Moosung Jae

May 4, 2015

Quantitative Reliability Analysis

System Reliability Analysis

System reliability analysis is conducted in terms of probabilities

The probabilities of events can be modelled as logical

combinations or logical outcomes of other random events

Graphical methods include

Failure Modes and Effects Analysis (FMEA)

Reliability Block Diagrams (RBD)

Master Logic Diagrams etc

Two main methods used include:

Fault tree analysis

Event tree analysis

Failure Modes and Effects Analysis

Failure modes and effects analysis (FMEA) is a qualitative

technique for understanding the behaviour of components in

an engineered systems

The objective is to determine the influence of component

failure on other components, and on the system as a whole

It is often used as a preliminary system reliability analysis to

assist the development of a more quantitative event tree/fault

tree analysis

FMEA can also be used as a stand-alone procedure for

relative ranking of failure modes that screens them according

to risk

i.e., as a screening tool

FMEA (cont’d)

As a risk evaluation technique, FMEA treats risk in it true

sense as the combination of likelihood and consequences

However, strictly speaking, it is not a probabilistic method

because it does not generally use quantified probability

statements

Rather, failure mode occurrences are described using

qualitative statements of likelihood (e.g., rare vs. frequent etc.)

Consequences are also ranked qualitatively using levels or

categories

e.g., ranging from safe to catastrophic

FMEA uses a rank-ordered scale of likelihood with respect to

failure mode occurrence, so that together with the

consequence categories, a rank-ordered level of relative risk

can be derived for each failure mode

FMEA (cont’d)

FMEA consists of sequentially tabulating each component

with

all associated possible failure modes

impacts on other components and the system

consequence ranking

failure likelihood

detection methods

compensating provisions

Failure modes effect and criticality analysis (FMECA) is

similar to FMEA except that the criticality of failure is

analyzed in greater detail

Example Example: Consider the

following water heater system

used in a residential home.

The objective is to conduct

a failure modes and effects

analysis (FMEA) for the

system.

Solution (cont’d)

Define consequence categories as

I. Safe – no effect on system

II. Marginal – failure will degrade system to some extent but will

not cause major system damage or injury to personnel

III. Critical – failure will degrade system performance and/or

cause personnel injury, and if immediate action is not taken,

serious injuries or deaths to personnel and/or loss of system will

occur

IV. Catastrophic – failure will produce severe system

degradation causing loss of system and/or multiple deaths or

injuries

The FMEA is shown in the following table

Fundamentals of Reliability

© M. Pandey, University of Waterloo

Solution Component Failure

Mode

Effects on

other

components

Effects on

whole system

Consequence

Category

Failure

Likelihood

Detection

Method

Compensating

Provisions

Pressure

relief valve

Jammed

open

Increased gas

flow and

thermostat

operation

Loss of hot

water, more

cold water

input and gas

I - Safe Reasonably

probable

Observe at

pressure

relief valve

Shut off water

supply, reseal or

replace relief

valve

Jammed

closed

None None I - Safe Probable Manual

testing

No conseq.

unless

combined with

other failure

modes

Gas valve Jammed

open

Burner

continues to

operate,

pressure relief

valve opens

Water temp.

and pressure

increase; water

turns to steam

III - Critical Reasonably

probable

Water at

faucet too hot;

pressure

relief valve

open (obs.)

Open hot water

faucet to relieve

pres., shut off

gas; pressure

relief valve

compensates

Jammed

closed

Burner ceases

to operate

System fails to

produce hot

water

I - Safe Remote Observe at

faucet (cold

water)

Thermostat Fails to

react to

temp.

rise

Burner

continues to

operate,

pressure relief

valve opens

Water temp.

rises; water

turns to steam

III - Critical Remote Water at

faucet too hot

Open hot water

faucet to relieve

pressure;

pressure relief

valve

compensates

Fails to

react to

temp.

drop

Burner fails to

function

Water

temperature

too low

I - Safe Remote Observe at

faucet (cold

water)

Fundamentals of Reliability

© M. Pandey, University of Waterloo

Reliability Block Diagrams

Most systems are defined through a combination of both

series and parallel connections of subsystems

Reliability block diagrams (RBD) represent a system using

interconnected blocks arranged in combinations of series

and/or parallel configurations

They can be used to analyze the reliability of a system

quantitatively

Reliability block diagrams can consider active and stand-by

states to get estimates of reliability, and availability (or

unavailability) of the system

Reliability block diagrams may be difficult to construct for

very complex systems

Series Systems

Series systems are also referred to as weakest link or chain

systems

System failure is caused by the failure of any one component

Consider two components in series

Failure is defined as the union of the individual component

failures

For small failure probabilities

1 2

where Q denotes the

probability of failure

Series Systems (cont’d)

For n components in series, the probability of failure is then

Therefore, for a series system, the system probability of

failure is the sum of the individual component probabilities

In case the component probabilities are not small, the system

probability of failure can be expressed as

For n components in series

Series Systems (cont’d)

Reliability is the complement of the probability of failure

For the two components in series, the system reliability can

be expressed as

Assuming independence

For n components in series

Therefore, for a series system, the reliability of the system is

the product of the individual component reliabilities

Parallel Systems

Parallel systems are also referred to as redundant

The system fails only if all of the components fail

Consider two components in parallel

Failure is defined by the intersection of the individual

(component) failure events

Assuming independence

1

2

Parallel Systems

For n components in parallel, the probability of failure is then

Therefore, for a parallel system, the system probability of

failure is the product of the individual component

probabilities

The reliability of the parallel system is

For n components in parallel, the system reliability is

Example Problem

Solution:

First combine the parallel components 2 and 3

The probability of failure is

The reliability is

Example: Compute the reliability and probability of failure for the

following system. Assume the failure probabilities for the

components are Q1 = 0.01, Q2 = 0.02 and Q3 = 0.03.

2

3

1

Solution (cont’d)

Next, combine component 1 and the sub-system (2,3) in

series

The probability of failure for the system is then

The system reliability is

Solution (cont’d)

The system probability of failure is equal to

The system reliability is

which is also equal to RSYS = 1 – QSYS

As shown in this example, the system probability of failure

and reliability are dominated by the series component 1

i.e. a series system is as good as its weakest link

Things to Consider

Reliability block diagrams can also be used to assess

Voting systems (k-out-of-n logic)

Standby systems (load sharing or sequential operation)

Simple systems can be assessed by gradually reducing them

to equivalent series/parallel configurations

More complex systems would require the use of a more

comprehensive approach, such as conditional probabilities or

imaginary components

For complex systems, great effort is needed to identify the

ways in which the system fails or survives

Fault trees can be used to decompose the main failure event

into unions and intersections of sub-events

Event trees can be used to identify the possible sequence of

events (also failures)

Examples

A series system is one which operates if and only if all of its components operate.

The equivalent circuit diagram is

Let Ri = P(component i works) and Rsys = P(system works)

Then, if the components operate independently,

( ) (1 2 )

(1 ) (2 ) ( )

P system operates P works works n works

P works P works P n works

1 2sys nR R R R

Series systems

Parallel systems

A parallel system is one which operates if and only if any of its components operate.

1 ( )

( ) (1 )

(1 ) ( )

i iQ R P component i fails

unreliability of component i

P system fails P fails n fails

P fails P n fails

1 2sys nQ Q Q Q

1 21 (1 )(1 ) (1 )sys nR R R R

Series/parallel systems

Example Find the reliability of the system shown below, if all components have reliability 0.8.

Solution. The system can be broken down into subsystems that are

series or parallel.

Decomposition

2

1

1

0.2 0.04

1 0.04 0.96

C DQ Q Q

R

2 1

0.8 0.8 0.96

0.6144

A BR R R R

2

(1 0.6144) 0.2

0.077

0.923

sys E

sys

Q Q Q

R

Conditional probability method

Some complex systems cannot be broken down into series and parallel subsystems. There are several reliability analysis methods such as conditional probability, and cut

sets and fault trees.

Example. Find the reliability of the complex system shown below, if all components have reliability 0.8.

This keystone component is chosen carefully. In this case we choose component E. Using the law of total probability gives

( ) ( ) ( | )

( ) ( | )

P system works P E works P system works E works

P E fails P system works E fails

| |sys E sys E E sys ER R R Q R

Conditional probability method

20.2 0.04AB A B CDQ Q Q Q

| 0.96 0.96 0.9216sys E AB CDR R R

|sys ER

|sys ER

2

2

|

|

0.8 0.64

(1 0.64) 0.1296

1 0.1296 0.8704

AC A C B D

AC BDsys E

sys E

R R R R R

Q Q Q

R

| |

0.8 0.9216 0.2 0.8704 0.91136

sys E sys E E sys ER R R Q R

Cut set method

Find the reliability of the complex system shown below, if all components have reliability 0.8.

A cut set is a subset of the components with the property that if all components in set

fail, then the system fails. For example, {A, B, E} is a cut set in the system above.

Definition. A minimal cut set is a cut set for which no subset is a cut set.

For example, in the system above, {A, B, E} is not a minimal cut set, but {A, B} is a minimal cut set.

The list of all minimal cut sets for the system above is

Define C1 to be the event “all components in cut set C1 fail ”, etc. Then

1 2 3 4, , , , , , , , ,C A B C C D C A D E C B C E

1 2 3 4( ) ( )sysP system fails Q P C C C C

Standby systems

One example of a standby system is of the power supply to a hospital. The primary

supply is from the electricity grid. The backup might be a diesel generator. A standby

system differs from a parallel system in two ways:

• the backup component is not in use while the primary component is operational

(and therefore is not susceptible to failure)

• there is a switching mechanism, which detects failure of the primary component

and activates the backup component. This switching mechanism may fail to operate

(e.g., the diesel generator may fail to start).

There may be more than one backup component, as shown below.

We show the general case, with failure rates

Note that we need to distinguish here between two cases. If then we will end

up with the result of the Example. Here we consider only the case , which gives

which, after some manipulation,

1 2 and

1 1 2

1 2 2 1

1 1 20

( )

10

( )

10

( ) ( ) ( ) ( )

t

syss

tt t t s

s

tt t s

s

R t R t sf s R t s d

e e e ds

e e e ds

1 2

1 2

2 1

1 2

2 1

1 2

( )

1

2 1 0

( )

1

2 1

( )

1

ts

t t

sys

s

tt t

eR t e e

ee e

1 2

12

2 1

t t

e e

To Calculate the reliability function of a standby system in which, the primary

component has a constant hazard rate of 0.01 per hour, while the backup component

has hazard rate 0.02 per hour

Example of Standby systems The primary component has a constant hazard rate of 0.01 per hour, while the backup

component has hazard rate 0.02 per hour. Compare the mean times to failure if these

components are operated (a) in parallel, (b) in standby mode.

(a) For the parallel system,

(b) For the standby system, we have

Thus the standby system gives the longer MTTF. This agrees with common sense,

since in the standby system the standby component begins its life later.

Suppose that the primary and backup components both have hazard rate 0.01 per

hour. Compare reliability functions if these components are operated (a) in parallel, (b)

in standby mode.

1 2 1 2

1 1 1 1 1 1116.67

0.01 0.02 0.03sysE T hr

1 2

1 2

1 1 1 1150

0.01 0.02sysE T E T E T hr

2( ) 1 (1 )(1 ) 2 0.991t t t t

sysR t e e e e

0.1

( ) ( 1) (2 )

( 1) 1.1 0.995

sys sysR t P T P nd failure occurs after time t

P N e

Review on Reliability and Failure Rate

ET and FT

IE Sys-A Sys-B Result

10-1/yr

Success

Failure

Success

Failure

OK

OK

CD CDF=1.1X10-7/yr

Sys-AFailure

Pump-1Failure

Pump-2Failure

Sys-BFailure

Pump-1Failure

Pump-2Failure

10-4/yr

10-2/yr 10-2/yr

1.1x10-2/yr

10-2/yr 10-3/yr

Emergency Electric Power System (비상전력계통)

E1 E2

G1 G3G2

At least 60KVA

30KVA 30KVA 30KVA

The probability that a device will perform successfully for

the period of time intended under the operating

conditions.

Emergency Diesel Generators

An Example

기본사건 (Basic Event) 기기고장률 (Demand Failure)

E1 3.18E-03

E2 3.18E-03

G1 7.72E-06

G2 7.72E-06

G3 7.72E-06

Emergency Electric Power System (비상전력계통)

E1 E2

G1 G3G2

At least 60KVA

30KVA 30KVA 30KVA

FT in KIRAP

최소단절집합(Minimal Cut Set) – 총 8 가지 {G1,G2} {G2,G3} {G1,G3} {E1,E2} {E1,G2} {E1,G3} {E2,G2} {E2,G1}

An Example Results

KCUT Version 4.8a(20) 1999.4.9 + Uncertainty

Boolean Equation Reduction Program + Uncertainty

Copyright Han, S.H. KAERI

Fri Feb 06 18:00:50 2004

>

>

LEVEL ( 0.000e+000 ) .

Reporting for FAIL

value = 1.053e-005

Final Cut Sets

no value f-v acc cut sets

1 1.043e-005 0.9905 0.9905 E1 E2

2 2.532e-008 0.0024 0.9929 E1 G3

3 2.532e-008 0.0024 0.9953 E1 G2

4 2.455e-008 0.0023 0.9977 G1 E2

5 2.455e-008 0.0023 1.0000 G2 E2

6 5.960e-011 0.0000 1.0000 G1 G2

7 5.960e-011 0.0000 1.0000 G1 G3

8 5.960e-011 0.0000 1.0000 G2 G3

Execution time 0 seconds (gen:0, exp:0, abs:0), Return Code = 1

End of CUT Run

시스템 이용불능도

KIRAP의 실행초기화면

KIRAP의 실행

KIRAP Menu

KIRAP 메뉴설명

이 름 의 미

Name 현재사건의 이름을 입력

Type 현재사건의 형태를 입력

Description 현재사건에 대한 자세한 설명

Mean, Cal. Type, Lambda, Tau Mean은 현재사건에 대한 신뢰도 값을 의미하며, 사용자가 직접 입력하는 것이 아니라 Lambda, Tau값들의 계산을 통해 얻어짐

EF 각 사건에 대해 주어진 Mean 값의 오차인자

Dist. Type 사건에 주어진 신뢰도 값에 대한 확률분포를 정해줌

Transfer 현재사건은 전이게이트(Transfer gate)로 만듬

Module 현재사건을 전이게이트에서 해제

Remark 현재사건에 대한 비고나 특기사항들을 기록

KIRAP 메뉴설명

Cal.

Type

Lambda

(λ)

Tau

(τ)

Mean의

계산 의 미

0 Demand

Failure Prob. - Mean = λ

: 고장확률을 바로 줄 경우 사용. 대부분의 demand failure가 이에 해당.

1 Running

Failure Rate

Mission

Time

Mean =

λx τ : 사고 후 주어진 시간 동안 운전하지 못하는 확률을 표현

2 Running

Failure Rate

Repair

Time

Mean =

λx τ : 항상 기기를 감시하다가 고장이 나면 바로 수리하는 경우에 이용 불능도를 표현

3 Standby

Failure Rate

Test

Interval

Mean =

λx τ/2 : 대기상태에 있으면서 정기적으로 점검하는 기기의 이용 불능도를 표현.

4 Failure

Rate - Mean = λ : 고장율을 단위로 가진 Event에 사용.

KIRAP 메뉴설명

KIRAP 메뉴설명

Tree Display Option

KIRAP 메뉴설명

계산수행 - 이용불능도 계산 - 최소단절군(MCS) 및 중요도 계산 etc