广义Z-numbers证据下考虑专家影响力和评价值一致性的应急决策方法

doi:10.16381/j.cnki.issn1003-207x.2022.1222

Abstract

Abstract:

Emergency decision-making is characterized by multi-attribute group decision-making， and in the complex and uncertain scenarios， the reliability of decision information with hesitation will affect the decision outcome. A novel emergency decision-making method considering the influence of expert and consistency of valuation is proposed， from the perspective of generalized Z-numbers evidence. Firstly， the emergency decision problem is modeled based on the generalized Z-numbers， and then the expert trust network is constructed to analyze the influence of expert. On this basis， a method is proposed to analyze the weight of valuation by integrating the influence of expert and the reliability of valuation. Secondly， according to the intersection of hesitant fuzzy evaluation values， different consistent situations are analyzed and the corresponding evidence generation methods are constructed. Then， the D-S evidence theory is used to fuse multiple evidence to make the optimal decision of response plans. Finally， the validity of the proposed method is verified through a decision example of medical waste disposal device with infectious diseases， and the results show the decision outcome are more reliable.

Key words: emergency decision-making, generalized Z-numbers, D-S evidence theory, expert influence, consistency

CLC Number:

C934

Lei Zhang, Keke Han, Xin Ye. Emergency Decision-making Method Considering the Influence of Expert and the Consistency of Valuation from the Perspective of Generalized Z-numbers Evidence[J]. Chinese Journal of Management Science, 2025, 33(6): 160-170.

Figures/Tables 10

输入： $O H i j, A H i j, r i j k k = 1,2, …, d$ 输出： $O H i j$ 和 $A H i j$ 的融合结果 $H C r$
Step 1：若 $∑ k = 1 d r i j k > 1$ ，执行式(7)，得到 $r i j k' k = a + 1, …, d$ ；
Step 2：若 $∑ k = 1 a r i j k > 1$ ，依次执行式(7)和式(6)；若 $∑ k = 1 a r i j k = 1$ ，执行式(6)；若 $∑ k = 1 a r i j k < 1$ 时，依次执行式(8)和式(6)，得到 $F H i j = H i j b, r i j b \| b = 1,2, …, f$ ；
Step 3：使用式(10)将 $F H i j$ 中的 $r i j b$ 进行修正，得到 $F H i j' = H i j b, r i j b' \| b = 1,2, …, f$ ；
Step 4：使用式(9)融合集合 $A H i j$ 和 $F H i j$ 的评价值，得到 $H C r$ ；
Step 5：若 $∑ k = 1 d r i j k = 1$ ，返回Step 2；
Step 6：若 $∑ k = 1 d r i j k < 1$ ，依次使用式(8)和式(6)融合 $O H i j$ 和 $A H i j$ ，得到 $H C r$ ；
Step 7：返回 $H C r$ 。

方案	$c 1$	$c 2$	$c 3$	$c 4$	$c 5$
$e 1$
$x 1$	$({H 3, H 4}, R 3)$	$({H 5, H 6}, R 5)$	$(H 3, {R 4, R 5})$	$(H 5, H 6, {R 3, R 4})$	$H 3, R 5$
$x 2$	$(H 4, {R 3, R 4})$	$({H 3, H 4}, R 4)$	$H 4 R 3, R 4$	$({H 3, H 4}, R 4)$	${H 5, H 6, R 5)}$
$x 3$	$H 3, R 4$	$(H 4, R 4)$	$({H 4, H 5}, R 3)$	$(H 3, {R 3, R 4})$	$H 3, R 4$
$x 4$	$H 5, R 4$	$({H 4, H 5}, R 4)$	$(H 4, R 3)$	$({H 4, H 5}, R 4)$	$H 4, H 5, R 4$
$e 2$
$x 1$	$H 3, R 4$	$(H 6, {R 5, R 6})$	$H 3, R 4$	$(H 6, {R 4, R 5})$	$({H 2, H 3}, {R 5, R 6})$
$x 2$	$({H 4, H 5}, {R 4, R 5})$	$H 3, R 5$	$({H 4, H 5}, R 4)$	$H 4, R 5$	$H 5, R 5$
$x 3$	$({H 3, H 4}, {R 4, R 5})$	$({H 3, H 4}, R 5)$	$H 5, R 5$	$H 3, R 4$	$({H 2, H 3}, R 5)$
$x 4$	$({H 4, H 5}, R 4)$	$H 4, R 5$	$(H 4, {R 4, R 5})$	$H 5, R 4$	$(H 5, {R 5, R 6})$
$e 3$
$x 1$	$(H 3, {R 5, R 6})$	$H 5, R 4$	$(H 3, {R 5, R 6})$	$H 6, R 5$	$H 3, R 5$
$x 2$	$H 5, R 5$	$({H 3, H 4}, R 4)$	$({H 4, H 5}, R 5)$	$H 3, R 5$	$H 5, R 4$
$x 3$	$({H 3, H 4}, {R 5, R 6})$	$({H 1, H 2}, R 4)$	$H 5, R 5$	$({H 2, H 3}, R 5)$	$H 3, R 4$
$x 4$	$H 6, R 5$	$H 4, R 5$	$({H 4, H 5}, R 5)$	$(H 5, H 6, {R 5, R 6}))$	$(H 5, {R 4, R 5})$
$e 4$
$x 1$	$(H 4, {R 4, R 5})$	$(H 5, H 6, {R 4, R 5})$	$({H 3, H 4}, R 4)$	$(H 6, R 5)$	$H 4, R 5$
$x 2$	$({H 4, H 5}, R 4)$	$({H 3, H 4}, R 4)$	$H 4, R 3$	$H 3, R 5$	$({H 4, H 5}, {R 4, R 5})$
$x 3$	$H 4, R 5$	$({H 3, H 4}, R 4)$	$(H 4, {R 3, R 4})$	$(H 3, {R 4, R 5})$	$(H 3, {R 4, R 5})$
$x 4$	$({H 4, H 5}, R 5)$	$({H 4, H 5}, {R 4, R 5})$	$({H 3, H 4}, R 4)$	$(H 5, H 6, R 5)$	$H 5, R 5$

方案	$c 1$	$c 2$	$c 3$	$c 4$	$c 5$
$e 1$
$x 1$	0.328	0.594	0.428	0.339	0.594
$x 2$	0.339	0.423	0.339	0.423	0.594
$x 3$	0.423	0.423	0.328	0.339	0.423
$x 4$	0.423	0.423	0.328	0.423	0.423
$e 2$
$x 1$	0.482	0.637	0.482	0.488	0.637
$x 2$	0.488	0.651	0.482	0.651	0.651
$x 3$	0.488	0.651	0.651	0.482	0.651
$x 4$	0.482	0.651	0.488	0.482	0.637
$e 3$
$x 1$	0.544	0.387	0.544	0.558	0.558
$x 2$	0.558	0.387	0.558	0.558	0.387
$x 3$	0.544	0.387	0.558	0.558	0.387
$x 4$	0.558	0.558	0.558	0.544	0.392
$e 4$
$x 1$	0.411	0.411	0.405	0.577	0.577
$x 2$	0.405	0.405	0.313	0.577	0.411
$x 3$	0.577	0.405	0.323	0.411	0.411
$x 4$	0.577	0.411	0.405	0.577	0.577

方案	$c 1$	$c 2$		$c 3$
$x 1$	${(H 3, 0.674), (H 4, 0.326)}$	${(H 5, 0.438), (H 6, 0.562)}$		${(H 3, 0.891), (H 4, 0.109)}$
$x 2$	${(H 4, 0.439), (H 5, 0.561)}$	${(H 3, 0.674), (H 4, 0.326)}$		${(H 4, 0.693), (H 5, 0.307)}$
$x 3$	${(H 3, 0.462), (H 3, 0.538)}$	${H 1, H 2, 0.208, (H 3, 0.283), (H 4, 0.51)}$		${(H 4, 0.262), (H 5, 0.738)}$
$x 4$	${(H 4, 0.26), (H 5, 0.467), (H 6, 0.274)}$	${(H 4, 0.769), (H 5, 0.204)}$		${(H 3, 0.057), (H 4, 0.843), (H 5, 0.1)}$
	$c 4$		$c 5$
$x 1$	${(H 5, 0.086), (H 6, 0.914)}$		${(H 2, 0.135), (H 3, 0.622), (H 4, 0.244)}$
$x 2$	${(H 3, 0.61), (H 4, 0.39)}$		${(H 4, 0.05), (H 5, 0.855), (H 6, 0.095)}$
$x 3$	${(H 2, 0.156), (H 3, 0.844)}$		${(H 2, 0.174), (H 3, 0.826)}$
$x 4$	${(H 4, 0.026), (H 5, 0.776), (H 6, 0.198)}$		${(H 4, 0.104), (H 5, 0.896)}$

方案	$c 1$	$c 2$		$c 3$
$x 1$	$m H 3 = 0.674, m (H 4) = 0.326$	$m (H 5) = 0.438, m (H 6) = 0.562$		$m H 3 = 0.888, m (H 4) = 0.112$
$x 2$	$m (H 4) = 0.439, m (H 5) = 0.561$	$m (H 3) = 0.674, m (H 4) = 0.326$		$m (H 4) = 0.75, m (H 5) = 0.25$
$x 3$	$m (H 3) = 0.462, m (H 4) = 0.538$	$m (H 1, H 2) = 0.208, m (H 3) = 0.283,$ $m (H 4) = 0.51$		$m (H 4) = 0.733, m (H 5) = 0.267$
$x 4$	$m (H 4) = 0.26, m (H 5) = 0.467,$ $m (H 6) = 0.274$	$m (H 4) = 0.769, m (H 5) = 0.204$		$m H 3 = 0.057, m (H 4) = 0.843,$ $m (H 5) = 0.1$
	$c 4$		$c 5$
$x 1$	$m (H 5) = 0.086, m (H 6) = 0.914$		$m (H 2) = 0.135, m (H 3) = 0.622, m (H 4) = 0.244$
$x 2$	$m (H 3) = 0.61, m (H 4) = 0.39$		$m (H 4) = 0.05, m (H 5) = 0.855, m (H 6) = 0.095$
$x 3$	$m H 2 = 0.156, m (H 3) = 0.844$		$m H 2 = 0.174, m (H 3) = 0.826$
$x 4$	$m (H 4) = 0.026, m (H 5) = 0.776, m (H 6) = 0.198$		$m (H 4) = 0.104, m (H 5) = 0.896$

属性	$c 1$	$c 2$	$c 3$	$c 4$	$c 5$	$λ j$
$c 1$	1	1	2	3	3	0.313
$c 2$	1	1	2	3	3	0.313
$c 3$	1/2	1/2	1	2	2	0.176
$c 4$	1/3	1/3	1/2	1	1	0.099
$c 5$	1/3	1/3	1/2	1	1	0.099

方案	证据理论集结结果
$x 1$	$m (H 2) = 0.006, m H 3 = 0.2675, m (H 4) = 0.0834, m (H 5) = 0.0852, m H 6 = 0.1542, m H 2, H 3, H 4, H 5, H 6 = 0.4038$
$x 2$	$m (H 3) = 0.1563, m (H 4) = 0.2605, m (H 5) = 0.1835, m H 6 = 0.0041, m (H 3, H 4, H 5, H 6) = 0.3956$
$x 3$	$m (H 2) = 0.0158, m (H 1, H 2) = 0.0372, m H 3 = 0.2475, m H 4 = 0.244, m (H 5) = 0.062, m (H 1, H 2, H 3, H 4, H 5) = 0.3936$
$x 4$	$m (H 3) = 0.0047, m (H 4) = 0.3171, m (H 5) = 0.2362, m H 6 = 0.0574, m (H 3, H 4, H 5, H 6) = 0.3847$

方法	$x 1$	$x 2$	$x 3$	$x 4$
HFLWA^[21]	0.5370	0.5103	0.4173	0.5989
本文方法	0.5267	0.5847	0.4322	0.6391

评价信息

本文方法

HFLWA方法^[21]

z 51

$m (H 1, H 2) = 0.208$

$m (H 3) = 0.283$

$m (H 4) = 0.51$

$m (H 2) = 0.023$

$m (H 3) = 0.755$

$m (H 4) = 0.222$

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Emergency Decision-making Method Considering the Influence of Expert and the Consistency of Valuation from the Perspective of Generalized Z-numbers Evidence

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