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2 Answers

1 votes
1 votes
80 is the normalized traffic in channel.(100-20).

20 is the normalized throughput(as given that 20 percent are idle )
1 votes
1 votes

Let,$G$ denote normalized order traffic load in the channel.

Given 20% slots are idle which implies frames will be successfully transmitted if we send in those 20% of idle slots.

Probability of successfull transmission =$P_{succ}$=0.2

we know , In slotted aloha probability of successfull transmission=$e^{-G}$

so ,

$e^{-G}=0.2$

$G=-\log_{e}0.2=1.609$

So, Normalized order traffic load=1.609.

Normalized Throughput =$G*P_{succ}=1.609*0.2=0.321$=32% 

Why in slotted aloha probability of successfull transmission = $e^{-G}$ ?

derivation :- https://wiki.eecs.yorku.ca/course_archive/2009-10/W/3213/_media/cse3213_14_randaccess1_w2010.pdf

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