Question

Customers arrive at a service station, manned by a single server who serves at an exponential rate \(\mu_{1}\), at a Poisson rate \(\lambda .\) After completion of service the customer then joins a second system where the server serves at an exponential rate \(\mu_{2} .\) Such a system is called a tandem or sequential queueing system. Assuming that \(\lambda<\mu_{i}\), \(i=1,2\), determine the limiting probabilities. Hint: Try a solution of the form \(P_{n, m}=C \alpha^{n} \beta^{m}\), and determine \(C, \alpha, \beta\).

Short answer

The limiting probability for a tandem queueing system with given arrival rate λ and service rates μ₁ and μ₂ is: \( P_{n, m} = \left[\left(1 - \frac{\lambda}{\mu_{2}}\right) \left(1 - \frac{\lambda}{\mu_{1}}\right)\right] \left(\frac{\lambda}{\mu_{2}}\right)^n \left(\frac{\lambda}{\mu_{1}}\right)^m \)

Step-by-step solution

Establish the balance equation

In order to find the limiting probability, we'll use a balance equation for the system. To do this, we need to consider the flow of customers in and out of each part of the system. Using the birth-death process theorem, we can establish that the balance equation is: \( \lambda P_{n-1, m} = \mu_{1} \: P_{n, m-1} + \mu_{2} \: P_{n, m} \) where \(P_{n, m}\) is the limiting probability of having n customers in the first queue and m customers in the second queue.

Set up the solution form

The exercise suggests that we try a solution of the form \(P_{n, m} = C \alpha^{n} \beta^{m}\). For the balance equation established in step 1, we plug in the suggested form: \( \lambda C \alpha^{n-1} \beta^{m} = \mu_{1} \: C \alpha^n \beta^{m-1} + \mu_{2} \: C \alpha^n \beta^m \)

Solve for α and β

We can rewrite the balance equation to isolate α and β, then divide both sides of the equation by \(\alpha^n \beta^m\): \( \frac{\lambda}{\alpha} = \mu_{1}\frac{\beta - 1}{\beta} + \mu_{2}\frac{\alpha - 1}{\alpha} \) This equation will give us the values of α and β. Using algebraic manipulation, we can find the values as follows: \( \alpha = \frac{\lambda}{\mu_{2}}, \quad \beta = \frac{\lambda}{\mu_{1}} \)

Solve for C

To find out the value of C, we need to incorporate the fact that in the limiting case, the total probability (that is, the sum of probabilities of all states in the queue) must equal to 1: \( \sum_{n=0}^{\infty} \sum_{m=0}^{\infty} C \alpha^n \beta^m = 1 \) This equation is a double geometric series, and thus can be solved as follows: \( C \frac{1}{(1 - \alpha)(1 - \beta)} = 1 \) By substituting α and β in terms of λ, μ₁, and μ₂, the value of C is: \( C = \left(1 - \frac{\lambda}{\mu_{2}}\right) \left(1 - \frac{\lambda}{\mu_{1}}\right) \)

Finalize the solution

Now we have determined the values of C, α, and β. The limiting probability is given by: \( P_{n, m} = \left[\left(1 - \frac{\lambda}{\mu_{2}}\right) \left(1 - \frac{\lambda}{\mu_{1}}\right)\right] \left(\frac{\lambda}{\mu_{2}}\right)^n \left(\frac{\lambda}{\mu_{1}}\right)^m \) This is the final solution for the limiting probabilities of a tandem queueing system with given arrival rate λ and service rates μ₁ and μ₂.