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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 a component fails, 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.

Short Answer

Expert verified
In order to analyze the given system as a continuous-time Markov chain, the state space is defined as the number of failed components, S = {0, 1, 2, ..., n}. Transition rates between states are determined for the scenarios of repair (decrease in faulty components), failure (increase in faulty components) and no change. By examining the detailed balance equations, we showed that the chain is time-reversible in the steady state. Limiting probabilities of having i failed components, π_i, were computed by normalizing the sum of all probabilities to equal 1. In conclusion, we demonstrated how to analyze the given system as a continuous-time Markov chain and compute its time-reversibility and limiting probabilities.

Step by step solution

01

Markov Chain Definition and Parameters

To analyze the given system as a continuous-time Markov chain, we will first define the states. The state space S can be defined as the number of failed components in the system. Therefore, S = {0, 1, 2, ..., n}, where 0 represents no failed components and n represents all components have failed. For the continuous-time Markov chain, we need to determine the transition rates between states. Let q(i, j) be the transition rate from state i to state j. The transition rate for i to j is given as the following: 1. Decrease in the number of faulty components (Repair of a component): If j = i - 1, then q(i, j) = i * α^i * μ. (where μ = Σμ_i for all components) 2. Increase in the number of faulty components (Failure of a component): If j = i + 1, then q(i, j) = λ_{j - i} (λ from working components) 3. No change in the number of faulty components : If j = i, then the transition rate remains 0, i.e., q(i, j) = 0.
02

Time Reversibility and Limiting Probabilities

Let π_i be the limiting probability that there are i failed components in the system. To show that the chain is time-reversible in the steady state, we need to prove that the detailed balance equations hold: π_i * q(i, i+1) = π_{i+1} * q(i+1, i) for all i in {0, 1, 2, ..., n-1}. Using the transition rates defined in the previous step, the detailed balance equations can be written as: π_i * λ_{i+1} = π_{i+1} * (i+1) * α^{i+1} * μ, which can be rewritten as: π_{i+1} = (λ_{i+1} / ((i+1) * α^{i+1} * μ)) * π_i. The limiting probabilities can be calculated as a product of π_0 and the ratios obtained above: π_1 = (λ_1 / (α * μ)) * π_0, π_2 = (λ_2 / (2 * α^2 * μ)) * π_1 = (λ_1 * λ_2 / (2 * α^3 * μ^2)) * π_0, π_3 = (λ_3 / (3 * α^3 * μ)) * π_2 = (λ_1 * λ_2 * λ_3 / (6 * α^6 * μ^3)) * π_0, and so on. Now, we have to normalize the probabilities to ensure that the sum of all probabilities equals 1: π_0 + π_1 + π_2 + ... + π_n = 1. Using the expressions for limiting probabilities obtained above, we can compute these probabilities using the equations: π_0 * (1 + (λ_1 / (α * μ)) + (λ_1 * λ_2 / (2 * α^3 * μ^2)) + ... + (λ_1 * λ_2 * ... * λ_n / (n! * α^(n^2) * μ^n))) = 1. In conclusion, we have shown that the given system can be analyzed as a continuous-time Markov chain, and its time-reversibility and limiting probabilities can be computed based on the chain parameters.

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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 crucial in modeling the time until an event occurs, particularly when the events happen independently and at a constant average rate. In the context of our exercise, the working time of each component is exponentially distributed with a specific rate parameter \( \lambda_i \). This rate parameter indicates the average number of times a component will fail per unit time.

A distinctive property of the exponential distribution is its memorylessness, meaning the probability of failure in the next instance is the same, irrespective of how long the component has been functioning. This aligns perfectly with the prerequisites for a continuous-time Markov chain (CTMC), where the future state only depends on the current state and not on the history.
Steady State
A steady state in the context of a continuous-time Markov chain represents a condition where the probabilities of being in any given state are constant over time. In other words, it's the scenario in which the system has 'settled down', and the state probabilities do not change. When a CTMC reaches a steady state, the likelihood of transition to different states balances out, resulting in a stable distribution of states known as the limiting probabilities. This concept ensures predictability in long-term behavior, which is why calculating the steady state is an essential step in understanding CTMCs.
Time Reversibility
Time reversibility is a property of some Markov chains where the process can be reversed without altering the statistical properties of the system. In time-reversible chains, the trajectory of the process in either time direction appears statistically the same. The detailed balance equations are instrumental in showcasing this property. A continuous-time Markov chain is time-reversible when, in the steady state, the rate of transitioning from state 'i' to state 'j' is equal to the rate of transitioning from state 'j' to state 'i', for all pairs of states. Our exercise involves proving this by ensuring the detailed balance equations are satisfied in steady state.
Limiting Probabilities
Limiting probabilities are the probabilities that a continuous-time Markov chain will be in a particular state after a long period. They essentially define the steady state distribution of the system. If we consider an infinite time horizon, the limiting probability for each state can be viewed as the proportion of time the system spends in that state. In the exercise, the calculation of limiting probabilities involves the determination of state probabilities in the steady state and requires normalization to ensure all probabilities sum up to one, reflecting the certainty that the system must be in one of the defined states.
Transition Rates
Transition rates are fundamental to continuous-time Markov chains, denoting the rate at which transitions occur between states. For our system with failed components, we consider two types of transitions—failures of components increasing the number of failed states, and repairs reducing these numbers. The exercise specifies that the rate of transition from state 'i' to state 'i+1' (failure) is given by the corresponding \( \lambda \) rate, while the transition from 'i' to 'i-1' (repair) depends on the current number of failed components and is calculated using \( \alpha \) and \( \mu \). Understanding and calculating these transition rates is essential for analyzing the behavior of the CTMC.
Detailed Balance Equations
The detailed balance equations are a series of equations that provide conditions under which a continuous-time Markov chain is in equilibrium. These equations correlate the flow of probability into each state with the flow out of that state, ensuring the net probability flow is zero in the steady state. This is a requirement for time reversibility. In the provided exercise, the detailed balance condition is used to relate the limiting probabilities of consecutive states, which subsequently aids in determining all the state probabilities in the steady state.
State Space
The state space of a continuous-time Markov chain represents all the possible states the system can be in. In our problem, the state space is defined by the number of failed components in the system, ranging from 0 (all components are functioning) to 'n' (all components have failed). This discrete set of potential outcomes forms the backbone of the CTMC analysis as it characterizes the entire set of scenarios that the system might experience over time and is integral for calculating transitions and probabilities within the system.
Rate Parameter
The rate parameter within the context of the exponential distribution and continuous-time Markov chains is a measure of how quickly or frequently transitions between states occur. In the exercise, each component has a failure rate parameter \( \lambda_i \) and a repair rate parameter that depends on \( \mu_i \) and the number of failed components. The rate parameter thus influences the transition rates between states in the Markov chain. It is a pivotal value in defining the dynamics of the system since it determines how often components are expected to fail and be repaired, ultimately shaping the steady state probabilities.

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Most popular questions from this chapter

Four workers share an office that contains four telephones. At any time, each 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 \(\mathrm{X}(t)=\left(X_{1}(t), X_{2}(t), X_{3}(t), X_{4}(t)\right)\) (b) Argue that \(\\{\mathrm{X}(t), t \geqslant 0\\}\) is a continuous-time Markov chain and give its infinitesimal rates. (c) Is \(\\{\mathrm{X}(t)\\}\) 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 them all being used begins a new "working" period. (d) What proportion of time are all workers "working"?

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?

(a) Show that Approximation 1 of Section \(6.8\) is equivalent to uniformizing the continuous-time Markov chain with a value \(v\) such that \(v t=n\) and then approximating \(P_{i j}(t)\) by \(P_{i j}^{* n}\). (b) Explain why the preceding should make a good approximation. Hint: What is the standard deviation of a Poisson random variable with mean \(n\) ?

A single repairperson looks after both machines 1 and \(2 .\) Each time it is repaired, machine \(i\) stays up for an exponential time with rate \(\lambda_{i}, i=1,2 .\) When machine \(i\) fails, it requires an exponentially distributed amount of work with rate \(\mu_{i}\) to complete its repair. The repairperson will always service machine 1 when it is down. For instance, if machine 1 fails while 2 is being repaired, then the repairperson will immediately stop work on machine 2 and start on \(1 .\) What proportion of time is machine 2 down?

Consider a taxi station where taxis and customers arrive in accordance with Poisson processes with respective rates of one and two per minute. A taxi will wait no matter how many other taxis are present. However, an arriving customer that does not find a taxi waiting leaves. Find (a) the average number of taxis waiting, and (b) the proportion of arriving customers that get taxis.

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