Reliability Engineering
- Part 1
Product Probability Law of Series
Components
n
R   Ri
i 1
If a system comprises a large number of components,
the system reliability may be rather low, even
though the individual components have high
reliabilities, e.g. V-1 missile in WW II.
Transition of Component States
Normal
state continues
Component fails
N
F
Component is repaired
Failed
state
continues
The Repair-to-Failure Process
Definitions of Reliability
• The probability that an item will
adequately perform its specified purpose for
a specified period of time under specified
environmental conditions.
• The ability of an item to perform an
required function, under given
environmental and operational conditions
and for a stated period of time. (ISO 8402)
Definition of Quality
The totality of features and characteristics of a
product or service that bear on its ability to satisfy
or implies needs (ISO 8402).
Quality denotes the conformity of the product to its
specification as manufactured, while reliability
denotes its ability to continue to comply with its
specification over its useful life. Reliability is
therefore an extension of quality into the time
domain.
REPAIR -TO-FAILURE PROCESS
MORTALITY DATA
t=age in years ; L(t) =number of living at age t
t
L(t)
0 1,023,102
1 1,000,000
2
994,230
3
990,114
4
986,767
5
983,817
10 971,804
t
L(t)
t
L(t)
t
L(t)
15
20
25
30
35
40
45
962,270
951,483
939,197
924,609
906,554
883,342
852,554
50
55
60
65
70
75
80
810,900
754,191
677,771
577,822
454,548
315,982
181,765
85
90
95
99
78,221
21,577
3,011
125
After Bompas-Smith. J.H. Mechanical Survival : The Use of Reliability
Data, McGraw-Hill Book Company, New York , 1971.
HUMAN RELIABILITY
t
Age in Years
0
1
2
3
4
5
10
15
20
25
30
40
45
50
55
60
65
70
75
80
85
90
95
99
100
L(t), Number Living at
Age t
R(t)=L(t)/N F(t)=1-R(t)
1,023,102
1,000,000
994,230
986,767
983,817
983,817
971,804
962,270
951,483
939,197
924,609
883,342
852,554
810,900
754,191
677,771
577,882
454,548
315,982
181,765
78,221
21,577
3,011
125
0
1.
0.9774
0.9718
0.9645
0.9616
0.9616
0.9499
0.9405
0.9300
0.9180
0.9037
0.8634
0.8333
0.7926
0.7372
0.6625
0.5648
0.4443
0.3088
0.1777
0.0765
0.0211
0.0029
0.0001
0.
0.
0.0226
0.0282
0.0322
0.0355
0.0384
0.0501
0.0595
0.0700
0.0820
0.0963
0.1139
0.1667
0.2074
0.2628
0.3375
0.4352
0.5557
0.6912
0.8223
0.9235
0.9789
0.9971
0.9999
1.
repair= birth
failure = death
Meaning of R(t):
(1) Prob. Of Survival (0.86)
of an individual of an
individual to age t (40)
(2) Proportion of a
population that is
expected to Survive to a
given age t.
1.0
P
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
10
20
30
40
50
60
70
80
90
100
Time to Failure
The time elapsing from when the unit is put into
operation until it fails for the first time, i.e. a random
variable T. It may also be measured by indirect time
concepts:
• The number of times a switch is operated
• The number of kilometers driven by a car
• The number of rotations of a bearing
• etc.
State Variable
The state of the unit at time t can be described by the
state variable
if the unit is functioning at time t
1
X (t )  
0 if the unit is in a failed state at time t
Reliability, R(t)
= probability of survival to (inclusive) age t
= the number of surviving at t divided by the
total sample
Unreliability, F(t)
= probability of death to age t (t is not included)
=the total number of death before age t divided
by the total population
Reliability - R(t)
• The probability that the component
experiences no failure during the the time
interval (0,t] or, equivalently, the probability
that the unit survives the time interval (0, t]
and is still functioning at time t.
R (t )  Pr(T  t )
lim R (t )  1
t 0
lim R (t )  0
t 
• Example: exponential distribution
R(t )  e
 t
Unreliability - F(t)
• The probability that the component
experiences the first failure during (0,t].
F (t )  Pr(T  t )
R(t )  F (t )  1
lim F (t )  0
t 0
lim F (t )  1
t 
• Example: exponential distribution
F (t )  1  e
 t
FALURE DENSITY FUNCTION f(t)
n(t  )  n(t )
f (t ) 
Age in Years
0
1
2
3
4
5
10
15
20
25
30
35
40
45
50
60
65
70
75
80
85
90
95
99
100
23,102
5,770
4,116
3,347
2,950
12,013
9,543
10,787
12,286
14,588
18,055
23,212
30,788
41,654
56,709
99,889
123,334
138,566
134,217
103,554
56,634
18,566
2,886
125
0
n(t   )  n(t )
N
0.02260
0.00564
0.00402
0.00327
0.00288
0.00235
0.00186
0.00211
0.00240
0.00285
0.00353
0.00454
0.00602
0.00814
0.01110
0.01500
0.01950
0.02410
0.02710
0.02620
0.02020
0.01110
0.00363
0.00071
000012
f (t ) 
dF (t )
dt
0.00540
0.00454
0.00284
0.00330
0.00287
0.00192
0.00198
0.00224
0.00259
0.00364
0.00393
0.00436
0.00637
0.00962
0.01367
0.01800
0.02200
0.02490
0.02610
0.02460
0.01950
0.00970
0.00210
-
140
Number of Deaths (thousands)
120
100
80
60
40
20
20
40
60
80
100
Age in Years (t)
0.14
0.12
Failure Density f (t)
0.10
0.8
0.6
0.4
0.2
20
40
60
80
100
Age in Years (t)
Failure Density - f(t)
f (t )t  Pr(t  T  t  t )
dF (t )
 t
 e
f (t ) 
dt
t
F (t )   f (u )du
0

R (t )   f (u )du
t
(exponential
distribution)
number of deaths during [t , t  ) f (t )
r (t ) 

number of survivals at age t
R(t )
CALCULATION OF FAILURE RATE r(t)
Age in Years
0
1
2
3
4
5
10
15
20
25
30
35
No. of Failures r(t)= f (t )
1  F (t )
(death)
23,102
5,770
4,116
3,347
2,950
12,013
9,534
10,787
12,286
14,588
18,055
23,212
0.02260
0.00570
0.00414
0.00338
0.00299
0.00244
0.00196
0.00224
0.00258
0.00311
0.00391
0.0512
Age in Years
No. of Failures
r(t)=
f (t )
1  F (t )
(death)
40
45
50
55
60
65
70
75
80
85
90
95
99
30,788
41,654
56,709
76,420
99,889
123,334
138,566
134,217
103,554
56,634
18,566
2,886
125
0.00697
0.00977
0.01400
0.02030
0.02950
0.04270
0.06100
0.08500
0.11400
0.14480
0.17200
0.24000
1.20000
Random failures
Early failures
Wearout failures
0.2
0.15
0.1
0.05
20
40
60
80
Failure rate r(t) versus t.
100
Failure Rate,

(faults/time)
Period of Approximately Constant
failure rate
Infant Mortality
Old Age
Time
Figure 11-2 A typical “bathtub” failure rate curve for process hardware. The
failure rate is approximately constant over the mid-life of the component.
Comments on Bathtub Curve
• Often units are tested before they are
distributed to the users. Thus, much of the
infant mortality will be removed before the
units are delivered for use.
• For the majority of mechanical units, the
failure rate will usually show a slightly
increasing tendency in the useful life period.
Failure Rate - r(t)
• The probability that the component fails per
unit time at time t, given that the component
has survived to time t.
f (t )
f (t )
r (t ) 

R(t ) 1  F (t )
• Example: r (t )  
The component with a constant failure rate is considered
as good as new, if it is functioning.
As Good As New?
Pr T  t  x e   (t  x )
Pr T  t  x | T  t 
  t  e   x  Pr T  x
Pr T  t
e
This implies that the probability that a unit will be
functioning at time t+x, given that it is functioning
at time t, is equal to the probability that a new unit
has a time to failure longer than x. Hence the
remaining life of a unit, functioning at time t, is
independent of t. The exponential distribution has
no “memory.”
Relation Between Reliability
and Failure Rate
t
-dR / dt
r (t ) 
 R(t )  exp [-  r (u ) du ]
0
R
t
f (t )  r (t )exp [-  r (u )du ]
0
r (t )  t  Pr(t  T  t  t | T  t )
Interpretation of Failure Rate
In actuarial statistics, the failure rate is called
the force of mortality (FOM). The failure
rate or FOM is a function of the life
distribution of a single unit and an
indication of the “proneness of failure” of
the unit after time t has elapsed.
Failure-Rate Experiment
Split the time interval (0, t) into disjoint intervals of
equal length dt. Then put n identical units into
operation at time t=0. When a unit fails, record
the time and leave that unit out. For each
interval, determine
1. The number of units n(i) that fail in interval i.
2. The functioning times of the individual units in
interval i. If a unit has failed before interval i, its
functioning time is zero.
Failure-Rate Experiment
T ji  functioning time of unit j in interval i
n
T
j 1
ji
r (i ) 
 total functioning time for all units in interval
n(i )
n
T
j 1
 failure rate
ji
let m(i ) denotes the number of units which are functioning
at the start of interval i.
n(i )
z (i ) 
m(i )t

n(i )
z (i )t 
m(i )
Mean Time to Failure - MTTF
MTTF   tf (t )dt    tR(t )dt


0
0


0
0
  tR(t ) 0   R(t )dt   R (t )dt


1

Variance of Time to Failure

Var T    (t  MTTF ) 2 f (t )dt
0

1
2
Failure Rate
Failure Density
Unreliability
1
f (t) Area = 1

F (t)
1
 f t dt
t
0
t
(a)
t
(b)
0
Reliability
t
(c)
R (t)
0
1 - F (t)
t
(d)
Figure 11-1 Typical plots of (a) the failure rate  (b) the failure density
f (t), (c) the unreliability F(t), and (d) the reliability R (t).
TABLE 11-1: FAILURE RATE DATA FOR
VARIOUS SELECTED PROCESS COMPONENTS1
Instrument
Fault/year
Controller
Control valve
Flow measurement (fluids)
0.29
0.60
1.14
Flow measurement (solids)
Flow switch
Gas - liquid chromatograph
3.75
1.12
30.6
Hand valve
Indicator lamp
Level measurement (liquids)
0.13
0.044
1.70
Level measurement (solids)
Oxygen analyzer
pH meter
6.86
5.65
5.88
Pressure measurement
Pressure relief valve
Pressure switch
1.41
0.022
0.14
Solenoid valve
Stepper motor
Strip chart recorder
0.42
0.044
0.22
Thermocouple temperature measurement
Thermometer temperature measurement
Valve positioner
0.52
0.027
0.44
1Selected
from Frank P. Lees, Loss Prevention in the Process Industries
(London: Butterworths, 1986), p. 343.
Example
Consider two independent components with failure
ratesλ1andλ2, respectively. Determine the probability
that component 1 fails before component 2.

Pr T2  T1   Pr T2  t | T1  t fT1 (t )dt
0

Similarly,
 e
 2t
0

1e
 1t

dt  1  e  ( 1  2 )t dt
0
1
1  2
Pr component j fails first among n components 
j
n

i 1
i
A System with n Components
in Parallel
• Unreliability
• Reliability
n
F   Fi
i 1
n
R  1  F  1   (1  Ri )
i 1
A System with n Components
in Series
• Reliability
• Unreliability
n
R   Ri
i 1
n
F  1  R  1   (1  Fi )
i 1
Upper Bound of Unreliability
for Systems with n Components
in Series
n
n
i 1
F   Fi   Fi Fj    (1)
i 1
n
  Fi
i 1
i  2 j 1
n 1
n
F
l
l 1
The Poisson Process
Assumptions of Homogeneous
Poisson Process (HPP)
Suppose we are studying the occurrence of a certain event A in the
course of a given time period. Let us assume
1. A can occur at any time in the interval. The probability of A
occurring in the interval (t , t  t ] is independent of t and
may be written as   t   (t ) where  is a positive constant.
2. The probability of more than one event A in this interval
is  (t ) , which is a function with property
 (t )
lim
t 0
3.
t
0
Let (t11,t12], (t21,t22],…be any sequence of disjoint
intervals in the time period in question. Then the events “A
occurs in (tj1,tj2],” j= 1, 2, …, are independent.
The process is said to have intensityλ
Probability of No Event Occurring
in (0, t]
Let N(t) denotes the number of times the event A occurs
during the period (0, t]. Let p(n, t )  Pr N (t )  n
p(0, t  t )  p(0, t ) 1  t   (t ) 
p(0, t  t )  p(0, t )
 (t )
  p(0, t ) 
p(0, t )
t
t
d
let t  0

p(0, t )   p(0, t )
dt
 t
From p(0, 0)  1  p(0, t )  e
for t  0
Exponential Distribution
Let T1 denotes the time point when A occurs for the
first time. T1 is a random variable and
FT1 (t )  Pr{T1  t}  1  Pr{T1  t}  1  p(0, t )
Thus,
1  e  t
for
t0
FT1 (t )  
otherwise
 0
 e   t
for
t0
fT1 (t )  
otherwise
 0

1
E[T1 ]   tfT1 (t )dt 

Thus, thee waiting time T between consecutive occurrences in a
HPP is exponentially distributed.
0
Probability of n Events Occurring
in (0, t]
p (n, t  t )  p (n, t ) 1  t   ( t )   p ( n  1, t )t
 p (n, t )  t  p (n  1, t )  p (n, t )  p (n, t ) (t )
p (n, t  t )  p (n, t )
 (t )
   p (n  1, t )  p( n, t )   p( n, t )
t
t
d
p (n, t )    p (n  1, t )  p (n, t )  ;
p( n, 0)  0 for n  1
dt
d
p (1, t )    p (0, t )  p (1, t )    e  t   p (1, t )
dt
 p (1, t )  te  t
( t ) n   t
 p ( n, t ) 
e
for n  0,1, 2,
n!
Poisson Distribution
Since p(n, t )  Pr N (t )  n
This distribution is called the Poisson distribution with
parameter t and random variable n
(t )  t
E  N (t )   n 
e  t
n!
n 0

n
Notice that, since the expected number of occurrences
of event A per unit time (t=1) is  ,  expresses the
intensity of the process.
Example 1
Suppose that exactly one event (failure) of a
HPP with intensityλis known to have
occurred in the interval (0, t0]. Determine
the distribution of the time T1 at which this
event occurred.
Example 1
Pr T1  t N (t0 )  1
Pr T1  t | N (t0 )  1 
Pr  N (t0 )  1
Pr 1 event in (0, t ] 0 events in (t , t0 ]

Pr  N (t0 )  1
Pr 1 event in (0, t ]}  Pr{0 events in (t , t0 ]

Pr  N (t0 )  1
te  t e   (t t ) t
=

 t
 t0 e
t0
0
0
for 0  t  t0
In other words, the time at which the first failure occurs in
uniformly distributed over (0,t0]. The expected time is thus
E T1 | N (t0 )  1  t0 / 2
Example 2
Suppose that the failure of a system are
occurring in accordance with a HPP. Some
failures develop into a consequence C, and
others do not. The probability of this
development is p and is assumed to be
constant for each failure. The failure
consequences are further assumed to be
independent of each other. Determine the
distribution of the consequences.
Example 2
M (t )  the number of C failures in (0,t]
n m
Pr M (t )  m | N (t )  n    p (1  p) n m
m
m  0,1, 2, , n
n m
n m  t 
e  t
Pr M (t )  m     p (1  p)
n!
nm  m 
( pt ) m  pt
e

m!

n
Thus, M(t) is also a HPP with intensity (pλ) .
The mean number of C consequences in (0,t] is (pλt).
Gamma Distribution
Consider a unit that is exposed to a series of shocks
which occur a HPP with intensity λ. The time
intervals T1, T2, T3, …, between consecutive shocks
are then independent and exponentially distributed
with parameter λ. Assume that the unit fails exactly
at the kth shock, and not earlier. The time to failure
of the unit
T  T1  T2 
is then gamma distributed (k, λ).
 Tk
Gamma Distribution
Consider a homogeneous Poisson distribution,
k 1
FTk (t )  Pr Tk  t  1  Pr Tk  t  1   p( j, t )
j 0
( t )   t
FTk (t )  1  
e
j!
j 0
k 1
fTk (t ) 
dFTk (t )
dt
j


(k  1)!
( t )
k 1  t
e


( k )
(t ) k 1 e t
The waiting time until the kth occurrence of event A in a
HPP with intensity λis gamma distributed. The gamma
distribution (1, λ) is an exponential distribution with
parameter λ.
Weibull Distribution
For majority of mechanical units, the failure
rates are slightly increasing in the useful life
period (not constant). A distribution often
used when r(t) is monotonic is the Weibull
distribution. The time to failure T of a unit
is said to be Weibull distributed with scale
parameterλand shape parameterα.
Weibull Distribution
1  e
F (t )  Pr T  t  
0

 (  t )
  t  e
d
f (t )  F (t )  
dt
0

 1
R(t )  Pr T  t  e
 (  t )
f (t )
 1
   t 
r (t ) 
R(t )
t 0
otherwise
 (  t )
t 0
otherwise
for t  0
for t  0
Weibull Distribution
1. α=1: The Weibull distribution reduces to
exponential distribution.
2. α>1: The failure rate is increasing.
3. α<1: The failure rate is decreasing.
Weibull Distribution
R t 1

1

e
Hence, the probability of time to failure larger than 1/λ
is independent of α. This quantity is called the
Characteristic Lifetime.
MTTF  

0
1 1 
R (t )dt     1
  
1  2 

2 1
Var T   2    1     1 
   
  
Three-Parameter Weibull
Distribution
1  e [  (t  )]
F (t )  Pr T  t  
0


   (t   ) 
d
f (t )  F (t )  
dt
0

 1
R(t )  Pr T  t  e
[  ( t  )]
f (t )
 1
    (t   ) 
r (t ) 
R(t )
t 
otherwise
e
[  ( t  )]
t 
otherwise
for t  
for t  
Normal Distribution
The normal distribution is sometimes used as a
lifetime distribution, even though it allows negative
TTF values with positive probability.
 (t   ) 2 
1
f (t ) 
exp  

2
2

 2


  t  
The probability density of standard normal
distribution is
 t2 
1
 (t ) 
exp   
2
 2
Normal Distribution
The unreliability, reliability and failure rate can be
written as
t 
t 
F (t )  Pr T  t    ( )d   


  
t 
R (t )  1  F (t )  1   




t 


R(t ) 1



r (t )  

R (t ) 
t 
1  

  

Normal Distribution, Left
Truncated at 0
  t 


 

R(t )  Pr T  t | T  0 

 
 
t 


R(t ) 1
 

r (t ) 

R(t ) 
t 
1  

  
for t  0
Log Normal Distribution
The time to failure T of a unit is said to be
lognormally distributed if Y=ln(T) is normally
distributed.
The lognormal distribution is commonly used as a
distribution for repair time. When modeling the
repair time, it is natural to assume that the repair
rate in increasing in a first phase. When the repair
has been going on for a rather long time, this
indicates serious problems. It is thus natural to
believe that the repair rate is decreasing after a
certain period of time.
Log Normal Distribution
2

1
1  log t    
f (t ) 
exp  
 
t 2
 
 2  
2

t1
1
1  log t    
F (t ) 
exp  
dt



 2 0 t
 
 2  
f (t )
r (t ) 
1  F (t )
1 2

MTTF  exp     
2 
