基于三边匹配理论的应急救援分配决策

doi:10.16265/j.cnki.issn1003-3033.2024.03.0625

摘要/Abstract

摘要：

为解决大型地质灾害背景下应急救援初期救援人员、应急救援装备与受灾地点的分配问题,提高应急救援的效率和质量,首先,将应急救援三方主体之间的选择过程描述为三边匹配问题;然后,标准化处理匹配三方主体的多属性偏好评价信息,构造多属性综合偏好得分矩阵,并以系统稳定性和总偏好效用最大化为目标,建立应急救援初期背景下的三边匹配满意稳定模型;最后,通过算例验证模型的可行性和有效性。结果表明:该模型考虑各主体偏好对于资源分配的影响,能实现应急救援背景下救援人员、应急救援装备与受灾地点三方主体的合理匹配,有助于缩短应急救援部署时间,提高应急救援的效率和质量。

关键词: 应急救援, 三边匹配, 分配决策, 偏好值, 满意稳定匹配模型

Abstract:

To address the issues of allocation of emergency rescuers, emergency rescue equipment, and disaster locations in the early stage of emergency rescue under the background of large-scale geological disasters, as well as improve the efficiency and quality of emergency rescue, an emergency rescue allocation decision-making method was proposed based on the three-sided matching theory. Firstly, the selection process among the three parties of emergency rescue was presented as a three-sided matching issue. Then, the multi-attribute preference evaluation information of the matching three parties was standardized, and a multi-attribute comprehensive preference score matrix was proposed. Furthermore, a three-sided matching stable and satisfactory model in the early stage of emergency rescue was developed to maximize system stability and total preference utility. Finally, the performance and accuracy of the proposed model were validated against cases. The results indicated that the proposed model reached reasonable matching between rescuers, emergency rescue equipment, and disaster locations in emergency rescue scenarios due to considering the effects of various parties' preferences on resource allocation. Furthermore, it can contribute to reducing the deployment time of emergency rescue operations and improve emergency efficiency and quality.

Key words: emergency rescue, three-sided matching, allocation decision, preference value, stable and satisfactory matching model

中图分类号:

X913.1安全运筹学

郭茜, 周伊可, 吴刚. 基于三边匹配理论的应急救援分配决策[J]. 中国安全科学学报, 2024, 34(3): 222-229.

GUO Qian, ZHOU Yike, WU Gang. Emergency rescue allocation decision-making based on three-sided matching theory[J]. China Safety Science Journal, 2024, 34(3): 222-229.

图/表 7

图1

表1

表2

表3

表4

救援人员组对救援装备组的标准化偏好值

$D ¯ p (a i, b j)$	b₁						b₂
$D ¯ p (a i, b j)$	D₁	D₂	D₃	D₄	D₅	D₆	D₁	D₂	D₃	D₄	D₅	D₆
a₁	0.167	1.000	0.667	1.000	0.333	0.500	0.333	0.000	1.000	0.250	0.000	0.000
a₂	0.667	1.000	0.500	0.333	0.000	1.000	1.000	0.000	0.000	1.000	1.000	0.750
a₃	0.000	0.500	1.000	0.500	0.000	0.500	1.000	0.500	1.000	0.000	0.667	0.000
a₄	0.500	0.750	1.000	1.000	1.000	0.000	0.000	1.000	0.500	0.000	0.000	1.000
$D ¯ p (a i, b j)$	b₃						b₄
$D ¯ p (a i, b j)$	D₁	D₂	D₃	D₄	D₅	D₆	D₁	D₂	D₃	D₄	D₅	D₆
a₁	1.000	1.000	0.000	0.250	1.000	1.000	0.000	1.000	0.667	0.000	0.000	0.500
a₂	0.000	0.500	0.667	0.000	0.667	0.250	0.667	0.000	1.000	0.667	0.000	0.000
a₃	0.250	1.000	0.000	0.500	1.000	0.500	0.500	0.000	1.000	1.000	0.667	1.000
a₄	1.000	1.000	0.750	0.833	0.333	0.333	0.500	0.000	0.000	0.333	0.333	0.667

表4

表5

救援人员组对受灾地点的标准化偏好值

$E q ¯ (a i, c k)$	c₁				c₂				c₃
$E q ¯ (a i, c k)$	E₁	E₂	E₃	E₄	E₁	E₂	E₃	E₄	E₁	E₂	E₃	E₄
a₁	1.000	0.000	0.400	1.000	0.750	0.714	0.000	0.667	1.000	0.714	0.400	0.000
a₂	0.250	1.000	0.000	1.000	0.000	0.600	1.000	0.833	0.750	0.400	0.800	0.000
a₃	1.000	0.333	0.667	0.667	0.000	0.000	1.000	1.000	0.000	1.000	0.000	0.167
a₄	0.500	0.000	0.667	0.000	0.750	1.000	1.000	0.000	1.000	0.667	0.000	0.000
$E q ¯ (a i, c k)$	c₄				c₅
$E q ¯ (a i, c k)$	E₁	E₂	E₃	E₄	E₁	E₂	E₃	E₄
a₁	0.000	1.000	1.000	0.667	0.500	0.714	0.400	0.000
a₂	1.000	0.400	0.000	1.000	1.000	0.000	0.800	0.667
a₃	0.000	1.000	1.000	0.000	1.000	0.000	0.000	0.500
a₄	1.000	0.667	0.667	0.667	0.000	0.333	0.333	1.000

表5

表6

救援装备组对受灾地点的标准化偏好值

$F t ¯ (b j, c k)$	c₁					c₂					c₃
$F t ¯ (b j, c k)$	F₁	F₂	F₃	F₄	F₅	F₁	F₂	F₃	F₄	F₅	F₁	F₂	F₃
b₁	1.000	0.200	0.333	1.000	0.000	0.667	0.800	1.000	0.667	1.000	0.667	0.800	0.167
b₂	0.800	0.000	1.000	0.750	0.250	0.000	1.000	0.000	0.250	0.500	0.000	0.000	0.800
b₃	0.000	1.000	1.000	0.000	0.857	0.750	0.000	0.400	0.250	0.000	1.000	0.000	0.000
b₄	0.000	0.667	0.000	0.000	0.400	1.000	0.000	0.000	1.000	1.000	1.000	1.000	0.000
$F t ¯ (b j, c k)$	c₃		c₄					c₅
$F t ¯ (b j, c k)$	F₄	F₅	F₁	F₂	F₃	F₄	F₅	F₁	F₂	F₃	F₄	F₅
b₁	0.333	0.750	0.000	0.000	0.667	0.000	0.250	0.000	1.000	0.000	0.333	1.000
b₂	0.000	0.000	0.000	0.333	0.200	1.000	1.000	1.000	0.000	1.000	0.250	0.250
b₃	1.000	0.571	1.000	0.000	0.800	0.000	0.857	0.000	0.000	0.400	0.000	1.000
b₄	0.500	0.000	1.000	0.333	1.000	1.000	0.000	0.000	1.000	0.000	1.000	0.400

表6

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