted by SAT don't ntroduce a delay of 2 k to the SAT propagaton tme, but they ntroduce only a delay of k slots At ths pont we have: SAT _ TIME S + ( l + k ) + 2 k k Τ Last( Τ) Τ where = Last( ) f = {,, } Τ Τ, The rght sde of the above equaton has an hgh value when the last term s very low; ths happen when Τ = then Last( Τ ) = (e ), and so the last term would assume ts hgher value - Otherwse f { } (sum of all the k quantty) leadng the mnmum value to the rght sde of the equatons UBLCS-98-5 28