Suppose that the flight of an aircraft can be regarded as a system having 3 comp
ID: 3217039 • Letter: S
Question
Suppose that the flight of an aircraft can be regarded as a system having 3 components A (aircraft), B (pilot) and C (airport). Suppose, furthermore, that component B can be regarded as a parallel subsystem consisting of B1 (captain) , B2 (first officer) and B3 (flight engineer); and C is a parallel subsystem consisting of C1 (scheduled airport) and C2(alternate airport) . Under given flight conditions, the reliabilities of components A, B1, B2 , B3, C1 and C2 ( deined as the probabilities that they can contribute to the successful completion of the flight) are, 0.9999, 0.9995, 0.999, 0.2,0.95 and 0.85 respectively.
(i) What is the reliability of the system?
(ii) What is the effect on the system reliabiliy of having a flight engineer who is also trained as a pilot, so increasing the reliability of B3 from 0.20 to 0.99?
(iii) if the flight crew did not have a first officer, what would be the effect of increasing the reliability of B3 from 0.20 to 0.99?
(iv) What is the effect of adding a second alternative landing point, C3, with reliability 0.80?
Explanation / Answer
Reliability of component A = 0.9999
Reliability of component B1 = 0.9995
Reliability of component B2 = 0.999
Reliability of component B3 = 0.2
Reliability of component C1 = 0.95
Reliability of component C2 = 0.85
Reliability of parallel B1 B2 B3 = Atleast one 1works= 1 - (all fails) = 1 - (0.0005 * 0.001 * 0.8) = 0.9999996
Reliability of parallel c1 C2 = 1 - (all fails) = 1 - (0.05 * 0.15) = 0.9925
Reliability of system = serien reliability of A B C = 0.9999 * 0.9999996 * 0.9925 = 0.9924003
b)
If reliability of B3 is increased from 0.20 to 0.99
Total B reliability = 1 - (0.0005 * 0.001 * 0.01) = 0.999999999995
Total reliability = 0.999999999995 * 0.9925 * 0.9999 = 0.99240074
Reliability increased
c)
Reliability of B in absence of first officer = 1 - ( 0.01 * 0.0005 ) = 0.99995
Reliability will decrease
d)
adding second landing point 0.8
total c reliabilty = 1 - (0.05 * 0.15 * 0.2 ) = 0.9985
As C reliability increases total reliability also increases
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