Question

Suppose users share a 3 Mbps link. Also suppose each user requires 150 kbps when transmitting, but each user transmits only 10 percent of the time. When circuit switching is used, how many users can be supported? For the remainder of this problem, suppose packet switching is used. Find the probability that a given user is transmitting. Suppose there are 120 users. Find the probability that at any given time, exactly n users are transmitting simultaneously. (Hint: Use the binomial distribution.) Find the probability that there are 21 or more users transmitting simultaneously.

Answer 1

Answer:

The probability that a given user is transmitting is 0.1 = 10%.

The probability that at any given time, exactly n users are transmitting simultaneously is $P(X = n) = C_{120,n}.(0.1)^{n}.(0.9)^{120-n}$.

0.0048 = 0.48% probability that there are 21 or more users transmitting simultaneously.

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$, $\sigma = \sqrt{V(X)}$.

Each user transmits only 10 percent of the time.

This means that $p = 0.1$

Find the probability that a given user is transmitting.

The probability that a given user is transmitting is 0.1 = 10%.

Suppose there are 120 users.

This means that $n = 120$

Find the probability that at any given time, exactly n users are transmitting simultaneously.

This is $P(X = n)$.

This n is different from the n of total number of users(120 in this case) from the standard binomial formula. This is the number of successes, which is the equivalent of x. So

$P(X = n) = C_{120,n}.(0.1)^{n}.(0.9)^{120-n}$

The probability that at any given time, exactly n users are transmitting simultaneously is $P(X = n) = C_{120,n}.(0.1)^{n}.(0.9)^{120-n}$

Find the probability that there are 21 or more users transmitting simultaneously.

Now we use the binomial approximation to the normal. We have that:

$\mu = E(X) = np = 120*0.1 = 12$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{120*0.1*0.9} = 3.2863$

Using continuity correction, this probability is $P(X \geq 21 - 0.5) = P(X \geq 20.5)$, which is 1 subtracted by the pvalue of Z when X = 20.5. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{20.5 - 12}{3.2863}$

$Z = 2.59$

$Z = 2.59$ has a pvalue of 0.9952

1 - 0.9952 = 0.0048

0.0048 = 0.48% probability that there are 21 or more users transmitting simultaneously.