8. Assessing Product Reliability
8.1. Introduction
8.1.8. How can you evaluate reliability from the "bottom-up" (component failure mode to system failure rate)?

8.1.8.4.

R out of N model

An \(r\) out of \(n\) model is a system that survives when at least \(r\) of its components are working (any \(r\))

An "\(r\) out of \(n\)" system contains both the series system model and the parallel system model as special cases. The system has \(n\) components that operate or fail independently of one another and as long as at least \(r\) of these components (any \(r\)) survive, the system survives. System failure occurs when the (\(n - r + \) 1)-th component failure occurs.

When \(r = n\), the \(r\) out of \(n\) model reduces to the series model. When \(r\) = 1, the \(r\) out of \(n\) model becomes the parallel model.

We treat here the simple case where all the components are identical.

Formulas and assumptions for the \(r\) out of \(n\) model (identical components):

1. All components have the identical reliability function \(R(t)\).
2. All components operate independently of one another (as far as failure is concerned).
3. The system can survive any (\(n - r\)) of the components failing. The system fails at the instant of the (\(n - r + \) 1)-th component failure.

Formula for an \(r\) out of \(n\) system where the components are identical

System reliability is given by adding the probability of exactly \(r\) components surviving to time \(t\) to the probability of exactly (\(r\) + 1) components surviving, and so on up to the probability of all components surviving to time \(t\). These are binomial probabilities (with \(p = R(t)\)), so the system reliability is given by: $$ R_S(t) = \sum_{i=r}^n \left( \begin{array}{c} n \\ i \end{array} \right) \left[ R(t) \right]^i \left[ 1 - R(t) \right] ^{n-i} \,\, . $$ Note: If we relax the assumption that all the components are identical, then \(R_S(t)\) would be the sum of probabilities evaluated for all possible terms that could be formed by picking at least \(r\) survivors and the corresponding failures. The probability for each term is evaluated as a product of \(R(t)\)'s and \(F(t)\)'s. For example, for \(n\) = 4 and \(r\) = 2, the system reliability would be (abbreviating the notation for \(R(t)\) and \(F(t)\) by using only \(R\) and \(F\)) $$ \begin{eqnarray} R_S & = & R_1 R_2 F_3 F_4 + R_1 R_3 F_2 F_4 + R_1 R_4 F_2 F_3 + R_2 R_3 F_1 F_4 \\ & + & R_2 R_4 F_1 F_3 + R_3 R_4 F_1 F_2 + R_1 R_2 R_3 F_4 + R_1 R_3 R_4 F_2 \\ & + & R_1 R_2 R_4 F_3 + R_2 R_3 R_4 F_1 + R_1 R_2 R_3 R_4 \,\, . \end{eqnarray} $$