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Chapter 3: Problem 1

A system with three independent components works correctly if at least one component is functioning properly. Failure rates of the individual components are \(\lambda_1=0.0001, \lambda_2=0.0002\), and \(\lambda_3=0.0004\) (assume exponential lifetime distributions). (a) Determine the probability that the system will work for \(1000 \mathrm{~h}\). (b) Determine the density function of the lifetime \(X\) of the system.

Short Answer

Expert verified
The probability that the system will work for 1000 hours is approximately 0.9943. The density function of the lifetime of the system is \[ f_X(t) = 0.0007 e^{-0.0007 t} \].

Step by step solution

01

Find the reliability of each component

The reliability of each component can be found using the exponential distribution formula: \( R_i(t) = e^{-\text{failure rate} \times \text{time}} \). For each component, we have:\[ R_1(t) = e^{-0.0001 \times 1000}, \ R_2(t) = e^{-0.0002 \times 1000}, \ R_3(t) = e^{-0.0004 \times 1000} \]
02

Calculate the reliability for each component at 1000 hours

Substitute the values to get:\[ R_1(1000) = e^{-0.1} \approx 0.9048 \], \[ R_2(1000) = e^{-0.2} \approx 0.8187 \], \[ R_3(1000) = e^{-0.4} \approx 0.6703 \]
03

Determine the probability that the system works

The system works if at least one component works. Use the formula:\[ R_{system}(t) = 1 - (1 - R_1(t))(1 - R_2(t))(1 - R_3(t)) \]. Substitute the values of reliabilities at 1000 hours:\[ R_{system}(1000) = 1 - (1 - 0.9048)(1 - 0.8187)(1 - 0.6703) \]
04

Compute the overall probability

\[ R_{system}(1000) = 1 - (0.0952)(0.1813)(0.3297) \approx 1 - 0.0057 \approx 0.9943 \]
05

Find the failure rate of the system

Since the system works as long as at least one component works, we need to find the combined failure rate:\[ \text{Combined failure rate} = \text{failure rate of component 1} + \text{failure rate of component 2} + \text{failure rate of component 3} \]. Thus, \( \text{Combined failure rate} = 0.0001 + 0.0002 + 0.0004 = 0.0007 \)
06

Determine the probability density function of the system's lifetime

The density function of the lifetime of the system follows an exponential distribution with the combined failure rate:\[ f_X(t) = \text{combined failure rate} \times e^{-\text{combined failure rate} \times t} \]. Therefore, \[ f_X(t) = 0.0007 \times e^{-0.0007 \times t} \]

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Key Concepts

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

exponential distribution
An exponential distribution is a continuous probability distribution used to model the time until an event occurs. Here, it is utilized to determine the likelihood of components lasting a certain duration without failure. The key parameter in the exponential distribution is the failure rate, denoted as \(\lambda\). The probability density function (PDF) of the exponential distribution is given by \ f(t) = \lambda e^{-\lambda t} \ where \(t\) is the time. For example, if the failure rate of a component is \(\lambda = 0.0001\), the PDF would be \ f(t) = 0.0001 e^{-0.0001 t} \.
failure rates
Failure rates represent the frequency at which a component or system is expected to fail within a certain timeframe. In this exercise, the failure rates for the three independent components are \(\lambda_1 = 0.0001\), \(\lambda_2 = 0.0002\), and \(\lambda_3 = 0.0004\). These values indicate the number of failures expected per hour of operation. For any time \(t\), the reliability \(R(t)\) of a component with failure rate \(\lambda\) is given by the reliability function \ R(t) = e^{-\lambda t} \
reliability function
The reliability function measures the probability that a system or component will perform its intended function without failure for a specific period. For an exponential distribution, the reliability function is \ R(t) = e^{-\lambda t} \ where \(R(t)\) is the probability that the component will not fail by time \(t\). For example, calculating \(R_1(1000) = e^{-0.0001 \times 1000} = e^{-0.1} \) gives \(R_1(1000) \approx 0.9048\), showing that there鈥檚 about a 90% chance of the first component operating properly for 1000 hours.
probability density function
The probability density function (PDF) provides the likelihood of the component failing at an exact time point. It is derived from the failure rate \(\lambda\) for an exponentially distributed lifetime. The PDF for such a distribution is given as \ f(t) = \lambda e^{-\lambda t} \. For instance, if the combined failure rate of the system is \(0.0007\), the PDF describing the system's lifetime is \ f_X(t) = 0.0007 e^{-0.0007 t} \.
system lifetime
System lifetime refers to the duration for which the entire system performs its function correctly. In our example, the system operates with at least one working component. Here, we calculated the system reliability as \(R_{system}(1000) \approx 0.9943\), meaning there's a 99.43% probability the system will work for 1000 hours. The system lifetime is modeled using the combined failure rate of individual components, resulting in an exponential distribution: \ f_X(t) = 0.0007 e^{-0.0007 t} \.

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Most popular questions from this chapter

Lifetimes of VLSI chips manufactured by a semiconductor manufacturer are approximately normally distributed with \(\mu=5 \times 10^6 \mathrm{~h}\) and \(\sigma=5 \times 10^5 \mathrm{~h}\). A computer manufacturer requires that at least \(95 \%\) of a batch should have a lifetime greater than \(4 \times 10^6 \mathrm{~h}\). Will the deal be made?

A series system has \(n\) independent components. For \(i=1,2, \ldots, n\), the lifetime \(X_i\) of the \(i\) th component is exponentially distributed with parameter \(\lambda_i\). Compute the probability that a given component \(j=(1,2, \ldots, n)\) is the cause of the system failure.

The CPU time requirement of a typical program measured in minutes is found to follow a three-stage Erlang distribution with \(\lambda=\frac{1}{2}\). What is the probability that the CPU demand of a program will exceed 1 min?

A calculator operates on two \(1.5-\mathrm{V}\) batteries (for a total of \(3 \mathrm{~V}\) ). The actual voltage of a battery is normally distributed with \(\mu=1.5\) and \(\sigma^2=0.45\). The tolerances in the design of the calculator are such that it will not operate satisfactorily if the total voltage falls outside the range \(2.70-3.30 \mathrm{~V}\). What is the probability that the calculator will function correctly?

A multiprocessor system has \(n\) processors. Service time of a process executing on the \(i\) th processor is exponentially distributed with parameter \(\mu_i(i=1,2, \ldots, n)\). Given that all \(n\) processors are active and that they are executing mutually independent processes, what is the distribution of time until a processor becomes idle?
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