Clustering-based maintainability demonstration for complex systems with a mixed maintenance time distribution

doi:10.21629/JSEE.2019.06.20

Journal of Systems Engineering and Electronics ›› 2019, Vol. 30 ›› Issue (6): 1260-1271.doi: 10.21629/JSEE.2019.06.20

• • 上一篇

收稿日期:2018-12-26 出版日期:2019-12-20 发布日期:2019-12-25

Clustering-based maintainability demonstration for complex systems with a mixed maintenance time distribution

Zhenya WU*(), Jianping HAO()

Department of Equipment Command and Management, Shijiazhuang Campus of Army Engineering University, Shijiazhuang 050003, China

Received:2018-12-26 Online:2019-12-20 Published:2019-12-25
Contact: Zhenya WU E-mail:wzy9012@163.com;hao001@sina.com
About author:WU Zhenya was born in 1990. He received his B.S. degree in missile engineering from Air Force Engineering University, Xi'an, Shaanxi, China, in 2013, and M.S. degree in equipment management from Air Force Engineering University, in 2016. He is currently pursuing his Ph.D. degree in armament science and technology at Shijiazhuang Campus of Army Engineering University. His current research interests include maintainability demonstration and statistics. E-mail: wzy9012@163.com|HAO Jianping was born in 1969. He is a professor of Army Engineering University. He received his M.S. and Ph.D. degrees from Mechanical Engineering College, Shijiazhuang, China. His research mainly focuses on maintainability and maintenance engineering. Since 2016, he has established a research group and has begun to study demonstration of complex system maintainability and theory of equipment general quality characteristics. E-mail: jianping hao001@sina.com
Supported by:
This work was supported by the National Defense Pre-research Funds(9140A27010215JB34422);This work was supported by the National Defense Pre-research Funds (9140A27010215JB34422)

摘要/Abstract

Abstract:

During maintainability demonstration, the maintenance time for complex systems consisting of mixed technologies generally conforms to a mixture distribution. However existing maintainability standards and guidance do not explain explicitly how to deal with this situation. This paper develops a comprehensive maintainability demonstration method for complex systems with a mixed maintenance time distribution. First of all, a K-means algorithm and an expectation-maximization (EM) algorithm are used to partition the maintenance time data for all possible clusters. The Bayesian information criterion (BIC) is then used to choose the optimal model. After this, the clustering results for equipment are obtained according to their degree of membership. The degree of similarity for the maintainability of different kinds of equipment is then determined using the projection method. By using a Bootstrap method, the prior distribution is obtained from the maintenance time data for the most similar equipment. Then, a test method based on Bayesian theory is outlined for the maintainability demonstration. Finally, the viability of the proposed approach is illustrated by means of an example.

Key words: complex system, maintainability demonstration, mixture distribution model, Bayesian test

. [J]. Journal of Systems Engineering and Electronics, 2019, 30(6): 1260-1271.

Zhenya WU, Jianping HAO. Clustering-based maintainability demonstration for complex systems with a mixed maintenance time distribution[J]. Journal of Systems Engineering and Electronics, 2019, 30(6): 1260-1271.

图/表 12

参考文献 36

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Feature	Number	Maintainability attribute
	1	Accessibility
	2	Disassembly/assembly
	3	Standardization
	4	Simplicity
Design	5	Identification
	6	Diagnosability
	7	Modularization
	8	Tribo-concepts
Personnel	9	Personnel including ergonomics
Personnel	10	System environment
Logistic	11	Tools and test equipment
Support	12	Documentation

Number	A	B	C	D	E	F	G	H	I
1	17(2.83)	35(3.56)	13(2.56)	16(2.77)	31(3.43)	20(3.00)	31(3.43)	19(2.94)	45(3.81)
2	9(2.19)	39(3.66)	21(3.04)	14(2.64)	28(3.33)	11(2.40)	16(2.77)	28(3.33)	47(3.85)
3	29(3.37)	21(3.04)	28(3.33)	15(2.71)	23(3.14)	18(2.89)	30(3.40)	23(3.14)	42(3.74)
4	11(2.40)	13(2.56)	19(2.94)	16(2.77)	33(3.50)	20(3.00)	16(2.77)	25(3.22)	41(3.71)
5	20(3.00)	11(2.40)	18(2.89)	15(2.71)	37(3.61)	8(2.08)	16(2.77)	28(3.33)	36(3.58)
6	10(2.30)	17(2.83)	33(3.50)	12(2.48)	29(3.37)	9(2.20)	37(3.61)	23(3.14)	41(3.71)
7	11(2.40)	19(2.94)	12(2.48)	16(2.77)	38(3.64)	13(2.56)	9(2.20)	25(3.22)	31(3.43)
8	25(3.22)	10(2.30)	19(2.94)	36(3.58)	29(3.37)	10(2.30)	20(3.00)	28(3.33)	32(3.47)
9	10(2.30)	15(2.71)	23(3.14)	19(2.94)	38(3.64)	32(3.47)	19(2.94)	33(3.50)	34(3.53)
10	15(2.71)	-	23(3.14)	-	39(3.66)	33(3.50)	22(3.09)	27(3.30)	34(3.53)
11	-	-	18(2.89)	-	39(3.66)	11(2.40)	-	36(3.58)	34(3.53)
12	-	-	17(2.83)	-	32(3.47)	18(2.89)	-	28(3.33)	-
13	-	-	20(3.00)	-	23(3.14)	-	-	-	-
Mean= 23.62 min(3.07); Variance ≈ 97.61 min²(0.20)

Model	Centroid				Boundary
Model	1	2	3	4	1	2	3	4
k = 2	2.68	3.44	–	–	[2.07 3.05]	[3.09 3.86]	–	–
k = 3	2.38	2.95	3.52	–	[2.07 2.64]	[2.70 3.22]	[3.29 3.86]	–
k = 4	2.36	2.79	3.06	3.52	[2.07 2.57]	[2.63 2.90]	[2.94 3.22]	[3.29 3.86]

Model	Component distribution
Model	1	2	3	4
k = 2	$N(2.68, 0.08), \widehat{\alpha}^0_1=0.49$	$N(3.44, 0.04), \widehat{\alpha}^0_2=0.51$	–	–
k = 3	$N(2.38, 0.02), \widehat{\alpha}^0_1=0.20$	$N(2.95, 0.02), \widehat{\alpha}^0_2=0.40$	$N(3.52, 0.02), \widehat{\alpha}^0_3=0.40$	–
k = 4	$N(2.36, 0.02), \widehat{\alpha}^0_1=0.18$	$N(2.79, 0.01), \widehat{\alpha}^0_2=0.18$	$N(3.06, 0.01), \widehat{\alpha}^0_3=0.24$	$N(3.49, 0.03), {\hat \alpha _4} = 0.44$

Model	Component distribution
Model	1	2	3	4
k = 2	$N(2.84, 0.14), \widehat{\alpha}_1=0.67$	$N(3.53, 0.02), \widehat{\alpha}_2=0.33$	–	–
k = 3	$N(2.34, 0.02), \widehat{\alpha}_1=0.16$	$N(2.91, 0.04), \widehat{\alpha}_2=0.42$	$N(3.50, 0.03), \widehat{\alpha}_3=0.42$	–
k = 4	$N(2.38, 0.02), \widehat{\alpha}_1=0.19$	$N(2.75, 0.01), \widehat{\alpha}_2=0.07$	$N(2.97, 0.02), \widehat{\alpha}_3=0.30$	$N(3.49, 0.03), \widehat{\alpha}_4=0.44$

Clustering-based maintainability demonstration for complex systems with a mixed maintenance time distribution

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参考文献 36

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Number	A	B	C	D	E	F	G	H	I
1	4.91^*	2.38	1.21	0.95	0	5.31^*	1.01	0	0
2	3.67	4.60^*	9.58^*	7.04^*	2.00	4.69	5.98$^{*}$	4.43	0.05
3	1.42	2.02	2.21	1.00	11.00^*	2.00	3.02	7.57^*	10.95^*

Model	k = 2	k = 3	k = 4
BIC	0.580 1	0.564 4^*	0.577 7