MIT16_36s09_lec20

# MIT16_36s09_lec20 - MIT OpenCourseWare http/ocw.mit.edu...

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Unformatted text preview: MIT OpenCourseWare http://ocw.mit.edu 16.36 Communication Systems Engineering Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms . 16.36: Communication Systems Engineering Lecture 20: Delay Models for Data Networks Part 2: Single Server Queues Eytan Modiano Eytan Modiano Slide 1 Single server queues buffer λ packet per second Server µ packet per second ➨ Service time = 1/ µ • M/M/1 – Poisson arrivals, exponential service times • M/G/1 – Poisson arrivals, general service times • M/D/1 – Poisson arrivals, deterministic service times (fixed) Eytan Modiano Slide 2 Markov Chain for M/M/1 system 0 1 2 k λδ λδ λδ λδ 1 −λδ µ δ µ δ µ δ µ δ • State k ⇒ k customers in the system • P(i,j) = probability of transition from state I to state j – As δ ⇒ 0, we get: P(0,0) = 1 - λδ , P(j,j+1) = λδ P(j,j) = 1 - λδ − µ δ P(j,j-1) = µ δ P(i,j) = 0 for all other values of I,j. • Birth-death chain: Transitions exist only between adjacent states – λδ ,µ δ are flow rates between states Eytan Modiano Slide 3 Equilibrium analysis • We want to obtain P(n) = the probability of being in state n • At equilibrium λ P(n) = µ P(n+1) for all n – P(n+1) = ( λ /µ )P(n) = ρ P(n), ρ = λ /µ • It follows: P(n) = ρ n P(0) • Now by axiom of probability: P ( n ) = 1 i = ! " # \$ n P (0) = P (0) 1 % \$ = 1 i = ! " # P (0) = 1 %\$ P ( n ) = \$ n (1 % \$ ) Eytan Modiano Slide 4 Average queue size N = nP ( n ) = n = ! " n # n (1 \$ # ) = # 1 \$ # n = ! " N = # 1 \$ # = % / μ 1 \$ % / μ = % μ \$ % • N = Average number of customers in the system • The average amount of time that a customer spends in the system can be obtained from Little’s formula (N=...
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MIT16_36s09_lec20 - MIT OpenCourseWare http/ocw.mit.edu...

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