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Chapter 9: Problem 30

Compute the expected system lifetime of a three-out-of-four system when the first two component lifetimes are uniform on \((0,1)\) and the second two are uniform on \((0,2)\).

Short Answer

Expert verified
The expected system lifetime of the three-out-of-four system, with first two component lifetimes uniform on (0,1) and second two component lifetimes uniform on (0,2), is \(\frac{5}{3}\).

Step by step solution

01

Find the pdf of the component lifetimes

First, we need to find the probability density functions (pdfs) of the individual component lifetimes. Since the first two components are uniformly distributed on (0,1), their pdfs can be given as: f_1(x) = f_2(x) = \[ \begin{cases} 1, & 0 < x \leq 1 \\ 0, & \mathrm{otherwise} \end{cases} \] For the second two components, which are uniformly distributed on (0,2), their pdfs can be given as: f_3(y) = f_4(y) = \[ \begin{cases} \frac{1}{2}, & 0 < y \leq 2 \\ 0, & \mathrm{otherwise} \end{cases} \]
02

Find the pdf for each case

Now, let's find the pdf for each of the three cases: Case 1: One of the first two components and two of the second two components fail. Using order statistics, we can calculate the pdf of the minimum of these three components as: h_1(x, y) = 2(1)(\frac{1}{2})^2 = y Case 2: Two of the first two components and one of the second two components fail. Using the same method, we can find the pdf of the minimum of these three components as: h_2(x, y) = \binom{2}{2}(\frac{1}{2})^2 = \frac{1}{4} Case 3: Two of the second two components fail and one of the first two components fails. The pdf of the minimum of these three components is: h_3(x, y) = 2(1)(\frac{1}{2})^2 = y
03

Compute the expected values for each case

Using the pdfs from the previous step, we'll now compute the expected values for each case through integration: E[Case 1] = \(\int_0^1 \int_0^2 xh_1(x, y) dy dx = \int_0^1 \int_0^2 xy dy dx = \frac{2}{3}\) E[Case 2] = \(\int_0^1 \int_0^2 xh_2(x, y) dy dx = \int_0^1 \int_0^2 \frac{1}{4}x dy dx = \frac{1}{3}\) E[Case 3] = \(\int_0^1 \int_0^2 xh_3(x, y) dy dx = \int_0^1 \int_0^2 xy dy dx = \frac{2}{3}\)
04

Calculate the overall expected system lifetime

Sum up the expected values for these three cases to get the expected system lifetime: Expected System Lifetime = E[Case 1] + E[Case 2] + E[Case 3] = \(\frac{2}{3} + \frac{1}{3} + \frac{2}{3}= \frac{5}{3}\) So the expected system lifetime of this three-out-of-four system is \(\frac{5}{3}\).

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Most popular questions from this chapter

Show that if each (independent) component of a series system has an IFR distribution, then the system lifetime is itself IFR by (a) showing that $$ \lambda_{F}(t)=\sum_{i} \lambda_{i}(t) $$ where \(\lambda_{F}(t)\) is the failure rate function of the system; and \(\lambda_{i}(t)\) the failure rate function of the lifetime of component \(i\). (b) using the definition of IFR given in Exercise 22 .

Consider a structure in which the minimal path sets are \(\\{1,2,3\\}\) and \(\\{3,4,5\\}\). (a) What are the minimal cut sets? (b) If the component lifetimes are independent uniform \((0,1)\) random variables, determine the probability that the system life will be less than \(\frac{1}{2}\).

Show that the mean lifetime of a parallel system of two components is $$ \frac{1}{\mu_{1}+\mu_{2}}+\frac{\mu_{1}}{\left(\mu_{1}+\mu_{2}\right) \mu_{2}}+\frac{\mu_{2}}{\left(\mu_{1}+\mu_{2}\right) \mu_{1}} $$ when the first component is exponentially distributed with mean \(1 / \mu_{1}\) and the second is exponential with mean \(1 / \mu_{2}\).

Let \(F\) be a continuous distribution function. For some positive \(\alpha\), define the distribution function \(G\) by $$ \bar{G}(t)=(\bar{F}(t))^{\alpha} $$ Find the relationship between \(\lambda_{G}(t)\) and \(\lambda_{F}(t)\), the respective failure rate functions of \(G\) and \(F\).

Component \(i\) is said to be relevant to the system if for some state vector \(\mathbf{x}\), $$ \phi\left(1_{i}, \mathbf{x}\right)=1, \quad \phi\left(0_{i}, \mathbf{x}\right)=0 $$ Otherwise, it is said to be irrelevant. (a) Explain in words what it means for a component to be irrelevant. (b) Let \(A_{1}, \ldots, A_{s}\) be the minimal path sets of a system, and let \(S\) denote the set of components. Show that \(S=\bigcup_{i=1}^{s} A_{i}\) if and only if all components are relevant. (c) Let \(C_{1}, \ldots, C_{k}\) denote the minimal cut sets. Show that \(S=\bigcup_{i=1}^{k} C_{i}\) if and only if all components are relevant.
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