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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.

- 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

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

- Mean or expected value of X
- Mean = E(X) =
- Median value of X
- X0.5 =
- Standard deviation

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.

Standard Normal Distribution Table of Probabilities

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

Enter table with

and find the

value of

f(z)

z

0

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?

- 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(,)

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)

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)

d. What is the probability that an elapsed time to

repair will exceed the mean elapsed time to repair?

- 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

Failure Occurs

Detection

Failure Confirmed

Preparation for

Maintenance

Active Maintenance Commences

Location and

Isolation

Faulty Item Identified

Disassembly

(Access)

Disassembly Complete

or

Repair of

Equipment

Removal of

Fault Item

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

and

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

where

where a1, a2, … an are real numbers

- 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

- 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

- 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

- 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.

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

Uniform Probability Integral Transformation

Remark: the cumulative probability distribution

function for any continuous random variable is

uniformly distributed over the interval (0, 1).

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

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=20and 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, σas 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.

Select a random value from

each of these distributions

Repeat

n

times

Calculate the value of system

performance for a system

composed of components with the

values obtained in the previous step.

Input 2: Relationship

between component

variables and system

performance

Output: Summarize and plot resulting

values of system performance. This

provides an approximation of the

distribution of system performance.

- 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

Systems Maintainability Analysis Examples

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

Solution:

MTTF = 400 h

MTTR = 2.5 h

MTTF = 250 h

MTTR = 1 h

MTTF = 100 h

MTTR = 0.5 h

MTTF = 500 h

MTTR = 2 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

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

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=

N :

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 thatΣCi=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 thatΣCi=1.

MMH/FHi=MFH/FH*Ci

Roll-up

Given component’s MMH/FHi or MTTRi , CSi andMTBMi

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 andMTBMi

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 andMTBMi

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 thatΣCi=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 thatΣCi=1.

MMH/FHi=MFH/FH*Ci

Roll-up

Given component’s MMH/FHi or MTTRi , CSi andMTBMi

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 andMTBMi

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

Estimation of the Mean - Normal Distribution

- X1, X2, …, Xn is a random sample of size n from
- N(, ), where both m & s 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

- 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

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

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