1 An Application of Probability to Reliability Modeling and Analysis Dr. Jerrell T. Stracener, SAE...

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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

• 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?

• 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

“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

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(F1dt)t(f

The Reliability Function

Probability Distribution Function• Weibull

• Exponential

e )t(R

Reliability Function

Reliability Functions

t is in multiples of

β=5.0

β=1.0

β=0.5

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

• Weibull

and, in particular

• Exponential

P p)-ln(1- t

t 632.0

p)-ln(1- Pt

Percentiles, tp

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

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

• Failure Rate

• Note:

Only for the Exponential Distribution

•Cumulative Failure

rate failure

Failure Rates - Exponential

• 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

Failure Rates - Weibull

Failure Rates

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

0 1.0 2.0

β=0.5

The Weibull Model - Distributions

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

Weibull

Exponential

1 MTBF

Mean Time Between Failure - MTBF

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

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

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?

The Weibull Model –Solution

9608.0

)200(2)1000/200(

/te)t(R

0004.0

2002)200( 2

failures per hour

The Weibull Model –Solution continued

From the Gamma Function table:

111000

23.886

5.11000

88623.0)5.1(

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

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

So that

And t0.5=3748

Mean life is

6.4227eeE(t) 8.34939μ 2

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

,861.0

• 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

• Reliability Block Diagram

•Element Time to Failure Distribution

with failure rate , for i=1, 2,…, n

• System reliability

Se)t(R

S θλ

is the system failure rate

• System mean time to failure

1iiS )t(

ii θE~T

E1 E2 En

Series Reliability Configuration with Exponential Distribution

• 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

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

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

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.

• Block Diagram

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

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

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

System Reliability Model - Parallel Configuration

• Product rule for unreliabilities

iiS tRtR

)(11)(

•Mean Time Between System Failures

SS (t)dtRMTBF

p=R(t)

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

e12)t(h

ttS eetR 22)(

θE~Ti θ

Parallel Reliability Configuration with Exponential Distribution

• System Mean Time Between Failures:

MTBFS = 1.5

Parallel Reliability Configuration with Exponential Distribution

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.

Example - 3 Configurations

Example - 3 Configurations - Solution

For the baseline system:

,etR λtS

,λthS

θ,MTBFS

θMTBMS

Example - 3 Configurations – Solution

For alternative A:

,e-e2tR λt2λtSA

e12λth

1.5θ.MTBFSA

0.5θ2

1MTBMSA

For alternative B:

2θMTBFSB

,eλt1tR λtSB

θMTBMSB

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

In terms of reliability,

λtλt

λt2λt

SA e-2e

In terms of failure rate,

0.01t0.01t

λe2e1

λλt1tλ

In terms of MTBF,

In terms of MTBM,

0 100 200 300 400 500 600 700

Rsa/Rs

Rsb/Rs

0 100 200 300 400 500 600 700

hsa/hs

hsb/hs

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

Example

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

Since Pi=e-λt for i=1,2,3,4,5

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

3λ-2λ-λ-

3λ-2λ-λ-λ2

5λ-4λ-3λ-2λ-

e5e8e36λ

)ee2e3(

λe5λe8λe36e

0.87θ θ30

6151045λ5

5λ-4λ-3λ-2λ-

s dttR

θ2.0λ5

1SMTBM

1 An Application of Probability to Reliability Modeling and Analysis Dr. Jerrell T. Stracener, SAE...

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