﻿ 配电变压器状态评价权重系数改进方法的研究

# 配电变压器状态评价权重系数改进方法的研究Method for Evaluating State of Distribution Transformer Based on Fuzzy Theory

Abstract: With the continuous development of the economy and society, the demand for electricity in all as-pects of society is gradually expanding, which requires the power system to be more secure and re-liable. In this paper, the related theoretical knowledge related to the state evaluation of distribution transformers is deeply studied. The fuzzy comprehensive evaluation method is introduced, and the general algorithm and improved algorithm for determining the weight coefficients are analyzed. The weight coefficients of the two algorithms are calculated by an example. It is concluded that the calculation of the state quantity weight coefficient in the distribution transformer state evaluation should be improved based on the rough set theory based weight coefficient determination method.

1. 引言

2. 模糊综合评判方法

2.1. 模糊集合的基本概念

2.2. 模糊综合评判的步骤

$B=W\circ R=W\circ \left[\begin{array}{c}{W}_{1}\circ {R}_{1}\\ {W}_{2}\circ {R}_{2}\\ ⋮\\ {W}_{n}\circ {R}_{n}\end{array}\right]=W\circ \left[\begin{array}{c}{B}_{1}\\ {B}_{2}\\ ⋮\\ {B}_{n}\end{array}\right]$ (1.1)

Figure 1. Fuzzy converter

1) 确定被评价的对象X。这里表示为需要评价的配电网某变压器。

2) 建立因素集U，也就是评价指标体系。

3) 建立表示评价结果等级的评价集V。

${v}_{i}$ 为评价等级，等级数为m，则评价集 $V=\left\{{v}_{1},{v}_{2},\cdots ,{v}_{m}\right\}$ 。本文中将配电变压器的运行状态和状态

$V=\left({v}_{1},{v}_{2},{v}_{3},{v}_{4}\right)=\left(正常,注意,异常,严重\right)$ (1.2)

4) 确定权重系数集W

$\underset{i=1}{\overset{n}{\sum }}{w}_{i}=1$ (1.3)

3. 基于粗糙集理论的权重系数确定方法的研究

3.1. 状态量权重系数的确定方法分类

3.2. 基于粗糙集理论权重系数的确定方法

3.2.1. 粗糙集理论确定权重系数的一般算法

1) 决策表的建立

2) 依赖度的计算

${\gamma }_{C}\left(D\right)=\frac{|po{s}_{C}\left(D\right)|}{|U|}$ (2.1)

${\gamma }_{C-{C}_{i}}\left(D\right)=\frac{|po{s}_{C-{C}_{i}}\left(D\right)|}{|U|}$ (2.2)

3) 重要度的计算

$sig\left({C}_{i}\right)={\gamma }_{C}\left(D\right)-{\gamma }_{C-{C}_{i}}\left(D\right)$ (2.3)

4) 权重系数的计算

$W\left({C}_{i}\right)=\frac{sig\left({C}_{i}\right)}{\underset{i=1}{\overset{m}{\sum }}sig\left({C}_{i}\right)}$ (2.4)

3.2.2. 改进后的权重系数的确定算法

1) 条件熵的计算

$I\left(D|C\right)=\underset{i=1}{\overset{m}{\sum }}\frac{{|{C}_{i}|}^{2}}{{|U|}^{2}}\underset{j=1}{\overset{m}{\sum }}\frac{|{D}_{j}\cap {C}_{i}|}{|{C}_{i}|}\left(1-\frac{|{D}_{j}\cap {C}_{i}|}{|{C}_{i}|}\right)$ (2.5)

2) 改进后重要度的计算

$sig\left({C}_{i}\right)=I\left(D|\left(C-{C}_{i}\right)\right)-I\left(D|C\right)+\frac{\underset{a\in C}{\sum }|a\left(x\right)|-\underset{a\in \left(C-{C}_{i}\right)}{\sum }|a\left(x\right)|}{\underset{a\in C}{\sum }|a\left(x\right)|}$ (2.6)

3) 改进后的权重系数的计算

$W\left({C}_{i}\right)=\frac{sig\left({C}_{i}\right)+I\left(D|\left\{{C}_{i}\right\}\right)}{\underset{a\in C}{\sum }\left\{sig\left(a\right)+I\left(D|\left\{a\right\}\right)\right\}}$ (2.7)

4. 实例分析

Table 1. Oil insulation fault diagnosis decision table

4.1. 决策表的知识分类

$U/IND\left({C}_{1}\right)=\left\{\left\{1,2,3\right\},\left\{4,5,6\right\}\right\}$

$U/IND\left({C}_{2}\right)=\left\{\left\{1,2,3,4,6\right\},\left\{5\right\}\right\}$

$U/IND\left({C}_{3}\right)=\left\{\left\{1,4\right\},\left\{2,5\right\},\left\{3,6\right\}\right\}$

$U/IND\left(C\right)=\left\{\left\{1\right\},\left\{2\right\},\left\{3\right\},\left\{4\right\},\left\{5\right\},\left\{6\right\}\right\}$

$U/IND\left(D\right)=\left\{\left\{1,4,5\right\},\left\{2,3,6\right\}\right\}$

$po{s}_{C}\left(D\right)=\left\{1\right\}\cup \left\{2\right\}\cup \cdots \cup \left\{6\right\}=\left\{1,2,\cdots ,6\right\}=U$

4.2. 基于一般算法确定权重系数

$U/IND\left(C-{C}_{1}\right)=\left\{\left\{1,4\right\},\left\{2\right\},\left\{3,6\right\},\left\{5\right\}\right\}$

$U/IND\left(C-{C}_{2}\right)=\left\{\left\{1\right\},\left\{2\right\},\left\{3\right\},\left\{4\right\},\left\{5\right\},\left\{6\right\}\right\}$

$U/IND\left(C-{C}_{3}\right)=\left\{\left\{1,2,3\right\},\left\{4,6\right\},\left\{5\right\}\right\}$

${\gamma }_{C}\left(D\right)=\frac{|po{s}_{C}\left(D\right)|}{|U|}=\frac{|U|}{|U|}=\frac{6}{6}=1$

$po{s}_{C-{C}_{1}}\left(D\right)=\left\{1,2,3,4,5,6\right\}=U$

$po{s}_{C-{C}_{2}}\left(D\right)=\left\{1,2,3,4,5,6\right\}=U$

$po{s}_{C-{C}_{3}}\left(D\right)=\left\{5\right\}$

${\gamma }_{C-{C}_{1}}\left(D\right)=\frac{|po{s}_{C-{C}_{1}}\left(D\right)|}{|U|}=\frac{|U|}{|U|}=1$

${\gamma }_{C-{C}_{2}}\left(D\right)=\frac{|po{s}_{C-{C}_{2}}\left(D\right)|}{|U|}=\frac{|U|}{|U|}=1$

${\gamma }_{C-{C}_{3}}\left(D\right)=\frac{|po{s}_{C-{C}_{3}}\left(D\right)|}{|U|}=\frac{1}{6}$

$sig\left({C}_{1}\right)={\gamma }_{C}\left(D\right)-{\gamma }_{C-{C}_{1}}\left(D\right)=0$

$sig\left({C}_{2}\right)={\gamma }_{C}\left(D\right)-{\gamma }_{C-{C}_{2}}\left(D\right)=0$

$sig\left({C}_{3}\right)={\gamma }_{C}\left(D\right)-{\gamma }_{C-{C}_{3}}\left(D\right)=1-\frac{1}{6}=\frac{5}{6}$

$W\left({C}_{1}\right)=\frac{sig\left({C}_{1}\right)}{sig\left({C}_{1}\right)+sig\left({C}_{2}\right)+sig\left({C}_{3}\right)}=0$

$W\left({C}_{2}\right)=\frac{sig\left({C}_{2}\right)}{sig\left({C}_{1}\right)+sig\left({C}_{2}\right)+sig\left({C}_{3}\right)}=0$

$W\left({C}_{3}\right)=\frac{sig\left({C}_{3}\right)}{sig\left({C}_{1}\right)+sig\left({C}_{2}\right)+sig\left({C}_{3}\right)}=1$

4.3. 基于改进算法确定权重系数

$\begin{array}{c}I\left(D|C\right)=\underset{i=1}{\overset{m}{\sum }}\frac{{|{C}_{i}|}^{2}}{{|U|}^{2}}\underset{j=1}{\overset{m}{\sum }}\frac{|{D}_{j}\cap {C}_{i}|}{|{C}_{i}|}\left(1-\frac{|{D}_{j}\cap {C}_{i}|}{|{C}_{i}|}\right)\\ ={\left(\frac{1}{6}\right)}^{2}\left[\frac{1}{1}\left(1-\frac{1}{1}\right)+\frac{0}{1}\left(1-\frac{0}{1}\right)\right]+{\left(\frac{1}{6}\right)}^{2}\left[\frac{0}{1}\left(1-\frac{0}{1}\right)+\frac{1}{1}\left(1-\frac{1}{1}\right)\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+{\left(\frac{1}{6}\right)}^{2}\left[\frac{0}{1}\left(1-\frac{0}{1}\right)+\frac{1}{1}\left(1-\frac{1}{1}\right)\right]+{\left(\frac{1}{6}\right)}^{2}\left[\frac{1}{1}\left(1-\frac{1}{1}\right)+\frac{0}{1}\left(1-\frac{0}{1}\right)\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+{\left(\frac{1}{6}\right)}^{2}\left[\frac{1}{1}\left(1-\frac{1}{1}\right)+\frac{0}{1}\left(1-\frac{0}{1}\right)\right]+{\left(\frac{1}{6}\right)}^{2}\left[\frac{0}{1}\left(1-\frac{0}{1}\right)+\frac{1}{1}\left(1-\frac{1}{1}\right)\right]\\ =0\end{array}$

$\begin{array}{c}I\left(D|\left\{{C}_{1}\right\}\right)=\underset{i=1}{\overset{m}{\sum }}\frac{{|{C}_{i}|}^{2}}{{|U|}^{2}}\underset{j=1}{\overset{m}{\sum }}\frac{|{D}_{j}\cap {C}_{i}|}{|{C}_{i}|}\left(1-\frac{|{D}_{j}\cap {C}_{i}|}{|{C}_{i}|}\right)\\ ={\left(\frac{3}{6}\right)}^{2}\left[\frac{1}{3}\left(1-\frac{1}{3}\right)+\frac{2}{3}\left(1-\frac{2}{3}\right)\right]+{\left(\frac{3}{6}\right)}^{2}\left[\frac{2}{3}\left(1-\frac{2}{3}\right)+\frac{1}{3}\left(1-\frac{1}{3}\right)\right]\\ =\frac{2}{9}\end{array}$

$\begin{array}{c}I\left(D|\left(C-\left\{{C}_{1}\right\}\right)\right)=\underset{i=1}{\overset{m}{\sum }}\frac{{|{C}_{i}|}^{2}}{{|U|}^{2}}\underset{j=1}{\overset{m}{\sum }}\frac{|{D}_{j}\cap {C}_{i}|}{|{C}_{i}|}\left(1-\frac{|{D}_{j}\cap {C}_{i}|}{|{C}_{i}|}\right)\\ ={\left(\frac{2}{6}\right)}^{2}\left[\frac{2}{2}\left(1-\frac{2}{2}\right)+\frac{0}{2}\left(1-\frac{0}{2}\right)\right]+{\left(\frac{1}{6}\right)}^{2}\left[\frac{0}{1}\left(1-\frac{0}{1}\right)+\frac{1}{1}\left(1-\frac{1}{1}\right)\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }+{\left(\frac{2}{6}\right)}^{2}\left[\frac{0}{2}\left(1-\frac{0}{2}\right)+\frac{2}{2}\left(1-\frac{2}{2}\right)\right]+{\left(\frac{1}{6}\right)}^{2}\left[\frac{1}{1}\left(1-\frac{1}{1}\right)+\frac{0}{1}\left(1-\frac{0}{1}\right)\right]\\ =0\end{array}$

$I\left(D|\left\{{C}_{2}\right\}\right)=\frac{1}{3}$

$I\left(D|\left(C-\left\{{C}_{2}\right\}\right)\right)=0$

$I\left(D|\left(C-\left\{{C}_{3}\right\}\right)\right)=\frac{1}{6}$

$\frac{\underset{a\in C}{\sum }|a\left(x\right)|-\underset{a\in \left(C-\left\{{C}_{i}\right\}\right)}{\sum }|a\left(x\right)|}{\underset{a\in C}{\sum }|a\left(x\right)|}=\frac{1}{3}$

$sig\left({C}_{1}\right)=I\left(D|\left(C-\left\{{C}_{1}\right\}\right)\right)-I\left(D|C\right)+\frac{\underset{a\in C}{\sum }|a\left(x\right)|-\underset{a\in \left(C-\left\{{C}_{i}\right\}\right)}{\sum }|a\left(x\right)|}{\underset{a\in C}{\sum }|a\left(x\right)|}=\frac{1}{3}$

$sig\left({C}_{3}\right)=I\left(D|\left(C-\left\{{C}_{3}\right\}\right)\right)-I\left(D|C\right)+\frac{\underset{a\in C}{\sum }|a\left(x\right)|-\underset{a\in \left(C-\left\{{C}_{i}\right\}\right)}{\sum }|a\left(x\right)|}{\underset{a\in C}{\sum }|a\left(x\right)|}=\frac{1}{2}$

$W\left({C}_{1}\right)=\frac{sig\left({C}_{1}\right)+I\left(D|\left\{{C}_{1}\right\}\right)}{\underset{a\in C}{\sum }\left\{sig\left(a\right)+I\left(D|\left\{a\right\}\right)\right\}}=\frac{5}{16}=0.3125$

$W\left({C}_{2}\right)=\frac{sig\left({C}_{2}\right)+I\left(D|\left\{{C}_{2}\right\}\right)}{\underset{a\in C}{\sum }\left\{sig\left(a\right)+I\left(D|\left\{a\right\}\right)\right\}}=\frac{3}{8}=0.375$

$W\left({C}_{3}\right)=\frac{sig\left({C}_{3}\right)+I\left(D|\left\{{C}_{3}\right\}\right)}{\underset{a\in C}{\sum }\left\{sig\left(a\right)+I\left(D|\left\{a\right\}\right)\right\}}=\frac{5}{16}=0.3125$

4.4. 对比分析

5. 总结

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