Question

Consider a single-server queue with Poisson arrivals and exponential service times having the following variation: Whenever a service is completed a departure occurs only with probability \(\alpha .\) With probability \(1-\alpha\) the customer, instead of leaving, joins the end of the queue. Note that a customer may be serviced more than once. (a) Set up the balance equations and solve for the steady-state probabilities, stating conditions for it to exist. (b) Find the expected waiting time of a customer from the time he arrives until he enters service for the first time. (c) What is the probability that a customer enters service exactly \(n\) times, \(n=\) \(1,2, \ldots ?\) (d) What is the expected amount of time that a customer spends in service (which does not include the time he spends waiting in line)? Hint: Use part (c). (e) What is the distribution of the total length of time a customer spends being served? Hint: Is it memoryless?

Short answer

In summary: (a) We set up balance equations and solved for steady-state probabilities, \(P_i\). They exist if \(\sum_{i=0}^{\infty} P_i = 1\). (b) The expected waiting time of a customer until entering service for the first time is \(W = \frac{P_1}{\lambda(1 - P_0)}\). (c) The probability that a customer enters service exactly \(n\) times is \(P(n) = (1 - \alpha)^{n-1}\alpha\). (d) The expected amount of time a customer spends in service is \(T_s = \frac{\alpha}{\mu}\). (e) The distribution of the total length of time a customer spends being served is exponential with parameter \(\frac{\alpha}{\mu}\) and is memoryless.

Step-by-step solution

Notations

Let \(P_i\) denote the steady-state probability that there are \(i\) customers in the system. Let \(\lambda\) be the arrival rate of customers and let \(\mu\) be the service rate.

Balance equations

For \(i \ge 0\), we can set up balance equations as follows: \(\lambda P_0 = \mu P_1\) \(\lambda P_1 = \mu P_2\) \(\lambda P_2 = \mu P_3 + (1-\alpha) \mu P_0\) \( \cdots \) \(\lambda P_{i} = \mu P_{i+1} + (1-\alpha) \mu P_{i-1}\) for \(i \ge 2\)

Solve steady-state probabilities

Using the balance equations, we can express the probabilities \(P_i\) as follows: \(P_1 = \frac{\lambda}{\mu} P_0\) \(P_2 = (\frac{\lambda}{\mu})^2 P_0 - (1 - \alpha) P_0\) For \(i \ge 2\), \(P_{i+1} = \frac{1}{\mu} (\lambda P_i - (1 - \alpha)\mu P_{i-1})\) To find the steady-state probabilities, we need to find the normalization constant \(P_0\). For the steady-state probabilities to exist, the following condition must hold: \(\sum_{i=0}^{\infty} P_i = 1\) (b) Expected waiting time until entering service

Expected waiting time

Let \(W\) be the expected waiting time before entering in service. Then, \(W = \frac{P_1}{\lambda(1 - P_0)}\) (c) Probability of entering service exactly n times

Probability of entering service n times

Let \(P(n)\) be the probability of entering service exactly n times, then \(P(n) = (1 - \alpha)^{n-1}\alpha\) (d) Expected amount of time in service

Expected time in service

Let \(T_s\) be the expected amount of time in service, then using part (c): \(T_s = \sum_{n=1}^{\infty} n (1 - \alpha)^{n-1}\alpha \cdot \frac{1}{\mu}\) \(T_s = \frac{\alpha}{\mu}\) (e) Distribution of total length of time in service

Distribution of total time in service

Since the expected time spent in service by a customer, \(T_s\) is equal to \(\frac{\alpha}{\mu}\) which is constant, the distribution of the total length of time a customer spends being served is exponential. The distribution has the parameter \(\frac{\alpha}{\mu}\) and is memoryless.

Key concepts

Poisson Arrivals

The concept of Poisson arrivals is fundamental to understanding many queueing theory models. Poisson arrivals occur when customers or particles come into a system at a continuous and steady average rate, with the time between arrivals following an exponential distribution. This assumption makes the Poisson process a vital component in modeling and analyzing queues where occurrences happen independently, such as customers entering a bank or calls coming into a call center.

In practical terms, if we say that the arrival of customers is a Poisson process with a rate of \(\(\lambda\)\), it means that the probability of a certain number of customers arriving in a fixed period only depends on the length of that period, not on the specific starting time. For instance, this makes the Poisson arrival process 'memoryless', meaning that the likelihood of an arrival happening in the next moment does not depend on when the previous arrival occurred.

Exponential Service Times

Exponential service times refer to another assumption in queueing theory, where the length of time it takes to serve a customer follows an exponential distribution. This implies that the likelihood of the service being completed at any given moment is the same, regardless of how long the service has already been in progress - another instance of the 'memoryless' property.

The service rate, denoted by \(\mu\), is the average rate at which customers are served by the system. This rate is the inverse of the expected time a customer spends being serviced. If, for instance, on average it takes 2 minutes to serve a customer, then the service rate would be \(\mu = 0.5\) services per minute. In our textbook example, the service times being exponentially distributed helps determine the expected amount of time a customer spends in service and the distribution of total service time.

Steady-State Probabilities

Steady-state probabilities are crucial for understanding the long-term behavior of a queueing system. They are the probabilities that a system will be found in a particular state, such as having a certain number of customers in the queue, after a long period of time when the transient effects of the initial conditions have worn off. To reach a steady state, a queueing system must be stable, meaning that it can handle incoming customers without growing indefinitely.

In our textbook problem, the balance equations were used to determine the steady-state probabilities \(P_i\) for each possible number of customers in the system. These probabilities must sum up to one for the system to be stable. If the system meets this condition, it allows us to predict the long-term average number of customers in the queue, among other characteristics.

Expected Waiting Time

The expected waiting time in a queueing system is a measure of how long a customer can expect to wait before receiving service upon arrival. This metric is important as it influences customer satisfaction and system efficiency. In the single-server queue scenario from our textbook, the expected waiting time is not straightforward due to the possibility of customers joining the end of the queue after being serviced, making calculations more complex.

To compute the expected waiting time, one must consider the probability of immediate service \(P_0\) and the steady-state probability of finding one customer in the system \(P_1\). Once these probabilities are determined through the balance equations, they are used to calculate the average waiting time for a customer to enter service for the first time. This figure is crucial for businesses in planning and improving their queue management strategies.