Transcript Application_reliability - Lyle School of Engineering

Systems Engineering Program
Department of Engineering Management, Information and Systems
EMIS 7370/5370 STAT 5340 :
PROBABILITY AND STATISTICS FOR SCIENTISTS AND ENGINEERS
An Application of Probability to
Reliability Modeling and Analysis
Dr. Jerrell T. Stracener, SAE Fellow
Leadership in Engineering
1
What is Reliability?
• Reliability is defined as the probability that an item will
perform its intended function for a specified interval
under stated conditions. In the simplest sense, reliability
means how long an item (such as a machine) will
perform its intended function without a breakdown.
• Reliability: the capability to operate as intended,
whenever used, for as long as needed.
Reliability is performance over time, probability
that something will work when you want it to.
2
Reliability Figures of Merit
• Basic or Logistic Reliability
MTBF - Mean Time Between Failures
measure of product support requirements
• Mission Reliability
Ps or R(t) - Probability of mission success
measure of product effectiveness
3
Reliability Humor: Statistics
“If I had only one day left to live,
I would live it in my statistics class -it would seem so much longer.”
From: Statistics A Fresh Approach
Donald H. Sanders
McGraw Hill, 4th Edition, 1990
4
The Reliability Function
The Reliability of an item is the probability that the item will
survive time t, given that it had not failed at time zero, when
used within specified conditions, i.e.,

Rt   P(T  t )   f ( t )dt  1  F( t )
t
5
Reliability Function
Probability Distribution Function
• Weibull

R (t)  e
t
 

• Exponential
R (t)  e

t

6
The Weibull Model - Distributions
Reliability Functions
β=5.0
1.0
0.8
R(t)
β=1.0
0.6
β=0.5
0.4
0.2
0
0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
t
t is in multiples of 
7
Percentiles, tp
• Weibull
t P   - ln(1 - p)
1

and, in particular
t 0.632  
• Exponential
t P   - ln(1 - p)
8
Relationship Between h(t), f(t), F(t) and R(t)
f t 
f t 
h t  

R t  1 - Ft 
Remark: The failure rate h(t) is a measure of proneness to
failure as a function of age, t.
9
Failure Rates - Exponential
• Failure Rate
1
h(t)   

• Note:
1
MTBF 
failure rate
Only for the Exponential Distribution
•Cumulative Failure
H( t )  
10
Failure Rates - Weibull
• Failure Rate
 -1
h(t)   t
 a decreasing function of t if  < 1
a constant if  = 1
an increasing function of t if  > 1
Notice that h(t) is
• Cumulative Failure Rate
-1
t
h(t)
H( t )   


• The Instantaneous and Cumulative Failure Rates, h(t)
and H(t), are straight lines on log-log paper.
11
The Weibull Model - Distributions
Failure Rates
3
β=5
h(t)
2
β=1
β=0.5
1
0
1.0
2.0
t
0
t is in multiples of  h(t) is in multiples of 1/ 
12
Mean Time to Failure and Mean Time Between Failures
Mean Time to Failure (or Between Failures) MTTF (or MTBF)
is the expected Time to Failure (or Between Failures)
Remarks:
MTBF provides a reliability figure of merit for expected failure
free operation
MTBF provides the basis for estimating the number of failures in
a given period of time
Even though an item may be discarded after failure and its mean
life characterized by MTTF, it may be meaningful to
characterize the system reliability in terms of MTBF if the
system is restored after item failure.
13
Mean Time Between Failure - MTBF
Weibull
1 
MTBF     1
 
Exponential
MTBF  
14
The Binomial Model - Example
Problem Four Engine Aircraft
Engine Reliability R(t) = p = 0.9
Mission success: At least two engines survive
Find RS(t)
15
The Binomial Model - Example
Solution –
1)
X1 = number of engines surviving in time t
Then X1 ~ B (4,0.9)
RS(t) = P(x1 ≥ 2) = b(2) + b(3) + b(4)
= 0.0486 + 0.2916 + 0.6561 = 0.9963
2)
X2 = number of engines failing in time t
Then X2 ~ B (4,0.1)
RS(t) = P(x2  2) = b(0) + b(1) + b(2)
= 0.6561 + 0.2916 + 0.0486 = 0.9963
16
The Weibull Model
Time to failure of an item follows a Weibull
distribution with  = 2 and  = 1000 hours.
(a) What is the reliability, R(t), for t = 200 hours?
(b) What is the hazard rate, h(t), (instantaneous
failure rate) at that time?
(c) What is the Mean Time To Failure?
17
The Weibull Model –Solution
(a)
R (t )  e
  t /  
R(200)  e
 0.9608
 ( 200/ 1000) 2
 1
(b)
t
h(t)  

2 1
2200
h(200) 
2
1000
 0.0004
failures per hour
18
The Weibull Model –Solution continued
(c)

1
MTTF  1  


1

 10001  
2

From the Gamma Function table:
(1.5)  0.88623
MTTF  10001.5
 886.23
19
Example – Pump life
The design life of a given type of pump for a given
operating environment has a Lognormal distribution. If t0.10
= 2000 hours and the median life is 3748 hours. What is the
mean life and the the 50th & 90th percentile? A system uses
five pumps of this type. What is the probability of at least
one of these pumps failing in 3000 hours?
20
Pump Life Solution
Given that t0.10 = 2000 hours and the median is
3748, we need to first find the values for  and .
Since the median life is
3748  eμ
=ln 3748
=8.2290
And since t0.1=
e
μ -1.28σ
=2000,
-1.28 = ln 2000
1.28  = 8.2290 – 7.6009
 = 0.4907
Pump Life Solution
Now t0.9 is the value of t for which
 ln t 0.90  8.229 
P(T  t 0.90 )  F(t 0.9 )  
  0.90
0.491


Ft 0.90   (1.28)  0.90
So that ln t 0.9  8.229  1.28,
0.491
But
and t0.9=7023.8
Probability of one pump failing within 3000 hours
 ln 3000  8.229 
F(3000)  P(T  3000)  Φ
  0.326
0.491


Pump Life - solution
Now t0.5 is the value of t for which
 ln t 0.50  μ 
P(T  t 0.50 )  F(t 0.5 )  
  0.50
σ


Ft 0.50   (0)  0.50
So that ln t 0.5  8.229
0
0.491
But
And t0.5=3748
Mean life is
E(t)  e
2
μ  σ2
 e8.34939  4227.6
Pump Life Solution
Probability of at least one pump failing in 3000
hours
Y = # of pumps failing in 3000 hours
y = 0, 1, 2, 3, 4, 5
Y has a binomial distribution with
n = 5 and p = 0.326
P(Y  1)  1  P(Y  0)
 5
0
5


 1   0.326 0.674
 0
 0.861,
or you could work this using a probability tree
Series Reliability Configuration
• Simplest and most common structure in reliability analysis.
• Functional operation of the system depends on the
successful operation of all system components Note: The
electrical or mechanical configuration may differ from the
reliability configuration
Reliability Block Diagram
• Series configuration with n elements: E1, E2, ..., En
E1
E2
En
• System Failure occurs upon the first element failure
25
Series Reliability Configuration with Exponential Distribution
• Reliability Block Diagram
E1
E2
En
•Element Time to Failure Distribution
1
Ti ~ Eθi  with failure rate λ i 
, for i=1, 2,…, n
θi
• System reliability
R S ( t )  e  St
n
where  S ( t )    i is the system failure rate
i 1
• System mean time to failure
1
MTTFS 
 θS
λS
26
Series Reliability Configuration
• System mean time to failure

MTTFS 
n
Note that /n is the expected time to the first failure,
E(T1), when n identical items are put into service
27
Parallel Reliability Configuration – Basic Concepts
• Definition - a system is said to have parallel reliability
configuration if the system function can be performed by any
one of two or more paths
• Reliability block diagram - for a parallel reliability configuration
consisting of n elements, E1, E2, ... En
E1
E2
En
28
Stracener_EMIS 7370/STAT 5340_Fall 08_09.30.08
Parallel Reliability Configuration
• Redundant reliability configuration - sometimes called a
redundant reliability configuration. Other times, the term
‘redundant’ is used only when the system is deliberately
changed to provide additional paths, in order to improve the
system reliability
• Basic assumptions
All elements are continuously energized starting at time t = 0
All elements are ‘up’ at time t = 0
The operation during time t of each element can be described
as either a success or a failure, i.e. Degraded operation or
performance is not considered
29
Stracener_EMIS 7370/STAT 5340_Fall 08_09.30.08
Parallel Reliability Configuration
System success - a system having a parallel reliability
configuration operates successfully for a period of time t if at least
one of the parallel elements operates for time t without failure.
Notice that element failure does not necessarily mean system
failure.
30
Stracener_EMIS 7370/STAT 5340_Fall 08_09.30.08
Parallel Reliability Configuration
• Block Diagram
E1
E2
En
• System reliability - for a system consisting of n elements, E1, E2, ... En
n
n
R S ( t )   R i ( t )   R i ( t )R j ( t ) 
i 1
ij
i j
n
n
 R (t ) R (t )R (t )  ...(1)  R (t )
n 1
i
ijk
i j k
j
k
i
i 1
if the n elements operate independently of each other and where
Ri(t) is the reliability of element i, for i=1,2,…,n
31
Stracener_EMIS 7370/STAT 5340_Fall 08_09.30.08
System Reliability Model - Parallel Configuration
• Product rule for unreliabilities
n
RS (t )  1   1  Ri (t ) 
i 1
•Mean Time Between System Failures

MTBFS   R S (t)dt
0
32
Stracener_EMIS 7370/STAT 5340_Fall 08_09.30.08
Parallel Reliability Configuration
s
p=R(t)
33
Parallel Reliability Configuration with Exponential Distribution
Element time to failure is exponential with failure rate 
• Reliability block diagram:
E1
E2
•Element Time to Failure Distribution
1
Ti ~ Eθ with failure rate λ 
for I=1,2.
θ
• System reliability
RS (t )  2e  t  e 2t
• System failure rate
 1  e  t 

h S ( t )  2
 t 
 2e 
34
Parallel Reliability Configuration with Exponential Distribution
• System Mean Time Between Failures:
MTBFS = 1.5 
35
Example - 3 Configurations
Compare the following reliability configurations I, II, and III in
terms of (a) system reliability, (b) system failure rate, (c) system
mean time between failures and (d) system mean time between
maintenance, assuming that a failure requires maintenance.
Element E has an exponential distribution of time to failure, T,
with failure rate  = 0.01. State all ground rules and
assumptions, show all work and present the results graphically
to convey your results.
I.
II.
III.
E
E
E
E
E
36
Example - 3 Configurations - Solution
For the baseline system:
 λt


RS t  e ,
E
h S t   λ,
MTBFS  θ,
and
MTBM S  θ
37
Example - 3 Configurations – Solution
E
For alternative A:
R SA t   2e  λt - e 2 λt ,

1 e
t   2λ
2  e
0.01t
h SA
0.01t
E
,

MTBFSA  1.5θ.
and
MTBM SA
1 θ

  0.5θ
2λ 2
38
Example - 3 Configurations – Solution
For alternative B:
E
E
 λt




R SB t  1  λt e ,
λ t
h SB t  
,
1  λt
MTBFSB  2θ
2
and
MTBM SB  θ
39
Example - 3 Configurations – Solution
Now we compare Alternatives A & B to the baseline system.
In terms of reliability,
R SA t  2e - e

R S t 
e  λt
 λt
2 λt
 2-e
 λt
and
R SB t  1  λt e

 λt
R S t 
e
 λt
 1  λt
40
Example - 3 Configurations – Solution
In terms of failure rate,

1 e
2λ
t   2  e
0.01t
h SA
h S t 
λ
 0.01t

  2 1  e
2  e
 0.01t
 0.01t


and
λ t
h SB t  1  λt
λt
1



h S t 
λ
1  λt 1  1
λt
2
41
Example - 3 Configurations – Solution
In terms of MTBF,
and
MTBFSA 1.5θ

 1.5
MTBFS
θ
MTBFSB 2θ

2
MTBFS
θ
In terms of MTBM, MTBM SA
MTBM S
and
0.5θ

 0.5
θ
MTBM SB θ
 1
MTBM S θ
42
Example - 3 Configurations – Solution
8
7
6
Rs
5
Rsa
4
Rsb
Rsa/Rs
3
Rsb/Rs
2
1
0
0
100
200
300
400
500
600
700
43
Example - 3 Configurations – Solution
1.2
1
0.8
hs
hsa
0.6
hsb
hsa/hs
hsb/hs
0.4
0.2
0
0
100
200
300
400
500
600
700
44
Example
A system consists of five components connected as shown.
Find the system reliability, failure rate, MTBF, and MTBM if
Ti~E(λ) for i=1,2,3,4,5
E2
E1
E3
E4
E5
45
Solution
This problem can be approached in several different ways. Here is
one approach:
There are 3 success paths, namely,
Success Path Event
E1E2
A
E1E3
B
E4E5
C
Then Rs(t)=Ps= P ( A  B  C )
=P(A)+P(B)+P(C)-P(AB)-P(AC)-P(BC)+P(ABC)
=P(A)+P(B)+P(C)-P(A)P(B)-P(A)P(C)-P(B)P(C)+
P(A)P(B)P(C)
=P1P2+P1P3+P4P5-P1P2P3-P1P2P4P5
-P1P3P4P5+P1P2P3P4P5
assuming independence and where Pi=P(Ei) for i=1, 2, 3, 4, 5 46
Since Pi=e-λt for i=1,2,3,4,5
Rs(t)
 (e -λt )(e -λt )  (e -λt )(e -λt )  (e -λt )(e -λt ) - (e -λt )(e -λt )(e -λt )
- (e -λt )(e -λt )(e -λt )(e -λt ) - (e -λt )(e -λt )(e -λt )(e -λt )
 (e -λt )(e -λt )(e -λt )(e -λt )(e -λt )
 3e -2λt  e -3λt  2e -4λt  e -5λt
 dtd Rs (t )

hs(t)
Rs (t )
6e -2λt  3λe-3λt  8λe-4λt  5λe-5λt

e  2 λt (3  e -λt  2e -2λt  e -3λt )
 6  3e -λt  8e -2λt  5e -3λt
 λ
- λt
- 2λt
-3λt
3

e

2
e

e




47

MTBFs   Rs (t )dt
0
 3e
e
e
e
  



3λ
2λ
5λ
 2λ
3
1
1
1

 

2λ 3λ 2λ 5λ
 45  10  15  6 

θ  0.87θ
30


- 2λt
1
MTBM S 
 0.2θ
5λ
-3λt
- 4λt
-5λt



0
48