Queueing Theory II. Summary. M/M/1 Output process Networks of Queue Method of Stages. General Distributions

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1 Queueig Theory II Suary M/M/1 Output process Networks of Queue Method of Stages Erlag Distributio Hyperexpoetial Distributio Geeral Distributios Ebedded Markov Chais 1

2 M/M/1 Output Process Burke s Theore: The Departure process of a stable M/M/1 queueig syste with arrival rate is also a Poisso process with rate. A k D k-1 I k Z k Z k D k D k-1 D k Y k Y k Iterdeparture Tie Burke s Theore Pr { Yk y X = } { X k ( A ) = k } + Pr { Yk y X ( A ) > 0} Pr{ X ( ) 0 k A > k } { Yk y} = ( A ) { ( k ) } { ( k ) } Pr 0 Pr 0 Pr X A = 0 = π0 = 1 ρ PASTA Pr X A > 0 = 1 π0 = ρ PASTA { Yk y X ( A k ) > } = Pr{ k } = 1 { Yk y X ( A k ) = } = { I Z y} Pr 0 Pr 0 y Z y e μ Pr k k + = μ y μy [ 1 e ] [ 1 e ] μ μ =

3 Two Queues i Series μ 1 μ Let the state of this syste be (X 1,X ) where X i is the uber of custoers i queue i. 0 1 μ μ μ 1 μ μ μ μ μ 1 μ 1 μ Two Queues i Series Balace Equatios π = μ π ( + μ ) π 0 μ π,1 π 1,0 1 = +, > 0 ( ) 0 0, 1 1 1, 1 + μ π = μ π + + μπ, > 0 ( + μ μ) π = μ μ π,, 0 1, + μ1π + 1, 1 + μ π, + 1 > μ 1 μ μ μ μ π = 1 = 0 = 0 μ 1 μ 1 μ

4 Product For Solutio Let ρ 1 =/μ 1, ρ =/μ = ( 1 ρ ) ( 1 ρ ) π ρ ρ Recall the M/M/1 Results π 1 1 ( 1 ) = ρ ρ Therefore the two queues ca be decoupled ad studied i isolatio. π = ππ 1 Jackso Networks The product for decopositio holds for all ope queueig etworks with Poisso iput processes that do ot iclude feedback Eve though custoer feedback causes the total iput process (exteral Poisso ad feedback) becoe o- Poisso, the product for solutio still holds! These types o etworks are referred to as Jackso Networks The total iput rate to each ote is give by Aggregate Rate Λ = + Λ i i i = 1 r Exteral Arrival Rate Routig Prob 4

5 Closed Networks The product for decopositio holds for also for closed etworks Aggregate Rate Λ = Λ r i i = 1 Routig Prob No-Poisso Processes Assue that the service tie distributio ca be approxiated by the su of iid expoetial rado variables with rate μ. That is = 1 ( y) Z = Y where Y ~ F = 1 e Y μ y I Hoework 4, you showed the desity of Z is give by ( ) 1 μ μt μt fz () t = e, t 0 ( 1)! Z is a Erlag rado variable with paraeters (, μ). 5

6 Erlag Distributio You ca show that the distributio fuctio of Z 1 μt ( μt) FZ () t = 1 e, t 0 = 0! The expected value of Z is give by 1 1 E[ Z ] = E Y = E Y = = = = 1 = 1 = 1 μ μ μ The variace of Z is give by 1 1 var [ Z ] = var Y = var Y = = = = 1 ( ) ( ) = 1 = 1 μ μ μ Note that the variace of a Erlag is always less tha or equal to the variace of the expoetial rado variable M/Er /1 Queueig Syste Meaig: Poisso Arrivals, Erlag distributed service ties (order ), sigle server ad ifiite capacity buffer. μ μ μ Stage 1 Stage Stage Erlag Server Oly a sigle custoer is allowed i the server at ay give tie. The custoer has to go through all stages before it is released. 6

7 M/Er /1 Queueig Syste The state of the syste eeds to take ito accout the stage that the custoer is i, so oe could use (x,s) where x=0,1, is the uber of custoers ad s=1,, is the stage that the custoer i service is curretly i. Alteratively, o ca use a sigle variable y to be the total uber of stages that eed to be copleted before the syste epties. Let = μ μ μ μ μ μ μ μ Er /Μ/1 Queueig Syste Siilar to the M/Er /1 Syste we ca let the state be the uber of arrival stages i the syste, so for exaple, for = μ μ μ μ μ μ μ 7

8 Hyperexpoetial Distributio Recall that the variace of the Erlag distributio is always less tha or equal tha the variace of the expoetial distributio. What if we eed service ties with higher variace? p 1 p μ 1 μ p μ Hyperexpoetial Server Hyperexpoetial Distributio Expected values if the Hyperexpoetial distributio p E[ Z ] = p Y E = μ The variace of Z is give by E [ ] Z = 1 = 1 p pe Y = 1 = 1 μ = = p [ ] p var Z = E [ Z ] ( E [ Z ]) = = 1 μ = 1 μ 8

9 M/H /1 Queueig Syste I this case, the state of the syste should iclude both, the uber of custoers i the syste ad the stage that the custoer i service is i. For exaple, for = p 0 11 μ pμ 1 pμ 1 pμ 1 pμ 1 (1-p)μ 1 (1-p)μ 1 (1-p)μ 1 μ pμ pμ pμ (1-p) 1 (1-p)μ (1-p)μ 3 (1-p)μ (1-p)μ M/G/1 Queueig Syste Whe the arrival ad service processes do ot possess the eoryless property, we are back to the GSMP fraework. I geeral, we will eed to keep track of the age or residual lifetie of each evet. Ebedded Markov Chais Soe ties it ay be possible to idetify specific poits such that the Markov property holds For exaple, for the M/G/1 syste, suppose that we choose to observe the state of the syste exactly after each custoer departure. For this proble, it does t atter how log ago the previous arrival occurred sice arrivals are geerated fro Poisso processes. Furtherore, at the poit of a departure, we kow that the age of the evet is always 0. 9

10 Ebedded Markov Chais So, right after each departure we observe the followig chai (oly the trasitios fro state are draw) Let a be the probability that arrivals will occur durig the iterval Y defied by two cosecutive departures a 1 a k a a 3 k a 0 Let N be a rado variable that idicates the uber of arrivals durig Y, the a = Pr{N = }. Ebedded Markov Chais Suppose we are give the desity of Y, f Y (y) the { } { } ( y) a = Pr N = = Pr N = Y = y f dy Sice we have a Poisso process 0 { } ( y) Pr N = Y = y = e! Y y Therefore ( y) y a = e fy ( y) dy 0! 10

11 State Iteratio Let X k be the state of the syste ust after the departure of the kth custoer A k X k-1 X k X k-1 X k D k-1 I k Z k Z k D k D k-1 D k Y k Y k X = X + N X = X 1 + N 1 k k 1 k k k k Arrivals Durig Y k X = X + N 1{ X > } k k 1 k k 1 0 M/G/1 Queueig Syste Pollaczek-Khichi (PK) Forula where E[ X ] = ρ ρ 1 μ σ 1 ρ 1 ρ X is the uber of custoers i the syste 1/μ is the average service tie σ is the variace of the service tie distributio is the Poisso arrival rate ρ=/μ is the traffic itesity ( ) ( ) 11

12 M/G/1 Exaple Cosider the queueig systes M/M/1 ad M/D/1 Copare their average uber i the syste ad average syste delay whe the arrival is Poisso with rate ad the ea service tie is 1/μ. For the M/M/1 syste, σ =1/μ, therefore E[ X ] ρ E[ X ] Μ / Μ /1 = 1 ρ For the M/D/1 syste, σ =0, therefore ρ ρ = 1 ( ρ ) Μ / D /1 1 1/ E [ S] = μ Μ / Μ /1 1 ρ 1/ μ ρ E [ S] Μ / D /1 = 1 1 ρ 1

Queueing Theory II. Summary. ! M/M/1 Output process. ! Networks of Queue! Method of Stages. ! General Distributions

Queueing Theory II. Summary. ! M/M/1 Output process. ! Networks of Queue! Method of Stages. ! General Distributions Queueing Theory II Summary! M/M/1 Output process! Networks of Queue! Method of Stages " Erlang Distribution " Hyperexponential Distribution! General Distributions " Embedded Markov Chains M/M/1 Output