1. Fault Tree Analysis Part 6 – Solutions of Fault Trees

2. Receiver explodes Air flow into receiver exceeds flow out at pressure danger level Pressure relief valve fails to give adequate discharge at pressure danger level Pressure reducing valve flow less than compressor flow Pressure control loop causes compressor to run Incorrect design Instrument air system pressure abnormally high See Subtree Instrument air system pressure abnormally low Other causes Air flow out of air system (demand + leakage) abnormal and exceeds pressure reducing valve capacity Air flow out air system normal but flow in abnormally low See Subtree Incorrect design Dirt Other causes T A B E B* F GHI

3. SUBTREE Pressure reducing valve Partially or completely Seized shut or blocked Dirt Other causes CD

4. CUT SETS T = (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)

5. 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.

6. [ Example ] T OR 1 AND OR 2 3 4 56 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) {1, 2, 3} and {1, 4, 5, 6} are the Minimum Path Sets

7. CUT SET TRANSFORMATION TOP OR 1 Component Cut Sets AND 2 Component Cut Sets AND 3 Component Cut Sets AND 4 Component Cut Sets AND 5 Component Cut Sets

8. 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.

9. [ Example ] TOP OR 1 2 G2 AND 3OR AND5 34 G3 G4 G5 OR 6AND G7 G2 3 G6 GATECUT 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).

10. 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.)

11. STEP 1 : Top Event Fault Event 1 Fault Event 2 Fault Event 3 BASIC EVENT 1 BASIC EVENT 2 BASIC EVENT 3 BASIC EVENT 2 BASIC EVENT 4 1 B A C 23 24 D

12. 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 :

13. 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.

14. STEP 2 : AA B D B 1 D C1D 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 OR

15. 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}

16. 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 }