Question

A group of \(n\) customers moves around among two servers. Upon completion of service, the served customer then joins the queue (or enters service if the server is free) at the other server. All service times are exponential with rate \(\mu .\) Find the proportion of time that there are \(j\) customers at server \(1, j=0, \ldots, n\).

Short answer

The proportion of time that there are \(j\) customers at server \(1\) for \(j = 0, 1, \ldots, n\) is given by: \[P_j = \begin{cases} \frac{2}{n+2}, &\text{if } j \text{ is even} \\ \frac{2}{n+2} \left(1 + \frac{n}{2}\right), &\text{if } j \text{ is odd} \end{cases}\]

Step-by-step solution

Understand the queueing model and the exponential distribution

In this problem, there are two servers, and \(n\) customers move back and forth between the servers. After being served by a server, a customer either enters service at the other server or joins the queue if the other server is busy. We are also told that service times at both servers follow an exponential distribution with parameter \(\mu\). The exponential distribution is a continuous probability distribution that describes the time between events in a Poisson point process. In this case, it describes the time a server spends serving a customer. The exponential distribution can be represented as \(f(t) = \mu e^{-\mu t}\), where \(t\) is the time and \(\mu\) is the rate parameter.

Set up the balance equation

To find the proportion of time that there are \(j\) customers at server \(1\), we can compute the stationary distribution of customers in the system. The stationary distribution is the probability distribution of the system’s state when time goes to infinity. Based on the given scenario, the balance equations can be defined as: For \(j=0\): \[P_0 \mu + P_{n-1} \mu = P_1 \mu\] For \(1 \leq j \leq n-1\): \[P_{j-1} \mu + P_{j+1} \mu = P_j (\mu + \mu)\] For \(j=1\), specifically: \[P_0 \mu + P_2 \mu = P_1 2\mu\]

Solve the balance equation

Solve the balance equations to find the probability, \(P_j\), of having \(j\) customers at server \(1\). From the equation for \(j=0\), we have: \[P_1 = P_0 + \frac{P_{n-1}}{2}\] From the equation for \(1 \leq j \leq n-1\), we observe that for any \(j\), \(P_{j+1} = P_{j-1}\). This means that for every non-negative integer \(k \leq n-2\), we have \[P_{2k+1} = P_{2k-1}\] Based on this, we can compute \(P_j\) for any \(j\) from \(P_0\): \[P_j = \begin{cases} P_0, &\text{if } j \text{ is even} \\ P_0 + \frac{P_{n-1}}{2}, &\text{if } j \text{ is odd} \end{cases}\] Notice that since \(P_0 + P_1 + \ldots + P_n = 1\) and \(P_{2k+1} = P_{2k-1}\) for every \(k\), \[P_0 \left(1 + \frac12 + \frac12 + \ldots + \frac12 \right) + \frac{P_{n-1}}{2} \left( 1 + 1 + \ldots + 1 \right) = 1\] \[P_0 \left( 1 + \frac{n}{2} \right) + \frac{P_{n-1}}{2} \left( \lfloor \frac{n+1}{2}\rfloor \right) = 1\] Now, we can compute \(P_0\) and \(P_{n-1}\): \[P_{0} = \frac{2}{n+2}\] \[P_{n-1} = 2P_0\] Finally, we can compute \(P_j\) as: \[P_j = \begin{cases} \frac{2}{n+2}, &\text{if } j \text{ is even} \\ \frac{2}{n+2} \left(1 + \frac{n}{2}\right), &\text{if } j \text{ is odd} \end{cases}\] These probabilities represent the proportion of time that there are \(j\) customers at server \(1\) for \(j = 0, 1, \ldots, n\).

Key concepts

Exponential Distribution

The exponential distribution is a staple concept in queueing theory, dealing with the time interval between successive events in a memoryless process. To put it simply, it’s used to model scenarios where events occur continuously and independently at a constant average rate.

Let’s consider the queueing problem at hand. The service times of customers at both servers are described by an exponential distribution with parameter \(\mu\). This parameter \(\mu\) represents the rate at which the servers work – or how many customers they serve in a unit of time, on average. Mathematically, the probability density function of an exponential distribution is denoted as \(f(t) = \mu e^{-\mu t}\), where \(t\) is the time.

In practical terms, if you know a server serves on average 2 customers per hour, the rate \(\mu\) would be 2. The memoryless property of the exponential distribution means that the likelihood of the server finishing with a customer is the same, no matter how long the service has already taken – reinforcing the idea that each customer's service time is independent and random.

Stationary Distribution

The concept of a stationary distribution is crucial in understanding the long-term behaviour of a queuing system. It represents the probability distribution of the number of customers in the system after it has been operating for a significant amount of time, such that the probabilities no longer change.

In regards to our example with two servers and \(n\) customers, we're interested in finding the stationary distribution for the number of customers at server 1. This stationary distribution will be able to tell us the proportions of time the server will have \(j\) customers. It's like taking a snapshot of the system at a random time in the future and seeing where customers are most likely to be.

To achieve this, we use the balance equations. They help us ensure that the rate at which customers enter a state (like being served by server 1) equals the rate at which they leave it, ultimately leading to the system finding an equilibrium. This equilibrium, or stationary state, provides us with the sought-after stationary distribution.

Balance Equations

The balance equations play a pivotal role in determining the stationary distribution in queuing systems. They are founded on the principle of conservation of flow in and out of a system's states, ensuring that in the long run, there is a balance. This balance is why the probabilities stay constant over time in a stationary distribution.

As we dissect the problem, we establish relationships between the probabilities of having different numbers of customers (\(j\)) at server 1. Crucially, the system has to be stable, which means, the rate of customers being served (exiting the queue) must equal the rate of customers arriving (entering the queue). These rates are captured by the balance equations set up in Step 2 of the solution.

For instance, the balance equation for when \(j=0\) customers are at server 1 translates to saying the rate at which customers leave state 0 to state 1 must be the same as when they go from having \(n-1\) customers to having no customers. Solving these equations, as demonstrated in the step-by-step solution, gives us the steady-state probabilities (\(P_0, P_1, ..., P_n\)), which reveal the average long-term behaviour of the queuing system.