土地资源评价层次分析法案例题分析
实验一 层次分析法(4学时)
一、实验原理 ⑴ 建立层次结构模型
⑵构造成对比较矩阵
两两对比分值标准
C1、C2、C3相对于G的权重:([1**********]016 李湘湘)
24
1
A1/213
1/41/31
0.5710.6150.5
0.2860.3080.375 0.1430.0770.125
1.686
0.9690.345
0.562
0.323=W 0.115
⑶一致性检验
CR
CIn1(AW)i
0.1 CI max CRn1niWi
240.5621.6681
AW=1/2130.323=0.949
1/41/310.1150.364
max
CI
1.6680.9490.3641(AW)i
=1/3( )=3.027 0.5620.3230.115niWi
n3.0273
==0.0135
n1
31
CI0.0135==0.023
CR
A1、A2、A3、A4相对于C1的权重:([1**********]006 丁凌)
11/21/31/4
11/221
C1
3111/242211/101/53/102/5
1/92/92/94/9
1/133/133/136/13
1/9
2/92/94/9
0.399
0.875
0.9751.75
0.1
0.2190.244=W 0.437
一致性检验
CR
CIn1(AW)i
0.1 CI max CRn1niWi
11/21/31/40.10.42111/20.2190.8815
C1W==
3111/20.2440.984542210.4371.763
max
1(AW)i0.40.88150.98151.763
=1/4()=4.02 niWi0.10.2190.2440.437
4.024
=0.0067 41
CI
n
n1
=
CR1
CI0.0067==0.007
0.9CR
A5、A6、A7相对于C2的权重:([1**********]032 杨雯)
11/51/3
3 C2=51
31/31
0.1110.1300.077
0.5560.6530.692 0.3330.2170.231
0.0.106 3181.
0.324
0.781
= W 0.089
(2) 一致性检验: CI=
maxn
n1
max=
1(AW)i
n iwi
C151/30.1060.165
2W=51
30.634=1.944 31/310.2600.819
max=130.1651.9440.8190.1060.634
0.260=3.0385 CI=
3.03853
31
=0.0193
CR2=CI0CR.0193
0.580.0332
CRCICR0.1
A8、A9相对于C3的权重:([1**********]052 李梦潇)
0.20.20.411/4
C31.6 0.80.841
0.2
W
0.8
构造矩阵为2阶,CR3=0,满足条件。
表 各指标的权重及组合权重表([1**********]024 王柄智)
CI1=CR1×RI1=0.007×0.90=0.0063 CI2=CR2×RI2=-0.0332×0.58=0.019256
c1CI1c2CI2cmCIm0.1CR
c1RI1c2RI2cmRIm
=(0.562×0.0063+0.323×0.019256)/(0.526×0.90+0.323×0.58) =0.009760288/0.66074 =0.[1**********]
Word编辑人 ([1**********]014 霍静宜)