Reliability Engineering Lec Notes #2

7/30/2019 Reliability Engineering Lec Notes #2

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INTRODUCTION

Definitions

Reliability is the probability that a component or a system will perform its required function (i.e.

not fail) for a given period of time when used under stated operating conditions. To determine

reliability in the operational sense:

1. Failure must be defined in an unambiguous and observable manner.

(if 1 brake fails, the system may not fail)

2. Period of time/function must be specified (e.g. 50,000 miles for the braking system)

3. Operating conditions must be specified (e.g. between given temperature limits for the brak-

ing system)

Maintainability is the probability that a failed component or a system will be restored to a speci-

fied condition (e.g. all wheel brakes functional) within a specified time (e.g. 1 day) when mainte-

nance is performed in accordance with prescribed procedures (e.g. replacing brake line).

Availability is the probability that a component or a system will perform its required function for

at given point time when used under stated operating conditions. If your car was in the shop last

year 3 days to repair a leaking oil pump (which had to be ordered) and 2 day to replace a seal

(which had to dry for 48 hours), its average availability over the year was (365-5)/365=0.99 or

99%.


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Why study reliability engineering?

Reliability is an important consideration in consumer purchasing. A 1985 Gallup poll con-

ducted in 1985 among 1000 individuals indicates that reliability as well as maintainability is

an important consideration in consumer purchasing:

Ten Most Important Product Attributes*

Attribute Average Score

Performance 9.5

Lasts a Long Time 9.0

Service 8.9

Easily Repaired 8.8

Warranty 8.4

Easy to Use 8.3

Appearance 7.7

Brand name 6.3

Packaging/Display 5.8

Latest Model 5.4

* Quality Progress, 18, 12-17 (1985)

Achieving high performance over the lifetime of a product requires considering reliability

and maintainability issues at the design and manufacturing stages. The following diamond

tree shows the interrelationships between reliability, maintainability and other design and

operational issues to achieve high performance. By looking downwards from the top of the

diamond tree, one can describe how various goals and subgoals are achieved. Looking

upward, one can identify why a goal needs to be achieved. As can be seen from the dia-

mond tree, reliability and maintainability assessment are not only important for the engi-

neering design process, but also necessary for:

o life-cycle costing

o cost-benefit analysis

o operational capability studies

o repair and facility resourcing

o inv entory and spare parts requirement determinations

o replacement decisions

o establishment of preventive maintenance programs.


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M. Modarres,Reliability and Risk Analysis, 3, Marcel Dekker, Inc, New York (1993)

Failures may lead to safety problems, possible damage, loss of life and/or property and sub-

sequent litigation.

Probabilistic vs. Deterministic Approach to Failure Prevention

The traditional approach to failure prevention is to design a high safety margin or high safety fac-

tor into the product (e.g. a brake line designed to withstand 4 times the expected load). Some

problems with this approach are:

Leads to overdesign and high cost.

The physics and chemistry of intrinsic failure mechanisms are not always accurately quan-

tifiable (e.g. crack formation and propagation, ionic contamination in semiconductors).

External events that can cause failures (e.g. earthquake, human error, excessive heat or

vibration) may appear to be occurring stochastically.

All failure mechanisms may not be known (e.g. TWA 800 fuel tank explosion in 1996 due

to spark ignition of the fuel vaporized by overheating of the HVAC system).

Cannot usually account for common cause failures (e.g. loss of power that supports both

operation and safety systems)


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Cannot be used to quantify the engineering or financial risk (probability of failure multi-

plied by consequence) associated with a process, product or mission.

A probabilistic approach to reliability engineering provides tools to solve these problems.

Overview of Reliability Engineering I

MODULE TOPIC SUBJECT MATTER HOURS

2INTRODUCTION Definition of reliability, availability and risk Overview of subject matterReliability, availability and maintainability

in life-cycle designProcess vs. product reliability

4PROBABILITY OVERVIEW Axioms of probabilityBayes theoremExpectationProbability density function (PDF)Cumulative distribution function (CDF)Failure rate, failure frequencyReliability function, mean-time-to-failure (MTTF)Common probability distributions

(binomial, Poisson, Erlang,exponential, gamma, normal,log-normal, Weibull, extreme value, composite)

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STATISTICS OVERVIEW Point and interval estimationMaximum likelihood estimatorsNon-parametric methodsCensored data

7SYSTEM RELIABILITY MODELINGAND PREDICTION - STATIC METHODS

Reliability block diagrams forseries, parallel, M-out-of-N systems(decomposition method, minimum cut-set method,minimum-tie-set method)

Failure-modes-and-effects analysisFault-tree/event-tree analysis,

concept of statistical importanceRoot cause analysisModeling of statistically dependent failures

(including common mode failures)New static methods(influence diagrams, digraph technique, GO-FLOW)

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SYSTEM RELIABILITY MODELINGAND PREDICTION - DYNAMIC METHODS

Back-up systems for full or partial function(standby systems including standby redundancy)

Monte-Carlo methodsMarkov models, state-space diagrams,

steady-state probabilitiesNew dynamic methods (DYLAM, cell-to-cell-mapping)

TAKE-HOME MIDTERM


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MODULE TOPIC SUBJECT MATTER HOURS

4MAINTAINABILITY AND AVAILABILITY Maintenance as a stochastic point process Avail-ability analysis of systems with repair Designtradeoff analysis Preventive maintenance

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DEVICE AND COMPONENT HAZARDRATES

Operating conditions Acceleration factors Hazardrate units Hazard rates, examples

2RELIABILITY OF PROGRAMMABLE

DEVICES

Software reliability factors and terminology

Severity of faults Current modeling techniquesCommunications software fault-tolerance Sys-tems design for fault detection

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HUMAN RELIABILITY SHARP methodHuman reliability analysis models (simulation models

expert judgement method, analytical methods)

3RELIABILITY OPTIMIZATION Reliability allocation (equal apportionment,ARINC apportionment, AGREE allocationmethod, effort minimization) Dynamic programmingOptimization in probabilistic design

2RISK ANALYSIS Risk perception and acceptability Determination of risk values Formalization of risk assessment Regula-tory requirements

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RELIABILITY AND RISK MANAGEMENT Business factors to consider in reliability and riskmanagement (cost, functionality, producibility, ser-viceability, program schedule) Organizational factors

to consider in reliability and risk management(resource allocation, training) Applicable standardsFailure analysis procedure,model reliability program

TAKE-HOME FINAL


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Proposition

If

P(A1A2) 0,

thenP(A1 + A2) = P(A1) + P(A2) P(A1A2)

Proof

First, recalling that "A1A2" means "A1 and A2" and "A1 + A2" means "A1 or A2", we can show

A1 = A1(A2 + A2) = A1A2 + A1A2

and

A1 + A2 = A1(A2 + A2) + A2(A1 + A1)

= A1A2 + A2A1 + A1A2.

Then

P(A1) = P[A1(A2 + A2)] = P(A1A2) + P(A1A2) => P(A1A2) = P(A1) P(A1A2)

and

P(A1 + A2) = P(A1A2 + A2A1 + A1A2)

= P(A1A2 + A2A1) + P(A1A2)

= P(A1A2) + P(A2A1) + P(A1A2)

= P(A1) P(A1A2) + P(A2) P(A1A2) + P(A1A2)

=> P(A1 + A2) = P(A1) + P(A2) P(A1A2)

Q.E.D.

Often the probability assigned to an event reflects a perceived state of knowledge rather than

having been deduced axiomatically. Such a probability is called subjective probability.


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Some Examples

Example 1

The outcome of 20 coin tosses is given below. Find the frequency of "tails" as a function

increasing number of tosses.

Toss# 1 2 3 4 5 6 7 8 9 10

Outcome h h t h t h t t t h

Toss# 11 12 13 14 15 16 17 18 19 20

Outcome h t t h t h t h h t

Solution

Frequency =Number of Desirable Outcomes

Number of Events

Toss# 1 2 3 4 5 6 7 8 9 10

Frequency 0 0 1/3 1/4 2/5 2/6 3/7 4/8 5/9 5/10

Toss# 11 12 13 14 15 16 17 18 19 20

Frequency 5/11 6/12 7/13 7/14 8/15 8/16 9/17 9/18 9/19 10/20


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Example 2

What is the probability of rolling 4 in single throw of 2 dice?

Solution

Possible combinations leading to the desired event and their respective probabilities are the fol-

lowing:

Event Die#1 Die#2 Probability of Event

A 1 3 1/36

B 2 2 1/36

C 3 1 1/36

Since the events are mutually exclusive

P(A + B + C) = P(A) + P(B) + P(C) = 3/36 = 1/12

Example 3

The turn signal circuit of an automobile consists of the turn signal lamp, the dashboard indicator

and a fuse. The turn signal does not work if any these circuit elements fail and each elements

can fail with a probability of 0.2 during 50,000 miles of driving. What is the probability that the

turn signal will fail during 50,000 miles of driving?

Solution

Define the events:

A: Turn signal lamp fails B: Dashboard indicator fails

C: Fuse fails D: Turn signal fails

Then

P(D) = P(A + B + C) = P(A + B) + P(C) P(A + B)P(C)

= P(A) + P(B) P(A)P(B) + P(C) P(C)[P(A) + P(B) P(A)P(B)]

= P(A) + P(B) + P(C) P(A)P(B) P(A)P(C) P(B)P(C) + P(A)P(B)P(C)

= 3(0. 2) 3(0. 2)2 + (0. 2)3 = 0. 488


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The Bayes Equation

From event decomposition rules recall that

P(Ai

B) = P(Ai

|B)P(B) = P(B|Ai

)P(Ai

)

and ifAi (i = 1, . . . , n) are mutually exclusive and exhaustive

P(B) = P(B|A1)P(A1) + P(B|A2)P(A2) + . . . + P(B|An)P(An) .

Then

P(Ai |B) =P(B|Ai)P(Ai)

n

i=1 P(B|Ai)P(Ai)

This equation is known as the Bayes Equation. The Bayes equation relates the a priori probabili-

ties P(B|Ai) to the a posteriori probabilities P(Ai |B).

Example 5

In a communication channel a 0 or 1 is transmitted correctly with probabilities P(X= 0) = P0and P(X= 1) = P1, respectively. Due to noise in the channel, a 0 can be received as 1 with prob-ability P(Y = 1|X= 0) = and a 1 can be received as 0 also with probability

P(Y = 0|X= 1) = . What is the probability that a 1 is transmitted if a 1 is received?

Solution

Note that the events X= 1 and X= 0 are exhaustive. Then

P(X= 1|Y = 1) =P(Y = 1 , X = 1)

P(Y = 1)

=

P(Y = 1|X= 1)P(X= 1)

P(Y = 1|X = 1)P(X= 1) + P(Y = 1|X= 0)P(X= 0)

=P1(1 )

P1(1 ) + P0

If = 0, then P(X= 1|Y = 1) = 1 and the channel is said to be noiseless.


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Example 6

Suppose there exists a fictitious test for instrumentation degradation which has the property

P(A|B) = P(A|B) = 0. 95 for P(B) = 0. 005,

where the events A, B, A and B are defined as

A The test states that the instrument is degraded

B Instrument is degraded

A The test states that the instrument is functioning properly

B The instrument is functioning properlyIs the test a good test?

Solution

In order to judge whether the test is good or not we need to know the likelihoods of the events

that:

the instrument is degraded when the test states that the instrument is degraded (i.e. P(B|A)),

and

the instrument is functioning properly when the test states that the instrument is functioning

properly, (i.e. P(B|A)).

From the Bayes equation

P(B|A) =P(A|B)P(B)

P(A|B)P(B) + P(A|B)P(B)

=(0. 95)(0. 005)

(0. 95)(0. 005) + (0. 05)(0. 995)= 0. 087156,

since P(A|B) + P(A|B) = 1 => P(A|B) = 1 0. 95, and

P(B|A) = P(A|B)P(B)P(A|B)P(B) + P(A|B)P(B)

=(0. 95)(1 0. 005)

(0. 95)(1 0. 005) + (0. 05)(0. 005)= 0. 999736

So the test is good for verifying that the instrument is functioning properly but not for detecting

the degradation. This conclusion can be also inferred from


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P(A|B) + P(A|B) = 1 => P(A|B) = 1 0. 95.

i.e., the probability that the test will state that the instrument is functioning properly when it is

degraded is 95%.

Probability Distribution Functions

Let p(x) dx Probability that the stochastic variable x is between x and x + dx

The p(x) is called the probability distribution function or probability density function (pdf) for

the variable x. Giv en p(x), the probability that x is within xmin and xmax can be found from

P(xmin x xmax) =

xmax

xmin

dx p(x)

This expression follows from the mutual exclusiveness of the events x = x1, x = x2, x = x3 etc.The P(xmin x xmax) PX(x) where X {x: xmin x xmax} is called the cumulative distri-bution function or Cdf ofx over X.

Example 7

For a certain type of resistors, the probability that the resistor resistance has the value rohms is

given by the distribution

p(r) = 12(200)2

exp 1

2

(r 1000)240, 000

a) What is the probability that the resistance of a resistor is between rmin = 900 and rmax =

1100 ohms?

b) What is the expected (or average) value ofr?

Solution

a) The Cdf we need to find is

PR P(rmin r rmax) =

rmax

rmin

dr p(r) = 12erf

rmax 10002002

erf

rmin 10002002

=1

2

erf

1100 10002002

erf

900 10002002


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=1

2{erf(0. 353553) erf(0. 353553)} = 0. 383

erf(z) =2

z

0e u

2

du

b) The expected value ofris

< r> =

dr rp(r) =

dr r2(200)2 exp

1

2

(r 1000)2

40, 000

= 1000 ohms.

Sometimes it is necessary to find the pdf of a given function y(x) of the stochastic variable x. In

this situation

p(y) = abs

dx

dy

p(x)x=x(y)

Example 8

Suppose

p(t) = e t , 0 t< .Then for

y = e t => ln(y) = t => absdtdy

=1y

,

p(y) =y

y= y1 , 0 y 1

Another practically important problem is to find the pdf pZ(z) ofz = x + y where x and y are sta-tistically independent stochastic variables with given pdfs pX(x) and pY(y) defined over

X {x: xmin x xmax} and Y {y: ymin y ymax}, respectively. First, we note that if

A: event that x is within x and x + dxand

B: event that y is within y and y + dy,

then

pXY(x, y)dxdy = Pr{AB} = pX(x)pY(y)dxdy => pXY(x, y) = pX(x)pY(y)

From the definition of Cdf


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PZ(z) =X

Y

dx dy pXY(x, y) =xmax

xmin

dx pX(x)zx

ymin

dy pY(y)

=> pZ(z) = PZ(z)

z=

xmax

xmindx pX(x)pY(z x) =

ymax

ymindy pY(y)pX(z y).

Example 9

Production Lines A and B in a factory produce shafts and bushings, respectively. In the finished

product, the bushings are supposed to fit over the shafts with a minimum clearance of 0.1 mm

and a maximum clearance of 0.5 mm. The probability that Line A will produce a shaft with

radius x mm is

pA(x) =1

2(0. 1)e

(x 10)2

0.02 mm1

and the probability that Line B will produce a bushing with inner radius y mm is

pB(y) =1

2(0. 1)e

(y 10.3)2

0.02 mm1.

Find the percent of finished products rejected.

Solution

Let z = y x. We want to find

1

0.5

0.1

dz p(z).Assume that Lines A and B are statistically independent. Then

p(z) =

dx pA(x)pB(z + x) =

5. 00

e (25.00z

2 15.00z + 2.25) mm1.

=>

0.5

0.1

dz p(z) = 0. 84Then the percent of finished products rejected is 100(1 0. 84) = 16 %


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9.5 9.6 9.7 9.8 9.9 10 10.1 10.2 10.3 10.4 10.50

0.2

0.4

0.6

0.8

1

1.2

1.4

x (mm)

pA(x)/mm

9.8 9.9 10 10.1 10.2 10.3 10.4 10.5 10.6 10.7 10.80

0.2

0.4

0.6

0.8

1

1.2

1.4

y (mm)

pB)(y)/m

m

0.2 0.1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.80

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

z(mm)

pz(z)/mm

Rejected Rejected

Reliability Engineering Lec Notes #2

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