Optimization of Two-Echelon Configuration of Maintenance Supply for Synthetic Forces based on Task Evolution Importance

doi:10.12382/bgxb.2021.0701

Abstract

Abstract:

Maintenance supply is an important material basis for the implementation of maintenance support, which directly affects the readiness of arms. In this paper, we analyze the operating mechanism of the two-echelon maintenance support system of the synthetic force and establish an importance analysis model based on task evolution structure. Then, the spare parts support degree is deemed as the objective to shorten the delay time and improve the availability, and a two-echelon dynamic configuration optimization model of equipment with multiple constraints on equipment volume, quality, importance, and cost is established. The Lagrange operator is introduced to convert multiple constraints into composite single factor constraints, and a dynamic update mechanism of constraint factors is established to meet the commander's different weight factor judgment rules timely. Finally, an improved margin analysis method based on MA-OD target decomposition is proposed. The numerical analysis shows that the proposed model provides effective support for the commander of the synthetic force to make the military equipment deployment plan in both peacetime and wartime, which is of great significance to improving the readiness.

Key words: maintenance supply, task evolution importance, multiple constraints, synthetci force, configuration optimization

CLC Number:

WU Weiyi, JIA Yunxian, SHI Xianming, LIU Bin, LIU Jie, YIN Shizhuang, ZHU Xi. Optimization of Two-Echelon Configuration of Maintenance Supply for Synthetic Forces based on Task Evolution Importance[J]. Acta Armamentarii, 2023, 44(2): 591-604.

Figures/Tables 11

References 24

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参数	装备1			装备2			装备3
体系贡献率	0.3			0.3			0.4
装备数量	3			3			3
部件编号	1	2	3	4	5	6	7	8	9	10
需求率/10^-2h^-1	0.269	0.0357	0.091	0.175	0.05	0.0625	0.194	0.0555	0.0833	0.0769
单装用数/个	1	1	1	2	1	2	1	2	1	2
本级修理率r_j	0	0.1	0.1	0	0.1	0	0.1	0	0.1	0
单价/万元	11.34	3.88	1.93	7.88	1.12	0.97	5.34	0.67	0.88	1.23
体积/m	5.6689	0.7719	0.3735	1.8011	0.3154	0.5478	0.3818	0.6972	0.2324	1.2533
质量/10²kg	6.0214	0.4998	1.5232	4.2126	0.4284	1.0948	1.8088	0.2142	0.5474	3.0464
报废率φ_j	0.1	0.2	0.2	0.1	0.2	0.1	0.2	0.1	0.1	0.1
器材运输工具载荷性能	5t通用运输车、额定车厢装载体积30m³、额定车厢承载质量5000kg
备件保障费用要求	有限费用下合理分配

参数	装备1			装备2			装备3
体系贡献率	0.3			0.3			0.4
装备数量	3			3			3
部件编号	1	2	3	4	5	6	7	8	9	10
需求率/10^-2h^-1	0.269	0.0357	0.091	0.175	0.05	0.0625	0.194	0.0555	0.0833	0.0769
单装用数/个	1	1	1	2	1	2	1	2	1	2
本级修理率r_j	0	0.1	0.1	0	0.1	0	0.1	0	0.1	0
单价/万元	11.34	3.88	1.93	7.88	1.12	0.97	5.34	0.67	0.88	1.23
体积/m	5.6689	0.7719	0.3735	1.8011	0.3154	0.5478	0.3818	0.6972	0.2324	1.2533
质量/10²kg	6.0214	0.4998	1.5232	4.2126	0.4284	1.0948	1.8088	0.2142	0.5474	3.0464
报废率φ_j	0.1	0.2	0.2	0.1	0.2	0.1	0.2	0.1	0.1	0.1
器材运输工具载荷性能	5t通用运输车、额定车厢装载体积30m³、额定车厢承载质量5000kg
备件保障费用要求	有限费用下合理分配

是否考虑体系贡献率	重要度	装备1			装备2			装备3
是否考虑体系贡献率	重要度	部件1	部件2	部件3	部件4	部件5	部件6	部件7	部件8	部件9	部件10
	重要度打分	0.584	0.9311	0.834	0.084	0.2606	0.9719	0.8753	0.6636	0.0865	0.0932
不考虑体系贡献率	重要度排序	6	2	4	10	7	1	3	5	9	8
	重要度归一	0.9584	0.9931	0.9834	0.9084	0.9261	0.9972	0.9875	0.9663	0.9086	0.9093
	重要度打分	0.1752	0.2793	0.2502	0.0252	0.0782	0.2915	0.3501	0.2654	0.0346	0.0372
考虑体系贡献率	重要度排序	6	3	5	10	7	2	1	4	9	8
	重要度归一	0.9175	0.9279	0.9250	0.9025	0.9078	0.9291	0.9350	0.9265	0.9034	0.9037

是否考虑体系贡献率	重要度	装备1			装备2			装备3
是否考虑体系贡献率	重要度	部件1	部件2	部件3	部件4	部件5	部件6	部件7	部件8	部件9	部件10
	重要度打分	0.584	0.9311	0.834	0.084	0.2606	0.9719	0.8753	0.6636	0.0865	0.0932
不考虑体系贡献率	重要度排序	6	2	4	10	7	1	3	5	9	8
	重要度归一	0.9584	0.9931	0.9834	0.9084	0.9261	0.9972	0.9875	0.9663	0.9086	0.9093
	重要度打分	0.1752	0.2793	0.2502	0.0252	0.0782	0.2915	0.3501	0.2654	0.0346	0.0372
考虑体系贡献率	重要度排序	6	3	5	10	7	2	1	4	9	8
	重要度归一	0.9175	0.9279	0.9250	0.9025	0.9078	0.9291	0.9350	0.9265	0.9034	0.9037

多维分析	单一约束	同一目标—综合约束目标:使用可用度		不同目标—综合约束目标:费用
部件编号	方案1 (体积) (S_j,S_o_,j)	方案2 (综合约束) (S_j,S_o_,j)	方案3 (考虑重要度) (S_j,S_o_,j)	方案4 (综合因素) (S_j,S_o_,j)	方案5 (考虑重要度) (S_j,S_o_,j)
1	3(1,2)	3(1,2)	3(1,2)	3(1,2)	3(1,2)
2	1(1,0)	1(1,0)	2(1,1)	0(0,0)	1(1,0)
3	2(2,0)	1(1,0)	1(1,0)	1(1,0)	1(1,0)
4	4(2,2)	4(2,2)	3(1,2)	4(1,3)	4(1,3)
5	1(1,0)	1(1,0)	1(1,0)	1(1,0)	0(0,0)
6	1(0,1)	1(0,1)	2(1,1)	2(1,1)	2(1,1)
7	5(2,3)	4(2,2)	5(2,3)	4(2,2)	4(2,2)
8	0(0,0)	1(0,1)	1(0,1)	0(0,0)	0(0,0)
9	0(0,0)	1(1,0)	1(1,0)	1(1,0)	1(1,0)
10	1(0,1)	1(0,1)	1(0,1)	1(0,1)	1(0,1)