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A large population of ALOHA users manages to generate 50 requests/sec, including both original and retransmission time is slotted in units of 40 msec. What is the change of success on the first attempt?

             A.  $e^-1$

            B.  $e^-2$

            C.  $ e^-0.5$

            D. $(1/2)*e^0.5$
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1 Answer

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Sourse: Tanenbaum

 

So, Success at first attempt = probability of zero frames = $e^{-G}$

Given, it generate 50 requests/sec i.e 50 requests in $10^{3}$ ms 

i.e in $10^{3}$ ms ${\Rightarrow }$ 50 frames generated 

So, in 40 ms ${\Rightarrow }$ $\frac{40 \times 50 }{10^{3}} = 2$ frames generated 

So G = 2

 

So, Success at first attempt  = $e^{-G}$ = $e^{-2}$

4 Comments

Okk ,, i think made easy answer is wrong
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yes probably.
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edited by

@AniMan_7 Success on the first attempt means we should let a single station to transmit; in that case the probability of x being 1 is 2 x e^-2 isn’t it? The term that you mentioned is, when no station transmit, the success probability will be e^-2.

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