Network of Markovian Queues. Lecture

Transcription

1 etwork of Markovan Queues etwork of Markovan Queues ETW09 20

2 etwork queue ed, G E ETW09 20 λ If the frst queue was not empty Then the tme tll the next arrval to the second queue wll be equal to the servce tme * D () s () queue s not empty = + s D *() s = What s the arrval rate to the second queue ( ρ) 2 If the frst queue was empty then we wll wat for the next arrval followed by ts servce tme D * () s λ + ρ λ + s + s + s () queue s empty λ = λ+ s + s 2

3 Burkes Theorm ed, G E ETW09 20 D * () s = ( ) λ ρ + ρ λ + s + s + s * λ D () s = ( ρ ) ρ + + s λ+ s ρs * ( ρ) λ+ ρ( λ + s) D () s = + s λ + s * s D () s λ+ = + s λ+ s ( \ ) * λ λ s + λ D () s = + s λ s + * λ + s D () s = + s λ+ s * λ D () s = λ + s Inter-departure tmes are exponentally dstrbuted wth- the same parameter as the nter-arrval tmes Burke's theorem states that the steady-state output of a stable M/M/m queue wth nput parameter λ and servce-tme parameter for each of the m channels s n fact a Posson pcess at the same rate.λ 3

4 Reversblty ed, G A random pcess s tme-reversble f the pcess, reversed n tme, has the same pbablstc ppertes as the forward pcess. E ETW

5 Departure Pcess fm the M/M/m Queue: Burke s Theorem Model ed, E ETW09 20 λ Inter-departure tmes are exponentally dstrbuted wth- the same parameter as the nter-arrval tmes Burke's theorem states that the steady-state output of a stable M/M/m queue wth nput parameter λ and servce-tme parameter for each of the m channels s n fact a Posson pcess at the same rate.λ 2 5

6 FEEDFORWARD ETWORKS (two queue example) λ 2 ed, G Pk (,2) k = Pq ( = k, q = k 2 2 Pk () = ( ρ ρ / ρ = λ / ) ρ k = λ Pk (2) = ( ρ ρ / 2) ρ k 2 2 = λ 2 Pk (,2) k = ( ρ ( ) ρ ρ ) ρ k k E ETW

7 FEEDFORWARD ETWORKS (two queue states example) ed, G E ETW ,0 0, 0,2 0,3 λ λ λ λ 2 2 2,0,,2,3 λ λ λ λ ,0 2, 2,2 2,3 λ λ λ λ ,0 3, 3,2 3,3

8 (two FEEDFORWARD ETWORKS queue states example) λ λ λ λ λ λ λ λ λ λ λ λ ETW09 20 Perf Model d, rsty n Ca

9 (n FEEDFORWARD ETWORKS ( queues s seres) λ 2 n n Pk (, k2, k3,..., kn ) = ( ρ ) ρ = n = k 9 ETW09 20 Perf Model d, rsty n Ca

10 ed, G E ETW09 20 Pduct Form Soluton for Queue P etworks x k ( k, k2, k3, k4,..., k ) = GK = β k GK ( ) k x k A = β( k) = ( ) ( ) ormalzaton Factor k! k < m β ( k ) = k! m m m k > m x s the soluton to the set o fequatons x = jxjrj, =,2,..., j= r j, s the Pbablty bl that a customer leav queue wll head to queue j 0

11 Pduct Form Soluton (specal Cases) ed, G E ETW09 20 Λ Λ P( k, k2, k3, k4,...., k ) = = e Λ P( k, k2, k3, k4,..., k ) = k! Λ k All queues use a sle server queue k All queues use = All queues use nfnte servers

12 Message Delay ed, G E ETW09 20 Average number of customers n the system T L= T λ = T λ = λ = = Average message delay L= Tλ Lttle's L = Formula = Tλ T D = λ = Average message delay (gnor ppagaton delay) Λ = I t t Input to queue 2

13 Open etworks λ 2 2 λ α, α 2, λ 4 4 α, α 3, λ 3 3 Λ2= α,λ+ λ2 Λ3= α,2λ+ λ3 Λ = α Λ + α Λ + λ 4 2 2, 2 3, Traffc Equatons 3 ETW09 20 Perf Model d, rsty n Ca

14 FeedBack λ 2 Model ed, ETW09 E 20 Feed Back destys the Posson characterstcs Consder the example when servce rate s much hgher than the arrval rate The result arrvals are not Posson nevertheless, t can be shown that the pduct form holds and the terms n the pduct are the same as f the flows were Posson. α 4

15 Closed etworks 2 ed, G E ETW09 20 umber of customers n the systems s fxed There s dependency between the queues snce the sum has to always be equal to a constant = k = K umber of possbltes K + 5

16 Example ETW09 20 Perf Model d, rsty n Ca

17 Three queue two customer example ETW09 20 Perf Model d, rsty n Ca