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DEVELOPMENT EVALUATIONS

Lecture Templates

Marko Nagode, Simon Oman, Domen Seruga

University of Ljubljana, Faculty of Mechanical Engineering

Askerceva 6, SI-1000 Ljubljana, Slovenia

25. Februry 2019

ii

Contents

1 Introduction 1

1.1 Concepts, Terms and Definitions . . . . . . . . . . . . . . . . . . . 3

1.2 Product Attributes . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.3 Basic Time Divisions . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.4 Effectiveness and Costs . . . . . . . . . . . . . . . . . . . . . . . . 7

2 Basic Reliability Models 9

2.1 The Failure Distribution . . . . . . . . . . . . . . . . . . . . . . . 9

2.2 Mean Time to Failure . . . . . . . . . . . . . . . . . . . . . . . . . 11

2.3 Hazard Rate Function . . . . . . . . . . . . . . . . . . . . . . . . 12

2.4 Bathtub Curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

2.5 Conditional Reliability . . . . . . . . . . . . . . . . . . . . . . . . 13

2.6 Probability Models . . . . . . . . . . . . . . . . . . . . . . . . . . 13

2.6.1 The Exponential Distribution . . . . . . . . . . . . . . . . 13

2.6.2 The Weibull Distribution . . . . . . . . . . . . . . . . . . . 14

2.6.3 The Normal Distribution . . . . . . . . . . . . . . . . . . . 15

2.6.4 The Log-normal Distribution . . . . . . . . . . . . . . . . 16

3 Reliability of Systems 17

3.1 Series and Parallel Configurations . . . . . . . . . . . . . . . . . . 18

3.2 Parallel configurations k out of n . . . . . . . . . . . . . . . . . . 18

3.3 Combined Series-Parallel Systems . . . . . . . . . . . . . . . . . . 19

3.4 Complex Configurations . . . . . . . . . . . . . . . . . . . . . . . 19

3.4.1 Decomposition . . . . . . . . . . . . . . . . . . . . . . . . 20

3.4.2 Enumeration . . . . . . . . . . . . . . . . . . . . . . . . . 21

3.4.3 The System Structure Function . . . . . . . . . . . . . . . 21

3.4.4 Coherent Systems . . . . . . . . . . . . . . . . . . . . . . . 22

3.4.5 Minimal Path and Cut Sets . . . . . . . . . . . . . . . . . 23

3.4.6 System Bounds . . . . . . . . . . . . . . . . . . . . . . . . 24

3.5 Low and High Level Redundancy . . . . . . . . . . . . . . . . . . 24

3.6 Three State Devices . . . . . . . . . . . . . . . . . . . . . . . . . . 25

3.6.1 Series Structure . . . . . . . . . . . . . . . . . . . . . . . . 25

3.6.2 Parallel Structure . . . . . . . . . . . . . . . . . . . . . . . 26

iii

CONTENTS

3.6.3 Low Level Redundancy . . . . . . . . . . . . . . . . . . . . 26

3.6.4 High Level Redundancy . . . . . . . . . . . . . . . . . . . 27

4 State Dependent Systems 29

4.1 Markov Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

4.2 Load Sharing Systems . . . . . . . . . . . . . . . . . . . . . . . . 32

4.3 Standby Systems or Passive Parallel Configurations . . . . . . . . 32

4.4 Passive Parallel System of Identical Components . . . . . . . . . . 33

4.5 Standby Systems with Switching Failure . . . . . . . . . . . . . . 34

4.6 Degraded Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

4.7 Three State Devices . . . . . . . . . . . . . . . . . . . . . . . . . . 35

5 Physical Reliability Models 37

5.1 Covariate Models . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

5.2 Static Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

5.3 Dynamic Models . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

5.3.1 Periodic Loads . . . . . . . . . . . . . . . . . . . . . . . . 40

5.3.2 Random Loads . . . . . . . . . . . . . . . . . . . . . . . . 41

5.3.3 Random Fixed Stress and Strength . . . . . . . . . . . . . 41

5.4 Physics of Failure Models . . . . . . . . . . . . . . . . . . . . . . 42

6 Design for Reliability 43

6.1 Reliability Objectives . . . . . . . . . . . . . . . . . . . . . . . . . 44

6.1.1 Life Cycle Costs . . . . . . . . . . . . . . . . . . . . . . . . 45

6.2 Reliability Allocation . . . . . . . . . . . . . . . . . . . . . . . . . 45

6.2.1 Exponential Case . . . . . . . . . . . . . . . . . . . . . . . 46

6.2.2 Optimal allocation . . . . . . . . . . . . . . . . . . . . . . 46

6.2.3 ARINC Method . . . . . . . . . . . . . . . . . . . . . . . . 47

6.2.4 AGREE Method . . . . . . . . . . . . . . . . . . . . . . . 47

6.2.5 Redundancies . . . . . . . . . . . . . . . . . . . . . . . . . 48

6.3 Design Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

6.3.1 Parts and Material Selection . . . . . . . . . . . . . . . . . 49

6.3.2 Derating . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

6.3.3 Stress Strength Analysis . . . . . . . . . . . . . . . . . . . 50

6.3.4 Complexity and Technology . . . . . . . . . . . . . . . . . 51

6.3.5 Redundancy . . . . . . . . . . . . . . . . . . . . . . . . . . 51

6.4 Failure and Effect Analysis . . . . . . . . . . . . . . . . . . . . . . 52

6.4.1 Review of the Process . . . . . . . . . . . . . . . . . . . . 53

6.4.2 Identification of the Potential Failure Modes . . . . . . . . 55

6.4.3 Identification of Potential Effects of Each Failure Mode . . 55

6.4.4 Assigning a Severity Rating for Each Failure Mode . . . . 55

6.4.5 Assigning an Occurrence Rating for Each Failure Mode . . 57

iv

CONTENTS

6.4.6 Assigning a Detection Rating for Each Failure Mode and/or

Effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

6.4.7 Calculation of the Risk Priority Numbe . . . . . . . . . . . 57

6.4.8 Prioritizing the Failure Modes for Action . . . . . . . . . . 58

6.4.9 Taking Action to Eliminate or Reduce the High Risk Fail-

ure Modes . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

6.4.10 Calculation of the Resulting RPN . . . . . . . . . . . . . . 58

6.5 System Safety and Fault Tree Analysis . . . . . . . . . . . . . . . 59

6.5.1 Fault Tree Analysis . . . . . . . . . . . . . . . . . . . . . . 59

6.5.2 Minimal Cut Sets . . . . . . . . . . . . . . . . . . . . . . . 61

6.5.3 Quantitative Analysis . . . . . . . . . . . . . . . . . . . . . 62

7 Maintainability 63

7.1 Analysis of Downtime . . . . . . . . . . . . . . . . . . . . . . . . . 63

7.2 The Repair Time Distribution . . . . . . . . . . . . . . . . . . . . 64

7.3 System Repair Time . . . . . . . . . . . . . . . . . . . . . . . . . 65

7.4 Reliability under Preventive Maintenance . . . . . . . . . . . . . . 66

7.5 Stochastic Point Processes . . . . . . . . . . . . . . . . . . . . . . 67

7.5.1 Renewal Process . . . . . . . . . . . . . . . . . . . . . . . 68

7.5.2 Minimal Repair process . . . . . . . . . . . . . . . . . . . 69

7.5.3 Overhaul . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

Note: The lecture templates are summarized after the book of Charles E. Ebeling,

An Introduction to Reliability and Maintainability Engineering, McGraw-Hill,

1997.

v

CONTENTS

For a reader who would like to have an in-depth knowledge of the subject, it is

recommended to read the entire book [Ebeling] content.

vi

Chapter 1

Introduction

Causes of the failures include:

• bad engineering design,

• faulty construction or manufacturing processes,

• improper use,

• human error,

• poor maintenance,

• inadequate testing and inspection,

• lack of protection against excessive environmental stress, etc.

The impact of product and system failures can be:

• so insignificant that the user does not even notice it,

• but they can also lead to a catastrophic consequences for humans and the

environment.

When evaluating on safety we conclude that the product will operate safely if

Ymin

Xmax

≥ SF

where SF is a safety factor. Because durability Y and load X are random variables

with probability density function fY (y) in fX(x), even safety factor SF� 1 does

not guarantee the operation without defects. Evaluation on safety is therefore

not sufficient because it does not take into account the scatter of the load and

the durability.

A modern development process, therefore, in addition to evaluations on:

• functionality,

1

1. INTRODUCTION

Xmax

fY(y)

fX(x)

Ymin

Figure 1.1. Circular cable car.

• safety,

• life cycle costs,

• ecological value,

• suitability of technology,

• aesthetic value,

• usefulness value,

• ergonomics,

• suitability for recycling, etc.

also includes evaluations on:

• reliability,

• maintainability,

• supportability,

• availability,

• effectiveness and

• product value.

2

1.1 Concepts, Terms and Definitions

Effectiveness

Readiness for

operationAvailability Capability

Maintainability SupportabilityReliability

Figure 1.2. Elements of the product effectiveness.

1.1 Concepts, Terms and Definitions

The ultimate objective of each product is the quality performance of the required

functions within the permissible deviations for a certain amount of time when

used under certain conditions of use, environmental and maintenance conditions,

at acceptable costs and minimizing environmental pollution. The function can be

described with the output characteristics of the product (the quality of message

transmission in communication systems, the load capacity of the conveyor or the

reliability of the brake) and the quality of the product is determined by reliability,

maintainability, supportability, availability, effectiveness, and product value.

Reliability is defined to be the probability that a component or system will

perform a required function within the permissible deviations for a given period

of time when used under stated operating and environmental conditions. Prior

to the determination of reliability, it is necessary to clearly define the failures, to

connect them with the functions of the product and to select the unit of time.

The specified time interval can be based on calendar or clock time, operating

hours or cycles. The cycle can be a turn of the engine, a load cycle or a block of

time load history. The terms of use and the environmental conditions in which

the product operates must meet the design requirements.

Maintainability is defined to be the probability that a failed component or

system will be restored or repaired to a specified condition within a period of

time when maintenance is performed in accordance with prescribed procedures.

Maintenance is usually a function of time and depends only on the time of repair.

The delay in logistical support, waiting time for maintenance personnel and parts

and administrative time are not included in the maintainability and are usually

treated separately.

Supportability is defined as the degree to which product characteristics and

planned support resources, including personnel, meet project requirements. Sup-

portability is a function of time and depends on the delay in logistic support and

3

1. INTRODUCTION

Product value

Efectiveness Life cycle costs Timetable Personnel

Figure 1.3. Elements of the product value.

the maintenance delay.

Reliability, maintainability and supportability are at a higher level linked to avail-

ability. Availability is defined as the probability that a component or system is

performing its required function at a given point in time or time interval when

used under stated operating, environmental and maintenance conditions. Avail-

ability is always greater or equal to reliability and is the preferred measure when

the system or component can be restored since it accounts for both failures (re-

liability) and repairs (maintainability).

In the broadest sense, product quality is determined by its effectiveness and prod-

uct value. Effectiveness is the probability that the product will meet project re-

quirements under certain operating, environmental and maintenance conditions,

depending on the readiness for operation, availability and capability. Readiness

for operation corresponds to a state vector at the first entry into operation,

capability however specifies the probability that the product will fulfill the re-

quired functions depending on the state in which it is located. The breakdown

of effectiveness is shown in Figure 1.2.

Product value links effectiveness, life cycle costs, timetables and personnel (Fig-

ure 1.3).Previous analyzes have shown that life cycle costs can be reduced by de-

voting enough attention to reliability, maintainability and supportability already

in the early stages of the product’s life cycle. If we want to increase the value,

we must increase the effectiveness at minimal costs in as short a time as possible

and with as little personnel as possible.

4

1.2 Product Attributes

1.2 Product Attributes

- effectiveness- acquisition costsoperations and- support costs- salvage value- timetables- personnel

reliability- selection of materials and parts- derating- stress analysis- strength analysis- complexity and technology- redundancymaintainability- failure isolation and diagnostics- standardization and parts - exchangeability- modularity and accessibility- repair or replacement- proactive maintenancesuportability- number of redundant components- number of spare parts- number of maintenance channelsavailability

operating- reach- speed- precision class- sensitivity- useful cargo- output power- ease of usephysical- volume and density- mass- shapefunctional- safety- fulfillment rate- functions

quality performance effectiveness product value

Tabel 1.1. Product attributes.

5

1. INTRODUCTION

costs share (%)

detailing andoptimization

testing

preparation of production

production

0 20 40 60 80 100

Pro

du

ct li

fe c

ycle

sta

ges

conceiving5

60

3

20

2

10

5

5

85

5

Figure 1.4. Cost breakdown by product life cycle stages.

1.3 Basic Time Divisions

The operating time may include:

• working hours,

• free time and time in standby mode.

Working hours is the time during which the product performs the required func-

tions within the permissible tolerances. In the free time there is no need for

operation, while the product performs only a limited number of functions or does

not operate during the standby time. The free time and standby time vary in the

speed of the product’s transition to the operating state. The non-operating time

is divided into:

• delay in logistics support,

• delay in maintenance and

• repair time.

t

free time

time in standby mode

working hours

time of operation down time

delay in logistics support

delay in maintenance

repair time

Figure 1.5. Time bus of the product status.

6

1.4 Effectiveness and Costs

1.4 Effectiveness and Costs

Epopt

pre-deliverycosts

after-deliverycosts

totalcosts

effectiveness

costs

Figure 1.6. Effectiveness and the manufacturer costs.

costs

effectiveness

Euopt

after-deliverycosts

purchasecosts

totalcosts

Figure 1.7. Effectiveness and the user costs.

7

1. INTRODUCTION

8

Chapter 2

Basic Reliability Models

We distinguish four characteristic functions that describe the reliability of the

product:

• reliability function,

• cumulative distribution function,

• probability density function and

• hazard rate function.

Each of these functions provides a unique and complete reliability description

from a different angle of view. If we know one, we can use it to express all others.

Knowing these functions we can also calculate:

• mean time to failure,

• failure distribution variance,

• median time to failure and,

• most likely observed failure time (mode).

2.1 The Failure Distribution

Reliability R(t) is the probability that the time to failure is going to be T ≥ t

R(t) = Pr{T ≥ t} =

∫ ∞t

f (t)dt (2.1)

Cumulative distribution function F (t) is the probability that a failure occurs

before time t

F (t) = Pr{T < t} = 1−R(t) =

∫ t

0

f (t)dt (2.2)

9

2. BASIC RELIABILITY MODELS

0 20 40 60 80 100

0.0

0.2

0.4

0.6

0.8

1.0

t

F(t),R(t)

R(t) F(t)

Figure 2.1. Reliability and cumulative distribution functions.

Probability density function is defined as

f (t) =dF (t)

dt= −dR(t)

dt(2.3)

Figure 2.2. Probability density function.

10

2.2 Mean Time to Failure

2.2 Mean Time to Failure

Expected mean time to failure MTTF is defined as

MTTF = E[T ] =

∫ ∞0

tf (t)dt (2.4)

It can also be shown that

MTTF =

∫ ∞0

R(t)dt (2.5)

Median time to failure tmed

R(tmed) = Pr{T ≥ tmed} = 0.5

Most likely observed failure time tmod

f (tmod) = max0≤t<∞

f (t)

Failure distribution variance

σ2 = E[(T −MTTF)2] =

∫ ∞0

(t−MTTF)2f (t)dt (2.6)

0 tmed 3 4

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

t

f(t)

MTTFtmod

area = 0.5

Figure 2.3. Comparison between mean time to failure, median and mode time.

11


2.3 Hazard Rate Function

Hazard rate function λ(t) provides an instantaneous (at time t) rate of failure

given that the system has survived to time t

λ(t) =f (t)

R(t)(2.7)

Reliability as a function of hazard rate function

R(t) = exp

{−∫ t

0

λ(t)dt

}(2.8)

Cumulative failure rate

L(t) =

∫ t

0

λ(t)dt (2.9)

Average failure rate defined between two times t1 ≤ T ≤ t2

AFR(t1, t2) =1

t2 − t1

∫ t2

t1

λ(t)dt (2.10)

2.4 Bathtub Curve

0 1000 2000 3000 4000 5000

0.000

0.005

0.010

0.015

0.020

t

l(t

)

randomfailures

wearoutfailures

earlyfailures

Figure 2.4. Bathtub curve.

12

2.5 Conditional Reliability

earlyfailures

randomfailures

wearoutfailures

Failuretype cause of failure prevention

burn-in testingscreeningquality controlacceptance testing

redundancysafety factor

deratingPreventive maintenaceParts replacemnettechnology

production failurepoor quality controlcontaminationpoor workmanship

operating conditionsenvironmental conditionshuman errorrandom events

material fatiguecorrosionagingwear

model

DFR

CFR

IFR

Tabel 2.1. Types and causes of failures and failure prevention.

2.5 Conditional Reliability

Conditional reliability is the probability that the product will be in operation

until the moment t provided that it was in the operating state after the burn-in

period or the warranty period T0.

R(t|T0) =R(t+ T0)

R(T0)(2.11)

Residual mean time to failure (MTTF taking into account T0)

MTTF(T0) =

∫ ∞0

R(t|T0)dt =1

R(T0)

∫ ∞T0

R(t)dt (2.12)

2.6 Probability Models

2.6.1 The Exponential Distribution

Reliability function

R(t) = exp

{−λ∫ t

0

dt

}= e−λt (2.13)

Cumulative distribution function of failures

F (t) = 1− e−λt (2.14)

Exponential probability density function

f (t) = −dR(t)

dt= λe−λt (2.15)

13


0 1 2 3 4 5

0

1

2

3

4

5

t

f(t)

λ = 5

λ = 1λ = 0.25

Figure 2.5. The exponential probability density function.

Mean time to failure

MTTF =

∫ ∞0

R(t)dt =

∫ ∞0

e−λt =e−λt

−λ

∣∣∣∣∞0

=1

λ(2.16)


σ2 =

∫ ∞0

(t−MTTF)2f (t)dt =

∫ ∞0

(t− 1/λ)2λe−λtdt =1

λ2(2.17)

The probability that the product (component) will be in operation until MTTF

is

R(MTTF) = e−MTTF/MTTF = e−1 = 0.367879

CFR model does not have a memory.

2.6.2 The Weibull Distribution

Hazard rate function

λ(t) =β

θ

(t

θ

)β−1

za θ > 0, β > 0 in t ≥ 0


R(t) = exp

{−∫ t

0

β

θ

(t

θ

)β−1

dt

}= e

−(tθ

)β(2.18)

Probability density function

f (t) =β

θ

(t

θ

)β−1

e−(tθ

)β(2.19)

14

2.6 Probability Models

0 1 2 3 4 5

0.0

0.3

0.6

0.9

1.2

1.5

t

f(t)

β = 0.5

β = 1.5 β = 2.0

β = 4.0

Figure 2.6. The Weibull probability density function for θ = 2.0.

0 1 2 3 4 5

0.0

0.5

1.0

1.5

2.0

t

f(t)

σ = 0.2

σ = 0.5

σ = 1.0

Figure 2.7. The Normal probability density function for µ = 2.0.


MTTF = θΓ[1 + 1/β ] (2.20)


σ2 = θ2(Γ[1 + 2/β ]− Γ2[1 + 1/β ]) (2.21)

2.6.3 The Normal Distribution


f (t) =1√2πσ

exp

{−(t− µ)2

2σ2

}(2.22)

15



R(t) =1√2πσ

∫ ∞t

exp

{−(t− µ)2

2σ2

}dt (2.23)


MTTF = µ (2.24)

2.6.4 The Log-normal Distribution

s = 0.1

0 1 2 3 4 5

0.0

0.5

1.0

1.5

2.0

t

f(t)

s = 2.0

s = 1.0

Figure 2.8. The Log-normal probability density function for tmed = 2.0.


f (t) =1√

2πtsexp

{−1

2

(ln(t/tmed))2

s2

}(2.25)

Reliability function expressed by Laplace function

R(t) = 1− Φ((ln t− ln tmed)/s) (2.26)


MTTF = tmed exp(s2/2) (2.27)


σ2 = t2med exp s2(exp s2 − 1) (2.28)

16

Chapter 3

Reliability of Systems

I2I1 I3 I5

I4

E1

S2S1 S3 S4

P2P1

P3

P5

P4system

product

assembly

sub-assembly

element

component

sub-component

Figure 3.1. Reliability block diagram of a system.

17

3. RELIABILITY OF SYSTEMS

21 n

1

2

n

(a) (b)

Figure 3.2. Reliability block diagram for n components in series (a) and in parallel

(b).

3.1 Series and Parallel Configurations

Reliability of a system made of n components in series

Rs(t) =n∏i=1

Ri(t) (3.1)

Reliability of a system made of n statistically independent components in parallel

Rs(t) = 1−n∏i=1

(1−Ri(t)) (3.2)

123456789

10

9.000000E-018.100000E-017.290000E-016.561000E-015.904900E-015.314410E-014.782969E-014.304672E-013.874205E-013.486784E-01

serialn

9.000000E-019.900000E-019.990000E-019.999000E-019.999900E-019.999990E-019.999999E-011.000000E+001.000000E+001.000000E+00

parallel

Tabel 3.1. The influence of component number n (reliability of each component

R = 0.9) on system reliability for serial and parallel configurations.

3.2 Parallel configurations k out of n

A generalization of n parallel components occurs when a requirement exists for

k out of n identical and independent components to function for the system to

function. Probability that at least k out of n statistically independent parallel

18

3.3 Combined Series-Parallel Systems

components will function is

Rs(t) =n∑i=k

(n

i

)R(t)i(1−R(t))n−i (3.3)

3.3 Combined Series-Parallel Systems

1

2

3

AB

4 5 6

CD

Figure 3.3. Combined series-parallel system.

3.4 Complex Configurations

For certain systems, the component configuration is such that the system relia-

bility cannot be simply decomposed into series and parallel relationship. Such

systems can be solved using the following procedures:

• decomposition,

• enumeration,

• minimal paths,

• minimal cuts and

• Monte Carlo symulations.

19


3.4.1 Decomposition

1

2

3

4

(b)

1

2

3

4

(c)

1

2

3

4

(a)

5

Figure 3.4. Decomposition of a linked network: Complex configuration (a), compo-

nent 5 does not fail (b) and component 5 fails (c).

20


3.4.2 Enumeration

123456789

1011121314151617181920212223242526272829303132

SFSSSSFSSSFFFSSSFSSFFSFFSFFSFFFF

1i

SSFSSSFFSSSSSFFSFFSSSFFFFSFFSFFF

2

SSSFSSSFFSFSSSSFFFFFSSSSFFFFFSFF

3

SSSSFSSSFFSFSFSSSFFFFFFSSSFFFFSF

4

SSSSSSFSFSSSSSSSFFFFFSFFFSFFFFFF

system

SSSSSFSSSFSSFSFFSSFSFFSFFFSFFFFF

5Pi

operates

0.5848200.0649800.0649800.0307800.0307800.146205

0.003420

0.0076950.0034200.0034200.0162450.0034200.0162450.007695

0.000855

0.000855

Pihas failed

0.007220

0.001620

0.0003800.0001800.0004050.0001800.000855

0.0003800.0018050.000855

0.0000200.0000450.0000450.0000950.0000950.000005

= 0.985815 = 0.014185

Figure 3.5. Enumeration method for complex configuration from Figure 3.4.

3.4.3 The System Structure Function

A very general approach for analyzing the reliability of complex systems is trough

the use od the system structure function. First component state need to be

defined by

Xi =

{1 if component operates

0 if component has failed(3.4)

21


If the system configuration consists of n components there are 2n possible states.

Each state is defined by the state vector

X = [X1, . . . , Xn] (3.5)

and the system structure function

Ψ(X) =

{1 if system operates

0 if system has failed(3.6)

Therefore for a series system the system structure function is defined by

Ψ(X) =n∏i=1

Xi = min{X1, . . . , Xn}

and for a parallel system by

Ψ(X) = 1−n∏i=1

(1−Xi) = max{X1, . . . , Xn}

Reliability of the system and the system structure function are interconnected

Rs = E[Ψ(X)] = 0 · Pr{Ψ(X) = 0}+ 1 · Pr{Ψ(X) = 1} (3.7)

From (3.4) and (3.7) it follows

Xni = Xi E[Xi] = Ri

For the system composed of n statistically independent series components it ap-

plies

Rs = Pr{Ψ(X) = 1}= Pr{min{X1, . . . , Xn} = 1}= Pr{X1 = 1, . . . , Xn = 1}= Pr{X1 = 1} · · ·Pr{Xn = 1}= R1 · · ·Rn

and for the system composed of n statistically independent parallel components

it applies

Rs = Pr{Ψ(X) = 1}= 1− Pr{X1 = 0, . . . , Xn = 0}= 1− Pr{X1 = 0} · · ·Pr{Xn = 0}= 1− (1−R1) · · · (1−Rn)

3.4.4 Coherent Systems

A system is coherent when a component reliability improvement does not de-

grade the system reliability. A coherent system has a structure function that is

22


monotonically increasing. That is if Y ≥ X then Ψ(Y) ≥ Ψ(X).

3.4.5 Minimal Path and Cut Sets

A path is a set of components whose functioning ensures that the system func-

tions. A minimal path Pi is one in which all the components within the set must

function for the system to function. The number of minimal paths is upside lim-

ited and denoted by p. A state vector X is a vector of a minimal path when

Ψ(X) = 1 and X < Y for each Y for which it applies Ψ(Y) = 1. The system

structure function is derived from the rules that apply to reliability calculation

Ψ(X) = 1−p∏i=1

(1−∏j∈Pi

Xj) (3.8)

123

p = 4

{1,3}{2,4}

{1,4,5}{2,3,5}

minimal path Pii

[1,0,1,0,0][0,1,0,1,0][1,0,0,1,1][0,1,1,0,1]

vector of minimal pathXi

Tabel 3.2. Minimal path sets for complex configuration from Figure 3.4.

1

2

4

3

2 4

1 3

5

5

Figure 3.6. Replacement block diagram formed from path sets.

A cut is a set of components whose failure will result in a system failure. A

minimal cut Ci is one in which all the components must fail in order for the

system to fail. The number of minimal paths is upside limited and denoted by

c. X is a vector of a minimal cut when Ψ(X) = 0 and X > Y for each Y for

which it applies Ψ(Y) = 0. The system structure function is also derived from

the rules that apply to reliability calculation

Ψ(X) =c∏i=1

(1−∏j∈Ci

(1−Xj)) (3.9)

23


123

c = 4

{1,2}{3,4}

{1,4,5}{2,3,5}

minimal cut Cii

[0,0,1,1,1][1,1,0,0,1][0,1,1,0,0][1,0,0,1,0]

vector of minimal cutXi

Tabel 3.3. Minimal cut sets for complex configuration from Figure 3.4.

1

2

3

4

4

5

3

5

1 2

Figure 3.7. Replacement block diagram formed from cut sets.

3.4.6 System Bounds

The rough upper-bound reliability is given by

Rsu = 1−n∏i=1

(1−Ri) (3.10)

The rough lower-bound reliability is given by

Rsl =n∏i=1

Ri (3.11)

The estimation of upper-bound reliability is given by

Rsu = 1−p∏i=1

(1−∏j∈Pi

Rj) (3.12)

The estimation of upper-bound reliability is given by

Rsl =c∏i=1

(1−∏j∈Ci

(1−Rj)) (3.13)

3.5 Low and High Level Redundancy

Reliability of a system with a low level redundancy

Rl = (1− (1−R)2)2 = R2(2−R)2

Reliability of a system with a high level redundancy

Rh = 1− (1−R2)2 = R2(2−R2)

24

3.6 Three State Devices

1

1

2

2

1

1

2

2

(b) (c)

1 2

(a)

Figure 3.8. Series configuration (a), low level redundancy (b)and high level redun-

dancy (c).


For three state devices it is typical that besides an operating state they have two

failure modes. For example failure at closing (short failure) and failure at opening

(open failure) the valve or electrical switch.

3.6.1 Series Structure

2

1(a)

2

1(d)

22

11(b) (c)

Figure 3.9. Series configuration. Failure ob both valves at closing (a), failure of valve

no. 1 at opening (b), failure of valve no. 2 at opening (c) and failure of both valves at

opening (d).

25


3.6.2 Parallel Structure

21(a) 21(b)

21(c) 21(d)

Figure 3.10. Parallel configuration. Failure ob both valves at opening (a), failure of

valve no. 1 at closing (b), failure of valve no. 2 at closing (c) and failure of both valves

at closing (d).

3.6.3 Low Level Redundancy

Reliability of a system with low level redundancy is calculated by

Rl = 1− (Fo − Fs)

In general it can be written

Rl =n∏i=1

(1− qmoi)−n∏i=1

(1− (1− qsi)m) (3.14)

2

2

2

1

1

1

n

n

n

m

Figure 3.11. Block diagram of a system composed of n assemblies in series where

each assembly has m three state components in parallel.

26


22

11

Figure 3.12. Low level redundancy for n = m = 2.

123456789

10111213141516

SFSSSFSSFFSFSFFF

1left

valvei

SSSFSSFFSFSFFFSF

SSFSSFFSSSFFFSFF

SSSSFSSFFSFSFFFF

1rightvalve

2left

valve

2rightvalve

systemat

closing

SSSSSSFSFFFFFFFF

SSSSSFSFSSSFFFFF

0.011475

0.0114750.0114750.0114750.0020250.0012750.0012750.0020250.000225

Psiat

closing

Fs = 0.052725

systemat

opening

0.002209

0.003249

0.0001410.0001710.0001710.0001410.000009

Poiat

opening

Fo = 0.006091

Tabel 3.4. Low level redundancy for n = m = 2, qo1 = 0.05, qo2 = 0.06, qs1 = 0.15,

qs2 = 0.10, Ps7 = (1 − qs1)qs1qs2(1 − qs2) = 0.011475, Po8 = (1 − qo1)2q2o2 = 0.003249

and Rl = 1− (Fo − Fs) = 0.941184.

3.6.4 High Level Redundancy

Reliability of a system with high level redundancy is calculated by

Rh =

(1−

m∏i=1

qsi

)n

−

(1−

m∏i=1

(1− qoi)

)n

(3.15)

27


21 m

21 m

1 2 m

n

Figure 3.13. Block diagram of a system composed of n assemblies in parallel where

each assembly has m three state components in series.

28

Chapter 4

State Dependent Systems

4.1 Markov Analysis

For a two component series system the system reliability is

Rs(t) = P1(t)

and for two component parallel system as follows

Rs(t) = P1(t) + P2(t) + P3(t)

Since the system must be in one of the states we can further write

P1(t) + P2(t) + P3(t) + P4(t) = 1 (4.1)

If the system at the time t is in state no. 1 then the probability that in time

t + ∆t will still be in the same state equals the probability P1(t) reduced by a

probability of transition to states no. 2 or 3. Probability that failure occurs in the

time interval t ≤ T ≤ t + ∆t with the condition that component no. 1 operates

until the time t is

Pr{t ≤ T ≤ t+ ∆t|T ≥ t} =R1(t)−R1(t+ ∆t)

R1(t)

Probability that the system will be in state no. 1 at time t and will go into the

state no. 2 during the time interval t ≤ T ≤ t+ ∆t corresponds to the associated

1234

SFSF

1state

SSFF

2

SSSF

systemparallelconf.

systemserialconf.

SFFF

Tabel 4.1. Possible states of the system.

29

4. STATE DEPENDENT SYSTEMS

probabilities of both events

R1(t)−R1(t+ ∆t)

R1(t)P1(t)

Probability that the system at time t+ ∆t will still be in state no. 1 therefore is

P1(t+ ∆t) = P1(t)− R1(t)−R1(t+ ∆t)

R1(t)P1(t)

− R2(t)−R2(t+ ∆t)

R2(t)P1(t)

(4.2)

Probability that the system will be in state no. 2 at time t + ∆t corresponds

to the probability that the system is in state no. 2 at time t increased by the

probability of its transition from the state no. 1 to the state no. 2 and decreased

by the probability of its transition from the state no. 2 to the state no. 4

P2(t+ ∆t) = P2(t) +R1(t)−R1(t+ ∆t)

R1(t)P1(t)

− R2(t)−R2(t+ ∆t)

R2(t)P2(t)

(4.3)

Similarly we can derive the probability that the system at time t + ∆t is going

to be in the state no. 3

P3(t+ ∆t) = P3(t) +R2(t)−R2(t+ ∆t)

R2(t)P1(t)

− R1(t)−R1(t+ ∆t)

R1(t)P3(t)

(4.4)

or in the state no. 4

P4(t+ ∆t) = P4(t) +R2(t)−R2(t+ ∆t)

R2(t)P2(t)

+R1(t)−R1(t+ ∆t)

R1(t)P3(t)

(4.5)

If we divide equation (4.2) with ∆t we get

P1(t+ ∆t)− P1(t)

∆t= −R1(t)−R1(t+ ∆t)

∆tR1(t)P1(t)

− R2(t)−R2(t+ ∆t)

∆tR2(t)P1(t)

Because it applies

lim∆t→0

P (t+ ∆t)− P (t)

∆t=dP (t)

dt= P ′(t)

and

lim∆t→0

−R(t+ ∆t)−R(t)

∆t= f (t)

30

4.1 Markov Analysis

the equations (4.2) to (4.5) forms the system of ordinary differential equations of

the first orderP ′1(t)

P ′2(t)

P ′3(t)

P ′4(t)

=

−λ1 − λ2 0 0 0

λ1 −λ2 0 0

λ2 0 −λ1 0

0 λ2 λ1 0

P1(t)

P2(t)

P3(t)

P4(t)

(4.6)

If failure rates are constant then the system of differential equations is analytically

solvable. Probability that the system will be in state no. 1 can be calculated by

an integration method for separating variables∫dP1(t)

P1(t)= −(λ1 + λ2)

∫dt+ C1

from there it follows

lnP1(t) = −(λ1 + λ2)t+ C1

and

P1(t) = e−(λ1+λ2)t+C1

Value of the integration constant C1 is calculated from the condition P1(0) = 1

which says that at time t = 0 the system is in state no. 1 with probability of 1.

From there it follows C1 = 0 and

P1(t) = e−(λ1+λ2)t (4.7)

Probability that the system at time t is in state no. 2 is a result of a linear

differential equation

P ′2(t) + λ2P2(t) = λ1e−(λ1+λ2)t

where

P2(t) = e−λ2∫dt

{λ1

∫e−(λ1+λ2)t+λ2

∫dtdt+ C2

}or

P2(t) = −e−(λ1+λ2)t + C2e−λ2t

Value of the integration constant C2 = 1 comes from boundary condition P2(0) =

0. Probability that the system at time t will be in state no. 2 is therefore

P2(t) = −e−(λ1+λ2)t + e−λ2t (4.8)

Similarly we can derive the following equation

P3(t) = e−λ1t − e−(λ1+λ2)t (4.9)

Probability that the system will be in state no. 4 is

P4(t) = 1− P1(t)− P2(t)− P3(t) (4.10)

31


Whilst the probability of a system composed of two components in series equals

Rs(t) = e−(λ1+λ2)t (4.11)

for a system with composed of two components in parallel applies

Rs(t) = e−λ1t + e−λ2t − e−(λ1+λ2)t (4.12)

4.2 Load Sharing Systems

1

2

1

3

4

2

λ 1

λ2

+

λ2

λ 1+

Figure 4.1. Block and rate diagrams for a two component load sharing system.

A system of differential equations for a load-sharing systemP ′1(t)

P ′2(t)

P ′3(t)

P ′4(t)

=

−λ1 − λ2 0 0 0

λ1 −λ+2 0 0

λ2 0 −λ+1 0

0 λ+2 λ+

1 0

P1(t)

P2(t)

P3(t)

P4(t)

(4.13)

Solution of a differential equation system

P1(t) = e−(λ1+λ2)t

P2(t) =λ1

λ1 + λ2 − λ+2

(e−λ+2 t − e−(λ1+λ2)t)

P3(t) =λ2

λ1 + λ2 − λ+1

(e−λ+1 t − e−(λ1+λ2)t)

(4.14)

4.3 Standby Systems or Passive Parallel

Configurations

1

3

4

2

λ 1

λ2

λ2 -

λ 1

p = 0

p = 0

1

2

Figure 4.2. Block and rate diagrams for a two component standby system with failures

in standby.

32

4.4 Passive Parallel System of Identical Components

A system of differential equations for a standby system without switching failure

p = 0P ′1(t)

P ′2(t)

P ′3(t)

P ′4(t)

=

−λ1 − λ−2 0 0 0

λ1 −λ2 0 0

λ−2 0 −λ1 0

0 λ2 λ1 0

P1(t)

P2(t)

P3(t)

P4(t)

(4.15)


P1(t) = e−(λ1+λ−2 )t

P2(t) =λ1

λ1 + λ−2 − λ2

(e−λ2t − e−(λ1+λ−2 )t)

P3(t) = e−λ1t − e−(λ1+λ−2 )t

(4.16)

System reliability

Rs(t) = e−λ1t +λ1

λ1 + λ−2 − λ2

(e−λ2t − e−(λ1+λ−2 )t) (4.17)

Mean time to failure MTTF

MTTF =1

λ1

+λ1

λ2(λ1 + λ−2 )(4.18)

4.4 Passive Parallel System of Identical

Components

A system of differential equations for a passive parallel system of identical com-

ponentsP ′1(t)

P ′2(t)

P ′3(t)

P ′4(t)

P ′5(t)

=

−λ 0 0 0 0

λ −λ 0 0 0

0 λ −λ 0 0

0 0 λ −λ 0

0 0 0 λ 0

P1(t)

P2(t)

P3(t)

P4(t)

P5(t)

(4.19)

12345

SFFFF

1state

SSFFF

2

SSSFF

3

SSSSF

4

SSSSF

system

Tabel 4.2. Possible states of the system.

33


1

2

3

4

5

λ

λ

λ

λ

1

1

1

1

Figure 4.3. Block and rate diagrams for a passive parallel system of identical compo-

nents.

Reliability of a passive parallel system of n identical components composed and

belonging MTTF are therefore

Rs(t) =n∑i=1

Pi(t) = e−λtn−1∑i=0

(λt)i

i!

MTTF =

∫ ∞0

Rs(t)dt =n−1∑i=0

∫ ∞0

(λt)ie−λt

i!dt =

n−1∑i=0

Γ[i+ 1]

λi!=n

λ

(4.20)

4.5 Standby Systems with Switching Failure

A system of differential equations for a standby system with switching failure

p > 0P ′1(t)

P ′2(t)

P ′3(t)

P ′4(t)

=

−qλ1 − pλ1 − λ−2 0 0 0

qλ1 −λ2 0 0

λ−2 0 −λ1 0

pλ1 λ2 λ1 0

P1(t)

P2(t)

P3(t)

P4(t)

(4.21)

1

3

4

2

qλ 1

λ2

λ2 -

λ 1

p > 0

p > 0

1

2

pλ 1

Figure 4.4. Block and rate diagrams for a two component standby system with failures

in standby and with switching failure.

34

4.6 Degraded Systems

where q = 1− p. Solution of a differential equation system (4.21)

P1(t) = e−(λ1+λ−2 )t

P2(t) =qλ1

λ1 + λ−2 − λ2

(e−λ2t − e−(λ1+λ−2 )t)

P3(t) = e−λ1t − e−(λ1+λ−2 )t

(4.22)

and system reliability

Rs(t) = e−λ1t +qλ1

λ1 + λ−2 − λ2

(e−λ2t − e−(λ1+λ−2 )t) (4.23)

4.6 Degraded Systems

(1) operating state(2) degraded state(3) state of failure

1

1

323

Figure 4.5. Block and rate diagrams for a degraded system.

A system of differential equations for a degraded systemP ′1(t)

P ′2(t)

P ′3(t)

=

−λ1 − λ2 0 0

λ2 −λ3 0

λ1 λ3 0

P1(t)

P2(t)

P3(t)

(4.24)


P1(t) = e−(λ1+λ2)t

P2(t) =λ2

λ1 + λ2 − λ3

(e−λ3t − e−(λ1+λ2)t)

P3(t) = 1− P1(t)− P2(t)

(4.25)


Rs(t) = P1(t) + P2(t) (4.26)


(1) operating state(2) failure at opening(3) failure at closing

1

1

32

Figure 4.6. Block and rate diagrams for a three state device.

35


A system of differential equations for a three state devicesP ′1(t)

P ′2(t)

P ′3(t)

=

−λ1 − λ2 0 0

λ1 0 0

λ2 0 0

P1(t)

P2(t)

P3(t)

(4.27)


P1(t) = e−(λ1+λ2)t

P2(t) =λ1

λ1 + λ2

(1− e−(λ1+λ2)t)

P3(t) = 1− P1(t)− P2(t)

(4.28)


Rs(t) = e−(λ1+λ2)t (4.29)

36

Chapter 5

Physical Reliability Models

In previous chapters, we dealt with reliability models in which the reliability

of a component or system was considered as a function of time only. Because

the operating and environmental conditions can change over time, reliability of-

ten also depends on other factors. Reliability models which in addition to time

also consider other parameters are called covariate models. Static stress-strength

models do not depend on time but on applied load and strength. Unlike static

stress-strength models, dynamic stress-strength models also take into account the

impact of load history. In addition to the static stress-strength reliability models

there are also so called physics of failure models.

5.1 Covariate Models

A simple example of a covariate model for the exponential probability density

function of failures is given by the equation

λ(x) = a0 +n∑i=1

aixi

The reliability of the component by taking into account additional parameters for

the case of the exponential probability density function of failures is calculated

by

R(t) = e−λ(x)t

37

5. PHYSICAL RELIABILITY MODELS

5.2 Static Models

Probability that the stress X will not be greater than x is calculated by

Pr{X < x} = FX(x) =

∫ x

0

fX(x)dx (5.1)

Probability that the strength Y will not exceed the value y can be calculated by

Pr{Y < y} = FY (y) =

∫ y

0

fY (y)dy (5.2)

If stress X is a random variable and strength a constant value Y = y (see Fig-

ure 5.1), then the reliability is

R = Pr{X < y} = FX(y) =

∫ y

0

fX(x)dx (5.3)

0 20 60 80 100

0.00

0.01

0.02

0.03

0.04

x

f X(x)

R

y

Figure 5.1. Reliability for random stress and constant strength.

If strength Y is a random variable and stress a constant value X = x see Fig-

ure 5.2), then the reliability is

R = Pr{Y ≥ x} = 1− FY (x) =

∫ ∞x

fY (y)dy (5.4)

If both stress X and strength Y are random variables, then the reliability is a

probability that the stress will not exceed the strength

R = Pr{X < Y } =

∫ ∞0

FX(y)fY (y)dy (5.5)

38

5.2 Static Models

0 20 60 80 100

0.00

0.01

0.02

0.03

0.04

y

f Y(y)

x

R

Figure 5.2. Reliability for constant stress and random strength.

or a probability that the strength will be greater than the stress

R = Pr{Y ≥ X} =

∫ ∞0

(1− FY (x))fX(x)dx (5.6)

0 20 60 80 100

0.00

0.01

0.02

0.03

0.04

x, y

f X(x),f Y(y)

FX(y)

fY(y)

fX(x)

y

Figure 5.3. Reliability for random stress and strength.

If the stress and strength belong to the exponential probability density function

fX(x) =1

µXe− xµX fY (y) =

1

µYe− yµY

with a mean value of stress µX and a mean value of strength µY then the reliability

39


1.0

0.50

0.9

0.53

0.8

0.56

0.7

0.59

0.6

0.63

0.5

0.67

0.4

0.71

0.3

0.77

0.2

0.83

0.1

0.91

µX/µY

R

Tabel 5.1. Reliability for the exponential probability density function of stress and

strength.

is

R =1

µX

∫ ∞0

exp

{−µX + µY

µXµY

}dx =

1

1 + µX/µY(5.7)

As obvious from Table 5.1,the value of µY must be at least 10x greater than the

value of µX in order for the reliability to exceed the value of 0.9.

5.3 Dynamic Models

Components for which the stress is changing over time requires separate treat-

ment in which dynamic reliability is calculated. Stress may appear at completely

random times or can be placed repetitively over time.

5.3.1 Periodic Loads

if we neglect the impact of aging and assume that in a moment ti the component

is loaded with a random load Xi having a probability density function of fX(x)

and assume that random strength Yi having a probability density function of

fY (y) is independent of time, then the dynamic reliability after n load cycles is

Rn = Pr{X1 < Y1, . . . , Xn < Yn}

Since the events X1, X2, . . . in Y1, Y2, . . . statistically independent it applies

Rn = Pr{X1 < Y1} · · ·Pr{Xn < Yn} = Rn (5.8)

If the time period ∆t is constant, t0 = 0 and ti = ti−1 + ∆t for i = 1, 2, . . ., then

the dynamic reliability can be calculated by

R(t) = Rt/∆t (5.9)

Y1

t1t2

tn

t

X1

Y2

X2

Yn

Xn

0 ∆t

Figure 5.4. The stress time history and strength for a periodic load.

40

5.3 Dynamic Models

5.3.2 Random Loads

Y1

t1

t2

t t

X1

Y2

X2

Yn

Xn

0

Figure 5.5. The stress time history and strength for a random load.

The probability that over the time t a component will be loaded i times correspond

to a binomial probability density function

fn(i) =

(n

i

)pi(1− p)n−i (5.10)

If number of events n� 1 and probability p very small then the equation (5.10)

can be approximated by Poisson probability density function

fn(i) = e−pn(pn)i

i!(5.11)

From mean period ∆t = t/n it follows pn = (p/∆t)t = αt where α is the mean

number of loads per unit of time. From here it follows

fi(t) = e−αt(αt)i

i!(5.12)

Dynamic reliability is a weighted sum of all possible events

R(t) =∞∑i=0

Rifi(t) = e−αt∞∑i=0

(αtR)i

i!(5.13)

Since

eαtR =∞∑i=0

(αtR)i

i!

the reliability can be written as

R(t) = e−(1−R)αt (5.14)

5.3.3 Random Fixed Stress and Strength

A different result is obtained if stress and strength are randomly determined once

and then fixed for each cycle. We assume that the first stress X1 and strength

Y1 are random variables and for all other realizations of stress and strength it

follows

Xi = X1 ∨ 0 Yi = Y1 i = 1, 2, . . .

41


Y1

Y2

Yn

t1

t2

t t

X1

X2

Xn

0

Figure 5.6. The load time history and strength for random fixed stress and strength.

From (5.13) the dynamic reliability is

R(t) = R0f0(t) +∞∑i=1

Rifi(t)

Because Ri = R for each i ≥ 1 it applies

R(t) = f0(t) +R(1− f0(t)) = e−αt +R(1− e−αt) (5.15)

5.4 Physics of Failure Models

An alternative to static reliability models is called physics of failure where time

to failure is expressed by a function

t = f (operating conditions,

environmental conditions,

material properties,

manufacturing technology,

geometry)

A typical representative of a physics of failure model is the calculation of expected

operating time of the roller bearing

t = a1a2a3

106

60n

(C

P

)pin hours where a1 is a reliability factor, a2 material factor, a3 factor of operating

conditions, n rotary speed in rev/min, p exponent, C dynamic load capacity and

P equivalent stress in N.

42

Chapter 6

Design for Reliability

definition of the reliability objectives

allocation of reliabilityby components

evaluation methods

failure mode andeffect analysisFMEA/FMECA

are goalsfulfilled

system effectiveness andlife cycle costs

system safety andfault tree analysis FTA

are goalsfulfilled

production

yes

yes

no

no

Figure 6.1. The reliability design process.

43

6. DESIGN FOR RELIABILITY

conceiving

detailingoptimizationandtesting

production

use andmaintenance

specification of reliabilityallocation of reliabilityevaluation methods

evaluation methodsFMEAtesting reliability growthsafety analysis and FTA

acceptance testingquality controlburn-in testing andmonitoring/screening

preventive maintenancepredictive maintenancemodificationsparts replacement

life cyclephases

activities relatedto reliability

Tabel 6.1. Reliability and product life cycle phases.

6.1 Reliability Objectives

Reliability indicators:

• mean time to failure MTTF,

• mean time between failures MTBF,

• target reliability after a certain time and,

• probability density function of failures.

Other activities:

• it is necessary to define the failures precisely,

• to eliminate a certain type of failure from the requirements,

• to monitor and record operation and non-operation times and,

• precisely define normal operating, environmental and maintenance condi-

tions.

44

6.2 Reliability Allocation

- R&D costs- production- distribution

operations costs- personnel- energy and fuelfailure costs- waranty- responsibility- repair or replacement- loss of trustsupport costs- repair resources- supply resources- support equipment- or facilities

- salvage value - disposal costs

acquisition costsoperations and support costs remaining costs

Tabel 6.2. Costs categories.

6.1.1 Life Cycle Costs

In general life cycle costs are

life cycle costs = acquisition costs

+ operations costs

+ failure costs

+ support costs

− net salvage value

(6.1)

where

net salvage value = salvage value

− disposal costs


Reliability allocation objective

Rs = h(R1(t), . . . , Rn(t)) ≥ R∗s (t) (6.2)

In case of serially related components

Rs =n∏i=1

Ri(t) ≥ R∗s (t) (6.3)

45


6.2.1 Exponential Case

If all components have constant failure rates, equation (6.3) can be written as

n∏i=1

e−λit ≥ e−λ∗s t

or

n∑i=1

λi ≤ λ∗s (6.4)

6.2.2 Optimal allocation

The objective of optimal allocation are minimal costs. We therefore need to solve

the following

minC =n∑i=1

Ci(xi) (6.5)

n∏i=1

(Ri + xi) ≥ R∗s (6.6)

at condition

0 < Ri + xi ≤ Bi < 1 i = 1, . . . , n (6.7)

The function of costs for reliability improvement Ci(x) can be approximated with

a second order polynomial equation

minC =n∑i=1

cix2i (6.8)

If we neglect (6.7) and equate both sides of the equation (6.6) we can determine

optimal values of xi using the Lagrange method

L(xi, θ) =n∑i=1

cix2i − θ

{n∏i=1

(Ri + xi)−R∗s

}(6.9)

Function extreme value L(xi, θ) is then calculated by solving the equation system

∂L(xi, θ)

∂xi= 2cixi − θ

n∏j=1j 6=i

(Rj + xj) = 0 i = 1, . . . , n

∂L(xi, θ)

∂θ=

n∏i=1

(Ri + xi)−R∗s = 0

(6.10)

46


By multiplying the above equation with (Ri + xi) and a small rearrangement we

get

θ

n∏i=1

(Ri + xi) = 2cixi(Ri + xi) = θR∗s

From here it follows

2cix2i + 2ciRixi − θR∗s = 0 (6.11)

Solution of the square equation (6.11) is its positive root

xi =−2ciRi +

√4c2iR

2i + 8ciθR

∗s

4ci(6.12)

6.2.3 ARINC Method

This method assumes that the components are in series, statistically indepen-

dent and have constant failure rates. If λi is the current failure rate of the ith

component and λ∗s is the target system failure rate, then from weights

wi =λi∑ni=1 λi

we can calculate new failure rate for each component

λi = wiλ∗s

6.2.4 AGREE Method

1

n2 = 3

2

3

4

6

5

n3 = 2n

1 = 1

n = 3, N = 6

Figure 6.2. Block diagram of a complex system suitable for solving with AGREE

method .

AGREE method (Advisory Group on Reliability of Electronic Equipment) as-

sumes that a system is comprised of n components (see Figure 6.2) each having

47


ni modules or sub-components. Let

t = system operating time

R∗s = system reliability target at time t

n = number of components

N =n∑i=1

ni = total number of modules in system

ti = operating time of ith component where ti ≤ t

λi = failure rate of the ith component

wi = importance index

Because it applies

n∏i=1

R∗ni/Ns = R∗s

we can write the following

wi(1− e−λiti) = 1−R∗ni/Ns

Where the left side of the equation is the joined probability that the ith compo-

nent fails and results in a system failure. The right side of the equation is the

failure probability allocated to the ith component. From this it follows

λi = − 1

tiln

{1− 1−R∗ni/Ns

wi

}i = 1, . . . , n (6.13)

Because the failure of a component does not necessarily cause a system failure

the following applies∏n

i=1 e−λiti ≤ R∗s .

6.2.5 Redundancies

1 4

3

2

23

Figure 6.3. Block diagram of a system.

48

6.3 Design Methods

6.3 Design Methods

6.3.1 Parts and Material Selection

• standard parts or parts manufacturing,

• material properties are a function of various factors.

6.3.2 Derating

0 20 40 60 80

0.0

0.2

0.4

0.6

0.8

T

λ

L/Ld = 0.9

L/Ld = 0.7

L/Ld = 0.5

Figure 6.4. The influence of temperature and stress on failure rate.

λ = λb

(s

sd

)0.7(L

Ld

)4.69(υd

υ

)0.54(c

cd

)0.67(T

Td

)3

49


where

λb = failure rate defined by the manufacturer

s = operating speed

sd = design speed

L = operating load

Ld = design load

υ = viscosity of lubricant used

υd = viscosity of specified lubricant

c = concentration of contaminants

cd = standard contamination level

T = operating temperature

Td = design temperature

6.3.3 Stress Strength Analysis

1 4 7 10

0.0

0.1

0.2

0.3

0.4

0.5

SF

Pr{SF<1}

sSF = 0.8

sSF = 1.2

sSF = 1

Figure 6.5. Failure probability and safety factor.

Safety factor SF and safety margin SM

SF =Y

XSM = Y −X (6.14)

Probability of a system failure as a function of the safety factor

Pr{SF < 1} = 1− Φ

(ln SF

sSF

)(6.15)

where

SF =my

mx

sSF =√s2x + s2

y

50

6.3 Design Methods

6.3.4 Complexity and Technology

• part count of the system,

• variability of used parts in the system,

• technology.

6.3.5 Redundancy

active

parallel k of n

load sharing

redundancy

combined passive

withoutswitching

failure

withswitching

failure

Figure 6.6. A classification of redundancy.

Optimization of active redundancy based on the costs

maxn∏i=1

{1− (1−Ri(t))ni} (6.16)

at the conditionn∑i=1

Cu,ini ≤ B +n∑i=1

Cu,i (6.17)

1 C := 0; ni := 1 za i := 1, . . . , n;

2 s := false; Calculate ∆i za i := 1, . . . , n;

3 while s := false do

4 k := maxarg {∆1, . . . ,∆n}; C := C + Cu,k;

5 if C < B then

6 nk := nk + 1; Calculate ∆k;

7 else

8 s := true; C := C − Cu,k;

9 end if

10 end while

Figure 6.7. Marginal analysis.

51


Optimal number of redundant components is determined using the marginal anal-

ysis which requires the separation of control variables ni. This is achieved by

logarithm of the equation (6.16)

maxn∑i=1

ln {1− (1−Ri(t))ni} (6.18)

Because logarithm is a monotonically increasing function, transformation ( ref

eq6:3:5:3) does not affect the position of the optimum. Marginal value

∆i =ln {1− (1−Ri(t))

ni+1} − ln {1− (1−Ri(t))ni}

Cu,i

is defined as the increase in the logarithm of the component reliability per dollar

investment for increasing the number of parallel components for one.

6.4 Failure and Effect Analysis

Failure mode and effect analysis FMEA is an iterative process that influences

design by identifying failure modes, assessing their probabilities of occurrence

and their effect on the system, isolating their causes and determining corrective

action or preventive measures. Main purpose of such analysis is to increase safety

and customer satisfaction. The company’s management needs to precisely define

the constraints within which the FMEA process will take place. There is a need

to answer a series of questions.

• Is the FMEA team responsible only for carrying out the analysis, or should

it also prepare proposals and/or implement improvements?

• What is the team budget?

• What other resources are available to the group?

• Is the date for carrying out the analysis determined and whether there are

other time constraints that need to be taken into account?

• What is the procedure if the group has to exceed the limits?

• How should the group inform the other entities of the company with the

results of the analysis?

FMEA of a product or a process consists of 10 steps

• review of the process,

• identification of the potential failure modes,

• identification of potential effects of each failure mode,

• assigning a severity rating for each failure mode,

52


1. Are all affected areas represented?

FMEA Analysis

FMEA number: 019 Date started:

Date completed:

05.03.02

Team leader: Martin M.

Action:Yes No

2. Are different levels and types of knowledge represented on the team?

3. Is the customer involved?

Yes No Action:

Yes No Action:

4. Who will take minutes and maintain records?

Sales and marketing represent the customer.

Andrej N.

FMEA team boundaries of freedom

5. What aspects of the FMEA is the team responsible for?.

Improvementrecommendations

Improvemntimplementation

6. What is the budget of the FMEA?

7. Does the project have a deadline? 15.04.02

8. Do team members have specific time constraints?

9. What is the procedure if the team needs to expand beyond boundaries ?

Revision of restrictions with the department head.

10. How should the ŽFMEA be communicated with others?

Final report.

11. What is the scope of the FMEA?

Perform the FMEA analysis for the new model of fire extinguisher X-1050.

The analysis must be completed before the 01.05.02.

Team members: Martin M. Andrej N. Marta L.

Boštjan V. Franc G.

5000 EUR

Figure 6.8. FMEA team start-up worksheet.

• assigning an occurrence rating for each failure mode,

• assigning a detection rating for each failure mode and/or effect,

• calculation of the risk priority number (RPN) for each effect,

• prioritizing the failure modes for action,

• taking action to eliminate or reduce the high risk failure modes and

• calculation of the resulting RPN as the failure modes are reduced or elimi-

nated.

6.4.1 Review of the Process

To ensure that everyone on the FMEA team has the same understanding of the

process that is being worked on, the team should review a blueprint of the product

53


8

Item and

function

Poten

tialca

use(s) o

ffailu

reO

Curren

t contro

ls

preven

tiond

ete

ction

D

RPN

Recom

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deda

ction

Respon

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an

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mp

letio

nA

ction

takenC

Poten

tial

failure

mo

de

Poten

tial

effect(s)of failure

S

tubecra

cks

Exposure to

high heator frost duringtran

sportation

- use

of

insu

latio

nm

ate

rials for

pa

cka

gin

g- tra

nsp

ortatio

nin

clima

tically

con

trolle

dco

nd

ition

s

it do

es n

ot

trigg

er a

ta

ctivatio

n

10

5

6

300

use oftem

peraturenon

-sensitivem

aterial

Martin M

.01

.04.02

replacemen

tof current tubew

ith thetem

peraturenon

-sensitive

tube

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ou

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tp

ressu

reD

amage of th

etube during th

eprodu

ction

8

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ob

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we

din

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pro

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ction

pro

ce

ss

4

256

protection o

ftub

es with

a Kevlar coating

Marta L.

15.04

.02pu

rchase of

protective

coatings

for tube

SO

D

RPN

10

2

6

120

8

5

4

160

clog

gin

gn

o o

utp

ut

10

foreign

object

in the tube

6

- inp

ut

con

trol

- test u

sin

gco

mp

resse

da

ir

3

180

cylinder

valvem

echan

ism

levelindicator

10

6

3

180

Team leader:

Martin M

.

fire extinguisher X-1050 F

MEA

team

FM

EA team

:

Subject (product / process):

fire extinguisher X-1050

019FM

EA num

ber:

FMEA

date (original):

(revised):

05

.03.02

01.05

.02

Pa

ge:

of

12

Figure

6.9.

FM

EA

ofa

pro

duct/p

rocess.

54


if they are conducting a product FMEA, or a detailed flowchart of the operation

if they are conducting a process FMEA.

6.4.2 Identification of the Potential Failure Modes

potentialfailure effect

potential cause offailure

potentialfailure mode

biological effectscyclical fatiguehuman errordegradationdepolimerizacijaevaporationexceptional load

chemical and electrolyticchangescontact failurecorrosioncreepmechanical loads

low-quality components

pollutiontemperature fluctuationfrictionmoisture

putrefactionfatigue damagedelayed reactionmagnetic degradationshort circuitfilament breakagelightning strike

electrolyticcorrosionbad contactcorrosin defectsdeformationsudden breakage

spillage of brakeliquidloss of contactfatigue damagewearsteaming up

degradationunstable driveairplane accidentdrop of permeabilitydrop of resistancelamp failureradio transmissionfailureloss of contact

relay does not workleaking of the tankbad guidancepower loss andincreased noiseno brake force

voltage droppressure dropLoses and noisebad visibility

Tabel 6.3. Potential causes of failure, failure modes and effects.

6.4.3 Identification of Potential Effects of Each Failure

Mode

In identifying the potential effects of the failure, we are wondering what might the

consequences of each failure mode be. This step must be thorough, because this

information will feed into the assignment of risk ratings for each of the failure.

6.4.4 Assigning a Severity Rating for Each Failure Mode

The severity rating S is based on a 10-point scale, with 1 being the lowest rating

and 10 being the highest. It is an estimation of how serious the effect would be if

a given failure did occur. Because each failure may have several different effects,

and each effect can have different level of severity, it is the effect, not the failure,

that is rated.

55


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t wo

uld

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ct on

the p

rocess o

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etectionratin

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56


6.4.5 Assigning an Occurrence Rating for Each Failure

Mode

The occurrence rating O is based on a 10-point scale and it specifies the proba-

bility for a certain failure mode to appear in the system or process life time. In

the assessment phase, the team tries to answer various questions.

• What is the frequency of failures of similar components that are still on the

market?

• Is the considered component similar to the previous generation of compo-

nents?

• What changes have been made when switching from previous to new gen-

eration?

• Is it a completely different or even completely new component?

• Has the operating conditions and/or environmental conditions changed?

• Has a statistical assessment of the probability of failure occurrence been

carried out?

6.4.6 Assigning a Detection Rating for Each Failure

Mode and/or Effect

The detection rating D is based on a 10-point scale and it looks how likely we

are to detect a failure or the effect of a failure. We start this step by identifying

current controls that may detect a failure or effect of a failure.

6.4.7 Calculation of the Risk Priority Numbe

The risk priority number (RPN)indexrisk priority number is simply calculated

by multiplying the severity rating S times the occurrence rating O times the

detection rating D for all of the items.

RPN = S ×O ×D (6.19)

it is a number between 1 and 1000. Because in the process of valuation of S, O

and D there are differences between the members of the group, it is worthwhile

to use one of the following techniques to achieve consensus

• vote in the group,

• inclusion of an expert,

• postponing the decision to one of the team members,

57


• classification of failures and/or effects by size according to S, O or D,

• prolonged debate or,

• voting for higher RPN value.

6.4.8 Prioritizing the Failure Modes for Action

The failure modes can now be prioritized by ranking them in order from the

highest RPN number to the smallest. A Pareto diagram is helpful to visualize

the differences between the various ratings. The team must now decide which

items to work on. Usually the team sets a cut-off RPN, where any failure modes

with a RPN above that point are attended to.

6.4.9 Taking Action to Eliminate or Reduce the High

Risk Failure Modes

Using an organized problem solving process the team identify and implement

actions to eliminate or reduce the high risk failure modes. The following are the

measures that can contribute to the reduction of the RPN index:

S Use of personal protective equipment (helmet, safety glasses, protective

gloves). Safety switches. Use of materials such as Safety glass that does

not cause such serious damage in case of failure.

O Increase the process performance index Cpk using test design and/or mod-

ification methods. Process of continuous improvement. Use of the mecha-

nisms to be activated in order for the product or process to work (example:

the garden lawnmower has a lever that needs to be pressed constantly dur-

ing operation).

D Statistical control of the process. Use of regularly calibrated measuring

devices. Preventive maintenance as a means of timely detection of potential

defects. Using encoding (colors and shapes) that informs the user about

what is right and what is not.

6.4.10 Calculation of the Resulting RPN

Once action has been taken to improve the product or process, new ratings for

severity, occurrence and detection should be determined and a resulting RPN

calculated.

58

6.5 System Safety and Fault Tree Analysis


Reliability and product safety are closely related. Safety is a state of the product

that does not lead to injuries, loss of life, or severe damage to equipment and

harmful consequences for the environment (MIL-STD-882). The safety analysis

is carried out primarily in cases where a certain type of failure could lead to

catastrophic consequences for humans or the environment. The aim of the anal-

ysis is to determine the types, probability of realization and corrective measures

for potential disasters with catastrophic consequences. Since the probability of

realization of these events is often very low, and because such failures are pre-

dominantly due to the sequence of several events, the fault tree analysis (FTA)

is used to detect and quantify them.

6.5.1 Fault Tree Analysis

AND gate. An output event occursonly if all incoming events occur.

Basic event. An independent basic event,which can no longer be broken down.

OR gate. An output event occurs,if at least one input event occurs.

Resultant event. Description of the failurethat arises from the logical combination ofother failures or description of the logicaldoor below it.

Incomplete event. An event that is notcompletely broken down due to a lack ofinformation.

Transfer-in and transfer-out. Used to linksections of the fault tree that are not contiguous.

Conditional event. A condition or restriction tied to a logic gate.

Normal event. A normally occurring eventthat is not a fault.

Figure 6.10. Symbols of fault tree analysis.

A fault tree analysis is a graphical design technique for detection of potential

failures that could lead to catastrophic consequences. There are four major steps

to a fault tree analysis:

• definition of the system, its boundaries and the top event,

59


observer

sensor 1

switch

Pissa tower

sensor 2alarm

auxiliary source of electricity

main source of electricity

Figure 6.11. Alarm device diagram.

Talarm failure

Hsensor 2failure

Gsensor 1 failure

Ihumanerror

Jsensor 1failure

Kswitchfailure

Apower failure D

secondary alarm failure

Bsensor failure C

alarmfailure

Eauxiliary sourcefailure

Fmain

sourcefailure

Figure 6.12. The alarm system fault tree.

• construction of the fault tree,

• qualitative evaluation by identifying those combinations of events that will

cause the top event and

• quantitative evaluation by assigning failure probabilities or unavailabilities

to the basic events and computing the probability of the top event.

Faults can be classified as primary, secondary and command. Faults may also be

classified as active or passive.

The top event T can be expressed as a function of basic and incomplete events

T = A ∪B ∪ C ∪D= (E ∩ F ) ∪ (G ∩H) ∪ C ∪D= (E ∩ F ) ∪ ((I ∪ J ∪K) ∩H) ∪ C ∪D

(6.20)

60


T

M2

or

Mk

E1

and

M1

E2 En1

and and

Figure 6.13. Equivalent tree of minimal cut sets.

T

BA

E F E A

C D

or or

and

or

T

A

or

E

Figure 6.14. A fault tree and an equivalent fault tree.

6.5.2 Minimal Cut Sets

Minimal cut set is a minimal set of basic events that cause the top event. If Mi

is ith minimal cut set the top event T can be expressed as union of all k minimal

cut sets

T = M1 ∪M2 ∪ · · · ∪Mk (6.21)

where Mi = E1 ∩ E2 ∩ · · · ∩ Eni and Ei are basic events.

E,FI,HJ,HK,HCD

iteration

ABCD

1 2

E,FG,HCD

3

iteration

AB

1 2

AC,D

3

AE,DF,D

AE,EE,AF,EF,A

4

AE

5

(a) (b)

Figure 6.15. Minimal cut sets examples: Figure 6.12 (a) and Figure 6.14 (b).

61


6.5.3 Quantitative Analysis

In the context of quantitative analysis, we estimate the probability of realization

of the top event under the condition that the probabilities of realizations of basic

and incomplete events are known. To help with the analysis we can use a tree of

minimal cut sets, since it applies

P (T ) = P (M1 ∪M2 ∪ · · · ∪Mk) (6.22)

If minimal cut sets represent unrelated events we can write

P (T ) = P (M1) + P (M2) + · · ·+ P (Mk) (6.23)

and if basic events Ei are statistically independent we can write

P (Mi) = P (E1 ∩ E2 ∩ · · · ∩ Eni) = P (E1)P (E2) · · ·P (Eni) (6.24)

Because the probabilities P (Mi) are small, the equation (6.23) gives satisfactory

results even in case of related events.

62

Chapter 7

Maintainability

We distinguish between:

• reactive,

• preventive and,

• predictive maintenance.

7.1 Analysis of Downtime

Delay time in the logistic support includes:

• waiting time for spare parts,

• initial administrative time,

• initial production or purchasing time,

total downtime

de

lay

in lo

gis

tics

sup

po

rt

mai

nte

na

nce

de

lay

tim

e

acce

ss t

ime

dia

gn

ost

ic t

ime

acti

ve t

ime

of

rep

air

or

rep

lace

men

t

tim

e o

f ve

rifi

cati

on

an

d

alig

nm

en

t

repair time

Figure 7.1. Breakdown of maintenance downtime.

63

7. MAINTAINABILITY

• maintenance time and,

• time of the transport.

Maintenance delay time includes:

• waiting time for available maintenance resources,

• waiting time for available maintenance facilities,

• administrative (notification) time and,

• travel time.

Maintenance resources includes:

• personnel,

• test and support equipment,

• tools,

• manuals or other technical data.

Maintenance facilities may be:

• repair dock,

• service bay,

• fixed test stand and,

• and other facilities.

7.2 The Repair Time Distribution

Let T be the continuous random variable representing the time to repair a failed

unit, having a probability density function of h(t) and cumulative distribution

function of

Pr{T < t} = H (t) =

∫ t

0

h(t)dt (7.1)

The mean time to repair may be found from

MTTR =

∫ ∞0

th(t)dt =

∫ ∞0

(1−H (t))dt (7.2)

and the variance of the repair distribution from

σ2 =

∫ ∞0

(t−MTTR)2h(t)dt (7.3)

64

7.3 System Repair Time

If the repair distribution function is exponential with h(t) = re−rt and the rate

of repair r = 1/MTTR then it applies

H (t) =

∫ t

0

e−t/MTTR

MTTRdt = 1− e−t/MTTR (7.4)

To represent the repair distribution often log-normal distribution is used

h(t) =1√

2πtsexp

{−1

2

(ln(t/tmed))2

s2

}(7.5)

where tmed is the median time to repair and s shape parameter. The probability

of a repair being finished in time t is found by utilizing the relationship between

the normal and log-normal distributions

Pr{T < t} = H (t) = Φ

(1

sln

t

tmed

)(7.6)

The mean time to repair is defined by

MTTR = tmedes2/2 (7.7)

7.3 System Repair Time

let MTTRi be the mean time to repair of the ith component, mumi be the expected

number of unexpected repairs of the ith component over the system design life

and qi the number of identical components of type i. Then the system mean time

to repair can be expressed as a weighted sum

MTTRs =

∑ni=1 qimumiMTTRi∑n

i=1 qimumi

(7.8)

The expected number of unexpected repairs of the ith component is

mumi =

{toi

MTBFifor renewal process∫ toi

0ρi(t)dt for minimal repair

Let us consider k from n redundant system. repair in such a case can be carried

out in various ways:

• repair of the component immediately after a failure,

• repair of one component after a failure of n− k + 1 n components,

• repair of one component while maintaining all other operating components

after a failure of n− k + 1 n components,

• repair of all faulty components while maintaining all other components after

a failure of n− k + 1 n components.

65

7. MAINTAINABILITY

7.4 Reliability under Preventive Maintenance

The reliability over the first and the second preventive maintenance intervals is

Rpm(t) = R(t) za 0 ≤ t < Tpm

Rpm(t) = R(Tpm)R(t− Tpm) za Tpm ≤ t < 2Tpm

where R(Tpm) is the probability of survival until the first preventive maintenance

and R(t − Tpm) is the probability to survive the additional time t − Tpm given

that the system was restored to its original condition at time Tpm. In general it

applies

Rpm(t) = Ri(Tpm)R(t− iTpm) za iTpm ≤ t < (i+ 1)Tpm (7.9)

where Ri(Tpm) is the probability of survival until the ith preventive maintenance

and R(t − iTpm) is the probability to survive the additional time t − iTpm. The

t

R(t)

Ri(Tpm)

Rpm(t)

R(t - iTpm)

0 Tpm

0.0

0.2

0.4

0.6

0.8

1.0

2Tpm 3Tpm 4Tpm 5Tpm 6Tpm

relia

bili

ty

Figure 7.2. The influence of preventive maintenance on reliability (increasing failure

rate).

MTTF under preventive maintenance can be calculated as

MTBF =

∫ ∞0

Rpm(t)dt =∞∑i=0

∫ (i+1)Tpm

iTpm

Rpm(t)dt

=∞∑i=0

Ri(Tpm)

∫ (i+1)Tpm

iTpm

R(t− iTpm)dt

=∞∑i=0

Ri(Tpm)

∫ Tpm

0

R(t)dt

In some preventive maintenance situations there is a possibility of a maintenance-

induced failure where p is the probability of a maintenance-induced failure during

66

7.5 Stochastic Point Processes

an individual preventive maintenance. The reliability after i preventive mainte-

nance procedures taking into account probability p is

Rpm(t) = Ri(Tpm)(1− p)iR(t− iTpm) za iTpm ≤ t < (i+ 1)Tpm

Ri(Tpm)

R(t - iTpm)

0 Tpm

t

R(t)

Rpm(t)

0.0

0.2

0.4

0.6

0.8

1.0

2Tpm 3Tpm 4Tpm 5Tpm 6Tpm

relia

bili

ty

Figure 7.3. The influence of preventive maintenance on reliability (decreasing failure

rate).


Relationships between individual random variables are given by equations

Yi = Xi + Si = Ti − Ti−1 Ti =i∑

j=1

Yj (7.10)

Let us also write the expected values of the operating times and down-times

E[Xi] = MTBF E[Si] = MTTR (7.11)

S1

T0

T2

T1

t

X1

Y1

Figure 7.4. An example of a time frame of a component state.

67

7. MAINTAINABILITY

7.5.1 Renewal Process

If Si ≈ 0 then random variable Xi is operating time between i− 1 in ith failure,

and random variable Ti operating time until ith failure. If random variables Xi

are statistically independent having the same probability density distribution of

failures f (t), then the probability distribution of operating times Ti for i → ∞converges to normal probability density function

fi(t) =1√

2πσiexp

{−(t− µi)2

2σ2i

}(7.12)

with the mean value of

µi = E[Ti] =i∑

j=1

E[Xj] = iE[Xi] = iMTBF (7.13)

and variance of

σ2i = E[(Ti − iMTBF)2]

= E[(i∑

j=1

(Xj −MTBF))2)]

=i∑

j=1

i∑k=1

E[(Xj −MTBF)(Xk −MTBF)]

=i∑

j=1

E[(Xj −MTBF)2] = iσ2

(7.14)

since it applies

E[(Xj −MTBF)(Xk −MTBF)] = 0 j 6= k

Stochastic point process can also be expressed by number of failure in time interval

[0, t].Let N (t) be a discrete random variable which gives a cumulative number of

failures in the interval [0, t]. Then it follows

Pr{N (t) = 0} = Pr{T1 > t}Pr{N (t) = i} = Pr{Ti ≤ t < Ti+1}

= Pr{Ti ≤ t} − Pr{Ti+1 ≤ t}= Fi(t)− Fi+1(t)

(7.15)

where cumulative distribution function is

F0(t) =

{0 za t ≤ 0

1 za t > 0

or

Fi(t) =

∫ t

0

Fi−1(t− u)f (u)du i = 1, 2, . . .

68


7.5.1.1 Homogeneous Poisson Process

If

f (t) = λe−λt (7.16)

with a failure rate of λ = 1/MTBF then it applies

F0(t) = 1

F1(t) = 1− e−λt

F2(t) = 1− e−λt − λte−λt

F3(t) = 1− e−λt − λte−λt − (λt)2

2e−λt

from which a general form of the cumulative distribution function can be gener-

ated

Fi(t) = 1−i−1∑j=0

(λt)j

j!e−λt i = 0, 1, . . . (7.17)

From here it follows

Pr{N (t) = i} =i∑

j=0

(λt)j

j!e−λt −

i−1∑j=0

(λt)j

j!e−λt =

(λt)i

i!e−λt (7.18)

7.5.1.2 Renewal Function

The renewal function provides the expected number of failures at time t

m(t) =∞∑i=1

iPr{N (t) = i} (7.19)

if in equation Pr{N (t) = i} is replaced by Fi(t)− Fi+1(t), we get

m(t) =∞∑i=1

Fi(t) (7.20)

7.5.2 Minimal Repair process

Aging of the product can be considered as a stochastic point process and it is

expressed with the intensity function

ρ(t) =dE[N (t)]

dt(7.21)

The intensity and renewal functions are related since it applies

m(t) = E[N (t)] =

∫ t

0

ρ(t)dt (7.22)

69

7. MAINTAINABILITY

7.5.3 Overhaul

If the probability density distribution of failures f (t) is known, then the expected

period of the overhaul is

E[Tov] = R(Tpm)E[t|t ≥ Tpm] + F (Tpm)E[t|t < Tpm]

= R(Tpm)Tpm + F (Tpm)

∫ ∞0

tf (t|t < Tpm)dt

Because from Bayes’s theorem it follows

f (t|t < Tpm) =

{0 za t ≥ Tpm

f (t)F (Tpm)

za t < Tpm

we can finally write

E[Tov] = R(Tpm)Tpm +

∫ Tpm

0

tf (t)dt =

∫ Tpm

0

R(t)dt (7.23)

Calculation of the expected period of the overhaul in case of the exponential

probability density distribution of the failure

E[Tov] =

∫ Tpm

0

e−λtdt =1

λ(1− e−λTpm)

with λ = 0.00001h−1 and overhaul period of Tpm = 10000h is shown in Figure 7.5.

0 5000 10000 15000

0.0E+00

2.0E-06

4.0E-06

6.0E-06

8.0E-06

1.0E-05

t

f(t) T

pmE[T

ov] = 9516h

R(Tpm)

Figure 7.5. Example of the overhaul period calculation.

70

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