Lecture 4: November 17, Part 1 Single Buffer Management

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1 Lecturer: Ad Rosén Algorthms for the anagement of Networs Fall Lecture 4: November 7, 2003 Scrbe: Guy Grebla Part Sngle Buffer anagement In the prevous lecture we taled about the Combned Input Output Queued (CIOQ) swtch, we dvded the swtch polcy nto two man parts:. Schedulng How we pass pacets from the swtch nputs to ts outputs. 2. Buffers anagement (When the sze of the buffers s fnte) acceptng/throwng pacets from the buffers. In ths part of the lecture we wll focus on the management of a sngle buffer, where the buffer has a fxed sze. Our goal s to maxmze the throughput of the swtch, that s, the number of pacets transmtted. In the more general case, each pacet has a weght/value, whch can stand for ts prorty (Dff Serv), and our goal s to maxmze the sum of weghts/values of pacets transmtted. 2 The odel We defne the model of a sngle buffer management n an OQ swtch as follows:. A buffer of sze B (pacets) 2. All pacets have fxed sze. 3. Every pacet p has a weght/value of vp ( ) 4. The buffer s FIFO (later we wll see ths s not a lmtaton, as every other buffer s algorthm can be smulated usng FIFO) 5. In each tme unt one pacet can be sent from the buffer. 6. Every pacet enterng the buffer can leave the buffer only when t s transmtted (non-preemptve). Our algorthm wll decde for each pacet, p, f p should be nserted to the buffer or not (dropped).

2 The goal of the algorthm s to maxmze the value of transmtted pacets, v( p ), or n other words to maxmze v( p). p transmtted 3 The Algorthm We assume that pvp, ( ) [, α ], and we dvde the range [, α ] to classes. Class contans pacets values n range[ e, e ). Defnton 3. Algorthm A s defned as follows:. Choose a random {,.., ln α }, s chosen wth unform dstrbuton. 2. From now on, a pacet p s accepted by the buffer only f the buffer has space left and v( p ) s n class. Defnton 3.2 A( σ ) - Sum of pacet values whch algorthm A transmtted after a seres of pacets σ s receved. OPT ( σ ) - Optmal number of pacets receved from class. OPT ( σ ) - Optmal weghts sum of pacets receved from class. Corollary 3.3 The Algorthm accepts an optmal number of pacets from class. Clam 3.4 The compettve rato of the Algorthm s e. Proof: A s a probablstc algorthm, let E( A( σ )) be the expectaton of A( σ ). Snce the weght of every pacet n class s at least e, and snce A devotes all ts tme to sendng pacets from class : EA e OPT () ( ( σ)) ( σ) = Corollary 3.5 The weght of every pacet n class s at most OPT ( σ) e OPT ( σ) e, so And from () and Corollary 3.5: 2

3 e OPT ( σ ) σ = = e EA ( ( σ)) e OPT( ) = OPT ( σ) OPT ( σ) ln α e = e 4 Smulatng an Arbtrary Polcy usng a FIFO ueue Clam 4. Gven an algorthm P, worng on a buffer of sze B, non-preemptve, and not necessarly a FIFO. There exsts an algorthm P ', worng on a buffer of sze B, nonpreemptve and FIFO, and whch for every seuence of pacets σ, P( σ) = P'( σ) Proof: We wll show that such algorthm, P ', exsts by defnng t. We smulate a run of P on σ n the bacground. Algorthm P ' : - If P accepts a pacet, P ' accepts t too. - When P transmts a pacet, P ' sends the pacet at the head of ts FIFO buffer. Corollary 4.2 Snce the buffer s non-preemptve, every pacet enterng the buffer s sent, therefore f P and P ' receve the same pacets, P( σ) = P'( σ). (even though the order of transmsson s not necessarly the same). Corollary 4.3 P ' has enough room n ts buffer to accept every pacet whch needs to be accepted by ts defnton. (It can be easly proved by nducton on the tme that the number of pacets at P ' and P s always the same). However, note that there mght be an addtve dfference of at most B α n the sum of weghts of the pacets sent by the two polces at any gven tme. The algorthm A defned earler (Defnton 3.), has a sgnfcant drawbac, even though we saw the expectaton s good. In practcal use, snce the probablty of droppng a pacet p s, many pacets are dropped, especally n case we have a burst of pacets not from class, the buffer mght be empty for a long tme, and we won t send any pacet. We wll show a determnstc algorthm A ' where ths drawbac s fxed. Note that from clam 4. we can now use a dfferent method than FIFO. 3

4 Defnton 4.4 Algorthm A ' s defned as follows: We dvde the buffer nto classes, each class s sze s B, as before, class contans values n range [ e, e ). A ' performs a round-robn n delvery (.e. n tme t a pacet s sent from class t mod + ) Corollary 4.5 A ' sends, for every class, at least OPT ( σ ) pacets. Clam 4.6 The compettve rato of the Algorthm A ' s e. Proof: Snce the weght of every pacet n class s at least e, and from corollary 4.5: A e OPT () ( σ) ( σ) = Corollary 4.7 The weght of every pacet n class s at most OPT ( σ) e OPT ( σ) e, so And from () and Corollary 4.5: e OPT ( σ ) σ = = e A( σ) e OPT( ) = OPT ( σ) OPT ( σ) ln α e = e 4

5 5 Part 2 Shared memory When modelng the swtch, we can tal about the buffer allocaton wthn the swtch. A smple scheme of our swtch loos le ths: nput ( pacets n memory) N outputs. In our model, n every tmestep pacet can leave on every output lne. Note that n ths dscusson the pacets don t have weghts/values. We manage N ueues and want to maxmze the number of pacets transmtted. There can be several strateges, for example:. Fxed partton assgn T cells for each output ueue. L - ueue length of. 2. Unform Fxed Partton specal case of Fxed partton where : T = N 3. Total Shared emory - every pacet s nserted to the swtch s memory (as long as there s space left) regardless of ts destnaton. 4. Defnng lower and upper bounds - R - lower bound on space we reserve for a ueue, T - upper bound on ueue sze. A new pacet s nserted to ueue f L T and max( L, R ). Note that f T > we need to prevent ueues from overlap. In the rest of ths lecture we wll defne and analyze a strategy called Harmonc partton. 6 Harmonc partton We wll start wth some defnton: Defnton 6. For N We defne constants B by: B = ln N + = s ()- the -th sze ueue at a specfc tme. (.e. numbers the ueues by ther sze - from = as the bggest ueue to =N as the smallest ueue). L - length of ueue s (). Defnton 6.2 The Harmonc partton polcy s: Accept a pacet to ts ueue (the ueue matchng ts output port) only f after acceptng = the pacet the followng condton holds:, N, L B Ths s an attempt to match the bounds on the ueues to the traffc at that tme. 5

6 Note that the bounds don t regard to a specfc ueue, but to a ueue number when sorted by sze. Defnton 6.3 We defne H as a shared memory swtch operatng under Harmonc buffer management polcy. H ( σ ) s the number of pacets the swtch transmts for seuence σ of pacets. Smlar to before, OPT s the optmal (off-lne) polcy whch transmts the maxmum possble number of pacets for any nput seuence. Theorem 6.4 The compettve rato of the Harmonc Algorthm s O(ln N ) It s possble to show Ω (ln N /(ln ln N)), but we won t see t here. Proof: Frst some defntons: Defnton 6.5 A pacet sent by OPT s called extra f at the tme step t s sent from ueue, H doesn t send a pacet from ueue. The above defnton of extra s of our nterest, snce the number of extra pacets euals to OPT( σ) H ( σ), and we actually needs to prove: OPT ( σ) H ( σ) O(ln N) H ( σ) So f we wll prove the number of extra pacets wth algorthm run on seuence σ O(ln N) H( σ ) we ve proved the theorem. Defnton 6.6 A pacet n the OPT ueue s called potentally extra at some tme step, f ts dstance from the OPT ueue head s bgger than the length of the same ueue for H. Observaton 6.7 If at a certan tme step a pacet s extra, then sometme n the past ths pacet was potentally extra. (.e. number of extra pacets number of potentally extra pacets). So, n order to prove theorem 6.4 t s suffcent to prove that the number of potentally extra pacet of algorthm run on seuence σ O(ln N) H( σ ). Theorem 6.8 Number of potentally extra pacets of the algorthm H s at most ln N + tmes the number of pacets that H accepts. Proof: We wll show that at every tme step of the algorthm we can have a mappng between potentally extra pacets and pacets that H has n the buffer at the same tme, or pacets already sent by H The mappng wll satsfy the followng: Every pacet of H s mapped to at most ln N + potentally extra pacets. 6

7 Every potentally extra pacet s mapped as long as t s potentally extra. If a potentally extra pacet, p, s mapped to a pacet then the dstance of p from the head of ts ueue (n H ) s eual or bgger than the dstance of from the head of ts ueue (n OPT ). (and when a pacet s sent from the swtch we consder her dstance from ts ueue s head as negatve). We wll now defne the mappng and show t satsfes the above propertes. The mappng: every tme step both H and OPT receve pacets. Afterwards changes and addtons are done to the mappng. Upon arrval of a new pacet:. If OPT ddn t accept the pacet, or f the pacet s not potentally extra, do nothng. 2. If the pacet s potentally extra: (a) If H too the pacet, then a pacet exsts at OPT n the same ueue, whch was potentally extra before (so t was mapped before), but now s not. The new potentally extra pacet taes the mappng of that pacet. (b) If H ddn t tae the pacet, we wll map the new potentally extra pacet to some pacet of H whch s closer to the head of ts ueue and has less than ln N + mappngs. Clam 6.9 Rule 2(b) s applcable,.e. we wll show that a pacet closer to the head of the ueue, wth less than ln N + mappngs, exsts. Note that f rule 2(b) s applcable, the defned mappng satsfes our demands. Proof: Snce H ddn t tae the pacet, we now that tang the pacet would have volated a condton, so f we too that pacet -, N, L > B, we defne to be the mnmal,.e. Therefore: L B, L > B = = = mn{ L > B }. = And so we obtan that f we too the pacet: L L L B B = = L = > = lnn + > ln N + = 7

8 Before tang the pacet - ueue s length). L lnn + (Tang the pacet adds at most one to ts Snce we now the condton L B s broen should we tae the pacet, we now = the pacet belongs to a ueue whose number s n range [, ] (otherwse the condton wouldn t have been volated). Let s count the number of pacets n ueues to, whose dstance from ther ueue s head s at most L. Queue of the arrved pacet Fgure As we can see from fgure, the number of such pacets s at least L = ln N + Snce the arrved pacet s potentally extra, ts dstance from the head of the ueue s at least L +, and therefore each one of the above pacets s sutable for the ln N + mappng. So far we have seen ln N + pacets whch are closer to the head of the ueue exsts. It s left to show that at least one of these pacets has less than ln N + mappngs. Ths s smple to see, snce every potentally extra pacet whch mapped to a pacet n the buffer of H, s stll wthn the buffer of OPT. 8

9 Before the new potentally extra pacet arrved to OPT, there were at most pacets at OPT (because OPT receved ths pacet, so he needed space for t). So the number of mappngs to the relevant pacets of H s. We have canddate pacets, and total mappng from these pacets s ln N +, therefore there exsts one pacet, whch s mapped to less than ln N + pacets. Ths pacet s mapped to our new potentally extra pacet. (n rule 2(b)). Concluson We showed at each tme step, a mappng between potentally extra pacets and pacets n the memory of H, and the pacet n H s mapped to at most ln N + potentally extra pacets. From the exstence of the mappng we conclude that: As a pacet of H leaves the swtch, there are at most ln N + pacets matched to t. After a pacet leaves, no new pacets are mapped to t. As a pacet of H leaves the swtch, all pacets matched to t n OPT are stll left n the buffer. For every pacet whch OPT sends from ueue, ether there s a pacet that H sends from at the same tme, or t s matched to a pacet H already sent. Therefore, the number of pacets that OPT has sent at tme t s at most O(ln N ) tmes the number of pacets H has sent at that same tme. References [] W. Aello, Y. ansour, S. Raagopolan, A. Rosén, Compettve Queue Polces for Dfferentated Servces. Proc. Of INFOCO 2000, pp [2] E. L. Hahne, A.Kesselman and Y.ansour, Compettve Buffer anagement for Shared-emory Swtches. Proceedngs of SPAA 200. [3] A.Kesselman and Y.ansour, Harmonc Buffer anagement Polcy for Shared emory Swtches. INFOCO