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

doi:10.21629/JSEE.2019.06.20

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

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.

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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	$N(2.38, 0.02), \widehat{\alpha}_1=0.19$	$N(2.75, 0.01), \widehat{\alpha}_2=0.07$	$N(2.97, 0.02), \widehat{\alpha}_3=0.30$	$N(3.49, 0.03), \widehat{\alpha}_4=0.44$