sol5 - − = − = − = − − 1 ) 1 1 ( lim = − ∞...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
EE3015 Tutorial 5 (Solution) Question 1 Max. Throughput = 18.4% (see lecture notes) = 10.304 kbps = 5.152 frames/second Question 2 dR Q QR d R Q Q + = + / The maximum throughput is Question 3 a) 1 ) 1 ( ) ( = N p Np p E 2 1 ) 1 )( 1 ( ) 1 ( ) ( ' = N N p N Np p N p E ) 1 ( ) 1 ( )) 1 ( ) 1 (( ) 1 ( 2 2 pN p N N p p p N N N = = N p p E 1 * 0 ) ( ' = = The case where p = 1 should be rejected, since when p = 1, the throughput is equal to zero. You should also check that the point is indeed a maximum. b) N N N N N N p E N N N 1 1 ) 1 1 ( ) 1 1 ( ) 1 1 ( 1 *) ( 1 1
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: − = − = − = − − 1 ) 1 1 ( lim = − ∞ → N N e N N N 1 ) 1 1 ( lim = − ∞ → Thus e p E N 1 *) ( lim = ∞ → . c) Slotted ALOHA has a higher efficiency because its probability of collision is lower. (Details can be found in the lecture notes.) Question 4 Throughput of A = 0.6 (1-0.3) = 0.42 Throughput of B = 0.3 (1-0.6) = 0.12...
View Full Document

This note was uploaded on 05/01/2011 for the course ELECTRICAL EE3015 taught by Professor Albertsun during the Spring '11 term at City University of Hong Kong.

Ask a homework question - tutors are online