Reliability Theory

What is reliability?Why it is important to study? Use of reliability in day to day life.

Reliability may be defined in several ways:

The idea that an item is fit for a purpose with respect to time. In the most discrete and practical sense: "Items that do not fail in use are reliable" and "Items that do fail in use are not reliable”.

Reliability of a device is denoted as and defined as

( )R t

( ) , 0.R t P T t t

Where “T” is a continuous random variable representing the life length of the device.

Some property of the reliability function :( )R t

( ) ( ) ( )t

i R t f u du

Where “f ” is probability density function (p.d.f) of “T”.

We assume( ) 0 0f t for t

( ) ( ) ( )ii R t f t

( ) ( ) is a nonincreasing function of iii R t t

( ) (0) 1 , ( ) 0iv R R

( ) ( ) 1 ( ) where ( ) is the cdf of v R t F t F t T

( ) ( ) 1 ( ) where ( ) is the cdf of v R t F t F t T

( )R t P T t

, ( ) 1or R t P T t

, ( ) 1 ( )or R t F t

Hence the results

( ) (0) 1 , ( ) 0( ) ( )iv R Rv iv

( ) ( ) is a nonincreasing function of ( ) ( )iii R t tiv iii

( ) ( ) ( )( ) ( )ii R t f tv ii

( ) ( ) 1 ( )( ) ( ) ( )

v R t F tR t F t f t

Hence the results

Conditional failure (hazard) rate function:

It is denoted by ( ) ( )Z t or h t

0

( ) limt

P t T t t T tZ t

t

For small

( )t

Z t t P t T t t T t

For small

( )t

Z t t P t T t t T t

For small , ( ) approximates the conditional probability that the device will fail during the interval , given that it has survived at the time

t Z t tt t t

t

In particular, if ( )Z t

Hazard rate is constant

In that case we can say that the device is as good as new at any time of its life.

( ) ( )Exerci .( )

se: f tShow that Z tR t

0

( ) limt

P t T t t T tZ t

t

0

( ) lim.t

P t T t t T tZ t

t P T t

0

( ) lim.t

P t T t tZ t

t P T t

0

( )( ) lim

. ( )

t t

tt

f u duZ t

t R t

0

( ) ( )( ) lim. ( )t

F t t F tZ tt R t

( ) ( )( ) .( ) ( )

F t f tZ tR t R t

Exercise :

Find the failure rate function for Weibull Distribution:

( ) 1 tF t e

( ) ( )( ) ( )( ) 1 ( )

f t F tZ t Z tR t F t

1( ) tF t e t

1

1( )t

t

e tZ t t

e

Failure Patterns

Theorem:

If T is the time to failure of a device with pdf f(t)and cdf F, and Z(t) is the failure rate function of thedevice, then :

0 0

( ) ( )

( ) , ( ) ( ) , for 0.

t t

Z s ds Z s ds

R t e f t Z t e t

( ) ( ) ( )( )( ) ( ) ( )

f s F s R sZ sR s R s R s

0

Integrating, Note that (0) 1

( )( )( )

t t

o

R

R sZ s ds dsR s

0

( )

ln ( ) ( ) ( )

tt Z s ds

o

R t Z s ds R t e

( )( )( )

f tZ tR t

( ) ( ) ( )f t Z t R t

0

( )

( ) ( )

t

Z s ds

f t Z t e

Theorem 6.3:

Expected system life time: Theorem 6.3 shows howmean/average lifetime of a system can bedetermined from a knowledge of the reliabilityfunction R(t)

0

( )

Where is the time to failure

E T R s ds

T

To prove that, it is sufficient to show that

0

E T P T t dt

0

. . ( )t

R H S f u du dt

0 0

( )u

f u dt du

0

( ) . .uf u du L H S

Example:

( ) /10 0 101 10

( )

F t t tt

Find E T

:( ) 1 ( )

1 /10 0 100 0

ClealyR t F t

t tt

0

( )E T R s ds

10

0

110sE T ds

102

0

10 5 520sE T s

Reliability of a system• Series System:- A system whose

components(say n components) are arranged in such a way that the system fails whenever any of its components fail is called a series system.

s1

s

i

R

R r e li a b i l i ty o f s e r ie s s y s te m R R e th e i th c o m p o n e n t

n

ii

R

w h e r el ia b i l i t y o f

Parallel System

• A system whose components are arranged in such a way that the system fails only if all its components fail is called a parallel system

s1

s

i

R ( ) 1 (1 ( ))

where R reliability of parallel system

R Reliability of ith component of the system

n

ii

t R t

Example 1: Consider a series system of three independentcomponents and the life length of each component followsuniformly distribution over (0,10).(i) Find the reliability of the system.(ii) Find the expected system life.Solution:

3 3

1 1

( ) ( ) 1 ( )s i ii i

R t R t F t

( ) /10 0 101 10

iF t t tt

3

3

( ) 1 ( )

1 /10 0 100 10

s iR t F t

t tt

0

( )E T R s ds

310

0

110sE T ds

52

E T

Example 2: Consider a system consisting of eight components . The system consists of five assemblies in series, where assembly 1 consists of component 1 with reliability 0.99, assembly II consists component 2 and 3 in parallel with reliabilities 0.95 and 0.95, assembly III consists component 4,5,6 in parallel with reliabilities 0.96, 0.92, 0.85, assembly IV and V consist of component 7 and 8 with reliabilities 0.95, 0.82 respectively. calculate system reliability.

Exercise VI.7 :Let Y be the life of an independent series system with ncomponent. Find the pdf and cdf of Y. Also prove that the hazardrate function of the system is the sum of the hazard functions of itscomponents.

Solution:

1

( ) ( )n

s ii

R t R t

1

( ) 1 ( ) 1 ( )n

Y s ii

F t R t R t

1

( ) ( ) ( )n

Y s ii

df t R t R tdt

1 1

, ( ) ( ) ( )nn

Y i ji j

j i

or f t R t R t

1 1

1

( ) ( )( )( )( ) ( )

nn

i ji j

j iYs n

si

i

R t R tf tZ tR t R t

1 2

1 2

( )( ) ( )( )( ) ( ) ( )

ns

n

R tR t R tZ tR t R t R t

1 2

1 2

( )( ) ( )( )( ) ( ) ( )

ns

n

f tf t f tZ tR t R t R t

1 2( ) ( ) ( ) ( )s nZ t Z t Z t Z t

1( ) ( )

n

s ii

Z t Z t

Exercise VI.8.Consider an n-component independent series system.Given that these components have constant failure rates

respectively. Find the(i) reliability of the system.(ii) hazard rate of the system.(iii) pdf of the life-length of the system.Also express the mean life of the system in terms ofmean life of its components.Solution:

1 1

( ) ( ) 1 ( )n n

s i ii i

R t R t F t

1 2, , , n

1

1

( )

n

iii

tnt

si

R t e e

1 2( ) ( ) ( ) ( )nZ t Z t Z t Z t

1 2 n

1

1( ) ( ( ))

n

ii

tn

s s ii

df t R t edt

The mean life length of the system is

where is the mean life-length of component.

1 2

11 1 1

n

1 2

1( ) sn

E T

i thi

Exercise VI.9.Let T be the life length of an independent parallelsystem with n-components.(i) Find the pdf and cdf of T(ii) Show that

Solution:

Hint: Use properties of reliability, f(t) and theirdefinitions.

0 0

( ) ( )

1

1 1

t t

iZ s ds Z s dsn

i

e e

Example: Probability that a device can survive after 0,1,2,3,4 or more shocks are 1,0.8,0.4,0.2, 0 respectively. If the arrival of shocks follow Poisson distribution with λ=0.15, find R(10).

Solution:

Put λ=0.15 and t=10 in the above.

2 3( ) ( )( ) 1 0.8( ) 0.4 0.22 6

t t tR t e t

CERTAIN LIFE MODELS

Exponential: In reliability, this distribution plays avital role. This is the distribution during phase II.

“Useful life phases” of the device, when failureoccurs due to an external causes, which may becalled “Shock/shocks”

Suppose that a device fails due to shocks, which occur in a Poisson processwith a rate . Assume that the device fails as soon as the first shock arrives.Then:

( ) 1 , 0tF t e t

,( ) , 0

0

t

orf t e t

else where

Thus the failure-time distribution is exponential distribution.

Let us assume, “p” is probability that even after the shock, no failure occurs.

( ) No failure occurs in the time interval (0, ]R t P t

0

( ) shocks arrive in time (0,t] device survives all shocksn

R t P n n

Using:

( ) ( ) BP A B P A P A

0

The device will survives all shocks( ) ttn

nR t P X n P X n

0

( ) nt

nR t P X n p

0

( )( ) , 0!

t nn

n

e tR t p tn

Where is the number of shocks which arrives in the time-interval(0, ]. Under the assumption that the shocks arrive in (0, ] is a Poissonprocess, then has a Poisson distribution with parameter

t

t

Xt t

X t

0

( )!

nt

n

p tR t e

n

(1 )( ) ...........(*)p tR t e

Differentiate (*) with respect to t we get

(1 )( ) ...........(*)p tR t e (1 )( ) ( ) (1 ) p tR t f t p e

(1 )( ) (1 ) , 0p tf t p e t

Thus has exponential distribution with parameter (1 )T p

Put p=0, we get

( ) tf t e

Hazard rate of exponential distribution

( ) tf t e ,( ) 1 , 0t

orF t e t

( ) ( )( )( ) 1 ( )

f t f tZ tR t F t

This is the phase-II in failure pattern graph

Gamma Distribution

Suppose that the shocks occur in a Poisson Process. And the device fails as soon as the r-th shock arrives

( ) ( )R t P T t

There will be no failure during the interval (0,t]P

There will be less than -shocks in the interval (0,t]P r

: be the random varible "Number of Shocks that arrive in (0,t]"t

LetX

Then Clearly distribution of follows poisson with parameter

Xt

( ) ( 1)tR t P X r

1

0 !

krt

k

te

k

Differentiating with respect to t we get

1

( ) , 0( 1)!

rtt

R t e tr

1

( ) , 0( )

rtt

f t e tr

Which is a Gamma distribution

Failure rate of Gamma distribution

1

1

( )( )( )

r t

r x

t

f t t eZ tR t

x e dx

1

( ) , 0( )

rtt

f t e tr

2

1L et .r x

t

D x e dx

2 1( 1 )( )

r t r x r t

t

t e r t x e dx t eZ t

D

1Now let (t)=( 1 ) r x r t

t

r t x e dx t e

2(t)= ( 1) , (using by parts)r x

t

r x e dx

Case I : If 0 1.r Then ( ) 0 for all 0.

( ) is an increasing function and increases from a negative value to zero.

( ) 0 for all 0.( ) for all 0.( ) is a decresing function of .

Thus Gamma distributio

t tt

t tZ t tZ t t

n for (0 1)

can be used as a life- model for any device when the device in phase-I.

r

C ase II : If 1.r 1,rlet t

1( ) 0 ( ) 0 rt Z t for t

1 , then (t)< 0 as r> 1 .

( ) is d ecreasin g fun ctio n an d decreases from a po sitive valu e to zero .

( ) 0 fo r a ll 0 .( ) 0 fo r a ll 0 .

( ) is an in cresing fu nction .T hu s G am m a d is tribu tion

rle t t

t

t tZ t tZ t

fo r ( 1)

can be u sed as a life- m odel fo r any d ev ice w h en the d evice in phase-III.

r

Case III : If 1.r

Then the Gamma distribution is changed to exponetial distribution with parameter .

( ) (constant function).Thus Gamma distribution for ( 1)can be used as a life- model for any device when the dev

Z tr

ice in phase-II.

Failure rate of Normal Distribution

Let T be the life length of any device which follows normal distribution with mean and variance

Assume that, the mean is so large.

is follows standard normal distribution.T

( 0) is neglegible, so that ( 0) 1.P T P T

2.

2

2

12

12

22

1 1 ( )2

( )(1 ( ))2

t

tte F t

eZ tF t

21

21Now let (t)= 1 ( )2

t te F t

Case I : If 0 .t

Then 0, and ( ) 0 .

( ) 0 for all 0 .( ) is an incresing function.

t t

Z t tZ t

C ase II : If .t 1T hen ( ) (1 ( )).

( ) 0 fo r all .( ) is a decreasing function and

decreases from a positive value to zero .( ) 0 fo r all .( ) 0 fo r all .

( ) is a increasing function o f 0 .

t F t

t tt

t tZ t tZ t t

T hus N orm al d istribu tion fo r ( large )can be used as a life- m odel fo r any device w hen the device in phase-III.

Statistical Methods of Determining Failure TimeDistribution

using correction for continuity.0.5ˆ ( ) ,iiF t

n

( )( ) ,i i i

d i i ii

n t n t tf t t t t t

n t

( )( ) ,

( )i i i

d i i ii i

n t n t tZ t t t t t

n t t

Data Analysis to check suitability of exponential distribution

Ex. V11:100 components were put on life-test. The test was terminated as soon as the 11th components failed. The life-length of those components which failed were as follows:8, 16.1, 24.3, 32.55, 40.9, 57.8, 66.4, 75.1, 83.9, 92.85. Guess which distribution fits the data. Estimate the parameters.

1110987Failure No.

49.340.932.5524.316.18Time tofailure

92.8583.975.166.457.8Time toFailure

654321Failure No.

Case-I:

0.0011183.9-92.850.0011932.55-40.90.0011375.1-83.90.0012124.3-32.55

0.0011657.8-66.40.001238-16.10.0011749.3-57.80.001250-8

0.0011940.9-49.3

0.0011466.4-75.10.0012116.1-24.3

Timeinterval

Timeinterval

( )df t ( )df t

Case-II:

0.110992.850.046040.90.099883.90.035632.55

0.078066.40.015116.10.067257.80.005018

0.056649.3

0.088875.10.025324.3

ˆln(1 ( ))iF t it it ˆln(1 ( ))iF t

Case-III:

0.0012483.9-92.850.0012432.55-40.90.0012575.1-83.90.0012624.3-32.55

0.0012557.8-66.40.001258-16.10.0012549.3-57.80.001250-8

0.0012540.9-49.3

0.0012566.4-75.10.0012416.1-24.3

Timeinterval

Timeinterval

( )dZ t ( )dZ t

Since is almost constant, we can assume exponential distribution fits the data. Now using least square method we can determine the parameter

Calculation for .

( )dZ t

.

1 1

1( )

ˆ 0 . 0 0 1 2 5 .1 1

ii

Z t

Data Analysis to check suitability of weibull distribution with α=0.

Weibull distribution with parameter

A random variable T is said to have Weibulldistribution if its density function is given by

( , , ).

1

,( )

0,

tt e tf t

t

Special case of Weibull distribution when

It can be proved that the hazard rate and reliability of the above Weibull distribution are

0.1

( ) , 0.ttf t e t

1

( ) , ( ) , 0.ttZ t R t e t

Consider the reliability . ( )R t

( ) , 0.t

R t e t

ln( ( )) ln( ( ))

1ln ln ln( ) ln( )

t tR t R t

tR t

Now plot

1ln ln ln( ) ln1 ( )

1ln ln ln( ) lnˆ1 ( ) ii

tF t

tF t

1l n ( ) , l n l n ˆ1 ( )ii

tF t

If the points

lies almost on a straight line, we can assume Weibull distribution with α=0 fits the data.

How to calculate the parameter and .Case-I (using graphical method)

slope of the above line

1l n ( ) , l n l n ˆ1 ( )ii

tF t

and can be obtained by putting any value for

in the equation given below:

1 1ˆln ln( ) ln lnˆ ˆ1 ( )ii

tF t

it

Case-ii (using least square method )

Least square methodSuppose given. We want to fit one

straight line to the above data. Least square

method estimates and by solving

(1)

( , ), 1, 2, ,i ix y i n

b x a

a b

2

( , ) 1min .

n

i ia b iy bx a

Using derivative concept, we can easily say that equation (1) gives minimum when

22ˆ

ˆˆ

i i i i

i i

i i

n x y x yb

n x x

y b xa

n

In our problem,

ln( )

1ln ln ˆ1 ( )

ln( )

i i

ii

x t

yF t

ba

Using least square method, we can estimate

ˆˆ

ˆˆ

ˆa

b

e

Consider the reliability . ( )Z t1

( ) , 0.tZ t t

ln( ( )) ln ( 1)ln( ) ( 1)ln

ln( ( )) ( 1)ln( ) ln ln

Z t t

Z t t

Now plot ln ( ), ln ( )i d it Z t

ln( ( )) ( 1) ln( ) ln lnd i iZ t t

If the points

lies almost on a straight line, we can assume Weibull distribution with α=0 fits the data.

How to calculate the parameter and .Case-I (using graphical method)

slope of the above line1+slope of the above line

ln ( ) , ln ( )i d it Z t

ˆ 1

and can be obtained by putting any value for

in the equation given below:

ˆ ˆln ( 1) ln( ) ln ( )ˆln ˆ

i d it Z t

it

Case-ii (using least square method )

Least square methodSuppose given. We want to fit one

straight line to the above data. Least square

method estimates and by solving

(1)

( , ), 1, 2, ,i ix y i n

b x a

a b

2

( , ) 1min .

n

i ia b iy bx a

Using derivative concept, we can easily say that equation (1) gives minimum when

22ˆ

ˆˆ

i i i i

i i

i i

n x y x yb

n x x

y b xa

n

Using least square method, we can estimate

ˆ ˆl nˆ

ˆˆ 1

ˆa

b

e

In our problem,

ln( )ln ( )

1ln ln

i i

i d i

x ty Z tba

Ex. V12:100 components were put on life-test. The test was terminated as soon as the 7th components failed. The life-length of those components which failed were 2, 3.3, 4.7, 5.9, 7, 8, 9. (a) Using hazard rate, check whether Weibull distribution fits

the data or not ? Estimate the parameters using (i) graphical method.(ii) least square method.

(b) Using cdf/reliability, check whether Weibull distribution fits the data or not ? Estimate the parameters using

(i) graphical method.(ii) least square method.

86

975.94.73.32Time to failure754321Failure No.

0.01057 - 80.01068 - 9

0.00945.9 – 70.00854.7 – 5.90.00723.3 – 4.70.00772 - 3.30.0050 – 2

Time interval ( )dZ t

Solution (a):

-4.552.078

-4.661.947-4.761.775.9

-4.851.193.3-5.290.692

-4.542.199

-4.871.544.7

ln( )it ln( ( ))d iZ tit

ln( ),ln ( )i d it Z tsince the points lies almost on a straight line, we can assume Weibulldistribution fits the data.

a(i). Calculation for parameters.

ˆ 1 1 0.01 1.01slope

ˆ ˆln ( 1 ) ln ( ) ln ( )ˆ

ln 1 .0 1 ( 0 .0 1) 0 .6 9 5 .2 91 .0 1

ˆ

1 8 8 .7 9

i d it Z t

e

e

a(ii). Calculation for parameters

7 7

1 17 7

2

1 1

ln (1 .4 4 ) 7 .1 31 .4 4

ln ( ) , ln ( ) , 1, 2 , , 7 .

1 1 .3 9 , 3 3 .5 2

2 0 .2 4 , 5 3 .7 4

ˆ ˆ0 .4 4 , 7 .1 3

ˆ ˆ1 .4 4 , 1 8 2 .3 2 .

i i i d i

i ii i

i i ii i

x t y Z t i

x y

x x y

b a

e