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Chapter 6: Problem 25

Customers artive at a service station, manned by a single servet 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

Expert verified
The limiting probabilities for the tandem queueing system, P(n, m), are given by: P(n, m) = \((1-\frac{位}{碌_1})(1-\frac{碌_1}{碌_2}) * (\frac{位}{碌_1})^n * (\frac{碌_1}{碌_2})^m\)

Step by step solution

01

Write the balance equations

Since the system is in steady-state, we can use the balance equations to describe the flow of customers between the servers. We have: 1. For the first server, the inflow rate is 位 and the outflow rate is 碌1. 2. For the second server, the inflow rate is 碌1 and the outflow rate is 碌2.
02

Define P(n, m) according to the hint

We define the limiting probability P(n, m) as follows: P(n, m) = C * 伪^n * 尾^m where n and m represent the number of customers in the first and second systems, respectively.
03

Apply the balance equations to P(n, m)

We can now apply the balance equations to the limiting probabilities: 1. For the first server: \(位P(n-1, m) = 碌_1P(n, m)\) (inflow rate = outflow rate for the first server) 2. For the second server: \(碌_1P(n, m-1) = 碌_2P(n, m)\) (inflow rate = outflow rate for the second server)
04

Find the relation between 伪 and 位, and 尾 and 碌2

From the balance equations in Step 3, we have: 1. \(位P(n-1, m) = 位C伪^{n-1}尾^m = 碌_1P(n, m) = 碌_1C伪^n尾^m\) 2. \(碌_1P(n, m-1) = 碌_1C伪^n尾^{m-1} = 碌_2P(n, m) = 碌_2C伪^n尾^m\) From these equations, we can find the relations between 伪, 位, 尾 and 碌2: 1. \(伪 = \frac{位}{碌_1}\) 2. \(尾 = \frac{碌_1}{碌_2}\)
05

Find the value for C, the normalization constant

To find the value of C, we make use of the fact that the summation of all probabilities equals 1: \(\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}P(n, m) = 1\) Substituting the given P(n, m) expression, we get: \(C\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}(伪^n尾^m) = 1\) We can then rewrite this as a double geometric series: \(C \frac{1}{(1-伪)(1-尾)} = 1\) Thus, the normalization constant C is: \(C = (1-伪)(1-尾)\)
06

Final solution for P(n, m)

Combining all the steps above, we can now write the final solution for P(n, m): P(n, m) = \((1-伪)(1-尾) *伪^n * 尾^m\) where \(伪 = \frac{位}{碌_1}\) \(尾 = \frac{碌_1}{碌_2}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Exponential Distribution
The exponential distribution is a fundamental concept in queueing theory, often used to model the time between events in a Poisson process. It is characterized by a constant rate, which makes it memoryless, meaning that the probability of an event occurring in the future is independent of any past events.
In the context of our exercise, the service times at the servers are exponentially distributed with rates \(\mu_1\) and \(\mu_2\). This means that the time it takes to serve a customer follows an exponential distribution with these respective rates.
Understanding this helps in setting up the balance equations since the inflow and outflow rates depend on these service times.
  • Exponential service means a randomly occurring event with a constant average rate.
  • Memoryless nature simplifies analysis in queueing scenarios.
  • Important for modeling because it leads to simple mathematical treatment.
Poisson Process
The Poisson process is used to model random arrivals in a queueing system. It describes how customers arrive at a service point, assuming arrivals follow a Poisson distribution at a constant average rate \(\lambda\).
In the example given, customers arrive at the first server in the system at a Poisson rate \(\lambda\). This random arrival process simplifies many aspects of queue analysis because it results in exponentially distributed inter-arrival times.
  • Constant arrival rate over time makes predictions feasible.
  • Inter-arrival times are independent, fitting well with exponential distributions.
  • Poisson processes and exponential distributions are often coupled in queueing models.
By understanding the Poisson process, we can justify the use of exponential distributions for service times, given their compatibility.
Steady-State Probabilities
Steady-state probabilities represent the long-term behavior of a queueing system where the state (like the number of customers) remains constant over time on average.
In our exercise, the goal is to find these steady-state probabilities, \(P(n,m)\), where \(n\) and \(m\) are the customers in the first and second servers, respectively.
Using the balance equations derived from the inflow and outflow rates of the servers, we calculate the probability of each state in the system. In this scenario:
  • Rates \(\alpha\) and \(\beta\) are determined by the relationships \(\alpha = \frac{\lambda}{\mu_1}\) and \(\beta = \frac{\mu_1}{\mu_2}\).
  • The normalization constant \(C\) ensures the sum of all probabilities equals 1.
  • This involves solving geometric series that sum up the probabilities of all possible states.
These probabilities help in understanding and predicting the performance of the queueing system, ensuring resource allocation is efficient.

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Most popular questions from this chapter

Consider two machines. Machine \(i\) operates for an exponential time with rate \(\lambda_{\ell}\) and then fails; its repair time is exponential with rate \(\mu_{1}, i=1,2\). The machines act independently of each other. Define a four- state continuous-time Markov chain which jointly describes the condition of the two machines. Use the assumed independence to compute the transition probabilities for this chain and then verify that these transition probabilities satisfy the forward and backward equations.

Consider a system of \(n\) components such that the working times of component \(i, i=1, \ldots, n\), are exponentially distributed with rate \(\lambda_{i} .\) When failed, however, the repair rate of component \(i\) depends on how many other components are down. Specifically, suppose that the instantaneous repair rate of component \(i, i=1, \ldots, n\), when there are a total of \(k\) failed components, is \(\alpha^{k} \mu_{i}\). (a) Explain how we can analyze the preceding as a continuous-time Markov chain. Define the states and give the parameters of the chain. (b) Show that, in steady state, the chain is time reversible and compute the limiting probabilities.

Consider a set of \(n\) machines and a single repair facility to service these machines. Suppose that when machine \(i, i=1, \ldots, n\), fails it requires an exponentially distributed amount of work with tate \(\mu_{1}\) to repair it. The repair facility divides its efforts equally among all failed machines in the sense that whenever there are \(k\) failed machines cach one receives work at a rate of \(1 / k\) per unit time. If there are a total of \(r\) working machines, including machine \(i\), then \(i\) fails at an instantaneous rate \(\lambda_{i} / r\). (a) Define an appropriate state space so as to be able to analyze the above system as a continuous-time Markov chain. (b) Give the instantaneous transition rates (that is, give the \(q_{j}\) ). (c) Write the time reversibility equations. (d) Find the limiting probabilities and show that the process is time reversible.

Four workers share an office that contains four telephones. At any time, cach worker is either "working" or "on the phone." Each "working" period of worker \(i\) lasts for an exponentially distributed time with rate \(\lambda_{i}\). and each "on the phone" period lasts for an exponentially distributed time with rate \(\mu_{i}, i=1,2,3,4\), (a) What proportion of time are all workers "working"? Let \(X_{i}(t)\) equal 1 if worker \(i\) is working at time \(t\), and let it be 0 otherwise. Let \(\mathbf{X}(t)=\left(X_{1}(t), X_{2}(t), X_{3}(t), X_{4}(t)\right)\). (b) Argue that \(\\{\mathbf{X}(t), t \geq 0]\) is a continuous-time Markov chain and give its infinitesimal rates. (c) Is \(\\{\mathbf{X}(t) \mid\) time reversible? Why or why not? Suppose now that one of the phones has broken down. Suppose that a worker who is about to use a phone but finds thern all being used begins a new "working'" period. (d) What proportion of time are all workers "working'r?

There are two machines, one of which is used as a spare. A working machine will function for an exponential time with rate \(\lambda\) and will then fail. Upon failure, it is immediately replaced by the other machine if that one is in working order, and it goes to the repair facility. The repair facility consists of a single person who takes an exponential time with rate \(\mu\) to repair a failed machine. At the repair facility, the newly failed machine enters service if the repairperson is free. If the repairperson is busy, it waits until the other machine is fixed. At that time, the newly repaired machine is put in service and repair begins on the other one. Starting with both machines in working condition, find (a) the expected value and (b) the variance of the time until both are in the repair facility. (c) In the long run, what proportion of time is there a working machine?
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