System Reliability

Study guide: Exam MAS-I

System Reliability is the probability that a system of components functions, when whether the system works is a fixed function of which components work — its structure function $\phi$ — and each component $i$ works independently with probability $p_i$ (Ross, ch. 9).

$\phi(\mathbf{x}) = \begin{cases} 1 & \text{if the system works} \\ 0 & \text{if it fails} \end{cases}$

$r(\mathbf{p}) = P\big(\phi(\mathbf{X}) = 1\big) = E\big[\phi(\mathbf{X})\big]$

$x_i = 1$ if component $i$ works and $0$ if not; $r(\mathbf{p})$ is the reliability function.
Series (works only if all work): $\phi = \prod x_i$ and $r = \prod p_i$ . Parallel (works if any works): $\phi = 1 - \prod(1 - x_i)$ and $r = 1 - \prod(1 - p_i)$ . A Joint Life status is a series system of lives; a last-survivor status is a parallel one.
$k$ -out-of- $n$ (works if at least $k$ work): with equal $p$ , $r = \sum_{i=k}^{n} \binom{n}{i} p^i (1-p)^{n-i}$ , a binomial tail. Series is $n$ -out-of- $n$ ; parallel is $1$ -out-of- $n$ .
A minimal path set is a set of components whose working alone makes the system work, with no proper subset doing so; a minimal cut set is a set whose failure alone makes it fail, again minimal. A $k$ -out-of- $n$ system has $\binom{n}{k}$ minimal path sets (every $k$ -subset) and $\binom{n}{n-k+1}$ minimal cut sets (every $(n-k+1)$ -subset) — the Combinatorics in the syllabus.
Bounds from minimal cut sets $C_j$ and minimal path sets $A_j$ , when the exact $r(\mathbf{p})$ is tedious:

$\prod_{j} \Big[1 - \prod_{i \in C_j} (1 - p_i)\Big] \;\leq\; r(\mathbf{p})$

$r(\mathbf{p}) \;\leq\; 1 - \prod_{j} \Big[1 - \prod_{i \in A_j} p_i\Big]$

Lifetimes: if component $i$ survives to $t$ with probability $\bar F_i(t)$ , the system does with probability $r(\bar{\mathbf{F}}(t))$ , and its expected life is $\int_0^\infty r(\bar{\mathbf{F}}(t))\,dt$ . A $k$ -out-of- $n$ system fails at the $(n-k+1)$ -th component failure, an order statistic.

Path and Cut Sets of a $k$ -out-of- $n$ System An insurer's claims platform runs on three servers and stays up while at least two are running. Each server is up independently with probability $0.9$ . List the minimal path and cut sets, find the reliability, and count the sets for a 3-out-of-5 design.
Answer Minimal path sets are the pairs $\{1,2\}, \{1,3\}, \{2,3\}$ ; minimal cut sets are also the pairs, since $n - k + 1 = 2$ failures bring it down.
 $\begin{align*} r &= 3p^2(1-p) + p^3 \\ &= 3(0.81)(0.1) + 0.729 \\ &= 0.972 \end{align*}$ 
For 3-out-of-5: $\binom{5}{3} = 10$ minimal path sets and $\binom{5}{5-3+1} = \binom{5}{3} = 10$ minimal cut sets.

Bounding a Bridge Structure A bridge system has components 1 and 2 leaving the source, 4 and 5 entering the sink, and 3 linking the two middle nodes. Its minimal path sets are $\{1,4\}, \{2,5\}, \{1,3,5\}, \{2,3,4\}$ and its minimal cut sets $\{1,2\}, \{4,5\}, \{1,3,5\}, \{2,3,4\}$ . Each component works with probability $0.9$ . Find $r$ exactly and compare with the bounds.
Answer Condition on component 3. If it works, the system is (1 or 2) in series with (4 or 5); if not, it is $\{1,4\}$ in parallel with $\{2,5\}$ :
 $\begin{align*} r &= 0.9\,\big[1 - 0.1^2\big]^2 + 0.1\,\big[1 - (1 - 0.81)^2\big] \\ &= 0.9(0.9801) + 0.1(0.9639) \\ &= 0.97848 \end{align*}$ 
Bounds: lower $= (1 - 0.1^2)^2 (1 - 0.1^3)^2 = 0.97814$ ; upper $= 1 - (1 - 0.81)^2 (1 - 0.729)^2 = 0.99735$ . The cut-set bound is very tight when components are reliable.

Expected Life of a 2-out-of-3 System Each server in the first example has an exponential lifetime with mean $10$ years. Find the system's expected life.
Answer With $p = e^{-t/10}$ , $r = 3p^2 - 2p^3$ :
 $\begin{align*} E[L] &= \int_0^\infty \left(3e^{-2t/10} - 2e^{-3t/10}\right) dt \\ &= 3(5) - 2\left(\tfrac{10}{3}\right) \\ &= 8.33 \text{ years} \end{align*}$ 
Shorter than one server's $10$ years: the system fails at the second failure, which on average comes sooner than a single server's failure. Redundancy here buys reliability early on, not a longer expected life.