Duplicate: https://gateoverflow.in/158532/Tanenbaum

Dark Mode

ajaysoni1924
asked
in Computer Networks
Mar 15, 2019

1,109 views
0 votes

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) 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?

both originals and retransmissions. Time is slotted in units of 40 msec.

(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?

0 votes

in 1 sec 50 requests are generated

In 40 ms(1 time slot) max number of requests that could be generated=$\frac{50*40}{1000}=2$

$\therefore$ Expected number of frames or mean=G=2 frames per time slot

Probability of successful transmission is given by Poisson distribution

$P_{r}=\frac{G^r*e^{-G}}{r!}$

a. Success on first attempt is achieved when no other frames is transmitted in the same time slot

$P_{0}=\frac{2^0*e^{-2}}{0!}$

$=e^{-2}$

b. Probability of unsuccessful transmission=$1-e^{-G}=1-e^{-2}$

Probability that there are k unsuccessful transmissions followed by a successful transmission=$(1-e^{-G})^k*e^{-G}=(1-e^{-2})^k*e^{-2}$

c. Expected number of transmissions

$\sum_{n=1}^{infinity}n*P_{n}=\sum_{n=1}^{infinity}n*(1-e^{-G})^{n-1}*e^{-G}$

$=e^{G}=e^2$

In 40 ms(1 time slot) max number of requests that could be generated=$\frac{50*40}{1000}=2$

$\therefore$ Expected number of frames or mean=G=2 frames per time slot

Probability of successful transmission is given by Poisson distribution

$P_{r}=\frac{G^r*e^{-G}}{r!}$

a. Success on first attempt is achieved when no other frames is transmitted in the same time slot

$P_{0}=\frac{2^0*e^{-2}}{0!}$

$=e^{-2}$

b. Probability of unsuccessful transmission=$1-e^{-G}=1-e^{-2}$

Probability that there are k unsuccessful transmissions followed by a successful transmission=$(1-e^{-G})^k*e^{-G}=(1-e^{-2})^k*e^{-2}$

c. Expected number of transmissions

$\sum_{n=1}^{infinity}n*P_{n}=\sum_{n=1}^{infinity}n*(1-e^{-G})^{n-1}*e^{-G}$

$=e^{G}=e^2$