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Chapter 6: Problem 8

Consider two machines, both of which have an exponential lifetime with mean \(1 / \lambda .\) There is a single repairman that can service machines at an exponential rate \(\mu .\) Set up the Kolmogorov backward equations; you need not solve them.

Short Answer

Expert verified
The Kolmogorov backward equations for the system are given by: 1. State 0 (Both machines working): \[ \frac{dP_0(t)}{dt} = \mu P_1(t) - 2\lambda P_0(t) \] 2. State 1 (One machine working, the other under repair): \[ \frac{dP_1(t)}{dt} = 2\lambda P_0(t) + \mu P_2(t) - (\lambda + \mu)P_1(t) \] 3. State 2 (Both machines failed, one under repair): \[ \frac{dP_2(t)}{dt} = \lambda P_1(t) - \mu P_2(t) \]

Step by step solution

01

State Definitions

First, let's define the possible states of the system: 1. State 0: Both machines are working. 2. State 1: One machine is working while the other is under repair. 3. State 2: Both machines are failed, and one of them is under repair.
02

Kolmogorov Backward Equations

To derive the backward differential equations for the probabilities of each state, we consider the rate at which the process is leaving each state and the rate at which it is entering each state. Let \( P_i(t) \) denote the probability of the system being in state \( i \) at time \( t \). We will now write down the Kolmogorov backward equations for each state.
03

State 0

State 0 is entered when there are two working machines and exited when one of the machines fails. Thus, the process can either enter state 0 by both machines finishing their repair or exit state 0 by one of the machines failure: \[ \frac{dP_0(t)}{dt} = \mu P_1(t) - 2\lambda P_0(t) \]
04

State 1

State 1 can be entered by either a failed machine in state 0 being fixed or a second failed machine in state 2 being fixed. State 1 is left when either a working machine in state 1 fails, or the repair finishes and the system returns to state 0 or state 2: \[ \frac{dP_1(t)}{dt} = 2\lambda P_0(t) + \mu P_2(t) - (\lambda + \mu)P_1(t) \]
05

State 2

State 2 is entered when one of the machines in state 1 fails, and exited when one of the two failed machines is repaired. In this case, the equation is given by: \[ \frac{dP_2(t)}{dt} = \lambda P_1(t) - \mu P_2(t) \] These three equations together represent the Kolmogorov backward equations for the given system characterized by the failure rates \( \lambda \) and repair rate \( \mu \).

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Most popular questions from this chapter

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 \(\mathbf{N}(t)=\left(N_{1}(t), N_{2}(t), \ldots, N_{k-1}(t)\right)\) (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 \(\mathbf{n}=\) \(\left(n_{1}, \ldots, n_{k-1}\right)\) give the possible next states along with their infinitesimal rates.

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