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On page $614$ of "Introduction to probability models" by Sheldon Ross ($9$th ed.), he describes the MTTR of a series system as a function of the failure rates ($\lambda_i$) and repair rates ($\mu_i$) of the individual components. The expression is:

$$\bar{D} = \frac{1}{\bar{\mu}} = \frac{1-\prod \left(\frac{\mu_i}{\mu_i+\lambda_i}\right)}{\prod \left(\frac{\mu_i}{\mu_i+\lambda_i}\right)} \frac{1}{\sum \lambda_i}$$

and for the MTBF ($\bar{U}$):

$$\bar{U} = \frac{1}{\bar{\lambda}} = \frac{1}{\sum \lambda_i}$$

Now, this formula should satisfy the associative property. For example, if you consider the first two components as a series system and find their combined failure and repair rates; $\bar{\lambda}$ and $\bar{\mu}$ and then consider this as a single component which is in series with the third component, finding the overall failure and repair rates; we should get the same answer as if we plugged all three into the formulas together.

More generally, all the different ways of combining the components into intermediate systems before getting to the final system (depicted in the figure below for three components) should yield the same result.

enter image description here

And this should hold for any number of components, with any number of ways of combining the components at various layers before coming up with the overall numbers (any way of taking the components as leaves and combining them into a tree that can have any number of leaves). I verified this for some particular cases, but not sure how to go about proving this in general.

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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.

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  • $\begingroup$ @RohitPandey: Many thanks for accepting my answer and granting the bounty. :-) $\endgroup$ Commented Mar 27, 2021 at 22:51

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