Question

Suppose a network access link of 1 Mbps capacity is being shared among 10 users through packet-switching and each user has 20% probability of using the link at any time. Assuming the link bandwidth is shared equally among its active users, what is the probability (in %) that each user is experiencing a bandwidth equal to or above 300 Kbps (Kilobits per second)?

Answer 1

Let the maximum active users allowed be n, for the required bandwidth requirements.

Then, we have :

Thus, the maximum number of users allowed are 1, 2, 3.

Now,

• probability that a user is active, p = 0.20
• probability that a user is not active = 1 - 0.20 = 0.80

Also, we consider each case 1 by 1.

1 user is online, and 9 users are offline :

We can select 1 out of 10 users, in ¹⁰C₁ ways.

Thus, the probability for this case

= ¹⁰C₁ * p * (1 - p)⁹

= 0.2684

2 users are online, and 8 users are offline :

We can select 2 out of 10 users, in ¹⁰C₂ ways.

Thus, the probability for this case

= ¹⁰C₂ * p² * (1 - p)⁸

= 0.3020

3 users are online, and 7 users are offline :

We can select 3 out of 10 users, in ¹⁰C₃ ways.

Thus, the probability for this case

= ¹⁰C₃ * p³ * (1 - p)⁷

= 0.2013

Thus, total probability

= 0.2684 + 0.3020 + .2013

= 0.7717

= 77.17 %