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Chapter 4: Problem 3

The failure rate for a certain type of component is \(\lambda(t)=a t(t \geq 0)\), where \(a>0\) and is constant. Find the component's reliability, and its expected life (or MTTF).

Short Answer

Expert verified
Reliability: \(R(t) = e^{-\frac{a t^2}{2}}\). MTTF: \(\sqrt{\frac{2}{a}}\).

Step by step solution

01

- Define Reliability Function

The reliability function, denoted as R(t), is given by\[R(t) = e^{-\int_0^t \lambda(s) \, ds}\]Substitute the given failure rate \(\lambda(t) = at\) into the integral.
02

- Evaluate the Integral

Compute the integral in the exponent:\[\int_0^t \lambda(s) \, ds = \int_0^t as \, ds = a \int_0^t s \, ds = a \left[ \frac{s^2}{2} \right]_0^t = \frac{a t^2}{2}\]
03

- Substitute and Simplify

Substitute \(\int_0^t \lambda(s) \, ds = \frac{a t^2}{2}\) back into the reliability function:\[R(t) = e^{-\frac{a t^2}{2}}\]
04

- Define Expected Lifetime (MTTF)

The expected lifetime (or Mean Time To Failure, MTTF) is given by the integral\[\text{MTTF} = \int_0^\infty R(t) \, dt\]Substitute the reliability function \(R(t)= e^{-\frac{a t^2}{2}}\) into the integral.
05

- Compute the MTTF Integral

Evaluate the integral for MTTF using the substitution \(u = \frac{a t^2}{2}\):\[MTTF = \int_0^\infty e^{-\frac{a t^2}{2}} \, dt = \int_0^\infty e^{-u} \frac{du}{a t} = \int_0^\infty \, \frac{e^{-u} \, du}{\sqrt{2a}} = \sqrt{\frac{2}{a}} \int_0^\infty e^{-u} \, du = \sqrt{\frac{2}{a}}\]

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Key Concepts

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

failure rate
Understanding this relationship helps engineers and scientists in planning maintenance schedules and improving the design of the component to enhance its longevity and performance.
reliability function
This equation tells us that as time progresses, the reliability function decreases. The decrease is exponential in nature, indicating that the component becomes significantly less reliable over time. Engineers use this function to predict the lifespan and usability of components in various applications.
Mean Time To Failure (MTTF)
MTTF provides valuable insights into how long a component is expected to perform reliably in its intended application. It helps in planning replacements and in understanding the factors that contribute to the longevity of the component.
expected lifetime
For consumers, understanding the expected lifetime of components they use can inform their maintenance schedules and help them make better purchasing decisions by comparing different products based on reliability metrics.

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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 \(X\).

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