Fault Tree Analysis
Part 6 – Solutions of Fault Trees
Receiverexplodes
Air flow into receiverexceeds flow out atpressure danger level
Pressure relief valve failsto give adequate dischargeat pressure danger level
Pressure reducingvalve flow lessthan compressor flow
Pressure controlloop causescompressor to run
Incorrectdesign
Instrumentair systempressureabnormallyhigh
See Subtree
Instrument airsystem pressureabnormally low
Other causes
Air flow outof air system(demand + leakage)abnormal and exceeds pressure reducingvalve capacity
Air flow out airsystem normal butflow in abnormally low
See Subtree
Incorrectdesign Dirt
Othercauses
T
A
B
E
B*F
G H I
SUBTREE
Pressure reducing valvePartially or completelySeized shut or blocked
DirtOthercauses
C D
CUT SETST = (A B C D) (B* F) (G H I)
Where “OR” “AND”
Note, B B* = ,
B* = C D E,
C C = C, D D = D,
A C, C D, C E, C F C
A D, D C, D E, D F D
Thus,
T = [(A B*) (C B*) (D B*) (A F) (B F) (C F) (D F)] (G H I)
= [(A C) (A D) (A E) C (C D) (C E) (D C) (D E) D (A F)
(B F) (C F) (D F)] (G H I)
= [(A E) (A F) (B F) C D] (G H I)
= [(A E G ) (A F G) (B F G ) (C G) (D G) (A E H) (A F H)
(B F H) (C H) (D H) (A E I) (A F I) (B F I) (C I) (D I)
DEFINITIONS:
Cut Set:
A cut set is a collection of basic events; if all these basic events occur, the top event is guaranteed to occur.
Minimal Cut Set
A minimal cut set is such that if any basic event is removed from the set, the remaining events collectively are no longer a cut set.
Path Set
A path set is a collection of basic events; if none of the events in the sets occur, the top event is guaranteed not to occur.
Minimal Path Set
A minimal path set is a path set such that if any basic event is removed from the set, the remaining events collectively are no longer a path set.
[ Example ] T
OR
1 AND
OR OR
2 OR
3
4 OR
5 6
Minimum Cut Sets
{1}, {2,4}, {2,5}, {2,6}, {3,4}, {3,5}, {3,6}
T = 1 (2 4) (2 5) (2 6) (3 4) (3 5) (3 6)
T 1 (2 4) (2 5) (2 6) (3 4) (3 5) (3 6)
= (1 2 3) (1 4 5 6)
{1, 2, 3} and {1, 4, 5, 6} are the Minimum Path Sets
CUT SET TRANSFORMATIONTOP
TOP
OR
1Component
Cut
Sets
AND
2Component
Cut
Sets
AND
3Component
Cut
Sets
AND
4Component
Cut
Sets
AND
5Component
Cut
Sets
PROCEDURE FOR FINDING CUT SETS
Method 1: Bottom Up
1) Form a table with two columns, one for gate numbers and the other for cut sets.
2) Search the tree for gates which have only primal events as inputs.
3) List the cut sets for each of these gates in the table.
4) Find a gate whose gate inputs already appear in the table.
5) If gate is an “OR” the cut set for it are gotten by forming the union of all cut sets for its inputs, with non-minimal sets removed.
6) If gate is an “AND”, its cut sets are formed by “anding” the cut sets for each input with those of all the inputs. After forming a given cut set, any duplicate members within it are removed and it is checked for minimality against all other cut sets formed to that point.
7) If the gate just developed is the TOP, its cut sets are those for the entire tree. If not, go to step 4.
[ Example ] TOP
OR
OR
1 2
G2 AND
3 OR
AND 5
3 4
G3
G4
G5
OR
6 ANDG7
G2 3
G6
GATE CUT SETS
2
5
4
7
3
6
1
(1) (2)
(3 , 4)
(3 , 4) (5)
(1 , 3) (2 , 3)
(3 , 4 , 3) (3 , 5)
(6) (1 , 3) (2 , 3)
(1) (2) (3 , 4) (3 , 5) (6) (1 , 3) (2 , 3)
Hence, the minimal cut sets for this tree are : (1) , (2) , (6) , (3 , 4) and (3 , 5).
PROCEDURE FOR FINDING DUT SETS
Method Two: Top Down
1) Uniquely identify all gates and basic events.
2) Resolve all gates into basic events.
3) Remove duplicate events within sets.
4) Delete all supersets (sets that contain another set as a subset.)
STEP 1 :
Top Event
Fault
Event 1
Fault
Event 2
Fault
Event 3BASIC
EVENT
1
BASIC
EVENT
2
BASIC
EVENT
3
BASIC
EVENT
2
BASIC
EVENT
4
1
B
A
C
2 3
2 4
D
•The second step is to resolve all the gates into BASIC events.
•This is done in a matrix format, beginning with the TOP event and proceeding through the matrix until all gates are resolved.
•The TOP event is always the first entry in the matrix and is entered in the first column of the first row.
•Gates are resolved by replacing them in the matrix with their inputs. There are two rules for entering the remaining information in the matrix : the OR – gate rule, and the AND-gate rule.
STEP 2 :
The OR-gate rule : When resolving an OR gate in the matrix, the first input of the OR gate replaces the gate identifier in the matrix, and all other inputs to the OR gate are inserted in the next available row, one input per row. The next available row means the next empty row of the matrix. In addition, if there are other entries in the row where the OR gate appeared, these entries must be entered (repeated) in all the rows that contain the gate’s inputs.
The AND-gate rule : When resolving an AND gate in the matrix, the first input to the AND gate replaces the gate identifier in the matrix, and the other inputs to the AND gate are inserted in the next available column, one input per column, on the same row that the AND gate appeared on.
STEP 2 :
A A B D
B 1 D C 1 D 2 C
1 4 C
1 2 C 2 1 4 C
1 2 3
1 2 2
1 4 C 2
1 2 3
1 4 3
(a) (b)
(c) (d)
(e) (f)
AND
ANDOR
OR
OR
STEP 3 :
Set 1 : 1 , 2 , 2 1 , 2
Set 2 : 1 , 2 , 4 1 , 2 , 4
Set 3 : 1 , 2 , 3 1 , 2 , 3
Set 4 : 1 , 3 , 4 1 , 3 , 4
STEP 4 :
Minimum cut sets :
{1 , 2} , {1 , 3 , 4}
PATH Sets :
A B
D
B 1
D
C
1
D 2 4
C
1
2 4
C 2 3
1
2 4
2 3
{ 1 } { 2 , 4 } { 2 , 3 }