Simulation with RGA Models

=Additional Tools=

Simulation
When analyzing developmental systems for reliability growth and conducting data analysis of fielded repairable systems, it is often useful to experiment with various what if scenarios or put together hypothetical analyses before having available test data in order to find the best way to analyze data when they become available. With that in mind, RGA 7 offers applications based on Monte Carlo simulation that can be used in order to: a) better understand reliability growth concepts. b) experiment with the impact of sample size, test time and growth parameters on analysis results. c) construct simulation-based confidence intervals. d) better understand concepts behind confidence intervals. e) design reliability demonstration tests. There are two applications of the Monte Carlo simulation in RGA 7. One is called Generate Monte Carlo Data and the other is called SimuMatic.

Generate Monte Carlo Data
Monte Carlo simulation is a computational algorithm in which we randomly generate input variables that follow a specified probability distribution. In the case of reliability growth and repairable system data analysis, we are interested in generating failure times for systems that we assume have specific characteristics. In our applications we want the inter-arrival times of the failures to follow a non-homogeneous poisson process with a Weibull failure intensity, as specified in the Crow-AMSAA (NHPP) model. The first time to failure, $${{t}_{1}},$$  is assumed to follow a Weibull distribution. It is obtained by solving for $${{t}_{1}}$$ :


 * $$R(t_1)=e^{(-\frac{t_1}{\eta})^\beta}= Uniform (0,1)$$

where:


 * $$\eta ={{\left( \frac{1}{\lambda } \right)}^{\tfrac{1}{\beta }}}$$

Solving Eqn. (R(t1) =uniform) for $${{t}_{1}}$$  yields:

The failure times are then obtained based on the conditional unreliability equation that describes the non-homogeneous poisson process (NHPP):

and then solving for $${{t}_{i}}$$  yields:

To access the data generation utility, either choose Tools > Generate Monte Carlo Data or click the Generate Monte Carlo Data icon on the Tools toolbar. There are different data types that can be generated with the Monte Carlo utility. For all of them, the basic parameters that are always specified are the beta $$(\beta )$$  and lambda  $$(\lambda )$$  parameters of the Crow-AMSAA (NHPP) model. That does not mean that the generated data can be analyzed only with the Crow-AMSAA (NHPP) model. Depending on the data type, the Duane, Crow Extended and Power Law models can also be used. They share the same growth patterns, which are based on the $$\beta $$  and  $$\lambda $$  parameters. In the case of the Duane model, $$\beta =1-\alpha $$, where  $$\alpha $$  is the growth parameter for the Duane model. Below we present the available data types that can generated with the Monte Carlo utility.

Failure Times
The data set is generated assuming a single system. There is a choice between a time terminated test, where the termination time needs to be specified, or a failure terminated test, where the number of failures needs to be specified. The generated failure times data can then be analyzed using the Duane, Crow-AMSAA (NHPP) or the Crow Extended model if classifications and modes are entered for the failures.

Grouped Failure Times
The data is generated assuming a single system. There is a choice between a time terminated test, where the termination time needs to be specified, or a failure terminated test, where the number of failures needs to be specified. In addition, constant or user defined intervals need to be specified for the grouping of the data. The generated grouped data can then be analyzed using the Duane, Crow-AMSAA (NHPP) or the Crow Extended model if classifications and modes are entered for the failures.

Multiple Systems - Concurrent
In this case, the number of systems needs to be specified. There is a choice between a time terminated test, where the termination time needs to be specified, or a failure terminated test, where the number of failures needs to be specified. The generated folio contains failure times for each of the systems. The data can then be analyzed using the Duane, Crow-AMSAA (NHPP) or the Crow Extended model if classifications and modes are entered for the failures.

Repairable Systems
In this case, the number of systems needs to be specified. There is a choice between a time terminated test, where the termination time needs to be specified, or a failure terminated test, where the number of failures needs to be specified. The generated folio contains failure times for each of the systems. The data can then be analyzed using the Power Law model or the Crow Extended model if classifications and modes are entered for the failures. Figure Montearlo shows the Monte Carlo utility and all the necessary user inputs. The seed determines the starting point from which the random numbers will be generated. The use of a seed forces the software to use the same sequence of random numbers, resulting in repeatability. In other words, the same failure times can be generated if the same seed, data type, parameters and number of points/systems are used. If no seed is provided, the computer's clock is used to initialize the random number generator and a different set of failure times will be generated at each new request.

Example 1
A reliability engineer wants to experiment with different testing scenarios as the reliability growth test of the company's new product is being prepared. From the reliability growth test data of a similar product that was developed previously, the beta and lambda parameters are $$\beta =0.5$$  and  $$\lambda =0.75.$$  Three systems are to be used to generate a representative data set of expected times-to-failure for the upcoming test. The purpose is to explore different test durations in order to demonstrate an MTBF of 200 hours.

Solution
In the Monte Carlo window, the parameters are set to $$\beta =0.5$$  and  $$\lambda =0.75.$$  Since we have three systems, the Multiple Systems - Concurrent option is selected and the number of systems is set to 3. Initially, the test is set to be time terminated with 2000 operating hours per system, for a total of 6000 operating hours. Figure concurrent monte carlo shows the Monte Carlo generation window for this example. $$$$

Figure folio monte carlo shows the generated failure times data. In this folio, the Advanced Systems View is used, so the data sheet shows the times-to-failure for system 2.

The data can then be analyzed just like a regular folio in RGA 7. In this case, we are interested in analyzing the data with the Crow-AMSAA (NHPP) model to calculate the demonstrated MTBF at the end of the test. In Figure folio monte carlo, in the results area of the folio, it can be seen that the demonstrated MTBF at the end of the test is 189.83 hours. Since that does not meet the requirement of demonstrating an MTBF of 200 hours, we can either generate a new Monte Carlo data set with different time termination settings, or access the Quick Calculation Pad in this folio and find the time for which the demonstrated (instantaneous) MTBF becomes 200 hours, as shown in Figure QCP monte carlo. From the figure it can be seen that, based on this specific data set, 6651.38 total operating hours are needed to show a demonstrated MTBF of 200 hours.

Note that since the Monte Carlo routine generates random input variables that follow the NHPP based on the specific $$\beta $$  and  $$\lambda $$, if the same seed is not used the failure times will be different the next time you run the Monte Carlo routine. Also note that since input variables are pulled from an NHPP with the expected values of $$\beta $$  and  $$\lambda ,$$  it should not be expected that the calculated parameters of the generated data set will match exactly the input parameters that were specified. In this example, the input parameters were set as $$\beta =0.5$$  and  $$\lambda =0.75$$  and the data set based on the Monte Carlo generated failure times yielded calculated Crow-AMSAA (NHPP) parameters of  $$\beta =0.4939$$  and  $$\lambda =0.8716.$$  The next time a data set is generated with a random seed, the calculated parameters will be slightly different, since we are essentially pulling input variables from a predefined distribution. The more simulations that are run, the more the calculated parameters will converge with the expected parameters. In RGA 7, the total number of generated failures with the Monte Carlo utility has to be less than 64,000.

SimuMatic
Reliability growth analysis using simulation can be a valuable tool for reliability practitioners. With this approach, reliability growth analyses are performed a large number of times on data sets that have been created using Monte Carlo simulation. RGA 7's SimuMatic utility generates calculated values of beta and lambda parameters, based on user specified input parameters of beta and lambda. SimuMatic essentially performs a user defined number of Monte Carlo simulations based on user defined required test time or failure termination settings, and then recalculates the beta and lambda parameters for each of the generated data sets. The number of times that the Monte Carlo data sets are generated and the parameters are re-calculated is also user defined. The final output presents the calculated values of beta and lambda and allows for various types of analysis. To access the SimuMatic utility, either choose Project > Add Other Tools > Add SimuMatic or click the SimuMatic icon on the Project toolbar. For all of the data sets, the basic parameters that are always specified are the beta $$(\beta )$$  and lambda  $$(\lambda )$$  parameters of the Crow-AMSAA (NHPP) model or the Power Law model. Failure Times The data set is generated assuming a single system. There is a choice between a time terminated test, where the termination time needs to be specified, or a failure terminated test, where the number of failures needs to be specified. SimuMatic will return the calculated values of $$\beta $$  and  $$\lambda $$  for a specified number of data sets. Grouped Failure Times The data set is generated assuming a single system. There is a choice between a time terminated test, where the termination time needs to be specified, or a failure terminated test, where the number of failures needs to be specified. In addition, constant or user defined intervals need to be specified for the grouping of the data. SimuMatic will return the calculated values of $$\beta $$  and  $$\lambda $$  for a specified number of data sets. Multiple Systems - Concurrent In this case, the number of systems needs to be specified. There is a choice between a time terminated test, where the termination time needs to be specified, or a failure terminated test, where the number of failures needs to be specified. SimuMatic will return the calculated values of $$\beta $$  and  $$\lambda $$  for a specified number of data sets. Repairable Systems In this case, the number of systems needs to be specified. There is a choice between a time terminated test, where the termination time needs to be specified, or a failure terminated test, where the number of failures needs to be specified. SimuMatic will return the calculated values of $$\beta $$  and  $$\lambda $$  for a specified number of data sets. Figure simumatic input shows the Main tab of the SimuMatic window where all the necessary user inputs for a multiple systems - concurrent data set have been entered. Figure simumatic output shows the generated results based on the input of Figure simumatic input. The sorted tab is shown, which allows us to extract conclusions about simulation generated confidence bounds, since the lambda and beta parameters and any other additional output are sorted by percentage. The Analysis tab allows the user to specify the confidence level for simulation generated confidence bounds as shown in Figure confidence bounds from simulation, where simulation confidence bounds are drawn for the cumulative number of failures based on the input parameters specified in Figure simumatic input. Extra results based on the calculated lambda and beta parameters can be generated in the Results tab, for example the instantaneous MTBF given a specific test time. This would be calculated as an additional result in addition to the lambda and beta parameters.

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Example 2
A manufacturer wants to design a reliability growth test for a redesigned product, in order to achieve an MTBF of 1,000 hours. Simulation is chosen to estimate the 1-sided 90% confidence bound on the required time to achieve the goal MTBF of 1,000 hours and the 1-sided 90% lower confidence bound on the MTBF at the end of the test time. The total test time is expected to be 15,000 hours. Based on historical data for the previous version, the expected beta and lambda parameters of the test are 0.5 and 0.3, respectively. Do the following: 1)	Generate 1,000 data sets using SimuMatic along with the required output. 2)	Plot the instantaneous MTBF vs. time with the 90% confidence bounds. 3)	Estimate the 1-sided 90% lower confidence bound on time for an MTBF of 1,000 hours. 4)	Estimate the 1-sided 90% lower confidence bound on the instantaneous MTBF at the end of the test.

Solution
1)	Figure Simumaticindow shows the SimuMatic window with all the appropriate inputs for creating the data sets.

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Figure Simumaticnalysis and Figure Simumaticesults show the settings in the Analysis and Results tab of the SimuMatic window in order to obtain the desired outputs.

$$$$ Figure Simumaticata displays the results of the simulation. The columns labeled Beta and Lambda contain the different parameters obtained by calculating each data set generated via simulation for the 1,000 data sets. The IMTBF(15,000) column contains the instantaneous MTBF at 15,000 hours (the end of test time) given the parameters obtained by calculating each data set generated via simulation. The Target DMTBF column contains the time required for the MTBF to reach 1,000 hours, given the parameters obtained from the simulation.

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2)	Figure SimumaticMTBF shows the plot of the instantaneous MTBF with the 90% confidence bounds.

3)	The 1-sided 90% lower confidence bound on time assuming MTBF = 1,000 hours can be obtained from the results of the simulation. On the Sorted sheet of the results, this is the target DMTBF value that corresponds to 10.00%, as shown in Figure Simumaticorted. Therefore the 1-sided 90% lower confidence bound on time is 12,642.21 hours.

1)	Figure SimumaticMTBFarget shows the 1-sided 90% lower confidence bound on time in the instantaneous MTBF plot. This is indicated by the target lines on the plot.

4)	The 1-sided 90% lower confidence bound on the instantaneous MTBF at the end of the test is again obtained from the Sorted sheet of the simulation results by looking at the value in the IMTBF(15,000) column that corresponds to 10.00%. As seen in Figure Simumaticorted, the 1-sided 90% lower confidence bound on the instantaneous MTBF at the end of the test is 605.93 hours.

Design of Reliability Tests for Repairable Systems
RGA 7 provides a new utility that allows you to design reliability demonstration tests for repairable systems. For example, you may want to design a test to demonstrate a specific cumulative MTBF for a specific operating period at a specific confidence level for a repairable system. The same applies for demonstration of an instantaneous MTBF or cumulative and instantaneous failure intensity at a given time $$t.$$

Underlying Theory
The failure process in a repairable system is considered to be a non-homogeneous poisson process (NHPP) with power law failure intensity. The instantaneous failure intensity at time $$t$$  is:


 * $${{\lambda }_{i}}\left( t \right)=\lambda \beta {{t}^{\beta -1}}={{\lambda }_{c}}\left( t \right)\beta $$

So, the cumulative failure intensity at time $$t$$  is:


 * $${{\lambda }_{c}}\left( t \right)=\lambda {{t}^{\beta -1}}=\frac{{{\lambda }_{i}}\left( t \right)}{\beta }$$

The instantaneous MTTF is:


 * $$\begin{align}

& MTT{{F}_{i}}\left( t \right)= & \frac{1} \\ & = & \frac{1}{\lambda \beta }{{t}^{1-\beta }} \\ & = & \frac{MTT{{F}_{c}}\left( t \right)}{\beta } \end{align}$$

The cumulative MTBF at time   is:


 * $$MTB{{F}_{c}}\left( \beta \right)=\frac{1}{{{\lambda }_{c}}\left( t \right)}=\frac{1}{\lambda }{{t}^{1-\beta }}=MTT{{F}_{i}}\left( t \right)\beta $$

The relation between confidence level, required test time, number of systems under test and the allowed total number of failures in the test is:


 * $$1-CL=\underset{i=0}{\overset{r}{\mathop \sum }}\,\frac{{{\left( m\lambda {{T}^{\beta }} \right)}^{i}}\exp (-m\lambda {{T}^{\beta }})}{i!}$$

where: •	 $$T$$ is the total test time for each system. •	 $$m$$ is the number of systems under test. •	 $$r$$ is the number of allowed failures in the test. •	 $$CL$$ is the confidence level. Given any three of the parameters, Eqn. (Cumulative poisson) can be solved for the fourth unknown parameter. Note that when $$\beta =1,$$  the number of failures is a homogeneous Poisson process and the time between failures is given by the exponential distribution.

Example 3
The objective is to design a test to demonstrate that the number of failures per system in five years is less than or equal to 10. In other words, demonstrate that the cumulative MTBF for a repairable system is less than or equal to 0.5 during a five year operating period, with 80% confidence level. Assume $$\beta =1.5$$, the number of systems for the test is  $$m=6$$  and the number of allowed failures in the test is  $$r=2.$$

Solution
Since the given requirement is number of failures, we transfer the requirement to cumulative MTBF or cumulative failure intensity.


 * $$MTBF_c=\frac{5}{10}=0.5 year$$

Then:


 * $${{\lambda }_{c}}=\frac{1}{MTB{{F}_{c}}}=2\text{ }failures/\ \ year$$

From Eqn. (lambda(c) (t)), we can then solve for $$\lambda $$ :


 * $${{\lambda }_{c}}\left( t \right)=\lambda {{t}^{\beta -1}}$$

For the five year period:


 * $${{\lambda }_{c}}\left( 5 \right)=\lambda \cdot {{5}^{\beta -1}}$$

Using the values of $$\lambda $$  and  $$\beta $$, we have:


 * $$2=\lambda \cdot {{5}^{1.5-1}}$$

Then solving for $$\lambda $$  yields:


 * $$\lambda =0.894$$

Using Eqn. (Cumulative poisson), we can then solve for the required test time, $$T,$$  for each system:


 * $$1-CL=\underset{i=0}{\overset{r}{\mathop \sum }}\,\frac{{{\left( m\lambda {{T}^{\beta }} \right)}^{i}}\exp (-m\lambda {{T}^{\beta }})}{i!}$$

or:


 * $$1-0.8=\underset{i=0}{\overset{2}{\mathop \sum }}\,\frac{{{\left( 6\cdot 0.894\cdot {{T}^{1.5}} \right)}^{i}}\exp (-6\cdot 0.894\cdot {{T}^{1.5}})}{i!}$$

or:


 * $$0.2=exp (-6\cdot 0.894\cdot T^1.5)+$$
 * $$\frac{(6\cdot 0.894\cdot T^1.5)^1 exp(-6\cdot 0.894\cdot T^1.5)}{1!}$$
 * $$+\frac{6\cdot 0.894\cdot T^1.5)^2 exp(-6\cdot 0.894\cdot T^1.5)}{2!}$$

Solving Eqn. (Example 1 equation poisson) numerically yields:


 * $$T=0.859$$

In other words, for this example we have to test for 0.859 years to demonstrate that the number of failures per system in five years is less than or equal to 10. The same result can be obtained in RGA 7, by using the Design of Reliability Tests (DRT) tool. Figure drt0 shows the calculated required test time per unit of 0.859 on the inputs of the example.

Example 4
At the end of a reliability growth testing program, a manufacturer wants to demonstrate that a new product has achieved an MTBF of 10,000 hours with 80% confidence. The available time for the demonstration test is 4,000 hours for each test unit. Assuming zero failures are allowed, what is the required number of units to be tested in order to demonstrate the desired MTBF?

Solution
Using the Design of Reliability Tests (DRT) tool in RGA 7, the required number of units can be obtained. Since this is a demonstration test then $$\beta =1$$, and no growth will be achieved. The results are shown in Figure DRTx2. It can be seen that in order to demonstrate a 10,000 hours MTBF with 80% confidence, 5 test units will be required.

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Figure DRT1x2 shows the different combinations of test time per unit and number of units in the test for different numbers of allowable failures. It helps to visually examine other possible test scenarios.