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Chapter 6: Problem 18

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?

Short Answer

Expert verified
The proportion of time the machine undergoes a phase i repair is: \[P_i = \frac{\mu_i}{\lambda + \sum_{i=1}^{k} \mu_i}\] The proportion of time the machine is working is: \[P_w = \frac{\lambda}{\lambda + \sum_{i=1}^{k} \mu_i}\]

Step by step solution

01

Determine overall rates

Let's first find out the overall rate for the whole system, which includes the machine's functioning time (rate λ) and the sum of repair processes' times (sum of rates μi). The overall rate is given by: \[\Omega = \lambda + \sum_{i=1}^{k} \mu_i\]
02

(a) Proportion of time spent on phase i repair

To find the proportion of time the machine undergoes a phase i repair, we can calculate it using the rate for that phase and the overall rate. The proportion of time spent on phase i repair is: \[P_i = \frac{\mu_i}{\Omega}\]
03

(b) Proportion of time the machine is working

Similarly, to find the proportion of time the machine is working, we need to divide the rate λ by the overall rate Ω. The proportion of time spent working is: \[P_w = \frac{\lambda}{\Omega}\] Now you can use these formulas to calculate the proportion of time spent on each repair phase and the proportion of time the machine is working, given the rate values λ and μi for the machine functioning and repair processes.

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Most popular questions from this chapter

Consider a Yule process starting with a single individual-that is, suppose \(X(0)=1\). Let \(T_{i}\) denote the time it takes the process to go from a population of size \(i\) to one of size \(i+1\) (a) Argue that \(T_{i}, i=1, \ldots, j\), are independent exponentials with respective rates i\lambda. (b) Let \(X_{1}, \ldots, X_{j}\) denote independent exponential random variables each having rate \(\lambda\), and interpret \(X_{i}\) as the lifetime of component \(i\). Argue that \(\max \left(X_{1}, \ldots, X_{j}\right)\) can be expressed as $$ \max \left(X_{1}, \ldots, X_{i}\right)=\varepsilon_{1}+\varepsilon_{2}+\cdots+\varepsilon_{j} $$ where \(\varepsilon_{1}, \varepsilon_{2}, \ldots, \varepsilon_{j}\) are independent exponentials with respective rates \(j \lambda\) \((j-1) \lambda, \ldots, \lambda\) Hint: Interpret \(\varepsilon_{i}\) as the time between the \(i-1\) and the ith failure. (c) Using (a) and (b) argue that $$ P\left[T_{1}+\cdots+T_{j} \leqslant t\right\\}=\left(1-e^{-\lambda t}\right)^{j} $$ (d) Use (c) to obtain $$ P_{1 j}(t)=\left(1-e^{-\lambda t}\right)^{j-1}-\left(1-e^{-\lambda t}\right)^{j}=e^{-\lambda t}\left(1-e^{-\lambda t}\right)^{j-1} $$ and hence, given \(X(0)=1, X(t)\) has a geometric distribution with parameter \(p=e^{-\lambda t}\) (e) Now conclude that $$ P_{i j}(t)=\left(\begin{array}{l} j-1 \\ i-1 \end{array}\right) e^{-\lambda t i}\left(1-e^{-\lambda t}\right)^{j-i} $$

Let \(O(t)\) be the occupation time for state 0 in the two-state continuous-time Markov chain. Find \(E[O(t) \mid X(0)=1]\).

A population of organisms consists of both male and female members. In a small colony any particular male is likely to mate with any particular female in any time interval of length \(h\), with probability \(\lambda h+o(h) .\) Each mating immediately produces one offspring, equally likely to be male or female. Let \(N_{1}(t)\) and \(N_{2}(t)\) denote the number of males and females in the population at \(t .\) Derive the parameters of the continuous-time Markov chain \(\left\\{N_{1}(t), N_{2}(t)\right\\}\), i.e., the \(v_{i}, P_{i j}\) of Section \(6.2\).

Let \(Y\) denote an exponential random variable with rate \(\lambda\) that is independent of the continuous-time Markov chain \(\\{X(t)\\}\) and let $$ \bar{P}_{i j}=P[X(Y)=j \mid X(0)=i\\} $$ (a) Show that $$ \bar{P}_{i j}=\frac{1}{v_{i}+\lambda} \sum_{k} q_{i k} \bar{P}_{k j}+\frac{\lambda}{v_{i}+\lambda} \delta_{i j} $$ where \(\delta_{i j}\) is 1 when \(i=j\) and 0 when \(i \neq j\) (b) Show that the solution of the preceding set of equations is given by $$ \overline{\mathbf{P}}=(\mathbf{I}-\mathbf{R} / \lambda)^{-1} $$ where \(\overline{\mathbf{P}}\) is the matrix of elements \(\bar{P}_{i j}, \mathbf{I}\) is the identity matrix, and \(\mathbf{R}\) the matrix specified in Section \(6.8\). (c) Suppose now that \(Y_{1}, \ldots, Y_{n}\) are independent exponentials with rate \(\lambda\) that are independent of \(\\{X(t)\\}\). Show that $$ P\left[X\left(Y_{1}+\cdots+Y_{n}\right)=j \mid X(0)=i\right\\} $$ is equal to the element in row \(i\), column \(j\) of the matrix \(\overline{\mathbf{p}}^{n}\). (d) Explain the relationship of the preceding to Approximation 2 of Section \(6.8\).

Individuals join a club in accordance with a Poisson process with rate \(\lambda\). Each new member must pass through \(k\) consecutive stages to become a full member of the club. The time it takes to pass through each stage is exponentially distributed with rate \(\mu\). Let \(N_{i}(t)\) denote the number of club members at time \(t\) who have passed through exactly \(i\) stages, \(i=1, \ldots, k-1 .\) Also, let \(\mathrm{N}(t)=\left(N_{1}(t), N_{2}(t), \ldots, N_{k-1}(t)\right)\) (a) Is \(\\{\mathbf{N}(t), t \geqslant 0\\}\) a continuous-time Markov chain? (b) If so, give the infinitesimal transition rates. That is, for any state \(\mathrm{n}=\) \(\left(n_{1}, \ldots, n_{k-1}\right)\) give the possible next states along with their infinitesimal rates.
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