System Maintainability Modeling & Analysis Leadership in Engineering Systems Engineering Program Department of Engineering Management, Information and Systems EMIS 7305/5305 Systems Reliability, Supportability and Availability Analysis System Maintainability Modeling & Analysis Dr. Jerrell T. Stracener, SAE Fellow Leadership in Engineering

Why Do Maintainability Modeling and Analysis? To identify the important issues To quantify and prioritize these issues To build better design and support systems 2

Bottoms Up Models Provide output to monitor design progress vs. requirements Provide input data for life cycle cost Provide trade-off capability Design features vs. maintainability requirements Performance vs. maintainability requirements Provide Justification for maintenance improvements perceived as the design progresses 3

Bottoms Up Models Provide the basis for maintainability guarantees/demonstration Provide inputs to warranty requirements Provide maintenance data for the logistic support analysis record Support post delivery design changes Inputs Task Time (MH) Task Frequency (MTBM) Number of Personnel-Elapsed Time (hours) For each repairable item 4

Bottoms Up Models Input Data Sources Task Frequency Reliability predictions de-rated to account for non-relevant failures Because many failures are repaired on equipment, the off equipment task frequency will be less than the task frequency for on equipment 5

Bottoms Up Models Input Data Sources (Continued) Task Time Touch time vs. total time That time expended by the technician to effect the repair Touch time is design controllable Total Time Includes the time that the technician expends in “Overhead” functions such as part procurement and paper work Are developed from industrial engineering data and analyst’s estimates 6

Task Analysis Model Task analysis modeling estimates repair time MIL-HDK-472 method V Spreadsheet template Allow parallel and multi-person tasks estimation Calculates elapsed time and staff hours Reports each task element and total repair time Sums staff hours by repairmen type Estimates impact of hard to reach/see tasks 7

When considering probability distributions in general, the time dependency between probability of repair and the time allocated for repair can be expected to produce one of the following probability distribution functions: Normal – Applies to relatively straightforward maintenance tasks and repair actions that consistently require a fixed amount of time to complete with little variation Exponential – Applies to maintenance tasks involving part substitution methods of failure isolation in large systems that result in a constant repair rate. Lognormal - Applies to Most maintenance tasks and repair actions comprised of several subsidiary tasks of unequal frequency and time duration.

The Exponential Model: Definition A random variable X is said to have the Exponential Distribution with parameters , where  > 0, if the probability density function of X is: , for x  0 , elsewhere

Properties of the Exponential Model: Probability Distribution Function for x < 0 for x  0 Note: the Exponential Distribution is said to be without memory, i.e. P(X > x1 + x2 | X > x1) = P(X > x2)

Properties of the Exponential Model: Mean or Expected Value Standard Deviation

Normal Distribution: A random variable X is said to have a normal (or Gaussian) distribution with parameters  and , where -  <  <  and  > 0, with probability density function -  < x <  where  = 3.14159… and e = 2.7183... f(x) x

Normal Distribution: Mean or expected value of X Mean = E(X) =  Median value of X X0.5 =  Standard deviation

Normal Distribution: Standard Normal Distribution If X ~ N(, ) and if , then Z ~ N(0, 1). A normal distribution with  = 0 and  = 1, is called the standard normal distribution.

 x z Normal Distribution: f(x) f(z)

Standard Normal Distribution Table of Probabilities http://www.smu.edu/~christ/stracener/cse7370/normaltable.html Enter table with and find the value of  z  z f(z)

Normal Distribution - Example: The time it takes a field engineer to restore a function in a logistics system can be modeled with a normal distribution having mean value 1.25 hours and standard deviation 0.46 hours. What is the probability that the time is between 1.00 and 1.75 hours? If we view 2 hours as a critically time, what is the probability that actual time to restore the function will exceed this value?

Normal Distribution - Example Solution:

Normal Distribution - Example Solution:

The Lognormal Model: Definition - A random variable X is said to have the Lognormal Distribution with parameters  and , where  > 0 and  > 0, if the probability density function of X is: , for x > 0 , for x  0

Properties of the Lognormal Distribution Probability Distribution Function where (z) is the cumulative probability distribution function of N(0,1) Rule: If T ~ LN(,), then Y = lnT ~ N(,)

Properties of the Lognormal Model: Mean or Expected Value Variance Median

Lognormal Model example The elapsed time (hours) to repair an item is a random variable. Based on analysis of data, elapsed time to repair can be modeled by a lognormal distribution with parameters  = 0.25 and  = 0.50. a. What is the probability that an elapsed time to repair will exceed 0.50 hours? b. What is the probability that an elapsed time to repair will be less than 1.2 hours? c. What is the median elapsed time to repair? d. What is the probability that an elapsed time to repair will exceed the mean elapsed time to repair? e. Sketch the cumulative probability distribution function.

Lognormal Model example - solution a. What is the probability that an elapsed time to repair will exceed 0.50 hours? X ~ LN(, ) where  = 0.25 and  = 0.50 note that: Y = lnX ~ N(, ) P(X > 0.50) = P(lnX > -0.693)

Lognormal Model example b. What is the probability that an elapsed time to repair will be less than 1.2 hours? P(X < 1.20) = P(lnX < ln1.20)

Lognormal Model example c. What is the median elapsed time to repair? P(X < x0.5) = 0.5 therefore

Lognormal Model example d. What is the probability that an elapsed time to repair will exceed the mean elapsed time to repair?

Lognormal Model example P(X > MTTR) = P(X > 1.455) = P(lnX > 0.375)

Lognormal Model example e. Sketch the cumulative probability distribution function. tmax

95th Percentile / MTTR Ratio If repair time, T, has a lognormal distribution with parameters μ and σ, then 95th percentile of time to repair Mean Time To Repair Ratio of 95th percentile to mean time to repair

95th Percentile / MTTR Ratio

95th Percentile / MTTR Ratio Since Then and

Analysis of Combination of Repair Times

Active Maintenance Commences Faulty Item Identified Corrective Maintenance Cycle Failure Occurs Detection Failure Confirmed Preparation for Maintenance Active Maintenance Commences Location and Isolation Faulty Item Identified Disassembly (Access) Disassembly Complete or Removal of Fault Item Repair of Equipment Installation of Spare/Repair Part Re-assembly Re-assembly Complete Alignment and Adjustment Condition Verification 34 Repair Completed

If X1, X2, ..., Xn are independent random variables with means 1, 2, ..., n and variances 12, 22, ..., n2, respectively, and if a1, a2, … an are real numbers then the random variable has mean and variance Linear Combinations of Random Variables

where a1, a2, … an are real numbers Linear Combinations of Random Variables If X1, X2, ..., Xn are independent random variables having Normal Distributions with means 1, 2, ..., n and variances 12, 22, ..., n2, respectively, and if then where and where a1, a2, … an are real numbers

Example

Generating Random Samples using Monte Carlo Simulation 38

Population vs. Sample Population the total of all possible values (measurement, counts, etc.) of a particular characteristic for a specific group of objects. Sample a part of a population selected according to some rule or plan. Why sample? - Population does not exist - Sampling and testing is destructive

Sampling Characteristics that distinguish one type of sample from another: the manner in which the sample was obtained the purpose for which the sample was obtained

Types of Samples Simple Random Sample The sample X1, X2, ... ,Xn is a random sample if X1, X2, ... , Xn are independent identically distributed random variables. Remark: Each value in the population has an equal and independent chance of being included in the sample. Stratified Random Sample The population is first subdivided into sub-populations for strata, and a simple random sample is drawn from each strata

Types of Samples (continued) Censored Samples Type I Censoring - Sample is terminated at a fixed time, t0. The sample consists of K times to failure plus the information that n-k items survived the fixed time of truncation. Type II Censoring - Sampling is terminated upon the Kth failure. The sample consists of K times to failure, plus information that n-k items survived the random time of truncation, tk. Progressive Censoring - Sampling is reduced in stage.

Uniform Probability Integral Transformation For any random variable Y with probability density function f(y), the variable is uniformly distributed over (0, 1), or F(y) has the probability density function

Uniform Probability Integral Transformation Remark: the cumulative probability distribution function for any continuous random variable is uniformly distributed over the interval (0, 1).

Generating Random Numbers f(y) F(y) y 1.0 0.8 0.6 0.4 0.2 ri yi

Generating Random Numbers Generating values of a random variable using the probability integral transformation to generate a random value y from a given probability density function f(y): 1. Generate a random value rU from a uniform distribution over (0, 1). 2. Set rU = F(y) 3. Solve the resulting expression for y.

Generating Random Numbers with Excel From the Tools menu, look for Data Analysis.

Generating Random Numbers with Excel If it is not there, you must install it. Generating Random Numbers with Excel

Generating Random Numbers with Excel Once you select Data Analysis, the following window will appear. Scroll down to “Random Number Generation” and select it, then press “OK”

Generating Random Numbers with Excel Choose which distribution you would like. Use uniform for an exponential or weibull distribution or normal for a normal or lognormal distribution

Generating Random Numbers with Excel Uniform Distribution, U(0, 1). Select “Uniform” under the “Distribution” menu. Type in “1” for number of variables and 10 for number of random numbers. Then press OK. 10 random numbers of uniform distribution will now appear on a new chart.

Generating Random Numbers with Excel Normal Distribution, N(μ, σ). Select “Normal” under the “Distribution” menu. Type in “1” for number of variables and 10 for number of random numbers. Enter the values for the mean (m) and standard deviation (s) then press OK. 10 random numbers of uniform distribution will now appear on a new chart.

Generating Random Values from an Exponential Distribution E() with Excel First generate n random variables, r1, r2, …, rn, from U(0, 1). Select “Uniform” under the “Distribution” menu. Type in “1” for number of variables and 10 for number of random numbers. Then press OK. 10 random numbers of uniform distribution will now appear on a new chart.

Generating Random Values from an Exponential Distribution E() with Excel Select a θ that you would like to use, we will use θ = 5. Type in the equation xi= -ln(1 - ri), with filling in θ as 5, and ri as cell A1 (=-5*LN(1-A1)). Now with that cell selected, place the cursor over the bottom right hand corner of the cell. A cross will appear, drag this cross down to B10. This will transfer that equation to the cells below. Now we have n random values from the exponential distribution with parameter θ=5 in cells B1 - B10.

Generating Random Values from an Weibull Distribution W(β, ) with Excel First generate n random variables, r1, r2, …, rn, from U(0, 1). Select “Uniform” under the “Distribution” menu. Type in “1” for number of variables and 10 for number of random numbers. Then press OK. 10 random numbers of uniform distribution will now appear on a new chart.

Generating Random Values from an Weibull Distribution W(β, ) with Excel Select a β and θ that you would like to use, we will use β =20, θ = 100. Type in the equation xi = [-ln(1 - ri)]1/, with filling in β as 20, θ as 100, and ri as cell A1 (=100*(-LN(1-A1))^(1/20)). Now transfer that equation to the cells below. Now we have n random variables from the Weibull distribution with parameters β =20 and θ =100 in cells B1 - B10.

Generating Random Values from an Lognormal Distribution LN(μ, σ) with Excel First generate n random variables, r1, r2, …, rn, from N(0, 1). Select “Normal” under the “Distribution” menu. Type in “1” for number of variables and 10 for number of random numbers. Enter 0 for the mean and 1 for standard deviation then press OK. 10 random numbers of uniform distribution will now appear on a new chart.

Generating Random Values from an Lognormal Distribution LN(μ, σ) with Excel Select a μ and s that you would like to use, we will use μ = 2, σ = 1. Type in the equation , with filling in μ as 2, σ as 1, and ri as cell A1 (=EXP(2+A1*1)). Now transfer that equation to the cells below. Now we have an Lognormal distribution in cells B1 - B10.

Flow Chart of Monte Carlo Simulation method Input 1: Statistical distribution for each component variable. Input 2: Relationship between component variables and system performance Select a random value from each of these distributions Calculate the value of system performance for a system composed of components with the values obtained in the previous step. Output: Summarize and plot resulting values of system performance. This provides an approximation of the distribution of system performance. Repeat n times

Sample and Size Error Bands Because Monte Carlo simulation involves randomly selected values, the results are subject to statistical fluctuations. Any estimate will not be exact but will have an associated error band. The larger the number of trials in the simulation, the more precise the final results. We can obtain as small an error as is desired by conducting sufficient trials In practice, the allowable error is generally specified, and this information is used to determine the required trials

Drawbacks of the Monte Carlo Simulation there is frequently no way of determining whether any of the variables are dominant or more important than others without making repeated simulations if a change is made in one variable, the entire simulation must be redone the method may require developing a complex computer program if a large number of trials are required, a great deal of computer time may be needed to obtain the necessary results

System Mean Time to Repair, MTTRS System without redundancy E1 E2 En 62

Systems Maintainability Analysis Examples Example 1: Compute the mean time to repair at the system level for the following system. Solution: MTTF = 500 h MTTR = 2 h MTTF = 400 h MTTR = 2.5 h MTTF = 250 h MTTR = 1 h MTTF = 100 h MTTR = 0.5 h 63

Maintainability Prediction Example 2: How does the MTTRs of the system in the previous example change if an active redundancy is introduced to the element with MTTF = 100h? MTTF = 100 h MTTR = 0.5 h MTTR = 0.5 h MTTF = 500 h MTTR = 2 h MTTF = 400 h MTTR = 2.5 h MTTF = 250 h MTTR = 1 h Solution: 64

Maintainability Math Task Time : MTTR ~ N(,σ)|LN(,σ) Task frequency : MTBM ~ E() Crew Size : CS is constant MMH/FH = CS*MTTR/MTBM T95%= N : +1.645σ LN : e+1.645σ T50%= N :  LN : e MTTR= LN : e+½σ^2 65

Maintainability Math: MTBM Given that MTBM is exponential MTBM = inverse of sum of inverse MTBMs Allocation Given a top level MTBM and a complexity factor for components Ci such that ΣCi =1. MTBMi=MTBM*Ci Roll-up Given components MTBMi MTBM=1/Σ(1/MTBMi) 66

Maintainability Math: MMH/FH MMH/FH = sum of MMH/FHs Allocation Given a top level MMH/FH and a complexity factor for component repairs Ci such that ΣCi =1. MMH/FHi=MFH/FH*Ci Roll-up Given component’s MMH/FHi or MTTRi , CSi and MTBMi MMH/FH=ΣMMH/FHi =ΣCSi*MTTRi/MTBMi 67

Maintainability Math: MTTR MTTR = weighted sum of MTTRs Weighting factor is frequency of maintenance = 1/MTBM Roll-up Given components MTTRi and MTBMi MTTR = MTBM*Σ(MTTRi /MTBMi) = (Σ(MTTRi /MTBMi)/Σ(1/MTBMi) 68

Maintainability Math: CS CS is computed from MMH/FM, MTBM and MTTR CS=MMH/FH*MTBM/MTTR Roll-up Given components MTTRi , CSi and MTBMi CSaverage= MMH/FH*MTBM/MTTR =(ΣCSi*MTTRi/MTBMi)/Σ(MTTRi /MTBMi) 69

Maintainability Math: MTBM Given that MTBM is exponential MTBM = inverse of sum of inverse MTBMs Allocation Given a top level MTBM and a complexity factor for components Ci such that ΣCi =1. MTBMi=MTBM*Ci Roll-up Given components MTBMi MTBM=1/Σ(1/MTBMi) 70

Maintainability Math: MMH/FH MMH/FH = sum of MMH/FHs Allocation Given a top level MMH/FH and a complexity factor for component repairs Ci such that ΣCi =1. MMH/FHi=MFH/FH*Ci Roll-up Given component’s MMH/FHi or MTTRi , CSi and MTBMi MMH/FH=ΣMMH/FHi =ΣCSi*MTTRi/MTBMi 71

Maintainability Math: MTTR MTTR = weighted sum of MTTRs Weighting factor is frequency of maintenance = 1/MTBM Roll-up Given components MTTRi and MTBMi MTTR = MTBM*Σ(MTTRi /MTBMi) = (Σ(MTTRi /MTBMi)/Σ(1/MTBMi) 72

Example: Roll-up Given the following R&M characteristics for the items comprising a subsystem, what are the subsystem R&M characteristics? 73

Solution: Roll-up MTBM =1/(Σ1/MTBMi) MMH/FH =ΣMMH/FHi MTTR =(ΣMTTRi/MTBMi)/(Σ1/MTBMi) CS =MMH/FH*MTBM/MTTR A B C D E F G H 1 2 3 4 5 74

Analysis and Model Selection Maintainability Data Analysis and Model Selection 75

Estimation of the Mean - Normal Distribution X1, X2, …, Xn is a random sample of size n from N(, ), where both µ & σ are unknown. Point Estimate of  Point Estimate of s ^ 76

Estimation of Lognormal Distribution Random sample of size n, X1, X2, ... , Xn from LN (, ) Let Yi = ln Xi for i = 1, 2, ..., n Treat Y1, Y2, ... , Yn as a random sample from N(, ) Estimate  and  using the Normal Distribution Methods 77

Estimation of the Mean of a Lognormal Distribution Mean or Expected value of Point Estimate of mean where and are point estimates of and respectively. ^ ^ 78

95th Percentile / MTTR Ratio If repair time, T, has a lognormal distribution with parameters μ and σ, then 95th percentile of time to repair Mean Time To Repair Ratio of 95th percentile to mean time to repair 79

Procedure for Prediction of 95th percentile time to repair using the predicted MTTR Obtain a random sample of n times to repair for a given subsystem, t1, t2, …, tn Utilize probability plotting on lognormal probability paper and/or use a statistical goodness of fit test to test the validity of the lognormal distribution Assuming the results indicate that the lognormal distribution provides a “good” fit to the data, estimate σ as follows: 80

Predict the 95th percentile repair time as follows: Procedure for Prediction of 95th percentile time to repair using the predicted MTTR Estimate r as follows: Predict the 95th percentile repair time as follows: 81