1 reliability

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Reliability Outline

Reliability Basic Concepts

Reliability Models

Reliability System Configurations Analysis of System Reliability

Application of Reliability in Aviation

Business



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Reliability in Logistics

Reliability Basic Concepts

± What is Reliability

± Reliability Definitions ± What Affects Reliability

± How is Reliability Used

± Goal of Reliability

± Bathtub Curve

± Reliability Function



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What is Reliability

± To the user of a product, reliability is problem

free operation

± Reliability is a function of stress

± To efficiently achieve reliability, rely on

analytical understanding of reliability and less

on understanding reliability through testing




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Reliability Definitions

± Reliability is a characteristic of an item,

expressed by the probability that the item will

perform its required function under given

conditions for a stated time interval.

(Prof. Dr. Alessandro Birolini, 1999)

± The probability that an item will perform arequired function without failure under stated

conditions for a stated period of time.

(Patrick D. T. O¶Connor, 1992)



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Reliability in Logistics ± The probability that an item will perform its

intended function for a specified interval under

stated conditions. (Reliability Analysis Center

and Rome Laboratory, 1993)

± The rigorous definition has four parts:

1. Reliability is the probability that a system

2. will demonstrate specified performance

3. For a stated period of time4. when operated under specified conditions.

(Daniel L. Babcock, 1996)



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Probability GivenTime

StatedConditions

Performanc

Reliability



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What Affects Reliability

± Redundancy (å

úüè)

± Design Simplicity

± Time

± Learning Curve

± Material Quality

± Experience

± Requirements




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How is Reliability Used

± It is used to define the longevity of a product

and the associated cost it incurs

± It helps identify risk of the product for both the

consumer and producer

± It incorporates statistics to better identify how

much ³give´ or ³take´ can go into a product or service - usually, the higher the reliability, the

higher the initial cost.




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± It predicts the likely hood of failure rates for a

given product or service

± Perform sensitivity analyses

Mission effectiveness

Supportability

Life cycle costs

Warranties




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Reliability Goals

1. Increase competitive position

2. Increase customer satisfaction3. Reduce customer support

requirements

4. Decrease cost of ownership




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Bathtub Curve


Infant

Mortality

Wear-outUseful Life

H a z a r d R a t e

P

F a i l u r e R a t

e s

System Life Cycle

Decreasing

Failure

Rate

Constant

Failure

Rate

Region

Increasing

Failure

Rate

Region

Exponential

Law applies

debuggingwearout



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Reliability Function

± If the random variable t has a density function

f(t), then Rs(t)

)(1)( t F t RS !

´g

!t

S dt t f t R )()(



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± If the time to failure is described by an

exponential density function, then

± and

U

U/1)( t et f !

´g !!t

t t

S edt et R U U

U//1)(



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± Mean life ( U) is the arithmetic average of the

lifetimes of all items considered5

± The mean life ( U) of the exponential function is

equivalent to the mean time between failure,hence

t S et R P!)(



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± Where P is the instantaneous failure rate

± Or

UP

1!

hoursoperatingtotalfailuresof number !P



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Weibull Distribution:

± Can be considered as a generalizationof the exponential

±Has three parameters

K = the time at which F(t) = 0 and is adatum parameter; i.e., failures start

occurring at time t W = the characteristic life and is a scale

parameter

F = shape parameter




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FP )(

)(t

S et R

!

1

1

)(

)()(

!

!

F

F

F

FP FPW

K FP

t t



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F FWK +! 1

MTT F



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H

a z a r d R a t e P

Infant

Mortality

Wear-outUseful Life

F<1 F>1

F=1

F=4

F=1/2

W0

F=1

t K



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Example

± Suppose ten

components were tested

under specifiedconditions

± Assume the following

data from the test

± Suppose thecomponents are non

repairable

± What is P?

Component Op Hours Failure

One 250 No

Two 85 Yes

Three 250 NoFour 95 Yes

Five 115 Yes

Six 250 No

Seven 75 Yes

Eight 250 No

Nine 105 YesTen 250 No

1725



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Solution:

0029.0

1725

5

hoursoperatingtotal

failuresof number

!

!

!

P

P

P



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Example

± Use the following figure to determine P and

MTBF

± Assume an exponential distribution

20 25 15 30 50 20 10

Operation

10 30 5 25 10 10

Failures



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Solution:

0353.0

170

6

!

!

!

P

P

Ptimeoperating total

failuresof number

33.28

0353.0

1

1

!

!

!

MTB F

MTB F

MTB F

P



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System Reliability Models

± The reliability definitions, concepts and

models presented apply at any level of asystem, from a single discretecomponent up

to and including the entire system



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± Systems reliability deals with the reliability of

the end-item system and is based on the

system configuration and component failure

rates as well intended service usage

± There are two basic types of reliability

configurations

Series

Parallel or Redundant



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Series Configuration

± 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 thereliability configuration



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± Series Reliability configuration with n

elements: E1, E2, ..., En Block Diagram:

± Since a single path exists, the failure of any

element in the system interrupts the path and

causes the system to fail

E1 E2 En




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± Ex ponential distributions of element time to

failure Ti ~ I( Ui) for i = 1, 2, ...n

System reliability

Where is the system failure rate

!

!!n

1i

i

t

S)t(R e)t(R S

P

§!

P!P

n

1i

iS )t(



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S S

S

MTT F 5!!P

1

System mean time to failure



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Parallel Configuration

± Definition - a system is said to have parallel

reliability configuration if the system function

can be performed by any one of two or morepaths

± Reliability block diagram - for a parallel

reliability configuration consisting of n

elements, E1, E2, ... En



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± Exponential distributions of element time to

failure Ti ~ I( Ui) for i= 1, 2, ... n

System reliability

? A ? A!!

!!n

1i

i

n

1i

t

S)t(R 11e11)t(R iP



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System mean time between failure MTBFS =

§§§§!

! P

¹¹ º

¸

©©ª

¨

PPP¹¹ º

¸

©©ª

¨

PPP n

1i

i

1n

k jik ji k ji

ji ji ji

n

1i i

1)1(...111



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± Redundant reliability configuration-

sometimes called a redundant reliability

configuration. Other times, the termµredundant¶ is used only when the system is

deliberately changed to provide additional

paths, in order to improve the system

reliability.



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± Basic assumptions

All elements are continuously energized starting at

time t = 0

All elements are µup¶ at time t = 0 The operation during time t of each element can

be described as either a success or a failure, i.e.,

degraded operation or performance is not

considered



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Rs(t) = 1 - (1-p)m

E1,1

m e

l e m e n

t s E1,2

E1,n

n elements

Rs(t) = 1 - (1-pn)m

E2,1 E2,2 E2,n

Em,n Em,n Em,n

E1

E2

Em



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E2

E1

Em

m e

l e m e n

t s

n elements

Rs(t) = [1 - (1-p)m ]n

E2

E1

Em

E2

E1

Em



éöè ÷öûû úâ å E åôâú E1

x E2 x E3

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÷öûû ã úú 1 ö è ûõýåôãè 1-å R øâö

1- (1-E1)(1-E2)(1-E3)

E1 E2 E3INPUT OUTPUT

E1

E2

E3

INPUT OUTPUT



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Configuration Considerations in Design

± Series Configuration - Relative to R edundant

Configuration

Simpler

Increases Basic Reliability

Reduces Support Resources

Decreases Mission Reliability




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± Redundant Configuration - Relative to S eries

Configuration

More Complex - Increases Weight

Requires More Testability

Increases Support Resources

Decreases Basic Reliability

Increases Mission Reliability




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Series Model Example:

± What is the reliability of the following system

given that:

E1 = 0.9400

E2 = 0.9500

E3 = 0.9800

E1 E2 E3



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Series Model Example:

± Solution:

Use the product rule

Rs(t) = E1 E2 E3

= (0.9400) (0.9500) (0.9800)

= 0.8751

0.94 0.95 0.98



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Parallel Model Example:


given that:

E1 = 0.9400

E2 = 0.9500

E3 = 0.9800E2

E1

E3



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Parallel Model Example:

± Solution:

Rs(t) = [1 - (1 - E1) (1 - E2) (1 - E3)]

= [1-(1-0.9400)(1-0.9500)(1-0.9800)]

= [1 - (0.0600)(0.0500)(0.0200)

= 0.9999

0.95

0.94

0.98



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Reliability Analysis

± Use to determine reliability of system

± Helps make informative decisions

± Illustrates faults in subsets of system

± Depicts through modeling increasing Rs(t)



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Bathtub Curve


Infant

Mortality

Wear-outUseful Life

H

a z a r d R a t e P



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Packer

QC

Pallets

Process Area

To Load



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To Store

ChooseVendor

CustomerPurchase

UnsatisfiedCustomer

To Shelf



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Exercise 1

± Determine the MTTF and MTTR for the given

data of a system (assume exp):

TTF TTR TTF TTR

(hr ) (hr ) (hr ) (hr )

125 1.0 58 1.0

44 1.0 53 0.8

27 9.8 36 0.5

53 1.0 25 1.7

8 1.2 106 3.646 0.2 200 6.0

5 3.0 159 1.5

20 0.3 4 2.5

15 3.1 79 0.3

12 1.5 27 3.8



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Exercise 1

± If the system operates for 8.0 consecutive

hours, what is the reliability of the system?



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Exercise 2

± Determine the MTTF and MTTR for the given

data of a system in the next slide (assume

exp).

± What is the reliability of the system if it

operates for three continuous hours?



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Reliability in LogisticsComponent Repair t Failure t Repair At

1 0.0 3.10 4.50

1 4.5 6.60 7.40

1 7.4 9.50

2 0.0 1.05 1.70

2 1.7 4.50 8.50

3 0.0 5.80 6.80

3 6.8 8.80

4 0.0 2.10 3.80

4 3.8 6.40 8.60

5 0.0 4.80 8.30

6 0.0 3.00 6.50

7 0.0 1.40 3.50

7 3.5 5.40 7.608 0.0 2.85 3.65

8 3.7 6.70 9.50

9 0.0 4.10 6.20

9 6.2 8.95

10 0.0 7.35



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Exercise 3


(assume all components exhibit an

exponential distribution)?

E3E1

E2E2 E3

E1

E1

E2 E3

E2E1

E3

E3

E3E3

E1 = 0.90E2 = 0.95E3 = 0.85



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Exercise 3

± If you could change only one of the

components, which one would you change?

Why? What is the reliability of the newsystem?



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Solution 1

± MTTF = 55.10

± MTTR = 2.19

± Rs(t) = 0.86

Solution 2

± MTTF = 3.05

± MTTR = 2.05

± Rs(t) = 0.37



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Solution 3

± Part I:

Rs(t) = 0.81

± Part II:

Change E3 on the output

In series with system ± failure at this junction

causes system failure

Rs(t) = 0.91



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1. Prof. Dr. Alessandro Birolini, R eliability E ngineering: Theory and

Practice, (Germany: Springer-Verlag Berlin Heidelberg, Third

Edition, 1999), p.2.

2. Patrick D. T. O¶Connor, Practical R eliability E ngineering , (New

York, New York:John Wiley & Sons, Third Edition, October 1992),p.3.

3. Reliability Analysis Center and Rome Laboratory, R eliability

Toolkit: Commercial Practices E dition, 1993, p.36.

4. Daniel L. Babcock, Managing E ngineering and Technology , (New

Jersey: Prentice-Hall, Inc., Second Edition, 1996), p. 204.

References



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5. Benjamin S. Blanchard, Logistics E ngineering and Management ,

(Upper Saddle River, New Jersey: Pearson/Prentice Hall, 2004),

p. 47.

6. Ernest J. Henley and Hiromistsu Kumamoto, R eliability

E ngineering and

R isk Assessment , (Englewood Cliffs, NewJersey: Prentice-Hall, 1981), p. 238.

References

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