数学建模层次分析法旅游景点选址举例.doc

假期到了, 某学生打算做一次旅游, 有四个地点可供选择, 假定他要考虑5个因素: 费用、景色、居住条件、饮食以及旅游条件. 由于该学生没有固定收入, 他对费用最为看重, 其次是旅游点的景色, 至于旅游条件、饮食, 差不多就行, 住什么地方就更无所谓了. 这四个旅游点没有一个具有明显的优势, 而是各有优劣. 该同学拿不定主意, 请用层次分析法帮助他找出最佳旅游点。正文：1、利用层次分析法构造层次分析模型：准则层目标层方案层旅游点44旅游点33旅游点22费用旅游点1景色居住条件旅游条件饮食景点选择择图1-12、利用成对比较法对准则层、方案层进行列表主要影响因素对比费用景色居住条件饮食旅游条件费用15393景色1/511/221/2居住条件1/32131饮食1/91/21/311/3旅游条件1/32131（表2-1）费用对比旅游点1旅游点2旅游点3旅游点4旅游点111/31/51/7旅游点2311/21/4旅游点35211/2旅游点41/7421（表2-2）景色对比旅游点1旅游点2旅游点3旅游点4旅游点111/243旅游点22155旅游点31/41/511旅游点41/31/511（表2-3）居住条件对比旅游点1旅游点2旅游点3旅游点4旅游点11658旅游点21/6112旅游点31/5117旅游点41/81/21/71（表2-4）饮食对比旅游点1旅游点2旅游点3旅游点4旅游点1111/31/3旅游点2111/21/5旅游点33211旅游点43511（表2-5）旅游条件对比旅游点1旅游点2旅游点3旅游点4旅游点11212旅游点21/211/21旅游点31212旅游点41/211/21（表2-6）2.构造成对比较判断矩阵(1) 建立准则层对目标层的成对比较判断矩阵(2) 建立方案层对准则层的成对比较判断矩阵 3.计算层次单排序权重向量并做一致性检验先利用Mathematica计算矩阵A的最大特征值及特征值所对应的特征向量.输入A=1.0,5,3,9,3,1/5,1,1/2,2,1/2,1/3,2,1,3,1,1/9,1/2,1/3,1,1/3,1/3,2,1,3,1T=Eigensystemj/Chop输出5.00974,-0.0048699+0.22084,-0.0048699-0.22084,0,0,0.88126,0.167913,0.304926,0.0960557,0.304926,0.742882,-0.223286-0.278709,-0.165421+0.346134,0.151384-0.057689,-0.165421+0.346134,0.742882,-0.223286+0.278709,-0.165421-0.346134,0.151384+0.057689,-0.165421-0.346134,-0.993367,0,0.0719207,0.0662245,0.0605282,0.884443,0,-0.380934,-0.0589629,0.263009得出A的最大特征值为 =5.00974,及其对应的特征向量x=0.88126,0.167913,0.304926,0.0960557,0.304926输入Clearx;x=T2,1;W1=x/ApplyPlus,x得到归一化之后的的特征向量=0.502119,0.0956728,0.173739,0.0547301,0.173739计算一致性指标， 故矩阵阶数表矩阵阶数1234567891011000.580.91.121.241.321.411.451.491.51（表3-1）查表（见表3-1）得到相应的随机一致性指标所以 通过了一致性检验,即认为的一致性程度在容许的范围之内, 可以用归一化后的特征向量作为排序权重向量.下面再求矩阵的最大特征值及特征值所对应的特征向量输入B1=1.0,1/3,1/5,1/7,3,1,1/2,1/4,5,2,1,1/2,1/7,4,2,1B2=1,1/2,4,3,2,1,5,5,1/4,1/5,1,1,1/3,1/5,1,1B3=1,6,5,8,1/6,1,1,2,1/5,1,1,7,1/8,1/2,1/7,1B4=1,1,1/3,1/3,1,1,1/2,1/5,3,2,1,1,3,5,1,1B5=1,2,1,2,1/2,1,1/2,1,1,2,1,2,1/2,1,1/2,1T1=EigensystemB1/ChopT2=EigensystemB2/ChopT3=EigensystemB3/ChopT4=EigensystemB4/ChopT5=EigensystemB5/Chop输出3.82325,0.0883772+0.544064,0.0883772-0.544064,0,0.111267,0.283002,0.536902,0.786934,-0.0248134-0.0681165,-0.141793+0.0729826,-0.154388+0.121345,0.964755,-0.0248134+0.0681165,-0.141793-0.0729826,-0.154388-0.121345 ,0.964755,0,0.299667,-0.832409,0.4661494.02113,-0.0105652+0.291301,-0.0105652-0.291301,0,0.495852,0.84036,0.149575,0.159851,-0.234515+0.517899,0.805208,-0.109665-0.110941,0.0407277 -0.0493071 ,-0.234515-0.517899 ,0.805208,-0.109665+0.110941 ,0.0407277 +0.0493071 ,0,-0.953463,-0.0953463,0.2860394.25551,-0.110262+1.03317,-0.110262-1.03317,-0.0349818,0.941183,0.179553,0.276018,0.0758271,0.898054,0.136097 +0.0728034 ,-0.309669+0.2519 ,-0.0331642-0.0960598,0.898054,0.136097-0.0728034,-0.309669-0.2519,-0.0331642+0.0960598,0.958653,-0.256222,0.123505,-0.009047724.08009,-0.0400469+0.570251,-0.0400469-0.570251,0,0.214349,0.214031,0.59059,0.747963,0.00228339-0.0861419,-0.0895045+0.220107,-0.388206-0.387638,0.796962,0.00228339+0.0861419,-0.0895045-0.220107,-0.388206+0.387638 ,0.796962,-0.424264,0,0.565685,0.7071074.,0,0,0,0.632456,0.316228,0.632456,0.316228,0.116296,0.629208,-0.687356,-0.343678,-0.92582,0.154303,0.308607,0.154303,-0.92582,0.154303,0.308607,0.154303分别得出其最大特征值=3.82325,= 4.02113,= 4.25551,= 4.08009,= 4，以及其特征向量如下：B1=（0.111267,0.283002,0.536902,0.786934）B2=（0.495852,0.84036,0.149575,0.159851）B3=（0.941183,0.179553,0.276018,0.0758271）B4=（0.214349,0.214031,0.59059,0.747963）B5=（0.632456,0.316228,0.632456,0.316228）其中 为求出归一化后的特征向量, 输入ClearB1,B2,B3,B4,B5;B1=T12,1;w1=B1/ApplyPlus,B1B2=T22,1;w2=B2/ApplyPlus,B2B3=T32,1;w3=B3/ApplyPlus,B3B4=T42,1;w4=B4/ApplyPlus,B4B5=T52,1;w5=B5/ApplyPlus,B5输出4.,0,0,0,0.632456,0.316228,0.632456,0.316228,0.116296,0.629208,-0.687356,-0.343678,-0.92582,0.154303,0.308607,0.154303,-0.92582,0.154303,0.308607,0.154303w1= 0.0647614,0.164718,0.312497,0.458024w2=0.301313,0.510659,0.0908919,0.0971363w3= 0.639138,0.121931,0.187438,0.0514926w4= 0.121311,0.121132,0.334246,0.423311w5= 0.333333,0.166667,0.333333,0.166667计算一致性指标,其中,输入lamda=T11,1,T21,1,T31,1,T41,1,T51,1CI=(lamda-4)/(4-1)/Chop则可以得到-0.0589181, 0.00704344,0.0851688,0.0266979,=0查表（见表3-1）得到相应的随机一致性指标计算一致性比率,输入CR=CI/0.90相应可得到因通过了一致性检验. 即认为的一致性程度在容许的范围之内, 可以用归一化后的特征向量作为其排序权重向量.4、计算层次总排序权重向量并做一致性检验购买个人电脑问题的第三层对第二层的排序权重计算结果列于表4-1k123450.06476140.3013130.6391380.1213110.3333330.1647180.5106590.1219310.1211320.1666670.3124970.09089190.1874380.3342460.3333330.4580240.09713630.05149260.4233110.1666673.823254.021134.255514.080094（表4-1）以矩阵表示第三层对第二层的排序权重计算结果为即是第三层对第二层的权重向量为列向量组成的矩阵. 最下层(第三层)对最上层(第一层)的总排序权向量为为了计算上式, 输入W2