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Chapter 6: Problem 10

Consider two machines. Machine \(i\) operates for an exponential time with rate \(\lambda_{i}\) and then fails; its repair time is exponential with rate \(\mu_{i}, i=1,2 .\) The machines act independently of each other. Define a four-state continuous-time Markov chain that 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.

Short Answer

Expert verified
To jointly model the condition of two machines, we create a four-state continuous-time Markov chain with states A (both operating), B (Machine 1 operating, Machine 2 failed), C (Machine 1 failed, Machine 2 operating), and D (both failed). The machines act independently, allowing us to compute the transition probabilities. For example, the transition rate from state A to B is given by \(\lambda_2\), while the rate from state B to A is \(\mu_2\). Using the transition rates, we form the forward and backward equations and verify that they hold for all possible transitions. Thus, the transition probabilities satisfy both forward and backward equations.

Step by step solution

01

Define the state space of the Markov chain

First, we define the state space of the continuous-time Markov chain. Since we have two machines and each machine can be in one of two states (operating or failed), there are a total of 4 possible states, which can be defined as follows: 1. State A: Both machines are operating 2. State B: Machine 1 is operating and Machine 2 is failed 3. State C: Machine 1 is failed and Machine 2 is operating 4. State D: Both machines are failed
02

Define the transition rates between these states

To find the transition rates between the states, we can use the fact that the machines act independently. The transition rates are given by: 1. Transition from state A to B: Machine 1 continues operating, and Machine 2 fails. Therefore, the rate is \(\lambda_{2}\). 2. Transition from state A to C: Machine 1 fails, and Machine 2 continues operating. Therefore, the rate is \(\lambda_{1}\). 3. Transition from state B to A: Machine 1 continues operating, and Machine 2 is repaired. Therefore, the rate is \(\mu_{2}\). 4. Transition from state B to D: Machine 1 fails, and Machine 2 remains failed. Therefore, the rate is \(\lambda_{1}\). 5. Transition from state C to A: Machine 2 continues operating, and Machine 1 is repaired. Therefore, the rate is \(\mu_{1}\). 6. Transition from state C to D: Machine 2 fails, and Machine 1 remains failed. Therefore, the rate is \(\lambda_{2}\). 7. Transition from state D to B: Machine 1 is repaired, and Machine 2 remains failed. Therefore, the rate is \(\mu_{1}\). 8. Transition from state D to C: Machine 2 is repaired, and Machine 1 remains failed. Therefore, the rate is \(\mu_{2}\).
03

Verify that the transition probabilities satisfy the forward and backward equations

Now we form the forward and backward equations to show that the system is consistent: Forward equations: \[\frac{dp_{AB}(t)}{dt} = -\lambda_{2} p_{AB}(t) + \mu_{2} p_{BA}(t)\] \[\frac{dp_{BA}(t)}{dt} = \lambda_{2} p_{AB}(t) - \mu_{2} p_{BA}(t)\] \[\frac{dp_{AC}(t)}{dt} = -\lambda_{1} p_{AC}(t) + \mu_{1} p_{CA}(t)\] \[\frac{dp_{CA}(t)}{dt} = \lambda_{1} p_{AC}(t) - \mu_{1} p_{CA}(t)\] Backward equations: \[\frac{dp_{BD}(t)}{dt} = -\lambda_{1} p_{BD}(t) + \mu_{1} p_{DB}(t)\] \[\frac{dp_{DB}(t)}{dt} = \lambda_{1} p_{BD}(t) - \mu_{1} p_{DB}(t)\] \[\frac{dp_{CD}(t)}{dt} = -\lambda_{2} p_{CD}(t) + \mu_{2} p_{DC}(t)\] \[\frac{dp_{DC}(t)}{dt} = \lambda_{2} p_{CD}(t) - \mu_{2} p_{DC}(t)\] Since the forward and backward equations hold for all possible transitions, the transition probabilities satisfy both the forward and backward equations.

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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 one of the key elements in understanding continuous-time Markov chains. It is a continuous probability distribution commonly used to model the time between events in a Poisson process.
The defining feature of an exponential distribution is its "memoryless" property, which means the probability of an event occurring is constant over time.
This is particularly useful when modeling the operation and failure times of machines. For a machine operating with a rate of \( \lambda \), we say that the time till the machine fails follows an exponential distribution with parameter \( \lambda \).
This essentially means that at each instant in time, the probability of the machine failing remains the same. This characteristic makes the exponential distribution ideal for modeling random processes such as machine failures in a Markov chain framework.
Transition Rates
Transition rates are a fundamental concept in continuous-time Markov chains. They tell us how likely it is to transition from one state to another in a small time interval.
In the case of our two machines, each machine has a specific rate at which it fails or gets repaired. For instance, from the given exercised machine 1 operates with a rate \( \lambda_1 \) and gets repaired with rate \( \mu_1 \).
These rates are used to form the transition rate matrix for the Markov chain, which defines the behavior of the system over time. This matrix is crucial as it helps determine how the system evolves and predicts long-term behavior.
Understanding and calculating these rates correctly is essential, as they directly influence the chain's probability distributions over different states.
State Space
The state space of a Markov chain defines all possible states that the process can be in. It is a comprehensive list or set of states the system can take on.
For the problem in the exercise, our state space consists of four distinct states:
  • State A: Both machines are operating.
  • State B: Machine 1 is operating while Machine 2 is failed.
  • State C: Machine 1 is failed while Machine 2 is operating.
  • State D: Both machines are failed.
The state space is crucial because it forms the foundation for modeling how the Markov chain behaves. Knowing all potential states allows us to map transitions between them correctly based on given transition rates.
Forward and Backward Equations
Forward and backward equations are used to ensure that the transition probabilities in a Markov chain are consistent over time.
These equations are differential equations that allow us to compute the rate of change of transition probabilities over an interval.
For example, the forward equations relate the change in probability of moving from one state to another over time.
The backward equations, on the other hand, allow tracing transitions backward in time, from future states to the current state.
In our two-machines scenario, these equations ensure that the computed probabilities of the state transitions remain consistent with expected behavior.
These equations can be solved to predict future behavior of the system and assure the internal consistency of the model.

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Most popular questions from this chapter

After being repaired, a machine functions for an exponential time with rate \(\lambda\) and then fails. Upon failure, a repair process begins. The repair process proceeds sequentially through \(k\) distinct phases. First a phase 1 repair must be performed, then a phase 2, and so on. The times to complete these phases are independent, with phase \(i\) taking an exponential time with rate \(\mu_{i}, i=1, \ldots, k\) (a) What proportion of time is the machine undergoing a phase \(i\) repair? (b) What proportion of time is the machine working?

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 \(C, \alpha, \beta\).

Potential customers arrive at a single-server station in accordance with a Poisson process with rate \(\lambda .\) However, if the arrival finds \(n\) customers already in the station, then he will enter the system with probability \(\alpha_{n}\). Assuming an exponential service rate \(\mu\), set this up as a birth and death process and determine the birth and death rates.

The following problem arises in molecular biology. The surface of a bacterium consists of several sites at which foreign molecules-some acceptable and some not-become attached. We consider a particular site and assume that molecules arrive at the site according to a Poisson process with parameter \(\lambda\). Among these molecules a proportion \(\alpha\) is acceptable. Unacceptable molecules stay at the site for a length of time that is exponentially distributed with parameter \(\mu_{1}\), whereas an acceptable molecule remains at the site for an exponential time with rate \(\mu_{2}\). An arriving molecule will become attached only if the site is free of other molecules. What percentage of time is the site occupied with an acceptable (unacceptable) molecule?

Consider a graph with nodes \(1,2, \ldots, n\) and the \(\left(\begin{array}{l}n \\\ 2\end{array}\right) \operatorname{arcs}(t, j), i \neq j, i, j,=1, \ldots, n\) (See Section 3.6.2 for appropriate definitions.) Suppose that a particle moves along this graph as follows: Events occur along the arcs \((i, j)\) according to independent Poisson processes with rates \(\lambda_{i j} .\) An event along arc \((i, j)\) causes that arc to become excited. If the particle is at node \(i\) at the moment that \((i, j)\) becomes excited, it instantaneously moves to node \(j, i, j=1, \ldots, n .\) Let \(P_{j}\) denote the proportion of time that the particle is at node \(j .\) Show that $$ P_{j}=\frac{1}{n} $$ Hint: Use time reversibility.
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