3. reliability modelling

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Reliability Modeling and E l ti

Indian Institute of Technology, Kharagpur

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Evaluation

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DEFINITIONS

System is a collection of componentsconfigured to realize a given task.

This can be described through:

• Structure Function (Mathematical)

L i Di (Vi l)


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i • Logic Diagram (Visual)

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TYPE OF SYSTEMS

Systems are of two types:

• Non-maintained (Reliability and associated metrics)

• Maintained (Availability and associated metrics)


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i metrics)

• Appropriate stochastic models are verymuch different for these two types ofsystems!

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SYSTEM RELIABILITY MODELLING

• In case of non-maintained systems, systemreliability or Mean Time To Failure ( MTTF) isgenerally the main criteria of systemperformance.

• On the other hand, in case of maintainedsystems, maintainability design (MD) goeshand in hand with its design for reliability


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i hand-in-hand with its design for reliability(DFR) and the system needs to be designed forboth reliability and maintainability, or forMTTF and MTTR, and

• Many a times, system availability is oftenoptimized within the resources available forsystem design.

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SYSTEM HIERARCHY

• All systems have a built-in hierarchy


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SYSTEM MODELLING

• One can model system in any of the followingways using failure/success criterion:

Black Box ApproachIn this approach, the state of system is described either interms of the two states (working/ failed) or more than twostates without linking them to the components of the


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system.

White Box ApproachIn this approach, one models the state of the systemspecifically in terms of the states of various components ofthe system.

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SYSTEM MODELLING

White Box Modeling: The first one is called Forward (or Bottom up) approach. The other

is called Backward (top-down) approach.

1. In the forward approach, one starts with events(success or failure) at item level and thenproceeds forward to system level to evaluate theconsequences of such events on system


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i consequences of such events on systemperformance.

2. In the Backward approach, one starts at systemlevel and proceeds downward to the item levelto link system performance to success/failures atitem level.

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SYSTEM MODELLING

Linking the system performance to thefailures/success at item level can be donequalitatively or quantitatively.

In the qualitative analysis, the focus is on causalrelationships that link the item level events to


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system events or vice-versa.

In quantitative analysis, one obtains variousmeasures of system performance (e.g., systemreliability) in terms of items performance. (e.g.,item reliabilities)

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GRAPHICAL REPRESENTATION

• There are two ways of graphically representing the functional relationship in a Qualitative analysis of a system. The first one is Block diagram or Reliability Logic Diagram (RLD) approach. An RLD indicates which components of the system must operate for


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successful operation of the system. The other approach involves indication of which components must failbefore the system fails. This graphical representation is done through Fault Tree Logic Diagram. Block Diagram approach is optimistic approach of modelling whereas the FTA is pessimistic approachof Modelling.

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BLOCK DIAGRAM OF A SYSTEM

• Since the system modeling is done on the basis offunctional interrelationship of the constituent itemsin relation to the overall functioning of the system,one has to keep the hierarchy of the system in view.

• Therefore, we must consider the success of thesystem vis à vis the success of the constituent


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i system vis-à-vis the success of the constituentsubsystems and then success of each subsystem isconsidered vis-a-vis the success of the constitutingcomponents. This exercise will eventually result ina Block Diagram of the system.

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RELIABILITY LOGIC/BLOCK/NETWORK DIAGRAM

• A diagram which shows the logical relationships of systemsuccess (or failure) with success (or failure) of itsconstituent parts is termed as RLD/RBD/NRLD.

• RLD indicates that which component(s) in a system mustoperate without any failure for successful operation of thesystem


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i system.

• To develop RLD and reliability analysis thereof requireunderstanding and depiction of operational relationships orlogical interrelationships amongst its constituent items(subsystems, units or elements or parts.)

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BLOCK DIAGRAM OF A JET ENGINE

• Jet Engine– Fuel Supply System

• Fuel Pump

• Fuel Filter

– Carburetor• Jet


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• Other Components

– Ignition System (2)• LV

• HV

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ASSUMPTION IN MODELING

• All elements and systems can be either in operatingor failed state, i.e., two state modeling.

• All elements are in operating state initially exceptpossibly in the case of redundancy.

• States of all elements are statistically independent.(Failure of one element does not affect the probability


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i (Failure of one element does not affect the probabilityof failure of other elements).

• The reliabilities of all constituent components areknown through some “reliability analysis”.

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INDEPENDENT FAILURES

• Failure times of components are influenced byenvironmental conditions. As the environmentgets harsher, the time to failure of componentdecreases.

• If the components of a system share the sameenvironment, failure times of components


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i environment, failure times of componentsbecome statistically dependent.

• However, if the dependence is weak, or thecomponents are located far apart, one canignore dependence and treat the failure times asbeing statistically independent.

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INDEPENDENT FAILURES

• The advantage of this assumption is thatthe failure times of a component can bemodelled separately using Univariatefailure distribution functions.

• If the dependence is significant


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i If the dependence is significant,multivariate failure distributions must beused and the analysis becomes much morecomplicated.

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Reliability Models


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VARIOUS SYSTEM MODELS

Now we will consider various systemmodels that can be built for a givensituation. Among these system models areseries, parallel, series-parallel andparallel-series, k-out-of-m:G, standby,


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partial standby , non-series parallel modeletc. We will consider these one by one andderive system reliability and system MTTFfor these configurations.

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Series & Parallel Systems

Series System Parallel System

non-redundant system(all components must work for system success)

fully redundant system(all components must fail for system failure)

TR1, 10 kVA TR1, 20 kVA


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,

TR2, 10 kVA

15 kVALoad

TR2, 20 kVA

15 kVALoad

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Let R = P [Success]

Q = P [Failure]

R + Q = 1

n

i

21s

R

RRR

Ss

RR1

R1Q

21

Series System


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1i

2121

21

21

QQQQ

)Q)(1Q-(1-1

RR1

product rule of reliability

S

R1 R2

S

Q1 Q2

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Example:

A system consists of 10 identical components, all ofwhich must work for system success. What is thesystem reliability if each component has a reliability of0.95?

Component reliability, R = 0.95

Number of components n = 10


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Number of components, n = 10

Using Product Rule of Reliability,

System Reliability, Rs = Rn

= (0.95)10 = 0.5987

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Series System

0.4

0.5

0.6

0.7

0.8

0.9

1.0

tem

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0.98

0.999

If each component has a reliability of 0.9.

Number of Components

Reliability

1 2 3 4 5

0.9 0.81 0.729 0.6561 0 59049


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0.0

0.1

0.2

0.3

0 10 20 30 40 50 60 70


Sys

t

0.9

System Reliability decreases as the number of components increases in a Series System. The number on the curve is the reliability of each component.

5 10 20 50

0.59049 0.348678 0.121577 0.005154

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• Most common and an important reliabilitymodel.

• Series system has no redundancy since afailure of any component causes failure ofthe entire system

Series Model


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i the entire system.

• The system reliability will be less than thereliability of least reliable component in themodel.

• System design would require high reliabilityand as few components.

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• Assuming Ei s refers to the events of units being functionally good.

Series Model

n

is

ns

n

pn

nR

R

321

13212121

321s

)P(E)...P(E)P(E)P(ER

then,tindependenarefailuresunittheIf

)E...EE/EP(E)...E/EP(E.)/EP(E.)P(E

)E...EEP(E

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• The above result can also be obtained using time to failure considerations

2323

i 1

n

iis

ns

ns

trtR

tttttttR

tttttttR

1

21

21

)()(

)Pr()...Pr()Pr()(

)()...()(Pr)(

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Parallel Redundant System

2

1

21

21

Ss

RRRR

)R)(1R-(1-1

QQ1

Q1R


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product rule of unreliability

S

Q1 Q2

S

R1 R2

2121 RRRR

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Example:

A system is to be designed with an overall reliability of 0.999 usingcomponents having individual reliabilities of 0.7. What is theminimum number of components that must be connected in parallel?

System Reliability, Rs = 0.999

Component reliability, R = 0.7

Number of components, n = ?


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System Unreliability, Qs = 1 - Rs = 1 – 0.999 = 0.001

Component Unreliability, Q = 1 – R = 1 – 0.7 = 0.3

Using Product Rule of Unreliability, Qs = Qn i.e. 0.001 = (0.3)n

therefore, n = ln (0.001) / ln(0.3) = 5.74

since n is an integer, n = 625

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Parallel System

Reliability of each component is 0.7


System Reliability

1 0.7000002 0 910000

0.9

1.0

elia

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2 0.9100003 0.9730004 0.9919005 0.9975706 0.9992717 0.999781 0.6

0.7

0.8

1 2 3 4 5 6 7 8


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PARALLEL MODEL

• System reliability will be more than any of theconstituent component.

• Most common and cheapest method of reliabilityimprovement, if feasible or permitted.

• An additional unit will improve the MTTF by 1.5times of the MTTF of a single unit


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i times of the MTTF of a single unit.

• There will not be much gain in system MTTF, ifnumber of units increased from 4 to 5 and so on.(Law of diminishing return-See in Next Slide).

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System MTTF for Various Configurations

PARALLEL MODEL


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Series/Parallel Systems

Network Reduction

Telcom. Repeater Station Power Supply System 4

2 31

Battery Bank

DG Rectifier

Cable


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TechniqueRX = R2.R3

QY = QX.Q4

RY = 1 - QX.Q4

= 1 – (1-RX).(1-R4)

RS = RY.R1

4

1X

1Y

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Example:

Derive a general expression for the unreliability of themodel shown below, and hence evaluate the unreliability ofthe system if all components have a reliability of 0.8.

input Output


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Reliability Network Model (Reliability Block Diagram) of the System

input p

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Ri = 0.8 for i = 1 to 5


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Q8 = Q7 . Q5

= (1 - R7). Q5

= (1 - R1 . R2 . R6). Q5

= [1 - R1 . R2 . (1 - Q6)]. Q5

= [1 - R1 . R2 . (1 - Q3 . Q4)]. Q5

Q8 = [1 – 0.8 x 0.8 (1 – 0.2 x 0.2)]x 0.2 = 0.07712

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m out of n Systems

partially redundant system

TR1, 10 kVA

TR2, 10 kVA

15 kVALoad

If components are identical then

(R + Q)3 = R3 + 3R2Q + 3RQ2 + Q3

Rs = R3 + 3R2Q

Q 3RQ2 Q3 1 R


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TR3, 10 kVA

Qs = 3RQ2 + Q3 = 1 - Rs

If components are non-identical then

R1 R2 R3

Rs = R1 R2 R3 + R1 R2 Q3 + R1 Q2 R3 + Q1 R2 R3

1

3

2

2/32 out 3 system

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Example:

Derive a general expression for the unreliability of the systemwhose reliability model is shown below. Consider the case inwhich all parallel branches of this system are fully redundantwith the exception of that consisting of components 4, 5 and 6 forwhich any 2 of the branches are required for system success.

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2/3


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Reliability Network Model (Reliability Block Diagram) of the System

input Output

1

7

2

3 6

5

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If Components 4, 5 and 6 are identical, each with reliability R, R9 = R3 + 3R2Q

Q8 = Q2 . Q3

1

7

2

3

4

6

5

2/3

1 8 9

7


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OtherwiseR9 = R4 R5 R6 + R4 R5 Q6 + R4 Q5 R6 + Q4 R5 R6

R10 = R1 R8 R9

7

10

7

11

Qs = Q10 . Q7

= (1 – R10). Q7

= (1 - R1 . R8 . R9). Q7

= [1 - R1 (1 – Q8).R9 ]. Q7

= [1 - R1 (1 – Q2 .Q3). R4 R5 R6 + R4 R5 Q6 + R4 Q5 R6 + Q4 R5 R6 ]. Q7

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2-out-of-3:G Model

The Reliability Logic Diagram and Fault Tree diagram for 2-out-of-3:G system are shown below:


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RELIABILITY OF 2-OUT-OF-3:G SYSTEM

Let us designate the paths in Block diagram asObviously,

where Ei is the event of unit i being good. Therefore, system reliability is given by:

., 321 TandTT

323312211 , EETandEETEET


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i system reliability is given by:

where,

3

1i ji kjikjijii TTTPTTPTP

.23

2

32

321323121

ppRor

pppppppppR

s

s

321Pr TTTRs

3

3321

3

1

3

2

EEEPTPRi

is

321312121 EEEPEEEEPTTP

321 EEEP

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For exponential failure distribution, we cancompute system MTTF of by:

where,

2-out-of-3:G Model (MTTF)

dttRMTTF s )(0


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and for exponential failure distribution, t

i etp )(

321323121

2111

MTTF

321323121 2 pppppppppRs

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Consider a 3 out of 4 system. The system is successful ifany of the path is good The success of each pathdepends on the success of two elements

1

2

3in

Out

1

1

1

2

2

3

34

4

Exercise 3-out-of-4:G Model (MTTF)


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3 4

Block Diagram Reliability logic Diagram

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...)AP(A)(PA)P(A)P(A)P(A

)AAAP(AR

yields,

EEEAEEEA

EEEA,EEEA

ji4321

4321s

43244313

42123211


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3939

)AAAP(A)AAP(A

...)()()()()(

4321

i j k

kji

i j

j3

Rs=P(E1E2E3)+P(E1E2E4)+P(E1E3E4)+P(E2E3E4)- 3P(E1E2E3E4)

If the elements are identical and their failures are independentthen

Rs=4P3-3p4

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Identical Units k-out of – M:G system (Using Binomial)

A system functions properly if any k out of m units function properly, the probability of exactly k successes out of m is given by,

unitsmor1,m1,kk,aslongasfunctionalremainssystemThe

p)(1pk

mp)m,B(k; kmk


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4040

kthanlargeris1)k(mtheifly,particularp)(1pi

m1R

or

p)(1pi

mR

by,givenissuccesssystemofThe.function

mi1k

0i

s

mim

ki

s

i

i

yprobabilit

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For identical units, for various

MTTF OF K-OUT-OF-M:G SYSTEM

m

kj jmkMTTF

11,


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combinations of k and m ,(for k m) , weprovide system MTTF for identical in thenext slide.

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k-out-of-m:G Model (MTTF)


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System MTTF for various values of k and m

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Example: System Reliability

Block A

ARR


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The system reliability can be calculated as:

3

110113

3211112

3

9876543

2110

)1()1(1

)1()(1)(11

,

RRCRRCR

RRRRRRRR

RRRwhereRRRRR

C

B

ACBAs

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System Reliability Evaluation Using Probability Distributions

Time dependent reliability – described by probability distributions

Reliability of Component 1 for a time period t, R1(t) =t

0

1 (t)dtλ

e

where (t) is the hazard rate of Component 1.

During useful life period of the component R1(t) =tλ1e


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where is constant, and is called the failure rate of Component 1.

g p p 1( )

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Series Systems

Rs = R1 . R2 product rule of reliability

Considering time dependent probabilities,

Rs(t) = R1(t). R2(t) =t

01 (t)dtλ

et

02 (t)dtλ

e

For ‘n’ component series system with hazard rates 1(t), 2(t), …, n(t),


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Rs(t) =

n

i 1

t

0i (t)dtλ

ealso applicable if failure distributions for different components are different

During useful life period when component failures are exponentially distributed

Rs(t) = = =

n

i 1

tλ ietλ i

1e

n

i tee where equivalent hazard rate, e = is also constant

n

ii

1

i.e. resulting distribution for the system is also exponentialne

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Parallel Systems

Qs = Q1 . Q2 product rule of unreliability

Considering time dependent probabilities,

Qs(t) = Q1(t). Q2(t) = [1- ].[1- ]t

01 (t)dtλ

et

02 (t)dtλ

e

For ‘n’ component parallel system with hazard rates 1(t), 2(t), …, n(t),

2

1


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Qs(t) = [1 - ]

n

i 1

t

0i (t)dtλ

e


Qs(t) = [1 - ]

n

i 1

tλ ie

Therefore, resulting distribution for the system is non-exponentiali.e. resulting hazard rate for the system is no longer constant, but a function of time

cannot obtain equivalent hazard rate for exponential distribution

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Partially Redundant (m out of n) Systems

[R(t) + Q(t)] n = nCr R(t) n-r Q(t) r

n

r 0


Apply Binomial Expansion


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R(t) = and Q(t) = 1 -λteλte

Let see some simple examples.

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Example:

A simple electronic circuit consists of 6 transistors each having a failure rate of 10-6 f/hr, 4 diodes each having a failure rate of 0.5 x 10-6 f/hr, 3 capacitors each having a failure rate of 0.2 x 10-6 f/hr, 10 resistors each having a failure rate of 5 x 10-6 f/hr and 2 switches each having a failure rate of 2 x 10-6 f/hr. Assuming connectors and wiring are 100% reliable, evaluate the equivalent failure rate of the system and the probability of the system surviving 1000 and 10000 hours if all components must operate for system success.

n


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i1

Equivalent failure rate of the system, e =

= 6(10-6 ) + 4(0.5 x 10-6 ) + 3(0.2 x 10-6 ) + 10(5 x 10-6 ) + 2(2 x 10-6 ) = 6.26 x 10-5

f/hr

Rs(1000 hr) = = = 0.9393tee 1000x1026.6 6

e x

Rs(10,000 hr) = = = 0.5347tee 10000x1026.6 6

e x

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Example:

Consider a system comprising of 4 identical units each having a failurerate of 0.1 f/yr. Evaluate the probability of the system surviving 0.5 yearsand 5 years if at least two units must operate successfully.

Using Binomial Expansion,

[R(t) + Q(t)] 4 = R4(t) + 4 R3(t)Q(t) + 6 R2(t)Q2(t) + 4 R(t)Q3(t) + Q4(t)

where, R(t) = and Q(t) = 1 -λte

λte


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For 2 out of 4 system,

Rs(t) = R4(t) + 4 R3(t)Q(t) + 6 R2(t)Q2(t)

= + 4 (1 - ) + 6 (1 - )2λt4e λt3e λte λt2e λte

For t = 0.5 years, t = 0.1 x 0.5 = 0.05 Rs(0.5) = 0.9996

For t = 5 years, t = 0.1 x 5 = 0.5 Rs(5) = 0.8282

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Standby Model

One or more units are in standby mode waiting to take overthe operation from the main operating unit as soon as thefailure of the operating unit takes place.

Decision switch senses the failure of the basic unit. It is assumed that the operation of failure sensing and

switching on to the standby unit is instantaneous, i.e.,uninterrupted system operation


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uninterrupted system operation. The standby system fails only when all the system units

including the standby units have failed. The reliability of such system will be higher than the

parallel system with equal number of active units.(Conditions?).

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STANDBY MODELS


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STANDBY MODELS: MISSILE SYSTEM


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Two Unit Standby ModelObviously, from the Fig (b), it

can be seen easily that the system success would be obtained:

Either (i) when the primary unit # 1 continues to work beyond the mission time t, i.e., t1 > t, where t1 is the time to failure

System Modelling


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i where t1 is the time to failure of unit # 1.

or (ii) unit # 1 fails at time t1 (t1< t), and the standby unit ( unit # 2) comes into operation immediately upon the failure of unit #1 and continues to work for t2 > (t-t1). The system reliability

therefore can be given by:

])}ttt(t){(tt)Pr[(t)t(R 1211s ne

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Two Unit Standby ModelTherefore, reliability for a two-unit standby system would be obtained as:

or

System Modelling

])}ttt(t){(tt)Pr[(t)t(R 1211s

t

dtttptftp 112111


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i 0

tetp 11

11111

tetf 1212

ttettp

1

0121

1211 dteeeRRtRt

tttts

21121

21

1

ttt

s

eeetR

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The function of decision switch (DS) is critical and thesuccess of a standby system to great extent dependsupon the reliability of this decision switch.

The reliability decision switch is driving the systemreliability.

STANDBY MODELS: IMPERFECT SWITCH


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Two-unit standby system ,

tds represents a random variable denoting the time to failure of the decision switch.

1121

0

11 )()()()( dtttptfptptRt

dsS

)}]()(){()[()( 12111 tttttttttPtR dss

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DS perfect but Degradation of standby unit in standby mode. Combining degradation of standby unit and

Imperfect switching

STANDBY MODELS: OTHER


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THREE-UNIT STANDBY MODEL

Similarly, we can determine the reliability of a three unit system as:

System Modelling

1

21

1

tt

ts

eeetR


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i 2112

1 s

32312321131221

321

ttt eee

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PARTIAL STANDBY MODEL

As shown in fig. there is another way to configure threeunits.

Two units are in operational mode and third unit maybe put in standby mode.

Actually both units are required in operational mode forsatisfactory operation of the system.

Thus are in series as far as reliability logic diagramconcerned.


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Third unit replaces any one failed from these twooperational units.

1

2

3

Units 1 and 2 are in active mode and unit 3 is in standby mode

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Standby ModelsSystem Modelling

)exp(!

)(1

0

ti

tR

m

i

i

s

Rs(t) = e-2( 1+ 2 t )


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Triple Modular Redundancy

System Modelling


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The triple modular redundancy (TMR) consists of threeparallel digital circuits – A, B, and C - all having the sameinput. The outputs of the three circuits are compared by avoter, which sides with the majority and gives the majorityopinion as the system output. If all three circuits areoperating properly, all outputs agree; thus the system outputi t

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System Modelling

Triple Modular RedundancyHowever, if one element has failed so that it has produced an

incorrect output, the voter chooses the output of the twogood elements as the system output because they bothagree; thus the system output is correct. If two elements havefailed, the voter agrees with the majority (the two that have


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failed); thus the system output is incorrect. The systemoutput is also incorrect if all three circuits have failed.Actually, circuits-A, B, and C- in most cases are identical andthey are three replications of the same design. Using thisassumption, and also assuming that the voter does not failand all digital circuits are independent and identical withprobability of success p, the system reliability is given by:

123

2

033

3s p1pp1pR

p23pp2p3 232

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SHARED LOAD MODEL

System Modelling

A model in which the load on a memberchanges and consequently the hazard rate ofthe member also changes, as one or moremember fail.

In this model all the units/sub-systems are


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connected in parallel and each unit shares theload equally.

After the failure of one or more units theremaining units share the load equally but witha higher value of hazard rate.

The levels of hazard rate change only when asubsystem failure occurs.

It is assumed that failure distribution of thesurvivor subsystem after failure of asubsystem does not depend on the interval of

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TWO UNIT SHARED LOAD MODEL

System Modelling

In this case, we will have two units sharing the given load equally however if one unit fails, the total load will have to be take over by the single unit as shown in the following figure.


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Half load

Half load

Full load

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System ModellingTWO UNIT SHARED LOAD MODEL

Let, fh(t) : be the p.d.f. for time to failure of a unit

under half load condition ff(t) : be the p.d.f. for time to failure of a unit

under full load condition( this occurs


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i under full load condition( this occurs when one unit fails and other unit is forced to take the full load )

Therefore the system is successful if,1. Both component remain operational until the

mission time. The probability of this is givenby:P[(t1>t)∩(t2>t)] = {ph(t)}2

Where, ph(t) is reliability of a unit under halfload

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System Modelling

2. The first unit fails and the second unit survives and the takes the full load (after the failure of the first unit) until the time t, The probability of this operation is given by:


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P[(t1≤t, under half load)∩(t2>t1, under half load)∩

( t2>t-t1, under full load)] =

3. This is identical to the case 2 the only difference is that here unit 2 fails and unit 1 take the full load( after failure of the second unit) and the probability of this operation can be found by replacing t1 by t2 .

Incase both the units are same the probabilities in case 2 and case 3 turns out to be same.

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System Modelling

Now the system reliability is found by adding theprobabilities of case 1 , case 2 and case 3 i.e.

If we assume the failure distribution to the


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i If we assume the failure distribution to theexponential distribution for both the units anddenote the failure rates at half and full load by λh

and λf respectively. Then the reliability expressioncan be written as,

][)(

)( t2

fh

tht2

sfh

f

h e12

e2etR

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THREE UNIT SHARED LOAD

SYSTEMS

System Modelling

Similarly the reliability expression for the threeunit shared load model can be derived, thesuccess probability versus time plot is shownbelow, E1∩E2∩E3#1

E ∩E ∩ESu


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#2

#3

E1∩E2∩E3

E1∩E2∩E3

ccess

Probability

Time

321 EEE

321 EEE

321 EEE

t1 t2 t

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SYSTEMS

System Modelling

1/3 load

1/3 load

The diagrammatic representation of the load shifting is shown below,


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Full load

1/2 load

1/2 load

1/3 load

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SYSTEMS

System Modelling

For three similar units the above possibilities can be described as,

1. All the three units operate until the mission time and share the load equally.

2. Unit #1 fails and the other two units survive


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i 2. Unit #1 fails and the other two units survive and share the load equally.

3. Unit #1 and Unit #2 fail and Unit #3 operates and is forced to take the full load.

Let t1, t2 and t3 be the time to failure of the first, second and third unit respectively then the probabilities of the above possibilities can be written as,P(#1) = P[(t1>t)∩(t2>t) ∩(t3>t)] = [po(t)]3

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SYSTEMS

System Modelling

P(#2) = 3P[(t1≤t,at 1/3 load)∩{(t2>t1) ∩(t3>t1), at 1/3load}∩ {(t2>t-t1) ∩(t3>t-t1), at ½ load}]

AndP(#3) = 6P[(t1≤t, at 1/3 load) ∩{(t2>t1) ∩(t3>t1), at 1/3 load}


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i }∩{(t2≤t, at ½ load) ∩(t3>t2, at ½ load)} ∩{(t3>t-t1-t2), at full load}

Now the reliability expression for the three unit shared load model canbe written as,

Rs(t) = P(#1) + P(#2) + P(#)

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SYSTEMS

System Modelling

In the integral form it can be written as,

Rs(t) = [po(t)]3 ++

12

1h2

1o1

t

0

o dtttptptf3 )}({}{

tt

2t

dtdttttttfttf61

)()()(}{


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i 1

0

221f2h2h2

1o1

0

o dtdttttptptftptf6 )()()(}{

Now if we take the failure distribution to be exponential distribution with failure rates λo, λh

and λf for the 1/3, ½ and full load respectively then the reliability expression can be written as,

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SYSTEMS

System Modelling

The expression comes out to be,

)()32(

3)( 233 tt

oh

ots

hoo eeetR


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])2)(3(

)2)(32()3)(23([6

)(23

hfof

t

fhoh

t

foho

t

ho

oh

f

ho

e

ee

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SYSTEMS

System Modelling

One very important thing to note is that,• If we put, λ1=2λh and λ2= λf in the two unit

standby reliability expression we get thereliability expression for the two unit

h d l d d l


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i shared load model.• Similarly if we put λ1=3λo, λ2=2λh and λ3=

λf in three unit standby reliabilityexpression we will end up with thereliability expression of three unit sharedload model.

Finally MTTF can be found byintegrating the reliability expressionbetween 0 and Infinity

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COMPARISON OF DIFFERENT MODELS

• Basis:– Assuming constant failure rate.

– All elements or subsystems are identical with reliability, p.


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SINGLE COMPONENT

0.5

0.6

0.7

0.8

0.9

1

R


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0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.1

0.2

0.3

0.4

p

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Three Component in S & P

0.5

0.6

0.7

0.8

0.9

1

R 0.5

0.6

0.7

0.8

0.9

1

R 0.5

0.6

0.7

0.8

0.9

1

R 0.5

0.6

0.7

0.8

0.9

1

R


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0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.1

0.2

0.3

0.4

p

Series

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.1

0.2

0.3

0.4

p

Series

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.1

0.2

0.3

0.4

p

Series

parallel

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.1

0.2

0.3

0.4

p

Series

parallel

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SP:(2-IN-P WITH 1-IN-S)

0 5

0.6

0.7

0.8

0.9

1

R


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0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.1

0.2

0.3

0.4

0.5

p

R

SeriesparallelSP

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PS:(2-IN-S WITH 1-P)

0 5

0.6

0.7

0.8

0.9

1

R


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0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.1

0.2

0.3

0.4

0.5

p

R

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2-OUT-OF-3:G

0.6

0.7

0.8

0.9

1


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0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.1

0.2

0.3

0.4

0.5

p

R

Series

parallelSPPS

2O3:G

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STANDBY WITH ONE ACTIVE AND 2-STNDBY, DS PERFECT, NO

DEGRADATION

0 5

0.6

0.7

0.8

0.9

1

R


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0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.1

0.2

0.3

0.4

0.5

p

R

SeriesparallelSPPS2O3:GStndby

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WITH DS HAS HALF THE FR OF UNITS’ FR, NO DEGRADATION

0 5

0.6

0.7

0.8

0.9

1

R


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0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.1

0.2

0.3

0.4

0.5

p

R

SeriesparallelSPPS2O3:GStndbyStndbyDSFRH

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0.6

0.7

0.8

0.9

1

With DS has same FR of units’ FR, No Degradation


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0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.1

0.2

0.3

0.4

0.5

p

R

Series

parallelSP

PS2O3:GStndby

StndbyDSFRHStndbyDSFRF

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PARTIAL STANDBY

0 5

0.6

0.7

0.8

0.9

1


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0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.1

0.2

0.3

0.4

0.5

p

R

SeriesparallelSPPS2O3:GStndbyStndbyDSFRHStndbyDSFRFPstndby

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Comparison of Various Models


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• Series and standby provides the lowest and highest bounding curves.

• Reliability of partial standby and k-out-of-m for some values of p(0<=p<=0.5) lie below the single component curve line and for highervalues of p lie above the single component line.

• This implies that reliability of such model is worse than the reliabilityof single component for some values of p close to 0 while same modelsoffers better reliability for higher values of p.


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• SP/PS also would also exhibit similar characteristics for higher ordersystems.

• k-out-of-m system invariably shows this characteristic.

• Therefore, it is necessary to know the breakeven point for which amulti-component system has the same reliability as that of a singlecomponent.

• The designer must ensure that the reliability of his system is better thanthis breakeven value while designing.

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• Although, standby model offers best possible reliability. However, inreality it may not be possible to introduce this kind of redundancy withall types of subsystems or elements.

• Besides, the advantage will level off if DS is imperfect and has highFR.

• SO BE CAREFUL IN DESIGNING A SYSTEMS.


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SHARED LOAD MODEL

This model is considered in which load on a member changes and consequently the hazard rate of the member changes, as one or more members fail.

In this model all the units or subsystem comprising the system are connected in parallel and each unit shares the load equally.

If any of subsystem fails the surviving subsystems will share the load


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If any of subsystem fails, the surviving subsystems will share the load equally but with higher value of hazard rate.

For example,

If three generators of 110MW each, connected in parallel electricity andthe load is 200MW, then each of these generators shares a load of66.67MW and at this load each generator has definite failure rate.

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If any one fails the remaining two will share the load of 100MW eachto meet requirement , obviously now failure rate of the survivinggenerator s will be higher than before.

Therefore in this model the levels of hazard rate changes only when a

subsystem failure occurs.


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Two units shared load Model:

Two units are sharing given load equally. However if a unit fails, the total load will have to be taken over by a single unit.

L f ( ) b h d f f

E1∩E2

E1∩E2

E1∩E2Success probabilities

#1

#2


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Let fh(t) be the p.d.f. for time to failure of a unit under half load condition.

Let ff(t)be the p.d.f. for time to failure of a unit under full load condition.

t1 t

Time

Success probabilities of two unit shared load system

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The system will be good if:

1. Both components remain in operational mode until the mission time.

The probability of this given by,

P[t1>t∩t2>t]={ph (t)}2


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Where, ph (t) is reliability of a unit at half load condition i.e.

t1 and t2 are time to failure of two units and the ‘t’ is mission time.

duuftpt

hh )()(

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2 Unit 1 is failed at time t1 , and prior to this the units are taking half of the load each. After time t1 ,unit 2 alone takes the full load until time t.

the probability of this operation ,

),(),[( 121 loadhalfunderttloadhalfunderttP


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)],( 12 loadfullunderttt

1111

0

)()()( dtttptptf fh

t

h

)()()( uduftpwheret

ff

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3 This is like 2 but instead of the first unit failing fist, the second unit is fails first and the other survives until time t.

the probability of this operation

)],(

),(),[(

21

212

loadfullunderttt

loadhalfunderttloadhalfunderttP


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System Reliability = [ sum of the probabilities of earlier three equations]

2222

0

)()()( dtttptptf fh

t

h

111

0

12 )()()(2)]([ dtttptptftpR fh

t

hhS

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3. reliability modelling

Technology

Transcript of 3. reliability modelling