Question

Consider a two-server parallel queuing system where customers arrive according to a Poisson process with rate \(\lambda\), and where the service times are exponential with rate \mu. Moreover, suppose that arrivals finding both servers busy immediately depart without receiving any service (such a customer is said to be lost), whereas those finding at least one free server immediately enter service and then depart when their service is completed. (a) If both servers are presently busy, find the expected time until the next customer enters the system. (b) Starting empty, find the expected time until both servers are busy. (c) Find the expected time between two successive lost customers.

Short answer

In summary, the expected time until the next customer enters the system when both servers are busy is \( \frac{1}{\mu} \), the expected time until both servers are busy starting from an empty system is \( \frac{2}{\lambda} \), and the expected time between two successive lost customers is \( \frac{2}{\lambda} + \frac{3}{2\mu} \).

Step-by-step solution

Set up the relevant probabilities and rates

Let's first define the relevant parameters: - Arrival rate (λ) - Service rate (μ) Now let's set up some of the probabilities and rates related to the parallel queuing system: 1. Probability that a server is available (P_available): This is the probability that there is at least one server available to provide service. 2. Probability that a server becomes available (P_next_service): This is the probability that the service of a customer is completed and a server becomes available for another customer.

(a) Calculate the expected time until the next customer enters the system when both servers are busy

When both servers are busy, the Poisson process and exponential distribution properties can help us determine the expected time until the next customer enters the system. - For a server to become available, at least one of them has to complete its service. - Let Z be the random variable representing the time until the next available server. We can calculate Z as the minimum of the service times X_1 and X_2 for the two servers: Z = min(X_1, X_2) Since X_1 and X_2 are independent exponential random variables with rate μ, their density functions are given by: \( f_{X_1}(x) = \mu e^{-\mu x} \) and \( f_{X_2}(x) = \mu e^{-\mu x} \) The probability density function for Z is given by: \( f_Z(z) = \mu^2 e^{-\mu z} (1 - e^{-\mu z}) \) The expected time until the next customer enters the system is then given by the expected value of Z: \( E[Z] = \int_0^{\infty} z f_Z(z) dz = \frac{1}{\mu} \)

(b) Calculate the expected time until both servers are busy starting empty

We need two customers to arrive and enter the system to make both servers busy. Waiting times between consecutive arrivals in a Poisson process (with rate λ) are also exponentially distributed with rate λ. Let Y_1 and Y_2 be the waiting times between the first and second customers, and the second and third customers, respectively. We calculate the expected value of Y_1 and Y_2 separately. \( E[Y_1] = \frac{1}{\lambda} \) and \( E[Y_2] = \frac{1}{\lambda} \) Since Y_1 and Y_2 are independent, the expected time until both servers are busy is the sum of their expected values: \( E[Y_1+Y_2] = E[Y_1] + E[Y_2] = \frac{1}{\lambda} + \frac{1}{\lambda} = \frac{2}{\lambda} \)

(c) Calculate the expected time between two successive lost customers

A lost customer occurs when both servers are busy and a new customer arrives. We evaluate the expected time between two successive lost customers as the sum of these three components: 1. The expected waiting time until both servers are busy starting from an empty system (calculated in step 3): \( \frac{2}{\lambda} \) 2. The expected time until the next customer enters the system when both servers are busy (calculated in step 2): \( \frac{1}{\mu} \) 3. The expected time until the customer is served once available: \( \frac{1}{2\mu} \) Adding these components, we get the expected time between two successive lost customers: \( \frac{2}{\lambda} + \frac{1}{\mu} + \frac{1}{2\mu} = \frac{2}{\lambda} + \frac{3}{2\mu} \)