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kotegsom [21]
3 years ago
5

A large population of ALOHA users manages to generate 50 requests/sec, including both originals and retransmissions. Time is slo

tted in units of 40 msec.
Required:
a. What is the chance of success on the first attempt?
b. What is the probability of exactly k collisions and then a success?
c. What is the expected number of transmission attempts needed?
Computers and Technology
1 answer:
krok68 [10]3 years ago
6 0

Answer:

The answer is below

Explanation:

Given that:

Frame transmission time (X) = 40 ms

Requests = 50 requests/sec, Therefore the arrival rate for frame (G) = 50 request * 40 ms = 2 request

a) Probability that there is success on the first attempt = e^{-G}G^k but k = 0, therefore Probability that there is success on the first attempt = e^{-G}=e^{-2}=0.135

b) probability of exactly k collisions and then a success = P(collisions in k attempts) × P(success in k+1 attempt)

P(collisions in k attempts) = [1-Probability that there is success on the first attempt]^k = [1-e^{-G}]^k=[1-0.135]^k=0.865^k

P(success in k+1 attempt) = e^{-G}=e^{-2}=0.135

Probability of exactly k collisions and then a success = 0.865^k0.135

c) Expected number of transmission attempts needed = probability of success in k transmission = e^{G}=e^{2}=7.389

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