Lectures10-14

Lectures10-14 - Paradox of Residual Life ECE-CSE 861:...

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1 ECE & CSE ECE-CSE 861: Introduction to Computer Communication Networks Ness B. Shroff ECE & CSE Lectures 10-14 ECE & CSE Paradox of Residual Life Formally: Let {N(t),t 0} be a Poisson Process with rate λ Let S 1 , S 2 ,… be successive arrival times, and S 0 0 The number of arrivals in [0,t] N(t) The time of the last arrival before t S N(t) The time until the next arrival after t = S N(t)+1 ECE & CSE Paradox of Residual Life To Prove: P{Y t u|N(s), s t} = 1-e - λ u , u 0 In other words what we want to prove is that the time until the next arrival is exponentially distributed with mean 1/ λ . Proof: For u 0: {Y t >u} = {S N(t)+1 -t > u} = {S N(t)+1 > u +t} = {N(t+u) – N(t) = 0} A t =age Y t residual lifetime S N(t) t S N(t)+1 ECE & CSE Paradox of Residual Life P{Y t >u | N(s), s t} = P{N(t+u) – N(t) = 0 | N(s), s t} = P{N(t+u)-N(t) = 0} (Ind. Increment) = P{N(u) =0} (Stationary Increment) = e - λ u , u 0 P {Y t u | N(s), s t} = 1 - e - λ u , u 0 Q.E.D. The Paradox of residual life is important in calculating the E(W) in the queueing system. Homework: Find the distribution of the age of the process. In other words find P{A t u} ? E[S N(t)+1 - S N(t) ] = ? as t E[S N(t)+1 - S N(t) ] 2/ λ as t
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2 ECE & CSE Reminder: Queueing Notation Notation: Infinite Buffer: A/B/C old notation A/B/C/ Single server Multiserver queue Inter-arrival time Service time number of servers ECE & CSE Reminder: Queueing Notation Finite Buffer: A/B/C/D Infinite Buffer: A/B/C Example: M/M/1 (Arrival Poisson, Service exponential, 1- server) M/G/1 Memoryless General ECE & CSE Reminder: Queueing Notation (c) M/M/m (d) M/M/m/N (e) G/M/1 , etc… # of servers = m # of servers = m Buffer size = N ECE & CSE M/M/1 Queue Possion arrivals and exponential service time. If τ n is the service time of the n th packet (customer), then P{ τ n s} = 1 - e - μ s , s 0 Given that a packet has been in service for t seconds, what is the probability that it will wait in service for an additional s seconds. In other words find P { τ n > s+t | τ n > t}. μ λ Poisson
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3 ECE & CSE M/M/1 system } { } { } { ) ( s P e e e t P t s P n s t t s n n > = = = > + > = + τ μ } { } , { } | { t P t t s P t t s P n n n n n > > + > = > + > For all s, t 0. ECE & CSE M/M/1 system Additional time a customer (packet) waits in service is independent of how long it has waited so far. Quantities of interest: o Queue length distribution: P n o Avg. Queue length = Σ n P n o Avg. time delay E(T) o Throughput γ P n is defined to be the steady state probability of having n customers in the system ECE & CSE M/M/1 system P n = lim t P n (t). P n (t+ Δ t) = f(P j (t)) Question: If at time t+ Δ t the queue is in state n, then what are the possible states the queue can be in at time t? Probability of n packets in system at time t ECE & CSE M/M/1 system Let n 1 a) At time t, Queue is in state n+2 or n-2 with probability o( Δ T) b) Queue is in state n at time t a) No arrivals or departure in (t, t+ Δ T] b) 1 arrival and a departure in (t, t+ Δ T] c) Any other scenario is o( Δ T) c)
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This note was uploaded on 03/18/2012 for the course ECE 861 taught by Professor Shroff during the Spring '11 term at Ohio State.

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Lectures10-14 - Paradox of Residual Life ECE-CSE 861:...

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