Question

DO NOT COPY AND PASTE FROM ALREADY POSTED ANSWERS! those answers are WRONG answer to the problem is posted BELOW. thank you!!! X(t), t 2 0, be the infinite server queue and suppose that initially there are x customers present. Compute the mean and variance of X(t). Let Var x) - (+x- Al ANSWE R

Answer 1

The problem depicts of a $M/M/\infty$ process,

Let
X(t),t>0
, be the infinite server queue and given that initially there are customers present.

To compute the mean and variance of .

X (t)

$\lambda$ is the rate at which people arrive in the queue, according to a homogeneous Poisson process. This implies that
λ(t-s)
people are expected to arrive in any time interval from to .

The parameter $\mu$ represents how long it takes to serve people after they arrive.

Suppose that anybody who arrives at time s will be served by time
t > s
with probability
F(t, s
and therefore will still be in the queue at time t with probability,

.

S(t, s) = 1-F(t, s)

Thus we have two kinds of people in this queue: the customers who are initially present and those who arrive later.

Let be the common survival function for all those initially present.

S(t, 0)

Now, an individual who was initially present is still in the queue at time t with probability .

S(t, 0)

Because everyone is served independently, the number of these individuals
Z(t)
still in the queue thus have a
Bin x, S(t, 0))
distribution.

Any individual arriving between nearby times and (which occurs with probability $\lambda(s)ds$) has a chance of still being in the queue at time t with probability $S(t,s)\lambda(s)ds$ . Hence the number of people in the queue at time t (say
Y(t)
) equivalent to the number in the interval under the operation of an inhomogeneous Poisson process with rate

$\gamma(t)=\int_{0}^{t}S(t,s)\lambda(s)ds$..................................(1)

Thus, the entire distribution of the number of people at any time t, is the sum of the Binomial distribution with parameters
(n:P(t)) = (x, S(t, 0))
and the Poisson distribution with parameter $\gamma(t)$.

Now, the mean of X(t) is given as,

$E[X(t)]=E[Y(t)+Z(t)]=E[Y(t)]+E[Z(t)]=\gamma(t)+np(t)$.......................................(2)

Whereas the variance of X(t) is given as,

.........................................(3)

Var. ( X ( t ) )-Va,.( Y ( t ) )+ Var(Z(t)) (t)+ np(t)(1-p(t))

Since Y(t) and Z(t) are independent

Now, is assumed to follow exponential distribution with rate $\mu$

................................................... (#)

S (t, s) = e-mt-s)

Thus by putting (#) in (1) we have,

$\gamma(t)=\int_{0}^{t}e^{-\mu(t-s)}\lambda ds = \frac{\lambda}{\mu}(1-e^{-\mu t})$

Now $p(t)=S(t,0)=e^{-\mu t}$ and

(n:P(t)) = (x, S(t, 0))

Inserting above 2 expressions in (1) and (2), we get,

$E[X(T)]= \frac{\lambda}{\mu}(1- e^{-\mu t}) + xe^{-\mu t}$

$Var(X(T))=(\frac{\lambda}{\mu}+x e^{-\mu t})(1-e^{-\mu t} ).$

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