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A system is composed of \(N\) identical components; each independently operates a random length of time until failure. Suppose the failure time distribution is exponential with parameter \(\lambda\). When a component fails it undergoes repair. The repair time is random, with distribution function exponential with parameter \(\mu .\) The system is said to be in state \(n\) at time \(t\) if there are exactiy \(n\) components under repair at time \(t .\) This process is a birth and death process. Determine its infinitesimal parameters.

Short Answer

Expert verified
The infinitesimal parameters for the birth and death process representing the number of components under repair in a system with N identical components are given by: \(q_{n, n+1} = (N-n)\lambda\) \(q_{n, n-1} = n\mu\) where \(q_{n, n+1}\) and \(q_{n, n-1}\) represent the jump rates from state n to n+1 (failure of one more component) and from state n to n-1 (repair of one component), respectively. \(\lambda\) is the failure rate, which is exponentially distributed, and \(\mu\) is the repair rate, which is also exponentially distributed.

Step by step solution

01

Understand the Birth and Death Process

The birth and death process is a type of continuous-time Markov chain, where the process transitions between states representing the number of components under repair. In our case, state n represents the situation where exactly n components are under repair. The process can only transition between adjacent states, i.e., from state n to n+1 or n-1.
02

Calculate the Jump Rates from State to State

To find the infinitesimal parameters of this birth and death process, let's first calculate the jump rates from state to state. The rate of transition from state n to n+1 represents the failure of one more component, with rate λ for each functioning component. Similarly, the rate of transition from state n to n-1 represents the repair of one component, with rate μ for each repairing component. There are (N-n) functioning components and n repairing components. - Transition from state n to n+1 : jump rate = (N-n)λ - Transition from state n to n-1 : jump rate = nμ
03

Infinitesimal Parameters

Now that we know the jump rates for each state transition, we can write the infinitesimal parameters for the birth and death process as follows: q_{n,n+1} = (N-n)λ q_{n,n-1} = nμ These are the infinitesimal parameters for the birth and death process, which describe the rates of transitions between the different states of the system, based on the number of components under repair.

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Key Concepts

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

Understanding Continuous-Time Markov Chains
A continuous-time Markov chain (CTMC) is a stochastic model that describes a sequence of events occurring continuously over time, with the key property that the future state depends only on the current state and not on the sequence of events that preceded it. This 'memoryless' property is called the Markov property. In the context of our exercise, the system's state at any given time is the number of components currently under repair, making it a classic example of a CTMC.

In a CTMC, time between state transitions does not need to be fixed but is instead determined by the probability of transitioning from one state to another. For the system described, each state change is triggered either by the failure of a non-repairing component, or the repair completion of a currently repairing component, with these events happening randomly over continuous time.

CTMCs are widely used across various fields, from queueing theory in operations research to population genetics in biology. Understanding the mechanism of CTMCs is fundamental in predicting the behavior of systems over time and making informed decisions based on probabilistic models.
Exponential Distribution and Its Significance
The exponential distribution is a continuous probability distribution commonly used to model the time until an event occurs, such as the lifespan of an electronic component or the time between arrivals in a queue. The key characteristic of the exponential distribution is that it is memoryless, meaning that no matter how much time has elapsed, the probability of the event occurring in the next instant remains constant.

This property aligns perfectly with the Markov property of CTMCs, hence its use in our exercise where the time until a component fails (or is repaired) is exponentially distributed with parameters \(\lambda\) and \(\mu\), respectively. The parameter \(\lambda\), the failure rate, dictates how quickly components fail on average, while \(\mu\) determines the speed of repairs. Modeling with exponential distribution greatly simplifies analysis and calculations in stochastic processes like the described birth and death process because it reduces complex time-related dependencies to straightforward rates.
Infinitesimal Parameters in Birth and Death Processes
In the birth and death processes, infinitesimal parameters are the fundamental building blocks used to describe the rate at which transitions between states occur. These parameters \( q_{n,n+1} \) and \( q_{n,n-1} \) are referred to as 'birth rates' and 'death rates,' respectively, in our exercise context. They define the instantaneous rate at which the system will 'give birth to' (add to) or 'witness the death of' (subtract from) the number of components being repaired.

The 'birth rate' \( q_{n,n+1} \) reflects the likelihood of a component failing and joining the repair queue, effectively increasing the number of components under repair. Conversely, the 'death rate' \( q_{n,n-1} \) refers to the chance of a component’s repair being completed, thus decreasing the number of components under repair.

Understanding these rates is crucial for determining the system's long-term behaviour and stability. By analyzing the infinitesimal parameters, we can make probabilistic predictions about future states, such as the average number of components under repair at any given time or the likelihood of all components functioning correctly.

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Most popular questions from this chapter

A system is composed of \(N\) machines. At most \(M \leq N\) can be operating at any one time; the rest are "spares". When a machine is operating, it operates a random length of time until failure. Suppose this failure time is exponentially distributed with parameter \(\mu\). When a machine fails it undergoes repair. At most \(R\) machines can be "in repair" at any one time. The repair time is exponentially distributed with parameter \(\lambda\). Thus a machine can be in any of four states: (i) Operating, (ii) "Up", but not operating, i.e., a spare, (iii) In repair, (iv) Waiting for repair. There are a total of \(N\) machines in the system. At most \(M\) can be operating. At most \(R\) can be in repair. Let \(X(t)\) be the number of machines "up" at time \(t\), either operating or spare. Then, (we assume) the number operating is min \(\\{X(t), M\\}\) and the number of spares is max \(\\{0, X(t)-M\\}\). Let \(Y(t)=N-X(t)\) be the number of machines " down". Then the number in repair is \(\min \\{Y(t), R\\}\) and the number waiting for repair is max \(\\{0, Y(t)-R\\}\). The above formulas permit to determine the number of machines in any category, once \(X(t)\) is known. \(X(t)\) is a birth and death process. (a) Determine the birth and death parameters, \(\lambda_{l}\) and \(\mu_{i}, i=0, \ldots, N\). (b) In the following special cases, determine \(\pi_{j}\), the stationary probability that \(X(t)=j\). (a) \(R=M=N\). (b) \(R=1, M=N\).

Let \(\mathscr{R}\) be a continuous time birth and death process where \(\lambda_{n}=\lambda>0\), \(n \geq 0, \mu_{0}=0, \mu_{n}>0, n \geq 1 .\) Let \(\pi=\sum_{n} \pi_{n}<\infty\), where \(\pi_{n}=\lambda^{n} /\left(\mu_{1} \mu_{2} \cdots \cdots \mu_{n}\right)\) so that \(\pi_{V} / \pi\) is the stationary distribution of the process. Suppose the initial state is a r.v. whose distribution is the stationary distribution of the process. Prove that the number of deaths in \([0, t]\) has a Poisson distribution with parameter \(\lambda t\).

Consider a birth and death process with infinitesimal parameter \(\lambda_{n}, \mu_{n}\). Show that the expected length of time for reaching state \(r+1\) starting from state 0 is $$ \sum_{n=0}^{r} \frac{1}{\lambda_{n} \pi_{n}} \sum_{k=0}^{n} \pi_{k} $$ For the definition of \(\pi_{n}\) see Eq. (4.5).

Consider the following continuous version of the Ehrenfest model (see pago 51, Chapter 2). We have \(2 N\) balls labeled \(1,2,3, \ldots, 2 N\). At time 0 each Iull is equally likely to be placed in one of two urns. Subsequently, the balls independently undergo displacement randomly in time from one urn to the other by the following rules. A ball has a probability \(\frac{1}{2} h+o(h)\) of changing urns during the time interval \((t, t+h)\) and probability \(1-(h / 2)+o(h)\) of remaining in the same urn during that interval. The movements over disjoint intervals of time are independent. Let \(X(t)\) denote the number of balls in urn I at time \(t .\) Set $$ P_{j k}(t)=\operatorname{Pr}\\{X(t)=k \mid X(0)=j\\}, \quad j, k=0,1, \ldots, 2 N $$ Establish the formula $$ g(t, s)=\sum_{k=0}^{2 N} P_{J k}(t) s^{k}=2^{-2 N}\left[1-e^{-t}+\left(1+e^{-t}\right) s\right]^{j}\left[1+e^{-t}+\left(1-e^{-t}\right) s\right]^{2 N-j} $$

Let \(\left\\{X_{i}(t) ; t \geq 0\right\\} \quad i=1,2\) be two independent Poisson processes with parameters \(\lambda_{1}\) and \(\lambda_{2}\) respectively. Let \(X_{1}(0)=m, X_{2}(0)=N-1\), and \(m

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