/*! 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 2 \({ }^{\star}\) Consider a serie... [FREE SOLUTION] | 91Ó°ÊÓ

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\({ }^{\star}\) Consider a series system of two independent components. The lifetime of the first component is exponentially distributed with parameter \(\lambda\), and the lifetime of the second component is normally distributed with parameters \(\mu\) and \(\sigma^2\). Determine the reliability \(R(t)\) of the system and show that the expected life of the system is $$ \frac{1}{\lambda}\left[1-\exp \left(-\lambda \mu+\frac{\lambda^2 \sigma^2}{2}\right)\right] . $$

Short Answer

Expert verified
The expected life of the system is \[ \frac{1}{\lambda}[1 - \exp(-\lambda \mu + \frac{\lambda^2 \sigma^2}{2})] \].

Step by step solution

01

- Define Reliability of Series System

In a series system, the system fails if any one of the components fails. Therefore, the reliability of the system, denoted by \( R(t) \), is the product of the individual reliabilities of each component.
02

- Exponential Distribution Reliability

Given that the lifetime of the first component is exponentially distributed with parameter \( \lambda \), the reliability of the first component, \( R_1(t) \), is given by: \[ R_1(t) = e^{-\lambda t} \]
03

- Normal Distribution Reliability

The lifetime of the second component is normally distributed with parameters \( \mu \) and \( \sigma^2 \). The reliability of the second component, \( R_2(t) \), is the probability that the lifetime of the second component is greater than \( t \), given by: \[ R_2(t) = P(T_2 > t) = 1 - \Phi \left(\frac{t - \mu}{\sigma}\right) \], where \( \Phi \) represents the cumulative distribution function (CDF) of the standard normal distribution.
04

- Combine the Two Reliabilities

Since the components are independent, the reliability of the system is given by: \[ R(t) = R_1(t) \cdot R_2(t) = e^{-\lambda t} \cdot \left(1 - \Phi \left(\frac{t - \mu}{\sigma}\right) \right) \]
05

- Expected Lifetime Calculation

To find the expected lifetime of the system \( E[T] \), integrate the reliability function \( R(t) \) over all time: \[ E[T] = \int_0^{\infty} R(t) \, dt \] Substitute \( R(t) \) obtained in previous step:\[ E[T] = \int_0^{\infty} e^{-\lambda t} \cdot \left(1 - \Phi \left(\frac{t - \mu}{\sigma}\right) \right) \, dt \]
06

- Simplify the Integral

Use integration techniques for the exponential function and standard normal distribution properties. Recognize that the integral involves the convolution of an exponential and normal distribution, where the expected lifetime can be derived by evaluation.The integral evaluates to: \[ E[T] = \frac{1}{\lambda} \left[1 - \exp \left(-\lambda \mu + \frac{\lambda^2 \sigma^2}{2}\right)\right] \]

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Key Concepts

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

Series System
In a series system, all components must function properly for the system to be operational. If one component fails, the entire system fails. Therefore, the overall reliability of the system depends on each individual component's reliability. In mathematical terms, for a series system with components having reliabilities \(R_1(t)\) and \(R_2(t)\), the combined reliability \(R(t)\) is calculated using the product of individual reliabilities: \[ R(t) = R_1(t) \times R_2(t) \] This relationship indicates that even if one component has high reliability, the overall system reliability might decrease significantly if another component has low reliability. Understanding this concept is vital for analyzing and improving the reliability of systems with multiple parts.
Exponential Distribution
The exponential distribution is often used to model the time until an event, such as a failure, happens. It's characterized by a constant failure rate, which makes it memoryless. This means the probability of the component failing within the next time unit is the same, regardless of how long it has already been operating. The parameter \lambda (lambda) represents the failure rate. The reliability function, which gives the probability that a component will function for a time \(t\) without failing, is given by: \[R(t) = e^{-\lambda t}\] Here, \lambda is crucial in determining how quickly reliability declines over time. For instance, a higher \lambda value means a faster decline in reliability, indicating a more failure-prone component.
Normal Distribution
The normal distribution is another common probability distribution used to model the lifetime of components. Unlike the exponential distribution, the normal distribution describes a variable measured around a mean value \mu (mu) with a spread characterized by the variance \sigma^2 (sigma squared). The reliability function for a normally distributed component's lifetime is obtained from the probability that the component's lifetime \((T_2)\) is greater than a given time \(t\): \[R_2(t) = 1 - \Phi \left(\frac{t - \mu}{\sigma}\right)\]Here, \Phi denotes the cumulative distribution function (CDF) of the standard normal distribution. This function gives the probability that the lifetime exceeds \(t\), thus providing the reliability at that time.
Reliability Function
The reliability function is central to understanding system performance over time. For a series system with components having different distributions, the reliability function is the product of the individual reliabilities. Given an exponentially distributed first component and a normally distributed second component, the system reliability at time \(t\) is given by: \[ R(t) = e^{-\lambda t} \times \left(1 - \Phi\left(\frac{t - \mu}{\sigma}\right) \right)\]This shows how the system's probability of functioning properly until time \(t\) is influenced by the characteristics of both distributions. By multiplying the individual reliabilities, the overall reliability accounts for the independent probability of each component's performance over time.
Expected Lifetime Calculation
To determine the expected lifetime \(E[T]\) of the series system, integrate the reliability function over all time. This means summing the probabilities of the system surviving each small interval over its entire life. The integral is expressed as: \[ E[T] = \int_0^{\infty} R(t) \, dt \]Substituting the reliability function: \[ E[T] = \int_0^{\infty} e^{-\lambda t} \times \left(1 - \Phi\left(\frac{t - \mu}{\sigma}\right)\right) \, dt\] Solving this integral, using the properties of exponential and normal distributions, results in: \[ E[T] = \frac{1}{\lambda} \left[1 - e^{-\lambda \mu + \frac{\lambda^2 \sigma^2}{2}}\right]\] This formula shows the expected lifetime in terms of the failure rate \lambda, mean \mu and variance \sigma^2 of the components. Understanding this helps in predicting and enhancing the lifetime of complex systems.

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Most popular questions from this chapter

The CPU time requirement \(X\) of a typical job can be modeled by the following hyperexponential distribution: $$ P(X \leq t)=\alpha\left(1-e^{-\lambda_1 t}\right)+(1-\alpha)\left(1-e^{-\lambda_2 t}\right), $$ where \(\alpha=0.6, \lambda_1=10\), and \(\lambda_2=1\). Compute (a) the probability density function of \(X\), (b) the mean service time \(E[X]\), (c) the variance of service time \(\operatorname{Var}[X]\), and (d) the coefficient of variation. Plot the distribution and the density function of \(X\).

Consider a discrete random variable \(X\) with the following pmf: $$ p_X(x)= \begin{cases}\frac{1}{x(x+1)}, & x=1,2, \ldots \\ 0, & \text { otherwise }\end{cases} $$ cant copy graph Figure 4.P.1. An alternative method of computing \(E[X]\) Show that the function defined satisfies the properties of a pmf. Show that the formula (4.1) of expectation does not converge in this case and hence \(E[X]\) is undefined. [Hint: Rewrite \(1 / x(x+1)\) as \(1 / x-1 /(x+1) \cdot]\)

\({ }^*\) The problem of dynamic storage allocation in the main memory of a computer system [WOLM 1965] can be simplified by choosing a fixed node size, \(k\), for allocation. Out of the \(k\) units (bytes, say) of storage allocated to a node, only \(k-b\) bytes are available to the user, since \(b\) bytes are required for control information. Let the random variable \(L\) denote the length in bytes of a user request. Thus \(\lceil L /(k-b)\rceil\) nodes must be allocated to satisfy the user request. Thus the total number of bytes allocated is \(X=k\lceil L /(k-b)\rceil\). Find \(E[X]\) as a function of \(k\) and \(E[L]\). Then, by differentiating \(E[X]\) with respect to \(k\), show that the optimal value of \(k\) is approximately \(b+\sqrt{2 b E[L]}\).

The time to failure distribution of Tandem software was found to be captured well by a two-phase hyperexponential distribution with the following pdf: $$ f(t)=\alpha_1 \lambda_1 e^{-\lambda_1 t}+\alpha_2 \lambda_2 e^{-\lambda_2 t}, $$ with \(\alpha_1=0.87, \alpha_2=0.13, \lambda_1=0.10, \lambda_2=2.78\) [LEE 1993]. Find the mean and variance of the time to failure.

Consider a parallel redundant system of two independent components with the lifetime of \(i\) th component \(X_i \sim \operatorname{EXP}\left(\lambda_i\right)\). Show that system MTTF is given by $$ \operatorname{MTTF}=\frac{1}{\lambda_1}+\frac{1}{\lambda_2}-\frac{1}{\lambda_1+\lambda_2} . $$ Generalize to the case of \(n\) components. Next consider a standby redundant system consisting these two components. Assuming that the component in the spare status does not fail, obtain the reliability and the MTTF of the system.

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