# Queueing Systems

This exam covers customer queuing systems including arrival rates, Poisson processes, probability density functions, steady states, operator utilisation, machine efficiency, and exponentially distributed service times.

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#### Solution Preview

a) This question can be solved using Bayes' Theorem. The theorem states that, given f(x|y) (the conditional density function of x given y) and the density function f(y),

inf.

f(x) = integral f(x|y)f(y)dy

y=0

Let's see how to use this theorem to solve this problem. We are asked to find the probability function of the number of customers that arrive during an interval t of time, where t is a random variable with density function f(t). Since customers arrive according to a Poisson process, we have that, given t,

P(An = k | t) = (lambda*t)^k * exp(-lambda*t)

-------------------------------------

k!

(that's the Poisson probability function) This would be the f(x|y) in Bayes' Theorem. Since we know that t follows the distribution f(t) then, using the theorem, we arrive at:

inf.

P(An = k) = integral (lambda*t)^k * exp(-lambda*t)

t=0 -------------------------------------- f(t)dt

k!

We can take lambda^k and k! out of the integral (since they don't depend on t) to get:

inf.

P(An = k) = lambda^k * integral t^k * exp(-lambda*t) * f(t)dt

------------ t=0

k!

That's the probability function we were looking for.

The second part of this question can be easily solved using the conditional probability theorem that states that:

E(X) = E[ E(X|Y) ]

Again, given the interval of time t, we have that:

E(An | t) = lambda*t

This is simply the expected value of the Poisson process given t. So, using the theorem:

E(An) = E[ E(An | t) ] = E(lambda*t) = lambda*E(t)

(since lambda is a constant, it can be ...

#### Solution Summary

This solution set includes complete calculations and explanations. 900 words.