A channel with capacity 150 Gbps is serving n users. Each user has 1.5
Gbps connection and is ON 10% of the time. Find the maximum number of
users that can be served if the probability of congestion (bit rate of
the n users is larger than the capacity) is to be less than 0.0014.

Hi smeski!
We can find the answer to this question using the binomial distribution.

Binomial Distribution - MathWorld
http://mathworld.wolfram.com/BinomialDistribution.html

Since each user is online 10% of the time, it's clear that, at any
given point in time, a user has a 10% chance (probability 0.1) of
being online. We can interpret each user as being a "trial" in an
experiment that has a 10% chance of succeeding (success = user is
online), and that this "trial" is repeated n (the number of users)
times. Under  this interpretation, the number of users online at any
given time follows a binomial distribution.

The binomial distribution has 2 parameters: N and p. N is the number
of times the experiment is repeated (the total number of users in this
case) and it's the value we want to find; while p is the probability
that the experiment succeeds (user is online) on each trial. So we
know that p=0.1.

Let's call X to the number of users that are online at any given time.
We know that X has a binomial distribution with parameters n unknown
and p=0.1. Since the channel has capacity for up to 100 users (150
Gbps, while each user uses 1.5 Gbps when online), we need to solve the
following equation:

Prob(X > 100) = 0.0014
(the probability of the numbers of users online at any given time
being greater than 100 -congestion- is 0.0014)

We know that N>=100, because the channel has capacity for up to 100
users. So the channel can serve at least 100 users. Now, since N is
large and p is too close to 0% or 100%, we can use the normal
distribution instead of the binomial distribution, as the normal
distribution is easier to handle when calculating probabilities.

Normal Approximation to Binomial
http://www.ruf.rice.edu/~lane/stat_sim/binom_demo.html
http://www.mis.coventry.ac.uk/~nhunt/binomial/normal.html

As a rule of a thumb, the normal approximation can be used when Np>5
and N(1-p)>5. Since N>=100, then Np>=10, so we'll use the normal
approximation. The parameters of the normal distribution are the mean
and the std. dev, so we need to know these. The mean of the binomial
distribution is N*p (N*0.1 in your case), while the variance is
N*p*(1-p) (N*0.1*0.9=N*0.09 in your case). Therefore, we can
approximate the distribution of X with a normal distribution with mean
N*0.1 and standard deviation sqrt(N*0.09) [sqrt() means square root
of]

Now, recall that the equation we had was:

Prob(X > 100) = 0.0014

Let's subtract the mean of X and divide by the std. dev.

Prob( (X-0.1N)/sqrt(0.09N) > (100-0.1N)/sqrt(0.09N) ) = 0.0014

The rationale for doing that step was the following. If we have a
normally distributed random variable, then subtract its mean and
divide the result by its std. dev., we get a standrd normally
distributed random variable. The standard normal distribution is just
the normal distribution with mean 0 and std. dev. 1. We need to this
in order to use the probability tables that are found in Statistics
books or in the following link, which gives the probabilities for the
standard normal distribution.

Distribution Tables
http://www.math.unb.ca/~knight/utility/NormTble.htm

Renaming Z the standard normal distribution we have in the left-hand
side of the equation, we have:

Prob( Z > (100-0.1N)/sqrt(0.09N) ) = 0.0014

Using the table, we find that

Prob( Z > 3 ) = 0.0013

which is the nearest value to 0.0014 we can get from the table. [If
you don't understand how to use the table, please request
clarification]

Therefore, we now have to find N from the following equation

(100-0.1N)/sqrt(0.09N) = 3

So, using basic algebra

100 - 0.1N = 3*sqrt(0.09N)
(100 - 0.1N)^2 = 9*0.09N
10000 - 20*N + 0.01*N^2 = 0.81*N
10000 - 20.81*N + 0.01*N^2 = 0

The two solutions (rounded) to this quadratic equation are N=753 and
N=1327. However, the N=1327 solution is invalid, it appeared
artificially when we squared both sides of the equation in the
previous steps. To check this, plug N=1327 in the equation:

100 - 0.1N = 3*sqrt(0.09N)

Clearly, the left-hand side will be negative, while the right-hand
side will be positive. So the answer to your question is that the
channel can serve a maximum of 753 users so that the congestion
probability will be 0.0014.

[Actually, it could hold a bit more, since with this N the probability
of congestion is actually 0.0013]


Google search terms
normal distribution table
://www.google.com/search?hl=en&lr=&q=normal+distribution+table&btnG=Search
normal approximation binomial
://www.google.com/search?hl=en&lr=&q=normal+approximation+binomial&btnG=Search


I hope this helps! If you have any questions regarding my answer,
please don't hesitate to request a clarification. Otherwise I await
your rating and final comments.

Best wishes!
elmarto

Thanks much! Makes sense!! Thanks for explaining it this way. Much
better than the book!!