notes5_EE549_Spring08mod

notes5_EE549_Spring08mod - UNIVERSITY OF SOUTHERN...

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UNIVERSITY OF SOUTHERN CALIFORNIA, SPRING 2008 1 Lecture Notes 5 EE 549 — Queueing Theory Instructor: Michael Neely I. BACKLOG AND DELAY BOUNDS FOR LEAKY BUCKET ( r, σ ) INPUTS Recall that X ( t ) ( r, σ ) if X [ t 1 , t 2 ] r ( t 2 - t 1 ) + σ for all intervals [ t 1 , t 2 ] . Consider a single server work conserving queue, and assume: X ( t ) ( r, σ ) input stream single server queue constant server rate μ initially empty system ( U (0) = 0) Fact 1: (Queue Bound) If r μ , then U ( t ) σ for all time t . Proof: Fix a particular time t . We want to show that U ( t ) σ . If U ( t ) = 0 , then we are done. Else, U ( t ) > 0 ¡ and so we are in a busy period that started at some time t b , where t b t . Thus: U ( t ) = X [ t b , t ] - μ · ( t - t b ) r ( t - t b ) + σ - μ · ( t - t b ) σ Fig. 1. An illustration of the unfinished work bound for leaky bucket inputs. Fig. 1 illustrates the bound on unfinished work. Corollary 1: (Delay Bound) If service is FIFO then delay σ μ . Proof: Take any packet i . Let t i be the time of the packet arrival, and let U ( t i ) be the unfinished work at this time (which includes all bits that were in the queue just before the packet arrived, plus the total number of bits in packet i ). Then under FIFO, the delay of packet i is exactly U ( t i ) , which is less than or equal to σ/μ . Fact 2: (Summing Leaky Bucket Inputs) If X 1 ( t ) ( r 1 , σ 1 ) and X 2 ( t ) ( r 2 , σ 2 ) , then X 1 ( t ) + X 2 ( t ) ( r 1 + r 2 , σ 1 + σ 2 ) . Proof: X [ t, t + T ] = X 1 [ t, t + T ] + X 2 [ t, t + T ] ( r 1 T + σ 1 ) + ( r 2 T + σ 2 ) = ( r 1 + r 2 ) T + ( σ 1 + σ 2 )
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UNIVERSITY OF SOUTHERN CALIFORNIA, SPRING 2008 2 Fact 3: (Input-Output Parameter Invariance) Consider a leaky bucket input stream X ( t ) with parameters ( r, σ ) entering a single server work conserving queue with constant processing rate μ bits/sec. Assume r μ , and let Y ( t ) represent the departure process. Then Y ( t ) is leaky bucket with parameters ( r, σ ) . Proof: We want to show that for any time t and any interval duration T 0 , we have: Y [ t, t + T ] rT + σ To show this, fix any time t and any T 0 , and consider the interval [ t, t + T ] . Note that the maximum amount of bits that depart over this interval is no more than U ( t ) plus the total new bits that enter during ( t, t + T ] : Y [ t, t + T ] U ( t ) + X ( t, t + T ] (1) We now have two cases: Case 1: U ( t ) = 0 . In this case, we have from (1): Y [ t, t + T ] 0 + X ( t, t + T ] rT + σ and we are done. Case 2: U ( t ) > 0 . In this case, there is a time t b t (at which the current busy period started) such that: U ( t ) = X [ t b , t ] - μ · ( t - t b ) (2) Using (1) and (2) Y [ t, t + T ] X [ t b , t ] - μ · ( t - t b ) + X ( t, t + T ] = X [ t b , t + T ] - μ · ( t - t b ) r ( t + T - t b ) + σ - μ · ( t - t b ) = rT + σ - ( μ - r ) · ( t - t b ) rT
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notes5_EE549_Spring08mod - UNIVERSITY OF SOUTHERN...

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