3 Constant Failure Rate Models

7/21/2019 3 Constant Failure Rate Models

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Constant Failure Rate Models KKKP6324 Quality Systems and Reliability

2

3.1 Exponential Reliability Function

3.2 Failure Modes

3.3 Poisson Process

3.4 Redundancy




3

3.1 Exponential Reliability Function

• A failure distribution that has a constant failure rate (CFR) is called an

exponential probability distribution.

• It is one of the most common failure distributions in reliability engineering

that is referred to the exponential, or CFR, model.

• It should dominate during the useful life of a system or a component.

! Refer to the bathtub curve.

• Assuming that ! (t ) = ! , t ! 0, ! > 0, from Eq. 1.14:

R t ( ) = exp ! ! "t ( ) d "t 0

t

# $%&

'()= e

!! t , t * 0 (3.1)

and,

F t ( ) =1! e!! t




4

Then

f t ( ) =!dR t ( )

dt

= ! e!! t

• From these relations:

MTTF =! e!! t

dt 0

"

# =

e!! t

!! 0

"

=

1

! (3.2)

! 2= ! t !

1

"

"

#$

%

&'" e

!" t dt

0

(

) =

1

" 2 (3.3)

Hence, the standard deviation is 1 / ! = MTTF.




5

0

0.2

0.4

0.6

0.8

1.0

1.2

0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0

R(t )

t

0.25 1.0 2.5 5.0

0

0.2

0.4

0.6

0.8

1.0

1.2

0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0

F (t )

t

0.25 1.0 2.5 5.0

0

1.0

2.0

3.0

4.0

5.0

6.0

0 0.5 1 1.5 2

f (t )

t

0.25 1.0 2.5 5.0




6

• For a given reliability R:

R t R( ) = e

!! t R= R

Then, t R = !

1

! ln R (3.4)

When R = 0.5, t R = t

med (3.5)

•

The CFR model is memoryless, i.e., o It is not dependent on how long the component has been operating,

o There is no aging and wear-out effect.

• In other words, a burn-in period has no subsequent effect on reliability and

will not improve the component’s reliability.




7

• Example 3.1:

A microwave transmitter has exhibited a constant failure rate of

0.00034 failure per operating hour. Calculate:

(a) MTTF,

(b) median of the distribution,

(c) reliability over 30 days of continuous operation,

(d) design life for a reliability of 0.95.

• Find (a) MTTF; (b) t med; (c) R(30!24 hr); (d) t 0.95.

( Ans: 2941 hr; 2039 hr; 0.7829; 150.9 hr)




8

3.2 Failure Modes

• Complex systems will fail through various causes, which can be

categorised based on failure modes.

• If Ri(t ) is the reliability function for the ith failure mode, then, assuming

independence among the failure modes, the system reliability is

R t

( )= R

i t

( )i=1

n

! (3.6)




9

• Let ! i(t ) be the failure rate function for the ith failure mode. Then,

Ri t ( ) = exp ! !

i "t ( ) d "t

0

t

#

$

%&

'

()

and R t ( ) = exp ! ! i "t ( ) d "t

0

t

# $%&

'()

i=1

n

*

= exp ! ! i "t ( )

i=1

n

# d "t

0

t

$

%

&'

(

)*

= exp ! ! "t ( ) d "t 0

t

$ %&'

()*

where ! t ( ) = ! i t ( )

i=1

n

! (3.7)




10

• For the CFR model, if a system consists of n independent, serially related

components each having a constant failure rate ! i, then

! t ( ) = ! = ! i

i=1

n

!

and R t ( ) = exp ! ! d "t 0

t

# $%&

'()= e

!! t

where MTTF =

1

! =

1

! i

i=1

n

!=

1

1 MTTFi

i=1

n

!, MTTFi

=

1

! i

(3.8)

• If the components are also all identical, i.e., ! i = ! 1 for i = 1, 2, …, n, then

! = n! 1

and MTTFi =

1

n! 1

(3.10)




11

• Example 3.2:

An aircraft engine consists of three modules having constant

failure rate of ! 1 = 0.002, ! 2 = 0.015 and ! 3 = 0.0025 failure peroperating hour. Calculate the corresponding MTTF.

( Ans: 51.28 hr)




12

• Certain components that operate on a cyclical basis may fail on demand.

Hence, the failure rate in failures per clock unit of time is

! eff = t I

t I + t

O

! I + t O

t I + t

O

! O + p

t I + t

O

(3.10)

where ! I = the average failure rate while idle;

! O = the average failure rate while operating;

p = the probability of failure on demand;

t I = average length of the idle time period per cycle;t O = average length of the operating time per cycle.

• A renewal process, in which a failed component is immediately replaced

with a new one, will cause the system to reach a steady CFR state.




13

• Example 3.3:

An air conditioning compressor operates once for an average time

of 20 minutes each hour. While operating, it has experienced afailure rate of 0.01 failure per operating hour, and while idle has

experienced a dormant failure date of 0.0002 failure per idle hour.

The probability that the compressor fails on demand is 0.03.

Calculate the probability that the compressor will not fail over a

24-hour period.

• Find ! eff . Then, find R(24).

( Ans: 0.9092)




14

• If a failure will never occur prior to some specified time t 0, then t 0 is a

minimum or threshold time, also known as the guaranteed lifetime. Hence,

f t ( ) = ! dR t ( )dt

= ! e!! t !t 0( ) , 0 < t

0 " t <# (3.11)

R t ( ) = e!! t !t 0( )

, t > t 0 (3.12)

• From Eq. 3.11 and Eq. 3.12, the failure rate is still ! , but the mean of the

distribution has shifted to a distance t 0. Hence,

MTTF = ! te!! t !t

0( )dt

t 0

!

" = t 0 +1

! (3.13)

R t med( ) = e

!! t med!t 0( )= 0.5 (3.14)

t med

= t 0 +

ln0.5

!! , t

R = t

0 +

ln R

!! (3.15)




15

• Example 3.4:

Let ! = 0.001 and t 0 = 200. Calculate the MTTR, t med, t 0.95 and " .

( Ans: 1200; 893.15; 251.3; 1000)




16

3.3 Poisson Process

• If a component having a constant failure rate ! is immediately repaired or

replaced upon failing, the number of failures observed over a time period t has a Poisson distribution.

• The probability of observing n failures in time t is given by the Poisson

probability mass function pn(t ):

pn t ( ) =

e!! t

! t ( )n

n!n = 0, 1, 2,… (3.16)

• The Poisson process is often used in inventory analysis to determine the

number of spare components when the time between failures is

exponential.




17

• If S spare components are available to support a continuous operation over

a time period t , then the cumulative probability of S or fewer failures

occurring during time t is

RS t ( ) = p

n t ( )

n=0

S

! (3.17)




18

• Example 3.5:

A specially designed welding machine has a non-repairable motor

with a constant failure rate of 0.05 failures per year. The companyhas purchased two spare motors. If the design life of the welding

machine is 10 years, what is the probability that the two spares

will be adequate?

• Find Pr{ X " 2} for ! = 0.05.( Ans: 0.9856)




19

3.4 Redundancy

• Consider the case of two independent and redundant components each

having the same constant failure rate ! , where the system failure will occuronly when both components have failed.

• Since the probability that both components will fail by time t is 1! e!! t ( )

2

(see Topic 5), the system reliability is given by,

RS t ( ) =1! 1! e

!! t ( )2

=1! 1!2e!! t

+ e!2! t ( )

RS t ( ) = 2e!! t

! e!2! t

(3.18)




20

• Thus, the hazard rate in this case is (which is not constant ),

! S

t

( )=

f S t ( )

RS t ( ) =

2! e!! t

!2! e!2! t

2e!! t !e!2! t

! S t ( ) =

! 1! e!! t ( )

1! 1

2 e

!! t (3.19)

•

The system MTTF can be determined as follows,

MTTFS = R

S t ( ) dt

0

!

" = 2e#! t # e

#2! t ( ) dt 0

!

" =

2

! # 1

2! =

3

2!

MTTFS =

1.5

! (3.20)




21

• Example 3.6:

For the microwave transmitter described in Example 3.1, a

second redundant transmitter is added. Calculate:

(a) MTTF,

(b) reliability over 30 days of continuous operation,

(c) design life for a reliability of 0.95.

• Derive for the system RS (t ). Then, find (a) MTTFS ;(b) RS (30!24 hr); (c) t 0.95.

( Ans: 4412 hr; 0.9529; 744.4 hr)

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3 Constant Failure Rate Models

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