Communication Networks II Contents

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1 6 / -- Communcaton etworks II (Görg) -- Communcaton etworks II Contents Fundamentals of probablty theory 2 Traffc n communcaton networks 3 Stochastc & Markovan Processes (SP & MP) 4 Fnte State Markovan Processes 5 Analyss of Markovan servce systems 6 Queueng networks for modelng communcaton networks 7 M/G/ model 8 The model M/G//FCFS/OPRE 9 The model M/G//FCFS/PRE

2 6 / 2 -- Communcaton etworks II (Görg) Queueng networks for modelng communcaton networks Queueng networks are nterconnected servce statons [Bo ], that are decoupled through buffers. Open System: System Closed System: System

3 ar rv als nter upted j obs general wat ng s y st em q pr oc es or l mt ed r ate of c ent ral sy s tem sourc es outfl ow 6 / 3 -- Communcaton etworks II (Görg) -- Closed system (no arrvals, no departures): E.g. constant number of actve termnals q, q jobs n system, one per termnal, each job s ether n the termnal (thnkng), n the queue or s beng served n the computer. The central system here can also be seen as a substtute for a so-called open system, whch s a complcated queueng system wth several queues and servers. In general: The processors n queueng networks are characterzed through In general: The processors n queueng networks are characterzed through (servng) statons. Each such staton s accompaned by a queue. In some cases more than one staton shares the same queue.

4 6 / 4 -- Communcaton etworks II (Görg) -- Open system (arrvals from outsde / departures to outsde) Fgure 6.2: General open queung network

5 sourc e0 λ 0 λ 0 = p 0 p 0 0 p λ 0 λ 0 λ λ p p queue queue queue p p µ µ µ p p (+ ) (+ ) ( + ) s wal + 6 / 5 -- Communcaton etworks II (Görg) -- Precondtons for the computablty of open queung networks n statonary operaton: the th staton can contan one or more servers ndvdual transton probabltes can be equal to zero ndvdual servce rates per staton (equal rates n case of parallel servers) selected, robust servce strateges The followng notatons wll be used: ε n µ = n ε p j λ j = p j servce rate of a server n staton number of servers n staton servce rate of all the servers n staton transton probablty from staton to staton j total arrval rate at staton (= throughput) probablty p ( + ) for leavng the open network after staton

6 6 / 6 -- Communcaton etworks II (Görg) -- Closed queung systems E.g. short crcut the source and the snk. Then there are exactly K crculatng jobs. Thus no queue needs more than K places. Open networks can be computed more easly than closed ones. Closed networks are often better suted for modelng of real systems, as they allow the followng: mappng of manyfold ndependent resources sequental usage of resources by the jobs smultaneous usage of dfferent resources through dfferent users

7 6 / 7 -- Communcaton etworks II (Görg) -- Job classes n queung networks The selecton of job paths n a queueng network wth statons s done dependng on the transton probabltes. It leads to a stochastc routng matrx, that defnes the transton probabltes from staton to staton j. p p2 L p p2 p22 L p2 P= M M O M p p L p 2 L (6.0.) It s allowed to group the jobs, so that each group has ts own routng matrx. Such groups are called classes or chans. Each class can be descrbed by a Markov chan. All the classes combned together also buld a Markov chan. Classes are qute sutable, for example, to model jobs wth dfferent behavor, e.g. nteractve and non-nteractve type jobs. Queung networks can contan some classes, whch are open, whle other classes can be closed.

8 6 / 8 -- Communcaton etworks II (Görg) Open queung networks: Tradtonal Soluton method If the queung network can be modeled as a dscrete Markov process wth known state transton probabltes, then the statonary state probablty dstrbuton can be obtaned by formulatng and solvng a lnear system of equatons. Each state transton here corresponds to a job transfer from one staton to the other, a new arrval or a departure. If k = k, k 2,Kk denotes the state vector of a system wth server statons, where k ( =,2, K, ) s the number of jobs at staton, then we can ntroduce the followng notatons P(k) the statonary dstrbuton of jobs n the system k a state of the system and P(k) the correspondng state probablty, where each state s denoted through a state vector k.

9 6 / 9 -- Communcaton etworks II (Görg) -- Such processes satsfy the global equlbrum condton at the statonary stuaton; see also equaton (4.3a) k : P( k) (rate from k) = k' k P( k') (rate from k' to k) (6..2) along wth the total probablty condton P( k) = k (6..3) Ths system of equatons has a unque soluton for the statonary state probabltes (provded that the system s rreducble, aperodc and recurrent non-null). From ths result we can derve the performance measures such as: mean number of jobs n each staton load factor at each of the statons mean watng tme and the throughput of each staton. The soluton of ths system of equatons s only for lmted sze of state space wth less than states numercally possble (Gauss-Sedel-Method).

10 6 / 0 -- Communcaton etworks II (Görg) -- Jackson found n 963 that for a specfc class of these networks the soluton can be presented n product form: P( k) p ( k ) = = p ( k ) wth =,2, K, and k = 0,,2,K (6..4) s the boundary probablty for havng exactly jobs n staton. etworks havng the followng propertes are dentfed as open exponental networks or so-called Jackson etworks: all statons have M-dstrbuton for servce tme and FCFS strategy the network s open wth Posson arrvals (at the rate λ o ), see fgure 6.3 only one sngle class of jobs unlmted buffer space between statons unsaturated statons: λ < ε n k λ = λ0 + pjλj wth, j =,2, K, j= (6..5)

11 6 / -- Communcaton etworks II (Görg) -- A sngle staton has followng structure: Fgure 6.3: A staton of a Jackson network Thus we get; see equaton (5.6a): (6..6) = n k k n k n n p n k k n p k p for! (0) for! ) ( (0) ) ( ρ ρ

12 6 / 2 -- Communcaton etworks II (Görg) -- wth: n λ ε µ = n ε number of server at staton, each wth the rate total arrval rate at staton servce rate of a sngle server at staton total servce rate of all the servers n staton ε k ρ number of jobs at staton load at staton, Where t holds, see equaton (5.3) λ ρ = = n ε A n A = λ ε (6..7) and also as pre-condton for statonary operaton: ρ λ = <, A = ρ n µ (6..8)

13 6 / 3 -- Communcaton etworks II (Görg) -- In so-called Jackson networks, the ndvdual statons behave under gven assumptons n such a way, that each of them functons as an solated M/M/n-model. Thus t holds from equaton (5.6b): p ( 0) = n A n A n!( n A ) j! + n j 0 = j (6..9) Other performance measures can be derved from statonary dstrbutons. Thus we get for the mean queue length (wthout the jobs that are beng processed): n n LQ = k n p k p ( ) ( ) = ( ) ( ρ ) ρ 0 2 n!( ρ ) k = n If we start agan from the model M/M/n wth n=, we get the known equaton (calculaton exercse!): (6..0) L Q 2 ( n = ) = ρ ρ (6..)

14 6 / 4 -- Communcaton etworks II (Görg) --

15 6 / 5 -- Communcaton etworks II (Görg) --

16 6 / 6 -- Communcaton etworks II (Görg) -- For the mean watng tme W of all jobs and for the mean system tme V we get from Lttle s law, see equaton (5.): L L = λ Q W = and V = W + or V (6..2) λ ε The mean number of jobs n the th staton L s gven by: L= n ρ+ LQ (6..3) For the case n = we agan get: ρ L = ρ ρ (6..4) For the total mean response tme of the network (sojourn tme n the network) can be obtaned by summng up the ndvdual values per staton L ρ V = = : λ λ ( ρ ) V where t holds: λ = = = λ λ 0 = L = λ = ρ ρ for total throughput (6..5) (6..6)

17 6 / 7 -- Communcaton etworks II (Görg) -- For the watng tme dstrbuton functon we get; see equatons (5.7), (5.8) P( T W n ( n ρ ) t) = p (0) e n!( ρ ) For n = we get: n ε ( ρ ) t for t 0 (6..7) P( T t) = e W ( ) ρ ε ρ t (6..8) e The relatve frequency of vsts to staton of an arbtrary job s gven λ 0 = p by e and equaton (6..5) wth, see equaton (6..6): 0 = λ λ λ e = p + e p 0 j j j= (6..9)

18 6 / 8 -- Communcaton etworks II (Görg) -- Example 6. Fgure 6.4 gves a system wth relevant parameters ndcated n the fgure. µ 2 µ 4 µ. Fgure 6.4: Open queung network =4 statons wth a sngle server per staton.e. servce tme s ndependent and M-dstrbuted wth the rate FCFS strategy for all statons throughput (neg. exp. arrval process) λ= s 4 n = µ 3 µ µ = 004. s, µ = 003. s, µ = 006. s, µ = 005. s 2 3 4

19 6 / 9 -- Communcaton etworks II (Görg) -- From λ = λ p 2 λ = λ 0+ p 2 + λ p 3 j= + λ p Ths system of equatons leads to the soluton: It further follows wth 3 λ = λ p λ = λ p the result. 4 j λ 4 j, ρ λ = µ follows : ρ = 08., ρ = 03., ρ = 06., ρ = , 3 3, λ = λ 4 λ = 20, λ 2 = 0, λ 3 = 0, λ 4 = 4 [ s ] Assume that we have to calculate the statonary state probabltes k = 3, k = 2, k = 4, k =. Because = k can take any value and ( ) P( k) = p( k, k, Kk ) = p ( k ) 2 = we get here. p( 3,2,4, ) = p (3) p2 (2) p3 (4) p4 ()

20 6 / Communcaton etworks II (Görg) -- k Wth n = we get p ( k ) = ( ρ ) ρ from equaton (6..6) and thus p 3) = 0.3, p (2) = 0.06, p (4) = 0.05, p () 0.6 ( = Thus, t holds: p(3,2,4,) = = ρ The mean number of jobs n staton s wth L = : ρ L = 4, L = 043., L = 5., L = The mean response tme s from V L = : λ V = 02., V = 0043., V = 05., V = For the total mean response tme, we get from V = Σ : λ L V =54. s

21 6 / 2 -- Communcaton etworks II (Görg) -- From e = p + e p 0 j j j= we can calculate the mean number of vsts made to staton by a job. As jobs arrve at rate λ from the outsde and result n a mean number of vsts to each of the statons, the frequency at whch staton s vsted s gven by λ e. Ths leads to: e = λ λ Thus, we get: e = 5, e = 25., e = 25., e = 2 3 4

22 6 / Communcaton etworks II (Görg) -- WnPEPSY-QS s a tool for calculatng performance measures of queueng networks: (download:

23 6 / Communcaton etworks II (Görg) Closed networks Gordon and ewell have shown n 967, that the product form soluton can also be extended for closed networks. All assumptons relevant to Jackson networks contnue to hold, and n addton to that t s assumed that the number of jobs n the network s constant: = k = K The throughput at staton s now, see also equaton(6..5) λ = λ p j j j= because there are no external arrvals. (6.2.20) (6.2.2) We can treat any closed network as an open one, by takng the arrval rate to vary n such a way, that the number of jobs n the network reman constant (= K). The number of dfferent states n a closed system becomes equal to the number of possbltes to dstrbute K jobs among statons, and ths number s gven by

24 6 / Communcaton etworks II (Görg) -- n s + K = (6.2.22) By puttng e = λ n equaton (6.2.2) we get the lnear system of equatons: λ e = j= e p j j (see equaton (6..9)) (6.2.23) From the theory of Gordon and ewell (967), the statonary dstrbuton at equlbrum has the unque soluton: p( k, k2, K, k) = G( K) = x k ( ) b k wth x e = ε (6.2.26) where G(K) s the normalzaton constant, that comes from the condton that all the state probabltes should sum up to one (condton for total probablty). G( K) = k K = = = x k b ( k ) (6.2.27)

25 6 / Communcaton etworks II (Görg) --

26 6 / Communcaton etworks II (Görg) -- Ths constant also has the product form. However, t s not the product of state probabltes of ndvdual statons. Because of the constant number of jobs n the network, the number of jobs n ndvdual statons are now nterdependent. We state the assumptons made, before gvng the soluton for closed networks. Frst, we defne a help functon : b( k ) k! k n b k n n k ( ) = n! k > n n = b ( 0) = For n =, the soluton smplfes to: p( k, k2, K, k) = G( K) wth: G( K)= x k k = K = = = x k (6.2.28) (6.2.29) (6.2.30)

27 6 / Communcaton etworks II (Görg) -- Example 6.2 It holds here: =3, K=3 and n = (full mesh) Fgure 6.5: Closed network.

28 6 / Communcaton etworks II (Görg) -- The followng transton matrx results: p = 06. p2 = 03. p3 = 0. P= p2 = 02. p22 = 03. p23 = 05. p = 0.4 p = 0. p = Each staton has a server wth M-servce tme dstrbuton: µ = 08. s, µ 2 = 06. s, µ 3 = 04. s, ( ε = µ ) The statonary dstrbuton at equlbrum for ths system works out to from (6.2.29) p( k, k2, k3) = G wth G ( 3) ( 3) = 3 k + k + k = 3 = k x = k x

29 6 / Communcaton etworks II (Görg) -- From the equlbrum equatons, see equaton (6.2.23) we get Wth vstng frequency e = t follows: e = e p + e p + e p = e = j = ej pj e = e p + e p + e p = e = e p + e p + e p = Determnaton of : x / b ( k ) k Due to = we get b ( k ) n = x = ; x = 25. ; x = ; x = 953. ; x = ; x = x = ; x = ; x = ; x = 833. ; x = 336. ; x = 662. ;

30 6 / Communcaton etworks II (Görg) -- There are exactly 5 possbltes to dstrbute 3 jobs among 3 statons, 2.e. 0 states. They are: k k k All these states are summed up to G(3). ( 3) G = x x x + x x x + x x x + x x x + x x x + x x x x 0 x2 3 x3 0 + x 0 x2 2 x3 + x 0 x 2x3 2 + x 0 x2 0 x3 3 = As, we get b k = for all k =, 2,K K. ow, we can calculate the state probabltes: n = ( ) p( k, k2, k3) = G 3 x k = ( 3)

31 6 / 3 -- Communcaton etworks II (Görg) -- ( ) ( ) ( ) p 30,, 0 = p,, = p 20,, = ( ) ( ) ( ) p 0,2, = p,2, 0 = p 0,,2 = 07. ( ) ( ) ( ) p 030,, = p 0,,2 = 02. p 20,, = 06. p( 003,, ) = The boundary probabltes p k - thus, f we consder only staton we get: ( ) p (0) = p(0,3,0) + p(0,2,) + p(0,,2) + p(0,0,3) = p () = p(,2,0) + p(,,) + p(,0,2) = p (2) = p(2,,0) + p(2,0,) = 0.72 p (3) = p(3,0,0) = The boundary probabltes of other statons are obtaned smlarly through summaton of the correspondng state probabltes.

32 6 / Communcaton etworks II (Görg) -- The mean number of jobs n staton can be calculated as n open networks: L = ( ) k p k = k 3 Here as an example for staton : [ ( ) ( ) ( )] ( ) ( ) [ ] ( ) L = p,, + p,2, 0 + p 0,,2 + 2 p 20,, + p 2, 0, + 3 p 30,, 0 = L 2 = 054. ; L 3 = 585. ; The load or probablty for havng at least one job at staton works out to ρ ρ ρ 3 = p, j, k wth k, j = ( ) ( 0,,2 ) 3 2= p j= (, j, k) wth, k ( 0,,2 ) 3 3= p k= (, j, k) wth, j ( 0,,2 ) λ ρ= p ( 0 ) = (holds only for n =, p s the dle probablty) µ ( 0) (6.2.3) (6.2.32)

33 6 / Communcaton etworks II (Görg) -- e.g. for staton to: ( ) ( ) ( ) ( ) ( ) ( ) ρ = p + p + p + p + p + p p,,,2, 0 0,,2 20,, 2, 0, 30,, 0 = - ( 0) = Analogously, for the other statons: ρ = ρ = Indvdual throughputs work out to λ = nρ ε; n = ε = µ ; λ = ; λ 2 = ; λ 3 = 039. ; The mean sojourn tme (response tme) n staton s obtaned usng Lttle s law V L (6.2.33) = λ e.g. V = ; V = ; V = ; 2 3

34 6 / Communcaton etworks II (Görg) -- The normalzaton constant G(K) A number of algorthms are avalable to calculate normalzaton constants [Bo ]. Buzen s algorthm (97/73) Here x ( =, L, ) are gven through e x = (6.2.34) ε where e denotes the mean number of vsts made to the -th staton and ε denotes the servce rate of a server at the -th staton, see equaton (6..9). As there are only - ndependent equatons n closed networks, t s assumed e =. We defne the followng help functon: g( k, n) = n n = k = k = k x b ( k ) for n=, L, k 0 where b ( k ) can be determned from the equaton (6.2.28). (6.2.35)

35 6 / Communcaton etworks II (Görg) -- For < n < t holds g(k,n) k j= 0 n k = k = & k n = j n x k k x n j n k x = = b k b ( k ) b j n ( j) ( ) = = 0 n = k = k j = & k n = 0 (6.2.36) k x = g( k j, n ) b ( j) j= 0 n n j From equaton (6.2.35) t drectly follows k x g( k,) = b ( k) g(0, n) = for for k =, L, K n=, L, (6.2.37) The equatons (6.2.36) and (6.2.37) completely defne the algorthm, whch s schematcally presented n table 6..

37 6 / Communcaton etworks II (Görg) -- The teratve equaton (6.2.36) along wth the ntal condton n equaton (6.2.37) completely defne the algorthm for the calculaton of G(K). The goal of ths algorthm s to calculate the bottom value n the last column, because ths value g(k,) s equal to the normalzaton constant G(K). The values g(k,) = G(k) n the last column are also mportant, as they can be used to calculate performance measures bypassng the calculaton of state probabltes. Wth the help of normalzaton constant G(K) t s possble to calculate frst the boundary probabltes p and then all the performance measures. The probablty for exactly k jobs to be n the -th staton results from: k x p ( k = k) = p ( k) = [ G( K k) x G( K k ) ] G( K) (6.2.38) where x can be obtaned from the equaton (6.2.34).

39 6 / Communcaton etworks II (Görg) n- n x b ( k) 0 g( 0, n ) + x g(, n ) + 2 g( 2, n ) + b b n n n n k n k ( k ) x n k 2 ( k 2) n k-2 g( k 2, n ) + n k- g( k, n ) + xn k g( k, n ) + b n b 0 x n b 2 ( 2) x n ( 0) ( ) g( k, n) K K x b( K) g( K, ) Table 6.: Buzen s Algorthm ( b n ( ) s n the numerator)

41 6 / 4 -- Communcaton etworks II (Görg) -- The throughput results from λ λ K e G ( K ) = ( ) = G( K) =, L, (6.2.39) where e can be determned from equaton (6..9). ρ λ µ The load factor of a staton can be derved from the relaton = /. Puttng from equaton (6.2.39) n ths formula, we get λ ρ = K x G ( ) G( K) (6.2.40) The mean number of jobs n the -th staton s gven through K K L = k p ( k) = k= k= x k G( K k) G( K) for =, L, (6.2.4)

43 6 / Communcaton etworks II (Görg) -- The r-th moment of the number of jobs n the -th staton s r r [ k ( k ) ] for =, L K ( r) k G( K k) L = x, G( K) k= (6.2.42) The mean sojourn tme can be calculated usng Lttle s law V = G( K k) G ( K ) e K k x k = for =, L, (6.2.43) The mean watng tme results from the known relatonshp: W = V / ε The mean queue length can also be obtaned usng Lttle s law: LQ =λ W

45 6 / Communcaton etworks II (Görg) -- Example 6.2a: In ths example we calculate the performance measures wth the help of the normalzaton constant. Frst we calculate e from equaton (6.2.23): e = e p + e p + e p = e = e p + e p + e p = e = e p + e p + e p = From equaton (6.2.34) we get x (=,2,3) x = 25. x = x =

47 6 / Communcaton etworks II (Görg) -- The table 6.2 gves us g( k, n) : K= The normalzaton constant GK ( ) = g ( K, ) = and Gk ( ) = g ( k, ), k =02,,, 3 The probablty for havng, for example, exactly 2 jobs n staton s calculated from the equaton (6.2.38): x 2 p( k = 2) = [ G( ) x G( 0) ] = 072. G( 3)

49 6 / Communcaton etworks II (Görg) -- The throughput λ ( =,2, 3) results from the equaton (6.2.39) λ = G(2) e G(3) = jobs/sec. λ 2 = e 2 G(2) G(3) = jobs/sec. λ 3 = e 3 G(2) = 0.38 jobs/sec. G(3) The load factor ρ ( =,2, 3 ) s calculated from the equaton (6.2.40) ρ = ρ2 = ρ 3 = The mean number of jobs n the -th staton (=,2,3) from equaton (6.2.4) L L Smlarly x G ( 2) x G ( ) x G ( 0) = + + = G( 3) G( 3) G( 3) 2 3 x G ( 2) x G ( ) x G ( 0) = + + = G( 3) G( 3) G( 3) L 3 = Thus we have obtaned the same set of results as n example 6.2 usng a dfferent approach.