Imperfect Maintenance Policy for a Two-Machine One-Buffer System Based on Markov Decision Process

doi:10.16183/j.cnki.jsjtu.2019.270

Abstract

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

CLC Number:

TH17

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.

Figures/Tables 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)