91Ó°ÊÓ

A system consists of five identical components connected in series as shown: As soon as one component fails, the entire system will fail. Suppose each component has a lifetime that is exponentially distributed with \(\lambda=.01\) and that components fail independently of one another. Define events \(A_{i}=\) \(\\{i\) th component lasts at least \(t\) hours \(\\}, i=1, \ldots, 5\), so that the \(A_{i} \mathrm{~s}\) are independent events. Let \(X=\) the time at which the system fails-that is, the shortest (minimum) lifetime among the five components. a. The event \(\\{X \geq t\\}\) is equivalent to what event involving \(A_{1}, \ldots, A_{5}\) ? b. Using the independence of the \(A_{i}{ }^{\prime}\) s, compute \(P(X \geq t)\). Then obtain \(F(t)=P(X \leq t)\) and the pdf of \(X\). What type of distribution does \(X\) have? c. Suppose there are \(n\) components, each having exponential lifetime with parameter \(\lambda\). What type of distribution does \(X\) have?

Step by step solution

01

Define the Event in Terms of Components

We need to express the event \( \{ X \geq t \} \), which means the system lasts at least \( t \) hours. Since the system fails when any component fails, this is equivalent to all components lasting at least \( t \) hours. Therefore, \( \{ X \geq t \} \) is equivalent to \( A_1 \cap A_2 \cap A_3 \cap A_4 \cap A_5 \).

02

Probability of X ≥ t

The probability that each component lasts at least \( t \) hours is \( P(A_i) = e^{-\lambda t} \), given the exponential distribution with rate \( \lambda = 0.01 \). Since the components are independent, this gives \[ P(X \geq t) = P(A_1 \cap A_2 \cap A_3 \cap A_4 \cap A_5) = (e^{-\lambda t})^5 = e^{-5\lambda t}. \]

03

Calculate F(t) and PDF of X

The cumulative distribution function (CDF) for failure time is \( F(t) = P(X \leq t) = 1 - P(X \geq t) = 1 - e^{-5\lambda t} \). The probability density function (PDF) is the derivative of the CDF, \[ f(t) = \frac{d}{dt}(1 - e^{-5\lambda t}) = 5\lambda e^{-5\lambda t}. \] The distribution of \( X \) is exponential with rate \( 5\lambda = 0.05 \).

04

Generalize for n Components

With \( n \) components, each having an exponential lifetime with parameter \( \lambda \), the minimum lifetimes \( X \) also follow an exponential distribution. The rate parameter for \( X \) in this case is \( n\lambda \), given by \[ f(t) = n\lambda e^{-n\lambda t}. \] This confirms that for \( n \) independent exponentially distributed components, \( X \) follows an exponential distribution with parameter \( n\lambda \).

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

System Reliability

When referring to system reliability, we often look at how dependable a system is over time, especially in terms of its probability to perform successfully without failure. For systems with components in series, just like in the given exercise, understanding system reliability is crucial. Here, the system fails as soon as any one component fails, which is a common setup in many real-world applications.
In such a system, the reliability is directly linked to the reliability of individual components. If one fails, the whole system is considered to have failed. Given the exponential nature of each component’s lifetime, we use probabilistic methods to estimate system reliability.
More generally:

• The reliability of a system in series configuration is less than that of any individual component.
• This emphasizes the importance of high reliability at the component level to ensure overall system reliability.

Independent Events

In probability theory, independent events are those whose outcomes do not affect each other. In the context of the problem, each component's failure is independent of the others due to the given exponential lifetime distribution.
For independent components, the probability of all components lasting at least a certain time can be calculated by multiplying the probabilities for each component. This beneficial property simplifies calculations like those in the exercise, where we calculate the joint probability of all components lasting a certain period.
Key points:

• Independence allows us to multiply probabilities for joint events.
• This is crucial for calculating overall system probability outcomes when components act independently.

Probability Density Function

The probability density function (PDF) gives insights into the likelihood of a random variable taking on a particular value. With exponential distributions, the PDF describes how the lifetime of a component is distributed.
In the exercise, the PDF is derived from the cumulative distribution function (CDF). It gives us the probability density of the system failing at a precise time, offering essential insights into both expected system behavior and risks involved.
Essential elements:

• The PDF for the system's failure time is a derivative of its CDF.
• Understanding the PDF allows us to predict how often the system might fail within a certain time frame.

Minimum Lifetime

The minimum lifetime in this context refers to the shortest time to failure among all system components. It is the driving factor in determining when the entire system will stop functioning.
For components with exponential distributions, the minimum lifetime also follows an exponential distribution. The rate parameter of this new distribution is adjusted by multiplying the original rate by the number of components, as shown in the exercise.
These insights highlight:

• The exponential nature of the minimum lifetime's distribution, demonstrating how independent components affect overall system behavior.
• By understanding the minimum lifetime, you can assess and enhance system design for improved durability.