Reliability engineers often work with systems having components connected in par
ID: 3040262 • Letter: R
Question
Reliability engineers often work with systems having components connected in parallel. In this problem, we will interpret the phrase “in parallel” as follows: The system is reliable (or, functioning) if at least one of the components is functioning. As a frame of reference, consider a two-engine aircraft.
-If both engines are functioning, the aircraft is functioning (at least in lieu of non-engine related problems).
-If only one engine is functioning, the aircraft still functions (although losing one engine may warrant an immediate landing.)
-If both engines are not functioning, the aircraft is not functioning.In this problem, we will denote by n the number of components in a parallel system.
1.Suppose n = 2. If the two components are functioning independently, each with probability p, show that the system reliability r2 is given by r2 =1(1p)2(hint: Let A1 = {component 1 is functioning} and A2 = {component 2 is functioning}. Then, r2 = P(A1 A2).)
2.Generalize the result in part (a) to consider a parallel system with n components (each finctioning independently with probability p). That is, show that the system reliability is rn =1(1p)n (hint: rn = P(A1 A2 ...An) = 1P(A1 A2 ...An). Then, use the fact that the components are independent.)
3.I have a parallel system with n = 5 components (each functioning independently with probability p). How unreliable can the individual components be and still have a system with reliability r5 = 0.9999?
Explanation / Answer
(a)
System reliability r2 = Probability that atleast one component is functioning out of 2 components
= 1 - Probability that no component is functioning
= 1 - (1-p) * (1-p)
= 1 - (1-p)2
(b)
System reliability rn = Probability that atleast one component is functioning out of n components
= 1 - Probability that none of 'n' component is functioning
= 1 - (1-p) * (1-p) * (1-p) * ..... n times
= 1 - (1-p)n
(c)
For parallel system with n = 5 components,
Reliability , r5 = 1 - (1-p)5
=> 0.9999 = 1 - (1-p)5
=> (1-p)5 = 1 - 0.9999
=> (1-p)5 = 0.0001
=> (1-p) = 0.1585
So, the individual components can be unreliable with probability 0.1585
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