Question

Suppose users share a 1 Mbps link. Also suppose each user

requires100

kbps when transmitting, but each user transmits only 10 percent

ofthe time.

(See the discussion Packet Switching Versus Circuit Switching

inSection 1.3).

A. When circuit switching is used, how many users can

besupported?

B. For the remainder of this problem, suppose packetswitching is

used.

Find the probability that a given user is transmitting.

C. (Suppose that there are 50 users. Find the probability that

atany given

time exactly n users are transmitting simultaneously (Hint: use

thebinomial

distribution).

D. Find the probability that there are 11 or more

userstransmitting

simultaneously. (Hint: use the central limit theorem.)

E. Solve D for 20 or more users.

Answer 1

The answer is below

Step-by-step explanation:

A) The total bandwidth = 1 Mbps = 1000 Kbps

The requirement for each user = 100 Kbps

Total number of users = Total bandwidth / requirement for each user

Total number of users = 1000 Kbps / 100 Kbps = 10 users

B) Since each user transmits only 10 percent of the time, probability that a given user is transmitting is:

Probability = 0.1

C) Binomial distribution formula is given by:

`P(x)= C(n,x)p^x(1-p)^(n-x)\\\\n = number\ of \ trials,x=number\ of\ success, p=probability\ of\ success \ for\ one\ trial`Hence:

Probability=
`(50\ choose\ n)p^n(1-p)^(50-n)`

`P=C(50,n)p^n(1-p)^(50-n)`

D) The probability that there are 11 or more is:

`Probability=1-\Sigma\limits^(20)_0 {(50\ choose\ n)p^n(1-p)^(50-n)}`