/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 19 A single repairperson looks afte... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 19

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?

Short Answer

Expert verified
The proportion of time machine 2 is down can be calculated using the formula: \(p_{2\_down} = \frac{\lambda_{2}(\mu_{1} + \lambda_{1})}{(\lambda_{1} + \lambda_{2})(\mu_{1} + \mu_{2} - \lambda_{2})}\). Plug in the given values for 位鈧�, 位鈧�, 渭鈧�, and 渭鈧� to find the proportion of downtime for machine 2.

Step by step solution

01

Identify possible system states

In this problem, there are four possible scenarios in which machines 1 and 2 can be down or up: 1. Both machines are up (State UU). 2. Machine 1 is down and machine 2 is up (State DU). 3. Machine 1 is up and machine 2 is down (State UD). 4. Both machines are down (State DD).
02

Set up rate equations for each state

The balance of flow into and out of each state can be described by rate equations. In these equations, p_UU, p_DU, p_UD, and p_DD represent the steady-state probabilities of states UU, DU, UD, and DD, respectively. Rate equations: 1. State UU: \( (\lambda_{1} + \lambda_{2}) p_{UU} = \mu_{1} p_{DU} \) 2. State DU: \( \lambda_{1} p_{DU} = \mu_{1} p_{UU} + \mu_{2} p_{DD} \) 3. State UD: \( \lambda_{2} p_{UD} = \mu_{2} p_{UU} \) 4. State DD: \( \lambda_{1} p_{DD} = \mu_{1} p_{UD} + \lambda_{2} p_{DU} \)
03

Solve the rate equations

Solving these rate equations, we can get the steady-state probabilities of each state: 1. State UU: \(p_{UU} = \frac{\mu_{1} \mu_{2}}{(\lambda_{1} + \lambda_{2})(\mu_{1} + \mu_{2} - \lambda_{2})}\) 2. State DU: \(p_{DU} = \frac{\lambda_{1} \mu_{2}}{(\lambda_{1} + \lambda_{2})(\mu_{1} + \mu_{2} - \lambda_{2})}\) 3. State UD: \(p_{UD} = \frac{\lambda_{2} (\mu_{1} - \lambda_{2})}{(\lambda_{1} + \lambda_{2})(\mu_{1} + \mu_{2} - \lambda_{2})}\) 4. State DD: \(p_{DD} = \frac{\lambda_{1} \lambda_{2}}{(\lambda_{1} + \lambda_{2})(\mu_{1} + \mu_{2} - \lambda_{2})}\)
04

Calculate the proportion of time machine 2 is down

Machine 2 is down either when both machines are down (State DD) or when machine 1 is up and machine 2 is down (State UD). Hence, the proportion of time machine 2 is down can be given by: Proportion of downtime for machine 2: \(p_{2\_down} = p_{UD} + p_{DD}\) Replace p_UD and p_DD with their values from Step 3: \(p_{2\_down} = \frac{\lambda_{2} (\mu_{1} - \lambda_{2})}{(\lambda_{1} + \lambda_{2})(\mu_{1} + \mu_{2} - \lambda_{2})} + \frac{\lambda_{1} \lambda_{2}}{(\lambda_{1} + \lambda_{2})(\mu_{1} + \mu_{2} - \lambda_{2})}\) Combining the terms: \(p_{2\_down} = \frac{\lambda_{2}(\mu_{1} + \lambda_{1})}{(\lambda_{1} + \lambda_{2})(\mu_{1} + \mu_{2} - \lambda_{2})}\) Thus, the proportion of time machine 2 is down can be calculated by plugging in the given values for 位鈧�, 位鈧�, 渭鈧�, and 渭鈧� into the above formula.

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Key Concepts

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

Exponential Distribution in Repair Times and Failures
In the world of modeling systems like machine repairs, exponential distribution plays a crucial role. It helps describe time durations between events, such as failures and repairs. In this exercise, the time a machine stays operational before breaking down follows an exponential distribution. This time duration is characterized by a rate, known as \( \lambda \), which tells us how frequently events (machine failures) occur.

Similarly, the time needed to repair a machine also follows an exponential distribution with a different rate, \( \mu \). This rate indicates how efficiently repairs are completed. Both rates, \( \lambda_i \) for failure and \( \mu_i \) for repair, help us model the continuous and random nature of these real-world processes.
  • Rate \( \lambda_i \) - Frequency of machine failures.
  • Rate \( \mu_i \) - Speed of completing repairs.
Machines tend to fail unpredictably, and repairs can vary in length. Using exponential distribution captures this stochastic behavior effectively, making it a powerful tool in queueing theory and reliability engineering.
Using Rate Equations to Model System Behavior
Rate equations are vital in capturing the movement between different states in a continuous-time Markov chain. They essentially provide a balance between the flow in and out of states. In our example, the states represent different combinations of machines either being up or down.

For instance, one of the rate equations is \( (\lambda_{1} + \lambda_{2}) p_{UU} = \mu_{1} p_{DU} \), where the left part indicates the total rate with which the system moves out of the state where both machines are up (\( p_{UU} \)), and the right part shows the movement into it. Solving these equations allows us to determine the probabilities of each state being occupied.

These probabilities help in understanding how frequently the system is expected to be found in each state, thereby aiding in assessing performance metrics such as machine downtimes. Rate equations are fundamental to steady-state analysis where systems are presumed to reach an equilibrium.
Understanding Steady-State Probabilities
Steady-state probabilities reflect the long-term behavior of a system when it reaches equilibrium, meaning the likelihood of finding the system in a particular state stabilizes. In our machine repair scenario, they reveal how often each state occurs over an infinite amount of time.

Each state (e.g., both machines down (\( p_{DD} \))) has an associated steady-state probability. Solving the rate equations, we find these probabilities:
  • \( p_{UU} = \frac{\mu_{1} \mu_{2}}{(\lambda_{1} + \lambda_{2})(\mu_{1} + \mu_{2} - \lambda_{2})} \) - both machines up.
  • \( p_{DU} = \frac{\lambda_{1} \mu_{2}}{(\lambda_{1} + \lambda_{2})(\mu_{1} + \mu_{2} - \lambda_{2})} \) - machine 1 down, machine 2 up.
  • \( p_{UD} = \frac{\lambda_{2} (\mu_{1} - \lambda_{2})}{(\lambda_{1} + \lambda_{2})(\mu_{1} + \mu_{2} - \lambda_{2})} \) - machine 2 down, machine 1 up.
  • \( p_{DD} = \frac{\lambda_{1} \lambda_{2}}{(\lambda_{1} + \lambda_{2})(\mu_{1} + \mu_{2} - \lambda_{2})} \) - both machines down.
By observing these probabilities, one can deduce performance metrics like downtime proportions. For instance, the proportion of time machine 2 is down is the sum of \( p_{UD} \) and \( p_{DD} \), offering a practical viewpoint of machine availability and reliability over time.

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Most popular questions from this chapter

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 \(6.8\).

Consider two machines that are maintained by a single repairman. Machine \(i\) functions for an exponential time with rate \(\mu_{i}\) before breaking down, \(i=1,2 .\) The repair times (for either machine) are exponential with rate \(\mu .\) Can we analyze this as a birth and death process? If so, what are the parameters? If not, how can we analyze it?

Each time a machine is repaired it remains up for an exponentially distributed time with rate \(\lambda\). It then fails, and its failure is either of two types. If it is a type 1 failure, then the time to repair the machine is exponential with rate \(\mu_{1}\); if it is a type 2 failure, then the repair time is exponential with rate \(\mu_{2} .\) Each failure is, independently of the time it took the machine to fail, a type 1 failure with probability \(p\) and a type 2 failure with probability \(1-p\). What proportion of time is the machine down due to a type 1 failure? What proportion of time is it down due to a type 2 failure? What proportion of time is it up?

Each individual in a biological population is assumed to give birth at an exponential rate \(\lambda\), and to die at an exponential rate \(\mu .\) In addition, there is an exponential rate of increase \(\theta\) due to immigration. However, immigration is not allowed when the population size is \(N\) or larger. (a) Set this up as a birth and death model. (b) If \(N=3,1=\theta=\lambda, \mu=2\), determine the proportion of time that immigration is restricted.

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.
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