Lecture20

Lecture20 - 1 I I t E k p t = ∴ = = − = Purdue...

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Purdue University ECE ECE - - 547 547 Introduction to Computer Introduction to Computer Communication Networks Communication Networks Instructor: Instructor: Xiaojun Xiaojun Lin Lin Lecture 20 Lecture 20
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Purdue University Max. Throughput analysis of Max. Throughput analysis of selective repeat selective repeat ¾ Some assumption as in stop and wait and go back n: ¾ Infinite reordering buffer at receiver ¾ p = P{receiving a frame in error} ¾ P{successful transmission in first attempt} = 1-p ¾ P{successful transmission in nth attempt} = p n-1 (1-p) ¾ Let k = # of transmissions required for a successful transmission 1 1 1 () ( 1 ) 1 tranmission time of one frame n n I Ek np p p t = =− = =
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Purdue University Max. throughput analysis of Max. throughput analysis of selective repeat selective repeat I max 1 Saturation: # frames/ unit time being transmitted t Ave # of packets/unit time Ave # of tranmissions required 1 ()
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Unformatted text preview: 1 I I t E k p t λ = ∴ = = − = Purdue University Compare the schemes Compare the schemes ¾ Let ¾ Throughput comparison( normalized by t I ) Stop and wait 1-p a Go back n 1-p 1+(a-1)p Selective repeat (1-p) 1 > ∴ + = = a t t t t t a I I out I T Purdue University Compare the schemes Compare the schemes ¾ Note: 1. If a is small (ie. a ~ 1), all schemes are basically the same. This means transmission time dominates propagation delay. 2. If a is large but p is very small such that ap << 1 i. Stop and wait is bad but, ii. Go back n and selective repeat have comparable performance 3. If ap ~ 1 or ap > 1, then selective repeat is much better than go back n ¾ Identify regimes in which each scheme is the best ¾ Selective repeat is best in terms of throughput ¾ Selective repeat is most complex to implement...
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This note was uploaded on 01/06/2010 for the course ECE 547 taught by Professor Xiaojunlin during the Spring '09 term at Purdue.

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Lecture20 - 1 I I t E k p t = ∴ = = − = Purdue...

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