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An Application of Probability toReliability Modeling and Analysis

Dr. Jerrell T. Stracener, SAE Fellow

Leadership in Engineering

EMIS 7370/5370 STAT 5340 : PROBABILITY AND STATISTICS FOR SCIENTISTS AND ENGINEERS

Systems Engineering ProgramDepartment of Engineering Management, Information and Systems

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• Reliability is defined as the probability that an item will perform its intended function for a specified interval under stated conditions. In the simplest sense, reliability means how long an item (such as a machine) will perform its intended function without a breakdown.

• Reliability: the capability to operate as intended, whenever used, for as long as needed.

Reliability is performance over time, probability that something will work when you want it to.

What is Reliability?

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• Basic or Logistic Reliability

MTBF - Mean Time Between Failures

measure of product support requirements

• Mission Reliability

Ps or R(t) - Probability of mission success

measure of product effectiveness

Reliability Figures of Merit

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“If I had only one day left to live, I would live it in my statistics class --it would seem so much longer.”

From: Statistics A Fresh ApproachDonald H. SandersMcGraw Hill, 4th Edition, 1990

Reliability Humor: Statistics

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The Reliability of an item is the probability that the item willsurvive time t, given that it had not failed at time zero, when used within specified conditions, i.e.,

)tT(PtR

t

)t(F1dt)t(f

The Reliability Function

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Probability Distribution Function• Weibull

• Exponential

t

e )t(R

t

e )t(R

Reliability Function

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Reliability Functions

R(t)

t

t is in multiples of

β=5.0

β=1.0

β=0.5

1.0

0.8

0.6

0.4

0.2

00 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0

The Weibull Model - Distributions

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• Weibull

and, in particular

• Exponential

1

P p)-ln(1- t

t 632.0

p)-ln(1- Pt

Percentiles, tp

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Remark: The failure rate h(t) is a measure of proneness to failure as a function of age, t.

tF-1

tf

tR

tfth

Relationship Between h(t), f(t), F(t) and R(t)

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• Failure Rate

• Note:

Only for the Exponential Distribution

•Cumulative Failure

1

)t(h

)t(H

rate failure

1MTBF

Failure Rates - Exponential

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• Failure Rate

a decreasing function of t if < 1Notice that h(t) is a constant if = 1

an increasing function of t if > 1

• Cumulative Failure Rate

• The Instantaneous and Cumulative Failure Rates, h(t) and H(t), are straight lines on log-log paper.

1-t )t(h

)t(ht )t(H

1-

Failure Rates - Weibull

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Failure Rates

h(t)

t

t is in multiples of h(t) is in multiples of 1/

3

2

1

0

0 1.0 2.0

β=5

β=1

β=0.5

The Weibull Model - Distributions

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Mean Time to Failure (or Between Failures) MTTF (or MTBF)is the expected Time to Failure (or Between Failures)

Remarks:

MTBF provides a reliability figure of merit for expected failure free operation

MTBF provides the basis for estimating the number of failures ina given period of time

Even though an item may be discarded after failure and its mean life characterized by MTTF, it may be meaningful tocharacterize the system reliability in terms of MTBF if thesystem is restored after item failure.

Mean Time to Failure and Mean Time Between Failures

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Weibull

Exponential

MTBF

1

1 MTBF

Mean Time Between Failure - MTBF

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Problem -

Four Engine Aircraft

Engine Reliability R(t) = p = 0.9

Mission success: At least two engines survive

Find RS(t)

The Binomial Model - Example

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Solution –

1) X1 = number of engines surviving in time t

Then X1 ~ B (4,0.9)

RS(t) = P(x1 ≥ 2) = b(2) + b(3) + b(4)

= 0.0486 + 0.2916 + 0.6561 = 0.9963

2) X2 = number of engines failing in time t

Then X2 ~ B (4,0.1)

RS(t) = P(x2 2) = b(0) + b(1) + b(2)

= 0.6561 + 0.2916 + 0.0486 = 0.9963

The Binomial Model - Example

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The Weibull Model

Time to failure of an item follows a Weibull distribution with = 2 and = 1000 hours.

(a) What is the reliability, R(t), for t = 200 hours?

(b) What is the hazard rate, h(t), (instantaneous failure rate) at that time?

(c) What is the Mean Time To Failure?

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The Weibull Model –Solution

(a)

(b)

9608.0

)200(2)1000/200(

eR

1t

)t(h

/te)t(R

0004.0

1000

2002)200( 2

12

h

failures per hour

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The Weibull Model –Solution continued

(c)

From the Gamma Function table:

2

111000

11

MTTF

23.886

5.11000

MTTF

88623.0)5.1(

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The design life of a given type of pump for a given operating environment has a Lognormal distribution. If t0.10 = 2000 hours and the median life is 3748 hours. What is the mean life and the the 50th & 90th percentile? A system uses five pumps of this type. What is the probability of at least one of these pumps failing in 3000 hours?

Example – Pump life

Pump Life Solution

Given that t0.10 = 2000 hours and the median is 3748, we need to first find the values for and .

Since the median life is

=ln 3748

=8.2290

And since t0.1= =2000,

-1.28 = ln 2000

1.28 = 8.2290 – 7.6009

= 0.4907

μe3748

1.28σ-μe

Pump Life Solution

Now t0.9 is the value of t for which

But

So that

and t0.9=7023.8

Probability of one pump failing within 3000 hours

326.00.491

229.83000ln Φ3000)P(TF(3000)

90.00.491

229.8ln t)F(t)tP(T 0.90

0.90.90

90.0)28.1(tF 0.90

1.28,0.491

229.8ln t0.9

Pump Life - solution

Now t0.5 is the value of t for which

But

So that

And t0.5=3748

Mean life is

6.4227eeE(t) 8.34939μ 2

2σ

50.0σ

μln t)F(t)tP(T 0.50

0.50.50

50.0)0(tF 0.50

00.491

229.8ln t0.5

Pump Life Solution

Probability of at least one pump failing in 3000 hours

Y = # of pumps failing in 3000 hoursy = 0, 1, 2, 3, 4, 5

Y has a binomial distribution withn = 5 and p = 0.326

or you could work this using a probability tree

)0(1)1( YPYP

50 674.0326.00

51

,861.0

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• Simplest and most common structure in reliability analysis.

• Functional operation of the system depends on the successful operation of all system components Note: The electrical or mechanical configuration may differ from the reliability configuration

Reliability Block Diagram

• Series configuration with n elements: E1, E2, ..., En

• System Failure occurs upon the first element failure

E1 E2 En

Series Reliability Configuration

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• Reliability Block Diagram

•Element Time to Failure Distribution

with failure rate , for i=1, 2,…, n

• System reliability

where

tS

Se)t(R

SS

S θλ

1MTTF

is the system failure rate

• System mean time to failure

n

1iiS )t(

ii θE~T

E1 E2 En

ii θ

1λ

Series Reliability Configuration with Exponential Distribution

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• System mean time to failure

Note that /n is the expected time to the first failure, E(T1), when n identical items are put into service

nMTTFS

Series Reliability Configuration

Stracener_EMIS 7370/STAT 5340_Fall 08_09.30.08

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Parallel Reliability Configuration – Basic Concepts

• Definition - a system is said to have parallel reliability configuration if the system function can be performed by any one of two or more paths

• Reliability block diagram - for a parallel reliability configuration consisting of n elements, E1, E2, ... En

E1

E2

En

Stracener_EMIS 7370/STAT 5340_Fall 08_09.30.08

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Parallel Reliability Configuration

• Redundant reliability configuration - sometimes called a redundant reliability configuration. Other times, the term ‘redundant’ is used only when the system is deliberately changed to provide additional paths, in order to improve the system reliability

• Basic assumptions

All elements are continuously energized starting at time t = 0

All elements are ‘up’ at time t = 0

The operation during time t of each element can be describedas either a success or a failure, i.e. Degraded operation orperformance is not considered

Stracener_EMIS 7370/STAT 5340_Fall 08_09.30.08

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Parallel Reliability Configuration

System success - a system having a parallel reliability configuration operates successfully for a period of time t if at least one of the parallel elements operates for time t without failure. Notice that element failure does not necessarily mean system failure.

Stracener_EMIS 7370/STAT 5340_Fall 08_09.30.08

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Parallel Reliability Configuration

• Block Diagram

• System reliability - for a system consisting of n elements, E1, E2, ... En

n

jiij

ji

n

1iiS )t(R)t(R)t(R)t(R

n

ii

nk

n

kjiijk

ji tRtRtRtR1

1 )()1...()()()(

if the n elements operate independently of each other and where Ri(t) is the reliability of element i, for i=1,2,…,n

E1

E2

En

Stracener_EMIS 7370/STAT 5340_Fall 08_09.30.08

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System Reliability Model - Parallel Configuration

• Product rule for unreliabilities

n

iiS tRtR

1

)(11)(

•Mean Time Between System Failures

0

SS (t)dtRMTBF

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Parallel Reliability Configuration

s

p=R(t)

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Element time to failure is exponential with failure rate

• Reliability block diagram:

•Element Time to Failure Distribution

with failure rate for I=1,2.

• System reliability

• System failure rate

t

t

S e2

e12)t(h

ttS eetR 22)(

E1

E2

θE~Ti θ

1λ

Parallel Reliability Configuration with Exponential Distribution

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• System Mean Time Between Failures:

MTBFS = 1.5

Parallel Reliability Configuration with Exponential Distribution

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Compare the following reliability configurations I, II, and III in terms of (a) system reliability, (b) system failure rate, (c) system mean time between failures and (d) system mean time between maintenance, assuming that a failure requires maintenance.Element E has an exponential distribution of time to failure, T, with failure rate = 0.01. State all ground rules and assumptions, show all work and present the results graphically to convey your results.

I. II. III.

EE

E

E

E

Example - 3 Configurations

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Example - 3 Configurations - Solution

For the baseline system:

and

,etR λtS

,λthS

θ,MTBFS

θMTBMS

E

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Example - 3 Configurations – Solution

For alternative A:

and

E

E

,e-e2tR λt2λtSA

,e2

e12λth

0.01t

0.01t

SA

1.5θ.MTBFSA

0.5θ2

θ

2λ

1MTBMSA

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Example - 3 Configurations – Solution

For alternative B:

and

E

E

,λt1

tλth

2

SB

2θMTBFSB

,eλt1tR λtSB

θMTBMSB

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Now we compare Alternatives A & B to the baseline system.

In terms of reliability,

and

λtλt

λt2λt

S

SA e-2e

e-2e

tR

tR

λt1

e

eλt1

tR

tRλt

λt

S

SB

Example - 3 Configurations – Solution

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In terms of failure rate,

and

0.01t

0.01t0.01t

0.01t

S

SA

e2

e12

λe2e1

2λ

th

th

λt1

1

1

λt1

λt

λλt1tλ

th

th

2

S

SB

Example - 3 Configurations – Solution

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In terms of MTBF,

and

In terms of MTBM,

and

1.5θ

1.5θ

MTBF

MTBF

S

SA

2θ

2θ

MTBF

MTBF

S

SB

5.0θ

.5θ0

MTBM

MTBM

S

SA

1θ

θ

MTBM

MTBM

S

SB

Example - 3 Configurations – Solution

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0

1

2

3

4

5

6

7

8

0 100 200 300 400 500 600 700

Rs

Rsa

Rsb

Rsa/Rs

Rsb/Rs

Example - 3 Configurations – Solution

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0

0.2

0.4

0.6

0.8

1

1.2

0 100 200 300 400 500 600 700

hs

hsa

hsb

hsa/hs

hsb/hs

Example - 3 Configurations – Solution

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A system consists of five components connected as shown.Find the system reliability, failure rate, MTBF, and MTBM if Ti~E(λ) for i=1,2,3,4,5

E1

E2

E3

E4 E5

Example

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This problem can be approached in several different ways. Here is one approach:There are 3 success paths, namely,Success Path EventE1E2 AE1E3 BE4E5 C

Then Rs(t)=Ps= =P(A)+P(B)+P(C)-P(AB)-P(AC)-P(BC)+P(ABC) =P(A)+P(B)+P(C)-P(A)P(B)-P(A)P(C)-P(B)P(C)+

P(A)P(B)P(C) =P1P2+P1P3+P4P5-P1P2P3-P1P2P4P5

-P1P3P4P5+P1P2P3P4P5

assuming independence and where Pi=P(Ei) for i=1, 2, 3, 4, 5

)( CBAP

Solution

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Since Pi=e-λt for i=1,2,3,4,5

Rs(t)

hs(t)

tttt

ttttt

tttttttt

ttttttttt

5λ-4λ-3λ-2λ-

λ-λ-λ-λ-λ-

λ-λ-λ-λ-λ-λ-λ-λ-

-λ-λ-λ-λ-λ-λ-λ-λ-λ

ee2e3e

))(e)(e)(e)(e(e

))(e)(e)(e(e-))(e)(e)(e(e-

))(e)(e(e-))(e(e))(e(e))(e(e

ttt

ttt

tttt

tttt

s

sdtd

e

tR

tR

3λ-2λ-λ-

3λ-2λ-λ-

3λ-2λ-λ-λ2

5λ-4λ-3λ-2λ-

ee2e3

e5e8e36λ

)ee2e3(

λe5λe8λe36e

)(

)(

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MTBFs

0.87θ θ30

6151045λ5

1

λ2

1

λ3

1

λ2

3

λ5

e

λ2

e

λ3

e

λ2

3e

)(

0

5λ-4λ-3λ-2λ-

0

tttt

s dttR

θ2.0λ5

1SMTBM