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A large population of ALOHA users manages to generate 50 requests/sec, including
both originals and retransmissions. Time is slotted in units of 40 msec.
a. (a) What is the chance of success on the first attempt?
b. (b) What is the probability of exactly k collisions and then a success?
c. (c) What is the expected number of transmission attempts needed?
For the a) Chance of success ,it shud be 1/e or 1/e^2  ??

+1 vote

We know , here Poisson distribution is followed .So

P(k)   =   e-G . (G)k / k!   where G is the number of frames generated in one frame transmission time(or frame slot time)

Here to find G we have to find number of frames that are generated in 1 slot time which is given as 40 ms.

Given , in  1000 ms  no of frames generated   =   50

So , no of frames generated in 40 ms       =   (50 / 1000) * 40

=   2

Hence the value of G obtained                  =   2

Now we solve the given parts one by one.

Solution to part a) :

P(success at first attempt)         =      P(0)

=      e-G

=      e-2

Solution to part b) :

P(success after k collisions)      =    [ P(failure) ]k  P(success)

=    (1 - e-G)k e-G

=    (1 - e-2)k e-2

Solution to part c) :

Expected Number of attempts    =   1 . P(1)  + 2 . P(2) ............... infinite terms

=    Σ k . P(k)  [ Where P(k) is the probability of success at kth attempt ]

=    Σ k . e-G (1 - e-G)k-1

This is an infinite arithmetico geometric progression.So finding this sum , we get :

Number of attempts(expected or mean number)    =   eG

=   e2

=   7.39

= 8 [As number of attempts should be an integer and hence we need to round it to higher integer in this case ]

answered by Veteran (89k points) 15 58 294
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Thank you so much @Habibkhan Sir

This is for slotted aloha right?? What wud be the Poisson distribution formula for Pure aloha??
For pure ALOHA we take -2G instead of G as vulnerable time there is twice of slotted ALOHA..Vulnerable time means time in which collision of frames may occur which is the twice the slot time for pure ALOHA..
Thanku so much sir