Availability Modelling

8/12/2019 Availability Modelling

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RELIABILITY ENGINEERING UNIT

ASST4403

Lecture 27-28 AVAILABILITY MODELLING


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earn ng outcomes

xp a n e ramewor o ava a y re a e o

reliability, maintainability and maintenance Interpret and analyse different times for availability

and downtime

Understand mathematical basis for availability

measures

rt cu ate t e system ava a ty assessmentmethods

Predict availability of simple systems


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reliability, maintainability and


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Dependability, framework of reliability,availabilit maintainabilit etc

AS IEC 60300.12004


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ome e n ons

availability performance and its influencing factors:

reliability performance, maintainability performance andmaintenance support performance

given conditions of use, to be retained in, or restored to astate in which it can perform a required function, whenmaintenance is performed under given conditions and

using stated procedures and resources Maintenance support performance is the ability of a

maintenance organization, under given conditions, toprovide upon demand, the resources required tomaintain an item, under a given maintenance policy


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MTBF

t

etR

)(

yearst 30


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v w y


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Review of Reliability System A

Modem POD A

Rin Pair A1

0.659

0.9500.950

0.950 0.942

0.741 0.8610.861

0.861

0.6590.942

0.741 0.950Ring Pair A2

0.9500.741 0.861 0.659

0.659

0.950 0.9420.861

SCM(16

Wires)

Junction(incl

Conrs)System B

ep0.9500.861 0.942

Modem POD B

POD B

0.659

0.950

0.950 0.942

0.861

0.861

0.942

Ring Pair B1

0.741

.. .

0.950Ring Pair B2

0.9500.741 0.861

.

0.6590.950 0.9420.861

.

0.9500.861 0.942


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Review of Reliability


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Step 6

ys em

0.689SCM(16

Wires)

Junction(incl

Conrs)


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Reliability and Confidence

Reliability is the probability, at a specified confidence level, that a deviceor s stem will erform its intend function for a iven interval of timeunder specified operating conditions.

What is the relationship between confidence and reliability?


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Eg Haul Pack Operational Capability

Task Haul Pack required to travel into operational area return and dump its load

Success Criteria - Haul Pack successfully travels to the AO return and dump its load

Mission Phases:

Haul Pac

Available

tart

Haul Pack

Transits toHaul Pac

Dumps Load


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Eg Haul Pack Operational Capability

Pr(HPA) = Probability that HP is Available = 0.7

Pr(HPS) = Probability that HP Starts = 0.95 Pr(HPT) = Probability that HP Transits = 0.9

Pr(HPD) = Probability that HP Dumps Load = 0.8

HPTransits to &

HP HP HP

from AO

(HPT)

(HPA) (HPS)

(HPD)

Pr(Mission Success) = Pr(HPA) x Pr(HPS) x Pr(HPT) x Pr(HPD)


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Eg - HP Operational Capability


Pr(HPS) = Probability that HP Starts = 0.95 Pr(HPT) = Probability that HP Transits to AO = 0.9

Pr(HPD) = Probability that HP Dumps Load = 0.8

Pr(HPA)Pr(HPT) Pr(HPT)Pr(HPG) Pr(Mission Success)0.700 0.950 0.900 0.800 0.479

.

0.000

Pr(Mission Success) = Pr(HPA) x Pr(HPS) x Pr(HPT) x Pr(HPD)= .


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What if the probabilities of each event are increased?

What is the impact on the mission success ?

Pr(HPA) = Probability that HP is Available from 0.7 to 0.8 Pr(HPS) = Probability that HP Starts from 0.95 to 0.975 Pr HPT = Probabilit that HP Transits from 0.9 to 0.95

Pr(HPD) = Probability that HP Dumps from 0.8 to 0.9


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Pr(HPS) = Probability that HP Starts = 0.975 Pr(HPT) = Probability that HP Transits = 0.95

Pr(HPD) = Probability that HP Dumps = 0.9


. . . . .

0.000

Pr(Mission Success) = Pr(HPA) x Pr(HPS) x Pr(HPT) x Pr(HPD)= .


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Case 1 - 47.9% of missions succeed

Case 2 - 66.7% of missions succeed


. . . . .

Increase in Capability 39.29%


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Question?

What is the effect of Improved Reliability and Maintainability on Cost and

Data:

Require Haul Packs for four (4) Mine Sites 12 HPs are re uired to be available er Mine Site

Each HP costs $2 million


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MTBF = 10 hours

MTTR = 5 hours Inherent Availability = MTBF/(MTBF+MTTR)

Inherent Availability of Each HP = 67%

= = .

Total Cost of Task Forces No of HPs x # per Mine Site x Unit Cost

== $144M


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MTBF = 20 hours (double original Baseline)

MTTR = 5 hours (same as original Baseline) Inherent Availability of Each HP = 80%

HPs required for each Mine Site = 12/ 0.8 = 15 Total Cost of Task Forces

No of HPs x # per Mine Site x Unit Cost=15 x 4 x $2M= $120M


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MTBF = 20 hours (double original Baseline)

MTTR = 2.5 hours (half the original Baseline)

Inherent Availability of Each HP = 88.8%

HPs required for each Mine Site = 12/ 0.888 = 14

No of HPs # per Mine Site x Unit Cost=14 x 4 x $2M= $112M


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g - pera ona apa y

Capability Cost Comparison

Baseline

System

Case 1 Case 2

MTBF hrs

MTTR (hrs) 5 5 2.5Availability 67% 80% 88.8%

Data

Ca abilito o s

$144M $120M $112MCOST($M)

A $32M saving and this only includes procurement costs


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Types of availability to be discussed

Inherent availability, Ai

Achieved availability, Aa

, o


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Different times for availabilit and downtime


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What is time

All approaches to availability are time related

, OT=operating time per given total calendar time

ST=standby time (not operating but assumed operable) TCMT=Total corrective maintenance time TPMT=TCMT=Total preventive maintenance time =

Adapted from the Defence Reliability Management Course, 2/2005


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Breakdown of downtime

Supply delay: total delay time in

components for the repair external factors,not art of the

Maintenance delay: time spent waitingfor maintenance resources or facilities

system

Repair time: sum of the following

Access time

n erent repa rtime

Diagnosis Repair or replacement


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Factors influencing downtime

Main factors are equipment design and maintenance

Active repair is determined by the design

philosophy

Ke desi n areas: access ad ustment built-in testequipment, display & indicators, handling & ergonomics,

Interchangeability, least replaceable assembly (LRA),mounting, redundancy, test points

Maintenance strate ies


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Some mathematical basics for availability


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A note before we head on

Some of the slides that follow in this topiccontain quite a few mathematic expressionsand formulas. These are intended from the

author to be reference material for the easeof the participants


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systems is working at time t

The (average) interval or mission availability in the time

atwor ngssystem tPt

interval (t1, t2) is

2

)(1

),( 21t

tav dttAttA

which can be interpreted as the mean proportion of time in

12

.t1=0, t2=, we have

0

)()( dttAA av


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v y The long run availability of system is

0

)(1

lim dttAA av

which can be interpreted as the average proportion of along period of time where the system is able to function

The limiting or steady-state availability is , when the limitexists,lim tAt

riodMission

downtimeunplannedtotalmeandowntimeplannedtotalmean1

opA

av

The operational availability is the mean proportion of amission period the system is able to perform its intended

function


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Long run average availability

A failed item is replaced to an as good as new

Up-times T1, T2, , Tn are independent and identicallydistributed (iid) with mean time to failure MTTF

Down-times D , D , , D are inde endent andidentically distributed (iid) with mean downtime MDT

,

MTTFTEAav

)(


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Inherent availability

MTTF

tAAinh )(lim

Inherent availability is based solely on the failure distribution and-

Equipment design parameter, based on which trade-offs betweenreliability and maintainability can be made


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component

Assuming constant failure rate (exponential timeo a ure an cons an repa r ra e exponen atime to repair), where =1/(MTTR)

Steady-state availability MTTRMTTFA

Instantaneous availability teA )(


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ome most common y app e ava a tymeasures


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If we are only concerned with correctivemaintenance



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Inherent availabilit

Ai is availability when we are only concerned with

,maintenance, no administrative & logistic delay time

TCMOTA

MTTRMTBFA

ii

timeofin termsor

where OT=operating time, TCM=total correctivemaintenance

Ai is primarily a function of design



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Exam le of inherent s stem availabilit

Assume the system had been running for two yearsan you a een mon or ng e a ures.

If you had 20 failures the MTBF would be?

What would the Inherent Availability be if the meantime to repair was 4 hours?

MTBF

TTRMTBFinh inh ____

Reproduced with courtesy from Mark Mackenzie


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If we are concerned with both corrective andreventive maintenance


A hi d il bili


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Achieved availability

Aa is availability when we are concerned with both,

logistic delay time

TPMTCMOTA

MMTMTBMA aa

timeofin termsor

where MTBM=mean time between maintenance MMT=mean maintenance time OT=operating time,

TPM=total preventive maintenance

A is now both a function of desi n and reventivemaintenance (may also be partly a function of design)



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Example of achieved availability

A generator runs non-stop for 3 months and fails 3 times.

=once which takes 5 hrs.

e ac eve ava a y s

24313 OT.

52524313

TPMTCMOTa



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Is more frequent preventive maintenancer f r v il ili ?

C.E. Ebeling (1997), Introduction to reliability and maintainability engineering, McGraw-Hill, Boston


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administration and logistics


Ope ational a ailabilit


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Operational availability

Ao is availability when we are concerned with corrective& preventive maintenance, also administrative & logisticdelay time

timeofin termsorMALDTMMTMTBM

Ao

OT

where MTBM=mean time between maintenance

TALDTTPMTCMOT

o

MMT=mean maintenance time MALDT,TALDT = mean/total adm. and logistic delay time

, TCM, TPM=total corrective/preventive maintenance

organisational effectivenessAdapted from the Defence Reliability Management Course, 2/2005


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What do the factors mean for operationalavailabilit

What can we do about each of these?



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What about the administrative logistics down time, a e ea me or pump par s was one

week? mean down time (MDT) = 172hrs

What if there was preventative maintenance orscheduled maintenance? MTBM = 504hrs

MTBM

MDTMTBMo o ____

Reproduced with courtesy from Mark Mackenzie


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Most common approaches

RBD

FTA

Flow networks

Petri Nets

on e ar o s mu a on


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Series systems

Assuming constant failure

rate (exponential time to

R1 R2

failure) and constant repairrate (exponential time to

=

Steady-state availability 21122121

21

A

Generally for n components n

iA


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Parallel systems



R1


=

R2

Steady-state availability *) 22 22

2

A

Generally for n components n

iA 1

* 1 2 . , . .

team is assumed


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Standby systems Unit 1



Sensor


=

Unit 2

Steady-state availability *) 22

A

* 1 2 .


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- - - Unit 1 Assuming constant failure

rate (exponential time to Unit 2

rate (exponential time torepair), where =1/(MTTR) Unit 3

Steady-state availability *),

e.g. n=3, k=23223

223

63 A

k1 For general n and k ini

in i

A

0)(

1

*) Assuming 1= 2 = 3 = .


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Example: ship missile system

.

Q: find steady-state availability of the system excluding themissiles and disregard switching failure

Exam le: shi missile s stem


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Exam le: shi missile s stem

A: Availability of the radar system is

999996.0

001.05.05.0

22

2

22

2

radarA

The availabilit of the launch and uidance s stem

....

9974.00013.05.0

5.0

LGA

The system availability is

... LGradarsys


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Monte Carlo simulation

Benefits


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Benefits

The designer can be confident that the system hasspec e re a ty or t e r t o componentcharacteristics, provided all the analytical results are

It is suitable for computerized design;

Any probability distribution is simulated;

No complex mathematical treatments are needed.

An advantage with Monte Carlo simulation is that theevents in the RBD do not have to be combinedanalytically since the simulation itself takes into account

whether each block is failed or functional


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Key elements

Identification of the probability distribution for each

Identification of random variable generation for designparame ers ase on e g ven pro a y s r u on y

computer;

Identification of the probability distribution, its mean andvariance of system performance by simulation.


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Limitations

Mathematical models for simulation are required;

All the system components need to be included inorder to obtain reasonable analytical results;

A large number of replicas of the system aresimulated.


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Example:

repair logicfor a typicalsystem


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Simulation results of depot stock levels


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A type of state-space analysis technique

probability of state transitions from failed state too eratin state and vice versa

A component in a system is assumed to be in either

Probability of failure and of returning to an available state

are o n eres s

Particularly useful to maintained systems for which RBDcan be not directly applicable


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Example (1-out-of-2 active redundantsys em 11 5

When the two componentsare identical


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The solution

The

availability, A S0 (t) is

The unavailabilitfor some specificand is shown


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SUMMARY


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Inherent

Achieved

Documents

Availability Modelling