#### 期刊菜单

Reliability Model of Three Parts Series Repairable System
DOI: 10.12677/AAM.2019.81009, PDF, HTML, XML, 下载: 1,103  浏览: 1,414  科研立项经费支持

Abstract: This paper studies the components that do not obey the exponential distribution, establishes the series repairable system model of three different types of components by supplementary variable method, and solves the instantaneous availability and steady availability by Laplace transform. The results of single component and double components are verified, and extended to n components series repairable systems and verified with special cases. Hope to be able to solve the relevant calculation of different component systems in real life.

1. 引言

2. 系统假设与状态描述

2.1. 系统假设

Figure 1. Logic block diagram of series system of three different types of components

2.2. 系统状态

$S\left(t\right)$ 表示系统状态，

Figure 2. Three-component state transition diagram

3. 模型建立

t为系统开始工作到目前状态的时间，显然 $t\ge 0$ ；记 ${x}_{i}\left(t\right)$ 表示 $S\left(t\right)=i$ 时，立即开始维修且持续维修时间为x， $0\le x\le t$ 。过程 $\left\{S\left(t\right),x\left(t\right);t\ge 0\right\}$ 是一个连续时间下的广义马尔科夫过程 [1] 。由于指数分布的无记忆性，可知t时刻前的过程与t时刻后的过程无关。令 ${P}_{i}\left(t,x\right)$ 表示t时刻系统因第i个部件发生故障而使系统驻留的时间为x的概率。记 ${G}_{i}\left(x\right)=1-{\text{e}}^{-{\mu }_{i}\left(x\right)t}$ 为部件i的维修度， ${g}_{i}\left(x\right)={{G}^{\prime }}_{i}\left(x\right)$ 为维修度的概率密度函数。

${P}_{0}\left(t+\Delta t\right)={P}_{0}\left(t\right)\left(1-{\lambda }_{1}\Delta t-{\lambda }_{2}\Delta t-{\lambda }_{3}\Delta t\right)+\underset{i=1}{\overset{3}{\sum }}{\int }_{0}^{t}{\mu }_{i}\left(x\right){P}_{i}\left(t,x\right)\text{d}x\cdot \Delta t+ο\left(\Delta t\right)$

$\frac{\text{d}}{\text{d}t}{P}_{0}\left(t\right)=\left(-{\lambda }_{1}-{\lambda }_{2}-{\lambda }_{3}\right){P}_{0}\left(t\right)+\underset{i=1}{\overset{3}{\sum }}{\int }_{0}^{t}{P}_{i}\left(t,x\right){\mu }_{i}\left(x\right)\text{d}x$ (1)

$\begin{array}{l}{P}_{i}\left(t+\Delta t,x+\Delta x\right)-{P}_{i}\left(t,x\right)=-{\mu }_{i}\left(x\right)\Delta x{P}_{i}\left(t,x\right)+ο\left(\rho \right)\\ ⇒\frac{\partial }{\partial t}{P}_{i}\left(t,x\right)+\frac{\partial }{\partial x}{P}_{i}\left(t,x\right)=-{\mu }_{i}\left(x\right){P}_{i}\left(t,x\right)\end{array}$ (2)

$\rho =\sqrt{{\left(\Delta x\right)}^{2}+{\left(\Delta t\right)}^{2}}=\sqrt{2}\cdot \Delta t$ , $\Delta x=x\left(t+\Delta t\right)-x\left(t\right)=\Delta t$ .

${P}_{0}\left(0\right)=1;{P}_{i}\left(0,x\right)=0,i=1,2,3\left(x\ge 0\right)$ (3)

${P}_{i}\left(t,0\right)={\lambda }_{i}{P}_{0}\left(t\right),i=1,2,3$ (4)

$x\ge t$${P}_{i}\left(t,x\right)=0$ ，有 ${\int }_{0}^{t}{P}_{i}\left(t,x\right){\mu }_{i}\left(x\right)\text{d}x={\int }_{0}^{+\infty }{P}_{i}\left(t,x\right){\mu }_{i}\left(x\right)\text{d}x$

$\left\{\begin{array}{l}\frac{\text{d}}{\text{d}t}{P}_{0}\left(t\right)=-\left({\lambda }_{1}+{\lambda }_{2}+{\lambda }_{3}\right){P}_{0}\left(t\right)+\underset{i=1}{\overset{3}{\sum }}{\int }_{0}^{+\infty }{P}_{i}\left(t,x\right){\mu }_{i}\left(x\right)\text{d}x\\ \frac{\partial }{\partial t}{P}_{1}\left(t,x\right)+\frac{\partial }{\partial x}{P}_{1}\left(t,x\right)=-{\mu }_{1}\left(x\right){P}_{1}\left(t,x\right)\\ \frac{\partial }{\partial t}{P}_{2}\left(t,x\right)+\frac{\partial }{\partial x}{P}_{2}\left(t,x\right)=-{\mu }_{2}\left(x\right){P}_{2}\left(t,x\right)\\ \frac{\partial }{\partial t}{P}_{3}\left(t,x\right)+\frac{\partial }{\partial x}{P}_{3}\left(t,x\right)=-{\mu }_{3}\left(x\right){P}_{3}\left(t,x\right)\\ {P}_{0}\left(0\right)=1;\text{\hspace{0.17em}}{P}_{1}\left(0,x\right)={P}_{2}\left(0,x\right)={P}_{3}\left(0,x\right)=0\\ {P}_{1}\left(t,0\right)={\lambda }_{1}{P}_{0}\left(t\right),{P}_{2}\left(t,0\right)={\lambda }_{2}{P}_{0}\left(t\right),{P}_{3}\left(t,0\right)={\lambda }_{3}{P}_{0}\left(t\right)\end{array}$ (5)

4. 模型求解

$L\left[{P}_{i}\left(t,x\right)\right]={P}_{i}^{\text{*}}\left(s,x\right),i=1,2,3$ ，得：

$s{P}_{\text{0}}^{\text{*}}\left(s\right)-1=-\left({\lambda }_{1}+{\lambda }_{2}+{\lambda }_{3}\right){P}_{0}^{*}\left(s\right)+\underset{i=1}{\overset{3}{\sum }}{\int }_{0}^{+\infty }{P}_{i}^{*}\left(s,x\right){\mu }_{i}\left(x\right)\text{d}x$ (6)

$s{P}_{1}^{*}\left(s,x\right)+\frac{\partial }{\partial x}{P}_{1}^{*}\left(s,x\right)=-{\mu }_{1}\left(x\right){P}_{1}^{*}\left(s,x\right)$ (7)

$s{P}_{2}^{*}\left(s,x\right)+\frac{\partial }{\partial x}{P}_{2}^{*}\left(s,x\right)=-{\mu }_{2}\left(x\right){P}_{2}^{*}\left(s,x\right)$ (8)

$s{P}_{3}^{*}\left(s,x\right)+\frac{\partial }{\partial x}{P}_{3}^{*}\left(s,x\right)=-{\mu }_{3}\left(x\right){P}_{3}^{*}\left(s,x\right)$ (9)

${P}_{0}^{*}\left(0\right)=\frac{1}{s};\text{\hspace{0.17em}}{P}_{1}^{*}\left(0,x\right)={P}_{2}^{\text{*}}\left(0,x\right)={P}_{\text{3}}^{\text{*}}\left(0,x\right)=0$ (10)

${P}_{1}^{*}\left(s,0\right)={\lambda }_{1}{P}_{0}^{*}\left(s\right),{P}_{2}^{*}\left(s,0\right)={\lambda }_{2}{P}_{0}^{*}\left(s\right),{P}_{3}^{*}\left(s,0\right)={\lambda }_{3}{P}_{0}^{*}\left(s\right)$ (11)

$-sx+\mathrm{ln}|1-{G}_{1}\left(x\right)|=\mathrm{ln}|{P}_{1}^{*}\left(s,x\right)|-\mathrm{ln}|{P}_{1}^{*}\left(s,0\right)|⇒{P}_{1}^{*}\left(s,x\right)={\text{e}}^{-sx}\left(1-{G}_{1}\left(x\right)\right){P}_{1}^{*}\left(s,0\right)$

${P}_{1}^{*}\left(s,x\right)={\text{e}}^{-sx}\left(1-{G}_{1}\left(x\right)\right){\lambda }_{1}{P}_{0}^{*}\left(s\right)$ (12)

${P}_{2}^{*}\left(s,x\right)={\text{e}}^{-sx}\left(1-{G}_{2}\left(x\right)\right){\lambda }_{2}{P}_{0}^{*}\left(s\right)$ (13)

${P}_{3}^{*}\left(s,x\right)={\text{e}}^{-sx}\left(1-{G}_{3}\left(x\right)\right){\lambda }_{3}{P}_{0}^{*}\left(s\right)$ (14)

$s{P}_{\text{0}}^{\text{*}}\left(s\right)-1=-\left({\lambda }_{1}+{\lambda }_{2}+{\lambda }_{3}\right){P}_{0}^{*}\left(s\right)+\underset{i=1}{\overset{3}{\sum }}{\int }_{0}^{\infty }{\text{e}}^{-sx}{\lambda }_{i}{P}_{0}^{*}\left(s\right){g}_{i}\left(x\right)\text{d}x$ (15)

$s{P}_{\text{0}}^{\text{*}}\left(s\right)-1=-\left({\lambda }_{1}+{\lambda }_{2}+{\lambda }_{3}\right){P}_{0}^{*}\left(s\right)+\underset{i=1}{\overset{3}{\sum }}{\lambda }_{i}{P}_{0}^{*}\left(s\right){g}_{i}^{*}\left(s\right)$

${P}_{\text{0}}^{\text{*}}\left(s\right)=\frac{1}{s+{\lambda }_{1}+{\lambda }_{2}+{\lambda }_{3}-\underset{i=1}{\overset{3}{\sum }}{\lambda }_{i}{g}_{i}^{*}\left(s\right)}$ (16)

${P}_{1}^{*}\left(s,x\right)=\frac{{\text{e}}^{-sx}\left(1-{G}_{1}\left(x\right)\right){\lambda }_{1}}{s+{\lambda }_{1}+{\lambda }_{2}+{\lambda }_{3}-\underset{i=1}{\overset{3}{\sum }}{\lambda }_{i}{g}_{i}^{*}\left(s\right)}$ (17)

${P}_{2}^{*}\left(s,x\right)$${P}_{3}^{*}\left(s,x\right)$ 也可类似得出。

$\left\{\begin{array}{l}{P}_{\text{0}}^{\text{*}}\left(s\right)=\frac{1}{s+{\lambda }_{1}+{\lambda }_{2}+{\lambda }_{3}-\underset{i=1}{\overset{3}{\sum }}{\lambda }_{i}{g}_{i}^{*}\left(s\right)}\\ {P}_{1}^{*}\left(s,x\right)=\frac{{\text{e}}^{-sx}\left(1-{G}_{1}\left(x\right)\right){\lambda }_{1}}{s+{\lambda }_{1}+{\lambda }_{2}+{\lambda }_{3}-\underset{i=1}{\overset{3}{\sum }}{\lambda }_{i}{g}_{i}^{*}\left(s\right)}\\ {P}_{2}^{*}\left(s,x\right)=\frac{{\text{e}}^{-sx}\left(1-{G}_{2}\left(x\right)\right){\lambda }_{2}}{s+{\lambda }_{1}+{\lambda }_{2}+{\lambda }_{3}-\underset{i=1}{\overset{3}{\sum }}{\lambda }_{i}{g}_{i}^{*}\left(s\right)}\\ {P}_{3}^{*}\left(s,x\right)=\frac{{\text{e}}^{-sx}\left(1-{G}_{3}\left(x\right)\right){\lambda }_{3}}{s+{\lambda }_{1}+{\lambda }_{2}+{\lambda }_{3}-\underset{i=1}{\overset{3}{\sum }}{\lambda }_{i}{g}_{i}^{*}\left(s\right)}\end{array}$ (18)

${A}^{*}\left(s\right)={P}_{0}^{*}\left(s\right)=\frac{1}{s+{\lambda }_{1}+{\lambda }_{2}+{\lambda }_{3}-\underset{i=1}{\overset{3}{\sum }}{\lambda }_{i}{g}_{i}^{*}\left(s\right)}$

$A=\frac{1}{1+\underset{i=1}{\overset{3}{\sum }}\frac{{\lambda }_{i}}{{\mu }_{i}}}$

5. 验证与推广

$\left\{\begin{array}{l}\frac{\text{d}}{\text{d}t}{P}_{0}\left(t\right)=-{\lambda }_{1}{P}_{0}\left(t\right)+{\int }_{0}^{t}{P}_{\text{1}}\left(t,x\right){\mu }_{\text{1}}\left(x\right)\text{d}x\\ \frac{\partial }{\partial t}{P}_{1}\left(t,x\right)+\frac{\partial }{\partial x}{P}_{1}\left(t,x\right)=-{\mu }_{1}\left(x\right){P}_{1}\left(t,x\right)\\ {P}_{0}\left(0\right)=1;\text{\hspace{0.17em}}{P}_{1}\left(0,x\right)=0\\ {P}_{1}\left(t,0\right)={\lambda }_{1}{P}_{0}\left(t\right)\end{array}$

${A}^{*}\left(s\right)={P}_{0}^{*}\left(s\right)=\frac{1}{s+{\lambda }_{1}-{\lambda }_{1}{g}_{1}^{*}\left(s\right)}$

$A=\frac{1}{1+\frac{{\lambda }_{1}}{{\mu }_{1}}}$

${\lambda }_{1}=\lambda$${\mu }_{1}=\mu$ 时，与参考文献 [6] 推论1所得出的稳态可用度结果相同。

$\left\{\begin{array}{l}\frac{\text{d}}{\text{d}t}{P}_{0}\left(t\right)=-\left({\lambda }_{1}+{\lambda }_{2}\right){P}_{0}\left(t\right)+\underset{i=1}{\overset{2}{\sum }}{\int }_{0}^{t}{P}_{i}\left(t,x\right){\mu }_{i}\left(x\right)\text{d}x\\ \frac{\partial }{\partial t}{P}_{i}\left(t,x\right)+\frac{\partial }{\partial x}{P}_{i}\left(t,x\right)=-{\mu }_{i}\left(x\right){P}_{i}\left(t,x\right),i=1,2\\ {P}_{0}\left(0\right)=1;\text{\hspace{0.17em}}{P}_{1}\left(0,x\right)={P}_{2}\left(0,x\right)=0\\ {P}_{1}\left(t,0\right)={\lambda }_{1}{P}_{0}\left(t\right),{P}_{2}\left(t,0\right)={\lambda }_{2}{P}_{0}\left(t\right)\end{array}$

${A}^{*}\left(s\right)={P}_{0}^{*}\left(s\right)=\frac{1}{s+{\lambda }_{1}+{\lambda }_{2}-\underset{i=1}{\overset{2}{\sum }}{\lambda }_{i}{g}_{i}^{*}\left(s\right)}$

$A=\frac{1}{1+\underset{i=1}{\overset{2}{\sum }}\frac{{\lambda }_{i}}{{\mu }_{i}}}$

${\lambda }_{1}={\lambda }_{2}=\lambda$${\mu }_{1}={\mu }_{2}=\mu$ 时与参考文献 [8] 所得出的稳态可用度结果相同。

$\left\{\begin{array}{l}\frac{\text{d}}{\text{d}t}{P}_{0}\left(t\right)=-\left({\lambda }_{1}+{\lambda }_{2}+\cdots +{\lambda }_{n}\right){P}_{0}\left(t\right)+\underset{i=1}{\overset{n}{\sum }}{\int }_{0}^{t}{P}_{i}\left(t,x\right){\mu }_{i}\left(x\right)\text{d}x\\ \frac{\partial }{\partial t}{P}_{i}\left(t,x\right)+\frac{\partial }{\partial x}{P}_{i}\left(t,x\right)=-{\mu }_{i}\left(x\right){P}_{i}\left(t,x\right),i=1,2,\cdots ,n\\ {P}_{0}\left(0\right)=1;\text{\hspace{0.17em}}{P}_{1}\left(0,x\right)={P}_{2}\left(0,x\right)=\cdots ={P}_{n}\left(0,x\right)\\ {P}_{i}\left(t,0\right)={\lambda }_{i}{P}_{0}\left(t\right),i=1,2,\cdots ,n\end{array}$

${A}^{*}\left(s\right)={P}_{0}^{*}\left(s\right)=\frac{1}{s+{\lambda }_{1}+{\lambda }_{2}+\cdots +{\lambda }_{n}-\underset{i=1}{\overset{n}{\sum }}{\lambda }_{i}{g}_{i}^{*}\left(s\right)}$

$A=\frac{1}{1+\underset{i=1}{\overset{n}{\sum }}\frac{{\lambda }_{i}}{{\mu }_{i}}}$

${\lambda }_{1}={\lambda }_{2}=\cdots ={\lambda }_{n}=\lambda$${\mu }_{1}={\mu }_{2}=\cdots ={\mu }_{n}=\mu$ 时，其稳态可用度为

$A=\frac{\mu }{\mu +n\lambda }$

6. 小结

 [1] 曹晋华, 程侃. 可靠性数学引论[M]. 北京: 高等教育出版社, 2006. [2] 张欣欣, 孟宪云, 陈胜强, 等. 修理工单重休假且两部件相依一部件温贮备的三部件退化可修系统[J]. 黑龙江大学自然科学学报, 2014, 31(4): 425-436. [3] 陈燕, 朱翼隽, 陈佩树. 单重休假且有优先修理权的三部件系统[J]. 江苏大学学报(自然科学版), 2014, 35(6): 738-744. [4] 胡林敏, 田瑞玲, 吴俊波, 等. 具有休假的三部件串并联可修系统的可靠性分析[J]. 燕山大学学报, 2007, 31(4): 299-303. [5] 刘东旭, 司文艺, 袁玉娇. 一类单部件可修复系统的稳定性即可靠性分析[J]. 延边大学(自然科学版), 2014(40): 15-19. [6] 唐应辉, 刘晓云. 一种新型的单部件可修系统[J]. 系统工程理论与实践, 2003, 23(7): 106-111. [7] 陶有德, 乔兴, 于景元, 朱广田. 一类可修复计算机系统的数学模型[J]. 信阳师范学院学报(自然科学版), 2011(3): 330-332. [8] 陶有德, 于景元, 朱广田. 一类可修复计算机系统的稳定性和可靠性[J]. 信阳师范学院学报(自然科学版), 2011(1): 18-21. [9] 宋保维. 系统可靠性设计与分析[M]. 西安: 西北工业大学出版社, 2000.