fault tree analysis · fault tree = set of logical gates (algebraic equations) x. t = Φ (x. 1, x....

Fault tree analysis

Piero Baraldi

FailureProbabilty

Assessment

AccidentScenarios

Identification

Evaluation ofthe

consequences

Riskevaluation

RISK = Si, pi, xi4321

pi/xi A B C D

International StandardsBest PracticesLessons learnt

Expert judgmentsFlow and transport codes

Finite Element MethodsDC/AC power flows, etc.

ALARP = as low as reasonably practicable

FTAETA

Markov Models

Hazard Analysis

Hazop

FMECA

Monte Carlo Simulation

Classical Techniques for PRA

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Classical Techniques for Risk Analysis

• Hazard identification: Failure Mode and Effect Analysis (FMEA) …

• Accident Scenarios Identification: Fault Tree Analysis (FTA) Event Tree Analysis (ETA) …

• System Failure Probability Assessment: Fault Tree Analysis (FTA) Event Tree Analysis (ETA) …

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Fault Tree Analysis (FTA)

• Systematic and quantitative• Deductive (search for root causes)

Objectives:1. Decompose the system failure in elementary

failure events of constituent components 2. Compute the system failure probability,

from constituent component failure probabilities

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Fault Treeconstruction

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FT construction: procedure steps

Electrical generating system

E1 E2

G1 G2 G3

E1, E2 = enginesG1, G2, G3 = generators, each one

is rated at 30 KVA

1. Define top event: typically a (sub)system failure

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1. Define top event (system failure)2. Decompose (top) event by identifying sub-events

which can cause it

No direct contributions to failure:At least two out of the three generators do not work


E1 E2

G1 G2 G3

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which can cause it


Identification of the contributing event categories that may directly cause the top event to occur. At least four categories exist:

1. No input to the device2. Primary failure of the device (under operation in the

design envelope, random, due to aging or fatigue)3. Human error in actuating or installing the device4. Secondary failure of the device (due to present or

past stresses caused by neighboring components or the environments: e.g. common cause failure, external causes such as earthquakes, etc.)

If these events are considered to be indeed contributing to the system fault, then they are connected to the top event logically via an OR function and graphically through the OR gate

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which can cause it3. Decompose each sub-event in more elementary sub-

events which can cause it


E1 E2

G1 G2 G3

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events which can cause it4. Stop decomposition when sub-event probability data

are available (resolution limit): sub-event = basic or primary event

basic event

E1 E2

G1 G2 G3


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E1 E2

G1 G2 G3



events which can cause it4. Stop decomposition when sub-event probability data

are available (resolution limit): sub-event = basic or primary event

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FT Example 1

E1 E2

G1 G2 G3

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FT gate symbols

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FT event symbols

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Exercise 1: electric circuit breaker

HydraulicControl A

HydraulicControl B

Actuators

Linkage

• Draw the fault tree for the top event: “latch does not trip on demand”

Moving part fordisconnecting anelectrical circuit

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Exercise 1: Solution

HydraulicControl A

HydraulicControl B

ActuatorsLinkage

No input to linkage

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FT qualitative analysis

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FT qualitative analysis Let us introduce:

Xi = binomial indicator variable of i-th component state (basic event)

Xi =1 → failure event true

0 → failure event false

Fault Tree Gate: boolean algebraic equation (one for each gate) based on Boolean Algebra basic operators:

∧ = AND ∨ = OR NOT

(The dual definition can also be used, as long as we are coherent)

XT = binomial indicator variable of the FT top event

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Exercise 2A

E1 E2

D

• Write the Boolean equations linking the event at the top of the gate with those at the bottom in the following two cases:

𝑥𝑥𝐴𝐴 = 𝑓𝑓(𝑥𝑥𝐸𝐸1, 𝑥𝑥𝐺𝐺1) 𝑥𝑥𝐷𝐷 = 𝑓𝑓(𝑥𝑥𝐸𝐸1, 𝑥𝑥𝐸𝐸2)

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∨ = OR gate)1)(1(1

11

1111

GE

GEGEA

XX

XXXXX

−−−=

=−+=

∧ = AND gateE1 E2

D

21 EED X X X =

If X’s only assume integer values 0 and 1, then ordinary algebra can be used to express operator results

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Exercise 3• Write the Boolean equations linking the event at the top of the

gate (T1) with those at the bottom (E1,E2,G1,G2).

𝑥𝑥𝑇𝑇1 = 𝛷𝛷 (𝑥𝑥𝐸𝐸1, 𝑥𝑥𝐸𝐸2, 𝑥𝑥𝐺𝐺1, 𝑥𝑥𝐺𝐺2)D

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1) Commutative Law: (a) X YY X ∧=∧ (b) X YY X ∨=∨

2) Associative Law

(a) ( ) ( ) ZYX ZYX ∧∧=∧∧

(b) ( ) ( ) ZYX ZYX ∨∨=∨∨

3) Idempotent Law

(a) XX X =∧ (b) XX X =∨

4) Absorption Law

(a) ( ) X YX X =∨∧

(b) ( ) X YXX =∧∨

Most important laws of boolean algebra5) Distributive Law

(a) ( ) ( ) ( )ZXYX ZYX ∧∨∧=∨∧

(b) ( ) ( ) ( )ZYXZXYX ∧∨=∨∧∨

6) Complementation*

(a) =∧ XX ∅ (b) Ω=∨ XX

(c) XX =

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== BAT XX X1

=+−−−

+++−+=

=−+−+=

21211212112121

121212212121

2212121111

GGEEGEEGGEGGEE

GEEGGGEEEEGE

GEEEEGGEGE

XXXXXXXXXXXXXX

XXXXXXXXXXXX

)XXXXXX)(XXXX(

XXXXXXXXXXXX211221212121 GGEGEEGGEEGE −−++=

),,(2 2111 GGEET XX,XXX Φ=

D

Piero Baraldi

FT qualitative analysis Let us introduce:

Xi = binomial indicator variable of i-th component state (basic event)

Xi =1 → failure event true

0 → failure event false

Fault Tree = SET of Logical Gates(algebraic equations)

XT = Φ (X1 , X2 , …, Xn)

Structure (switching) function Φ

(The dual definition can also be used, as long as we are coherent)

XT = binomial indicator variable of the FT top event

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Exercise 4

HydraulicControl A

HydraulicControl B

Actuators

Linkage

• Find the structure function for the TE: “latch does not trip on demand”


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Exercise 4: Solution• Find the structure function for the TE: “latch does not

trip on demand”

))XXX)(XXXX(X)(1X(11X HBBHBBHAAHAALT −+−+−−−=

HydraulicControl A

HydraulicControl B

ActuatorsLinkage

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Fundamental Products

Theorem

A structure function can be written uniquely as the union of all those fundamental products which render the structure

function true (i.e., Φ = 1)

Canonical expansion or disjunctive normal form of Φ

Fundamental product = Product containing all of the n input variables

e.g. 𝒙𝒙𝑻𝑻 = 𝚽𝚽 𝒙𝒙𝟏𝟏,𝒙𝒙𝟐𝟐,𝒙𝒙𝟑𝟑 → 𝒙𝒙𝟏𝟏 𝒙𝒙𝟐𝟐 𝒙𝒙𝟑𝟑, 𝒙𝒙𝟏𝟏 𝒙𝒙𝟐𝟐 𝒙𝒙𝟑𝟑, … ,𝒙𝒙𝟏𝟏 𝒙𝒙𝟐𝟐 𝒙𝒙𝟑𝟑

• Number of fundamental products for a fault tree with n basic events= 2𝑛𝑛• A fundamental product is 1 (true) if all its variables are 1 (𝒙𝒙𝟏𝟏 𝒙𝒙𝟐𝟐 𝒙𝒙𝟑𝟑 𝒙𝒙𝟏𝟏=1, 𝒙𝒙𝟐𝟐 = 1,𝒙𝒙𝟑𝟑 = 𝟎𝟎)

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Exercise 5: fundamental productsConsider a series of 2 components

You are required to:• List the fundamental products• Write the structure function

1 2

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( ) ( )2121

212121

212121T

XXXXX1XXX1XX

XXXXXXX

−+==−+−+=

=++=

21212121 XXXXXXXX

• Series of 2 components

• List of the fundamental products

1 2

Which fundamental products are such that, when equal to 1, the structure function is also true (i.e., Φ = 1)?


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Coherent structure functions

Coherent structure function = function monotonically increasing in each input variable:

1. Φ(1) = 1 (if all components are failed, the system is failed)2. Φ(0) = 0 (if all components are working, the system is working)3. Φ(X)≤ Φ(Y) if X≤Y (each element of X is lower or equal than the

corresponding element of Y)

Improving the performance of a component (= replacing a failedcomponent by a functioning one) does not cause the system tochange from the success to the failed state

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Coherent structure functions

Coherent structure function = function monotonically increasing in each input variable:

1. Φ(1) = 1 (if all components are failed, the system is failed)2. Φ(0) = 0 (if all components are working, the system is working)3. Φ(X)≤ Φ(Y) if X≤Y (each element of X is lower or equal than the

corresponding element of Y)

Improving the performance of a component (= replacing a failedcomponent by a functioning one) does not cause the system tochange from the success to the failed state

It is possible to show that:• coherent structure functions do not contain complemented variables

1 2 𝑥𝑥𝑇𝑇 = 𝛷𝛷 𝑥𝑥1, 𝑥𝑥2 = 𝑥𝑥1 + 𝑥𝑥2 − 𝑥𝑥1𝑥𝑥2

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FT qualitative analysis: cut sets

• Cut sets = logic combinations of primary (basic) events which cause the top event to be true (system failure): 𝑋𝑋: 𝛷𝛷 𝑋𝑋 = 1

1 2

Cut sets:𝑥𝑥1𝑥𝑥2

𝑥𝑥1, 𝑥𝑥2

Set of components whose failure causes the failure of the system

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FT qualitative analysis: minimal cut sets

• Cut sets = logic combinations of primary (basic) events which cause the top event to be true (system failure): 𝑋𝑋: 𝛷𝛷 𝑋𝑋 = 1

• Minimal cut sets = cut sets such that if one of the events (failures) is not verified, the top event is not verified

1 2Cut sets:

𝑥𝑥1𝑥𝑥2

𝑥𝑥1, 𝑥𝑥2

A component is repaired The system starts working

1 2Minimal cut sets:

𝑥𝑥1𝑥𝑥2

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Exercise 6: minimal cut sets

HydraulicControl A

HydraulicControl B

Actuators

Linkage

• Identify the minimal cut sets for the TE: “latch does not trip on demand”


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Exercise 6: Solution• Identify the minimal cut sets for the TE: “latch does

not trip on demand”HydraulicControl A

HydraulicControl B

ActuatorsLinkage

5 minimal cut sets:

M1=XLM2=XAXBM3=XAXHBM4=XHAXB

M5=XHAXHB

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• Coherent structure functions can be expressed in reduced expresions in terms of minimal cut sets:• Unique and irreducible form of the structure function: union of all

the cut sets

TE

𝑀𝑀1 𝑀𝑀𝑚𝑚𝑚𝑚𝑚𝑚𝑀𝑀2 …

FT qualitative analysis: structure function and minimal cut sets

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• Coherent structure functions can be expressed in reduced expresions in terms of minimal cut sets:• Unique and irreducible form of the structure function: union of all

the minimal cut sets:

𝑇𝑇 = 𝑀𝑀1 ∪𝑀𝑀2 ∪⋯ ,𝑀𝑀𝑚𝑚𝑚𝑚𝑚𝑚

𝑥𝑥𝑇𝑇 = 1 − 1 − 𝑥𝑥1𝑀𝑀1𝑥𝑥2

𝑀𝑀1 … 1 − 𝑥𝑥1𝑀𝑀2𝑥𝑥2

𝑀𝑀2 … … (1 − 𝑥𝑥1𝑀𝑀𝑚𝑚𝑚𝑚𝑚𝑚𝑥𝑥2

𝑀𝑀𝑚𝑚𝑚𝑚𝑚𝑚 … )

FT qualitative analysis: structure function and minimal cut sets

𝑀𝑀1 = 𝑥𝑥1𝑀𝑀1𝑥𝑥2

𝑀𝑀1 …𝑀𝑀2 = 𝑥𝑥1

𝑀𝑀2𝑥𝑥2𝑀𝑀2 …

…𝑀𝑀𝑚𝑚𝑚𝑚𝑚𝑚 = 𝑥𝑥1

𝑀𝑀𝑚𝑚𝑚𝑚𝑚𝑚𝑥𝑥2𝑀𝑀𝑚𝑚𝑚𝑚𝑚𝑚 …

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Example• Build the structure function from the minimal cut sets for the

circuit breaker system

𝑇𝑇 = 𝐿𝐿 ∪ 𝐴𝐴 𝐵𝐵 ∪ 𝐴𝐴 𝐻𝐻𝐵𝐵 ∪ 𝐻𝐻𝐴𝐴 𝐵𝐵 ∪ 𝐻𝐻𝐴𝐴 𝐻𝐻𝐵𝐵

𝑥𝑥𝑇𝑇 = 1 − 1 − 𝑥𝑥𝐿𝐿 1 − 𝑥𝑥𝐴𝐴𝑥𝑥𝐵𝐵 1 − 𝑥𝑥𝐴𝐴𝑥𝑥𝐻𝐻𝐵𝐵 1 − 𝑥𝑥𝐻𝐻𝐴𝐴𝑥𝑥𝐵𝐵 (1 − 𝑥𝑥𝐻𝐻𝐴𝐴𝑥𝑥𝐻𝐻𝐵𝐵)

HydraulicControl A

HydraulicControl B

ActuatorsLinkage

M1=XLM2=XAXBM3=XAXHBM4=XHAXB

M5=XHAXHB

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Example• Verify whether the structure function obtained in Exercise 4

is the same of that obtained above from the minimal cut set:

From Ex. 4

From minimal cut sets:

𝑥𝑥𝑇𝑇 = 1 − 1 − 𝑥𝑥𝐿𝐿 1 − 𝑥𝑥𝐴𝐴𝑥𝑥𝐵𝐵 1 − 𝑥𝑥𝐴𝐴𝑥𝑥𝐻𝐻𝐵𝐵 1 − 𝑥𝑥𝐻𝐻𝐴𝐴𝑥𝑥𝐵𝐵 (1 − 𝑥𝑥𝐻𝐻𝐴𝐴𝑥𝑥𝐻𝐻𝐵𝐵)

𝑥𝑥𝑇𝑇 = 1 − 1 − 𝑥𝑥𝐿𝐿 𝑥𝑥𝐴𝐴 + 𝑥𝑥𝐻𝐻𝐴𝐴 − 𝑥𝑥𝐴𝐴𝑥𝑥𝐻𝐻𝐴𝐴 𝑥𝑥𝐵𝐵 + 𝑥𝑥𝐻𝐻𝐵𝐵 − 𝑥𝑥𝐵𝐵𝑥𝑥𝐻𝐻𝐵𝐵It is possible to show that the two terms are equal by

performing the products and using the rulesof boolean algebra

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XT = Φ (X1 , X2 , …, Xn)

• Boolean algebra to solve FT equations

How to find the minimal cut sets?


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XT = Φ (X1 , X2 , …, Xn)




𝑥𝑥𝑇𝑇1 = 𝑥𝑥𝐸𝐸1𝑥𝑥𝐺𝐺2 + 𝑥𝑥𝐸𝐸1𝑥𝑥𝐸𝐸2 + 𝑥𝑥𝐺𝐺1𝑥𝑥𝐺𝐺2 −−𝑥𝑥𝐸𝐸1𝑥𝑥𝐸𝐸2𝑥𝑥𝐺𝐺2 − 𝑥𝑥𝐸𝐸1𝑥𝑥𝐺𝐺1𝑥𝑥𝐺𝐺2

E1 E2

G1 G2

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== BAT XX X1

XT = Φ (X1 , X2 , …, Xn)





cut sets

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== BAT XX X1




cut set

minimal cut set

minimal cut set

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mcs: Example 1

2112212121211 GGEGEEGGEEGET XXXXXXXXXXXX X −−++=

=+−−−−=

=+−−−+−−−=

=−++−+−−−=

=−+++−−−−=

=++−−−−=

)]XXXXXXXX1)(XX1[(1

)]XXXXXXXX1(XXXXXXXXXX1[1

]XXXXXXXXXXXXXXXXXXXX1[1

]XXXXXXXXXXXXXXXXXXXX1[1

]XXXXXXXXXXXX1[1

2121212121

212121212121212121

21212112212121212121

21212121211221212121

211221212121

GGEEGGEEGE

GGEEGGEEGEGGEEGGEE

GGEEGGEGEEGEGGEEGGEE

GGEEGGEEGGEGEEGGEEGE

GGEGEEGGEEGE

)]XX1)(XX1)(XX1[(1212121 GGEEGE −−−−=

3 minimal cut sets:

211 GEM =

212 EEM =

213 GGM =

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FT qualitative analysis: results

1. Mcs identify the component basic failure events which contribute to system failure

2. Qualitative component criticality: those components appearing in low order mcs’ or in many mcs’ are the most critical

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FT quantitative analysis

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Compute system failure probability from primary events probabilities by:1. Using the laws of probability theory at the fault tree gates

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Exercise 8

Compute system failure probability for the circuit breaker of Exercise 1 using the laws of probability theory at the fault tree gates

HydraulicControl A

HydraulicControl B

ActuatorsLinkage

p = 0.1 p = 0.1 p = 0.1 p = 0.1

p ≅ 0.01

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p = 0.1 p = 0.1 p = 0.1 p = 0.1

p ≅ 0.2 p ≅ 0.2

p ≅ 0.04

p ≅ 0.05

Compute system failure probability for the circuit breaker of Exercise 1 using the laws of probability theory at the fault tree gates

p ≅ 0.01HydraulicControl A

HydraulicControl B

ActuatorsLinkage

Attention:This method may easily lead to errors and should not be used in case of basic events shared by more branches!

No approximation = 0.0457

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1. Consider a system whose 2 minimal cut sets are:2. 𝑀𝑀1 = 𝑥𝑥𝐴𝐴𝑥𝑥𝐵𝐵3. 𝑀𝑀2 = 𝑥𝑥𝐴𝐴𝑥𝑥𝐶𝐶4. You are required to compute the probability of the top

event

Exercise 9

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Compute system failure probability from primary events probabilities by:1. Using the laws of probability theory at the fault tree gates2. Using the mcs found from the qualitative analysis


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1. 𝑻𝑻 = 𝑴𝑴𝟏𝟏 ∪𝑴𝑴𝟐𝟐

𝑷𝑷 𝑻𝑻 = 𝑷𝑷(𝑴𝑴𝟏𝟏) + 𝑷𝑷 𝑴𝑴𝟐𝟐 − 𝑷𝑷(𝑴𝑴𝟏𝟏 ∩𝑴𝑴𝟐𝟐)


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1. 𝑻𝑻 = 𝑴𝑴𝟏𝟏 ∪𝑴𝑴𝟐𝟐

𝑷𝑷 𝑻𝑻 = 𝑷𝑷(𝑴𝑴𝟏𝟏) + 𝑷𝑷 𝑴𝑴𝟐𝟐 − 𝑷𝑷(𝑴𝑴𝟏𝟏 ∩𝑴𝑴𝟐𝟐)

Ex 9 Solution

𝑷𝑷 𝑻𝑻 = 𝑷𝑷 𝑨𝑨 𝑷𝑷(𝑩𝑩) + 𝑷𝑷 𝑨𝑨 𝑷𝑷(𝑪𝑪) − 𝐏𝐏(𝑨𝑨𝑩𝑩𝑨𝑨𝑪𝑪)

𝑷𝑷 𝑻𝑻 = 𝑷𝑷 𝑨𝑨 𝑷𝑷(𝑩𝑩) + 𝑷𝑷 𝑨𝑨 𝑷𝑷(𝑪𝑪) − 𝐏𝐏(𝑨𝑨𝑩𝑩𝑪𝑪)

𝑷𝑷 𝑻𝑻 = 𝑷𝑷 𝑨𝑨 𝑷𝑷(𝑩𝑩) + 𝑷𝑷 𝑨𝑨 𝑷𝑷(𝑪𝑪) − 𝐏𝐏 𝑨𝑨 𝑷𝑷 𝑩𝑩 𝑷𝑷(𝑪𝑪)

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Compute system failure probability from primary events probabilities by:1. Using the laws of probability theory at the fault tree gates2. Using the mcs found from the qualitative analysis

It can be shown that:

∑ ∑ ∏∑−

= += =

+

=

−++−==1

1 1 1

1

1][)1(][][]1)([

mcs

i

mcs

ij

mcs

jj

mcsji

mcs

jj MPMMPMPXP Φ

∑∑ ∑∑=

−

= +==

≤=≤−mcs

jj

mcs

i

mcs

ijji

mcs

jj MPXPMMPMP

1

1

1 11][]1)([][][ Φ


“Rare event” approximation

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Exercise 10

Compute system failure probability from the mcs

HydraulicControl A

HydraulicControl B

ActuatorsLinkage

p = 0.1 p = 0.1 p = 0.1 p = 0.1

p = 0.01

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Exercise 10 Solution

5 mcs:P(M1) =P(XL=1)=0.01P(M2)=P(XAXB=1)=0.1·0.1=0.01P(M3)=P(XAXHB) =0.1·0.1=0.01 P(M4)= P(XHAXB=1)=0.1·0.1=0.01P(M5) =P(XHAXHB)=0.1·0.1=0.01

0464.0][][]1)([

05.0][]1)([

1

1 11

1

=−≥=

=≤=

∑ ∑∑

∑−

= +==

=

mcs

i

mcs

ijji

mcs

jj

mcs

jj

MMPMPXP

MPXP

Φ

Φ

HydraulicControl A

HydraulicControl B

Actuators

LinkLatch

No approximation = 0.0457

0.0454

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Compute system failure probability from primary events probabilities by:1. Using the laws of probability theory at the fault tree gates2. Using the mcs found from the qualitative analysis3. Reducing (if necessary) the structure function by

boolean algebra rules and directly applying the expected value operator:


))XXX)(XXXX(X)(1X(11X HBBHBBHAAHAALT −+−+−−−=

( ) [ ] [ ][ ] [ ] [ ]))XXX(XE)XXX(XE)(1XE(11

))XXX)(XXXX(X)(1X(11EXETP

HBBHBBHAAHAAL

HBBHBBHAAHAALT

−+−+−−−=−+−+−−−==

= 0.0457

595959Piero Baraldi

59

Exercise 11: Alarm of pressure excess

When the pressure in the reactor reaches a value above the acceptable an alarm is activated manually by the supervisor or automatically by an automatic sensor. The alarm is fed by a line of main energy or, in case of general energy failure, by a stand-by power generator.

You are required to compute the probability of the event: “failure of the alarm system”

Component Failureprobability

Manometer 0.08Supervisor 0.14Switch 0.04Alarm 0.05Stand by Power Generator

0.12

Main Electricpower

0.10

AutomaticSensor

0.14

606060Piero Baraldi


60p = 0.14 p = 0.08

p = 0.05

p = 0.14

p = 0.1 p = 0.12

p = 0.012

p = 0.24

p = 0.0336

p = 0.0929

Main powerfailure

Stand-bypowerfailure

Supervisorfailure

SwitchFailure

(p = 0.04)

Manometerfailure

Failure of the

automaticsensor

Alarmfailure

Power failure Signal does notreach to alarm

Failure of the alarmsystem

Failure on manualactivation

Correct! Branches areINDEPENDENT

Piero Baraldi

FT: comments

1. Straightforward modeling via few, simple logic operators

2. Focus on one top event of interest at a time

3. Providing a graphical communication tool whose analysis is transparent

4. Providing an insight into system behavior

5. Minimal cut sets are a synthetic result which identifies the critical components

fault tree analysis · fault tree = set of logical gates (algebraic equations) x. t = Φ (x. 1, x....

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