基于马尔科夫决策过程的带缓存双机系统不完美维护策略

doi:10.16183/j.cnki.jsjtu.2019.270

摘要/Abstract

摘要：

为有效解决退化系统的设备维护优化问题,对带缓存双机系统的预防性维护策略进行研究.首先,考虑两设备的随机退化,针对不完美的小修和预防性维护,通过马尔科夫过程描述系统状态.然后,基于收入和成本,建立生产系统的利润模型.最后,通过最大化长期期望利润,确定不同系统状态下的预防性维护决策.采用值迭代方法对模型进行求解,数值分析结果表明,设备的维护决策不仅取决于自身状态,还受其他设备状态和缓存量的影响.

关键词: 不完美维护, 缓存, 双机系统, 马尔科夫决策过程

Abstract:

To solve the problem of machine maintenance optimization of degraded systems, the preventive maintenance policy of a two-machine one-buffer production system is studied. First, random degradations of both machines are considered. For imperfect minimal repair and preventive maintenance, a Markov process is used to describe the system state. Then, a profit model for the production system is established based on cost and income. Finally, preventive maintenance decisions in different system states are determined by maximizing long-term expected profits. The model is solved by value iteration, and numerical analysis results show that the maintenance decision of the machine depends not only on its own state, but also on the state of other machines and the amount of the buffer.

Key words: imperfect maintenance, buffer, two-machine system, Markov decision process

中图分类号:

TH17

田雪雁, 王孟雅, 潘尔顺. 基于马尔科夫决策过程的带缓存双机系统不完美维护策略[J]. 上海交通大学学报, 2021, 55(4): 480-488.

TIAN Xueyan, WANG Mengya, PAN Ershun. Imperfect Maintenance Policy for a Two-Machine One-Buffer System Based on Markov Decision Process[J]. Journal of Shanghai Jiao Tong University, 2021, 55(4): 480-488.

图/表 10

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M2状态		κ=0	κ=1	κ=2	κ=3	κ=4	κ=5
运行	1	(-,-,-)	(-,-,-)	(6,5,5)	(6,5,5)	(4,3,3)	(3,3,3)
	2	(-,-,-)	(-,-,-)	(6,5,5)	(6,5,5)	(4,3,3)	(3,3,3)
	3	(-,-,-)	(-,-,-)	(6,5,4)	(6,5,4)	(3,2,2)	(3,3,3)
	4	(-,-,6)	(-,-,6)	(5,5,4)	(5,4,3)	(3,2,2)	(3,3,3)
	5	(-,5)	(-,5)	(5,3)	(5,3)	(3,4)	(3,4)
故障小修	7	(6,6,6)	(6,5,6)	(4,3,3)	(3,3,3)	(3,4,3)	(3,4,4)
	8	(6,5,5)	(6,5,5)	(4,3,3)	(3,3,3)	(3,4,3)	(3,4,4)
	9	(6,5,5)	(6,5,5)	(4,3,3)	(3,3,3)	(3,4,4)	(3,4,4)
	10	(6,5,4)	(6,5,3)	(3,2,2)	(3,3,3)	(3,4,4)	(4,4,4)
	11	(5,3)	(5,2)	(3,2)	(3,3)	(3,3)	(4,4)
预防性维护	13	(6,6,6)	(6,6,6)	(5,3,3)	(3,3,3)	(3,4,3)	(3,4,4)
	14	(6,6,6)	(6,6,6)	(4,3,3)	(3,3,3)	(3,4,3)	(3,4,4)
	15	(6,6,6)	(6,6,6)	(4,3,3)	(3,3,3)	(3,4,3)	(3,4,4)
	16	(6,6,5)	(6,5,5)	(4,3,3)	(3,3,3)	(3,4,4)	(3,4,4)
	17	(6,5,5)	(6,5,5)	(4,3,3)	(3,3,3)	(3,4,4)	(3,4,4)
	18	(6,5,5)	(6,5,5)	(4,3,3)	(3,3,3)	(3,4,3)	(3,4,4)

M1状态		κ=0	κ=1	κ=2	κ=3	κ=4	κ=5
运行	1	(6,5,6)	(6,5,6)	(6,5,6)	(6,5,6)	(6,5,6)	(6,5,6)
	2	(6,5,6)	(6,5,6)	(6,5,6)	(6,5,6)	(6,5,6)	(6,5,6)
	3	(6,5,6)	(6,5,6)	(6,5,6)	(6,5,6)	(6,5,6)	(6,5,6)
	4	(6,5,6)	(6,5,6)	(6,5,6)	(6,5,6)	(6,5,6)	(6,5,6)
	5	(6,5,6)	(6,5,6)	(6,5,6)	(6,5,6)	(6,5,6)	(6,5,6)
	6	(6,4,6)	(6,4,6)	(6,5,6)	(6,5,6)	(6,5,6)	(6,5,6)
故障小修	7	(1,1,1)	(5,2,5)	(6,5,6)	(6,5,6)	(6,5,6)	(6,5,6)
	8	(1,1,1)	(5,3,5)	(6,5,6)	(6,5,6)	(6,5,6)	(6,5,6)
	9	(1,1,1)	(5,3,5)	(6,5,6)	(6,5,6)	(6,5,6)	(6,5,6)
	10	(1,1,1)	(5,3,5)	(6,5,6)	(6,5,6)	(6,5,6)	(6,5,6)
	11	(1,1,1)	(5,3,5)	(6,5,6)	(6,5,6)	(6,5,6)	(6,5,6)
	12	(1,1,1)	(5,3,5)	(6,5,6)	(6,5,6)	(6,5,6)	(6,5,6)
预防性维护	13	(1,1,1)	(4,2,4)	(6,5,6)	(6,5,6)	(6,5,6)	(6,5,6)
	14	(1,1,1)	(4,2,4)	(6,5,6)	(6,5,6)	(6,5,6)	(6,5,6)
	15	(1,1,1)	(4,2,4)	(6,5,6)	(6,5,6)	(6,5,6)	(6,5,6)
	16	(1,1,1)	(4,2,4)	(6,5,6)	(6,5,6)	(6,5,6)	(6,5,6)
	17	(1,1,1)	(4,2,4)	(6,5,6)	(6,5,6)	(6,5,6)	(6,5,6)
	18	(1,1,1)	(4,2,4)	(6,5,6)	(6,5,6)	(6,5,6)	(6,5,6)