﻿ 基于AHP-FUZZY施工现场安全评价研究

# 基于AHP-FUZZY施工现场安全评价研究Research on Safety Evaluation of Construction Site Based on AHP-FUZZY

Abstract: Aiming at the problems of safety accidents and the seriousness of losses in the construction industry, this paper introduces the fuzzy analytic hierarchy process (AHP-FUZZY) and establishes a safety evaluation model based on AHP-FUZZY. The weight of each evaluation index of the safety evaluation model is determined by AHP method, and the comprehensive safety evaluation of the construction site of Qichun Hexi Industrial Park is carried out in combination with the fuzzy mathematics theory. The construction site is in a basically normal safety state, which is consistent with the actual situation. This method is of great significance for improving the level of enterprise safety management.

1. 引言

2. AHP-FUZZY安全评价模型

2.1. 建立安全评价指标体系

2.2. 确定评语集

Figure 1. Qichun Hexi Industrial Park safety evaluation index system

Table 1. Construction site safety and safety evaluation index classification

2.3. 确定评价指标模糊隶属度

$R=\left(\begin{array}{cccc}{r}_{11}& {r}_{12}& \cdots & {r}_{1n}\\ {r}_{21}& {r}_{22}& \cdots & {r}_{2n}\\ ⋮& ⋮& \ddots & ⋮\\ {r}_{m1}& {r}_{m2}& \cdots & {r}_{mn}\end{array}\right)$

2.4. 构造判断矩阵

2.5. 确定指标权重

Table 2. Judgment matrix scale meaning

Table 3. RI value

Figure 2. Indicator weight determination process

3. 实例分析

3.1. 工程概况

3.2. 构建判断矩阵

Table 4. U-Ui, Ui-Uij judgment matrix

3.3. 确定评价指标权重

Table 5. Evaluation index weight

3.4. 确定评价指标模糊隶属度

3.5. 模糊综合评价

${B}_{1}=\left(0.101,0.267,0.119,0.513\right)×{R}_{1}=\left(0.266,0.442,0.241,0.051,0\right)$

${B}_{2}=\left(0.059,0.431,0.399,0.111\right)×{R}_{2}=\left(0.336,0.463,0.145,0.056,0\right)$

${B}_{3}=\left(0.833,0.167\right)×{R}_{3}=\left(0.517,0.350,0.108,0.025,0\right)$

${B}_{4}=\left(0.144,0.053,0.157,0.297,0.311,0.038\right)×{R}_{4}=\left(0.484,0.453,0.059,0.004,0\right)$

$B=\left({B}_{1},{B}_{2},{B}_{3},{B}_{4}\right)=\left(\begin{array}{ccccc}0.266& 0.442& 0.241& 0.051& 0.000\\ 0.336& 0.463& 0.145& 0.056& 0.000\\ 0.517& 0.350& 0.108& 0.025& 0.000\\ 0.484& 0.453& 0.059& 0.004& 0.000\end{array}\right)$

$S=\left(0.293,0.137,0.048,0.523\right)×B=\left(0.401,0.447,0.127,0.025,0\right)$

4. 结论

NOTES

*通讯作者。

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