Question

Consider an infinite server queuing system in which customers arrive in accordance with a Poisson process with rate \(\lambda\), and where the service distribution is exponential with rate \(\mu\). Let \(X(t)\) denote the number of customers in the system at time \(t\). Find (a) \(E[X(t+s) \mid X(s)=n] ;\) (b) \(\operatorname{Var}[X(t+s) \mid X(s)=n]\). Hint: Divide the customers in the system at time \(t+s\) into two groups, one consisting of "old" customers and the other of "new" customers. (c) Consider an infinite server queuing system in which customers arrive according to a Poisson process with rate \(\lambda\), and where the service times are all exponential random variables with rate \(\mu .\) If there is currently a single customer in the system, find the probability that the system becomes empty when that customer departs.

Short answer

The expected number of customers in the system at time \(t+s\) given that there are \(n\) customers at time \(s\) is: \[ E[X(t+s) \mid X(s)=n] = ne^{-\mu t} + \lambda t\] The variance of the number of customers in the system at time \(t+s\) given that there are \(n\) customers at time \(s\) is: \[\operatorname{Var}[X(t+s) \mid X(s)=n] = ne^{-\mu t} - n^2e^{-2\mu t} + \lambda t\] To find the probability that the system becomes empty when a single customer departs, it is necessary that \(E[Y(t)]\geq 0\), which gives us the inequality \(e^{-\mu t}+\lambda t\geq 1\). The specific solution will depend on the values of \(\lambda\) and \(\mu\).

Step-by-step solution

Define Known Quantities

We have the following quantities: - Poisson arrival process with rate \(\lambda\) - Exponential service distribution with rate \(\mu\) - \(X(t)\), the number of customers in the system at time \(t\) Let \(N(t)\) be the number of new customers arriving to the system during the time interval \((s, t+s]\). This is a Poisson random variable with mean \(\lambda t\). Let \(M(t)\) be the number of old customers remaining in the system at \(t+s\). Now, we will find the mean and variance for both \(M(t)\) and \(N(t)\).

Find mean and variance of new customers, \(N(t)\)

Since \(N(t)\) is a Poisson random variable, its mean and variance can be found easily. Mean of \(N(t)\): \[ E[N(t)] = \lambda t \] Variance of \(N(t)\): \[ Var[N(t)] = \lambda t \]

Find mean and variance of old customers, \(M(t)\)

Now we need to find the mean and variance of old customers still in the system at time \(t+s\). For an individual customer, the probability of still being in the system is given by the survival function of the exponential distribution, which is: \(S(t) = e^{-\mu t}\). Mean of \(M(t)\): \[ E[M(t)] = nS(t) = ne^{-\mu t} \] For the variance, we need to find \(E[M(t)^2]\) first: \[E[M(t)^2] = \sum_{k=0}^n {n\choose k} S(t)^k (1-S(t))^{n-k} k^2 = n(ne^{-\mu t} + e^{-2\mu t} - ne^{-\mu t})\] Now we can find the variance: Variance of \(M(t)\): \[\operatorname{Var}[M(t)] = E[M(t)^2] - [E[M(t)]]^2 = ne^{-\mu t} - n^2e^{-2\mu t}\]

Find expectation and variance of \(X(t+s)\) given \(X(s)=n\)

Since the number of old customers and new customers are independent, we can now find the expected value and variance of \(X(t+s)\) given \(X(s)=n\): (a) Expectation: \[ E[X(t+s) \mid X(s)=n] = E[M(t)] + E[N(t)] = ne^{-\mu t} + \lambda t\] (b) Variance: \[\operatorname{Var}[X(t+s) \mid X(s)=n] = \operatorname{Var}[M(t)] + \operatorname{Var}[N(t)] = ne^{-\mu t} - n^2e^{-2\mu t} + \lambda t\]

Find the probability that the system becomes empty with a single customer

Now, let's find the probability that the system becomes empty when a single customer departs. Let \(Y(t)\) denote the number of customers in the system when a single customer departs, then the probability we need to find is \(P(Y(t)=0)\). According to the previous steps, \(E[Y(t)]=e^{-\mu t}+\lambda t-1\). We know the number of customers has to be an integer, so the expected number of customers in the system should be a non-negative integer. Therefore, for \(P(Y(t)=0)\) to be positive, it is necessary that \(E[Y(t)]\geq 0\), which gives us the inequality \(e^{-\mu t}+\lambda t\geq 1\). To find a solution for this inequality, we can try different methods, such as graphical or numerical methods. The specific solution will depend on the values of \(\lambda\) and \(\mu\).