Question

Consider a single-server exponential system in which ordinary customers arrive at a rate \(\lambda\) and have service rate \(\mu .\) In addition, there is a special customer who has a service rate \(\mu_{1}\). Whenever this special customer arrives, she goes directly into service (if anyone else is in service, then this person is bumped back into queue). When the special customer is not being serviced, she spends an exponential amount of time (with mean \(1 / \theta\) ) out of the system. (a) What is the average arrival rate of the special customer? (b) Define an appropriate state space and set up balance equations. (c) Find the probability that an ordinary customer is bumped \(n\) times.

Short answer

The probability of an ordinary customer being bumped \(n\) times in this single-server exponential system is given by: \(P_{n} = \left(\frac{\mu_1(\lambda+\theta)}{(\mu+\mu_1)(\mu+\theta)}\right)^{n}\)

Step-by-step solution

(Step 1: Calculate the average arrival rate of the special customer)

Since the special customer spends an exponential amount of time with mean \(1/\theta\) out of the system, we know that the time between arrivals of this special customer follows an exponential distribution with a rate of \(\theta\). Therefore, the average arrival rate of the special customer is \(\theta\).

(Step 2: Define state spaces and set up balance equations)

Let's define the state space as follows: State \(n\) represents the number of ordinary customers in the system, from \(n = 0, 1, 2, ...\) Now, let's set up balance equations for this system. For \(n = 0\) (state 0), the balance equation becomes: Balance In: \(q_{1}(\lambda + \theta) = q_{0}(\mu + \theta)\) Balance Out: \(q_{0}\lambda = q_{1}(\mu + \mu_1)\) For \(n \geq 1\) (state n), the balance equation becomes: \(q_{n-1}\mu + q_{n+1}(\lambda+\theta) = q_{n}(\mu_n+\lambda+\theta)\) Where \(\mu_n = \mu + \mu_1\) if the special customer is in service with the ordinary customer at state \(n\).

(Step 3: Find the probability that an ordinary customer is bumped n times)

Let \(P_{n}\) be the probability that an ordinary customer is bumped \(n\) times. When a special customer arrives, she bumps the ordinary customer with a probability of \(\frac{\mu_1}{\mu+\mu_1}\). Using the balance equations from step 2, we can write the probability \(P_{n}\) as: \(P_{n} = \frac{\mu_1(\lambda+\theta)}{(\mu+\mu_1)(\mu+\theta)} P_{n-1}\) Now, to find the probability \(P_{n}\), we will use the initial condition \(P_{0} = 1\) and iterate the above equation as follows: \(P_{n} = \left(\frac{\mu_1(\lambda+\theta)}{(\mu+\mu_1)(\mu+\theta)}\right)^{n} P_{0}\) Therefore, the probability of an ordinary customer being bumped \(n\) times is: \(P_{n} = \left(\frac{\mu_1(\lambda+\theta)}{(\mu+\mu_1)(\mu+\theta)}\right)^{n}\)