EEE 522 - RELIABILITY ENGINEERING (2 C)
Course Lecturer: Dr Lanre Olatomiwa
Course Content
1. Basic reliability concept: definition, need for reliability, reliability programme plan.
2. Continuous theoretical probability distribution; Exponential, Log-Normal and
Weibull.
3. Reliability measures; failure function, reliability function, hazard rate function (bathtub shape), mean time to failure (MTTF), mean operating time between failures
(MTBF).
4. Reliability block diagram; series, parallel, series-parallel, redundant systems and
complex.
5. Reliability assessment at design phase: Failure Modes Effects and Critical Analysis
(FMECA), Fault Tree Analysis (FTA).
6. Reliability tests.
7. Maintainability and Availability.
8. Economics of Reliability.
Intended Learning Outcome
By the end of this course, every student is expect to:
1) Apply engineering knowledge and specialist techniques to prevent or to reduce the
likelihood or frequency of failures.
2) Identify and correct the causes of failures that do occur, despite the efforts to prevent
them.
3) Determine ways of coping with failures that do occur, if their causes have not been
corrected.
4) Apply methods for estimating the likely reliability of new designs, and for analysing
reliability data.
1
1.1 Defination of Terms
1. Quality: The totality of features and characteristics of a product or service that bear on its
ability to satisfy stated or implied need.
N.B: The quality of a products is characterized not only by its conformity to specification at
the time of its supplied to the user, but also its ability to meet those specification over its entire
lifetime. However, according to common usage, quality denoted 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.
Parameter of Quality







Reliability
Availability
Maintainability ( Ability to be maintain)
Reparability (availability of repair expert)
Compatibility
Durability
Interchangeability of parts
R t 
2. Reliability,
,: The probability that an item/product will perform its intended function
(prescribe duty) without failure for specified period of time, when operated correctily in a
specified environment.
N.B: Reliability is the extension of quality into the time domain
Example:
Suppose we start a test at
t 0 with N 0 devices. After some time, N f devices of the total have
 N0  N f  Ns  .
failed and N s will have survived
The Reliability at any time t is given by;
R t  
Ns
N0
3. Reliability definition in term of Electric power system
Reliability definition of electric power system can be classify into parts:
a) Adequacy: This is the ability of electric power system to supply the aggregate
electrical demand and energy requirements of consumer at all time taking into
consideration the schedule and forced outages of the system components.
b) Security: The ability of electric power system to withstand sudden disturbances such
as electric short circuits or unanticipated loss of system components or elements.
4. Durability: This is the ability of an item to withstand the effects of time dependent mechanisms
such as fatigue, wear, corrosion, electrical parameter change, and so on. Durability is usually
2
expressed as a minimum time before the occurrence of wearout failures. In repairable systems it
often characterizes the ability of the product to function after repairs.
5. Availability: It is the ability of an item to perform its intended function at a specified instant of
time or over a stated period of time
6. Maintainability: The ability of an item under specified conditions of use, to be retained in, or
restored to a state in which it can perform its required functions when maintenance is performed
under stated conditions and using prescribed procedure and resources.
N.B: Maintainability is a main factor determining the availability of items.
7. Failure: The termination of the ability of the product to perform its intended function.
8. Failure Rate,
time.
 t 
  t    MTBF 

: The ratio of number of failures within a sample to the cumulative operating
1
1
MTBF
Example:
Q1: Samples of 1000 transistors are tested for a week, and two (2) of them failed. Assuming they
failed at the end of the week, what is the Failure Rate?
Solution:
 t  
Failure rate :
2 failures
1000* 24*7 hrs
 1.19e 5 failures / hr
h t 
9. Hazard Rate,
: The instantaneous probability of failure of an item given that it has survived
until that time. This is also called as “Instantaneous Failure Rate”.
10. Mean Time Between Failure (MTBF): for repairable systems, it is the ratio of the
cumulative operating time to the number of failures for that item.
i. For a repairable system, MTBF is the average time in service between
failures. Note
by the system.
that,
this
does
not
include
the
time
spent
at
repair
facility
Examples:
Q1: A motor is repaired and returned to service six (6) times during its life and provides
45,000 hours of service. Calculate MTBF.
Solution:
MTBF 
total operating time 45000

 7,500 hrs
no. of failures
6
Q2: If MTBF for a motor is 7,500 hours, what is the probability that it will operate for 30 days
without failure?
3
Solution 2:
1


*total operating time 

 MTBF

R  e*t  e
 30*24 hrs 


 7,500 hrs 
e
 0.908
11. Mean Time To Failure (MTTF):, for non-repairable items, it is the total number of life-units
of an item population divided by the number of failures within that population, during a particular
measurement interval under-stated conditions.
i. MTTF is the average life of a non-repairable system.
ii. For a repairable system, MTTF represents the average time before the
first failure.
1.2 Need for Reliability/ The objective reliability engineering study
Failure is inevitable for everything in the real world, and engineering systems are no exception. The
impact of failures varies from minor inconvenience and costs to personal injury, significant economic
loss, environmental impact, and deaths. Causes of failure include, bad engineering design, faulty
manufacturing, inadequate testing, human error, poor maintenance, improper use and lack of
protection against excessive stress. Designers, manufacturers and end users strive to minimize the
occurrence and recurrence of failures. In order to minimize failures in engineering systems, it is
essential to understand ‘why’ and ‘how’ failures occur. It is also important to know how often such
failures may occur.
Reliability deals with the failure concept where as the safety deals with the consequences after the
failure. Inherent safety systems/measures ensure the consequences of failures are minimal. Reliability
and safety engineering provides a quantitative measure of performance, identifies important
contributors, gives insights to improve system performance such as how to reduce likelihood of
failures and risky consequences, measures for recovery, and safety management. The objectives of
reliability engineering, in the order of priority, are:
5) To apply engineering knowledge and specialist techniques to prevent or to reduce the
likelihood or frequency of failures.
6) To identify and correct the causes of failures that do occur, despite the efforts to prevent them.
7) To determine ways of coping with failures that do occur, if their causes have not been
corrected.
8) To apply methods for estimating the likely reliability of new designs, and for analysing
reliability data.
1.3.
Application Area or Reliability
One main objective of reliability study is to provide information as a basis for decision making.
Reliability technology has a potentially wide range of application area. Some of these area includes:
1) Risk Analysis
Risk = (Probability of failure) x (exposure) x (consequence)
4
Causal
analysis
(b)
Consequence
analysis
Accidental Event
(a)
Methods
(b) Causal Analysis
(a) Accidental Event
- Fault Tree
- Checklist
analysis
- Preliminary hazard
- Reliability
analysis
Block Diagram
- FMECA
- FMECA
- Event data sources
- Reliability data
sources
(c)
(c) Consequence Analysis
- Event tree Analysis
- Consequence Model
- Reliability
assessments
- Simulation
Fig.1: Main steps of Risk Analysis with main method
The main steps in Quantitative Risk Analysis (QRA) is illustrated bellow:
(a) Identification and description of potential accidental events (e.g, gas lieak in an oil/gas
processing plants
(b) The potential causes of each accidental events are identified by a causal analysis using
e.g fault tree analysis
(c) The consequence analysis is usually carried out by event-tree analysis
2) Environmental Protection
Reliability studies may be used to improve the design and operational regularity of antipollution system, such as gas/water cleaning system
3) Quality
Reliability is one of the major parameter of quality of a product. Reliability is used in Total
Quality Management (TQM)
4) Optimization of Maintenance and Operation in Industry
The prime objective of maintenance is to maintain or improve the system reliability and
production/operation regularity. Maintenance is carried out to prevent system failures and
to restore the system function when a failure has occurred. Reliability Centered Maintenance
(RCM) approach is the main tools to improve the cost effectiveness and control of
maintenances in many industries, and hence to improve availability and safety.
5) Engineering Design
5
Reliability is considered to be one of the most important quality characteristics of technical
products. Reliability assurance should therefore be an important topic during the engineering
design process.
1.4.
Product Life-Cycle
Conceptualization
Custotomer
Design
After Sales
Services
Procurement/
Purchasing
Sales
Manufacturing/
Fabrication
Inspection and
Testing
Production
Fig 2. Product Life-Cycle
1.5.
Reliability Programme Plan
What are the actions that managers and engineers can take to influence reliability?
One obvious activity is quality assurance (QA), the whole range of functions designed to ensure that
delivered products are compliant with the design. For many products, QA is sufficient to ensure high
reliability, and we would not expect a company mass-producing simple diecastings for non-critical
applications to employ reliability staff. In such cases the designs are simple and well proven, the
environments in which the products will operate are well understood and the very occasional failure
has no significant financial or operational effect. QA, together with craftsmanship, can provide
adequate assurance for simple products or when the risks are known to be very low. Risks are low
when safety margins can be made very large, as in most structural engineering. Reliability
engineering disciplines may justifiably be absent in many types of product development and
manufacture. QA disciplines are, however, essential elements of any integrated reliability
programme.
A formal reliability programme is necessary whenever the risks or costs of failure are not low. Risks
of failure usually increase in proportion to the number of components in a system, so reliability
programmes are required for any product whose complexity leads to an appreciable risk.
6
An effective reliability programme should be based on the conventional wisdom of responsibility and
authority being vested in one person. Let us call him or her the reliability programme manager. The
responsibility must relate to a defined objective, which may be a maximum warranty cost figure, an
MTBF to be demonstrated or a requirement that failure will not occur. Having an objective and the
authority, how does the reliability programme manager set about his or her task, faced as he or she is
with a responsibility based on uncertainties? Brief outline is given below.






1.6.
The reliability programme must begin at the earliest, conceptual phase of the project. It is at
this stage that fundamental decisions are made, which can significantly affect reliability.
These are decisions related to the risks involved in the specification (performance,
complexity, cost, producibility, etc.), development time-scale, resources applied to evaluation
and test, skills available, and other factors.
The shorter the project time-scale, the more important is this need, particularly if there will be
few opportunities for an iterative approach. The activities appropriate to this phase are an
involvement in the assessment of these trade-offs and the generation of reliability objectives.
The reliability staff can perform these functions effectively only if they are competent to
contribute to the give-and-take inherent in the trade-off negotiations, which may be conducted
between designers and staff from manufacturing, marketing, finance, support and customer
representatives.
As the project proceeds from initial study to detail design, the reliability risks are controlled
by a formal, documented approach to the review of design and to the imposition of design
rules relating to selection of components, materials and processes, stress protection,
tolerancing, and so on. The objectives at this stage are to ensure that known good practices
are applied, that deviations are detected and corrected, and that areas of uncertainty are
highlighted for further action.
The programme continues through the initial hardware manufacturing and test stages, by
planning and executing tests to show up design weaknesses and to demonstrate achievement
of specified requirements and by collecting, analysing and acting upon test and failure data.
During production, QA activities ensure that the proven design is repeated, and further testing
may be applied to eliminate weak items and to maintain confidence. The data collection,
analysis and action process continues through the production and in-use phases.
Throughout the product life-cycle, therefore, the reliability is assessed, first by initial
predictions based upon past experience in order to determine feasibility and to set objectives,
then by refining the predictions as detail design proceeds and subsequently by recording
performance during the test, production and in-use phases. This performance is fed back to
generate corrective action, and to provide data and guidelines for future products.
Reliability Programme Activities
1) Define the system― Define the prescribed duty of the product i.e the goals of the product.
Here we also define the failure mode (i.e what are the diffident ways that the particular
products can fail to meet its intended purpose. At this stage we also define the appropriate
measure of reliability.
2) Decide on overall reliability target/objectives
7
3) Reliability apportionment― Apportion reliability to each component of the system in
order to determine the overall reliability of the system. This can be done through two
ways, namely
a) Similar-familiar technique― this getting information from similar or familiar
product that is currently in existence in the market to apportion the reliability to
each component.
b) Factor of influence― this is obtaining the information on the reliability by
looking at the following critical factors; weight, size, complexity, criticality and
state-of-the-art.
4) Reliability modelling and evaluation
5) Reliability improvement
6) Designs reviews and quality assurance
7) Component testing and data collection
8) Monitoring field performance
9) Writing reliability report
1.7.
Function of Reliability Engineer
1) Reliability estimation
2) Reliability prediction
3) Reliability growth plan
4) Reliability apportionment
5) Participation in all design reviews
6) Conducting reliability test
7) Perform statistical analysis on test data
8) Maintenance of relevant data system
9) Provision of assistance to production, quality assurance and purchasing department
10) Writing reliability specification
2. Continuous theoretical probability distribution
It is necessary to emphasize that all theoretical distributions represent the family of distributions
defined by a common rule through unspecified constants known as parameters of distribution. The
particular member of the family is defined by fixing numerical values for the parameters, which define
the distribution. The probability distributions most frequently used in reliability, maintainability and
supportability engineering are examined below:
Note: Parameters of a distribution can be classified in the following three categories (note that not
all distributions will have all the three parameters, many distributions may have either one or two
parameters):
1. Scale parameter, which controls the range of the distribution on the horizontal scale.
2. Shape parameter, which controls the shape of the distribution curves.
3. Source parameter or Location parameter, which defines the origin or the minimum value
which random variable, can have. Location parameter also refers to the point on the horizontal
axis where the distribution is located.
8
2.1.
Normal (Gaussian) Distribution
This is the most frequently used and most extensively covered theoretical distribution in the literature.
The Normal Distribution is continuous for all values of X between -∞ and + ∞. It has a characteristic
symmetrical shape, which means that the mean, the median and the mode have the same numerical
value. The normal data distribution pattern occurs in many natural phenomena, such as human
heights, weather patterns etc. The mathematical expression for normal distribution probability density
function (pdf) is given by:
( )=
(
) ⁄
exp −
where is the scale parameter, equal to the standard deviation (SD), while
is the location
parameter, equal to mean. The mode and the median are coincident with the mean as the pdf is
symmetrical. The influence of the parameter on the location of the distribution on the horizontal
axis is shown in Figure 1, where the values for parameter are constant
Figure 1: Probability density of normal distribution for different values of
An important reason for the wide use of normal distribution is the fact that whenever several random
variables are added together, the resulting sum tends to normal regardless of the distribution of the
variable being added. This is known as central limit theorem. It justifies the use of normal distribution
in many engineering applications including, quality control. Normal distribution is a close fit to most
quality control and some reliability observations, such as size of machined parts and the lives of items
subject to wearout failure.
The pdf of a normal distribution with parameter and can be calculated using Excel as ( ) =
( , , ,
) and the reliability as ( ) = 1 −
( , , ,
).
9
Advantages of Normal Distribution
1) It is widely use because resulting composite distribution of many additive sources of
variations approaches normal distribution
2) It is use to model various physical, mechanical, chemical or electrical properties of system
(e.g variation of electrical condenser, electrical power in a given area, Generator output
voltage, electrical resistance etc.)
2.2.
Log-Normal Distribution
When working with continuous distributions that show positive skewness, calculations using the
Gaussian (normal) distribution prove inadequate. When this situation occurs, and all the values are
(or transformed to be) > 0, taking the natural logs of the values may result in a normal distribution.
For the log-normal distribution:
f  x 
1
x l 2
e
 1  ln x   2 
l
 
 
 2 l  


, x0
The mean and variance of log-normal distribution are:
mean     e
1 2

 l   l 
2 

 
variance  2  e
 2 l  l2  e l2  1


Where l  mean of the natural logs of individuals
 l2  variance of natural logs of individuals
Note: Log-normal distribution is more versatile than normal distribution as it has wild range of
shapes, and therefore, is often a better fit to reliability data, such as for populations with ware out
characteristics. The reliability of a system following log-normal distribution with parameter
( , , ).
can also be calculated using Excel functions: ( ) = 1 −
Examples:
Q1: Twenty five measurements are taken of time to failure of a component in hours. The
natural logarithms are found to be normally distributed with an estimated mean
l  3.5
and
variance
 l2  1.3
(note
that
l
natural log values). Find the untransformed mean and variance in hours.
10
and
 l2
are
for
Solution:
m ean 
1 .3 

 3 .5 

2 

e
v a ria n c e  e 
 6 3 .4 3 4
2  3 .5   1 .3 
 e 1 .3  1 


 1 0 7 4 0 .9 1
Advantages of Lognormal Distribution
1)
2)
3)
4)
It is more versatile than normal distribution
It better fit population with wear out characteristics than normal distribution
It can be used for electrical installation life
It can be used to model situation where large occurrence are connected at the tail end of the
range e.g ( amount of electricity used by different consumers, downtimes of a system, timeto-repair, light intensity of a bulb etc).
2.3.
Exponential Distribution
Exponential distribution of Figure 2 is the most commonly used distribution in reliability, and
is generally used to predict the probability of surviving at a (t) time.
Figure 2: Exponential distribution is the most commonly used in reliability.
The probability density function (pdf) of the exponential distribution is:
11
f  t    e  t ,
t0
or
f  t   1   e t ,
where
t0
MTBF  
  rate of failure 
R  t   e  t
1

,
or et 
t  time
where
t0
F  t   Unreliability  Failure  1  R  t 
The hazard function for the exponential distribution is  , and is constant throughout the function.
Therefore, the exponential distribution should be used for reliability prediction during the rate of
constant failure or at random cause of failure or period of operation.
Some unique failures to the exponential distribution include:
1. The mathematical mean and standard deviation are equal.
2. Of all the values 63.21% fall below the mean value, which translates into only a 36.79%
probability of surviving past the time period of one MTBF.
3. The Reliability R(t) as the time t approaches zero, approaches to one as a limit.
Advantage of Exponential reliability
1)
2)
3)
4)
It is a very simple function,
Algebraic manipulation is simple since the failure rate is constant
It is very good for a reparable system
It can only be use at CFR phase of the bathtub curve.
Examples:
Q1: An equipment in a manufacturing plant has an MTBF
hr. What is the probability of operating for a period of 750 hours without failure?
Solution:
  0.000666, t  750
 0.000666  750 
R 750   e  t  e 
 e 0.5  0.6065
12
of
1500
Note: MTBF and λ do not need to be a function of time in hours. The characteristic of “time” or
usage can be such units as cycles instead of hours. In this case, MCBF (Mean Cycles Between
Failures) could be the appropriate measure.
Q2: One cycle of a machine completes the assembly of 20 units. A study of this machine predicted
an MCBF of 14,000 cycles (λ = 0.00007143/cycle). What is the probability of operating 15,000
cycles without failure?
Solution:
a.
R  t   e t 
or
e c  ; where c  cycles,
R  t   e  t
or
e c
or
b.
Using the first equation:
R15,000  e15,000 14000  e1.0714  0.3425
Q3: A component failure rate is constant and found to be 0.02per thousand hours. Calculate the
probability that the component will survive 10,000 hours.
Solution
=
.
= 2 × 10
ℎ
, t = 10,000
R(t) =
=
( ×
×
13
,
)
= .
2.3.1 Exponential Reliability of Series components
Consider a number of components arrange in series and each component with different failure rate
( ) as shown below. The reliability of the entire components connected in series is thus:
1
n
2
R1(t)
R2(t)
Rn(t)
( ) = 1( ) × 2( ) × − − − −×
( )=
×
× − − − −×
( )=
ℎ
( )
=
Example 1
An electronic component has an exponential life distribution with constant failure rate of 0.0002.
(a) What is the probability that the component will survive the first 300hrs of operation?
(b) What is the 300hrs reliability of the system comprising 4 of such components connected in
series?
Solution
= 0.0002 failure per hours, and t = 300hrs
(a)
(b)
( )=
(300) =
( )=
( .
ℎ
×
)
= 0.9418
=
Therefore, (300) =
=
( .
×
+
+
+
= 4(0.0002) = 0.0008
= 0.7866
Example 2 ( Non- censored experiment)
Ten electrical switches were put to test and the number of cycle (on and off of operation of a
switch) to failure counted. The number of cycles of failure are; 560,685, 820, 956, 1024,
1150,1689,1850,1900, 1956. Assuming constant failure rate, calculate the following:
(a)
(b)
(c)
(d)
Mean number of cycle to failure (MCTF)
Failure rate
Reliability at 300 cycles
Number of cycle at which the reliability will be 90%
14
Solution
(a)
=
=
560 + 685 + 820 + 956 + 1024 + 1150 + 1689 + 1850 + 1900 + 1956 12590
=
10
10
= 1259
(b) Failure rate,
=
=
= 7.94 × 10
/
(c) Reliability at 300 cycles
.
( )=
=
(d) No. of cycle at which reliability is 90%
×
×
= 0.7880
( )=
Take natural log of both sides:
( )=−
Therefore, =
( )
( . )
=
.
×
= 132.67 = 133
(
)
Example 3 (Censored experiment)
Fifteen (15) automobile tyres were tested to destruction and the distance to the first 8 failures (in
1,000km) at 10,000km were 2.92, 3.26, 5.61, 6.09, 7.71, 8.36, 9.55, and 9.65. Calculate:
(a)
(b)
(c)
(d)
Mean distance to failure (MDTF)
Failure rate
Reliability at 5000km
No. of distance at which the reliability is 50%
Note: In many cases when life data are analyzed, all of the component in the sample may not have
failed (i.e, the event of interest was not observed) or the exact time to failure of all the components
are not known. This is sometimes done for economic or other reasons. In censored test, all items are
activated at time t= 0, and follow until failure or until time (t), when the experiment is terminated.
Solution
=
(a) For a censored test,
Where,
9.65) × 1000km =
And
,
= (2.92 + 3.26 + 5.61 + 6.09 + 7.71 + 8.36 + 9.55 +
m
= 7 × 1000
= 70,000
N.B: The remaining 7tyres survived the test after 10,000km distance when the test is
terminated.
15
Therefore,
=
(b) Failure rate,
=
=
,
.
53,150 + 70,000
= 15,393.75
8
= 6.5 × 10
/
(c) Reliability at 5000km
. ×
×
( )=
=
(d) No. of distance at which the reliability is 50%
( )=
Take natural log of both sides
( )=−
Therefore, =
( )
=
( . )
. ×
= 0.7227
= 10,663.8
2.3.2 Exponential Reliability of Parallel components
Consider a number of components arrange in parallel and each component with different failure
rate, as shown below. The reliability of the entire component connected in parallel is thus:
R1(t)
R2(t)
Rn(t)
( ) = 1−
( ) = 1− 1−
(1 −
1−
( ))
− − − − − (1 −
Using the approx. formula:
=
1+ + +−−−+
(1)
Assumption: Assume that all parallel component have equal reliablity (i.e ):
( )=
Therefore,
( )=
=
But, ( ) =
16
( )
)
( )=−
Take natural log of both sides,
Therefore,
=
( )
(2)
Substituting eqn(2) in eqn(1) and find the reciprocal of
parallel system.
to obtain the failure rate of the entire
Example 1
The figure below shows the reliability diagram of a system consisting of 4 different units. The
reliability value for 12hrs of operation is as indicated. Each unit is assumed to have an exponential
life distribution model.
(a) Calculate the failure rate of each unit
(b) Obtain the total system reliability after 12hours of operation.
A
(0.92)
B
(0.92)
D
(0.95)
C
(0.92
Solution
Reliability of the each parallel system units are:
( )= ( )=
( ) = 0.95
While the reliability of the series units is
(a) Since the failure rate
( ) = 0.92
( )
=
, Therefore, for each of the parallel system the failure rate will
be,
=
Substituting for
=
=
( )
=
−
=
(0.92)
= 0.00695/ℎ
−12
in eqn (1):
1
=
1
1 1
1
1+ + +−−−+
2 3
Since n =3,
1
1
=
=
1
1+
1
6.95 × 10
1 1
+
2 3
11
= 263.79
6
17
Therefore,
=
=
.
= 3.79 × 10 /ℎ
( )
=
Likewise for the series components,
( .
=
)
= 4.27 × 10 /ℎ
(b) To obtain the total system reliablity after 12hours of operation, we first obtain the overall
for the entire system.
=
+
= 3.79 × 10 + 4.27 × 10 = 8.06 × 10
Total system reliablity after 12hours of operation,
(12) =
=
2.4.
( .
×
×
)
= .
Weibull Distribution
The Weibull distribution is arguably the most popular statistical distribution used by reliability
engineers. It has the great advantage in reliability work that by adjusting the distribution parameters,
it can be made to fit many life distributions.
The Weibull pdf is (in terms of time t):
  t  
f t   
   
 1
  t    
exp   
 
    
where  ,  ,   0 .
β(beta) = shape parameter, which determines distribution shape
η(eta) = scale parameter; 63.21% of the values fall below this parameter
γ(gamma) =location parameter
If failures start at time t  0 , then   0 and the pdf of the Weibull distribution becomes;
t
f t    
  
 1
  t  
exp     
    
By altering the shape parameter , the Weibull distribution takes different shapes. For example, when = 3.4
the Weibull approximates to the normal distribution; when =1, it is identical to the exponential distribution.
The cumulative distribution functions for the Weibull distribution is:
  t    
F  t   1  exp   
  for   0
    
18
  t  
F  t   1  exp      for   0
    
The corresponding reliability function is;
  t  
R  t   exp     
    
The hazard rate is;
t
h t    
  
 1
The mean or MTTF is;
1

 1
 
   
2 
1 
Standard deviation:       1     1
 
 
2
  x  = gamma function of a variable ( x ). Values of   x  are listed in the gamma
function tables.
Advantage of Weibull Distribution
1) It is two-parameter model hence more flexible than exponential model that has only one
parameter ( ).
2). It describe well the weakest link in data
3) It is amendable to graphical analysis
4) It can be used for life-test (i.e from early stage to aging period)
5) The value of obtain in Weibull distribution model can be used to determine the stage of the
product (i.e DFR, CRF, or IFR). If:



< 1,
= 1,
> 1,
(
(
(
19
Weibull Distribution Graphical Analysis
Hazard rate, ℎ( )
( )=
------------------------------(1)
While, Reliability, ( ) =
---------------------------------(2)
But the probability density function (pdf), ( ) = ( ) ( )
=
×
---------------------- (3)
From eqn.(2), if =
Therefore, ( ) = ( ) =
= 0.3679
and the failure probability, ( ) = 1 − ( )
= 1 − 0.3679 = 0.6321
Also, , ( ) = 1 − ( )
Inversing both side,
1
1
=
( ) 1− ( )
Therefore,
( )
=
( )
=
Taking double log of both side (i.e loglog);
1
1− ( )
This can be compare to equation of a straight line,
Where y is the
( )
log( ) −
=
=
on vertical axis, m is
and c (the intercept) =−
20
+
(the slope),
log( ) on horizontal axis
Example 1
100 electric lamps are put to life test, and the failure in days for the first 10 that fail are as follows;
12, 25, 36, 46, 54, 66, 70, 82, 88, and 95. On plotting these data on Weibull graph paper, the
estimate of Weibull distribution parameters i.e scale parameter ( ) and shape parameter ( ) were
obtain to be 550 days and 1.25 respectively.
(i)
(ii)
Determine the failure probability at 1000days
What is the reliability value at 200days
Solution
= 1.25 , = 550
Given :
Failure probability ( ) = 1 − ( )
(i)
But, ( ) =
At t = 1000days,
.
( )=1−
(ii)
=
.
Reliability at 200 days
( )=
.
At t = 200days, ( ) =
= .
Example 2
A unit was tested and the following were the results of the test; = 25, 000, : = 2.0. Calculate (i)
Weibull mean (MTTF), (ii) the standard deviation and (iii) the probability of survival for
10,000hours.
Solution
i.
Estimate of Weibull mean (MTTF),
1

 1
 
   
= 25,000
Note:
(ii)
=
+ ) =
√
Weibull standard deviation
21
,
=
,
.
2 
1 
      1     1
 
 
2
= 25,000
( )−
= 25,0000
(iii)
+ ) −
Reliability at 10,000hrs
( )=
At t = 10, 000hrs, ( ) =
= .
,
,
22
+ )
=
,
.