ASSIGNMENT | SYSTEM RELIABILITY

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System Reliability
Name
Institution Affiliation

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1. The figure below gives a fault tree for a top event marked with the
probabilities for the basic events.
a. Write down the
logical expressions for the outputs from the gates A, B and C are as
follows and hence derive the minimal cuts sets for the system (Birolini,
2013).
P ( C ) =P ( 3 )+ P ( 4 ) + P ( 5 )−P ( 3 ) P ( 4 )−P ( 3 ) P ( 5 ) −P ( 4 ) P (5 )+ P ( 3 ) P ( 4 ) P ( 5 )
(1∈3 [ ¿ ] logic tree for success)
P ( B ) =P ( 2 ) P ( C ) =P(2 C) (1 in 2 [AND] logic tree for failure)
P ( B ) =P ( 2 ) { P ( 3 ) + P ( 4 ) + P ( 5 ) −P ( 3 ) P ( 4 ) −P ( 3 ) P ( 5 ) −P ( 4 ) P ( 5 ) + P ( 3 ) P ( 4 ) P ( 5 ) }
P ( A ) = P(1 ∨ B) = P (1) + P ( B ) −P(1 ∧ B) (1 in 2 (OR)logic tree for success ¿
P ( A ) = P (1) +
P ( 2 ) {P ( 3 ) + P ( 4 ) + P ( 5 )−P ( 3 ) P ( 4 ) −P ( 3 ) P ( 5 )−P ( 4 ) P ( 5 )+ P ( 3 ) P ( 4 ) P ( 5 ) }−P (1) ∧ P ( 2 ) { P ( 3 ) + P ( 4 )+ P ( 5 ) −P ( 3 )
Minimal cuts.
A=1+ B

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B=¿2. C
C=3+ 4+5
b. Using the minimal cut set derived, use the inclusion –exclusion theorem
to calculate the upper and lower bounds for the probability of the top
event, T (Smith, 2011).
∑
i=1
N
P( C¿¿ i)−∑
i<1
P(Ci Λ C j )≤ P(T ) ≤∑
i=1
N
P(C¿ ¿i)¿ ¿
∑
i=1
N
P(C¿¿ i)= ( B1 ) + ( B2 X B3 ) + ( B2 X B4 ) + ( B2 X B5 ) ¿
( 0.4 ) + ( 0.2 X 0.3 ) + ( 0.2 X 0.2 )+ ( 0.2 X 0.4 )=0.58
∑
i< 1
P ( Ci Λ C j ) =0.58−P ( C1 Λ C2 ) + P ( C2 Λ C3 ) + P ( C3 Λ C4 ) + P ( C1 Λ C2 ) + p ¿
C1= ( 1 )−B1 ,1=0.4
C2= ( 2,3 ) −B1 , 2=0.2 , B2 , 2=0.3
C2= ( 2,3 ) −B1 , 2=0.2 , B2 , 2=0.3
C4= ( 2,5 ) −B1 ,4 =0.2 , B2 , 4=0.4
P ( C1 Λ C2 ) =B1,1 X ( B1,2 X B2,2 )=0.4 X 0.2 X 0.3=0.024
P ( C1 Λ C3 ) =B1,1 X ( B1,3 X B2,3 ) =0.4 X 0.2 X 0.2=0.016
P ( C1 Λ C4 )=B1,1 X ( B1,4 X B2,4 )=0.4 X 0.2 X 0.4=0.032
P ( C2 Λ C3 ) = ( B1,2 X B2,2 ) X ( B1,3 X B2,3 )=0.2 X 0.3 X 0.2=0.012
P ( C2 Λ C4 )= ( B1,2 X B2,2 ) X ( B1,4 X B2,4 )=0.2 X 0.3 X 0.4=0.024
P ( C3 Λ C4 )= ( B1,3 X B2,3 ) X ( B1,4 X B2,4 ) =0.2 X 0.2 X 0.4=0.016
∑
i< 1
P ( Ci Λ C j ) =0.024+ 0.016+0.032+0.012+0.024+ 0.016=0.124

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∑
i=1
N
P(C¿¿ i)−∑
i<1
P ( Ci Λ C j )=0.58−0.124=0.456 ¿
0.456 ≤ P(T ) ≤0.58
c. Calculate the exact value of the probability of the top event. Comment
on how closely this compares with limits found in part (b) (Nachlas,
2017).
P ( A ) = P (1) +
P ( 2 ) { P ( 3 ) + P ( 4 ) + P ( 5 ) −P ( 3 ) P ( 4 ) −P ( 3 ) P ( 5 ) −P ( 4 ) P ( 5 ) + P ( 3 ) P ( 4 ) P ( 5 ) } −P (1) ∧ P ( 2 ) { P ( 3 ) + P ( 4 ) + P ( 5
¿ ( 0.4 )+ ( 0.2 ) { ( 0.3 ) + ( 0.2 ) + ( 0.4 )− ( 0.3 X 0.2 ) − ( 0.3 X 0.4 ) − ( 0.2 X 0.4 )−(0.3 X 0.2 X 0.4) }−(0.4 X 0.2) { ( 0.3 ) + ( 0.2 ) + ( 0
¿ 0.6 { ( 0.9 )− ( 0.06 )− ( 0.12 )−(0.08)− ( 0.024 ) }−(0.8) {0.9− ( 0.06 )− ( 0.12 ) −0.8+0.024 }
P ( A )=0.4797
The exact value of the probability is within therange obtained ∈ part ( b ) above
CITATION Lie 17 ¿ 1033(Lienig ,2017) .
2. The figure below shows the power output from a batch of experimental
geothermal power production units based in Northern Alaska. The
geothermal heat input is essentially constant, but the harsh conditions
have rendered the units somewhat unreliable, necessitating frequent
repair to return them to normal production