Transcript Systems Maintainability Models & analysis

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
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Why Do Maintainability Modeling and Analysis?
• To identify the important issues
• To quantify and prioritize these issues
• To build better design and support systems
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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
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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
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Bottoms Up Models
• Input Data Sources
– Task Frequency
 Reliability predictions de-rated to account for nonrelevant failures
 Because many failures are repaired on equipment,
the off equipment task frequency will be less than
the task frequency for on equipment
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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
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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
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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.
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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:
1
f (x)  e

0

x

,
for x  0
,
elsewhere
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Properties of the Exponential Model:
• Probability Distribution Function
F (x)  PX  x  
for x < 0
0
1- e

x

for x  0
Note: the Exponential Distribution is said to be
without memory, i.e.
• P(X > x1 + x2 | X > x1) = P(X > x2)
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Properties of the Exponential Model:
• Mean or Expected Value
  E (X )  
• Standard Deviation
 
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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
1
f (x) 
e
 2
where

 = 3.14159…
1
2
2

x



2
-<x<
and
e = 2.7183...
f(x)
x
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Normal Distribution:
• Mean or expected value of X
Mean = E(X) = 
• Median value of X
X0.5 = 
• Standard deviation
Var(X )  
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Normal Distribution:
Standard Normal Distribution
If X ~ N(, ) and if Z 
X 

, then Z ~ N(0, 1).
A normal distribution with  = 0 and  = 1, is called
the standard normal distribution.
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Normal Distribution:
f(z)
f(x)
x

Z
z
x
0

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Normal Distribution:
Standard Normal Distribution Table of Probabilities
http://www.smu.edu/~christ/stracener/cse7370/normaltable.html
Enter table with
Z
f(z)
x

and find the
value of 

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z
z
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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?
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Normal Distribution - Example Solution:
P1.00  X  1.75
1.75  1.25 
 1.00  1.25
 P
X

0.46 
 0.46
 P 0.54  X  1.09
 1.09  0.54
 0.8621  0.2946  0.5675
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Normal Distribution - Example Solution:
2  1.25 

P  X  2   P Z 

0.46 

 PZ  1.63  1  1.63
 0.0516
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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:
f (x) 
1
x 2
0

e
1
2
2

ln
x



2
,
for x > 0
,
for x  0
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Properties of the Lognormal Distribution
Probability Distribution Function
 ln x   
F( x )  

  
where (z) is the cumulative probability distribution
function of N(0,1)
Rule:
If T ~ LN(,), then Y = lnT ~ N(,)
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Properties of the Lognormal Model:
• Mean or Expected Value
E ( X)  e
1 2
 
2
• Median
x0.5  e

• Variance
Var (X)  e
2   2
e
2

1
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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.
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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)
 lnX  μ  0.693  0.25 
 P


0.50
 σ

 PZ  1.89
 0.9716
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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)
 lnX  μ 0.182  0.25 
 P


0.50
 σ

 PZ  0.136
 0.4404
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Lognormal Model example
c. What is the median elapsed time to repair?
P(X < x0.5) = 0.5
ln x0.5   

 P Z 




 PZ  0
therefore
 0.5
ln x0.5  
0

ln x0.5    0.25
x0.5  e   e0.25  1.284
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Lognormal Model example
d. What is the probability that an elapsed time to
repair will exceed the mean elapsed time to repair?
MTTR  e
e
σ2
μ
2

0.502
0.25
2
 e 0.375
 1.455
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Lognormal Model example
P(X > MTTR) = P(X > 1.455)
= P(lnX > 0.375)
 lnX  μ 0.375  0.25 
 P


0.50
 σ

 PZ  0.25
 0.4013
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Lognormal Model example
e. Sketch the cumulative probability distribution
function.
Cumulative Probability Distribution Function
1
P(t<x)
0.8
0.6
0.4
0.2
0
0
1
2
tmax
3
4
5
6
time to repair
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95th Percentile / MTTR Ratio
If repair time, T, has a lognormal distribution with parameters
μ and σ, then
• 95th percentile of time to repair
t 0.95  e
μ 1.645σ
•Mean Time To Repair
MTTR  e
σ2
μ
2
•Ratio of 95th percentile to mean time to repair
σ2
1.645σ 2
t 0.95
r
e
MTTR
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95th Percentile / MTTR Ratio
4
95th percentile / MTTR ratio
3.5
3
2.5
2
1.5
1
0.5
0
0
1
2
3
4
5
6
Sigma
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95th Percentile / MTTR Ratio
Since
t 0.95
e
MTTR
σ2
 1.645σ -  2
σ2
1.645σ 2
re
Then
1.645 
2
2
2
2
 ln r
 1.645  ln r  0
 2  3.29  2 * ln r  0
and
3.29  10.84  8 ln r

2
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Analysis of Combination of Repair Times
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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
Alignment
and Adjustment
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Re-assembly Complete
Condition
Verification
Repair Completed
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Linear Combinations of Random Variables
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
n
Y   ai X i
i 1
has mean
n
Y   ai  i
i 1
and variance
n
 Y2   ai2 i2
i 1
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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
n
Y   ai X i
then
Y ~ N Y ,  Y 
i 1
where
n
n
Y   ai  i
and
i 1
where a1, a2, … an are real numbers
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  a 
2
Y
i 1
2
i
2
i
Example
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Generating Random Samples
using Monte Carlo Simulation
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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
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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
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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
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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.
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Uniform Probability Integral Transformation
For any random variable Y with probability density
function f(y), the variable
y
F ( y) 
 f ( x)dx

is uniformly distributed over (0, 1), or F(y) has the
probability density function
gF ( y)  1
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for 0  y  1
Uniform Probability Integral Transformation
Remark: the cumulative probability distribution
function for any continuous random variable is
uniformly distributed over the interval (0, 1).
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Generating Random Numbers
f(y)
y
F(y)
ri
1.0
0.8
0.6
0.4
0.2
0
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y
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.
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Generating Random Numbers with Excel
From the Tools menu, look for Data Analysis.
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Generating Random Numbers with Excel
If it is not there, you must install it.
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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”
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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
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
  ri
x

e
Type in the equation , i
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.
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Flow Chart of Monte Carlo Simulation method
Input 1: Statistical distribution
for each component variable.
Select a random value from
each of these distributions
Input 2: Relationship
between component
variables and system
performance
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.
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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
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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
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System Mean Time to Repair, MTTRS
System without redundancy
E1
E2
En
n
n
MTTR i
λ i MTTR i


i 1 MTBFi
MTTRs  n
 i 1 n
1
λi


i 1 MTBFi
i 1
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Systems Maintainability Analysis Examples
• Example 1: Compute the mean time to repair at the system
level for the following system.
MTTF = 500 h
MTTR =
2h
•
MTTF = 400 h
MTTR = 2.5 h
MTTF = 250 h
MTTR =
1h
MTTF = 100 h
MTTR = 0.5 h
Solution:
2h
2.5h
1h
0.5h



0.01925
500
h
400
h
250
h
100
h
MTTRs 

 1.04h
1
1
1
1
1
0.0185h



500h 400h 250h 100h
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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 =
2h
MTTF = 400 h
MTTR = 2.5 h
Solution:
MTTF = 250 h
MTTR =
1h
MTTF = 100 h
MTTR = 0.5 h
2h
2.5h
1h
0.5h 0.5h




0.02425
500
h
400
h
250
h
100
h
100
h
MTTRs 

 0.85h
1
1
1
1
1
1
0.0285h




500h 400h 250h 100h 100h
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Maintainability Math
•
•
•
•
•
Task Time
Task frequency
Crew Size
MMH/FH =
T95%=
– N
– LN
• T50%=
– N
– LN
• MTTR=
– N
– LN
:
:
:
MTTR ~ N(,σ)|LN(,σ)
MTBM ~ E()
CS is constant
CS*MTTR/MTBM
:
:
+1.645σ
e+1.645σ
:
:

e
:
:

e+½σ^2
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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)
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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
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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)
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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)
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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)
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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
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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)
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Example: Roll-up
• Given the following R&M characteristics for the items
comprising a subsystem, what are the subsystem R&M
characteristics?
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Solution: Roll-up
(1)
(2)
(3)
(4)
A
MTBM =1/(Σ1/MTBMi)
MMH/FH
=ΣMMH/FHi
MTTR
=(ΣMTTRi/MTBMi)/(Σ1/MTBMi)
CS
=MMH/FH*MTBM/MTTR
B
C
D
E
MTBM MTTR
CS
MMH/FH
Item 1
13009
1.51
1.00
0.000116
Item 2
71872
1.36
1.54
0.000029
Item 3
285714
0.52
1.00
0.000002
Total
10606
1.45
1.08
0.000147
Equation: =1/G5 =H5/G5 =E5/C5*B5 =SUM(E2:E4)
F
G
H
MTTR/MTBM
1 1/MTBM
0.000077
0.000116
2
0.000014
0.000019
3
0.000004
0.000002
4
0.000094
0.000137
5
=SUM(G2:G4) =SUM(H2:H4)
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Maintainability Data
Analysis and Model Selection
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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 
n
1
μ^   X i  X
n i 1
• Point Estimate of s

1 n

Xi  X

n i 1
^
s

2
n 1
n
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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
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Estimation of the Mean of a Lognormal Distribution
• Mean or Expected value of
mean  E(x)  e
σ2
μ
2
• Point Estimate of mean
^
^
mean  E(x)  e
σ̂ 2
μ̂ 
2
where μ̂ and σ̂ are point estimates of μ and σ respectively.
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95th Percentile / MTTR Ratio
If repair time, T, has a lognormal distribution with parameters μ
and σ, then
• 95th percentile of time to repair
t 0.95  e
μ 1.645σ
•Mean Time To Repair
MTTR  e
σ2
μ
2
•Ratio of 95th percentile to mean
time to repair
2
t 0.95
r
e
MTTR
1.645σ -
Stracener_EMIS 7305/5305_Spr08_03.25.08
σ
2
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:


2
n ln ti     ln ti 
i 1
 i 1

ˆ 
n(n  1)
n
n
2
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Procedure for Prediction of 95th percentile
time to repair using the predicted MTTR
• Estimate r as follows:
1.645ˆ 
rˆ  e
ˆ 2
2
• Predict the 95th percentile repair time as follows:
tˆ0.95  rˆ  MTTR
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