reliability design example

To do so, first approximate the expected value and variance of prior system reliability [math]R_{0}\,\![/math]. This generally means ensuring that things continue to conform to requirements in the face of real world conditions. The benchmark study will help you fill in gaps by identifying existing internal best practices and techniques to yield the desired results. This includes: Readers may also be interested in test design methods for quantitative accelerated life tests. Reliability Testing can be categorized into three segments, 1. If the expected test duration can be estimated prior to the test, test resources can be better allocated. We have to either increase the sample size or the test duration. So, if we duplicate the devices at each stage then the reliability of the system can be increased. The split-half method assesses the internal consistency of a test, such as psychometric tests and questionnaires. By substituting [math]f=0\,\! This example solved in Weibull++ is shown next. In the above scenario, we know that we have the testing facilities available for [math]t=48\,\! }{i!\cdot (n-i)! The SimuMatic utility in Weibull++ can be used for this purpose. [/math] are then calculated as before: For each subsystem i, from the beta distribution, we can calculate the expected value and the variance of the subsystem’s reliability [math]R_{i}\,\! \end{align}\,\! [/math], [math]\beta_{0}=\left(1-E\left(R_{0}\right)\right)\left[\frac{E\left(R_{0}\right)-E^{2}\left(R_{0}\right)}{Var\left(R_{0}\right)}-1\right]\,\! Languages: Assume that there are two design options for a new product. But for this data to be of any use, the tests must possess certain properties like reliability and validity, that ensure unbiased, accurate, and authentic results. The Weibull reliability equation is: Since we know the values of [math]{{t}_{DEMO}}\,\! [/math], [math] \alpha\,\!=\alpha\,\!_{0}+s=146.07943\,\! But this maximization should be considered along with the cost. [/math], and must determine the test time, [math]{{t}_{TEST}}\,\![/math]. The Concepts of Reliability and Validity Explained With Examples All research is conducted via the use of scientific tests and measures, which yield certain observations and data. [/math] and [math]\theta\,\! [/math] and [math]f\,\! [/math], [math]1-CL=\underset{i=0}{\overset{f}{\mathop \sum }}\,\frac{n! » Contact us Based on previous experiments, they assume the underlying failure distribution is a Weibull distribution with [math]\beta = 2\,\! » Kotlin Are we designing the system with reliability and maintenance in mind? The demonstrated reliability is 68.98% as shown below. For example, a design should require the minimal possible amount of non-value-added manual work and assembly. [/math], [math]{{t}_{TEST}}\,\! Solved programs: [/math] from the binomial equation with Weibull distribution. [/math], [math]E\left(R_{i}\right)=\frac{s_{i}}{n_{i}+1}\,\! » SQL After analyzing the data set with the MLE and FM analysis options, we can now calculate the B10 life and its interval in the QCP, as shown next. That topic is discussed in the Accelerated Life Testing Reference. In analytical methods, both Fisher bounds and likelihood ratio bounds need to use assumptions. From the above results, we can see the upper bound of the last failure is about 955 hours. 17 Examples of Reliability posted by John Spacey, January 26, 2016 updated on February 06, 2017. The following are reliability engineering techniques and considerations. » SEO We have already determined the value of the scale parameter, [math]\eta \,\! [/math] have already been calculated or specified, so it merely remains to solve the binomial equation for [math]n\,\![/math]. In this article, we will learn about the concept of reliability design problem. E\left(R_{i}\right)=\frac{n_{i}-r_{i}}{n_{i}+1} [/math] are already known, and it is just a matter of plugging these values into the appropriate reliability equation. It can be said that multiple copies of the same device type are connected in parallel through the use of switching circuits. In cases like this, it is useful to have a "carpet plot" that shows the possibilities of how a certain specification can be met. Design Situation 1: One Variable Load Design Situation 2: Two Variable Loads Check Design Situation Structural Steel, etc. [/math], which is the reliability that is going to be incorporated into the actual test calculation. This value is [math]n=85.4994\,\! [/math] units, since the fractional value must be rounded up to the next integer value. [/math] known, the above beta distribution equation can now be used to calculate a quantity of interest. Then the parameters in the posterior beta distribution for R are calculated as: Finally, from this posterior distribution, the system reliability R at a confidence level of CL=0.9 is solved as: Given R = 0.85, n = 20, and r = 1, using the above prior information on system reliability to solve for CL. Using the estimated median rank for each failure and the assumed underlying failure distribution, we can calculate the expected time for each failure. During this correct operation, no repair is required or performed, and the system adequately follows the defined performance specifications. Let c is the maximum allowable cost and ci be the cost of each unit of device i. In reliability design, the problem is to design a system that is composed of several devices connected in series.. [/math], [math]Var\left(R_{0}\right)=\prod_{i=1}^{k}\left[E^{2}\left(R_{i}\right)+Var\left(R_{i}\right)\right]-\prod_{i=1}^{k}\left[E^{2}\left(R_{i}\right)\right]\,\! For the initial setup, set the sample size for each design to 20, and use two test durations of 3,000 and 5,000 hours. [/math], [math]\begin{align} If 11 samples are used and one failure is observed by the end of the test, then the demonstrated reliability will be less than required. » C++ [/math], in the previous example. [/math] is 2117.2592 hours. This methodology requires the use of the cumulative binomial distribution in addition to the assumed distribution of the product's lifetimes. We now incorporate a form of the cumulative binomial distribution in order to solve for the required number of units. [/math], [math]{{R}_{TEST}}=g({{t}_{TEST}};\theta ,\phi )\,\! Example: Suppose a questionnaire is distributed among a group of people to check the quality of a skincare product and repeated the same questionnaire with many groups. In this example, we will use the parametric binomial method to design a test to demonstrate a reliability of 90% at [math]{{t}_{DEMO}}=100\,\! [/math], [math]{{R}_{TEST}}={{e}^{-{{({{t}_{TEST}}/\eta )}^{\beta }}}}={{e}^{-{{(60/83.1)}^{1.5}}}}=0.541=54.1%\,\! }\cdot {{(1-{{R}_{TEST}})}^{i}}\cdot R_{TEST}^{(n-i)}\,\! [/math] is calculated by: The last step is to substitute the appropriate values into the cumulative binomial equation, which for the Weibull distribution appears as: The values of [math]CL\,\! Join our Blogging forum. It will also help define a set of reliability practices to move defec… [/math] is calculated as: The last step is to substitute the appropriate values into the cumulative binomial equation. [/math], [math]f\,\! [/math] depending on the type of prior information available. The process steps each include a slightly different focus and set of tools. This example solved in Weibull++ is shown next. Reliability is the probability that a product will continue to work normally over a specified interval of time, under specified conditions. [/math], [math]{{R}_{TEST}}={{e}^{-{{({{t}_{TEST}}/\eta )}^{\beta }}}}={{e}^{-{{(48/448.3)}^{1.5}}}}=0.966=96.6%\,\! [/math] are required inputs to the process and [math]{{R}_{TEST}}\,\! [/math], [math]\eta =\frac{MTTF}{\Gamma (1+\tfrac{1}{\beta })}\,\! If we imagine that r1 is the reliability of the device. [/math], https://www.reliawiki.com/index.php?title=Reliability_Test_Design&oldid=61749. The calculated Q is given in the figure below: In this example you will use the Expected Failure Times plot to estimate the duration of a planned reliability test. » C » C This is done by comparing the results of one half of a test with the results from the other half. For Design 1, its shape parameter [math]\beta = 3\,\! [/math], [math] R=\text{BetaINV}\left(1-CL,\alpha\,\!,\beta\,\!\right)=0.838374 \,\! Given the above subsystem test information, in order to demonstrate the system reliability of 0.9 at a confidence level of 0.8, how many samples are needed in the test? Related terms: Reliability Analysis; Power Device Run-length encoding (find/print frequency of letters in a string), Sort an array of 0's, 1's and 2's in linear time complexity, Checking Anagrams (check whether two string is anagrams or not), Find the level in a binary tree with given sum K, Check whether a Binary Tree is BST (Binary Search Tree) or not, Capitalize first and last letter of each word in a line, Greedy Strategy to solve major algorithm problems. [/math], [math]Var\left(R_{i}\right)=\frac{s_{i}\left(n_{i}+1-s_{i}\right)}{\left(n_{i}+1\right)^{2}\left(n_{i}+2\right)}\,\! Reliability engineering is a well-developed discipline closely related to statistics and probability theory. [/math] and [math]\phi\,\! In this example, you will use the Difference Detection Matrix to choose the suitable sample size and duration for a reliability test. The values of [math]\alpha_{0}\,\! : The probability that a PC in a store is up and running for eight hours without crashing is 99%; this is referred as reliability. A value of 0 means the difference cannot be detected through the test, 1 means the difference can be detected if the test duration is 5,000 hours, and 2 means the difference can be detected if the test duration is 3,000 hours. [/math], [math] For example, given n = 4, r = 2 and CL = 0.5, the calculated Q is 0.385728. [/math] hours with a 95% confidence if no failure occur during the test. » C This chapter discusses several methods for designing reliability tests. These represent the true exponential distribution confidence bounds referred to in The Exponential Distribution. Aptitude que. If the reliability of the system is less than or equal to 80%, the chance of passing this test is 1-CL = 0.1, which is the Type II error. Several methods have been designed to help engineers: Cumulative Binomial, Non-Parametric Binomial, Exponential Chi-Squared and Non-Parametric Bayesian. [/math] can be calculated. » Java [math]\alpha_{0}\,\! [/math], [math]\beta_{0}=\left(1-E \left(R_{0}\right)\right)\left[\frac{E\left(R_{0}\right)-E^{2}\left(R_{0}\right)}{Var \left(R_{0}\right)}-1\right]=5.448499634\,\! The binomial equation used in non-parametric demonstration test design is the base for predicting expected failure times. Example: The levels of employee satisfaction of ABC Company may be assessed with questionnaires, in-depth interviews and focus groups and results can be compared. [/math], we can substitute these in the equation and solve for [math]\eta \,\! [/math], [math]f\,\! [/math], and the confidence level, [math]CL\,\![/math]. » DBMS [/math], [math]1-CL=\underset{i=0}{\overset{r}{\mathop \sum }}\,\frac{n! » DOS first half and second half, or by odd and even numbers. » CS Basics In other words, in cases where the available test time is equal to the demonstration time, the following non-parametric binomial equation is widely used in practice: where [math]CL\,\! Using this value and the assumed Weibull distribution, the median value of the failure time of the second failure is calculated as: Its bounds and other failure times can be calculated in a similar way. » News/Updates, ABOUT SECTION [/math], [math]{{R}_{DEMO}}\,\! [/math], [math]\begin{align} We want to determine the number of units to test for [math]{{t}_{TEST}}=60\,\! From this result, we can see that the estimated B10 life and its confidence intervals are the same as the results displayed in the Difference Detection Matrix. In this example, we will use the parametric binomial method to design a test that will demonstrate [math]MTTF=75\,\! The test is time-terminated and the termination time is set to T. Using the method given in Expected Failure Times Plots, we can generate the failure times. » Machine learning }{i!\cdot (n-i)! This method only returns the necessary accumulated test time for a demonstrated reliability or [math]MTTF\,\! » C++ [/math] and [math]\beta_{0}\,\! Figure 7.2 Design for reliability (DfR) activities flow, from Practical Reliability Engineering, outlines the basic stages or elements of a product generation process. This requires knowledge of the lowest possible reliability, the most likely possible reliability and the highest possible reliability of the system. [/math], [math] CL=\text{Beta}\left(R,\alpha,\beta\right)=0.81011 \,\! » Data Structure Reliability is the ability of things to perform over time in a variety of expected conditions. [/math] value. Design for reliability (or RBDO) includes two distinct categories of analysis, namely (1) design for variability (or variability-based design optimization), which focuses on the variations at a given moment in time in the product life; From: Diesel Engine System Design, 2013. In this case, [math]{{R}_{TEST}}\,\! The questions are how many samples and how long should the test be conducted in order to detect a certain amount of difference. [/math], [math]{{T}_{a}}=\frac{\tfrac{{{t}_{DEMO}}}{-ln(R)}\cdot \chi _{1-CL;2f+2}^{2}}{2}\,\! & ans. [/math] for the Weibull distribution is: where [math]\Gamma (x)\,\! Design modifications might be necessary to improve robustness. Achieving reliability, however, requires thoughtful planning and execution. This means, at the time when the second failure occurs, the estimated system probability of failure is 0.385728. [/math], [math]{{t}_{DEMO}}\,\! By running the simulations you can assess whether the planned test design can achieve the reliability target. [/math] for the Weibull distribution using the Quick Parameter Estimator tool, as shown next. Then they make use of such devices at each stage, that result is increase in reliability at each stage. [/math], [math]\text{Var}\left(R_{0}\right)=0.003546663\,\! Since required inputs to the process include [math]{{R}_{DEMO}}\,\! [/math] and [math]\beta_{0}\,\! Another advantage of using the simulation method is that it is straightforward and results can be visually displayed in SimuMatic. [/math], [math]f\,\! For example, suppose you wanted to know the reliability of a system and you had the following prior knowledge of the system: This information can be used to approximate the expected value and the variance of the prior system reliability. One of the key factors in asset/system performance is its reliability- “inherent reliability” or designed in reliability. With this information, the next step involves solving the binomial equation for [math]{{R}_{TEST}}\,\![/math]. This can be rearranged in terms of [math]\eta\,\! [/math], [math]1-CL=\underset{i=0}{\overset{f}{\mathop \sum }}\,\frac{n! » O.S. For details, see the Weibull++ SimuMatic chapter. The engineers need to design a test that compares the reliability performance of these two options. The median failure times are used to estimate the failure distribution. & Q=1-{{e}^-{{{\left( \frac{t}{\eta } \right)}^{\beta }}}}\Rightarrow \\ [/math], [math]{{T}_{a}}=\frac{MTTF\cdot \chi _{1-CL;2f+2}^{2}}{2}\,\! [/math], [math]\eta =\frac{{{t}_{DEMO}}}{{{(-\text{ln}({{R}_{DEMO}}))}^{\tfrac{1}{\beta }}}}\,\! » DS There, it measures the extent to which all parts of the test contribute equally to what is being measured. For the above example, if we set CL=0.9, from the calculated Q we can get the upper bound of the time for each failure. [/math] and [math]\eta \,\! By testing 20 samples each for 3,000 hours, the difference of their B10 lives probably can be detected. For cell (1000, 2000), Design 1's B10 life is 1,000 and the assumed [math]\beta\,\! Ad: Author: Andrew Taylor BSc MA FRSA - Art and Engineering in Product Design Design for Reliability What is Product Reliability? When sample size is small or test duration is short, these assumptions may not be accurate enough. You can specify various factors of the design, such as the test duration (for a time-terminated test), number of failures (for a failure-terminated test) and sample size. In this case, we will assume that we have 20 units to test, [math]n=20\,\! For example, the mouse on your computer In reliability design, we try to use device duplication to maximize reliability. Then the reliability of the function can be given by πr1. Usually the engineer designing the test will have to study the financial trade-offs between the number of units and the amount of test time needed to demonstrate the desired goal. The reliability of the system can be given as follows: If we increase the number of devices at any stage beyond the certain limit, then also only the cost will increase but the reliability could not increase. » DBMS Depending on the results, you can modify the design by adjusting these factors and repeating the simulation process—in effect, simulating a modified test design—until you arrive at a modified design that is capable of demonstrating the target reliability within the available time and sample size constraints. [/math] are calculated as: With [math]\alpha_{0}\,\! }{i!\cdot (n-i)! [/math] equation, and following the previously described methodology to determine [math]{{t}_{TEST}}\,\! These values can then be used to find the prior system reliability and its variance: From the above two values, the parameters of the prior distribution of the system reliability can be calculated by: With this prior distribution, we now can design a system reliability demonstration test by calculating system reliability R, confidence level CL, number of units n or number of failures r, as needed. The figure below shows the result from Weibull++. [/math], [math]{{t}_{TEST}}\,\! With the exception of the exponential distribution (and ignoring the location parameter for the time being), this reliability is going to be a function of time, a shape parameter and a scale parameter. Then the reliability of the function can be given by πr1. The results of these calculations are given in the table below. Assume the failure distribution is Weibull, then we know: Using the above equation, for a given Q, we can get the corresponding time t. The above calculation gives the median of each failure time for CL = 0.5. » C++ [/math], [math] \alpha\,\!_{0}=E\left(R_{0}\right)\left[\frac{E\left(R_{0}\right)-E^{2}\left(R_{0}\right)}{Var\left(R_{0}\right)}-1\right]=127.0794\,\! [/math], the number of units that need to be tested. [/math] and [math]\beta_{0}\,\![/math]. » Java 1-CL=\sum_{i=0}^{f}\binom{n}{i}(1-{{R}_{TEST}})^{i}{{R}_{TEST}}^{n-i} [/math] have already been calculated or specified. You can use the non-parametric Bayesian method to design a test for a system using information from tests on its subsystems. [/math], [math]Var({{R}_{0}})={{\left( \frac{c-a}{6} \right)}^{2}}=0.000803 \,\! [/math], [math]CL\,\! The product's reliability should be reevaluated in light of these additional variables. [/math] has already been calculated, it merely remains to solve the cumulative binomial equation for [math]n\,\! For Reliability Design with Example in Hindi Follow: https://www.youtube.com/watch?v=HAFjqjuUUQQ See the Worked out example starts at 00:04:00. Reliability engineering is the design, production and operation of things to retain their quality over time. [/math], [math]{{T}_{a}}=\frac{\tfrac{500}{-ln(0.85)}\cdot 10.6446}{2}=16,374\text{ hours}\,\! This example solved in Weibull++ is shown next. As we know, with 4 samples, the median rank for the second failure is 0.385728. If the two halves of th… If we assume the system reliability follows a beta distribution, the values of system reliability, R, confidence level, CL, number of units tested, n, and number of failures, r, are related by the following equation: where [math]Beta\,\! It’s important to consider reliability and validity when you are creating your research design, planning your methods, and writing up your results, especially in quantitative research. Therefore, the non-parametric binomial equation determines the sample size by controlling for the Type II error. Again, the above beta distribution equation for the system reliability can be utilized. For example, the confidence bounds of reliability from SimuMatic are purely based on simulation results. Assume the allowed number of failures is 1. 4 units were allocated for the test, and the test engineers want to know how long the test will last if all the units are tested to failure. However, there are difficulties with applying the traditional DOE analysis methods, such as ANOVA or … This page was last edited on 10 December 2015, at 21:22. The simulation method usually does not require any assumptions. [/math] have already been calculated or specified, so it merely remains to solve the equation for [math]n\,\![/math]. Submitted by Shivangi Jain, on August 21, 2018. [/math] and [math]{{\beta}_{0}} \gt 0\,\! With this value known, one can use the appropriate reliability equation to back out the value of [math]{{t}_{TEST}}\,\! [/math], [math] \beta\,\!=\beta\,\!_{0}+r=21.40153\,\! [/math], and [math]{{R}_{TEST}}\,\! The regular non-parametric analyses performed based on either the binomial or the chi-squared equation were performed with only the direct system test data. » Java Var\left(R_{i}\right)=\frac{\left(n_{i}-r_{i}\right)\left(r_{i}+1\right)}{\left(n_{i}+1\right)^{2}\left(n_{i}+2\right)} [/math] from the [math]MTTF\,\! In this case, the last failure is a suspension with a suspension time of 3,000 hours. [/math] hours. Assume we want to compare the B10 lives (or mean lives) of two designs. Given the value of the [math]MTTF\,\! Then the maximization problem can be given as follows: Here, Øi (mi) denotes the reliability of the stage i. And the reliability of the stage I becomes (1 – (1 - ri) ^mi). Reliability is about the consistency of a measure, and validity is about the accuracy of a measure. [/math], [math]E\left(R_{0}\right)=0.846831227\,\! Their B10 lives may range from 500 to 3,000 hours. Note that since the test duration is set to 3,000 hours, any failures that occur after 3,000 are treated as suspensions. [/math], [math]\alpha_{0}=E\left(R_{0}\right)\left[\frac{E\left(R_{0}\right)-E^{2}\left(R_{0}\right)}{\text{Var}\left(R_{0}\right)}-1\right]=30.12337003\,\! [/math], [math] \alpha\,\!=\alpha\,\!_{0}+s=146.0794\,\! [/math] is the test time. How this calculation is performed depends on whether one is attempting to solve for the number of units to be tested in an available amount of time, or attempting to determine how long to test an available number of test units. Use Prior Expert Opinion on System Reliability, Use Prior Information from Subsystem Tests, [math]{{R}_{DEMO}}=g({{t}_{DEMO}};\theta ,\phi )\,\! You can use the non-parametric Bayesian method to design a test using prior knowledge about a system's reliability. Measurement 3. The following picture shows the complete control panel setup and the results of the analysis. If r1 = 0.99 and n = 10 that n devices are set in a series, 1 <= i <= 10, then reliability of the whole system πri can be given as: Πri = 0.904. Submitted by Shivangi Jain, on August 21, 2018 . More Resources: Weibull++ Examples Collection, Download Reference Book: Life Data Analysis (*.pdf), Generate Reference Book: File may be more up-to-date. \end{align}\,\! Monte Carlo simulation provides another useful tool for test design.

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