Question

Think of the differences between circuit-switching and packet-switching paradigms in the Internet core design. Assume an Internet Service Provider (ISP) has an access link with 1500 Mbps capacity and, at any given time, only 1% of its customers are active. However, during that active period, users generate data at a constant rate of 25 Mbps. Assume that there is a total of 100 customers of this ISP. Please answer the following questions. Show your work.
i. What is the probability that more than 1 users will be active?
ii. What is the probability that more than 2 users will be active?

Answer 1

Answer:

0.264 ; 0.079

Explanation:

Given that:

Sample size, n = 100

Probability of being active, p = 1% = 1/100 = 0.01

Using the binomial probability relation :

P(x =x) = nCx * p^x * (1 - p)^(n - x)

Probability that more than 1 user will be active

P(x > 1) = 1 - [p(x=0) + p(x = 1)]

P(x = 0) = 100C0 * 0.01^0 * 0.99^100 = 0.366

P(x = 1) = 100C1 * 0.01^1 * 0.99^99 = 0.370

P(x > 1) = 1 - [0.366 + 0.370]

P(x > 1) = 0.264

2.)

Probability that more than 2 user will be active

P(x > 2) = 1 - [p(x=0) + p(x = 1) + p(x = 2)]

P(x = 0) = 100C0 * 0.01^0 * 0.99^100 = 0.366

P(x = 1) = 100C1 * 0.01^1 * 0.99^99 = 0.370

P(x = 2) = 100C2 * 0.01^2 * 0.99^98 = 0.185

P(x > 1) = 1 - [0.366 + 0.370 + 0.185]

P(x > 1) = 0.079