Probability of system failure, as it requires two adjacent component to fail

probabilityreliability

I am stuck with a problem in probability. It says that the system consists of 16 components distributed in corners and intersections as the following:

  s---s---s---s 
  |   |   |   | 
  s---s---s---s 
  |   |   |   | 
  s---s---s---s 
  |   |   |   | 
  s---s---s---s

Failure of two adjacent components will lead to system failure. The probability of failure for each component is 0.0005.

I tried different ways but doesn't work, and I was thinking to do it as a simulation but I didn't have any idea

Best Answer

It's easy to see that the number of pairs of two adjacent components is 8*3=24 (each large edge has 3 pairs and there are 8 edges). Because the probability of failure is so small, we can ignore the case where three or more fail simultaneously. For failure, two have to fail simultaneously and the two have to be adjacent. The probability becomes -

$$p = {16 \choose 2}(0.0005)^2 (1-0.0005)^{14} \frac{24}{{16 \choose 2}}$$

This is a lower bound since we didn't consider 3 failing, 4 failing and so on. But those probabilities are extremely low.

If you want to include 3 failing as well your estimate becomes (24*14 because two adjacent need to fail and choose any of the remaining 14 for the third component that failed):

$$p = {16 \choose 2}(0.0005)^2 (1-0.0005)^{14} \frac{24}{{16 \choose 2}} + {16 \choose 3}(0.0005)^3 (1-0.0005)^{13} \frac{24 \times 14}{{16 \choose 3}}$$

4 onwards becomes complicated with negligible addition to the answer.

Related Solutions

Two components with different probabilities of failure.

Taking the average is incorrect, but taking the sum is correct. You have correctly calculated the probabilities that exactly one component of either type fails. These two events are disjoint, so their probabilities must simply add to give the correct answer of $$\binom{4}{1} 0.9^3 \cdot 0.1 \cdot 0.8^7 + \binom{7}{1} 0.9^4 \cdot 0.8^6 \cdot 0.2 $$ for the probability that exactly one component fails.

MTTR and MTBF formulas for series system must be associative

We can use induction on $n\geq 3$ to prove associativity of the system repair rate $\overline{\mu}$. The associativity of the system failure rate $\overline{\lambda}=\lambda_1+\cdots\lambda_n$ is already given, since addition is associative.

At first we note that for $n\geq 1$ the expressions \begin{align*} \mu_{1,2,\ldots,n}&:=\left(\sum_{j=1}^n\lambda_j\right) \frac{\prod_{k=1}^n\frac{\mu_k}{\mu_k+\lambda_k}}{1-\prod_{l=1}^k\frac{\mu_l}{\mu_l+\lambda_l}}\\ \lambda_{1,2,\ldots,n}&:=\sum_{j=1}^n\lambda_j \end{align*} are both symmetric in $1,2,\ldots,n$, so that each index-permutation $\sigma(1),\sigma(2),\ldots,\sigma(n)$ yields the same result. Therefore we consider wlog \begin{align*} \mu_{n^{\star}}&:=\mu_{1,2,\ldots,n}\\ \lambda_{n^{\star}}&:=\lambda_{1,2,\ldots,n} \end{align*} with index-tuple $(1,2,\ldots,n)$ always in this order.

Base step: $n=3$

We have \begin{align*} \mu_{2^{\star}}&=\left(\lambda_1+\lambda_2\right) \frac{\frac{\mu_1}{\mu_1+\lambda_1}\,\frac{\mu_2}{\mu_2+\lambda_2}} {1-\frac{\mu_1}{\mu_1+\lambda_1}\,\frac{\mu_2}{\mu_2+\lambda_2}}\tag{2}\\ \lambda_{2^{\star}}&=\lambda_1+\lambda_2\\ \\ \mu_{3^{\star}}&=\left(\lambda_1+\lambda_2+\lambda_3\right) \frac{\frac{\mu_1}{\mu_1+\lambda_1}\,\frac{\mu_2}{\mu_2+\lambda_2}\,\frac{\mu_3}{\mu_3+\lambda_3}} {1-\frac{\mu_1}{\mu_1+\lambda_1}\,\frac{\mu_2}{\mu_2+\lambda_2}\,\frac{\mu_3}{\mu_3+\lambda_3}}\tag{3}\\ \lambda_{3^{\star}}&=\lambda_1+\lambda_2+\lambda_3\\ \end{align*}

In the base step it is sufficient due to symmetry to show \begin{align*} \mu_{2^{\star},3}=\mu_{3^{\star}}\tag{4} \end{align*} We obtain from (2) and (4) \begin{align*} \mu_{2^{\star},3}&=\left(\lambda_{2^{\star}}+\lambda_3\right) \frac{\frac{\mu_{2^{\star}}}{\mu_{2^{\star}}+\lambda_{2^{\star}}}\,\frac{\mu_3}{\mu_3+\lambda_3}} {1-\frac{\mu_{2^{\star}}}{\mu_{2^{\star}}+\lambda_{2^{\star}}}\,\frac{\mu_3}{\mu_3+\lambda_3}}\tag{5} \end{align*}

Since \begin{align*} \color{blue}{\frac{\mu_{2^{\star}}}{\mu_{2^{\star}}+\lambda_{2^{\star}}}} &=\frac{\left(\lambda_1+\lambda_2\right) \frac{\frac{\mu_1}{\mu_1+\lambda_1}\,\frac{\mu_2}{\mu_2+\lambda_2}} {1-\frac{\mu_1}{\mu_1+\lambda_1}\,\frac{\mu_2}{\mu_2+\lambda_2}}} {\left(\lambda_1+\lambda_2\right) \frac{\frac{\mu_1}{\mu_1+\lambda_1}\,\frac{\mu_2}{\mu_2+\lambda_2}} {1-\frac{\mu_1}{\mu_1+\lambda_1}\,\frac{\mu_2}{\mu_2+\lambda_2}}+\left(\lambda_1+\lambda_2\right)}\\ &=\frac{ \frac{\frac{\mu_1}{\mu_1+\lambda_1}\,\frac{\mu_2}{\mu_2+\lambda_2}} {1-\frac{\mu_1}{\mu_1+\lambda_1}\,\frac{\mu_2}{\mu_2+\lambda_2}}} {\frac{\frac{\mu_1}{\mu_1+\lambda_1}\,\frac{\mu_2}{\mu_2+\lambda_2}} {1-\frac{\mu_1}{\mu_1+\lambda_1}\,\frac{\mu_2}{\mu_2+\lambda_2}}+1}\\ &=\frac{\frac{\mu_1}{\mu_1+\lambda_1}\,\frac{\mu_2}{\mu_2+\lambda_2}} {\frac{\mu_1}{\mu_1+\lambda_1}\,\frac{\mu_2}{\mu_2+\lambda_2}+\left(1-\frac{\mu_1}{\mu_1+\lambda_1}\,\frac{\mu_2}{\mu_2+\lambda_2}\right)}\\ &\,\,\color{blue}{=\frac{\mu_1}{\mu_1+\lambda_1}\,\frac{\mu_2}{\mu_2+\lambda_2}}\tag{6} \end{align*} we obtain by putting (6) into (5)

\begin{align*} \color{blue}{\mu_{2^{\star},3}} &=\left(\lambda_{2^{\star}}+\lambda_3\right) \frac{\frac{\mu_{2^{\star}}}{\mu_{2^{\star}}+\lambda_{2^{\star}}}\,\frac{\mu_3}{\mu_3+\lambda_3}} {1-\frac{\mu_{2^{\star}}}{\mu_{2^{\star}}+\lambda_{2^{\star}}}\,\frac{\mu_3}{\mu_3+\lambda_3}}\\ &=\left(\lambda_1+\lambda_2+\lambda_3\right) \frac{\frac{\mu_1}{\mu_1+\lambda_1}\,\frac{\mu_2}{\mu_2+\lambda_2}\,\frac{\mu_3}{\mu_3+\lambda_3}} {1-\frac{\mu_1}{\mu_1+\lambda_1}\,\frac{\mu_2}{\mu_2+\lambda_2}\,\frac{\mu_3}{\mu_3+\lambda_3}}\\ &\,\,\color{blue}{=\mu_{3^{\star}}} \end{align*} and the base step follows.

Induction hypothesis: $n=N-1$

We assume the claim is valid for $n=N-1$, i.e. we have \begin{align*} \mu_{(N-1)^{\star},N}&=\left(\lambda_{(N-1)^{\star}}+\lambda_N\right) \frac{\frac{\mu_{(N-1)^{\star}}}{\mu_{(N-1)^{\star}}+\lambda_{(N-1)^{\star}}}\,\frac{\mu_N}{\mu_N+\lambda_N}} {1-\frac{\mu_{(N-1)^{\star}}}{\mu_{(N-1)^{\star}}+\lambda_{(N-1)^{\star}}}\,\frac{\mu_N}{\mu_N+\lambda_N}}\\ &=\left(\sum_{j=1}^N\lambda_j\right)\, \frac{\prod_{k=1}^N\frac{\mu_k}{\mu_k+\lambda_k}}{1-\prod_{l=1}^N\frac{\mu_l}{\mu_l+\lambda_l}}\tag{7}\\ &=\mu_{N^{\star}} \end{align*}

Induction step: $n=N$

We have \begin{align*} \mu_{N^{\star},N+1}&=\left(\lambda_{N^{\star}}+\lambda_{N+1}\right) \frac{\frac{\mu_{N^{\star}}}{\mu_{N^{\star}}+\lambda_{N^{\star}}}\,\frac{\mu_{N+1}}{\mu_{N+1}+\lambda_{N+1}}} {1-\frac{\mu_{N^{\star}}}{\mu_{N^{\star}}+\lambda_{N^{\star}}}\,\frac{\mu_{N+1}}{\mu_{N+1}+\lambda_{N+1}}}\tag{8}\\ \end{align*}

Since according to the induction hypothesis \begin{align*} \color{blue}{\frac{\mu_{N^{\star}}}{\mu_{N^{\star}}+\lambda_{N^{\star}}}} &=\frac{\left(\sum_{j=1}^N\lambda_j\right)\, \frac{\prod_{k=1}^N\frac{\mu_k}{\mu_k+\lambda_k}}{1-\prod_{l=1}^N\frac{\mu_l}{\mu_l+\lambda_l}}} {\left(\sum_{j=1}^N\lambda_j\right)\, \frac{\prod_{k=1}^N\frac{\mu_k}{\mu_k+\lambda_k}}{1-\prod_{l=1}^N\frac{\mu_l}{\mu_l+\lambda_l}}+\left(\sum_{j=1}^N\lambda_j\right)}\\ &=\frac{\frac{\prod_{k=1}^N\frac{\mu_k}{\mu_k+\lambda_k}}{1-\prod_{l=1}^N\frac{\mu_l}{\mu_l+\lambda_l}}} {\frac{\prod_{k=1}^N\frac{\mu_k}{\mu_k+\lambda_k}}{1-\prod_{l=1}^N\frac{\mu_l}{\mu_l+\lambda_l}}+1}\\ &=\frac{\prod_{k=1}^N\frac{\mu_k}{\mu_k+\lambda_k}} {\prod_{k=1}^N\frac{\mu_k}{\mu_k+\lambda_k}+\left(1-\prod_{k=1}^N\frac{\mu_k}{\mu_k+\lambda_k}\right)}\\ &\,\,\color{blue}{=\prod_{k=1}^N\frac{\mu_k}{\mu_k+\lambda_k}}\tag{9} \end{align*} we obtain by putting (9) into (8)

\begin{align*} \color{blue}{\mu_{N^{\star},N+1}}&=\left(\lambda_{N^{\star}}+\lambda_{N+1}\right) \frac{\frac{\mu_{N^{\star}}}{\mu_{N^{\star}}+\lambda_{N^{\star}}}\,\frac{\mu_{N+1}}{\mu_{N+1}+\lambda_{N+1}}} {1-\frac{\mu_{N^{\star}}}{\mu_{N^{\star}}+\lambda_{N^{\star}}}\,\frac{\mu_{N+1}}{\mu_{N+1}+\lambda_{N+1}}}\\ &=\left(\sum_{j=1}^{N+1}\lambda_j\right) \frac{\prod_{k=1}^{N+1}\frac{\mu_k}{\mu_k+\lambda_k}} {1-\prod_{k=1}^{N+1}\frac{\mu_k}{\mu_k+\lambda_k}}\\ &\,\,\color{blue}{=\mu_{(N+1)^{\star}}} \end{align*} and the claim follows.

Best Answer

Related Solutions

Two components with different probabilities of failure.

MTTR and MTBF formulas for series system must be associative

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