Reliability Engineering- Part 1

Product Probability Law of Series Components

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.

n

iiRR

1

Transition of Component States

N F

Component fails

Component is repaired

Failedstatecontinues

Normalstate 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) t L(t) t L(t) t L(t)

0 1,023,102 15 962,270 50 810,900 85 78,221

1 1,000,000 20 951,483 55 754,191 90 21,577

2 994,230 25 939,197 60 677,771 95 3,011

3 990,114 30 924,609 65 577,822 99 125

4 986,767 35 906,554 70 454,548

5 983,817 40 883,342 75 315,982

10 971,804 45 852,554 80 181,765

After Bompas-Smith. J.H. Mechanical Survival : The Use of Reliability

Data, McGraw-Hill Book Company, New York , 1971.

HUMAN RELIABILITY

t L(t), Number Living at

Age in Years Age t R(t)=L(t)/N F(t)=1-R(t)

0 1,023,102 1. 0. 1 1,000,000 0.9774 0.0226 2 994,230 0.9718 0.0282 3 986,767 0.9645 0.0322 4 983,817 0.9616 0.0355 5 983,817 0.9616 0.0384 10 971,804 0.9499 0.0501 15 962,270 0.9405 0.0595 20 951,483 0.9300 0.0700 25 939,197 0.9180 0.0820 30 924,609 0.9037 0.0963 40 883,342 0.8634 0.1139 45 852,554 0.8333 0.1667 50 810,900 0.7926 0.2074 55 754,191 0.7372 0.2628 60 677,771 0.6625 0.3375 65 577,882 0.5648 0.4352 70 454,548 0.4443 0.5557 75 315,982 0.3088 0.6912 80 181,765 0.1777 0.8223 85 78,221 0.0765 0.9235 90 21,577 0.0211 0.9789 95 3,011 0.0029 0.9971 99 125 0.0001 0.9999 100 0 0. 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.

0 10 20 30 40 50 60 70 80 90 100

1.0

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

PSurvival distribution

Failur

e dist

ributio

n

Pro

babi

lity

of

Sur

viva

l R(t

) an

d D

eath

F(t

)

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

1 if the unit is functioning at time t( )

0 if the unit is in a failed state at time tX 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.

• Example: exponential distributiontetR )(

0

( ) Pr( )

lim ( ) 1

lim ( ) 0t

t

R t T t

R t

R t

Unreliability - F(t)

• The probability that the component experiences the first failure during (0,t].

• Example: exponential distribution

( ) Pr( )

( ) ( ) 1

F t T t

R t F t

1)(lim

0)(lim0

tF

tF

t

t

tetF 1)(

FALURE DENSITY FUNCTION f(t)

Age in Years No. of Failures (death)

N

)()()(

tntntf

dt

tdFtf

)()(

012345101520253035404550606570758085909599100

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 99,889123,334138,566134,217103,554 56,634 18,566 2,886 125 0

0.022600.005640.004020.003270.002880.002350.001860.002110.002400.002850.003530.004540.006020.008140.011100.015000.019500.024100.027100.026200.020200.011100.003630.00071000012

0.005400.004540.002840.003300.002870.001920.001980.002240.002590.003640.003930.004360.006370.009620.013670.018000.022000.024900.026100.024600.019500.009700.00210 - - -

)()( tntn

20 40 60

20

40

60

80

80

100

120

140

100

Nu

mb

er o

f D

eath

s (t

hou

san

ds)

Age in Years (t)

20 40 60

0.2

0.4

0.6

80

0.8

0.10

0.12

0.14

100 Age in Years (t)

Fai

lure

Den

sity

f (

t)

Failure Density - f(t)

0

( ) Pr( )

( )( )

( ) ( )

( ) ( )

t

t

t

f t t t T t t

dF tf t e

dt

F t f u du

R t f u du

(exponential distribution)

CALCULATION OF FAILURE RATE r(t)

Age in Years No. of Failures

(death)

r(t)=)(1

)(

tF

tf

Age in Years No. of Failures

(death)

r(t)=)(1

)(

tF

tf

0 1 2 3 4 5101520253035

23,102 5,770 4,116 3,347 2,95012,013 9,53410,78712,28614,58818,05523,212

0.022600.005700.004140.003380.002990.002440.001960.002240.002580.003110.003910.0512

40455055606570758085909599

30,788 41,654 56,709 76,420 99,889123,334138,566134,217103,554 56,634 18,566 2,886 125

0.006970.009770.014000.020300.029500.042700.061000.085000.114000.144800.172000.240001.20000

)(

)(

ageat survivals ofnumber

),[ during deaths ofnumber )(

tR

tf

t

tttr

20 40 60 80 100

0.05

0.1

0.15

0.2Early failures

Random failures

Wearout failuresF

ailu

re R

ate

r(t)

Failure rate r(t) versus t.

Period of Approximately Constant failure rate

Infant Mortality Old Age

Time

Failure Rate,

(faults/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.

• Example:

)(1

)(

)(

)()(

tF

tf

tR

tftr

)(trThe component with a constant failure rate is considered as good as new, if it is functioning.

As Good As New?

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.”

( )PrPr | Pr

Pr

t xx

t

T t x eT t x T t e T x

T t e

Relation Between Reliability and Failure Rate

- /( ) ( ) exp [- ( ) ]

0

( ) ( )exp [- ( ) ]0

( ) Pr( | )

tdR dtr t R t r u du

R

tf t r t r u du

r t t 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

1

1

functioning time of unit in interval

total functioning time for all units in interval

( )( ) failure rate

let ( ) denotes the number of units which are functioning

ji

n

jij

n

jij

T j i

T

n ir i

T

m i

at the start of interval i.

( ) ( )( ) ( )

( ) ( )

n i n iz i z i t

m i t m i

Mean Time to Failure - MTTF

Variance of Time to Failure

0 0

0 0 0

( ) ( )

( ) ( ) ( )

1

MTTF tf t dt tR t dt

tR t R t dt R t dt

2

0

2

( ) ( )

1

Var T t MTTF f t dt

Failure Rate Failure Density Unreliability Reliability

t(a)

t(b)

t(c)

t(d)

f (t) F (t) R (t)

1 1

0 0

Area = 1

t

0dttf

1 - F (t)

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 0.29Control valve 0.60Flow measurement (fluids) 1.14

Flow measurement (solids) 3.75Flow switch 1.12Gas - liquid chromatograph 30.6

Hand valve 0.13Indicator lamp 0.044Level measurement (liquids) 1.70

Level measurement (solids) 6.86Oxygen analyzer 5.65pH meter 5.88

Pressure measurement 1.41Pressure relief valve 0.022Pressure switch 0.14

Solenoid valve 0.42Stepper motor 0.044Strip chart recorder 0.22

Thermocouple temperature measurement 0.52Thermometer temperature measurement 0.027Valve positioner 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.

Similarly,

1

2 1 1 2

2 1 2 10

( )1 10 0

1

1 2

Pr Pr | ( )

T

t t t

T T T t T t f t dt

e e dt e dt

1

Pr component j fails first among n components j

n

ii

A System with n Components in Parallel

• Unreliability

• Reliability

n

iiFF

1

n

iiRFR

1

)1(11

A System with n Components in Series

• Reliability

• Unreliability

n

iiRR

1

n

iiFRF

1

)1(11

Upper Bound of Unreliability for Systems with n Components

in Series

n

ll

nj

n

i

i

ji

n

ii FFFFF

1

1

2

1

11

)1(

n

iiF

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 is independent of t and may be written as where is a positive constant.

2. The probability of more than one event A in this interval is , which is a function with property

3. 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.

( , ]t t t

( )t t

( )t 0

( )lim 0

t

t

t

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

(0, ) (0, ) 1 ( )

(0, ) (0, ) ( )(0, ) (0, )

let 0 (0, ) (0, )

From (0,0) 1 (0, ) for 0t

p t t p t t t

p t t p t tp t p t

t td

t p t p tdt

p p t e t

( , ) Pr ( )p n t N t n

Exponential DistributionLet T1 denotes the time point when A occurs for the

first time. T1 is a random variable and

1

1

1

1

1 1

1 0

( ) Pr{ } 1 Pr{ } 1 (0, )

Thus,

1 for 0( )

0 otherwise

for 0( )

0 otherwise

1[ ] ( )

T

t

T

t

T

T

F t T t T t p t

e tF t

e tf t

E T tf t dt

Thus, thee waiting time T between consecutive occurrences in aHPP is exponentially distributed.

Probability of n Events Occurring in (0, t]

( , ) ( , ) 1 ( ) ( 1, )

( , ) ( 1, ) ( , ) ( , ) ( )

( , ) ( , ) ( )( 1, ) ( , ) ( , )

( , ) ( 1, ) ( , ) ; ( ,0) 0 for 1

(1,

p n t t p n t t t p n t t

p n t t p n t p n t p n t t

p n t t p n t tp n t p n t p n t

t td

p n t p n t p n t p n ndtd

pdt

) (0, ) (1, ) (1, )

(1, )

( )( , ) for 0,1,2,

!

t

t

nt

t p t p t e p t

p t te

tp n t e n

n

Poisson DistributionSince

This distribution is called the Poisson distribution with parameter and random variable n

Notice that, since the expected number of occurrences of event A per unit time (t=1) is , expresses the intensity of the process.

( , ) Pr ( )p n t N t n

t

0

( )( )

!

nt

n

tE N t n e t

n

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

0

0

1 01 0

0

0

0

0

0

( )

00 0

Pr ( ) 1Pr | ( ) 1

Pr ( ) 1

Pr 1 event in (0, ] 0 events in ( , ]

Pr ( ) 1

Pr 1 event in (0, ]} Pr{0 events in ( , ]

Pr ( ) 1

= for 0t tt

t

T t N tT t N t

N t

t t t

N t

t t t

N t

te e tt t

t e t

In other words, the time at which the first failure occurs inuniformly distributed over (0,t0]. The expected time is thus

1 0 0| ( ) 1 / 2E T N t t

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

( ) the number of C failures in (0,t]

Pr ( ) | ( ) (1 )

0,1,2, ,

Pr ( ) (1 )!

( )

!

m n m

n

m n m t

n m

mp t

M t

nM t m N t n p p

m

m n

n tM t m p p e

m n

p te

m

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

is then gamma distributed (k, λ).1 2 kT T T T

Gamma DistributionConsider a homogeneous Poisson distribution,

1

0

( ) Pr 1 Pr 1 ( , )k

k

T k kj

F t T t T t p j t

1

0

1 1

( )( ) 1

!

( )( ) ( ) ( )

( 1)! ( )

k

k

k

jkt

Tj

T k t k tT

tF t e

j

dF tf t t e t e

dt k k

The waiting time until the kth occurrence of event A in aHPP 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 ( )

( )

1

1 0( ) Pr

0 otherwise

0( ) ( )

0 otherwise

( ) Pr for 0

( )( ) for 0

( )

t

t

t

e tF t T t

d t e tf t F t

dt

R t T t e t

f tr t t t

R t

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

11

tR

e

Hence, the probability of time to failure larger than 1/λis independent of α. This quantity is called the

Characteristic Lifetime.

0

22

1 1( ) 1

1 2 11 1

MTTF R t dt

Var T

Three-Parameter Weibull Distribution

[ ( )]

1 [ ( )]

[ ( )]

1

1( ) Pr

0 otherwise

( )( ) ( )

0 otherwise

( ) Pr for

( )( ) ( ) for

( )

t

t

t

e tF t T t

d t e tf t F t

dt

R t T t e t

f tr t t t

R t

Normal DistributionThe normal distribution is sometimes used as a

lifetime distribution, even though it allows negative TTF values with positive probability.

The probability density of standard normal distribution is

2

2

1 ( )( ) exp

22

tf t

t

21( ) exp

22

tt

Normal Distribution

The unreliability, reliability and failure rate can be written as

( ) Pr ( )

( ) 1 ( ) 1

( ) 1( )

( ) 1

t tF t T t d

tR t F t

tR t

r ttR t

Normal Distribution, Left Truncated at 0

( ) Pr | 0 for 0

( ) 1( )

( ) 1

t

R t T t T t

tR t

r ttR t

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

2

0

2

2

1exp

)(1

)()(

log

2

1exp

1

2

1)(

log

2

1exp

2

1)(

MTTF

tF

tftr

dtt

ttF

t

ttf

t