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DOE++ Examples

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DOE Reference Examples

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Xfmea Risk Discovery Analysis Example

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Weibull++ Reliability Growth Module Examples

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RGA Reference Examples

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BlockSim Reference Examples

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Weibull++ Accelerated Life Testing Module Examples

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ALTA Reference Examples

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Reliability Allocation

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In the process of developing a new product, the engineer is often faced with the task of designing a system that conforms to a set of reliability specifications. The engineer is given the goal for the system and must then develop a design that will achieve the desired reliability of the system, while performing all of the system's intended functions at a minimum cost. This involves a balancing act of determining how to allocate reliability to the components in the system so the system will meet its reliability goal while at the same time ensuring that the system meets all of the other associated performance specifications.

BlockSim provide three allocation methods: equal allocation, weighted reliability allocation and cost optimzation allocation. In these three methods, the simplest method is equal reliability allocation, which distributes the reliabilities uniformly among all components. For example, suppose a system with five components in series has a reliability objective of 90% for a given operating time. The uniform allocation of the objective to all components would require each component to have a reliability of 98% for the specified operating time, since <math>{{0.98}^{5}}\tilde{=}0.90\,\!</math>. While this manner of allocation is easy to calculate, it is generally not the best way to allocate reliability for a system. The optimum method of allocating reliability would take into account the cost or relative difficulty of improving the reliability of different subsystems or components.

The reliability optimization process begins with the development of a model that represents the entire system. This is accomplished with the construction of a system reliability block diagram that represents the reliability relationships of the components in the system. From this model, the system reliability impact of different component modifications can be estimated and considered alongside the costs that would be incurred in the process of making those modifications. It is then possible to perform an optimization analysis for this problem, finding the best combination of component reliability improvements that meet or exceed the performance goals at the lowest cost.

==Improving Reliability==
Reliability engineers are very often called upon to make decisions as to whether to improve a certain component or components in order to achieve a minimum required system reliability. There are two approaches to improving the reliability of a system: fault avoidance and fault tolerance. Fault avoidance is achieved by using high-quality and high-reliability components and is usually less expensive than fault tolerance. Fault tolerance, on the other hand, is achieved by redundancy. Redundancy can result in increased design complexity and increased costs through additional weight, space, etc.

Before deciding whether to improve the reliability of a system by fault tolerance or fault avoidance, a reliability assessment for each component in the system should be made. Once the reliability values for the components have been quantified, an analysis can be performed in order to determine if that system's reliability goal will be met. If it becomes apparent that the system's reliability will not be adequate to meet the desired goal at the specified mission duration, steps can be taken to determine the best way to improve the system's reliability so that it will reach the desired target.

Consider a system with three components connected reliability-wise in series. The reliabilities for each component for a given time are: <math>{{R}_{1}} = 70%,\,\!</math> <math>{{R}_{2}} = 80%\,\!</math>, and <math>{{R}_{3}} = 90%\,\!</math>. A reliability goal, <math>{{R}_{G}} = 85%\,\!</math> is required for this system.

The current reliability of the system is:

::<math>{{R}_{s}}={{R}_{1}}\cdot {{R}_{2}}\cdot {{R}_{3}}=50.4%\,\!</math>

Obviously, this is far short of the system's required reliability performance. It is apparent that the reliability of the system's constituent components will need to be increased in order for the system to meet its goal. First, we will try increasing the reliability of one component at a time to see whether the reliability goal can be achieved.

The following figure shows that even by raising the individual component reliability to a hypothetical value of 1 (100% reliability, which implies that the component will never fail), the overall system reliability goal will not be met by improving the reliability of just one component. The next logical step would be to try to increase the reliability of two components. The question now becomes: which two? One might also suggest increasing the reliability of all three components. A basis for making such decisions needs to be found in order to avoid the ''trial and error'' aspect of altering the system's components randomly in an attempt to achieve the system reliability goal.

[[Image:BS6.9.png|center|500px|Change in system reliability of a three-unit series system due to increasing the reliability of just one component.|link=]]

As we have seen, the reliability goal for the preceding example could not be achieved by increasing the reliability of just one component. There are cases, however, where increasing the reliability of one component results in achieving the system reliability goal. Consider, for example, a system with three components connected reliability-wise in parallel. The reliabilities for each component for a given time are: <math>{{R}_{1}} = 60%,\,\!</math> <math>{{R}_{2}} = 70%\,\!</math> and <math>{{R}_{3}} = 80%\,\!</math>. A reliability goal, <math>{{R}_{G}} = 99%,\,\!</math> is required for this system. The initial system reliability is:

::<math>{{R}_{S}}=1-(1-0.6)\cdot (1-0.7)\cdot (1-0.8)=0.976\,\!</math>

The current system reliability is inadequate to meet the goal. Once again, we can try to meet the system reliability goal by raising the reliability of just one of the three components in the system.

[[Image:6_10.png|center|500px|Meeting a reliability goal requirement by increasing a component's reliability|link=]]

From the above figure, it can be seen that the reliability goal can be reached by improving Component 1, Component 2 or Component 3. The reliability engineer is now faced with another dilemma: which component's reliability should be improved? This presents a new aspect to the problem of allocating the reliability of the system. Since we know that the system reliability goal can be achieved by increasing at least one unit, the question becomes one of how to do this most efficiently and cost effectively. We will need more information to make an informed decision as to how to go about improving the system's reliability. How much does each component need to be improved for the system to meet its goal? How feasible is it to improve the reliability of each component? Would it actually be more efficient to slightly raise the reliability of two or three components rather than radically improving only one?

In order to answer these questions, we must introduce another variable into the problem: cost. Cost does not necessarily have to be in dollars. It could be described in terms of non-monetary resources, such as time. By associating cost values to the reliabilities of the system's components, we can find an optimum design that will provide the required reliability at a minimum cost.

==Cost/Penalty Function==
There is always a cost associated with changing a design due to change of vendors, use of higher-quality materials, retooling costs, administrative fees, etc. The cost as a function of the reliability for each component must be quantified before attempting to improve the reliability. Otherwise, the design changes may result in a system that is needlessly expensive or overdesigned. Developing the "cost of reliability" relationship will give the engineer an understanding of which components to improve and how to best concentrate the effort and allocate resources in doing so. The first step will be to obtain a relationship between the cost of improvement and reliability.

The preferred approach would be to formulate the cost function from actual cost data. This can be done from past experience. If a reliability growth program is in place, the costs associated with each stage of improvement can also be quantified. Defining the different costs associated with different vendors or different component models is also useful in formulating a model of component cost as a function of reliability.

However, there are many cases where no such information is available. For this reason, a general (default) behavior model of the cost versus the component's reliability was developed for performing reliability optimization in BlockSim. The objective of this function is to model an overall cost behavior for all types of components. Of course, it is impossible to formulate a model that will be precisely applicable to every situation; but the proposed relationship is general enough to cover most applications. In addition to the default model formulation, BlockSim does allow the definition of user-defined cost models.

===Quantifying the Cost/Penalty Function===
One needs to quantify a cost function for each component, <math>{{C}_{i}}\,\!</math>, in terms of the reliability, <math>{{R}_{i}}\,\!</math>, of each component, or:

::<math>\begin{align}
{{C}_{i}}=f({{R}_{i}})
\end{align}\,\!</math>

This function should:

*Look at the current reliability of the component, <math>{{R}_{Current}}\,\!</math>.<br>
*Look at the maximum possible reliability of the component, <math>{{R}_{Max}}\,\!</math>.<br>
*Allow for different levels of difficulty (or cost) in increasing the reliability of each component. It can take into account:<br>
:*design issues.<br>
:*supplier issues.<br>
:*state of technology.<br>
:*time-to-market issues, etc.<br>

Thus, for the cost function to comply with these needs, the following conditions should be adhered to:

*The function should be constrained by the minimum and maximum reliabilities of each component (i.e., reliability must be less than one and greater than the current reliability of the component or at least greater than zero).<br>
*The function should not be linear, but rather quantify the fact that it is incrementally harder to improve reliability. For example, it is considerably easier to increase the reliability from 90% to 91% than to increase it from 99.99% to 99.999%, even though the increase is larger in the first case.<br>
*The function should be asymptotic to the maximum achievable reliability.<br>

The following default cost function (also used in BlockSim) adheres to all of these conditions and acts like a penalty function for increasing a component's reliability. Furthermore, an exponential behavior for the cost is assumed since it should get exponentially more difficult to increase the reliability. See Mettas [[Appendix_B:_References | [21]]].

::<math>{{C}_{i}}({{R}_{i}})={{e}^{(1-f)\cdot \tfrac{{{R}_{i}}-{{R}_{\min ,i}}}{{{R}_{\max ,i}}-{{R}_{i}}}}}\ \,\!</math>

where:

*<math>{{C}_{i}}({{R}_{i}})\,\!</math> is the penalty (or cost) function as a function of component reliability.<br>
*<math>f\,\!</math> is the feasibility (or cost index) of improving a component's reliability relative to the other components in the system.<br>
*<math>{{R}_{min,i}}\,\!</math> is the current reliability at the time at which the optimization is to be performed.<br>
*<math>{{R}_{max,i}}\,\!</math> is the maximum achievable reliability at the time at which the optimization is to be performed.<br>

Note that this penalty function is dimensionless. It essentially acts as a weighting factor that describes the difficulty in increasing the component reliability from its current value, relative to the other components.

Examining the cost function given by equation above, the following observations can be made:

*The cost increases as the allocated reliability departs from the minimum or current value of reliability. It is assumed that the reliabilities for the components will not take values any lower than they already have. Depending on the optimization, a component's reliability may not need to be increased from its current value but it will not drop any lower.<br>
*The cost increases as the allocated reliability approaches the maximum achievable reliability. This is a reliability value that is approached asymptotically as the cost increases but is never actually reached.<br>
*The cost is a function of the range of improvement, which is the difference between the component's initial reliability and the corresponding maximum achievable reliability.<br>
*The exponent in the above equation approaches infinity as the component's reliability approaches its maximum achievable value. This means that it is easier to increase the reliability of a component from a lower initial value. For example, it is easier to increase a component's reliability from 70% to 75% than to increase its reliability from 90% to 95%.

===The Feasibility Term, <math>f\,\!</math>===
The feasibility term in <math>{{C}_{i}}({{R}_{i}})={{e}^{(1-f)\cdot \tfrac{{{R}_{i}}-{{R}_{\min ,i}}}{{{R}_{\max ,i}}-{{R}_{i}}}}}\ \,\!</math> is a constant (or an equation parameter) that represents the difficulty in increasing a component's reliability relative to the rest of the components in the system. Depending on the design complexity, technological limitations, etc., certain components can be very hard to improve. Clearly, the more difficult it is to improve the reliability of the component, the greater the cost. The following figure illustrates the behavior of the function defined in the above equation for different values of <math>f\,\!</math>. It can be seen that the lower the feasibility value, the more rapidly the cost function approaches infinity.

[[Image:6_11.png|center|650px|Behavior of the cost function for different feasibility values.|link=]]

Several methods can be used to obtain a feasibility value. Weighting factors for allocating reliability have been proposed by many authors and can be used to quantify feasibility. These weights depend on certain factors of influence, such as the complexity of the component, the state of the art, the operational profile, the criticality, etc. Engineering judgment based on past experience, supplier quality, supplier availability and other factors can also be used in determining a feasibility value. Overall, the assignment of a feasibility value is going to be a subjective process. Of course, this problem is negated if the relationship between the cost and the reliability for each component is known because one can use regression methods to estimate the parameter value.

===Maximum Achievable Reliability===
For the purposes of reliability optimization, we also need to define a limiting reliability that a component will approach, but not reach. The costs near the maximum achievable reliability are very high and the actual value for the maximum reliability is usually dictated by technological or financial constraints. In deciding on a value to use for the maximum achievable reliability, the current state of the art of the component in question and other similar factors will have to be considered. In the end, a realistic estimation based on engineering judgment and experience will be necessary to assign a value to this input.

Note that the time associated with this maximum achievable reliability is the same as that of the overall system reliability goal. Almost any component can achieve a very high reliability value, provided the mission time is short enough. For example, a component with an exponential distribution and a failure rate of one failure per hour has a reliability that drops below 1% for missions greater than five hours. However, it can achieve a reliability of 99.9% as long as the mission is no longer than four seconds. For the purposes of optimization in BlockSim, the reliability values of the components are associated with the time for which the system reliability goal is specified. For example, if the problem is to achieve a system goal of 99% reliability at 1,000 hours, the maximum achievable reliability values entered for the individual components would be the maximum reliability that each component could attain for a mission of 1,000 hours.

As the component reliability, <math>{{R}_{i}}\,\!</math>, approaches the maximum achievable reliability, <math>{{R}_{i,max}}\,\!</math>, the cost function approaches infinity. The maximum achievable reliability acts as a scale parameter for the cost function. By decreasing <math>{{R}_{i,max}}\,\!</math>, the cost function is compressed between <math>{{R}_{i,min}}\,\!</math> and <math>{{R}_{i,max}}\,\!</math>, as shown in the figure below.

[[Image:6_12.png|center|650px|Effect of the maximum achievable reliability on the cost function.|link=]]

===Cost Function===
Once the cost functions for the individual components have been determined, it becomes necessary to develop an expression for the overall system cost. This takes the form of:

::<math>\begin{align}
{{C}_{s}}({{R}_{G}})={{C}_{1}}({{R}_{1}})+{{C}_{2}}({{R}_{2}})+...+{{C}_{n}}({{R}_{n}}),i=1,2,...,n
\end{align}\,\!</math>

In other words, the cost of the system is simply the sum of the costs of its components. This is regardless of the form of the individual component cost functions. They can be of the general behavior model in BlockSim or they can be user-defined. Once the overall cost function for the system has been defined, the problem becomes one of minimizing the cost function while remaining within the constraints defined by the target system reliability and the reliability ranges for the components. The latter constraints in this case are defined by the minimum and maximum reliability values for the individual components.

BlockSim employs a nonlinear programming technique to minimize the system cost function. The system has a minimum (current) and theoretical maximum reliability value that is defined by the minimum and maximum reliabilities of the components and by the way the system is configured. That is, the structural properties of the system are accounted for in the determination of the optimum solution. For example, the optimization for a system of three units in series will be different from the optimization for a system consisting of the same three units in parallel. The optimization occurs by varying the reliability values of the components within their respective constraints of maximum and minimum reliability in a way that the overall system goal is achieved. Obviously, there can be any number of different combinations of component reliability values that might achieve the system goal. The optimization routine essentially finds the combination that results in the lowest overall system cost.

==Determining the Optimum Allocation Scheme==
To determine the optimum reliability allocation, the analyst first determines the system reliability equation (the objective function). As an example, and again for a trivial system with three components in series, this would be:

::<math>{{R}_{_{S}}}={{R}_{1}}\cdot {{R}_{2}}\cdot {{R}_{3}}\ \,\!</math>

If a target reliability of 90% is sought, then the equation above is recast as:

::<math>0.90={{R}_{1}}\cdot {{R}_{2}}\cdot {{R}_{3}}\ \,\!</math>

The objective now is to solve for <math>{{R}_{1}}\,\!</math>, <math>{{R}_{2}}\,\!</math> and <math>{{R}_{3}}\,\!</math> so that the equality in the equation above is satisfied. To obtain an optimum solution, we also need to use our cost functions (i.e., define the total allocation costs) as:

::<math>{{C}_{T}}={{C}_{1}}({{R}_{1}})+{{C}_{2}}({{R}_{2}})+{{C}_{3}}({{R}_{3}})\ \,\!</math>

With the cost equation defined, then the optimum values for <math>{{R}_{1}}\,\!</math>, <math>{{R}_{2}}\,\!</math> and <math>{{R}_{3}}\,\!</math> are the values that satisfy the reliability requirement, the second equation above, at the minimum cost, the last equation above. BlockSim uses this methodology during the optimization task.

===Defining a Feasibility Policy in BlockSim===
In BlockSim, you can choose to use the default feasibility function, as defined by <math>{{C}_{i}}({{R}_{i}})={{e}^{(1-f)\cdot \tfrac{{{R}_{i}}-{{R}_{\min ,i}}}{{{R}_{\max ,i}}-{{R}_{i}}}}}\ \,\!</math>, or use your own function. The first picture below illustrates the use of the default values using the slider control. The second figure shows the use of an associated feasibility policy to create a user-defined cost function. When defining your own cost function, you should be aware of/adhere to the following guidelines:

*Because the cost functions are evaluated relative to each other, they should be correlated. In other words, if one function evaluates to 10, <math>{{C}_{i}}({{R}_{i}})=10\,\!</math> for one block and 20 for another, <math>{{C}_{i}}({{R}_{i}})=20\,\!</math>, then the implication is that there is a 1 to 2 cost relation. <br>
*Do not mix your own function with the software's default functions unless you have verified that your cost functions are defined and correlated to the default cost functions, as defined by:

::<math>\begin{align}
{{C}_{i}}({{R}_{i}})={{e}^{(1-f)\cdot \tfrac{{{R}_{i}}-{{R}_{\min ,i}}}{{{R}_{\max ,i}}-{{R}_{i}}}}}
\end{align}\,\!</math>

*Your function should adhere to the guidelines presented earlier.
*Lastly, and since the evaluation is relative, it is preferable to use the predefined functions unless you have a compelling reason (or data) to do otherwise. The last section in this chapter describes cases where user-defined functions are preferred.

[[Image:6.13.png|center|700px|thumb|Setting the default feasibility function in BlockSim with the feasibility slider. Note that the feasibility slider displays values, ''SV'', from 0.1 to 9.9 when moved by the user, with SV=9.9 being the hardest. The relationship between ''f'' and ''SV'' is "f = (1 - SV/10)").|link= ]]

[[Image:6.14.png|center|700px|thumb|Setting a user-defined feasibility function in BlockSim. Any user-defined equation can be entered as a function of ''R.''|link=]]

==Implementing the Optimization==
As was mentioned earlier, there are two different methods of implementing the changes suggested by the reliability optimization routine: fault tolerance and fault avoidance. When the optimized component reliabilities have been determined, it does not matter which of the two methods is employed to realize the optimum reliability for the component in question. For example, suppose we have determined that a component must have its reliability for a certain mission time raised from 50% to 75%. The engineer must now decide how to go about implementing the increase in reliability. If the engineer decides to do this via fault avoidance, another component must be found (or the existing component must be redesigned) so that it will perform the same function with a higher reliability. On the other hand, if the engineer decides to go the fault tolerance route, the optimized reliability can be achieved merely by placing a second identical component in parallel with the first one.

Obviously, the method of implementing the reliability optimization is going to be related to the cost function and this is something the reliability engineer must take into account when deciding on what type of cost function is used for the optimization. In fact, if we take a closer look at the fault tolerance scheme, we can see some parallels with the general behavior cost model included in BlockSim. For example, consider a system that consists of a single unit. The cost of that unit, including all associated mounting and hardware costs, is one dollar. The reliability of this unit for a given mission time is 30%. It has been determined that this is inadequate and that a second component is to be added in parallel to increase the reliability. Thus, the reliability for the two-unit parallel system is:

::<math>\begin{align}
{{R}_{S}}=1-{{(1-0.3)}^{2}}=0.51\text{ or }51%
\end{align}\,\!</math>

So, the reliability has increased by a value of 21% and the cost has increased by one dollar. In a similar fashion, we can continue to add more units in parallel, thus increasing the reliability and the cost. We now have an array of reliability values and the associated costs that we can use to develop a cost function for this fault tolerance scheme. The next figure shows the relationship between cost and reliability for this example.

[[Image:6.15.png|center|650px|Cost function for redundant parallel units.|link=]]

As can be seen, this looks quite similar to the general behavior cost model presented earlier. In fact, a standard regression analysis available in Weibull++ indicates that an exponential model fits this cost model quite well. The function is given by the following equation, where <math>C\,\!</math> is the cost in dollars and <math>R\,\!</math> is the fractional reliability value.

::<math>C(R)=0.3756\cdot {{e}^{3.1972\cdot R}}\,\!</math>

==Reliability Allocation Examples==

{{:BlockSim_Allocation_Analysis_Example}}

==Setting Specifications==
This methodology could also be used to arrive at initial specifications for a set of components. In the prior examples, we assumed a current reliability for the components. One could repeat these steps by choosing an arbitrary (lower) initial reliability for each component, thus allowing the algorithm to travel up to the target. When doing this, it is important to keep in mind the fact that both the distance from the target (the distance from the initial arbitrary value and the target value) for each component is also a significant contributor to the final results, as presented in the prior example. If one wishes to arrive at the results using only the cost functions then it may be advantageous to set equal initial reliabilities for all components.

==Other Notes on User-Defined Cost Functions==
The optimization method in BlockSim is a very powerful tool for allocating reliability to the components of a system while minimizing an overall cost of improvement. The default cost function in BlockSim was derived in order to model a general relationship between the cost and the component reliability. However, if actual cost information is available, then one can use the cost data instead of using the default function. Additionally, one can also view the feasibility in the default function as a measure of the difficulty in increasing the reliability of the component relative to the rest of the components to be optimized, assuming that they also follow the same cost function with the corresponding feasibility values. If fault tolerance is a viable option, a reliability cost function for adding parallel units can be developed as demonstrated previously.

Another method for developing a reliability cost function would be to obtain different samples of components from different suppliers and test the samples to determine the reliability of each sample type. From this data, a curve could be fitted through standard regression techniques and an equation defining the cost as a function of reliability could be developed. The following figure shows such a curve.

[[Image:6.21.png|center|650px|Typical reliability growth curve generated using ReliaSoft's Reliability Growth software.|link=]]

Lastly, and in cases where a reliability growth program is in place, the simplest way of obtaining a relationship between cost and reliability is by associating a cost to each development stage of the growth process. Reliability growth models such as the Crow (AMSAA), Duane, Gompertz and Logistic models can be used to describe the cost as a function of reliability.

If a reliability growth model has been successfully implemented, the development costs over the respective development time stages can be applied to the growth model, resulting in equations that describe reliability/cost relationships. These equations can then be entered into BlockSim as user-defined cost functions (feasibility policies). The only potential drawback to using growth model data is the lack of flexibility in applying the optimum results. Making the cost projection for future stages of the project would require the assumption that development costs will be accrued at a similar rate in the future, which may not always be a valid assumption. Also, if the optimization result suggests using a high reliability value for a component, it may take more time than is allotted for that project to attain the required reliability given the current reliability growth of the project.

RCM++ Examples

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Introduction to Confidence Bounds

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<noinclude>{{Navigation box}}[[Category: Shared Articles]]
''This article also appears in the [https://help.reliasoft.com/reference/life_data_analysis Life data analysis reference] and [https://help.reliasoft.com/reference/accelerated_life_testing_data_analysis Accelerated life testing reference] books.'' </noinclude>

== What Are Confidence Bounds?==

One of the most confusing concepts to a novice reliability engineer is estimating the precision of an estimate. This is an important concept in the field of reliability engineering, leading to the use of confidence intervals (or bounds). In this section, we will try to briefly present the concept in relatively simple terms but based on solid common sense.

=== The Black and White Marbles ===
To illustrate, consider the case where there are millions of perfectly mixed black and white marbles in a rather large swimming pool and our job is to estimate the percentage of black marbles. The only way to be absolutely certain about the exact percentage of marbles in the pool is to accurately count every last marble and calculate the percentage. However, this is too time- and resource-intensive to be a viable option, so we need to come up with a way of estimating the percentage of black marbles in the pool. In order to do this, we would take a relatively small sample of marbles from the pool and then count how many black marbles are in the sample.

'''Taking a Small Sample of Marbles'''

First, pick out a small sample of marbles and count the black ones. Say you picked out ten marbles and counted four black marbles. Based on this, your estimate would be that 40% of the marbles are black.

[[Image:estimation.png|center|200px]]

If you put the ten marbles back in the pool and repeat this example again, you might get six black marbles, changing your estimate to 60% black marbles. Which of the two is correct? Both estimates are correct! As you repeat this experiment over and over again, you might find out that this estimate is usually between <math>{{X}_{1}}%\,\!</math> and <math>{{X}_{2}}%\,\!</math>, and you can assign a percentage to the number of times your estimate falls between these limits. For example, you notice that 90% of the time this estimate is between <math>{{X}_{1}}%\,\!</math> and <math>{{X}_{2}}%.\,\!</math>

''' Taking a Larger Sample of Marbles '''

If you now repeat the experiment and pick out 1,000 marbles, you might get results for the number of black marbles such as 545, 570, 530, etc., for each trial. The range of the estimates in this case will be much narrower than before. For example, you observe that 90% of the time, the number of black marbles will now be from <math>{{Y}_{1}}%\,\!</math> to <math>{{Y}_{2}}%\,\!</math>, where <math>{{X}_{1}}%<{{Y}_{1}}%\,\!</math> and <math>{{X}_{2}}%>{{Y}_{2}}%\,\!</math>, thus giving you a more narrow estimate interval. The same principle is true for confidence intervals; the larger the sample size, the more narrow the confidence intervals.

'''Back to Reliability'''

We will now look at how this phenomenon relates to reliability. Overall, the reliability engineer's task is to determine the probability of failure, or reliability, of the population of units in question. However, one will never know the exact reliability value of the population unless one is able to obtain and analyze the failure data for every single unit in the population. Since this usually is not a realistic situation, the task then is to estimate the reliability based on a sample, much like estimating the number of black marbles in the pool. If we perform ten different reliability tests for our units, and analyze the results, we will obtain slightly different parameters for the distribution each time, and thus slightly different reliability results. However, by employing confidence bounds, we obtain a range within which these reliability values are likely to occur a certain percentage of the time. This helps us gauge the utility of the data and the accuracy of the resulting estimates. Plus, it is always useful to remember that each parameter is an estimate of the true parameter, one that is unknown to us. This range of plausible values is called a confidence interval.

=== One-Sided and Two-Sided Confidence Bounds ===
Confidence bounds are generally described as being one-sided or two-sided.

'''Two-Sided Bounds'''

[[Image:two sided bounds.png|center|350px]]

When we use two-sided confidence bounds (or intervals), we are looking at a closed interval where a certain percentage of the population is likely to lie. That is, we determine the values, or bounds, between which lies a specified percentage of the population. For example, when dealing with 90% two-sided confidence bounds of <math>(X,Y)\,\!</math>, we are saying that 90% of the population lies between <math>X\,\!</math> and <math>Y\,\!</math> with 5% less than <math>X\,\!</math> and 5% greater than <math>Y\,\!</math>.

'''One-Sided Bounds'''

One-sided confidence bounds are essentially an open-ended version of two-sided bounds. A one-sided bound defines the point where a certain percentage of the population is either higher or lower than the defined point. This means that there are two types of one-sided bounds: upper and lower. An upper one-sided bound defines a point that a certain percentage of the population is less than. Conversely, a lower one-sided bound defines a point that a specified percentage of the population is greater than.

[[Image:one sided bounds.png|center|350px]]

For example, if <math>X\,\!</math> is a 95% upper one-sided bound, this would imply that 95% of the population is less than <math>X\,\!</math>. If <math>X\,\!</math> is a 95% lower one-sided bound, this would indicate that 95% of the population is greater than <math>X\,\!</math>. Care must be taken to differentiate between one- and two-sided confidence bounds, as these bounds can take on identical values at different percentage levels. For example, in the figures above, we see bounds on a hypothetical distribution. Assuming that this is the same distribution in all of the figures, we see that <math>X\,\!</math> marks the spot below which 5% of the distribution's population lies. Similarly, <math>Y\,\!</math> represents the point above which 5% of the population lies. Therefore, <math>X\,\!</math> and <math>Y\,\!</math> represent the 90% two-sided bounds, since 90% of the population lies between the two points. However, <math>X\,\!</math> also represents the lower one-sided 95% confidence bound, since 95% of the population lies above that point; and <math>Y\,\!</math> represents the upper one-sided 95% confidence bound, since 95% of the population is below <math>Y\,\!</math>. It is important to be sure of the type of bounds you are dealing with, particularly as both one-sided bounds can be displayed simultaneously in Weibull++. In Weibull++, we use upper to represent the higher limit and lower to represent the lower limit, regardless of their position, but based on the value of the results. So if obtaining the confidence bounds on the reliability, we would identify the lower value of reliability as the lower limit and the higher value of reliability as the higher limit. If obtaining the confidence bounds on probability of failure we will again identify the lower numeric value for the probability of failure as the lower limit and the higher value as the higher limit.

== Fisher Matrix Confidence Bounds ==
This section presents an overview of the theory on obtaining approximate confidence bounds on suspended (multiple censored) data. The methodology used is the so-called Fisher matrix bounds (FM), described in Nelson [[Appendix:_Life_Data_Analysis_References|[30]]] and Lloyd and Lipow [[Appendix:_Life_Data_Analysis_References|[24]]]. These bounds are employed in most other commercial statistical applications. In general, these bounds tend to be more optimistic than the non-parametric rank based bounds. This may be a concern, particularly when dealing with small sample sizes. Some statisticians feel that the Fisher matrix bounds are too optimistic when dealing with small sample sizes and prefer to use other techniques for calculating confidence bounds, such as the likelihood ratio bounds. <br>

=== Approximate Estimates of the Mean and Variance of a Function ===
In utilizing FM bounds for functions, one must first determine the mean and variance of the function in question (i.e., reliability function, failure rate function, etc.). An example of the methodology and assumptions for an arbitrary function <math>G\,\!</math> is presented next.

'''Single Parameter Case'''

For simplicity, consider a one-parameter distribution represented by a general function <math>G,\,\!</math> which is a function of one parameter estimator, say <math>G(\widehat{\theta }).\,\!</math> For example, the mean of the exponential distribution is a function of the parameter <math>\lambda: G(\lambda) = 1 / \lambda = \mu\,\!</math>. Then, in general, the expected value of <math>G\left( \widehat{\theta } \right)\,\!</math> can be found by:

::<math>E\left( G\left( \widehat{\theta } \right) \right)=G(\theta )+O\left( \frac{1}{n} \right)\,\!</math>

where <math>G(\theta)\,\!</math> is some function of <math>\theta\,\!</math>, such as the reliability function, and <math>\theta\,\!</math> is the population parameter where <math>E\left( \widehat{\theta } \right)=\theta \,\!</math> as <math>n\to \infty \,\!</math>. The term <math>O\left( \tfrac{1}{n} \right)\,\!</math> is a function of <math>n\,\!</math>, the sample size, and tends to zero, as fast as <math>\tfrac{1}{n},\,\!</math> as <math>n\to \infty .\,\!</math> For example, in the case of <math>\widehat{\theta }=1/\overline{x}\,\!</math> and <math>G(x) = 1 / x\,\!</math>, then <math>E(G(\widehat{\theta }))=\overline{x}+O\left( \tfrac{1}{n} \right)\,\!</math> where <math>O\left( \tfrac{1}{n} \right)=\tfrac{{{\sigma }^{2}}}{n}\,\!</math>. Thus as <math>n\to \infty \,\!</math>, <math>E(G(\widehat{\theta }))=\mu \,\!</math> where <math>\mu\,\!</math> and <math>\sigma\,\!</math> are the mean and standard deviation, respectively. Using the same one-parameter distribution, the variance of the function <math>G\left( \widehat{\theta } \right)\,\!</math> can then be estimated by:

::<math>Var\left( G\left( \widehat{\theta } \right) \right)=\left( \frac{\partial G}{\partial \widehat{\theta }} \right)_{\widehat{\theta }=\theta }^{2}Var\left( \widehat{\theta } \right)+O\left( \frac{1}{{{n}^{\tfrac{3}{2}}}} \right)\,\!</math>

'''Two-Parameter Case'''

Consider a Weibull distribution with two parameters <math>\beta\,\!</math> and <math>\eta\,\!</math>. For a given value of <math>t\,\!</math>, <math>R(t)=G(\beta ,\eta )={{e}^{-{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}\,\!</math>. Repeating the previous method for the case of a two-parameter distribution, it is generally true that for a function <math>G\,\!</math>, which is a function of two parameter estimators, say <math>G\left( {{\widehat{\theta }}_{1}},{{\widehat{\theta }}_{2}} \right)\,\!</math>, that:

::<math>E\left( G\left( {{\widehat{\theta }}_{1}},{{\widehat{\theta }}_{2}} \right) \right)=G\left( {{\theta }_{1}},{{\theta }_{2}} \right)+O\left( \frac{1}{n} \right)\,\!</math>

and:

::<math>\begin{align}
Var( G( {{\widehat{\theta }}_{1}},{{\widehat{\theta }}_{2}}))= &{\left(\frac{\partial G}{\partial {{\widehat{\theta }}_{1}}}\right)^2}_{{\widehat{\theta_{1}}}={\theta_{1}}}Var(\widehat{\theta_{1}})+{\left(\frac{\partial G}{\partial {{\widehat{\theta }}_{2}}}\right)^2}_{{\widehat{\theta_{2}}}={\theta_{2}}}Var(\widehat{\theta_{2}})\\
& +2{\left(\frac{\partial G}{\partial {{\widehat{\theta }}_{1}}}\right)}_{{\widehat{\theta_{1}}}={\theta_{1}}}{\left(\frac{\partial G}{\partial {{\widehat{\theta }}_{2}}}\right)}_{{\widehat{\theta_{2}}}={\theta_{2}}}Cov(\widehat{\theta_{1}},\widehat{\theta_{2}}) \\
& +O\left(\frac{1}{n^{\tfrac{3}{2}}}\right)
\end{align}
\,\!</math>

Note that the derivatives of the above equation are evaluated at <math>{{\widehat{\theta }}_{1}}={{\theta }_{1}}\,\!</math> and <math>{{\widehat{\theta }}_{2}}={{\theta }_{2}},\,\!</math> where E <math>\left( {{\widehat{\theta }}_{1}} \right)\simeq {{\theta }_{1}}\,\!</math> and E <math>\left( {{\widehat{\theta }}_{2}} \right)\simeq {{\theta }_{2}}.\,\!</math>

'''Parameter Variance and Covariance Determination'''

The determination of the variance and covariance of the parameters is accomplished via the use of the Fisher information matrix. For a two-parameter distribution, and using maximum likelihood estimates (MLE), the log-likelihood function for censored data is given by:

::<math>\begin{align}
\ln [L]= & \Lambda =\underset{i=1}{\overset{R}{\mathop \sum }}\,\ln [f({{T}_{i}};{{\theta }_{1}},{{\theta }_{2}})] \\
& \text{ }+\underset{j=1}{\overset{M}{\mathop \sum }}\,\ln [1-F({{S}_{j}};{{\theta }_{1}},{{\theta }_{2}})] \\
& \text{ }+\underset{l=1}{\overset{P}{\mathop \sum }}\,\ln \left\{ F({{I}_{{{l}_{U}}}};{{\theta }_{1}},{{\theta }_{2}})-F({{I}_{{{l}_{L}}}};{{\theta }_{1}},{{\theta }_{2}}) \right\}
\end{align}\,\!</math>

In the equation above, the first summation is for ''complete data'', the second summation is for ''right censored data'' and the third summation is for ''interval or left censored data''.

Then the Fisher information matrix is given by:

::<math>{{F}_{0}}=\left[ \begin{matrix}
{{E}_{0}}{{\left[ -\tfrac{{{\partial }^{2}}\Lambda }{\partial \theta _{1}^{2}} \right]}_{0}} & {} & {{E}_{0}}{{\left[ -\tfrac{{{\partial }^{2}}\Lambda }{\partial {{\theta }_{1}}\partial {{\theta }_{2}}} \right]}_{0}} \\
{} & {} & {} \\
{{E}_{0}}{{\left[ -\tfrac{{{\partial }^{2}}\Lambda }{\partial {{\theta }_{2}}\partial {{\theta }_{1}}} \right]}_{0}} & {} & {{E}_{0}}{{\left[ -\tfrac{{{\partial }^{2}}\Lambda }{\partial \theta _{2}^{2}} \right]}_{0}} \\
\end{matrix} \right]\,\!</math>

The subscript 0 indicates that the quantity is evaluated at <math>{{\theta }_{1}}={{\theta }_{{{1}_{0}}}}\,\!</math> and <math>{{\theta }_{2}}={{\theta }_{{{2}_{0}}}},\,\!</math> the true values of the parameters.

So for a sample of <math>N\,\!</math> units where <math>R\,\!</math> units have failed, <math>S\,\!</math> have been suspended, and <math>P\,\!</math> have failed within a time interval, and <math>N = R + M + P,\,\!</math> one could obtain the sample local information matrix by:

::<math>F={{\left[ \begin{matrix}
-\tfrac{{{\partial }^{2}}\Lambda }{\partial \theta _{1}^{2}} & {} & -\tfrac{{{\partial }^{2}}\Lambda }{\partial {{\theta }_{1}}\partial {{\theta }_{2}}} \\
{} & {} & {} \\
-\tfrac{{{\partial }^{2}}\Lambda }{\partial {{\theta }_{2}}\partial {{\theta }_{1}}} & {} & -\tfrac{{{\partial }^{2}}\Lambda }{\partial \theta _{2}^{2}} \\
\end{matrix} \right]}^{}}\,\!</math>

Substituting the values of the estimated parameters, in this case <math>{{\widehat{\theta }}_{1}}\,\!</math> and <math>{{\widehat{\theta }}_{2}}\,\!</math>, and then inverting the matrix, one can then obtain the local estimate of the covariance matrix or:

::<math>\left[ \begin{matrix}
\widehat{Var}\left( {{\widehat{\theta }}_{1}} \right) & {} & \widehat{Cov}\left( {{\widehat{\theta }}_{1}},{{\widehat{\theta }}_{2}} \right) \\
{} & {} & {} \\
\widehat{Cov}\left( {{\widehat{\theta }}_{1}},{{\widehat{\theta }}_{2}} \right) & {} & \widehat{Var}\left( {{\widehat{\theta }}_{2}} \right) \\
\end{matrix} \right]={{\left[ \begin{matrix}
-\tfrac{{{\partial }^{2}}\Lambda }{\partial \theta _{1}^{2}} & {} & -\tfrac{{{\partial }^{2}}\Lambda }{\partial {{\theta }_{1}}\partial {{\theta }_{2}}} \\
{} & {} & {} \\
-\tfrac{{{\partial }^{2}}\Lambda }{\partial {{\theta }_{2}}\partial {{\theta }_{1}}} & {} & -\tfrac{{{\partial }^{2}}\Lambda }{\partial \theta _{2}^{2}} \\
\end{matrix} \right]}^{-1}}\,\!</math>

Then the variance of a function (<math>Var(G)\,\!</math>) can be estimated using equation for the variance. Values for the variance and covariance of the parameters are obtained from Fisher Matrix equation. Once they have been obtained, the approximate confidence bounds on the function are given as:

::<math>C{{B}_{R}}=E(G)\pm {{z}_{\alpha }}\sqrt{Var(G)}\,\!</math>

which is the estimated value plus or minus a certain number of standard deviations. We address finding <math>{{z}_{\alpha}}\,\!</math> next.

=== Approximate Confidence Intervals on the Parameters ===
In general, MLE estimates of the parameters are asymptotically normal, meaning that for large sample sizes, a distribution of parameter estimates from the same population would be very close to the normal distribution. Thus if <math>\widehat{\theta }\,\!</math> is the MLE estimator for <math>\theta\,\!</math>, in the case of a single parameter distribution estimated from a large sample of <math>n\,\!</math> units, then:

::<math>z\equiv \frac{\widehat{\theta }-\theta }{\sqrt{Var\left( \widehat{\theta } \right)}}\,\!</math>

follows an approximating normal distribution. That is

::<math>P\left( x\le z \right)\to \Phi \left( z \right)=\frac{1}{\sqrt{2\pi }}\int_{-\infty }^{z}{{e}^{-\tfrac{{{t}^{2}}}{2}}}dt\,\!</math>

for large <math>n\,\!</math>. We now place confidence bounds on <math>\theta,\,\!</math> at some confidence level <math>\delta\,\!</math>, bounded by the two end points <math>{{C}_{1}}\,\!</math> and <math>{{C}_{2}}\,\!</math> where:

::<math>P\left( {{C}_{1}}<\theta <{{C}_{2}} \right)=\delta \,\!</math>

From the above equation:

::<math>P\left( -{{K}_{\tfrac{1-\delta }{2}}}<\frac{\widehat{\theta }-\theta }{\sqrt{Var\left( \widehat{\theta } \right)}}<{{K}_{\tfrac{1-\delta }{2}}} \right)\simeq \delta \,\!</math>

where <math>{{K}_{\alpha}}\,\!</math> is defined by:

::<math>\alpha =\frac{1}{\sqrt{2\pi }}\int_{{{K}_{\alpha }}}^{\infty }{{e}^{-\tfrac{{{t}^{2}}}{2}}}dt=1-\Phi \left( {{K}_{\alpha }} \right)\,\!</math>

Now by simplifying the equation for the confidence level, one can obtain the approximate two-sided confidence bounds on the parameter <math>\theta\,\!,\,\!</math> at a confidence level <math>\delta,\,\!</math> or:

::<math>\left( \widehat{\theta }-{{K}_{\tfrac{1-\delta }{2}}}\cdot \sqrt{Var\left( \widehat{\theta } \right)}<\theta <\widehat{\theta }+{{K}_{\tfrac{1-\delta }{2}}}\cdot \sqrt{Var\left( \widehat{\theta } \right)} \right)\,\!</math>

The upper one-sided bounds are given by:

::<math>\theta <\widehat{\theta }+{{K}_{1-\delta }}\sqrt{Var(\widehat{\theta })}\,\!</math>

while the lower one-sided bounds are given by:

::<math>\theta >\widehat{\theta }-{{K}_{1-\delta }}\sqrt{Var(\widehat{\theta })}\,\!</math>

If <math>\widehat{\theta }\,\!</math> must be positive, then <math>\ln \widehat{\theta }\,\!</math> is treated as normally distributed. The two-sided approximate confidence bounds on the parameter <math>\theta\,\!</math>, at confidence level <math>\delta\,\!</math>, then become:

::<math>\begin{align}
& {{\theta }_{U}}= & \widehat{\theta }\cdot {{e}^{\tfrac{{{K}_{\tfrac{1-\delta }{2}}}\sqrt{Var\left( \widehat{\theta } \right)}}{\widehat{\theta }}}}\text{ (Two-sided upper)} \\
& {{\theta }_{L}}= & \frac{\widehat{\theta }}{{{e}^{\tfrac{{{K}_{\tfrac{1-\delta }{2}}}\sqrt{Var\left( \widehat{\theta } \right)}}{\widehat{\theta }}}}}\text{ (Two-sided lower)}
\end{align}\,\!</math>

The one-sided approximate confidence bounds on the parameter <math>\theta\,\!</math>, at confidence level <math>\delta,\,\!</math> can be found from:

::<math>\begin{align}
& {{\theta }_{U}}= & \widehat{\theta }\cdot {{e}^{\tfrac{{{K}_{1-\delta }}\sqrt{Var\left( \widehat{\theta } \right)}}{\widehat{\theta }}}}\text{ (One-sided upper)} \\
& {{\theta }_{L}}= & \frac{\widehat{\theta }}{{{e}^{\tfrac{{{K}_{1-\delta }}\sqrt{Var\left( \widehat{\theta } \right)}}{\widehat{\theta }}}}}\text{ (One-sided lower)}
\end{align}\,\!</math>

The same procedure can be extended for the case of a two or more parameter distribution. Lloyd and Lipow [[Appendix:_Life_Data_Analysis_References|[24]]] further elaborate on this procedure.

=== Confidence Bounds on Time (Type 1) ===
Type 1 confidence bounds are confidence bounds around time for a given reliability. For example, when using the one-parameter exponential distribution, the corresponding time for a given exponential percentile (i.e., y-ordinate or unreliability, <math>Q = 1 - R)\,\!</math> is determined by solving the unreliability function for the time, <math>T\,\!</math>, or:

::<math>\begin{align}\widehat{T}(Q)= &-\frac{1}{\widehat{\lambda }}
\ln (1-Q)= & -\frac{1}{\widehat{\lambda }}\ln (R)
\end{align}\,\!</math>

Bounds on time (Type 1) return the confidence bounds around this time value by determining the confidence intervals around <math>\widehat{\lambda }\,\!</math> and substituting these values into the above equation. The bounds on <math>\widehat{\lambda }\,\!</math> are determined using the method for the bounds on parameters, with its variance obtained from the Fisher Matrix. Note that the procedure is slightly more complicated for distributions with more than one parameter.

=== Confidence Bounds on Reliability (Type 2) ===
Type 2 confidence bounds are confidence bounds around reliability. For example, when using the two-parameter exponential distribution, the reliability function is:

::<math>\widehat{R}(T)={{e}^{-\widehat{\lambda }\cdot T}}\,\!</math>

Reliability bounds (Type 2) return the confidence bounds by determining the confidence intervals around <math>\widehat{\lambda }\,\!</math> and substituting these values into the above equation. The bounds on <math>\widehat{\lambda }\,\!</math> are determined using the method for the bounds on parameters, with its variance obtained from the Fisher Matrix. Once again, the procedure is more complicated for distributions with more than one parameter.

== Beta Binomial Confidence Bounds ==
Another less mathematically intensive method of calculating confidence bounds involves a procedure similar to that used in calculating median ranks (see [[Parameter Estimation]]). This is a non-parametric approach to confidence interval calculations that involves the use of rank tables and is commonly known as beta-binomial bounds (BB). By non-parametric, we mean that no underlying distribution is assumed. (Parametric implies that an underlying distribution, with parameters, is assumed.) In other words, this method can be used for any distribution, without having to make adjustments in the underlying equations based on the assumed distribution. Recall from the discussion on the median ranks that we used the binomial equation to compute the ranks at the 50% confidence level (or median ranks) by solving the cumulative binomial distribution for <math>Z\,\!</math> (rank for the <math>{{j}^{th}}\,\!</math> failure):

::<math>P=\underset{k=j}{\overset{N}{\mathop \sum }}\,\left( \begin{matrix}
N \\
k \\
\end{matrix} \right){{Z}^{k}}{{\left( 1-Z \right)}^{N-k}}\,\!</math>

where <math>N\,\!</math> is the sample size and <math>j\,\!</math> is the order number.

The median rank was obtained by solving the following equation for <math>Z\,\!</math>:

::<math>0.50=\underset{k=j}{\overset{N}{\mathop \sum }}\,\left( \begin{matrix}
N \\
k \\
\end{matrix} \right){{Z}^{k}}{{\left( 1-Z \right)}^{N-k}}\,\!</math>

The same methodology can then be repeated by changing <math>P\,\!</math> for 0.50 (50%) to our desired confidence level. For <math>P = 90%\,\!</math> one would formulate the equation as

::<math>0.90=\underset{k=j}{\overset{N}{\mathop \sum }}\,\left( \begin{matrix}
N \\
k \\
\end{matrix} \right){{Z}^{k}}{{\left( 1-Z \right)}^{N-k}}\,\!</math>

Keep in mind that one must be careful to select the appropriate values for <math>P\,\!</math> based on the type of confidence bounds desired. For example, if two-sided 80% confidence bounds are to be calculated, one must solve the equation twice (once with <math>P = 0.1\,\!</math> and once with <math>P = 0.9\,\!</math>) in order to place the bounds around 80% of the population. <br><br>Using this methodology, the appropriate ranks are obtained and plotted based on the desired confidence level. These points are then joined by a smooth curve to obtain the corresponding confidence bound.

In Weibull++, this non-parametric methodology is used only when plotting bounds on the mixed Weibull distribution. Full details on this methodology can be found in Kececioglu [[Appendix:_Life_Data_Analysis_References|[20]]]. These binomial equations can again be transformed using the beta and F distributions, thus the name beta binomial confidence bounds.

== Likelihood Ratio Confidence Bounds ==
Another method for calculating confidence bounds is the likelihood ratio bounds (LRB) method. Conceptually, this method is a great deal simpler than that of the Fisher matrix, although that does not mean that the results are of any less value. In fact, the LRB method is often preferred over the FM method in situations where there are smaller sample sizes.

Likelihood ratio confidence bounds are based on the following likelihood ratio equation:

::<math>-2\cdot \text{ln}\left( \frac{L(\theta )}{L(\widehat{\theta })} \right)\ge \chi _{\alpha ;k}^{2}\,\!</math>

where:

:*<math>L(\theta)\,\!</math> is the likelihood function for the unknown parameter vector <math>\theta\,\!</math>
:*<math>L(\widehat{\theta })\,\!</math> is the likelihood function calculated at the estimated vector <math>\widehat{\theta }\,\!</math>
:*<math>\chi _{\alpha ;k}^{2}\,\!</math> is the chi-squared statistic with probability <math>\alpha\,\!</math> and <math>k\,\!</math> degrees of freedom, where <math>k\,\!</math> is the number of quantities jointly estimated

If <math>\delta\,\!</math> is the confidence level, then <math>\alpha = \delta\,\!</math> for two-sided bounds and <math>\alpha = (2\delta - 1)\,\!</math> for one-sided. Recall from the [[Brief Statistical Background]] chapter that if <math>x\,\!</math> is a continuous random variable with ''pdf'':

::<math>\begin{align}
f(x;{{\theta}_{1}},{{\theta}_{2}},...,{{\theta}_{k}})
\end{align}\,\!</math>

where <math>{{\theta}_{1}},{{\theta}_{2}},...,{{\theta}_{k}}\,\!</math> are <math>k\,\!</math> unknown constant parameters that need to be estimated, one can conduct an experiment and obtain <math>R\,\!</math> independent observations, <math>{{x}_{1}},{{x}_{2}},...,{{x}_{R}}\,\!</math>, which correspond in the case of life data analysis to failure times. The likelihood function is given by:

::<math>L({{x}_{1}},{{x}_{2}},...,{{x}_{R}}|{{\theta }_{1}},{{\theta }_{2}},...,{{\theta }_{k}})=L=\underset{i=1}{\overset{R}{\mathop \prod }}\,f({{x}_{i}};{{\theta }_{1}},{{\theta }_{2}},...,{{\theta }_{k}})\,\!</math>

::<math>\begin{align}
i = 1,2,...,R
\end{align}\,\!</math>

The maximum likelihood estimators (MLE) of <math>{{\theta}_{1}},{{\theta}_{2}},...,{{\theta}_{k}}\,\!</math> are <math>k\,\!</math> are obtained by maximizing <math>L.\,\!</math> These are represented by the <math>L(\widehat{\theta })\,\!</math> term in the denominator of the ratio in the likelihood ratio equation. Since the values of the data points are known, and the values of the parameter estimates <math>\widehat{\theta }\,\!</math> have been calculated using MLE methods, the only unknown term in the likelihood ratio equation is the <math>L(\theta)\,\!</math> term in the numerator of the ratio. It remains to find the values of the unknown parameter vector <math>\theta\,\!</math> that satisfy the likelihood ratio equation. For distributions that have two parameters, the values of these two parameters can be varied in order to satisfy the likelihood ratio equation. The values of the parameters that satisfy this equation will change based on the desired confidence level <math>\delta;\,\!</math> but at a given value of <math>\delta\,\!</math> there is only a certain region of values for <math>{{\theta}_{1}}\,\!</math> and <math>{{\theta}_{2}}\,\!</math> for which the likelihood ratio equation holds true. This region can be represented graphically as a contour plot, an example of which is given in the following graphic.

[[Image:Examplecontourplot.png|center|450px]]

The region of the contour plot essentially represents a cross-section of the likelihood function surface that satisfies the conditions of the likelihood ratio equation.

'''Note on Contour Plots in Weibull++ and ALTA'''

Contour plots can be used for comparing data sets. Consider two data sets, one for an old product design and another for a new design. The engineer would like to determine if the two designs are significantly different and at what confidence. By plotting the contour plots of each data set in an overlay plot (the same distribution must be fitted to each data set), one can determine the confidence at which the two sets are significantly different. If, for example, there is no overlap (i.e., the two plots do not intersect) between the two 90% contours, then the two data sets are significantly different with a 90% confidence. If the two 95% contours overlap, then the two designs are NOT significantly different at the 95% confidence level. An example of non-intersecting contours is shown next. For details on comparing data sets, see [[Comparing Life Data Sets]].

<br>[[Image:Contourplot.png|center|450px]]

=== Confidence Bounds on the Parameters ===
The bounds on the parameters are calculated by finding the extreme values of the contour plot on each axis for a given confidence level. Since each axis represents the possible values of a given parameter, the boundaries of the contour plot represent the extreme values of the parameters that satisfy the following:

::<math>-2\cdot \text{ln}\left( \frac{L({{\theta }_{1}},{{\theta }_{2}})}{L({{\widehat{\theta }}_{1}},{{\widehat{\theta }}_{2}})} \right)=\chi _{\alpha ;1}^{2}\,\!</math>

This equation can be rewritten as:

::<math>L({{\theta }_{1}},{{\theta }_{2}})=L({{\widehat{\theta }}_{1}},{{\widehat{\theta }}_{2}})\cdot {{e}^{\tfrac{-\chi _{\alpha ;1}^{2}}{2}}}\,\!</math>

The task now is to find the values of the parameters <math>{{\theta}_{1}}\,\!</math> and <math>{{\theta}_{2}}\,\!</math> so that the equality in the likelihood ratio equation shown above is satisfied. Unfortunately, there is no closed-form solution; therefore, these values must be arrived at numerically. One way to do this is to hold one parameter constant and iterate on the other until an acceptable solution is reached. This can prove to be rather tricky, since there will be two solutions for one parameter if the other is held constant. In situations such as these, it is best to begin the iterative calculations with values close to those of the MLE values, so as to ensure that one is not attempting to perform calculations outside of the region of the contour plot where no solution exists.

<br>'''Example 1:''' {{Example: Likelihood Ratio Bounds on Parameters }}

=== Confidence Bounds on Time (Type 1) ===
The manner in which the bounds on the time estimate for a given reliability are calculated is much the same as the manner in which the bounds on the parameters are calculated. The difference lies in the form of the likelihood functions that comprise the likelihood ratio. In the preceding section, we used the standard form of the likelihood function, which was in terms of the parameters <math>{{\theta}_{1}}\,\!</math> and <math>{{\theta}_{2}}\,\!</math>. In order to calculate the bounds on a time estimate, the likelihood function needs to be rewritten in terms of one parameter and time, so that the maximum and minimum values of the time can be observed as the parameter is varied. This process is best illustrated with an example.

<br>'''Example 2:''' {{Example: Likelihood Ratio Bounds on Time (Type I)}}

=== Confidence Bounds on Reliability (Type 2) ===
The likelihood ratio bounds on a reliability estimate for a given time value are calculated in the same manner as were the bounds on time. The only difference is that the likelihood function must now be considered in terms of <math>\beta\,\!</math> and <math>R\,\!</math>. The likelihood function is once again altered in the same way as before, only now <math>R\,\!</math> is considered to be a parameter instead of <math>t\,\!</math>, since the value of <math>t\,\!</math> must be specified in advance. Once again, this process is best illustrated with an example.

<br>'''Example 3:''' {{Example: Likelihood Ratio Bounds on Reliability (Type 2)}}

== Bayesian Confidence Bounds ==
A fourth method of estimating confidence bounds is based on the Bayes theorem. This type of confidence bounds relies on a different school of thought in statistical analysis, where prior information is combined with sample data in order to make inferences on model parameters and their functions. An introduction to Bayesian methods is given in the [[Parameter Estimation]] chapter. Bayesian confidence bounds are derived from Bayes's rule, which states that:

::<math>f(\theta |Data)=\frac{L(Data|\theta )\varphi (\theta )}{\underset{\varsigma }{\int{\mathop{}_{}^{}}}\,L(Data|\theta )\varphi (\theta )d\theta }\,\!</math>

where:
:*<math>f (\theta | Data\,\!</math>) is the ''posterior pdf'' of <math>\theta\,\!</math>
:*<math>\theta\,\!</math> is the parameter vector of the chosen distribution (i.e., Weibull, lognormal, etc.)
:*<math>L(\bullet )\,\!</math> is the likelihood function
:*<math>\varphi (\theta )\,\!</math> is the ''prior pdf'' of the parameter vector <math>\theta\,\!</math>
:*<math>\varsigma \,\!</math> is the range of <math>\theta\,\!</math>.

In other words, the prior knowledge is provided in the form of the prior ''pdf'' of the parameters, which in turn is combined with the sample data in order to obtain the posterior ''pdf''. Different forms of prior information exist, such as past data, expert opinion or non-informative (refer to [[Parameter Estimation]]). It can be seen from the above Bayes's rule formula that we are now dealing with distributions of parameters rather than single value parameters. For example, consider a one-parameter distribution with a positive parameter <math>{{\theta}_{1}}\,\!</math>. Given a set of sample data, and a prior distribution for <math>{{\theta}_{1}},\,\!</math> <math>\varphi ({{\theta }_{1}}),\,\!</math> the above Bayes's rule formula can be written as:

::<math>f({{\theta }_{1}}|Data)=\frac{L(Data|{{\theta }_{1}})\varphi ({{\theta }_{1}})}{\int_{0}^{\infty }L(Data|{{\theta }_{1}})\varphi ({{\theta }_{1}})d{{\theta }_{1}}}\,\!</math>

In other words, we now have the distribution of <math>{{\theta}_{1}}\,\!</math> and we can now make statistical inferences on this parameter, such as calculating probabilities. Specifically, the probability that <math>{{\theta}_{1}}\,\!</math> is less than or equal to a value <math>x,\,\!</math> <math>P({{\theta }_{1}}\le x)\,\!</math> can be obtained by integrating the posterior probability density function (''pdf''), or:

::<math>P({{\theta }_{1}}\le x)=\int_{0}^{x}f({{\theta }_{1}}|Data)d{{\theta }_{1}}\,\!</math>

The above equation is the posterior ''cdf'', which essentially calculates a confidence bound on the parameter, where <math>P({{\theta }_{1}}\le x)\,\!</math> is the confidence level and <math>x\,\!</math> is the confidence bound. Substituting the posterior ''pdf'' into the above posterior ''cdf'' yields:

::<math>CL=\frac{\int_{0}^{x}L(Data|{{\theta }_{1}})\varphi ({{\theta }_{1}})d{{\theta }_{1}}}{\int_{0}^{\infty }L(Data|{{\theta }_{1}})\varphi ({{\theta }_{1}})d{{\theta }_{1}}}\,\!</math>

The only question at this point is, what do we use as a prior distribution of <math>{{\theta}_{1}}\,\!</math>? For the confidence bounds calculation application, non-informative prior distributions are utilized. Non-informative prior distributions are distributions that have no population basis and play a minimal role in the posterior distribution. The idea behind the use of non-informative prior distributions is to make inferences that are not affected by external information, or when external information is not available. In the general case of calculating confidence bounds using Bayesian methods, the method should be independent of external information and it should only rely on the current data. Therefore, non-informative priors are used. Specifically, the uniform distribution is used as a prior distribution for the different parameters of the selected fitted distribution. For example, if the Weibull distribution is fitted to the data, the prior distributions for beta and eta are assumed to be uniform. The above equation can be generalized for any distribution having a vector of parameters <math>\theta,\,\!</math> yielding the general equation for calculating Bayesian confidence bounds:

::<math>CL=\frac{\underset{\xi }{\int{\mathop{}_{}^{}}}\,L(Data|\theta )\varphi (\theta )d\theta }{\underset{\varsigma }{\int{\mathop{}_{}^{}}}\,L(Data|\theta )\varphi (\theta )d\theta }\,\!</math>

where:

:*<math>CL\,\!</math> is the confidence level
:*<math>\theta\,\!</math> is the parameter vector
:*<math>L(\bullet )\,\!</math> is the likelihood function
:*<math>\varphi (\theta )\,\!</math> is the prior ''pdf'' of the parameter vector <math>\theta\,\!</math>
:*<math>\varsigma \,\!</math> is the range of <math>\theta\,\!</math>
:*<math>\xi\,\!</math> is the range in which <math>\theta\,\!</math> changes from <math>\Psi(T,R)\,\!</math> till <math>\theta\,\!</math> 's maximum value, or from <math>\theta\,\!</math> 's minimum value till <math>\Psi(T,R)\,\!</math>
:*<math>\Psi(T,R)\,\!</math> is a function such that if <math>T\,\!</math> is given, then the bounds are calculated for <math>R\,\!</math>. If <math>R\,\!</math> is given, then the bounds are calculated for <math>T\,\!</math>.

If <math>T\,\!</math> is given, then from the above equation and <math>\Psi\,\!</math> and for a given <math>CL\,\!</math>, the bounds on <math>R\,\!</math> are calculated. If <math>R\,\!</math> is given, then from the above equation and <math>\Psi\,\!</math> and for a given <math>C\,\!</math> the bounds on <math>T\,\!</math> are calculated.

=== Confidence Bounds on Time (Type 1) ===
For a given failure time distribution and a given reliability <math>R\,\!</math>, <math>T(R)\,\!</math> is a function of <math>R\,\!</math> and the distribution parameters. To illustrate the procedure for obtaining confidence bounds, the two-parameter Weibull distribution is used as an example. The bounds in other types of distributions can be obtained in similar fashion. For the two-parameter Weibull distribution:

::<math>T(R)=\eta \exp (\frac{\ln (-\ln R)}{\beta })\,\!</math>

For a given reliability, the Bayesian one-sided upper bound estimate for <math>T(R)\,\!</math> is:

::<math>CL=\underset{}{\overset{}{\mathop{\Pr }}}\,(T\le {{T}_{U}})=\int_{0}^{{{T}_{U}}(R)}f(T|Data,R)dT\,\!</math>

where <math>f(T | Data, R)\,\!</math> is the posterior distribution of Time <math>T.\,\!</math> Using the above equation, we have the following:

::<math>CL=\underset{}{\overset{}{\mathop{\Pr }}}\,(T\le {{T}_{U}})=\underset{}{\overset{}{\mathop{\Pr }}}\,(\eta \exp (\frac{\ln (-\ln R)}{\beta })\le {{T}_{U}})\,\!</math>

The above equation can be rewritten in terms of <math>\eta\,\!</math> as:

::<math>CL=\underset{}{\overset{}{\mathop{\Pr }}}\,(\eta \le {{T}_{U}}\exp (-\frac{\ln (-\ln R)}{\beta }))\,\!</math>

Applying the Bayes's rule by assuming that the priors of <math>\beta\,\!</math> and <math>\eta\,\!</math> are independent, we then obtain the following relationship:

::<math>CL=\frac{\int_{0}^{\infty }\int_{0}^{{{T}_{U}}\exp (-\frac{\ln (-\ln R)}{\beta })}L(\beta ,\eta )\varphi (\beta )\varphi (\eta )d\eta d\beta }{\int_{0}^{\infty }\int_{0}^{\infty }L(\beta ,\eta )\varphi (\beta )\varphi (\eta )d\eta d\beta }\,\!</math>

The above equation can be solved for <math>{{T}_{U}}(R)\,\!</math>, where:

:*<math>CL\,\!</math> is the confidence level,
:*<math>\varphi (\beta )\,\!</math> is the prior ''pdf'' of the parameter <math>\beta\,\!</math>. For non-informative prior distribution, <math>\varphi (\beta )=\tfrac{1}{\beta }.\,\!</math>
:*<math>\varphi (\eta )\,\!</math> is the prior ''pdf'' of the parameter <math>\eta.\,\!</math> For non-informative prior distribution, <math>\varphi (\eta )=\tfrac{1}{\eta }.\,\!</math>
:*<math>L(\bullet )\,\!</math> is the likelihood function.

The same method can be used to get the one-sided lower bound of <math>T(R)\,\!</math> from:

::<math>CL=\frac{\int_{0}^{\infty }\int_{{{T}_{L}}\exp (\frac{-\ln (-\ln R)}{\beta })}^{\infty }L(\beta ,\eta )\varphi (\beta )\varphi (\eta )d\eta d\beta }{\int_{0}^{\infty }\int_{0}^{\infty }L(\beta ,\eta )\varphi (\beta )\varphi (\eta )d\eta d\beta }\,\!</math>

The above equation can be solved to get <math>{{T}_{L}}(R)\,\!</math>.

The Bayesian two-sided bounds estimate for <math>T(R)\,\!</math> is:

::<math>CL=\int_{{{T}_{L}}(R)}^{{{T}_{U}}(R)}f(T|Data,R)dT\,\!</math>

which is equivalent to:

::<math>(1+CL)/2=\int_{0}^{{{T}_{U}}(R)}f(T|Data,R)dT\,\!</math>

and:

::<math>(1-CL)/2=\int_{0}^{{{T}_{L}}(R)}f(T|Data,R)dT\,\!</math>

Using the same method for the one-sided bounds, <math>{{T}_{U}}(R)\,\!</math> and <math>{{T}_{L}}(R)\,\!</math> can be solved.

=== Confidence Bounds on Reliability (Type 2) ===
For a given failure time distribution and a given time <math>T\,\!</math>, <math>R(T)\,\!</math> is a function of <math>T\,\!</math> and the distribution parameters. To illustrate the procedure for obtaining confidence bounds, the two-parameter Weibull distribution is used as an example. The bounds in other types of distributions can be obtained in similar fashion. For example, for two parameter Weibull distribution:

::<math>R=\exp (-{{(\frac{T}{\eta })}^{\beta }})\,\!</math>

The Bayesian one-sided upper bound estimate for <math>R(T)\,\!</math> is:

::<math>CL=\int_{0}^{{{R}_{U}}(T)}f(R|Data,T)dR\,\!</math>

Similar to the bounds on Time, the following is obtained:

::<math>CL=\frac{\int_{0}^{\infty }\int_{0}^{T\exp (-\frac{\ln (-\ln {{R}_{U}})}{\beta })}L(\beta ,\eta )\varphi (\beta )\varphi (\eta )d\eta d\beta }{\int_{0}^{\infty }\int_{0}^{\infty }L(\beta ,\eta )\varphi (\beta )\varphi (\eta )d\eta d\beta }\,\!</math>

The above equation can be solved to get <math>{{R}_{U}}(T)\,\!</math>.

The Bayesian one-sided lower bound estimate for R(T) is:

::<math>1-CL=\int_{0}^{{{R}_{L}}(T)}f(R|Data,T)dR\,\!</math>

Using the posterior distribution, the following is obtained:

::<math>CL=\frac{\int_{0}^{\infty }\int_{T\exp (-\frac{\ln (-\ln {{R}_{L}})}{\beta })}^{\infty }L(\beta ,\eta )\varphi (\beta )\varphi (\eta )d\eta d\beta }{\int_{0}^{\infty }\int_{0}^{\infty }L(\beta ,\eta )\varphi (\beta )\varphi (\eta )d\eta d\beta }\,\!</math>

The above equation can be solved to get <math>{{R}_{L}}(T)\,\!</math>.

The Bayesian two-sided bounds estimate for <math>R(T)\,\!</math> is:

::<math>CL=\int_{{{R}_{L}}(T)}^{{{R}_{U}}(T)}f(R|Data,T)dR\,\!</math>

which is equivalent to:

::<math>\int_{0}^{{{R}_{U}}(T)}f(R|Data,T)dR=(1+CL)/2\,\!</math>

and

::<math>\int_{0}^{{{R}_{L}}(T)}f(R|Data,T)dR=(1-CL)/2\,\!</math>

Using the same method for one-sided bounds, <math>{{R}_{U}}(T)\,\!</math> and <math>{{R}_{L}}(T)\,\!</math> can be solved.

== Simulation Based Bounds ==
The SimuMatic tool in Weibull++ can be used to perform a large number of reliability analyses on data sets that have been created using Monte Carlo simulation. This utility can assist the analyst to a) better understand life data analysis concepts, b) experiment with the influences of sample sizes and censoring schemes on analysis methods, c) construct simulation-based confidence intervals, d) better understand the concepts behind confidence intervals and e) design reliability tests. This section describes how to use simulation for estimating confidence bounds. <br>

SimuMatic generates confidence bounds and assists in visualizing and understanding them. In addition, it allows one to determine the adequacy of certain parameter estimation methods (such as rank regression on X, rank regression on Y and maximum likelihood estimation) and to visualize the effects of different data censoring schemes on the confidence bounds.

'''Example:'''
{{:Simulation_Based_Bounds_Example}}

Parametric Recurrent Event Data Analysis

2023-09-18T21:47:52Z

Lisa Hacker:

<noinclude>{{Banner Weibull Articles}}
''This article appears in the [https://help.reliasoft.com/reference/life_data_analysis Life data analysis reference].''

{{Navigation box}}
</noinclude>Weibull++'s parametric RDA folio is a tool for modeling recurrent event data. It can capture the trend, estimate the rate and predict the total number of recurrences. The failure and repair data of a repairable system can be treated as one type of recurrence data. Past and current repairs may affect the future failure process. For most recurrent events, time (distance, cycles, etc.) is a key factor. With time, the recurrence rate may remain constant, increase or decrease. For other recurrent events, not only the time, but also the number of events can affect the recurrence process (e.g., the debugging process in software development).

The parametric analysis approach utilizes the General Renewal Process (GRP) model, as discussed in Mettas and Zhao [[Appendix:_Life_Data_Analysis_References|[28]]]. In this model, the repair time is assumed to be negligible so that the processes can be viewed as point processes. This model provides a way to describe the rate of occurrence of events over time, such as in the case of data obtained from a repairable system. This model is particularly useful in modeling the failure behavior of a specific system and understanding the effects of the repairs on the age of that system. For example, consider a system that is repaired after a failure, where the repair does not bring the system to an ''as-good-as-new'' or an ''as-bad-as-old'' condition. In other words, the system is partially rejuvenated after the repair. Traditionally, in as-bad-as-old repairs, also known as ''minimal repairs'', the failure data from such a system would have been modeled using a homogeneous or non-homogeneous Poisson process (NHPP). On rare occasions, a Weibull distribution has been used as well in cases where the system is almost as-good-as-new after the repair, also known as a ''perfect renewal process'' (PRP). However, for the intermediate states after the repair, there has not been a commercially available model, even though many models have been proposed in literature. In Weibull++, the GRP model provides the capability to model systems with partial renewal (''general repair'' or ''imperfect repair/maintenance'') and allows for a variety of predictions such as reliability, expected failures, etc.

== The GRP Model ==
In this model, the concept of virtual age is introduced. Let <math>{{t}_{1}},{{t}_{2}},\cdots ,{{t}_{n}}\,\!</math> represent the successive failure times and let <math>{{x}_{1}},{{x}_{2}},\cdots ,{{x}_{n}}\,\!</math> represent the time between failures ( <math>{{t}_{i}}=\sum_{j=1}^{i}{{x}_{j}})\,\!</math>. Assume that after each event, actions are taken to improve the system performance. Let <math>q\,\!</math> be the action effectiveness factor. There are two GRP models:

Type I:

::<math>\begin{align}
v_{i}=v_{i-1}+qx_{i}=qt_{i}
\end{align}\,\!</math>

Type II:

::<math>{{v}_{i}}=q({{v}_{i-1}}+{{x}_{i}})={{q}^{i}}{{x}_{1}}+{{q}^{i-1}}{{x}_{2}}+\cdots +{{q}{x}_{i}}\,\!</math>

where <math>{{v}_{i}}\,\!</math> is the virtual age of the system right after <math>i\,\!</math>th repair. The Type I model assumes that the <math>i\,\!</math>th repair cannot remove the damage incurred before the <math>(i-1)\,\!</math> th repair. It can only reduce the additional age <math>{{x}_{i}}\,\!</math> to <math>{{qx}_{i}}\,\!</math>. The Type II model assumes that at the <math>i\,\!</math>th repair, the virtual age has been accumulated to <math>v_{i-1} + {{x}_{i}}\,\!</math>. The <math>i\,\!</math>th repair will remove the cumulative damage from both current and previous failures by reducing the virtual age to <math>q(v_{i-1} + x_{i})\,\!</math>.

The power law function is used to model the rate of recurrence, which is:

::<math>\begin{align}
\lambda(t)=\lambda \beta t^{\beta -1}
\end{align}\,\!</math>

The conditional ''pdf'' is:

::<math>f({{t}_{i}}|{{t}_{i-1}})=\lambda \beta {{({{x}_{i}}+{{v}_{i-1}})}^{\beta -1}}{{e}^{-\lambda \left[ {{\left( {{x}_{i}}+{{v}_{i-1}} \right)}^{\beta }}-v_{i-1}^{\beta } \right]}}\,\!</math>

MLE method is used to estimate the model parameters. The log likelihood function is discussed in Mettas and Zhao [[Appendix:_Life_Data_Analysis_References|[28]]]:

::<math>\begin{align}
& \ln (L)= n(\ln \lambda +\ln \beta )-\lambda \left[ {{\left( T-{{t}_{n}}+{{v}_{n}} \right)}^{\beta }}-v_{n}^{\beta } \right] \\
& -\lambda \underset{i=1}{\overset{n}{\mathop \sum }}\,\left[ {{\left( {{x}_{i}}+{{v}_{i-1}} \right)}^{\beta }}-v_{i-1}^{\beta } \right]+(\beta -1)\underset{i=1}{\overset{n}{\mathop \sum }}\,\ln ({{x}_{i}}+{{v}_{i-1}})
\end{align}\,\!</math>

where <math>n\,\!</math> is the total number of events during the entire observation period. <math>T\,\!</math> is the stop time of the observation. <math>T = t_{n}\,\!</math> if the observation stops right after the last event.

== Confidence Bounds ==
In general, in order to obtain the virtual age, the exact occurrence time of each event (failure) should be available (see equations for Type I and Type II models). However, the times are unknown until the corresponding events occur. For this reason, there are no closed-form expressions for total failure number and failure intensity, which are functions of failure times and virtual age. Therefore, in Weibull++, a Monte Carlo simulation is used to predict values of virtual time, failure number, MTBF and failure rate. The approximate confidence bounds obtained from simulation are provided. The uncertainty of model parameters is also considered in the bounds.

=== Bounds on Cumulative Failure (Event) Numbers ===
The variance of the cumulative failure number <math>N(t)\,\!</math> is:

::<math>Var[N(t)]=Var\left[ E(N(t)|\lambda ,\beta ,q) \right]+E\left[ Var(N(t)|\lambda ,\beta ,q) \right]\,\!</math>

The first term accounts for the uncertainty of the parameter estimation. The second term considers the uncertainty caused by the renewal process even when model parameters are fixed. However, unless <math>q = 1\,\!</math> , <math>Var\left[ E(N(t)|\lambda ,\beta ,q) \right]\,\!</math> cannot be calculated because <math>E(N(t))\,\!</math> cannot be expressed as a closed-form function of <math>\lambda,\beta\,\,</math>, and <math>q\,\!</math>. In order to consider the uncertainty of the parameter estimation, <math>Var\left[ E(N(t)|\lambda ,\beta ,q) \right]\,\!</math> is approximated by:

::<math>Var\left[ E(N(t)|\lambda ,\beta ,q) \right]=Var[E(N({{v}_{t}})|\lambda ,\beta )]=Var[\lambda v_{t}^{\beta }]\,\!</math>

where <math>v_{t}\,\!</math> is the expected virtual age at time <math>t\,\!</math> and <math>Var[\lambda v_{t}^{\beta }]\,\!</math> is:

::<math>\begin{align}
& Var[\lambda v_{t}^{\beta }]= & {{\left( \frac{\partial (\lambda v_{t}^{\beta })}{\partial \beta } \right)}^{2}}Var(\hat{\beta })+{{\left( \frac{\partial (\lambda v_{t}^{\beta })}{\partial \lambda } \right)}^{2}}Var(\hat{\lambda }) \\
& +2\frac{\partial (\lambda v_{t}^{\beta })}{\partial \beta }\frac{\partial (\lambda v_{t}^{\beta })}{\partial \lambda }Cov(\hat{\beta },\hat{\lambda })
\end{align}\,\!</math>

By conducting this approximation, the uncertainty of <math>\lambda\,\!</math> and <math>\beta\,\!</math> are considered. The value of <math>v_{t}\,\!</math> and the value of the second term in the equation for the variance of number of failures are obtained through the Monte Carlo simulation using parameters <math>\hat{\lambda },\hat{\beta },\hat{q},\,\!</math> which are the ML estimators. The same simulation is used to estimate the cumulative number of failures <math>\hat{N}(t)=E(N(t)|\hat{\lambda },\hat{\beta },\hat{q})\,\!</math>.

Once the variance and the expected value of <math>N(t)\,\!</math> have been obtained, the bounds can be calculated by assuming that <math>N(t)\,\!</math> is lognormally distributed as:

::<math>\frac{\ln N(t)-\ln \hat{N}(t)}{\sqrt{Var(\ln N(t))}}\tilde{\ }N(0,1)\,\!</math>

The upper and lower bounds for a given confidence level <math>\alpha\,\!</math> can be calculated by:

::<math>N{{(t)}_{U,L}}=\hat{N}(t){{e}^{\pm {{z}_{a}}\sqrt{Var(N(t))}/\hat{N}(t)}}\,\!</math>

where <math>z_{a}\,\!</math> is the standard normal distribution.

If <math>N(t)\,\!</math> is assumed to be normally distributed, the bounds can be calculated by:

::<math>N{{(t)}_{U}}=\hat{N}(t)+{{z}_{a}}\sqrt{Var(N(t))}\,\!</math>

::<math>N{{(t)}_{L}}=\hat{N}(t)-{{z}_{a}}\sqrt{Var(N(t))}\,\!</math>

In Weibull++, the <math>N(t)_{U}\,\!</math> is the smaller of the upper bounds obtained from lognormal and normal distribution appoximation. The <math>N(t)_{L}\,\!</math> is set to the largest of the lower bounds obtained from lognormal and normal distribution appoximation. This combined method can prevent the out-of-range values of bounds for some small <math>t\,\!</math> values.

=== Bounds of Cumulative Failure Intensity and MTBF ===
For a given time <math>t\,\!</math> , the expected value of cumulative MTBF <math>m_{c}(t)\,\!</math> and cumulative failure intensity <math>\lambda_{c}(t)\,\!</math> can be calculated using the following equations:

::<math>{{\hat{\lambda }}_{c}}(t)=\frac{\hat{N}(t)}{t};{{\hat{m}}_{c}}(t)=\frac{t}{\hat{N}(t)}\,\!</math>

The bounds can be easily obtained from the corresponding bounds of <math>N(t)\,\!</math>.

::<math>\begin{align}
& {{{\hat{\lambda }}}_{c}}{{(t)}_{L}}= & \frac{\hat{N}{{(t)}_{L}}}{t};\text{ }{{{\hat{\lambda }}}_{c}}{{(t)}_{L}}=\frac{\hat{N}{{(t)}_{L}}}{t};\text{ } \\
& {{{\hat{m}}}_{c}}{{(t)}_{L}}= & \frac{t}{\hat{N}{{(t)}_{U}}};\text{ }{{{\hat{m}}}_{c}}{{(t)}_{U}}=\frac{t}{\hat{N}{{(t)}_{L}}}
\end{align}\,\!</math>

=== Bounds on Instantaneous Failure Intensity and MTBF ===
The instantaneous failure intensity is given by:

::<math>{{\lambda }_{i}}(t)=\lambda \beta v_{t}^{\beta -1}\,\!</math>

where <math>v_{t}\,\!</math> is the virtual age at time <math>t\,\!</math>. When <math>q\ne 1,\,\!</math> it is obtained from simulation. When <math>q = 1\,\!</math>, <math>v_{t} = t\,\!</math> from model Type I and Type II.

The variance of instantaneous failure intensity can be calculated by:

::<math>\begin{align}
& Var({{\lambda }_{i}}(t))= {{\left( \frac{\partial {{\lambda }_{i}}(t)}{\partial \beta } \right)}^{2}}Var(\hat{\beta })+{{\left( \frac{\partial {{\lambda }_{i}}(t)}{\partial \lambda } \right)}^{2}}Var(\hat{\lambda }) \\
& +2\frac{\partial {{\lambda }_{i}}(t)}{\partial \beta }\frac{\partial {{\lambda }_{i}}(t)}{\partial \lambda }Cov(\hat{\beta },\hat{\lambda })+{{\left( \frac{\partial {{\lambda }_{i}}(t)}{\partial v(t)} \right)}^{2}}Var({{{\hat{v}}}_{t}})
\end{align}\,\!</math>

The expected value and variance of <math>v_{t}\,\!</math> are obtained from the Monte Carlo simulation with parameters <math>\hat{\lambda },\hat{\beta },\hat{q}.\,\!</math> Because of the simulation accuracy and the convergence problem in calculation of <math>Var(\hat{\beta }),Var(\hat{\lambda })\,\!</math> and <math>Cov(\hat{\beta },\hat{\lambda }),\,\!</math> <math>Var(\lambda_{i}(t))\,\!</math> can be a negative value at some time points. When this case happens, the bounds of instantaneous failure intensity are not provided.

Once the variance and the expected value of <math>\lambda_{i}(t)\,\!</math> are obtained, the bounds can be calculated by assuming that  <math>\lambda_{i}(t)\,\!</math> is lognormally distributed as:

::<math>\frac{\ln {{\lambda }_{i}}(t)-\ln {{{\hat{\lambda }}}_{i}}(t)}{\sqrt{Var(\ln {{\lambda }_{i}}(t))}}\tilde{\ }N(0,1)\,\!</math>

The upper and lower bounds for a given confidence level <math>\alpha\,\!</math> can be calculated by:

::<math>{{\lambda }_{i}}(t)={{\hat{\lambda }}_{i}}(t){{e}^{\pm {{z}_{a}}\sqrt{Var({{\lambda }_{i}}(t))}/{{{\hat{\lambda }}}_{i}}(t)}}\,\!</math>

where <math>z_{a}\,\!</math> is the standard normal distribution.

If <math>\lambda_{i}(t)\,\!</math> is assumed to be normally distributed, the bounds can be calculated by:

::<math>{{\lambda }_{i}}{{(t)}_{U}}={{\hat{\lambda }}_{i}}(t)+{{z}_{a}}\sqrt{Var(N(t))}\,\!</math>

::<math>{{\lambda }_{i}}{{(t)}_{L}}={{\hat{\lambda }}_{i}}(t)-{{z}_{a}}\sqrt{Var(N(t))}\,\!</math>

In Weibull++, <math>\lambda_{i}(t)_{U}\,\!</math> is set to the smaller of the two upper bounds obtained from the above lognormal and normal distribution appoximation. <math>\lambda_{i}(t)_{L}\,\!</math> is set to the largest of the two lower bounds obtained from the above lognormal and normal distribution appoximation. This combination method can prevent the out of range values of bounds when <math>t\,\!</math> values are small.

For a given time <math>t\,\!</math>, the expected value of cumulative MTBF <math>m_{i}(t)\,\!</math> is:

::<math>{{\hat{m}}_{i}}(t)=\frac{1}{{{{\hat{\lambda }}}_{i}}(t)}\text{ }\,\!</math>

The upper and lower bounds can be easily obtained from the corresponding bounds of <math>\lambda_{i}(t)\,\!</math>:

::<math>{{\hat{m}}_{i}}{{(t)}_{U}}=\frac{1}{{{{\hat{\lambda }}}_{i}}{{(t)}_{L}}}\,\!</math>

::<math>{{\hat{m}}_{i}}{{(t)}_{L}}=\frac{1}{{{{\hat{\lambda }}}_{i}}{{(t)}_{U}}}\,\!</math>

=== Bounds on Conditional Reliability ===
Given mission start time <math>t_{0}\,\!</math> and mission time <math>T\,\!</math>, the conditional reliability can be calculated by:

::<math>R(T|{{t}_{0}})=\frac{R(T+{{v}_{0}})}{R({{v}_{0}})}={{e}^{-\lambda [{{({{v}_{0}}+T)}^{\beta }}-{{v}_{0}}]}}\,\!</math>

<math>v_{0}\,\!</math> is the virtual age corresponding to time <math>t_{0}\,\!</math>. The expected value and the variance of <math>v_{0}\,\!</math> are obtained from Monte Carlo simulation. The variance of the conditional reliability <math>R(T|t_{0})\,\!</math> is:

::<math>\begin{align}
& Var(R)= {{\left( \frac{\partial R}{\partial \beta } \right)}^{2}}Var(\hat{\beta })+{{\left( \frac{\partial R}{\partial \lambda } \right)}^{2}}Var(\hat{\lambda }) \\
& +2\frac{\partial R}{\partial \beta }\frac{\partial R}{\partial \lambda }Cov(\hat{\beta },\hat{\lambda })+{{\left( \frac{\partial R}{\partial {{v}_{0}}} \right)}^{2}}Var({{{\hat{v}}}_{0}})
\end{align}\,\!</math>

Because of the simulation accuracy and the convergence problem in calculation of <math>Var(\hat{\beta }),Var(\hat{\lambda })\,\!</math> and <math>Cov(\hat{\beta },\hat{\lambda }),\,\!</math> <math>Var(R)\,\!</math> can be a negative value at some time points. When this case happens, the bounds are not provided.

The bounds are based on:

::<math>\log \text{it}(\hat{R}(T))\tilde{\ }N(0,1)\,\!</math>

::<math>\log \text{it}(\hat{R}(T))=\ln \left\{ \frac{\hat{R}(T)}{1-\hat{R}(T)} \right\}\,\!</math>

The confidence bounds on reliability are given by:

::<math>R=\frac{{\hat{R}}}{\hat{R}+(1-\hat{R}){{e}^{\pm \sqrt{Var(R)}/[\hat{R}(1-\hat{R})]}}}\,\!</math>

It will be compared with the bounds obtained from:

::<math>R=\hat{R}{{e}^{\pm {{z}_{a}}\sqrt{Var(R)}/\hat{R}}}\,\!</math>

The smaller of the two upper bounds will be the final upper bound and the larger of the two lower bounds will be the final lower bound.

==Example: Air Condition Unit==

{{:Example:_Parametric_RDA_-_Air_Condition_Unit}}

ReliaSoft's Alternate Ranking Method

2023-09-18T21:47:38Z

Lisa Hacker:

<noinclude>{{Banner Weibull Articles}}
''This article appears in the [https://help.reliasoft.com/reference/life_data_analysis Life data analysis reference].''

</noinclude>
In probability plotting or rank regression analysis of '''interval''' or '''left censored''' data, difficulties arise when attempting to estimate the exact time within the interval when the failure actually occurs, especially when an overlap on the intervals is present. In this case, the ''standard ranking method'' (SRM) is not applicable when dealing with interval data; thus, ReliaSoft has formulated a more sophisticated methodology to allow for more accurate probability plotting and regression analysis of data sets with interval or left censored data. This method utilizes the traditional rank regression method and iteratively improves upon the computed ranks by parametrically recomputing new ranks and the most probable failure time for interval data.

In the traditional method for plotting or rank regression analysis of '''right censored''' data, the effect of the exact censoring time is not considered. One example of this can be seen at the [[Parameter_Estimation#Shortfalls_of_the_Rank_Adjustment_Method|parameter estimation]] chapter. The ReliaSoft ranking method also can be used to overcome this shortfall of the standard ranking method.

The following step-by-step example illustrates the ReliaSoft ranking method (RRM), which is an iterative improvement on the standard ranking method (SRM). Although this method is illustrated by the use of the two-parameter Weibull distribution, it can be easily generalized for other models.

Consider the following test data:

{|border="1" align="center" style="border-collapse: collapse;" cellpadding="5" cellspacing="5"
|-
|colspan="4" style="text-align:center"|Table B.1- The Test Data
|-
!Number of Items
!Type
!Last Inspection
!Time
|-align="center"
|1||Exact Failure|| ||10
|-align="center"
|1||Right Censored|| ||20
|- align="center"
|2||Left Censored||0||30
|-align="center"
|2||Exact Failure|| ||40
|-align="center"
|1||Exact Failure|| ||50
|- align="center"
|1||Right Censored|| ||60
|- align="center"
|1||Left Censored||0||70
|- align="center"
|2||Interval Failure||20||80
|- align="center"
|1||Interval Failure||10||85
|- align="center"
|1||Left Censored||0||100
|}

=== Initial Parameter Estimation===
As a preliminary step, we need to provide a crude estimate of the Weibull parameters for this data. To begin, we will extract the exact times-to-failure (10, 40, and 50) and the midpoints of the interval failures. The midpoints are 50 (for the interval of 20 to 80) and 47.5 (for the interval of 10 to 85). Next, we group together the items that have the same failure times, as shown in Table B.2.

Using the traditional rank regression, we obtain the first initial estimates:

::<math>\begin{align}
& {{\widehat{\beta }}_{0}}= & 1.91367089 \\
& {{\widehat{\eta }}_{0}}= & 43.91657736
\end{align}\,\!</math>

{|border="1" align="center" style="border-collapse: collapse;" cellpadding="5" cellspacing="5"
|colspan="4" style="text-align:center"|Table B.2- The Union of Exact Times-to-Failure with the "Midpoint" of the Interval Failures
|-
!Number of Items
!Type
!Last Inspection
!Time
|-
|1||Exact Failure|| ||10
|-
|2||Exact Failure|| ||40
|-
|1||Exact Failure|| ||47.5
|-
|3||Exact Failure|| ||50
|}

'''Step 1'''

For all intervals, we obtain a weighted ''midpoint'' using:

::<math>\begin{align}
{{{\hat{t}}}_{m}}\left( \hat{\beta },\hat{\eta } \right)= & \frac{\int_{LI}^{TF}t\text{ }f(t;\hat{\beta },\hat{\eta })dt}{\int_{LI}^{TF}f(t;\hat{\beta },\hat{\eta })dt}, \\
= & \frac{\int_{LI}^{TF}t\tfrac{{\hat{\beta }}}{{\hat{\eta }}}{{\left( \tfrac{t}{{\hat{\eta }}} \right)}^{\hat{\beta }-1}}{{e}^{-{{\left( \tfrac{t}{{\hat{\eta }}} \right)}^{{\hat{\beta }}}}}}dt}{\int_{LI}^{TF}\tfrac{{\hat{\beta }}}{{\hat{\eta }}}{{\left( \tfrac{t}{{\hat{\eta }}} \right)}^{\hat{\beta }-1}}{{e}^{-{{\left( \tfrac{t}{{\hat{\eta }}} \right)}^{{\hat{\beta }}}}}}dt}
\end{align}\,\!</math>

This transforms our data into the format in Table B.3.

{|border="1" align="center" style="border-collapse: collapse;" cellpadding="5" cellspacing="5"
|colspan="5" style="text-align:center"|Table B.3- The Union of Exact Times-to-Failure with the "Midpoint" of the Interval Failures, Based upon the Parameters <math>\beta\,\!</math> and <math>\eta\,\!</math>.
|-
!Number of Items
!Type
!Last Inspection
!Time
!Weighted "Midpoint"
|- align="center"
|1||Exact Failure|| ||10 ||
|- align="center"
|2||Exact Failure|| ||40||
|-align="center"
|1||Exact Failure|| || 50||
|-align="center"
|2||Interval Failure||20||80||42.837
|-align="center"
|1||Interval Failure||10||85||39.169
|}

'''Step 2'''

Now we arrange the data as in Table B.4.

{|border="1" align="center" style="border-collapse: collapse;" cellpadding="5" cellspacing="5"
|colspan="2"|Table B.4- The Union of Exact Times-to-Failure with the "Midpoint" of the Interval Failures, in Ascending Order.
|-align="center"
!Number of Items
!Time
|- align="center"
|1||10
|- align="center"
|1||39.169
|- align="center"
|2||40
|- align="center"
|2||42.837
|- align="center"
|1||50
|}

'''Step 3'''

We now consider the left and right censored data, as in Table B.5.

{|border="1" align="center" style="border-collapse: collapse;" cellpadding="5" cellspacing="5"
|-
|colspan="7" style="text-align:center"|Table B.5- Computation of Increments in a Matrix Format for Computing a Revised Mean Order Number.
|-
!Number of items
!Time of Failure
!2 Left Censored ''t'' = 30
!1 Left Censored ''t'' = 70
!1 Left Censored ''t'' = 100
!1 Right Censored ''t'' = 20
!1 Right Censored ''t'' = 60
|-
|1||10||<math>2 \frac{\int_0^{10} f_0(t)dt}{F_0 (30)-F_0 (0)}\,\!</math> ||<math>\frac{\int_0^{10} f_0 (t)dt}{F_0(70)-F_0(0)}\,\!</math> || <math>\frac{\int_0^{10} f_0(t)dt}{F_0(100)-F_0(0)}\,\!</math> || 0||0
|-
|1||39.169||<math>2 \frac{\int_{10}^{30} f_0(t)dt}{F_0(30)-F_0(0)}\,\!</math> ||<math>\frac{\int_{10}^{39.169} f_0(t)dt}{F_0(70)-F_0(0)}\,\!</math> ||<math>\frac{\int_{10}^{39.169} f_0(t)dt}{F_0(100)-F_0(0)}\,\!</math> || <math>\frac{\int_{20}^{39.169} f_0(t)dt}{F_0(\infty)-F_0(20)}\,\!</math>||0
|-
|2||40||0||<math>\frac{\int_{39.169}^{40} f_0(t)dt}{F_0(70)-F_0(0)}\,\!</math> || <math>\frac{\int_{39.169}^{40} f_0(t)dt}{F_0(100)-F_0(0)}\,\!</math> ||<math>\frac{\int_{39.169}^{40} f_0(t)dt}{F_0(\infty)-F_0(20)}\,\!</math> ||0
|-
|2||42.837||0|| <math>\frac{\int_{40}^{42.837} f_0(t)dt}{F_0(70)-F_0(0)}\,\!</math> || <math>\frac{\int_{40}^{42.837} f_0(t)dt}{F_0(100)-F_0(0)}\,\!</math>|| <math>\frac{\int_{40}^{42.837} f_0(t)dt}{F_0(\infty)-F_0(0)}\,\!</math>||0
|-
|1||50||0||<math>\frac{\int_{42.837}^{50} f_0(t)dt}{F_0(70)-F_0(0)}\,\!</math> ||<math>\frac{\int_{42.837}^{50} f_0(t)dt}{F_0(100)-F_0(0)}\,\!</math> || <math>\frac{\int_{42.837}^{50} f_0(t)dt}{F_0(\infty)-F_0(0)}\,\!</math>||0
|}

In general, for left censored data:

:• The increment term for <math>n\,\!</math> left censored items at time <math>={{t}_{0}},\,\!</math> with a time-to-failure of <math>{{t}_{i}}\,\!</math> when <math>{{t}_{0}}\le {{t}_{i-1}}\,\!</math> is zero.
:• When <math>{{t}_{0}}>{{t}_{i-1}},\,\!</math> the contribution is:

::<math>\frac{n}{{{F}_{0}}({{t}_{0}})-{{F}_{0}}(0)}\underset{{{t}_{i-1}}}{\overset{MIN({{t}_{i}},{{t}_{0}})}{\mathop \int }}\,{{f}_{0}}\left( t \right)dt\,\!</math>

:or:

::<math>n\frac{{{F}_{0}}(MIN({{t}_{i}},{{t}_{0}}))-{{F}_{0}}({{t}_{i-1}})}{{{F}_{0}}({{t}_{0}})-{{F}_{0}}(0)}\,\!</math>

where <math>{{t}_{i-1}}\,\!</math> is the time-to-failure previous to the <math>{{t}_{i}}\,\!</math> time-to-failure and <math>n\,\!</math> is the number of units associated with that time-to-failure (or units in the group).

In general, for right censored data:
:• The increment term for <math>n\,\!</math> right censored at time <math>={{t}_{0}},\,\!</math> with a time-to-failure of <math>{{t}_{i}}\,\!</math>, when <math>{{t}_{0}}\ge {{t}_{i}}\,\!</math> is zero.
:• When <math>{{t}_{0}}<{{t}_{i}},\,\!</math> the contribution is:

::<math>\frac{n}{{{F}_{0}}(\infty )-{{F}_{0}}({{t}_{0}})}\underset{MAX({{t}_{0}},{{t}_{i-1}})}{\overset{{{t}_{i}}}{\mathop \int }}\,{{f}_{0}}\left( t \right)dt\,\!</math>

:or:

::<math>n\frac{{{F}_{0}}({{t}_{i}})-{{F}_{0}}(MAX({{t}_{0}},{{t}_{i-1}}))}{{{F}_{0}}(\infty )-{{F}_{0}}({{t}_{0}})}\,\!</math>

where <math>{{t}_{i-1}}\,\!</math> is the time-to-failure previous to the <math>{{t}_{i}}\,\!</math> time-to-failure and <math>n\,\!</math> is the number of units associated with that time-to-failure (or units in the group).

'''Step 4'''

Sum up the increments (horizontally in rows), as in Table B.6.

{|border="1" align="center" style="border-collapse: collapse;" cellpadding="5" cellspacing="5"
|-
|colspan="8" style="text-align:center"|Table B.6- Increments Solved Numerically, Along with the Sum of Each Row.
|-
!Number of items
!Time of Failure
!2 Left Censored ''t''=30
!1 Left Censored ''t''=70
!1 Left Censored ''t''=100
!1 Right Censored ''t''=20
!1 Right Censored ''t''=60
!Sum of row(increment)
|-
|1||10||0.299065||0.062673||0.057673||0||0||0.419411
|-
|1||39.169||1.700935||0.542213||0.498959||0.440887||0||3.182994
|-
|2||40||0||0.015892||0.014625||0.018113||0||0.048630
|-
|2||42.831||0||0.052486||0.048299||0.059821||0||0.160606
|-
|1||50||0||0.118151||0.108726||0.134663||0||0.361540
|}

'''Step 5'''

Compute new mean order numbers (MON), as shown Table B.7, utilizing the increments obtained in Table B.6, by adding the ''number of items'' plus the ''previous MON'' plus the current increment.

{|border="1" align="center" style="border-collapse: collapse;" cellpadding="5" cellspacing="5"
|-
|colspan="4" style="text-align:center"|Table B.7- Mean Order Numbers (MON)
|-
!Number of items
!Time of Failure
!Sum of row(increment)
!Mean Order Number
|-
|1||10||0.419411||1.419411
|-
|1||39.169||3.182994||5.602405
|-
|2||40||0.048630||7.651035
|-
|2||42.837||0.160606||9.811641
|-
|1||50||0.361540||11.173181
|}

'''Step 6'''

Compute the median ranks based on these new MONs as shown in Table B.8.

{|border="1" align="center" style="border-collapse: collapse;" cellpadding="5" cellspacing="5"
|colspan="3" style="text-align:center"|Table B.8- Mean Order Numbers with Their Ranks for a Sample Size of 13 Units.
|-
!Time
!MON
!Ranks
|-
|10||1.419411||0.0825889
|-
|39.169||5.602405||0.3952894
|-
|40||7.651035||0.5487781
|-
|42.837||9.811641||0.7106217
|-
|50||11.173181||0.8124983
|}

'''Step 7'''

Compute new <math>\beta \,\!</math> and <math>\eta ,\,\!</math> using standard rank regression and based upon the data as shown in Table B.9.
<br>
{|border="1" align="center" style="border-collapse: collapse;" cellpadding="5" cellspacing="5"
!Time
!Ranks
|-
|10||0.0826889
|-
|39.169||0.3952894
|-
|40||0.5487781
|-
|42.837||0.7106217
|-
|50||0.8124983
|}

'''Step 8'''
Return and repeat the process from Step 1 until an acceptable convergence is reached on the parameters (i.e., the parameter values stabilize).

===Results===
The results of the first five iterations are shown in Table B.10.
Using Weibull++ with rank regression on X yields:
<br>
{|border="1" align="center" style="border-collapse: collapse;" cellpadding="5" cellspacing="5"
|colspan="3" style="text-align:center;"|Table B.10-The parameters after the first five iterations
|-align="center"
!''Iteration''
!<math>\beta\,\!</math>
!<math>\eta\,\!</math>
|-align="center"
|1||1.845638||42.576422
|-align="center"
|2||1.830621 ||42.039743
|-align="center"
|3||1.828010 ||41.830615
|-align="center"
|4||1.828030 ||41.749708
|-align="center"
|5||1.828383 ||41.717990
|}

::<math>{{\widehat{\beta }}_{RRX}}=1.82890,\text{ }{{\widehat{\eta }}_{RRX}}=41.69774\,\!</math>

The direct MLE solution yields:

::<math>{{\widehat{\beta }}_{MLE}}=2.10432,\text{ }{{\widehat{\eta }}_{MLE}}=42.31535\,\!</math>

Weibull Distribution Examples

2023-09-18T21:47:29Z

Lisa Hacker:

<noinclude>{{Banner Weibull Examples}}{{Navigation box}}
''These examples also appear in the [https://help.reliasoft.com/reference/life_data_analysis Life data analysis reference].''
</noinclude>
===Median Rank Plot Example===

In this example, we will determine the median rank value used for plotting the 6th failure from a sample size of 10. This example will use Weibull++'s Quick Statistical Reference (QSR) tool to show how the points in the plot of the following example are calculated.

First, open the Quick Statistical Reference tool and select the '''Inverse F-Distribution Values''' option.

In this example, n1 = 10, j = 6, m = 2(10 - 6 + 1) = 10, and n2 = 2 x 6 = 12.

Thus, from the F-distribution rank equation:

::<math>MR=\frac{1}{1+\left( \frac{10-6+1}{6} \right){{F}_{0.5;10;12}}}\,\!</math>

Use the QSR to calculate the value of F<sub>0.5;10;12</sub> = 0.9886, as shown next:

[[Image: F Inverse.png|center|550px]]

Consequently:

::<math>MR=\frac{1}{1+\left( \frac{5}{6} \right)\times 0.9886}=0.5483=54.83%\,\!</math>

Another method is to use the '''Median Ranks''' option directly, which yields MR(%) = 54.8305%, as shown next:
[[Image: MR.png|center|550px]]

===Complete Data Example===

Assume that 10 identical units (N = 10) are being reliability tested at the same application and operation stress levels. 6 of these units fail during this test after operating the following numbers of hours, <math>{T}_{j}\,\!</math>: 150, 105, 83, 123, 64 and 46. The test is stopped at the 6th failure. Find the parameters of the Weibull ''pdf'' that represents these data.

'''Solution'''

Create a new Weibull++ standard folio that is configured for grouped times-to-failure data with suspensions.

Enter the data in the appropriate columns. Note that there are 4 suspensions, as only 6 of the 10 units were tested to failure (the next figure shows the data as entered). Use the 3-parameter Weibull and MLE for the calculations.

[[Image: DataforExample_11.png.png|center|500px]]

Plot the data.

[[Image: Plot for Example 11.png|center|500px]]

Note that the original data points, on the curved line, were adjusted by subtracting 30.92 hours to yield a straight line as shown above.

===Suspension Data Example===

ACME company manufactures widgets, and it is currently engaged in reliability testing a new widget design. 19 units are being reliability tested, but due to the tremendous demand for widgets, units are removed from the test whenever the production cannot cover the demand. The test is terminated at the 67th day when the last widget is removed from the test. The following table contains the collected data.

{| {| border="1" align="center" style="border-collapse: collapse;" cellpadding="5" cellspacing="5"
|+ '''Widget Test Data'''
| align="center" style="background:#f0f0f0;"|'''Data Point Index'''
| align="center" style="background:#f0f0f0;"|'''State (F/S)'''
| align="center" style="background:#f0f0f0;"|'''Time to Failure'''
|-
| 1||F||2
|-
| 2||S||3
|-
| 3||F||5
|-
| 4||S||7
|-
| 5||F||11
|-
| 6||S||13
|-
| 7||S||17
|-
| 8||S||19
|-
| 9||F||23
|-
| 10||F||29
|-
| 11||S||31
|-
| 12||F||37
|-
| 13||S||41
|-
| 14||F||43
|-
| 15||S||47
|-
| 16||S||53
|-
| 17||F||59
|-
| 18||S||61
|-
| 19||S||67
|-
|
|}

'''Solution'''

In this example, we see that the number of failures is less than the number of suspensions. This is a very common situation, since reliability tests are often terminated before all units fail due to financial or time constraints. Furthermore, some suspensions will be recorded when a failure occurs that is not due to a legitimate failure mode, such as operator error. In cases such as this, a suspension is recorded, since the unit under test cannot be said to have had a legitimate failure.

Enter the data into a Weibull++ standard folio that is configured for times-to-failure data with suspensions. The folio will appear as shown next:

[[Image: Data Folio Example 13.png|center|550px]]

We will use the 2-parameter Weibull to solve this problem. The parameters using maximum likelihood are:

::<math>\begin{align}
& \hat{\beta }=1.145 \\
& \hat{\eta }=65.97 \\
\end{align}\,\!</math>

Using RRX:

::<math>\begin{align}
& \hat{\beta }=0.914\\
& \hat{\eta }=79.38 \\
\end{align}\,\!</math>

Using RRY:

::<math>\begin{align}
& \hat{\beta }=0.895\\
& \hat{\eta }=82.02 \\
\end{align}\,\!</math>

===Interval Data Example===

Suppose we have run an experiment with 8 units tested and the following is a table of their last inspection times and failure times:

{| border="1" align="center" style="border-collapse: collapse;" cellpadding="5" cellspacing="5"
| align="center" style="background:#f0f0f0;"|'''Data Point Index'''
| align="center" style="background:#f0f0f0;"|'''Last Inspection'''
| align="center" style="background:#f0f0f0;"|'''Failure Time'''
|-
| 1||30||32
|-
| 2||32||35
|-
| 3||35||37
|-
| 4||37||40
|-
| 5||42||42
|-
| 6||45||45
|-
| 7||50||50
|-
| 8||55||55
|-
|}

Analyze the data using several different parameter estimation techniques and compare the results.

'''Solution'''

Enter the data into a Weibull++ standard folio that is configured for interval data. The data is entered as follows:

[[Image: Data Folio.png|center|550px]]

The computed parameters using maximum likelihood are:

::<math>\begin{align}
& \hat{\beta }=5.76 \\
& \hat{\eta }=44.68 \\
\end{align}\,\!</math>

Using RRX or rank regression on X:

::<math>\begin{align}
& \hat{\beta }=5.70 \\
& \hat{\eta }=44.54 \\
\end{align}\,\!</math>

Using RRY or rank regression on Y:

::<math>\begin{align}
& \hat{\beta }=5.41 \\
& \hat{\eta }=44.76 \\
\end{align}\,\!</math>

The plot of the MLE solution with the two-sided 90% confidence bounds is:
[[Image: MLE Plot.png|center|550px]]

===Mixed Data Types Example===

From Dimitri Kececioglu, Reliability & Life Testing Handbook, Page 406. [[Appendix:_Life_Data_Analysis_References|[20]]].

Estimate the parameters for the 3-parameter Weibull, for a sample of 10 units that are all tested to failure. The recorded failure times are 200; 370; 500; 620; 730; 840; 950; 1,050; 1,160 and 1,400 hours.

'''Published Results:'''

Published results (using probability plotting):

::<math>{\widehat{\beta}} = 3.0\,\!</math>, <math>{\widehat{\eta}} = 1,220\,\!</math>, <math>{\widehat{\gamma}} = -300\,\!</math>

'''Computed Results in Weibull++'''

Weibull++ computed parameters for rank regression on X are:

::<math>{\widehat{\beta}} = 2.9013\,\!</math>, <math>{\widehat{\eta}} = 1195.5009\,\!</math>, <math>{\widehat{\gamma}} = -279.000\,\!</math>

The small difference between the published results and the ones obtained from Weibull++ are due to the difference in the estimation method. In the publication the parameters were estimated using probability plotting (i.e., the fitted line was "eye-balled"). In Weibull++, the parameters were estimated using non-linear regression (a more accurate, mathematically fitted line). Note that γ in this example is negative. This means that the unadjusted for γ line is concave up, as shown next.

[[Image:Weibull Distribution Example 19 Plot.png|center|450px]]

===Weibull Distribution RRX Example===

{{:Weibull_Distribution_RRX_Example}}

===Benchmark with Published Examples===

The following examples compare published results to computed results obtained with Weibull++.

{{Font|Complete Data RRY Example|11|Arial|bold|black}}

From Dimitri Kececioglu, Reliability & Life Testing Handbook, Page 418 [[Appendix:_Life_Data_Analysis_References|[20]]].

Sample of 10 units, all tested to failure. The failures were recorded at 16, 34, 53, 75, 93, 120, 150, 191, 240 and 339 hours.

'''Published Results'''

Published Results (using Rank Regression on Y):

::<math>\begin{align}
& \widehat{\beta }=1.20 \\
& \widehat{\eta} = 146.2 \\
& \hat{\rho }=0.998703\\
\end{align}\,\!</math>

'''Computed Results in Weibull++'''

This same data set can be entered into a Weibull++ standard data sheet. Use RRY for the estimation method.

Weibull++ computed parameters for RRY are:

::<math>\begin{align}
& \widehat{\beta }=1.1973 \\
& \widehat{\eta} = 146.2545 \\
& \hat{\rho }=0.9999\\
\end{align}\,\!</math>

The small difference between the published results and the ones obtained from Weibull++ is due to the difference in the median rank values between the two (in the publication, median ranks are obtained from tables to 3 decimal places, whereas in Weibull++ they are calculated and carried out up to the 15th decimal point).

You will also notice that in the examples that follow, a small difference may exist between the published results and the ones obtained from Weibull++. This can be attributed to the difference between the computer numerical precision employed by Weibull++ and the lower number of significant digits used by the original authors. In most of these publications, no information was given as to the numerical precision used.

{{Font|Suspension Data MLE Example|11|Arial|bold|black}}

From Wayne Nelson, Fan Example, Applied Life Data Analysis, page 317 [[Appendix:_Life_Data_Analysis_References|[30]]].

70 diesel engine fans accumulated 344,440 hours in service and 12 of them failed. A table of their life data is shown next (+ denotes non-failed units or suspensions, using Dr. Nelson's nomenclature). Evaluate the parameters with their two-sided 95% confidence bounds, using MLE for the 2-parameter Weibull distribution.

[[Image:example18table.png|center]]

'''Published Results:'''

Weibull parameters (2P-Weibull, MLE):

::<math>\begin{align}
& \widehat{\beta }=1.0584 \\
& \widehat{\eta} = 26,296 \\
\end{align}\,\!</math>

Published 95% FM confidence limits on the parameters:

::<math>\begin{align}
& \widehat{\beta }=\lbrace 0.6441, \text{ }1.7394\rbrace \\
& \widehat{\eta} = \lbrace 10,522, \text{ }65,532\rbrace \\
\end{align}\,\!</math>

Published variance/covariance matrix:

[[Image:example18formula3.png]]

Note that Nelson expresses the results as multiples of 1,000 (or = 26.297, etc.). The published results were adjusted by this factor to correlate with Weibull++ results.

'''Computed Results in Weibull++'''

This same data set can be entered into a Weibull++ standard folio, using 2-parameter Weibull and MLE to calculate the parameter estimates.

You can also enter the data as given in table without grouping them by opening a data sheet configured for suspension data. Then click the '''Group Data''' icon and chose '''Group exactly identical values'''.

[[Image:groupdataicon.png|center]]

[[Image:Weibull Distribution Example 18 Group Data.png|center|450px]]

The data will be automatically grouped and put into a new grouped data sheet.

Weibull++ computed parameters for maximum likelihood are:

::<math>\begin{align}
& \widehat{\beta }=1.0584 \\
& \widehat{\eta} = 26,297 \\
\end{align}\,\!</math>

Weibull++ computed 95% FM confidence limits on the parameters:

::<math>\begin{align}
& \widehat{\beta }=\lbrace 0.6441, \text{ }1.7394\rbrace \\
& \widehat{\eta} = \lbrace 10,522, \text{ }65,532\rbrace \\
\end{align}\,\!</math>

Weibull++ computed/variance covariance matrix:

[[Image:compexample18formula3.png]]

The two-sided 95% bounds on the parameters can be determined from the QCP. Calculate and then click '''Report''' to see the results.

[[Image: Weibull Distribution Example 18 QCP Parameter Bounds.png|center|550px]]

{{Font|Interval Data MLE Example|11|Arial|bold|black}}

From Wayne Nelson, Applied Life Data Analysis, Page 415 [[Appendix:_Life_Data_Analysis_References|[30]]]. 167 identical parts were inspected for cracks. The following is a table of their last inspection times and times-to-failure:

[[Image:example16table.png|center|800px]]

'''Published Results:'''

Published results (using MLE):

::<math>\begin{align}
& \widehat{\beta }=1.486 \\
& \widehat{\eta} = 71.687\\
\end{align}\,\!</math>

Published 95% FM confidence limits on the parameters:

::<math>\begin{align}
& \widehat{\beta }=\lbrace 1.224, \text{ }1.802\rbrace \\
& \widehat{\eta} = \lbrace 61.962, \text{ }82.938\rbrace \\
\end{align}\,\!</math>

Published variance/covariance matrix:

[[Image:example16formula3.png]]

'''Computed Results in Weibull++'''

This same data set can be entered into a Weibull++ standard folio that's configured for grouped times-to-failure data with suspensions and interval data.

Weibull++ computed parameters for maximum likelihood are:

::<math>\begin{align}
& \widehat{\beta }=1.485 \\
& \widehat{\eta} = 71.690\\
\end{align}\,\!</math>

Weibull++ computed 95% FM confidence limits on the parameters:

::<math>\begin{align}
& \widehat{\beta }=\lbrace 1.224, \text{ }1.802\rbrace \\
& \widehat{\eta} = \lbrace 61.961, \text{ }82.947\rbrace \\
\end{align}\,\!</math>

Weibull++ computed/variance covariance matrix:

[[Image:compexample16formula3.png]]

{{Font|Grouped Suspension MLE Example|11|Arial|bold|black}}

From Dallas R. Wingo, IEEE Transactions on Reliability Vol. R-22, No 2, June 1973, Pages 96-100.

Wingo uses the following times-to-failure: 37, 55, 64, 72, 74, 87, 88, 89, 91, 92, 94, 95, 97, 98, 100, 101, 102, 102, 105, 105, 107, 113, 117, 120, 120, 120, 122, 124, 126, 130, 135, 138, 182. In addition, the following suspensions are used: 4 at 70, 5 at 80, 4 at 99, 3 at 121 and 1 at 150.

'''Published Results (using MLE)'''

::<math>\begin{align}
& \widehat{\beta }=3.7596935\\
& \widehat{\eta} = 106.49758 \\
& \hat{\gamma }=14.451684\\
\end{align}\,\!</math>

'''Computed Results in Weibull++'''

::<math>\begin{align}
& \widehat{\beta }=3.7596935\\
& \widehat{\eta} = 106.49758 \\
& \hat{\gamma }=14.451684\\
\end{align}\,\!</math>

Note that you must select the '''Use True 3-P MLE'''option in the Weibull++ Application Setup to replicate these results.

{{Font|3-P Probability Plot Example|11|Arial|bold|black}}

Suppose we want to model a left censored, right censored, interval, and complete data set, consisting of 274 units under test of which 185 units fail. The following table contains the data.

{| {| border="1" align="center" style="border-collapse: collapse;" cellpadding="5" cellspacing="5"
|+ '''The Test Data'''
| align="center" style="background:#f0f0f0;"|'''Data Point Index'''
| align="center" style="background:#f0f0f0;"|'''Number in State'''
| align="center" style="background:#f0f0f0;"|'''Last Inspection'''
| align="center" style="background:#f0f0f0;"|'''State (S or F)'''
| align="center" style="background:#f0f0f0;"|'''State End Time'''
|-
| 1||2||5||F||5
|-
| 2||23||5||S||5
|-
| 3||28||0||F||7
|-
| 4||4||10||F||10
|-
| 5||7||15||F||15
|-
| 6||8||20||F||20
|-
| 7||29||20||S||20
|-
| 8||32||0||F||22
|-
| 9||6||25||F||25
|-
| 10||4||27||F||30
|-
| 11||8||30||F||35
|-
| 12||5||30||F||40
|-
| 13||9||27||F||45
|-
| 14||7||25||F||50
|-
| 15||5||20||F||55
|-
| 16||3||15||F||60
|-
| 17||6||10||F||65
|-
| 18||3||5||F||70
|-
| 19||37||100||S||100
|-
| 20||48||0||F||102
|-
|}

'''Solution'''

Since standard ranking methods for dealing with these different data types are inadequate, we will want to use the ReliaSoft ranking method. This option is the default in Weibull++ when dealing with interval data. The filled-out standard folio is shown next:

[[Image: Data Folio for Example 14.png|center|650px]]

The computed parameters using MLE are:

::<math>\hat{\beta }=0.748;\text{ }\hat{\eta }=44.38\,\!</math>

Using RRX:

::<math>\hat{\beta }=1.057;\text{ }\hat{\eta }=36.29\,\!</math>

Using RRY:

::<math>\hat{\beta }=0.998;\text{ }\hat{\eta }=37.16\,\!</math>

The plot with the two-sided 90% confidence bounds for the rank regression on X solution is:
[[Image: RRX Plot for Example 14.png|center|550px]]

Weibull Distribution Examples

2023-09-18T21:47:08Z

Lisa Hacker: Removed protection from "Weibull Distribution Examples"

<noinclude>{{Banner Weibull Examples}}{{Navigation box}}
''These examples also appear in the [[The_Weibull_Distribution#Weibull Distribution Examples|Life Data Analysis Reference book]].''
</noinclude>
===Median Rank Plot Example===

In this example, we will determine the median rank value used for plotting the 6th failure from a sample size of 10. This example will use Weibull++'s Quick Statistical Reference (QSR) tool to show how the points in the plot of the following example are calculated.

First, open the Quick Statistical Reference tool and select the '''Inverse F-Distribution Values''' option.

In this example, n1 = 10, j = 6, m = 2(10 - 6 + 1) = 10, and n2 = 2 x 6 = 12.

Thus, from the F-distribution rank equation:

::<math>MR=\frac{1}{1+\left( \frac{10-6+1}{6} \right){{F}_{0.5;10;12}}}\,\!</math>

Use the QSR to calculate the value of F<sub>0.5;10;12</sub> = 0.9886, as shown next:

[[Image: F Inverse.png|center|550px]]

Consequently:

::<math>MR=\frac{1}{1+\left( \frac{5}{6} \right)\times 0.9886}=0.5483=54.83%\,\!</math>

Another method is to use the '''Median Ranks''' option directly, which yields MR(%) = 54.8305%, as shown next:
[[Image: MR.png|center|550px]]

===Complete Data Example===

Assume that 10 identical units (N = 10) are being reliability tested at the same application and operation stress levels. 6 of these units fail during this test after operating the following numbers of hours, <math>{T}_{j}\,\!</math>: 150, 105, 83, 123, 64 and 46. The test is stopped at the 6th failure. Find the parameters of the Weibull ''pdf'' that represents these data.

'''Solution'''

Create a new Weibull++ standard folio that is configured for grouped times-to-failure data with suspensions.

Enter the data in the appropriate columns. Note that there are 4 suspensions, as only 6 of the 10 units were tested to failure (the next figure shows the data as entered). Use the 3-parameter Weibull and MLE for the calculations.

[[Image: DataforExample_11.png.png|center|500px]]

Plot the data.

[[Image: Plot for Example 11.png|center|500px]]

Note that the original data points, on the curved line, were adjusted by subtracting 30.92 hours to yield a straight line as shown above.

===Suspension Data Example===

ACME company manufactures widgets, and it is currently engaged in reliability testing a new widget design. 19 units are being reliability tested, but due to the tremendous demand for widgets, units are removed from the test whenever the production cannot cover the demand. The test is terminated at the 67th day when the last widget is removed from the test. The following table contains the collected data.

{| {| border="1" align="center" style="border-collapse: collapse;" cellpadding="5" cellspacing="5"
|+ '''Widget Test Data'''
| align="center" style="background:#f0f0f0;"|'''Data Point Index'''
| align="center" style="background:#f0f0f0;"|'''State (F/S)'''
| align="center" style="background:#f0f0f0;"|'''Time to Failure'''
|-
| 1||F||2
|-
| 2||S||3
|-
| 3||F||5
|-
| 4||S||7
|-
| 5||F||11
|-
| 6||S||13
|-
| 7||S||17
|-
| 8||S||19
|-
| 9||F||23
|-
| 10||F||29
|-
| 11||S||31
|-
| 12||F||37
|-
| 13||S||41
|-
| 14||F||43
|-
| 15||S||47
|-
| 16||S||53
|-
| 17||F||59
|-
| 18||S||61
|-
| 19||S||67
|-
|
|}

'''Solution'''

In this example, we see that the number of failures is less than the number of suspensions. This is a very common situation, since reliability tests are often terminated before all units fail due to financial or time constraints. Furthermore, some suspensions will be recorded when a failure occurs that is not due to a legitimate failure mode, such as operator error. In cases such as this, a suspension is recorded, since the unit under test cannot be said to have had a legitimate failure.

Enter the data into a Weibull++ standard folio that is configured for times-to-failure data with suspensions. The folio will appear as shown next:

[[Image: Data Folio Example 13.png|center|550px]]

We will use the 2-parameter Weibull to solve this problem. The parameters using maximum likelihood are:

::<math>\begin{align}
& \hat{\beta }=1.145 \\
& \hat{\eta }=65.97 \\
\end{align}\,\!</math>

Using RRX:

::<math>\begin{align}
& \hat{\beta }=0.914\\
& \hat{\eta }=79.38 \\
\end{align}\,\!</math>

Using RRY:

::<math>\begin{align}
& \hat{\beta }=0.895\\
& \hat{\eta }=82.02 \\
\end{align}\,\!</math>

===Interval Data Example===

Suppose we have run an experiment with 8 units tested and the following is a table of their last inspection times and failure times:

{| border="1" align="center" style="border-collapse: collapse;" cellpadding="5" cellspacing="5"
| align="center" style="background:#f0f0f0;"|'''Data Point Index'''
| align="center" style="background:#f0f0f0;"|'''Last Inspection'''
| align="center" style="background:#f0f0f0;"|'''Failure Time'''
|-
| 1||30||32
|-
| 2||32||35
|-
| 3||35||37
|-
| 4||37||40
|-
| 5||42||42
|-
| 6||45||45
|-
| 7||50||50
|-
| 8||55||55
|-
|}

Analyze the data using several different parameter estimation techniques and compare the results.

'''Solution'''

Enter the data into a Weibull++ standard folio that is configured for interval data. The data is entered as follows:

[[Image: Data Folio.png|center|550px]]

The computed parameters using maximum likelihood are:

::<math>\begin{align}
& \hat{\beta }=5.76 \\
& \hat{\eta }=44.68 \\
\end{align}\,\!</math>

Using RRX or rank regression on X:

::<math>\begin{align}
& \hat{\beta }=5.70 \\
& \hat{\eta }=44.54 \\
\end{align}\,\!</math>

Using RRY or rank regression on Y:

::<math>\begin{align}
& \hat{\beta }=5.41 \\
& \hat{\eta }=44.76 \\
\end{align}\,\!</math>

The plot of the MLE solution with the two-sided 90% confidence bounds is:
[[Image: MLE Plot.png|center|550px]]

===Mixed Data Types Example===

From Dimitri Kececioglu, Reliability & Life Testing Handbook, Page 406. [[Appendix:_Life_Data_Analysis_References|[20]]].

Estimate the parameters for the 3-parameter Weibull, for a sample of 10 units that are all tested to failure. The recorded failure times are 200; 370; 500; 620; 730; 840; 950; 1,050; 1,160 and 1,400 hours.

'''Published Results:'''

Published results (using probability plotting):

::<math>{\widehat{\beta}} = 3.0\,\!</math>, <math>{\widehat{\eta}} = 1,220\,\!</math>, <math>{\widehat{\gamma}} = -300\,\!</math>

'''Computed Results in Weibull++'''

Weibull++ computed parameters for rank regression on X are:

::<math>{\widehat{\beta}} = 2.9013\,\!</math>, <math>{\widehat{\eta}} = 1195.5009\,\!</math>, <math>{\widehat{\gamma}} = -279.000\,\!</math>

The small difference between the published results and the ones obtained from Weibull++ are due to the difference in the estimation method. In the publication the parameters were estimated using probability plotting (i.e., the fitted line was "eye-balled"). In Weibull++, the parameters were estimated using non-linear regression (a more accurate, mathematically fitted line). Note that γ in this example is negative. This means that the unadjusted for γ line is concave up, as shown next.

[[Image:Weibull Distribution Example 19 Plot.png|center|450px]]

===Weibull Distribution RRX Example===

{{:Weibull_Distribution_RRX_Example}}

===Benchmark with Published Examples===

The following examples compare published results to computed results obtained with Weibull++.

{{Font|Complete Data RRY Example|11|Arial|bold|black}}

From Dimitri Kececioglu, Reliability & Life Testing Handbook, Page 418 [[Appendix:_Life_Data_Analysis_References|[20]]].

Sample of 10 units, all tested to failure. The failures were recorded at 16, 34, 53, 75, 93, 120, 150, 191, 240 and 339 hours.

'''Published Results'''

Published Results (using Rank Regression on Y):

::<math>\begin{align}
& \widehat{\beta }=1.20 \\
& \widehat{\eta} = 146.2 \\
& \hat{\rho }=0.998703\\
\end{align}\,\!</math>

'''Computed Results in Weibull++'''

This same data set can be entered into a Weibull++ standard data sheet. Use RRY for the estimation method.

Weibull++ computed parameters for RRY are:

::<math>\begin{align}
& \widehat{\beta }=1.1973 \\
& \widehat{\eta} = 146.2545 \\
& \hat{\rho }=0.9999\\
\end{align}\,\!</math>

The small difference between the published results and the ones obtained from Weibull++ is due to the difference in the median rank values between the two (in the publication, median ranks are obtained from tables to 3 decimal places, whereas in Weibull++ they are calculated and carried out up to the 15th decimal point).

You will also notice that in the examples that follow, a small difference may exist between the published results and the ones obtained from Weibull++. This can be attributed to the difference between the computer numerical precision employed by Weibull++ and the lower number of significant digits used by the original authors. In most of these publications, no information was given as to the numerical precision used.

{{Font|Suspension Data MLE Example|11|Arial|bold|black}}

From Wayne Nelson, Fan Example, Applied Life Data Analysis, page 317 [[Appendix:_Life_Data_Analysis_References|[30]]].

70 diesel engine fans accumulated 344,440 hours in service and 12 of them failed. A table of their life data is shown next (+ denotes non-failed units or suspensions, using Dr. Nelson's nomenclature). Evaluate the parameters with their two-sided 95% confidence bounds, using MLE for the 2-parameter Weibull distribution.

[[Image:example18table.png|center]]

'''Published Results:'''

Weibull parameters (2P-Weibull, MLE):

::<math>\begin{align}
& \widehat{\beta }=1.0584 \\
& \widehat{\eta} = 26,296 \\
\end{align}\,\!</math>

Published 95% FM confidence limits on the parameters:

::<math>\begin{align}
& \widehat{\beta }=\lbrace 0.6441, \text{ }1.7394\rbrace \\
& \widehat{\eta} = \lbrace 10,522, \text{ }65,532\rbrace \\
\end{align}\,\!</math>

Published variance/covariance matrix:

[[Image:example18formula3.png]]

Note that Nelson expresses the results as multiples of 1,000 (or = 26.297, etc.). The published results were adjusted by this factor to correlate with Weibull++ results.

'''Computed Results in Weibull++'''

This same data set can be entered into a Weibull++ standard folio, using 2-parameter Weibull and MLE to calculate the parameter estimates.

You can also enter the data as given in table without grouping them by opening a data sheet configured for suspension data. Then click the '''Group Data''' icon and chose '''Group exactly identical values'''.

[[Image:groupdataicon.png|center]]

[[Image:Weibull Distribution Example 18 Group Data.png|center|450px]]

The data will be automatically grouped and put into a new grouped data sheet.

Weibull++ computed parameters for maximum likelihood are:

::<math>\begin{align}
& \widehat{\beta }=1.0584 \\
& \widehat{\eta} = 26,297 \\
\end{align}\,\!</math>

Weibull++ computed 95% FM confidence limits on the parameters:

::<math>\begin{align}
& \widehat{\beta }=\lbrace 0.6441, \text{ }1.7394\rbrace \\
& \widehat{\eta} = \lbrace 10,522, \text{ }65,532\rbrace \\
\end{align}\,\!</math>

Weibull++ computed/variance covariance matrix:

[[Image:compexample18formula3.png]]

The two-sided 95% bounds on the parameters can be determined from the QCP. Calculate and then click '''Report''' to see the results.

[[Image: Weibull Distribution Example 18 QCP Parameter Bounds.png|center|550px]]

{{Font|Interval Data MLE Example|11|Arial|bold|black}}

From Wayne Nelson, Applied Life Data Analysis, Page 415 [[Appendix:_Life_Data_Analysis_References|[30]]]. 167 identical parts were inspected for cracks. The following is a table of their last inspection times and times-to-failure:

[[Image:example16table.png|center|800px]]

'''Published Results:'''

Published results (using MLE):

::<math>\begin{align}
& \widehat{\beta }=1.486 \\
& \widehat{\eta} = 71.687\\
\end{align}\,\!</math>

Published 95% FM confidence limits on the parameters:

::<math>\begin{align}
& \widehat{\beta }=\lbrace 1.224, \text{ }1.802\rbrace \\
& \widehat{\eta} = \lbrace 61.962, \text{ }82.938\rbrace \\
\end{align}\,\!</math>

Published variance/covariance matrix:

[[Image:example16formula3.png]]

'''Computed Results in Weibull++'''

This same data set can be entered into a Weibull++ standard folio that's configured for grouped times-to-failure data with suspensions and interval data.

Weibull++ computed parameters for maximum likelihood are:

::<math>\begin{align}
& \widehat{\beta }=1.485 \\
& \widehat{\eta} = 71.690\\
\end{align}\,\!</math>

Weibull++ computed 95% FM confidence limits on the parameters:

::<math>\begin{align}
& \widehat{\beta }=\lbrace 1.224, \text{ }1.802\rbrace \\
& \widehat{\eta} = \lbrace 61.961, \text{ }82.947\rbrace \\
\end{align}\,\!</math>

Weibull++ computed/variance covariance matrix:

[[Image:compexample16formula3.png]]

{{Font|Grouped Suspension MLE Example|11|Arial|bold|black}}

From Dallas R. Wingo, IEEE Transactions on Reliability Vol. R-22, No 2, June 1973, Pages 96-100.

Wingo uses the following times-to-failure: 37, 55, 64, 72, 74, 87, 88, 89, 91, 92, 94, 95, 97, 98, 100, 101, 102, 102, 105, 105, 107, 113, 117, 120, 120, 120, 122, 124, 126, 130, 135, 138, 182. In addition, the following suspensions are used: 4 at 70, 5 at 80, 4 at 99, 3 at 121 and 1 at 150.

'''Published Results (using MLE)'''

::<math>\begin{align}
& \widehat{\beta }=3.7596935\\
& \widehat{\eta} = 106.49758 \\
& \hat{\gamma }=14.451684\\
\end{align}\,\!</math>

'''Computed Results in Weibull++'''

::<math>\begin{align}
& \widehat{\beta }=3.7596935\\
& \widehat{\eta} = 106.49758 \\
& \hat{\gamma }=14.451684\\
\end{align}\,\!</math>

Note that you must select the '''Use True 3-P MLE'''option in the Weibull++ Application Setup to replicate these results.

{{Font|3-P Probability Plot Example|11|Arial|bold|black}}

Suppose we want to model a left censored, right censored, interval, and complete data set, consisting of 274 units under test of which 185 units fail. The following table contains the data.

{| {| border="1" align="center" style="border-collapse: collapse;" cellpadding="5" cellspacing="5"
|+ '''The Test Data'''
| align="center" style="background:#f0f0f0;"|'''Data Point Index'''
| align="center" style="background:#f0f0f0;"|'''Number in State'''
| align="center" style="background:#f0f0f0;"|'''Last Inspection'''
| align="center" style="background:#f0f0f0;"|'''State (S or F)'''
| align="center" style="background:#f0f0f0;"|'''State End Time'''
|-
| 1||2||5||F||5
|-
| 2||23||5||S||5
|-
| 3||28||0||F||7
|-
| 4||4||10||F||10
|-
| 5||7||15||F||15
|-
| 6||8||20||F||20
|-
| 7||29||20||S||20
|-
| 8||32||0||F||22
|-
| 9||6||25||F||25
|-
| 10||4||27||F||30
|-
| 11||8||30||F||35
|-
| 12||5||30||F||40
|-
| 13||9||27||F||45
|-
| 14||7||25||F||50
|-
| 15||5||20||F||55
|-
| 16||3||15||F||60
|-
| 17||6||10||F||65
|-
| 18||3||5||F||70
|-
| 19||37||100||S||100
|-
| 20||48||0||F||102
|-
|}

'''Solution'''

Since standard ranking methods for dealing with these different data types are inadequate, we will want to use the ReliaSoft ranking method. This option is the default in Weibull++ when dealing with interval data. The filled-out standard folio is shown next:

[[Image: Data Folio for Example 14.png|center|650px]]

The computed parameters using MLE are:

::<math>\hat{\beta }=0.748;\text{ }\hat{\eta }=44.38\,\!</math>

Using RRX:

::<math>\hat{\beta }=1.057;\text{ }\hat{\eta }=36.29\,\!</math>

Using RRY:

::<math>\hat{\beta }=0.998;\text{ }\hat{\eta }=37.16\,\!</math>

The plot with the two-sided 90% confidence bounds for the rank regression on X solution is:
[[Image: RRX Plot for Example 14.png|center|550px]]

Lognormal Distribution Examples

2023-09-18T21:46:36Z

Lisa Hacker:

<noinclude>{{Banner Weibull Examples}}{{Navigation box}}
''These examples also appear in the [https://help.reliasoft.com/reference/life_data_analysis Life data analysis reference].''
</noinclude>
===Complete Data Example===

Determine the lognormal parameter estimates for the data given in the following table.
{|border="1" align="center" style="border-collapse: collapse;" cellpadding="5" cellspacing="5"
|-
|colspan="3" style="text-align:center"| '''Non-Grouped Times-to-Failure Data'''
|-
!Data point index
!State F or S
!State End Time
|-
|1 ||F||2
|-
|2 ||F||5
|-
|3 ||F||11
|-
|4 ||F||23
|-
|5 ||F||29
|-
|6 ||F||37
|-
|7||F||43
|-
|8||F||59
|}

'''Solution'''

Using Weibull++, the computed parameters for maximum likelihood are:

::<math>\begin{align}
& {{{\hat{\mu }}}^{\prime }}= & 2.83 \\
& {\hat{\sigma '}}= & 1.10
\end{align}\,\!</math>

For rank regression on <math>X\,\!</math>

::<math>\begin{align}
& {{{\hat{\mu }}}^{\prime }}= & 2.83 \\
& {{{\hat{\sigma' }}}}= & 1.24
\end{align}\,\!</math>

For rank regression on <math>Y:\,\!</math>

::<math>\begin{align}
& {{{\hat{\mu }}}^{\prime }}= & 2.83 \\
& {{{\hat{\sigma' }}}}= & 1.36
\end{align}\,\!</math>

===Complete Data RRX Example===

From Kececioglu [[Appendix:_Life_Data_Analysis_References|[20, p. 347]]]. 15 identical units were tested to failure and following is a table of their failure times:

<center>'''Times-to-Failure Data'''</center>

<center><math>\begin{matrix}
\text{Data Point Index} & \text{Failure Times (Hr)} \\
\text{1} & \text{62}\text{.5} \\
\text{2} & \text{91}\text{.9} \\
\text{3} & \text{100}\text{.3} \\
\text{4} & \text{117}\text{.4} \\
\text{5} & \text{141}\text{.1} \\
\text{6} & \text{146}\text{.8} \\
\text{7} & \text{172}\text{.7} \\
\text{8} & \text{192}\text{.5} \\
\text{9} & \text{201}\text{.6} \\
\text{10} & \text{235}\text{.8} \\
\text{11} & \text{249}\text{.2} \\
\text{12} & \text{297}\text{.5} \\
\text{13} & \text{318}\text{.3} \\
\text{14} & \text{410}\text{.6} \\
\text{15} & \text{550}\text{.5} \\
\end{matrix}\,\!</math></center>

'''Solution'''

Published results (using probability plotting):

::<math>\begin{matrix}
{{\widehat{\mu }}^{\prime }}=5.22575 \\
{{\widehat{\sigma' }}}=0.62048. \\
\end{matrix}\,\!</math>

Weibull++ computed parameters for rank regression on X are:

::<math>\begin{matrix}
{{\widehat{\mu }}^{\prime }}=5.2303 \\
{{\widehat{\sigma'}}}=0.6283. \\
\end{matrix}\,\!</math>

The small differences are due to the precision errors when fitting a line manually, whereas in Weibull++ the line was fitted mathematically.

===Complete Data Unbiased MLE Example===

From Kececioglu [[Appendix:_Life_Data_Analysis_References|[19, p. 406]]]. 9 identical units are tested continuously to failure and failure times were recorded at 30.4, 36.7, 53.3, 58.5, 74.0, 99.3, 114.3, 140.1 and 257.9 hours.

'''Solution'''

The results published were obtained by using the unbiased model.
Published Results (using MLE):

::<math>\begin{matrix}
{{\widehat{\mu }}^{\prime }}=4.3553 \\
{{\widehat{\sigma' }}}=0.67677 \\
\end{matrix}\,\!</math>

This same data set can be entered into Weibull++ by creating a data sheet capable of handling non-grouped time-to-failure data. Since the results shown above are unbiased, the Use Unbiased Std on Normal Data option in the User Setup must be selected in order to duplicate these results.
Weibull++ computed parameters for maximum likelihood are:

::<math>\begin{matrix}
{{\widehat{\mu }}^{\prime }}=4.3553 \\
{{\widehat{\sigma' }}}=0.6768 \\
\end{matrix}\,\!</math>

===Suspension Data Example===

From Nelson [[Appendix:_Life_Data_Analysis_References|[30, p. 324]]]. 96 locomotive controls were tested, 37 failed and 59 were suspended after running for 135,000 miles. The table below shows the failure and suspension times.

{|border="1" align="center" style="border-collapse: collapse;" cellpadding="5" cellspacing="5"
|-
|colspan="4" style="text-align:center"|'''Nelson's Locomotive Data'''
|-
!
!Number in State
!F or S
!Time
|-
|1||1||F||22.5
|-
|2||1||F||37.5
|-
|3||1||F||46
|-
|4||1||F||48.5
|-
|5||1||F||51.5
|-
|6||1||F||53
|-
|7||1||F||54.5
|-
|8||1||F||57.5
|-
|9||1||F||66.5
|-
|10||1||F||68
|-
|11||1||F||69.5
|-
|12||1||F||76.5
|-
|13||1||F||77
|-
|14||1||F||78.5
|-
|15||1||F||80
|-
|16||1||F||81.5
|-
|17||1||F||82
|-
|18||1||F||83
|-
|19||1||F||84
|-
|20||1||F||91.5
|-
|21||1||F||93.5
|-
|22||1||F||102.5
|-
|23||1||F||107
|-
|24||1||F||108.5
|-
|25||1||F||112.5
|-
|26||1||F||113.5
|-
|27||1||F||116
|-
|28||1||F||117
|-
|29||1||F||118.5
|-
|30||1||F||119
|-
|31||1||F||120
|-
|32||1||F||122.5
|-
|33||1||F||123
|-
|34||1||F||127.5
|-
|35||1||F||131
|-
|36||1||F||132.5
|-
|37||1||F||134
|-
|38||59||S||135
|}

'''Solution'''

The distribution used in the publication was the base-10 lognormal.
Published results (using MLE):

::<math>\begin{matrix}
{{\widehat{\mu }}^{\prime }}=2.2223 \\
{{\widehat{\sigma' }}}=0.3064 \\
\end{matrix}\,\!</math>

Published 95% confidence limits on the parameters:

::<math>\begin{matrix}
{{\widehat{\mu }}^{\prime }}=\left\{ 2.1336,2.3109 \right\} \\
{{\widehat{\sigma'}}}=\left\{ 0.2365,0.3970 \right\} \\
\end{matrix}\,\!</math>

Published variance/covariance matrix:

::<math>\left[ \begin{matrix}
\widehat{Var}\left( {{{\hat{\mu }}}^{\prime }} \right)=0.0020 & {} & \widehat{Cov}({{{\hat{\mu }}}^{\prime }},{{{\hat{\sigma' }}}})=0.001 \\
{} & {} & {} \\
\widehat{Cov}({{{\hat{\mu }}}^{\prime }},{{{\hat{\sigma' }}}})=0.001 & {} & \widehat{Var}\left( {{{\hat{\sigma '}}}} \right)=0.0016 \\
\end{matrix} \right]\,\!</math>

To replicate the published results (since Weibull++ uses a lognormal to the base <math>e\,\!</math> ), take the base-10 logarithm of the data and estimate the parameters using the normal distribution and MLE.

*Weibull++ computed parameters for maximum likelihood are:

::<math>\begin{matrix}
{{\widehat{\mu }}^{\prime }}=2.2223 \\
{{\widehat{\sigma' }}}=0.3064 \\
\end{matrix}\,\!</math>

*Weibull++ computed 95% confidence limits on the parameters:

::<math>\begin{matrix}
{{\widehat{\mu }}^{\prime }}=\left\{ 2.1364,2.3081 \right\} \\
{{\widehat{\sigma'}}}=\left\{ 0.2395,0.3920 \right\} \\
\end{matrix}\,\!</math>

*Weibull++ computed/variance covariance matrix:

::<math>\left[ \begin{matrix}
\widehat{Var}\left( {{{\hat{\mu }}}^{\prime }} \right)=0.0019 & {} & \widehat{Cov}({{{\hat{\mu }}}^{\prime }},{{{\hat{\sigma' }}}})=0.0009 \\
{} & {} & {} \\
\widehat{Cov}({\mu }',{{{\hat{\sigma' }}}})=0.0009 & {} & \widehat{Var}\left( {{{\hat{\sigma' }}}} \right)=0.0015 \\
\end{matrix} \right]\,\!</math>

===Interval Data Example===

Determine the lognormal parameter estimates for the data given in the table below.
{|border="1" align="center" style="border-collapse: collapse;" cellpadding="5" cellspacing="5"
|-
|colspan="3" style="text-align:center"| '''Non-Grouped Data Times-to-Failure with Intervals'''
|-
!Data point index
!Last Inspected
!State End Time
|-
|1 ||30||32
|-
|2 ||32||35
|-
|3 ||35||37
|-
|4 ||37||40
|-
|5 ||42||42
|-
|6 ||45||45
|-
|7||50||50
|-
|8||55||55
|}

'''Solution'''

This is a sequence of interval times-to-failure where the intervals vary substantially in length. Using Weibull++, the computed parameters for maximum likelihood are calculated to be:

::<math>\begin{align}
& {{{\hat{\mu }}}^{\prime }}= & 3.64 \\
& {{{\hat{\sigma' }}}}= & 0.18
\end{align}\,\!</math>

For rank regression on <math>X\ \,\!</math>:

::<math>\begin{align}
& {{{\hat{\mu }}}^{\prime }}= & 3.64 \\
& {{{\hat{\sigma' }}}}= & 0.17
\end{align}\,\!</math>

For rank regression on <math>Y\ \,\!</math>:

::<math>\begin{align}
& {{{\hat{\mu }}}^{\prime }}= & 3.64 \\
& {{{\hat{\sigma' }}}}= & 0.21
\end{align}\,\!</math>

Exponential Distribution for Grouped Data Example

2023-09-18T21:46:26Z

Lisa Hacker:

<noinclude>
{{Banner Weibull Examples}}
''This example appears in the [https://help.reliasoft.com/reference/life_data_analysis Life data analysis reference]''.
</noinclude>

20 units were reliability tested with the following results:

{| border="1" align="center" style="border-collapse: collapse;" cellpadding="5" cellspacing="5"
|-
|colspan="2" style="text-align:center"| '''Table - Life Test Data'''
|-
|-align="center"
!Number of Units in Group
!Time-to-Failure
|- align="center"
|7 ||100
|- align="center"
|5 ||200
|- align="center"
|3 ||300
|- align="center"
|2 ||400
|- align="center"
|1 ||500
|- align="center"
|2 ||600
|}

1. Assuming a 2-parameter exponential distribution, estimate the parameters by hand using the MLE analysis method.

2. Repeat the above using Weibull++. (Enter the data as grouped data to duplicate the results.)

3. Show the Probability plot for the analysis results.

4. Show the Reliability vs. Time plot for the results.

5. Show the ''pdf'' plot for the results.

6. Show the Failure Rate vs. Time plot for the results.

7. Estimate the parameters using the rank regression on Y (RRY) analysis method (and using grouped ranks).

'''Solution'''

1. For the 2-parameter exponential distribution and for <math>\hat{\gamma }=100\,\!</math> hours (first failure), the partial of the log-likelihood function, <math>\lambda\,\!</math>, becomes:

::<math> \begin{align}

\frac{\partial \Lambda }{\partial \lambda }= &\underset{i=1}{\overset{6}{\mathop \sum }}\,{N_i} \left[ \frac{1}{\lambda }-\left( {{T}_{i}}-100 \right) \right]=0\\

\Rightarrow & 7[\frac{1}{\lambda }-(100-100)]+5[\frac{1}{\lambda}-(200-100)]
+ \ldots +2[\frac{1}{\lambda}-(600-100)]\\
= & 0\\
\Rightarrow & \hat{\lambda}=\frac{20}{3100}=0.0065 \text{fr/hr}

\end{align}
\,\!</math>

2. Enter the data in a Weibull++ standard folio and calculate it as shown next.
<br>
<br>
[[Image:Exponential Distribution Example 8 Data.png|center|750px|]]

3. On the Plot page of the folio, the exponential Probability plot will appear as shown next.

[[Image:Exponential Distribution Example 8 Plot.png|center|650px|]]

4. View the Reliability vs. Time plot.

[[Image:Exponential Distribution Example 8 Rel Plot.png|center|650px|]]

5. View the ''pdf'' plot.

[[Image:Exponential Distribution Example 8 Pdf Plot.png|center|650px|]]

6. View the Failure Rate vs. Time plot.

[[Image:Exponential Distribution Example 8 Failure Rate Plot.png|center|650px|]]

Note that, as described at the beginning of this chapter, the failure rate for the exponential distribution is constant. Also note that the Failure Rate vs. Time plot does show values for times before the location parameter, <math>\gamma \,\!</math>, at 100 hours.

7. In the case of grouped data, one must be cautious when estimating the parameters using a rank regression method. This is because the median rank values are determined from the total number of failures observed by time <math>{{T}_{i}}\,\!</math> where <math>i\,\!</math> indicates the group number. In this example, the total number of groups is <math>N=6\,\!</math> and the total number of units is <math>{{N}_{T}}=20\,\!</math>. Thus, the median rank values will be estimated for 20 units and for the total failed units (<math>{{N}_{{{F}_{i}}}}\,\!</math>) up to the <math>{{i}^{th}}\,\!</math> group, for the <math>{{i}^{th}}\,\!</math> rank value. The median ranks values can be found from rank tables or they can be estimated using ReliaSoft's Quick Statistical Reference tool.

For example, the median rank value of the fourth group will be the <math>{{17}^{th}}\,\!</math> rank out of a sample size of twenty units (or 81.945%).

The following table is then constructed.

<center><math>\begin{matrix}
N & {{N}_{F}} & {{N}_{{{F}_{i}}}} & {{T}_{i}} & F({{T}_{i}}) & {{y}_{i}} & T_{i}^{2} & y_{i}^{2} & {{T}_{i}}{{y}_{i}} \\
\text{1} & \text{7} & \text{7} & \text{100} & \text{0}\text{.32795} & \text{-0}\text{.3974} & \text{10000} & \text{0}\text{.1579} & \text{-39}\text{.7426} \\
\text{2} & \text{5} & \text{12} & \text{200} & \text{0}\text{.57374} & \text{-0}\text{.8527} & \text{40000} & \text{0}\text{.7271} & \text{-170}\text{.5402} \\
\text{3} & \text{3} & \text{15} & \text{300} & \text{0}\text{.72120} & \text{-1}\text{.2772} & \text{90000} & \text{1}\text{.6313} & \text{-383}\text{.1728} \\
\text{4} & \text{2} & \text{17} & \text{400} & \text{0}\text{.81945} & \text{-1}\text{.7117} & \text{160000} & \text{2}\text{.9301} & \text{-684}\text{.6990} \\
\text{5} & \text{1} & \text{18} & \text{500} & \text{0}\text{.86853} & \text{-2}\text{.0289} & \text{250000} & \text{4}\text{.1166} & \text{-1014}\text{.4731} \\
\text{6} & \text{2} & \text{20} & \text{600} & \text{0}\text{.96594} & \text{-3}\text{.3795} & \text{360000} & \text{11}\text{.4211} & \text{-2027}\text{.7085} \\
\sum_{}^{} & {} & {} & \text{2100} & {} & \text{-9}\text{.6476} & \text{910000} & \text{20}\text{.9842} & \text{-4320}\text{.3362} \\
\end{matrix}\,\!</math></center>

Given the values in the table above, calculate <math>\hat{a}\,\!</math> and <math>\hat{b}\,\!</math>:

::<math>\begin{align}
& \hat{b}= & \frac{\underset{i=1}{\overset{6}{\mathop{\sum }}}\,{{t}_{i}}{{y}_{i}}-(\underset{i=1}{\overset{6}{\mathop{\sum }}}\,{{t}_{i}})(\underset{i=1}{\overset{6}{\mathop{\sum }}}\,{{y}_{i}})/6}{\underset{i=1}{\overset{6}{\mathop{\sum }}}\,t_{i}^{2}-{{(\underset{i=1}{\overset{6}{\mathop{\sum }}}\,{{t}_{i}})}^{2}}/6} \\
& & \\
& \hat{b}= & \frac{-4320.3362-(2100)(-9.6476)/6}{910,000-{{(2100)}^{2}}/6}
\end{align}\,\!</math>

or:

::<math>\hat{b}=-0.005392\,\!</math>

and:

::<math>\hat{a}=\overline{y}-\hat{b}\overline{t}=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}}}{N}-\hat{b}\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{t}_{i}}}{N}\,\!</math>

or:

::<math>\hat{a}=\frac{-9.6476}{6}-(-0.005392)\frac{2100}{6}=0.2793\,\!</math>

Therefore:

::<math>\hat{\lambda }=-\hat{b}=-(-0.005392)=0.05392\text{ failures/hour}\,\!</math>

and:

::<math>\hat{\gamma }=\frac{\hat{a}}{\hat{\lambda }}=\frac{0.2793}{0.005392}\,\!</math>

or:

::<math>\hat{\gamma }\simeq 51.8\text{ hours}\,\!</math>

Then:

::<math>f(T)=(0.005392){{e}^{-0.005392(T-51.8)}}\,\!</math>

Using Weibull++, the estimated parameters are:

::<math>\begin{align}
\hat{\lambda }= & 0.0054\text{ failures/hour} \\
\hat{\gamma }= & 51.82\text{ hours}
\end{align}\,\!</math>

The small difference in the values from Weibull++ is due to rounding. In the application, the calculations and the rank values are carried out up to the <math>15^{th}\,\!</math> decimal point.