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Component reliability

Jørn Vatn

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The state of a component is either “up” or “down”

T1, T2 and T3 are ”Uptimes”

D1 and D2 are “Downtimes”

State

Time

T1

D1 D2

T2 T3"Up"

"Down"

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Time to failure - TTF

The term ‘time to failure’ (TTF) denotes the time from a unit is put into service, until it fails That is TTF is equivalent to T1

In some situations we also use the term time to failure to denote T2, T3

It should however be denoted that the distribution of subsequent uptimes are not necessarily identical

A range of quantities relates to TTF Distribution function, F(t) = Pr(T t) Survivor function R(t) = 1 - F(t) = Pr(T > t) Hazard rate z(t) Mean Time To Failure (without maintenance), MTTF = MTTFWO

T1

D1 D2

T2 T3"Up"

"Down"

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Availability (A) and unavailability (U)

Availability is the probability that the component is able to perform its designated function Availability is thus the probability that the component is functioning

well when we consider an arbitrary point of time Availability may also be seen as the average “uptime”

Unavailability is the probability that the component is not able to perform its designated function Unavailability is thus the probability that the component is not

functioning well when we consider an arbitrary point of time Unavailability may also be seen as the average “downtime”

A+U = 1 = 100%

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Evident and hidden function

An evident function means that the a failure of the component immediately will be detected, i.e. when we go from “Up” to “Down will be known to the operator under normal operation procedures

A hidden function means that we do not continuously monitor the unit, thus we will not detect a failure before the unit is demanded, or we perform a functional test

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Unavailability for evident functions

The uptimes and downtimes are characterized by: Mean Time To Failure = MTTF = E(T1) = E(T2) = … Mean Down Time = MDT = E(D1) = E(D2) = …

Graphically we observe that the average “downtime” is given by

where = 1/MTTF = failure rate (assume constant failure rate) = 1/MDT = repair rate (assume constant repair rate)

1 2 3

1 1 2 2 3 3

MDTMDT /

( ) ( ) ( ) MTTF MDTn

D D DU

T D T D T D

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Hidden functions

A hidden function means that a component failure is not immediately detected, i.e. We do not know when we are going from “Up” to “Down” In this situation D1,D2,… represent the ”non detected” downtime +

the time of repair

In order to reduce the non-detected downtime, the component is tested (function test) periodically

Time between testing is denoted (test interval) If a failure occurs in a test period we will in average be

down half of the interval, i.e. /2

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Unavailability (probability of failure on demand)

State

Time

T1

D1 D2

T2 T3"Up"

"Dwon"

MDT / 2PFD

MTTF MDT MTTF / 2 2U

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Reliability block diagram (RBD)

Reliability block diagrams are valuable when we want to visualise the performance of a system comprised of several (binary) components

The interpretation of the diagram is The system is functioning if it is a connection between a and b, i.e.

it is a path of functioning components from a to b. The system is in a fault state (is not functioning) if it does not exist

a path of functioning components between a and b

1

2

3

a b

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Exercise

Define all combination of component states in the example RBD

List those combination that represent a system fault state

1

2

3

a b

1 2 3 System

Up Up Up Up

Down Up Up Down

Etc.

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Structure function

For components we have

For structures (systems) we have

denotes the structure function, and depends on the xi’s (x is a vector of all the xi’s).

1 if the component is functioning at time ( )

0 if the component is in a fault state at time

tx t

t

1 if the system is functioning at time ( , )

0 if the system is in a fault state (not functioning) at time

tt

t

x

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Structure function for simple structures

For a serial structure we have (x) = x1 x2 ... xn

For a parallel structure (redundancy) we have (x) = 1-(1-x1)(1- x2) …(1- xn)

For structures composed of serial- and parallel structureswe may combine the formulas above

1 2 3 na b

. . . .

1

2

3

n

a b

.

.

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Example

1

2

3

a b1

2

3

a b

I II

(x) = I II because I and II are in serialI = x1

II = 1-(1-x2)(1- x3) because 2 and 3 are in paralell(x) = x1(1-(1-x2)(1- x3))

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Pivotal decomposition

Some structure are more complex than just a collection of serial and parallel structures

We may then often use the rule of pivotal decomposition to establish the structure function Consider component number i in the structure The rule of pivotal decomposition now states: (x) = xi(x | xi=1) + (1-xi)(x | xi=0)

where (x | xi=1) is the structure function if component i is functioning (all the time), and (x | xi=0) is the structure function if component i is not functioning (is not there)

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Example

Component 5 is causing “problems” for us If component 5 is always functioning we have

(x | x5=1) = [1-(1-x1) (1-x3)] [1-(1-x2) (1-x4)]

If component 5 is never functioning we have (x | x5=0) = [1-(1-x1x2) (1-x3x4)]

In total (x) = x5(x | x5=1) + (1-x5)(x | x5=0)

= x1x2+x3x4-x1x2x3x4+x1x4x5-x1x2x4x5-x1x3x4x5+2x1x2x3x4x5+x2x3x5-x1x2x3x5-x2x3x4x5

1 2

3 4

5

1 2

3 4

1 2

3 4

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k-out-of-n structures

A k-out-of-n system is a system that functions if and only if at least k out of the n components in the system is functioning

We often write k oo n to denote a k out of n system, for example 2 oo 3

Example 1: We have three pumps which each has a 50% of the total required capacity Thus, it is sufficient that only two pumps is functioning to achieve

sufficient pump capacity 2 oo 3 system. Example 2: We have mounted 3 fire detectors in a process area

To avoid shut-down due to false alarms, we vote the detectors, and require 2 fire indications to shut-down and start fire fighting 2 oo 3 system

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Structure function for k oo n

1

2

3

n

a b

.

.

k oo n

n

ii

n

ii

kx

kx

1

1

if 0

if 1)(x

Example: 2 oo 3 (x) = x1x2+x1x3+x2x3-2x1x2x3

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Binary state variables

Note that the state variables are binary, and takes the values one or zero.

We note that 1n = 1 for n > 1 0n = 0 for n > 1

Thus we could replace xin with xi for n > 1

This simplifies the expansion of the terms in the structure function, and is also a prerequisite for system reliability assessments to come later in the course

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Exercise

Find the structure function for the following RBD

13

a b

45

2

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Solution

I(x) = 1-(1-x3)(1-x4) = x3+x4-x3x4

II(x) = x5I(x) = x3x5+x4x5-x3x4x5

III(x) = 1-(1-x2)(1- II(x)) = x2+x3x5+x4x5-x3x4x5-x2x3x5-x2x4x5+x2x3x4x5

(x) = x1III(x) = x1x2+x1x3x5+x1x4x5-x1x3x4x5-x1x2x3x5-x1x2x4x5+x1x2x3x4x5

13

a b

45

2

III

III

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Component reliability

We introduce pi = as the reliability of component i

Usually pi will be identical to the availability (A), and for repairable components we have

Since the Xi’s are binary values we have

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System reliability (h(p) = pS)

We introduce h(p) = probability that system is functioning where p is a vector of component reliabilities

Since the structure function, (x) also is a binary function, we have h(p) = Pr((X) = 1) = E((X))

Now, assume The Xi’s are independent (X) can be written as a sum of products (which we usually can)

) = where Sj is term j in the sum of products

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General procedure for calculating h(p)

1. Obtain the structure function (x)

2. Multiply out all terms in (x) (to get a sum of products)

3. Remove all exponents in powers of x, i.e. replace xin with

xi for n > 1. Denote the result M(x)

4. The system reliability is found by replacing the xi’s in M(x) with the corresponding pi’s, i.e.,

5. h(p) = pS = M(xx = p)

Note 1: All the Xi’s must be independent

Note 2: Step 2 could be extremely time consuming

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Exercise

Find the system reliability for the RBD:

When the following reliability parameters are given:

Component MTTF MDT1 20 000 82 4 000 243 1 000 484 1 000 485 4 000 12

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Solution

h(p) = pS = M(xx = p)= p1p2+p1p3p5+p1p4p5-p1p3p4p5-p1p2p3p5-p1p2p4p5+p1p2p3p4p5

= 0.99957

Component MTTF MDT1 20 000 8 0.999602 4 000 24 0.994043 1 000 48 0.954204 1 000 48 0.954205 4 000 12 0.99701