Fault Tree Analysis Part 3: Digraph-Based Fault Tree Synthesis Procedure (Tree and NFBL)
Fault tree analysis - LASAR · 2018-05-30 · Fault Tree Analysis (FTA) •Systematic and...
Transcript of Fault tree analysis - LASAR · 2018-05-30 · Fault Tree Analysis (FTA) •Systematic and...
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Fault tree analysis
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Piero Baraldi
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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Piero Baraldi
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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Piero Baraldi
Fault Tree
construction
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FT construction: procedure steps
Electrical generating system
E1 E2
G1 G2 G3
E1, E2 = engines
G1, 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
FT construction: procedure steps
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 device
2. 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 device
4. 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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Piero Baraldi
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
FT construction: procedure steps
E1 E2
G1 G2 G3
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Piero Baraldi
1. Define top event (system failure)
2. Decompose (top) event by identifying sub-events which can cause it
3. Decompose each sub-event in more elementary sub-events which can cause it
FT construction: procedure steps
E1 E2
G1 G2 G3
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Piero Baraldi
1. Define top event (system failure)
2. Decompose (top) event by identifying sub-events which can cause it
3. Decompose each sub-event in more elementary sub-events which can cause it
4. Stop decomposition when sub-event probability data are available (resolution limit): sub-event = basic or primary event
basic event
E1 E2
G1 G2 G3
FT construction: procedure steps
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Piero Baraldi
FT construction: procedure steps
E1 E2
G1 G2 G3
1. Define top event (system failure)
2. Decompose (top) event by identifying sub-events which can cause it
3. Decompose each sub-event in more elementary sub-events which can cause it
4. Stop decomposition when sub-event probability data are available (resolution limit): sub-event = basic or primary event
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Piero Baraldi
FT Example 1
E1 E2
G1 G2 G3
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Piero Baraldi
FT gate symbols
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Piero Baraldi
FT gate symbols
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FT event symbols
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Exercise 1: electric circuit breaker
Hydraulic
Control A
Hydraulic
Control B
Actuators
Linkage
• Draw the fault tree for the top event: “latch does not trip on demand”
Moving part for
disconnecting an
electrical circuit
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Piero Baraldi
Exercise 1: Solution
Hydraulic
Control A
Hydraulic
Control B
Actuators
Linkage
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Piero Baraldi
FT qualitative
analysis
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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 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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Piero Baraldi
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:
𝑥𝐴 = 𝑓(𝑥𝐸1, 𝑥𝐺1) 𝑥𝐷 = 𝑓(𝑥𝐸1, 𝑥𝐸2)
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Exercise 2: Solution
= OR gate)1)(1(1
11
1111
GE
GEGEA
XX
XXXXX
= AND gate
E1 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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Piero Baraldi
Exercise 3
• Write the Boolean equations linking the event at the top of the gate with those at the bottom
𝑥𝑇1 = 𝛷 (𝑥𝐸1, 𝑥𝐸2, 𝑥𝐺1, 𝑥𝐺2)
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 algebra
5) Distributive Law
(a) ZXYX ZYX
(b) ZYXZXYX
6) Complementation*
(a) XX
(b) XX
(c) XX
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Exercise 3: Solution
BAT XX X1
21211212112121
121212212121
2212121111
GGEEGEEGGEGGEE
GEEGGGEEEEGE
GEEEEGGEGE
XXXXXXXXXXXXXX
XXXXXXXXXXXX
)XXXXXX)(XXXX(
XXXXXXXXXXXX211221212121 GGEGEEGGEEGE
),,(2 2111 GGEET XX,XXX
D
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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 Gate: boolean algebraic equation (one for each gate) based on Boolean Algebra basic operators:
= AND = OR NOT
XT = (X1 , X2 , …, Xn)
Structure (switching) function
(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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Piero Baraldi
Exercise 4
Hydraulic
Control A
Hydraulic
Control B
Actuators
Linkage
• Find the structure function for the TE: “latch does not trip on demand”
Moving part for
disconnecting an
electrical circuit
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Piero Baraldi
Exercise 4: Solution
• Find the structure function for the TE: “latch does not trip on demand”
))XXX)(XXXX(X)(1X(11X HBBHBBHAAHAALT
Hydraulic
Control A
Hydraulic
Control B
ActuatorsLinkage
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Piero Baraldi
Fundamental Products
Theorem
A structure function can be written uniquely as the union of
the fundamental products which render the structure
function true (i.e., Φ = 1)
Canonical expansion or disjunctive normal form of
Fundamental product = Products containing all of the n input variables
• 𝒙𝑻 = 𝚽 𝒙𝟏, 𝒙𝟐, 𝒙𝟑 → 𝒙𝟏 ∙ 𝒙𝟐∙ 𝒙𝟑, 𝒙𝟏 ∙ 𝒙𝟐∙ 𝒙𝟑, … , 𝒙𝟏 ∙ 𝒙𝟐 ∙ 𝒙𝟑
Number of fundamental product for a fault tree with n basic events= 2𝑛
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Piero Baraldi
Exercise 5: fundamental products
Consider 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
XXXX
X1XXX1XX
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)?
Exercise 5: Solution
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Piero Baraldi
Coherent structure functions
Coherent structure functions = 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 XY
Improving the performance of a component (= replacing a failed
component by a functioning one) does not cause the system to
change from the success to the failed state
It is possible to show that:
• coherent structure functions do not contain complemented variables
1 2 𝑥𝑇 = 𝛷 𝑥1, 𝑥2 = 𝑥1 + 𝑥2 − 𝑥1𝑥2
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Piero Baraldi
FT qualitative analysis: cut sets
• Cut sets = logic combinations of primary (basic) events which cause
the top event to be true (system failure): 𝑋: 𝛷 𝑋 = 1
1 2
Cut sets:
𝑥1𝑥2
𝑥1, 𝑥2
Set of components whose failure causes the failure of the system
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Piero Baraldi
FT qualitative analysis: minimal cut sets
• Cut sets = logic combinations of primary (basic) events which cause
the top event to be true (system failure): 𝑋: 𝛷 𝑋 = 1
• Minimal cut sets = cut sets such that if one of the events is not
verified, the top event is not verified
1 2
Cut sets:
𝑥1𝑥2
𝑥1, 𝑥2
A component is repaired The system starts working
1 2
Minimal cut sets:
𝑥1𝑥2
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Piero Baraldi
Exercise 6: minimal cut sets
Hydraulic
Control A
Hydraulic
Control B
Actuators
Linkage
• Identify the minimal cut sets for the TE: “latch does not trip on demand”
Moving part for
disconnecting an
electrical circuit
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Piero Baraldi
Exercise 6: Solution
• Identify the minimal cut sets for the TE: “latch does not trip on demand”
Hydraulic
Control A
Hydraulic
Control B
ActuatorsLinkage
5 minimal cut sets:
M1=XL
M2=XAXB
M3=XAXHB
M4=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:
𝑇 = 𝑀1 ∪𝑀2 ∪⋯ ,𝑀𝑚𝑐𝑠
𝑥𝑇 = 1 − 1 − 𝑥1𝑀1𝑥2
𝑀1 … 𝑥1𝑀2𝑥2
𝑀2 … … (1 − 𝑥1𝑀𝑚𝑐𝑠𝑥2
𝑀𝑚𝑐𝑠 …)
FT qualitative analysis: structure function and minimal cut sets
𝑀1 = 𝑥1𝑀1𝑥2
𝑀1 …
𝑀2 = 𝑥1𝑀2𝑥2
𝑀2 …
…
𝑀𝑚𝑐𝑠 = 𝑥1𝑀𝑚𝑐𝑠𝑥2
𝑀𝑚𝑐𝑠 …
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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
𝑀1 𝑀𝑚𝑐𝑠𝑀2 …
FT qualitative analysis: structure function and minimal cut sets
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Exercise 7
• Build the structure function from the minimal cut sets for the circuit breaker system
• Verify whether the structure function obtained in Exercise 4 is the same of that obtained above from the minimal cut set:
Hydraulic
Control A
Hydraulic
Control B
ActuatorsLinkage
M1=XL
M2=XAXB
M3=XAXHB
M4=XHAXB
M5=XHAXHB
))XXX)(XXXX(X)(1X(11X HBBHBBHAAHAALT
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Piero Baraldi
Exercise 7: Solution
• Build the structure function from the minimal cut sets for the circuit breaker system
𝑇 = 𝐿 ∪ 𝐴 ∙ 𝐵 ∪ 𝐴 ∙ 𝐻𝐵 ∪ 𝐻𝐴 ∙ 𝐵 ∪ 𝐻𝐴 ∙ 𝐻𝐵
𝑥𝑇 = 1 − 1 − 𝑥𝐿 1 − 𝑥𝐴𝑥𝐵 1 − 𝑥𝐴𝑥𝐻𝐵 1 − 𝑥𝐻𝐴𝑥𝐵 (1 − 𝑥𝐻𝐴𝑥𝐻𝐵)
Hydraulic
Control A
Hydraulic
Control B
ActuatorsLinkage
M1=XL
M2=XAXB
M3=XAXHB
M4=XHAXB
M5=XHAXHB
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Exercise 7: Solution
• 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:
𝑥𝑇 = 1 − 1 − 𝑥𝐿 1 − 𝑥𝐴𝑥𝐵 1 − 𝑥𝐴𝑥𝐻𝐵 1 − 𝑥𝐻𝐴𝑥𝐵 (1 − 𝑥𝐻𝐴𝑥𝐻𝐵)
𝑥𝑇 = 1 − 1 − 𝑥𝐿 𝑥𝐴 + 𝑥𝐻𝐴 − 𝑥𝐴𝑥𝐻𝐴 𝑥𝐵 + 𝑥𝐻𝐵 − 𝑥𝐵𝑥𝐻𝐵
It is possible to show that the two terms are equal by
performing the products and using the rules
of boolean algebra
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Piero Baraldi
XT = (X1 , X2 , …, Xn)
• Boolean algebra to solve FT equations
How to find the minimal cut sets?
Structure (switching) function
![Page 41: Fault tree analysis - LASAR · 2018-05-30 · Fault Tree Analysis (FTA) •Systematic and quantitative •Deductive (search for root causes) Objectives: 1. Decompose the system failure](https://reader033.fdocuments.in/reader033/viewer/2022042622/5f9e41842c455d065a626335/html5/thumbnails/41.jpg)
Piero Baraldi
XT = (X1 , X2 , …, Xn)
• Boolean algebra to solve FT equations
How to find the minimal cut sets?
Structure (switching) function
𝑥𝑇 = 𝑥𝐸1𝑥𝐺2 + 𝑥𝐸1𝑥𝐸2 + 𝑥𝐺1𝑥𝐺2 −−𝑥𝐸1𝑥𝐸2𝑥𝐺2 − 𝑥𝐸1𝑥𝐺1𝑥𝐺2
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Piero Baraldi
BAT XX X1
XT = (X1 , X2 , …, Xn)
• Boolean algebra to solve FT equations
XXXXXXXXXXXX211221212121 GGEGEEGGEEGE
How to find the minimal cut sets?
Structure (switching) function
cut sets
![Page 43: Fault tree analysis - LASAR · 2018-05-30 · Fault Tree Analysis (FTA) •Systematic and quantitative •Deductive (search for root causes) Objectives: 1. Decompose the system failure](https://reader033.fdocuments.in/reader033/viewer/2022042622/5f9e41842c455d065a626335/html5/thumbnails/43.jpg)
Piero Baraldi
BAT XX X1
• Boolean algebra to solve FT equations
XXXXXXXXXXXX211221212121 GGEGEEGGEEGE
How to find the minimal cut sets?
E1 E2
G1 G2 G3
cut set
minimal cut set
minimal cut set
![Page 44: Fault tree analysis - LASAR · 2018-05-30 · Fault Tree Analysis (FTA) •Systematic and quantitative •Deductive (search for root causes) Objectives: 1. Decompose the system failure](https://reader033.fdocuments.in/reader033/viewer/2022042622/5f9e41842c455d065a626335/html5/thumbnails/44.jpg)
Piero Baraldi
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
E1 E2
G1 G2 G3
![Page 45: Fault tree analysis - LASAR · 2018-05-30 · Fault Tree Analysis (FTA) •Systematic and quantitative •Deductive (search for root causes) Objectives: 1. Decompose the system failure](https://reader033.fdocuments.in/reader033/viewer/2022042622/5f9e41842c455d065a626335/html5/thumbnails/45.jpg)
Piero Baraldi
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
![Page 46: Fault tree analysis - LASAR · 2018-05-30 · Fault Tree Analysis (FTA) •Systematic and quantitative •Deductive (search for root causes) Objectives: 1. Decompose the system failure](https://reader033.fdocuments.in/reader033/viewer/2022042622/5f9e41842c455d065a626335/html5/thumbnails/46.jpg)
Piero Baraldi
FT quantitative
analysis
![Page 47: Fault tree analysis - LASAR · 2018-05-30 · Fault Tree Analysis (FTA) •Systematic and quantitative •Deductive (search for root causes) Objectives: 1. Decompose the system failure](https://reader033.fdocuments.in/reader033/viewer/2022042622/5f9e41842c455d065a626335/html5/thumbnails/47.jpg)
Piero Baraldi
FT quantitative analysis
Compute system failure probability from primary events probabilities by:
1. Using the laws of probability theory at the fault tree gates
![Page 48: Fault tree analysis - LASAR · 2018-05-30 · Fault Tree Analysis (FTA) •Systematic and quantitative •Deductive (search for root causes) Objectives: 1. Decompose the system failure](https://reader033.fdocuments.in/reader033/viewer/2022042622/5f9e41842c455d065a626335/html5/thumbnails/48.jpg)
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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
Hydraulic
Control A
Hydraulic
Control B
Actuators
Linkage
p = 0.1 p = 0.1 p = 0.1 p = 0.1
p 0.01
![Page 49: Fault tree analysis - LASAR · 2018-05-30 · Fault Tree Analysis (FTA) •Systematic and quantitative •Deductive (search for root causes) Objectives: 1. Decompose the system failure](https://reader033.fdocuments.in/reader033/viewer/2022042622/5f9e41842c455d065a626335/html5/thumbnails/49.jpg)
Piero Baraldi
Exercise 8: Solution
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 0.01
Hydraulic
Control A
Hydraulic
Control B
Actuators
Linkage
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
![Page 50: Fault tree analysis - LASAR · 2018-05-30 · Fault Tree Analysis (FTA) •Systematic and quantitative •Deductive (search for root causes) Objectives: 1. Decompose the system failure](https://reader033.fdocuments.in/reader033/viewer/2022042622/5f9e41842c455d065a626335/html5/thumbnails/50.jpg)
Piero Baraldi
Compute system failure probability from primary events probabilities by:
1. Using the laws of probability theory at the fault tree gates
2. Using the mcs found from the qualitative analysis
FT quantitative analysis
![Page 51: Fault tree analysis - LASAR · 2018-05-30 · Fault Tree Analysis (FTA) •Systematic and quantitative •Deductive (search for root causes) Objectives: 1. Decompose the system failure](https://reader033.fdocuments.in/reader033/viewer/2022042622/5f9e41842c455d065a626335/html5/thumbnails/51.jpg)
Piero Baraldi
1. Consider a system whose 2 minimal cut sets are:
2. 𝑀1 = 𝑥𝐴𝑥𝐵3. 𝑀2 = 𝑥𝐴𝑥𝐶4. You are required to compute the probability of the top
event
Exercise 9
![Page 52: Fault tree analysis - LASAR · 2018-05-30 · Fault Tree Analysis (FTA) •Systematic and quantitative •Deductive (search for root causes) Objectives: 1. Decompose the system failure](https://reader033.fdocuments.in/reader033/viewer/2022042622/5f9e41842c455d065a626335/html5/thumbnails/52.jpg)
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1. 𝑻 = 𝑴𝟏 ∪𝑴𝟐
𝑷 𝑻 = 𝑷(𝑴𝟏) + 𝑷 𝑴𝟐 − 𝑷(𝑴𝟏 ∩𝑴𝟐)
Exercise 9: Solution
𝑷 𝑻 = 𝑷 𝑨 𝑷(𝑩) + 𝑷 𝑨 𝑷(𝑪) − 𝐏(𝑨𝑩𝑨𝑪)
𝑷 𝑻 = 𝑷 𝑨 𝑷(𝑩) + 𝑷 𝑨 𝑷(𝑪) − 𝐏(𝑨𝑩𝑪)
𝑷 𝑻 = 𝑷 𝑨 𝑷(𝑩) + 𝑷 𝑨 𝑷(𝑪) − 𝐏 𝑨 𝑷 𝑩 𝑷(𝑪)
![Page 53: Fault tree analysis - LASAR · 2018-05-30 · Fault Tree Analysis (FTA) •Systematic and quantitative •Deductive (search for root causes) Objectives: 1. Decompose the system failure](https://reader033.fdocuments.in/reader033/viewer/2022042622/5f9e41842c455d065a626335/html5/thumbnails/53.jpg)
Piero Baraldi
Compute system failure probability from primary events probabilities by:
1. Using the laws of probability theory at the fault tree gates
2. 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
j
j
mcs
ji
mcs
j
j MPMMPMPXP
mcs
j
j
mcs
i
mcs
ij
ji
mcs
j
j MPXPMMPMP1
1
1 11
][]1)([][][
FT quantitative analysis
“Rare event” approximation
![Page 54: Fault tree analysis - LASAR · 2018-05-30 · Fault Tree Analysis (FTA) •Systematic and quantitative •Deductive (search for root causes) Objectives: 1. Decompose the system failure](https://reader033.fdocuments.in/reader033/viewer/2022042622/5f9e41842c455d065a626335/html5/thumbnails/54.jpg)
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Exercise 10
Compute system failure probability from the mcs
Hydraulic
Control A
Hydraulic
Control B
Actuators
Linkage
p = 0.1 p = 0.1 p = 0.1 p = 0.1
p 0.01
![Page 55: Fault tree analysis - LASAR · 2018-05-30 · Fault Tree Analysis (FTA) •Systematic and quantitative •Deductive (search for root causes) Objectives: 1. Decompose the system failure](https://reader033.fdocuments.in/reader033/viewer/2022042622/5f9e41842c455d065a626335/html5/thumbnails/55.jpg)
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Exercise 10 Solution
5 mcs:
P(M1) =P(XL=1)=0.01
P(M2)=P(XAXB=1)=0.1·0.1=0.01
P(M3)=P(XAXHB) =0.1·0.1=0.01
P(M4)= P(XHAXB=1)=0.1·0.1=0.01
P(M5) =P(XHAXHB)=0.1·0.1=0.01
0464.0][][]1)([
05.0][]1)([
1
1 11
1
mcs
i
mcs
ij
ji
mcs
j
j
mcs
j
j
MMPMPXP
MPXP
Hydraulic
Control A
Hydraulic
Control B
Actuators
LinkLatch
No approximation = 0.0457
0.0454
![Page 56: Fault tree analysis - LASAR · 2018-05-30 · Fault Tree Analysis (FTA) •Systematic and quantitative •Deductive (search for root causes) Objectives: 1. Decompose the system failure](https://reader033.fdocuments.in/reader033/viewer/2022042622/5f9e41842c455d065a626335/html5/thumbnails/56.jpg)
Piero Baraldi
Compute system failure probability from primary events probabilities by:
1. Using the laws of probability theory at the fault tree gates
2. Using the mcs found from the qualitative analysis
3. Reducing (if necessary) the structure function by boolean algebra rules and directly applying the expected value operator:
FT quantitative analysis
))XXX)(XXXX(X)(1X(11X HBBHBBHAAHAALT
))XXX(XE)XXX(XE)(1XE(11
))XXX)(XXXX(X)(1X(11EXETP
HBBHBBHAAHAAL
HBBHBBHAAHAALT
= 0.0457
![Page 57: Fault tree analysis - LASAR · 2018-05-30 · Fault Tree Analysis (FTA) •Systematic and quantitative •Deductive (search for root causes) Objectives: 1. Decompose the system failure](https://reader033.fdocuments.in/reader033/viewer/2022042622/5f9e41842c455d065a626335/html5/thumbnails/57.jpg)
575757Piero Baraldi
57
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 0.08
Supervisor 0.14
Switch 0.04
Alarm 0.05
Stand by Power Generator
0.12
Main Electricpower
0.10
AutomaticSensor
0.14
![Page 58: Fault tree analysis - LASAR · 2018-05-30 · Fault Tree Analysis (FTA) •Systematic and quantitative •Deductive (search for root causes) Objectives: 1. Decompose the system failure](https://reader033.fdocuments.in/reader033/viewer/2022042622/5f9e41842c455d065a626335/html5/thumbnails/58.jpg)
585858Piero Baraldi
Exercise 11: Solution
58
p = 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 power
failure
Stand-by
power
failure
Supervisor
failure
Switch
Failure
(p = 0.04)
Manometer
failure
Failure
of the
automatic
sensor
Alarm
failure
Power failure Signal does not
reach to alarm
Failure of the alarm
system
Failure on manual
activation
Correct! Branches are
INDEPENDENT
![Page 59: Fault tree analysis - LASAR · 2018-05-30 · Fault Tree Analysis (FTA) •Systematic and quantitative •Deductive (search for root causes) Objectives: 1. Decompose the system failure](https://reader033.fdocuments.in/reader033/viewer/2022042622/5f9e41842c455d065a626335/html5/thumbnails/59.jpg)
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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