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Unformatted text preview: 1 ECE & CSE ECECSE 861: Introduction to Computer Communication Networks Ness B. Shroff ECE & CSE Lecture 11 ECE & CSE Paradox of Residual Life Formally: Let {N(t),t 0} be a Poisson Process with rate Let S 1 , S 2 , be successive arrival times, and S 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 uN(s), s t} = 1e 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/ . Proof: For u 0: {Y t >u} = {S N(t)+1t > 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 u , u 0 P {Y t u  N(s), s t} = 1  e 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/ as t 2 ECE & CSE Reminder: Queueing Notation Notation: Infinite Buffer: A/B/C old notation A/B/C/ Single server Multiserver queue Interarrival 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....
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 Winter '11
 shroff

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