(Statistical Multiplexing) In class, we saw that packet switching can make more efficient use of resources by taking advantage of the fact that only a fraction of potential senders are active at any time. In this problem, you will be asked to demonstrate this fact mathematically. Suppose we have a single link with capacity L bits per second and a population of users that generate data at r bits per second when busy. The probability that a user is busy generating data is p. a. What is the maximum number of users that can be supported using circuit switching? Call this value MC. b. Now suppose we use packet switching to support a population of MP users. Derive a formula (in terms of p, MP, N, L, and r) for the probability that more than N users are busy. [Hint: Start by using the binomial distribution to derive a formula for the probability that exactly N users are busy.]c. Plug in some numbers. Let L = 1 Mbps, r = 64 Kbps and p = 0.1. Give the value for MC. What is the probability that more than MC users are busy for MP = 2MC? What about MP = 4MC? [Note: Feel free to use a tool such as Mathematica or Excel, or even write a small program to compute these numerical values.]Note: 1 Kbps is 1, 000 bps, not 1, 024 bps. Same rule applies to 1Mbps, 1Mbps= 106 bps. Also note that if something is specified as Megabyte (8 ∗ 106 bits), Gigabyte (8 ∗ 109 bits), etc.