Question

Problem 3 Suppose users share a 9 Mbps link. Each user consumes 1.5 Mbps when transmitting, and only transmits with probability 0.2 How many users can be supported using circuit switching? Assume packet switching for the rest of the problem. What is the probability that 7 out of 10 users are transmitting? What is the maximum number of users, such that the probability of exceeding the link capacity is no more than 0.01? Hint: Gradually increase the number of users and evaluate the probability of exceeding the link capacity for each value of this number. a) b) c)

Answer 1

Answer:

We have given,

Users share link of 9Mbps, each user consumes 1.5Mbps, probability = 0.2

a. The number of users = 9Mbps/1.5Mbps = 6

Thus the number of users can be supported = 6 users

b. The probability that 7 out of 10 users are transmitting

= nCr P^r * (1-P)^(n - r) =  10C7 P^7 * (1-P)^(10 -7) = 10C7P^7*(1-P)^3

= 10! / (10-7)! *7! * (0.2)^7 *(1-0.2)^3

= 10! /3! *7! * (0.0000128) * (0.512)

10*9*8*7!/ 3! *7! *(0.0000128) * (0.512)

= 10*9*8/ 3*2 * (0.0000128) * (0.512)

= 720/6 *(0.0000128) * (0.512)= 120*(0.0000128) * (0.512)  

= 0.0007864

Hence the probability that 7 out of 10 users are transmitting = 0.0007864

c. We already got the values of P , (1 - P) and n . Now we will find the value of r ( number of users for which probability must be less than 0.1

That is , 10Cr P^r * (1-P)^10 - r < = 0.1

Now increase the value of r ( number of users ) from 1 to 10:

10Cr P^r * (1 - P)^10 - r

Now , r = 1 ---> 10C1P^1 * ( 1 -P)^10 - 1 = 10C1 (0.2)^1 * ( 0.8)^10 - 1 = 10C1 *(0.2) * (0.8)^9 = 0.2

r = 2 ----> 10C2P^2 ( 1 - P)^10-2 = 10C2c* (0.2)^2 *( 0.8)^10 - 2 = 10C2* (0.2)^2 * (0.8)^8 = 0.3

r = 3 ----> 10C3 P^3 * (1 - P)^10 - 3 = 10C3 * (0.2)^3 * (0.8)^10 -3 = 10C3 *(0.2)^3 *(0.8)^7 = 0.2

.....................

r = 6 -----> 10C6P^6 * (1-P)^10-6 = 10C6 *(0.2)^6 * ( 0.8)^10 -6 = 10C6 * (0.2)^6 * (0.8)^4 = 0.003 ( this is less than 0.01)

Thus the maximum users = 6 users

Rate the answer it it helps else let me know your doubts.

Thank you!!!