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 = [tex]e^{-G}G^k[/tex] but k = 0, therefore Probability that there is success on the first attempt = [tex]e^{-G}=e^{-2}=0.135[/tex]
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 = [tex][1-e^{-G}]^k=[1-0.135]^k=0.865^k[/tex]
P(success in k+1 attempt) = [tex]e^{-G}=e^{-2}=0.135[/tex]
Probability of exactly k collisions and then a success = [tex]0.865^k0.135[/tex]
c) Expected number of transmission attempts needed = probability of success in k transmission = [tex]e^{G}=e^{2}=7.389[/tex]
