Stracener_EMIS 7305/5305_Spr08_03.25.08 System Maintainability Modeling & Analysis Dr. Jerrell T. Stracener, SAE Fellow Leadership in Engineering EMIS

Stracener_EMIS 7305/5305_Spr08_03.25.08

System Maintainability Modeling & Analysis

Dr. Jerrell T. Stracener, SAE Fellow

Leadership in Engineering

EMIS 7305/5305Systems Reliability, Supportability and Availability Analysis

Systems Engineering ProgramDepartment of Engineering Management, Information and Systems


2

Why Do Maintainability Modeling and Analysis?

• To identify the important issues

• To quantify and prioritize these issues

• To build better design and support systems


3

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


4

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


5

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


6

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


7

Task Analysis Model

• Task analysis modeling estimates repair time– MIL-HDK-472 method V– Spreadsheet template

Allow parallel and multi-person tasks estimationCalculates elapsed time and staff hoursReports each task element and total repair timeSums staff hours by repairmen typeEstimates impact of hard to reach/see tasks


8

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.


9

The Exponential Model:

DefinitionA random variable X is said to have the ExponentialDistribution with parameters , where > 0, if the probability density function of X is:

, for x 0

, elsewhere

x

e1

)x(f

0


10

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)

xXP )(xFx

e

-1

0


11

Properties of the Exponential Model:

• Mean or Expected Value

• Standard Deviation

)(XE


12

Normal Distribution:

A random variable X is said to have a normal (orGaussian) distribution with parameters and ,where - < < and > 0, with probability density function

- < x <

where = 3.14159… and e = 2.7183...

22

x2

1

e2

1)x(f

f(x)

x


13


• Mean or expected value of X

Mean = E(X) =

• Median value of X

X0.5 =

• Standard deviation

)(XVar


14


Standard Normal Distribution

If X ~ N(, ) and if , then Z ~ N(0, 1).

A normal distribution with = 0 and = 1, is calledthe standard normal distribution.

X

Z


15

x

0

z

x

Z


f(x) f(z)


16


Standard Normal Distribution Table of Probabilities

http://www.smu.edu/~christ/stracener/cse7370/normaltable.html

Enter table with

and find thevalue of

z0

z

f(z)

x

Z


17

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?


18

Normal Distribution - Example Solution:

75.100.1 XP

46.0

25.175.1

46.0

25.100.1XP

09.154.0 XP

54.009.1

5675.02946.08621.0


19

Normal Distribution - Example Solution:

46.0

25.122 ZPXP

63.1163.1 ZP

0516.0


20

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

22

xln2

1

e2x

1 )x(f

0


21

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(,)

xln

)x(F


22

Properties of the Lognormal Model:

• Mean or Expected Value

2

2

1

e)X(E

1ee)X(Var222

• Variance

• Median

ex 5.0


23

Lognormal Model example

The elapsed time (hours) to repair an item is a random variable. Based on analysis of data, elapsedtime 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 torepair 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 torepair will exceed the mean elapsed time to repair?e. Sketch the cumulative probability distributionfunction.


24

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)

0.50

0.250.693

σ

μlnXP

89.1P Z

9716.0


25


b. What is the probability that an elapsed time torepair will be less than 1.2 hours?

P(X < 1.20) = P(lnX < ln1.20)

0.50

0.25182.0

σ

μlnXP

136.0P Z

4404.0


26


c. What is the median elapsed time to repair?

P(X < x0.5) = 0.5

therefore

5.0lnP

xZ

0P Z

5.0

0ln 5.0

x

25.0ln 5.0 x

284.125.05.0 eex


27


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

2

σμ

2

eMTTR

2

50.025.0

2

e

375.0e

455.1


28


P(X > MTTR) = P(X > 1.455)

= P(lnX > 0.375)

0.50

0.250.375

σ

μlnXP

0.25ZP

4013.0


29


e. Sketch the cumulative probability distributionfunction.

Cumulative Probability Distribution Function

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6

time to repair

P(t

<x)

tmax


30

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

1.645σμ0.95 et

2

σμ

2

eMTTR

2

σ-1.645σ

0.95

2

eMTTR

tr

95th Percentile / MTTR Ratio


31

0

0.5

1

1.5

2

2.5

3

3.5

4

0 1 2 3 4 5 6

Sigma

95th

per

cen

tile

/ M

TT

R r

atio



32


Since

Then

and

2

σ--1.645σ

0.95

2

eMTTR

t

2

σ-1.645σ

2

er

0ln*229.3

0ln645.12

ln2

645.1

2

2

2

r

r

r

2

ln884.1029.3 r


Analysis of Combination of Repair Times


34

Detection

Preparation forMaintenance

Location andIsolation

Disassembly(Access)

Removal ofFault Item

Re-assembly

Alignmentand Adjustment

ConditionVerification

or

Repair ofEquipment

Installation ofSpare/Repair Part

Failure Occurs

Failure Confirmed

Active Maintenance Commences

Faulty Item Identified

Disassembly Complete

Re-assembly Complete

Repair Completed

Corrective Maintenance Cycle


If X1, X2, ..., Xn are independent random variables with means 1, 2, ..., n and variances 1

2, 22, ...,

n2, respectively, and if a1, a2, … an are real numbers

then the random variable

has mean

and variance

n

iii XaY

1

n

iiiY a

1

n

iiiY a

1

222

Linear Combinations of Random Variables


If X1, X2, ..., Xn are independent random variables having Normal Distributions with means 1, 2, ..., n and variances 1

2, 22, ..., n

2, respectively, and if

n

iii XaY

1

then YYNY ,~

where

where a1, a2, … an are real numbers

n

iiiY a

1

n

iiiY a

1

222 and

Linear Combinations of Random Variables


Example


38

Generating Random Samplesusing Monte Carlo Simulation


Populationthe total of all possible values (measurement, counts, etc.) of a particular characteristic for aspecific group of objects.

Samplea part of a population selected according to some rule or plan.

Why sample?- Population does not exist- Sampling and testing is destructive

Population vs. Sample


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

Sampling


• Simple Random SampleThe 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 SampleThe population is first subdivided into sub-populations for strata, and a simple randomsample is drawn from each strata

Types of Samples


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.

Types of Samples (continued)


For any random variable Y with probability densityfunction f(y), the variable

is uniformly distributed over (0, 1), or F(y) has theprobability density function

y

dxxfyF )()(

1y0for 1)( yFg

Uniform Probability Integral Transformation


Remark: the cumulative probability distributionfunction for any continuous random variable isuniformly distributed over the interval (0, 1).

Uniform Probability Integral Transformation


f(y)

F(y)y

y

1.00.80.60.40.2 0

ri

yi

Generating Random Numbers


Generating values of a random variable using theprobability integral transformation to generate arandom value y from a given probability densityfunction f(y):

1. Generate a random value rU from a uniformdistribution over (0, 1).

2. Set rU = F(y)

3. Solve the resulting expression for y.

Generating Random Numbers


From the Tools menu, look for Data Analysis.

Generating Random Numbers with Excel


If it is not there, you must install it.



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



Choose which distribution you would like. Use uniform for an exponential or weibull distribution or normal for a normal or lognormal distribution



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.



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.



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 ExponentialDistribution 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 ExponentialDistribution 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 WeibullDistribution 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 WeibullDistribution W(β, ) 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 LognormalDistribution 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.

iri ex

Generating Random Values from an LognormalDistribution LN(μ, σ) with Excel


Flow Chart of Monte Carlo Simulation method

Input 1: Statistical distribution for each component variable.

Input 2: Relationshipbetween componentvariables and systemperformance

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 resultingvalues of system performance. Thisprovides an approximation of the distribution of system performance.

Repeatntimes


Because Monte Carlo simulation involves randomlyselected values, the results are subject to statisticalfluctuations.

• Any estimate will not be exact but will have anassociated 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 byconducting sufficient trials

• In practice, the allowable error is generally specified,and this information is used to determine the required trials

Sample and Size Error Bands


• 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 entiresimulation must be redone

• the method may require developing acomplex computer program

• if a large number of trials are required, a great deal of computer time may be needed to obtainthe necessary results

Drawbacks of the Monte Carlo Simulation


62

System Mean Time to Repair, MTTRS

System without redundancy

E1 E2 En

n

1ii

n

1iii

n

1i i

n

1i i

i

λ

MTTRλ

MTBF1

MTBFMTTR

MTTRs


63

Systems Maintainability Analysis Examples

• Example 1: Compute the mean time to repair at the system level for the following system.

• Solution:

MTTF = 500 hMTTR = 2 h

MTTF = 400 hMTTR = 2.5 h



h04.1h0185.0

01925.0

h1001

h2501

h4001

h5001

h100h5.0

h250h1

h400h5.2

h500h2

MTTRs1


64

h85.0h0285.0

02425.0

h1001

h1001

h2501

h4001

h5001

h100h5.0

h100h5.0

h250h1

h400h5.2

h500h2

MTTRs1

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?






Solution:


65

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=– N : – LN : e+½σ^2


66

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)


67

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


68

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)


69

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)


70

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)


71

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


72

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)


73

Example: Roll-up

• Given the following R&M characteristics for the items comprising a subsystem, what are the subsystem R&M characteristics?


74

Solution: Roll-up

MTBM MTTR CS MMH/FH 1/MTBM MTTR/MTBMItem 1 13009 1.51 1.00 0.000116 0.000077 0.000116Item 2 71872 1.36 1.54 0.000029 0.000014 0.000019Item 3 285714 0.52 1.00 0.000002 0.000004 0.000002Total 10606 1.45 1.08 0.000147 0.000094 0.000137Equation: =1/G5 =H5/G5 =E5/C5*B5 =SUM(E2:E4) =SUM(G2:G4) =SUM(H2:H4)

A B C D E F G H 1 2 3 4 5

(1) MTBM =1/(Σ1/MTBMi)

(2) MMH/FH =ΣMMH/FHi

(3) MTTR =(ΣMTTRi/MTBMi)/(Σ1/MTBMi)

(4) CS =MMH/FH*MTBM/MTTR


75

Maintainability Data Analysis and Model Selection


76

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

n

1ii XX

n

1μ̂

n

ii XX

n 1

2^ 1

n

ns

1


77

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 DistributionMethods


78

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.σ̂μ̂ μ σ

2

σ̂μ̂

2

eE(x)mean

^

2

σμ

2

eE(x)mean

^


79

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

1.645σμ0.95 et

2

σμ

2

eMTTR

2

σ-1.645σ

0.95

2

eMTTR

tr



80

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:

)1(

lnlnˆ

2

11

2

nn

ttnn

ii

n

ii


81

• Estimate r as follows:

• Predict the 95th percentile repair time as follows:

2

ˆˆ645.1

2

ˆ

er

MTTRrt ˆˆ95.0

Procedure for Prediction of 95th percentile time to repair using the predicted MTTR

Documents

Stracener_EMIS 7305/5305_Spr08_03.25.08 System Maintainability Modeling & Analysis Dr. Jerrell T. Stracener, SAE Fellow Leadership in Engineering EMIS