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PMHF計算に関する積分公式 (3) |
#223に示した理由により、本稿の議論は全て取り消します。
前稿の続きで、ISO 26262のPMHFの導出の場合、確率積分を実行する際に次の(226.1)が出てくるため、あらかじめ結果を導出しておき、後程積分公式として使用します。
$$
\img[-1.35em]{/images/withinseminar.png}
\tag{226.1}
$$
$t=i\tau+u, i=0,1,...,n-1, n:=\frac{T_\text{lifetime}}{\tau}$とおいて変数変換すれば、
$$
(226.1)=\frac{1}{T_\text{lifetime}}\sum_{i=0}^{n-1}\int_0^\tau(i\tau+u)e^{-\lambda_\text{IF}(i\tau+u)-\lambda_\text{SM}u}du\\
=\tau\sum_{i=0}^{n-1}i e^{-\lambda_\text{IF}i\tau}
\frac{1}{T_\text{lifetime}}\int_0^\tau e^{-(\lambda_\text{IF}+\lambda_\text{SM})u}du
+\sum_{i=0}^{n-1}e^{-\lambda_\text{IF}i\tau}\frac{1}{T_\text{lifetime}}\int_0^\tau ue^{-(\lambda_\text{IF}+\lambda_\text{SM})u}du\\
\tag{226.2}
$$
ここで、(226.2)右辺第1項の級数の和を求めるため和を$x$とおけば、
$$
x:=\sum_{i=0}^{n-1}i e^{-\lambda_\text{IF}i\tau}=e^{-\lambda_\text{IF}\tau}+2e^{-\lambda_\text{IF}2\tau}+...+(n-1)e^{-\lambda_\text{IF}(n-1)\tau}\tag{226.3}
$$
となり、
$$
e^{-\lambda_\text{IF}\tau}x=\sum_{i=0}^{n-1}i e^{-\lambda_\text{IF}(i+1)\tau}=e^{-\lambda_\text{IF}2\tau}+...+(n-2)e^{-\lambda_\text{IF}(n-1)\tau}+(n-1)e^{-\lambda_\text{IF}n\tau}\tag{226.4}
$$
よって、(226.3)-(226.4)より、
$$
x(1- e^{-\lambda_\text{IF}\tau})=e^{-\lambda_\text{IF}\tau}+e^{-\lambda_\text{IF}2\tau}+...+e^{-\lambda_\text{IF}(n-1)\tau}-(n-1)e^{-\lambda_\text{IF}n\tau}\\
=\underbrace{e^{-\lambda_\text{IF}\tau}+e^{-\lambda_\text{IF}2\tau}+...+e^{-\lambda_\text{IF}(n-1)\tau}+e^{-\lambda_\text{IF}n\tau}}_{\text{n terms}}-ne^{-\lambda_\text{IF}n\tau}\\
=e^{-\lambda_\text{IF}\tau}\frac{1-e^{-\lambda_\text{IF}T_\text{lifetime}}}{1-e^{-\lambda_\text{IF}\tau}}-n e^{-\lambda_\text{IF}T_\text{lifetime}}\tag{226.5}
$$
よって、Maclaurin展開の1次近似を用いれば、
$$
\require{cancel}
x\approx\frac{\bcancel{\lambda_\text{IF}}T_\text{lifetime}}{\lambda_\text{IF}^\bcancel{2}\tau^2}(1-\lambda_\text{IF}\tau)-\frac{n(1-\lambda_\text{IF}T_\text{lifetime})}{\lambda_\text{IF}\tau}\\
=\frac{T_\text{lifetime}(\bcancel{1}-\bcancel{\lambda_\text{IF}}\tau)-T_\text{lifetime}(\bcancel{1}-\bcancel{\lambda_\text{IF}}T_\text{lifetime})}{\bcancel{\lambda_\text{IF}}\tau^2}=\frac{T_\text{lifetime}(T_\text{lifetime}-\tau)}{\tau^2}\tag{226.6}
$$
次に、(226.2)右辺第2項の級数の和は、
$$
\sum_{i=0}^{n-1}e^{-\lambda_\text{IF}i\tau}=e^{-\lambda_\text{IF}\tau}+...+e^{-\lambda_\text{IF}(n-1)\tau}=\frac{1-e^{-\lambda_\text{IF}T_\text{lifetime}}}{1-e^{-\lambda_\text{IF}\tau}}
\approx\frac{\bcancel{\lambda_\text{IF}}T_\text{lifetime}}{\bcancel{\lambda_\text{IF}}\tau}
\tag{226.7}
$$
次に、(226.2)右辺第1項の定積分の値は、
$$
\int_0^\tau e^{-(\lambda_\text{IF}+\lambda_\text{SM})u}du
=\left[\frac{e^{-(\lambda_\text{IF}+\lambda_\text{SM})u}}{-(\lambda_\text{IF}+\lambda_\text{SM})}\right]_0^\tau
=\frac{e^{-(\lambda_\text{IF}+\lambda_\text{SM})\tau}-1}{-(\lambda_\text{IF}+\lambda_\text{SM})}\\
\approx\frac{1}{\bcancel{\lambda_\text{IF}+\lambda_\text{SM}}}\left(\bcancel{(\lambda_\text{IF}+\lambda_\text{SM})}\tau-\frac{1}{2}(\lambda_\text{IF}+\lambda_\text{SM})^\bcancel{2}\tau^2\right)
=\tau\left(1-\frac{1}{2}(\lambda_\text{IF}+\lambda_\text{SM})\tau\right)
\tag{226.8}
$$
以上から、$\color{red}{(226.5)}$と$\color{green}{(226.6)}$を(226.2)に適用し、$\color{blue}{(226.7})$と部分積分の結果$\color{purple}{(225.1)}$を用いれば、
$$
(226.2)=\bcancel{\tau}\color{red}{\left(\frac{\bcancel{T_\text{lifetime}}(T_\text{lifetime}-\tau)}{\bcancel{\tau^2}}\right)}\frac{1}{\bcancel{T_\text{lifetime}}}\color{blue}{\bcancel{\tau}\left(1-\frac{1}{2}(\lambda_\text{IF}+\lambda_\text{SM})\tau\right)}\\
+\color{green}{\frac{\bcancel{T_\text{lifetime}}}{\bcancel{\tau}}}\frac{1}{\bcancel{T_\text{lifetime}}}
\color{purple}{\left(\frac{\tau^\bcancel{2}}{2}-
\frac{(\lambda_\text{IF}+\lambda_\text{SM})\tau^{\bcancel{3}2}}{3}\right)}\\
=(T_\text{lifetime}-\tau)\left(1-\frac{1}{2}(\lambda_\text{IF}+\lambda_\text{SM})\tau\right)+\left(\frac{\tau}{2}-\frac{1}{3}(\lambda_\text{IF}+\lambda_\text{SM})\tau^2\right)\\
=\left(1-\frac{1}{2}(\lambda_\text{IF}+\lambda_\text{SM})\tau\right)T_\text{lifetime}-\frac{\tau}{2}+\frac{1}{6}(\lambda_\text{IF}+\lambda_\text{SM})\tau^2
\tag{226.9}
$$
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