Lecture20

Lecture20 - 1 I I t E k p t = = = = Purdue University...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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 = =− = =
Background image of page 2
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 ()
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 1 I I t E k p t = = = = Purdue University Compare the schemes Compare the schemes Let 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 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 Identify regimes in which each scheme is the best Selective repeat is best in terms of throughput Selective repeat is most complex to implement...
View Full Document

Page1 / 5

Lecture20 - 1 I I t E k p t = = = = Purdue University...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online