论文代写-MS 927
时间:2022-03-25
Minimal cut and path sets
MS 927
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Minimal Cut sets
Minimal cut set formula expresses the top event as a “collection of minimal
scenarios causing the top event to occur”
Calculating Minimal Cut Sets
• Express top event in terms of basic events using fault tree logic
• Multiply out all the brackets so simplified Boolean expression
• Simplify using the different laws, if necessary
• Final expression gives the minimal cut sets
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Question 1
1) Write the minimal cut set expression and find the minimal cut sets
for the figure on slide 4
2) Write the minimal path set expression and find the minimal path
sets for the same figure on slide 4
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Slide 4
Fault tree for Question 1
Minimal cut sets Question 1
T = G1+G2 – equation 1
G1 = G3*G4 – equation 2
G2 = G5*G6 – equation 3
Putting the values of G1 and G2 in equation 1
T = (G3*G4)+ (G5*G6) – equation 4
G3 = P*Q*R – equation 5
G4 = D*E*F – equation 6
G5 = P*Q*R – equation 7
G6 = D*E*F – equation 8
Putting the values of equations 5,6,7,8 in equation 4
T = (P*Q*R * D*E*F)+(P*Q*R * D*E*F ) – equation 9
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=> T = (PQR DEF)+(PQR DEF ) – equation 9
According to Idempotent law part1 A+A = A
Let PQR = A
Let DEF = B
T = (AB)+(AB) – equation 10
Applying Idempotent Law part1 AB+AB should be AB,
T = AB – equation 11
Putting back the original value of A and B in equation 11,
T = PQRDEF – equation 12 (expression for minimal cut set)
Minimal cut set = {PQRDEF} (since its multiplication we cannot further
break this down)
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Minimal path set
Minimal path set expresses the top event as a “collection of minimal
scenarios preventing the top event from occurring”
Calculating Minimal Path set
• Create the dual tree, the fault tree for NOT T by applying De Morgan’s laws:
1) Replace all events A by A’ (A NOT)
2) Replace AND gates by OR gates
3) Replace OR gates by AND gates
• Apply cut set calculation method
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Dual Tree for the fault tree in slide 4
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Minimal Path sets Question 1
T (NOT) = G1(NOT)*G2(NOT) – equation 1
G1(NOT) = G3(NOT)+G4(NOT) – equation 2
G2(NOT) = G5(NOT)+G6(NOT) – equation 3
Putting the values of G1(NOT) and G2(NOT) in T(NOT)
T(NOT) = (G3(NOT)+G4(NOT))*(G5(NOT)+G6(NOT)) – equation 4
G3(NOT) = P(NOT)+Q(NOT)+R(NOT) – equation 5
G4(NOT) = D(NOT)+E(NOT)+F(NOT) – equation 6
G5(NOT) = P(NOT)+Q(NOT)+R(NOT) – equation 7
G6(NOT) = D(NOT)+E(NOT)+F(NOT) – equation 8
Putting the values of equations 5,6,7,8 in equation 4
T(NOT)= (P(NOT)+Q(NOT)+R(NOT) +D(NOT)+E(NOT)+F(NOT))*(P(NOT)+Q(NOT)+R(NOT) +
D(NOT)+E(NOT)+F(NOT) – equation 9
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We can write A(NOT) as A’, P(NOT) as P’ and so on
=> T’= (P’+Q’+R’+D’+E’+F’)*(P’+Q’+R’ + D’+E’+F’) – equation 10 (rewriting
equation 9)
Let P’+Q’+R’ be A’
Let D’+E’+F’ be B’
T’ = (A’+B’)*(A’+B’ ) – equation 11
According to Idempotent law part2 AA=A
Hence (A’+B’)(A’+B’) should be (A’+B’)
T’ = A’+B’ – equation 12
Putting back the value of A’ and B’ in equation 12
T’ = P’+Q’+R’+D’+E’+F’ – equation 13 (minimal path set expression)
Minimal Path sets = {P’}, {Q’},{R’},{D’},{E’},{F’} (since there is + between them, we
can break them into multiple sets)
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Question 2
1) Write the minimal cut set expression and find the minimal cut sets
for the figure on slide 12
2) Write the minimal path set expression and find the minimal path
sets for the same figure on slide 12
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Slide 12
Fault tree for Question 2
Minimal Cut sets Question 2
T = G1+G2+F – equation 1
G1 = G3+G4+G5 – equation 2
G2 = D+E – equation 3
Putting the values of G1 and G2 in equation 1
T = (G3+G4+G5)+ (D+E) +F – equation 4
G3 = P*Q– equation 5
G4 = P*R – equation 6
G5 = Q*R – equation 7
Putting the values of equations 5,6,7 in equation 4
T = (P*Q+P*R+Q*R)+(D+E)+F – equation 8
Rearranging equation 8,
=> T = (PQ+QR+PR+D+E+F) – equation 9(Minimal cut set expression)
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Minimal cut sets are :
{D}, {E}, {PQ},{QR},{PR},{F}
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Dual Tree for the fault tree in slide 12
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Minimal Path sets Question 2
We can write A(NOT) as A’ and similarly for all variables
T(NOT) = G1(NOT)*G2(NOT)*F(NOT) – equation 1
G1(NOT) = G3(NOT)*G4(NOT)*G5(NOT) – equation 2
G2(NOT) = D(NOT)*E(NOT) – equation 3
Putting the values of G1(NOT) and G2(NOT) in equation 1
T’ = (G3’*G4’*G5’)*(D’*E’) * F’ – equation 4
G3’ = P’+Q’– equation 5
G4’ = P’+R’ – equation 6
G5’ = Q’+R’ – equation 7
Putting the values of equations 5,6,7 in equation 4
T’ = ((P’+Q’)*(P’+R’)*(Q’+R’))*(D’*E’)*F’ – equation 8
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T’ = ((P’Q’+P’R’+Q’*Q’+Q’R’)*(P’+R’)) *(D’E’F’) – equation 9
T’ = (P’Q’Q’+P’P’Q’+P’Q’R’+P’Q’R’+P’P’R’+P’R’R’+Q’Q’R’+Q’R’R’)*(D’E’F’) –
equation 10
According to Idempotent law part2 , AA=A
Hence applying Idempotent law part2 to equation 10
T’ = (P’Q’+P’Q’+P’Q’R’+P’Q’R’+P’R’+P’R’+Q’R’+Q’R’)*(D’E’F’) – equation 11
According to Idempotent law part1, A+A=A
Hence applying Idempotent law part1 to equation 11
T’ = (P’Q’+P’Q’R’+P’R’+Q’R’)*(D’E’F’) – equation 12
Let P’Q’ be A’ and R’ be B’, then P’Q’R = A’B’
Putting this in equation 12,
T’ = (A’+A’B’+P’B’+Q’B’)*(D’E’F’) – equation 13
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According to Absorption law A+AB = A
Hence applying absorption law in equation 13
T’ = (A’+P’B’+Q’B’)*(D’E’F’) – equation 14
Putting the value of A’, B’ back in equation 14
T’ = (P’Q’+P’R’+Q’R’)*(D’E’F’) – equation 15 (minimal path set
expression)
Minimal path set = {(P’Q’+P’R’+Q’R’)*(D’E’F’)}
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