Transcript Exponential & Weibull Distributions

Systems Engineering Program
Department of Engineering Management, Information and Systems
EMIS 7370/5370 STAT 5340 :
PROBABILITY AND STATISTICS FOR SCIENTISTS AND ENGINEERS
Special Continuous Probability Distributions
-Exponential Distribution
-Weibull Distribution
Dr. Jerrell T. Stracener, SAE Fellow
Leadership in Engineering
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Exponential Distribution
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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) 

0
e

x

,
for x 0
,
elsewhere
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Properties of the Exponential Model
• Probability Distribution Function
for x< 0
F (x)  P(X  x) 
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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Exponential Model - Example
Suppose the response time X at a certain on-line computer
terminal (the elapsed time between the end of a user’s
inquiry and the beginning of the system’s response to that
inquiry) has an exponential distribution with expected
response time equal to 5 sec. The E(X) = 5=θ,
so λ = 0.2.
(a) What is the probability that the response time is at most 10
seconds?
(b) What is the probability that the response time is between 5 and 10
seconds?
(c) What is the value of x for which the probability of exceeding that
value is 1%?
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Exponential Model - Example
The probability that the response time is at most 10 sec is:
P ( X  10)
 F (10,0.2)
 1  e (.2 )(10)
 1  0.135
 0.865
The probability that the response time is between 5 and 10 sec is:
P(5  X  10)
 F (10;0.2)  F (5;0.2)
 (1  e  2 )  (1  e 1 )
 0.233
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Exponential Model - Example
The value of x for which the probability of exceeding x is 1%:
P( X  x)  1  e  x  0.99
e x  0.01
 λx  ln( 0.01)
4.605
x
0 .2
x  23.025 sec
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Weibull Distribution
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The Weibull Probability Distribution Function
•Definition - A random variable X is said to have the
Weibull Probability Distribution with parameters  and ,
where  > 0 and  > 0, if the probability density function of x

is:
x
f ( x) 
  1   
x
e


,
0
,
forx  0
elsewhere
Where,  is the Shape Parameter,  is the Scale
Parameter. Note: If  = 1, the Weibull reduces to
the Exponential Distribution.
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The Weibull Probability Distribution Function
Probability Density Function
f(t)
1.8
β=5.0
1.6
1.4
1.2
β=0.5
β=3.44
β=1.0
β=2.5
1.0
0.8
0.6
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 2.2 2.4
t
t is in multiples of 
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The Weibull Probability Distribution Function
for x  0
F(x)  PX  x   1 - e
x
 
θ
β
F(t) for various  and  = 100
F(x)
probability, p
1
5
0.8
3
1
0.6
  0.5
0.4
0.2
0
0
50
100
x
Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
150
200
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Weibull Probability Paper (WPP)
• Derived from double logarithmic transformation of
the Weibull Distribution Function.
 ( t /  )
• Of the form
where
F(t )  1  e
y  ax  b




1
y  ln ln 


1

F
(
t
)




a   b   ln  x  ln t
•Any straight line on Weibull Probability paper is a Weibull
Probability Distribution Function with slope,  and intercept,
- ln , where the ordinate is ln{ln(1/[1-F(t)])} the abscissa is
ln t.
Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
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Weibull Probability Paper (WPP)
Weibull Probability Paper links
http://perso.easynet.fr/~philimar/graphpapeng.htm
http://www.weibull.com/GPaper/index.htm
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Use of Weibull Probability Paper
8 4 3
2
1.5
1.0
0.8
0.7
0.5

99.0
95.0
90.0
80.0
70.0
F(x)
in %
Cumulative probability in percent
50.0
40.0
30.0
1.8 in.  
20.0
1 in.
10.0
5.0
4.0
3.0
2.0
1.0
0.5
10
2
3
4 5 6 7 8 9 10
2

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3
4 5 6 7 8 9 1000
15
x
Properties of the Weibull Distribution
• 100pth Percentile
x p   - ln(1 - p)
1

and, in particular
x0.632  
• Mean or Expected Value
1 
  E(X)     1
 
Note: See the Gamma Function Table to
obtain values of (a)
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Properties of the Weibull Distribution
• Standard Deviation of X
  2  2  1 
     1     1
  
  
1
2
where
 (a)  (a)
2
2
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The Gamma Function 

(a )   e x dx
x
a 1
0
(a  1)  a(a )
Values of the
Gamma Function
y=a
1
1.01
1.02
1.03
1.04
1.05
1.06
1.07
1.08
1.09
1.1
1.11
1.12
1.13
1.14
1.15
1.16
1.17
1.18
1.19
1.2
1.21
1.22
1.23
1.24
 (a)
1
0.9943
0.9888
0.9836
0.9784
0.9735
0.9687
0.9642
0.9597
0.9555
0.9514
0.9474
0.9436
0.9399
0.9364
0.933
0.9298
0.9267
0.9237
0.9209
0.9182
0.9156
0.9131
0.9108
0.9085
a
1.25
1.26
1.27
1.28
1.29
1.3
1.31
1.32
1.33
1.34
1.35
1.36
1.37
1.38
1.39
1.4
1.41
1.42
1.43
1.44
1.45
1.46
1.47
1.48
1.49
 (a)
0.9064
0.9044
0.9025
0.9007
0.899
0.8975
0.896
0.8946
0.8934
0.8922
0.8912
0.8902
0.8893
0.8885
0.8879
0.8873
0.8868
0.8864
0.886
0.8858
0.8857
0.8856
0.8856
0.8858
0.886
a
1.5
1.51
1.52
1.53
1.54
1.55
1.56
1.57
1.58
1.59
1.6
1.61
1.62
1.63
1.64
1.65
1.66
1.67
1.68
1.69
1.7
1.71
1.72
1.73
1.74
 (a)
0.8862
0.8866
0.887
0.8876
0.8882
0.8889
0.8896
0.8905
0.8914
0.8924
0.8935
0.8947
0.8959
0.8972
0.8986
0.9001
0.9017
0.9033
0.905
0.9068
0.9086
0.9106
0.9126
0.9147
0.9168
a
1.75
1.76
1.77
1.78
1.79
1.8
1.81
1.82
1.83
1.84
1.85
1.86
1.87
1.88
1.89
1.9
1.91
1.92
1.93
1.94
1.95
1.96
1.97
1.98
1.99
2
 (a)
0.9191
0.9214
0.9238
0.9262
0.9288
0.9314
0.9341
0.9369
0.9397
0.9426
0.9456
0.9487
0.9518
0.9551
0.9584
0.9618
0.9652
0.9688
0.9724
0.9761
0.9799
0.9837
0.9877
0.9917
0.9958
1
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Properties of the Weibull Distribution
• Mode - The value of x for which the probability
density function is masimum
i.e.,
f x mode   max f ( x)
1
x mode   1  1  
f(x)
Max f(x)=f(xmode)
0
xmode
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x
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Weibull Distribution - Example
Let X = the ultimate tensile strength (ksi) at -200
degrees F of a type of steel that exhibits ‘cold
brittleness’ at low temperatures. Suppose X has a
Weibull distribution with parameters  = 20,
and  = 100. Find:
(a) P( X  105)
(b) P(98  X  102)
(c) the value of x such that P( X  x) = 0.10
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Weibull Distribution - Example Solution
(a)
P( X  105) = F(105; 20, 100)
 1 e
(b)
 (105/100) 20
 1  0.070  0.930
P(98  X  102)
= F(102; 20, 100) - F(98; 20, 100)
e
 ( 0.98) 20
e
 (1.02) 20
 0.513  0.226  0.287
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Weibull Distribution - Example Solution
(c)
P( X  x) = 0.10
P( X  x)
Then
e
 ( x / 100) 20
 1 e
 ( x /100) 20
 0.10
 0.90
( x / 100) 20   ln 0.90
( x / 100) 20   ln 0.90
x / 100   ln 0.90
1/ 20
x  100 ln 0.90
1/ 20
x  89.36
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Weibull Distribution - Example
The random variable X can modeled by a Weibull
distribution with  = ½ and  = 1000. The spec time limit is
set at x = 4000. What is the proportion of items not
meeting spec?
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Weibull Distribution - Example
The fraction of items not meeting spec is
PX  4000  1  P( X  4000)
 1  F(4000)
1/2
e
 4000 


1000


 e 2
 0.1353
That is, all but about 13.53% of the items will not meet
spec.
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An Application of Probability to
Reliability Modeling and Analysis
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Reliability Definitions and Concepts
• Figures of merit
• Failure densities and distributions
• The reliability function
• Failure rates
• The reliability functions in terms of the failure rate
• Mean time to failure (MTTF) and mean time between
failures (MTBF)
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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.
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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
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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
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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
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Reliability
Relationship between failure density and reliability
d
f t    R t 
dt
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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.
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The Reliability Function
The reliability of an item at time t may be expressed in terms
of its failure rate at time t as follows:

 0 h ( y ) dy
R ( t )  exp    h ( y)dy   e
 0

t
t
where h(y) is the failure rate
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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.
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Relationship Between MTTF and Failure Density
If T is the random time to failure of an item, the
mean time to failure, MTTF, of the item is

ET   MTTF   tf t dt
0
where f is the probability density function of time
to failure, iff this integral exists (as an improper
integral).
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Relationship Between MTTF and Reliability

MTBF  MTTF   R t dt
0
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Reliability “Bathtub Curve”
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Reliability Humor
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The Exponential Model: (Weibull Model with β = 1)
Definition
A random variable T is said to have the Exponential
Distribution with parameters , where  > 0, if the
failure density of T is:
1
f (t )  e

0

t

, for t  0
, elsewhere
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Probability Distribution Function
• Weibull W(, )
F( t )  1 - e
t
 


,
for t  0
Where F(t) is the population proportion failing in time t
• Exponential E() = W(1, )
F(t )  1 - e
t


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The Exponential Model
Remarks
The Exponential Model is most often used in
Reliability applications, partly because of mathematical
convenience due to a constant failure rate.
The Exponential Model is often referred to as the
Constant Failure Rate Model.
The Exponential Model is used during the ‘Useful Life’
period of an item’s life, i.e., after the ‘Infant Mortality’
period before Wearout begins.
The Exponential Model is most often associated with
electronic equipment.
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Reliability Function
Probability Distribution Function
• Weibull

R (t)  e
t
 

• Exponential
R (t)  e

t

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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 
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Mean Time Between Failure - MTBF
Weibull
1 
MTBF     1
 
Exponential
MTBF  
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The Gamma Function 

(a )   e x dx
x
a 1
0
(a  1)  a(a )
Values of the
Gamma Function
Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
y=a
1
1.01
1.02
1.03
1.04
1.05
1.06
1.07
1.08
1.09
1.1
1.11
1.12
1.13
1.14
1.15
1.16
1.17
1.18
1.19
1.2
1.21
1.22
1.23
1.24
 (a)
1
0.9943
0.9888
0.9836
0.9784
0.9735
0.9687
0.9642
0.9597
0.9555
0.9514
0.9474
0.9436
0.9399
0.9364
0.933
0.9298
0.9267
0.9237
0.9209
0.9182
0.9156
0.9131
0.9108
0.9085
a
1.25
1.26
1.27
1.28
1.29
1.3
1.31
1.32
1.33
1.34
1.35
1.36
1.37
1.38
1.39
1.4
1.41
1.42
1.43
1.44
1.45
1.46
1.47
1.48
1.49
 (a)
0.9064
0.9044
0.9025
0.9007
0.899
0.8975
0.896
0.8946
0.8934
0.8922
0.8912
0.8902
0.8893
0.8885
0.8879
0.8873
0.8868
0.8864
0.886
0.8858
0.8857
0.8856
0.8856
0.8858
0.886
a
1.5
1.51
1.52
1.53
1.54
1.55
1.56
1.57
1.58
1.59
1.6
1.61
1.62
1.63
1.64
1.65
1.66
1.67
1.68
1.69
1.7
1.71
1.72
1.73
1.74
 (a)
0.8862
0.8866
0.887
0.8876
0.8882
0.8889
0.8896
0.8905
0.8914
0.8924
0.8935
0.8947
0.8959
0.8972
0.8986
0.9001
0.9017
0.9033
0.905
0.9068
0.9086
0.9106
0.9126
0.9147
0.9168
a
1.75
1.76
1.77
1.78
1.79
1.8
1.81
1.82
1.83
1.84
1.85
1.86
1.87
1.88
1.89
1.9
1.91
1.92
1.93
1.94
1.95
1.96
1.97
1.98
1.99
2
 (a)
0.9191
0.9214
0.9238
0.9262
0.9288
0.9314
0.9341
0.9369
0.9397
0.9426
0.9456
0.9487
0.9518
0.9551
0.9584
0.9618
0.9652
0.9688
0.9724
0.9761
0.9799
0.9837
0.9877
0.9917
0.9958
46
1
Percentiles, tp
• Weibull
t P   - ln(1 - p)
1

and, in particular
t 0.632  
• Exponential
t P   - ln(1 - p)
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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.
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Failure Rates - Exponential
• Failure Rate
1
h(t)   

• Note:
1
MTBF 
failure rate
Only for the Exponential Distribution
•Cumulative Failure
H( t )  
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The Weibull Model - Distributions
Failure Rates
3
β=5
2
h(t)
β=1
1
β=0.5
0
0
1.0
2.0
t
t is in multiples of  h(t) is in multiples of 1/ 
50
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The Binomial Model - Example Application 1
Problem Four Engine Aircraft
Engine Unreliability Q(t) = p = 0.1
Mission success: At least two engines survive
Find RS(t)
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The Binomial Model - Example Application 1
Solution X = number of engines failing in time t
RS(t) = P(x  2) = b(0) + b(1) + b(2)
= 0.6561 + 0.2916 + 0.0486 = 0.9963
52
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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
Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
53
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
Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
54
Series Reliability Configuration
• Reliability Block Diagram
 Identical and independent Elements
 Exponential Distributions
E1
E2
En
• Element Time to Failure Distribution
1
with failure rate λ 
T ~ Eθ 
θ
• System reliability
R S ( t )  e  nt
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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
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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
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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
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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.
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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
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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
61
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Parallel Reliability Configuration
s
p=R(t)
62
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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 
Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
63
Parallel Reliability Configuration with Exponential Distribution
• System Mean Time Between Failures:
MTBFS = 1.5 
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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
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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 66
Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
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

Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08



67

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
Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
-3λt
- 4λt
-5λt



0
68