层次分析法选专业

层次分析法在专业选择中的应用.问题提出本人数学与应用数学专业在读，由于数学是一切自然科学与工程技术的基础，所以专业出路较广。我有意考研，现用层次分析法，考虑以下五个因素：对专业的执着兴趣，实现自我价值，解决该领域问题能力与发展潜力，就业形势与未来收入，与本科知识储备相关度。从基础数学、应用数学、计算机科学、经济金融、统计学、数学教育六个专业中选择一个作为硕士阶段研修专业。.模型建立1.建立层次结构模型在深入分析问题的基础上，将有关的各个因素按照不同的属性自上而下地分为若干层次。同一层的因素从属于上一层因素或对上层因素有影响，同时又支配下一层的因素或受到下层因素的作用，而同一层各因素之间尽量相互独立，最上层为目标层，通常只有一个因素，最下层通常为方案层或对象层，中间可以有一个或几个层次，通常称为准则或指标层，当准则过多时（比如多与九个）应进一步分解出子准则层。将决策问题分为3个层次，最上层为目标层，即选择专业，最下层为方案层，有六个专业供选择，中间层为准则层，各层间的联系用相连的直线表示。选择专业A1执着兴趣IA2A1执着兴趣IA2自我价值实现IA3解决问题能力发展潜力IA4就业形势未来收入IA5与本科知识曲备相关度应用数学计经[济机应用数学计经[济机金科融学数学教育通过相互比较个准则对于目标的权重，及各个方案对于每个准则的权重。这些权重通常在人的思维过程中是定性的，而在层次分析法中则要给出得到权重的定量方法。将方案层对准则层的权重及准则层对目标层的权重进行综合，最终确定方案层对目标层的权重。在层次分析法中要给出综合的计算方法。本文将定性分析和定量计算结合起来完成上述步骤，给出决策结果。2.构造成对比矩阵，计算权向量从层次结构模型的第二层开始，对于从属于（或影响及）上一层的每个因素的同一层诸因素，用成对比较法和1-9比较尺度构造成对比较阵，直至最下层。对于每一个成对比较矩阵计算最大特征根及对应特征向量，利用一致性指标、随机一致性指标和一致性比率做一致性检验。若检验通过，特征向量（归一化后）即为权向量；若不通过需要重新构造成对比较矩阵。五个因素对专业选择的重要性用矩阵A=aj55表示，其中％为准则相对于5A0上一次的重要性之比（A./A.）,矩阵A为：1.00003.00000.33331.00005.00001.00003.00000.33331.00005.00000.33331.00001.00001.00003.00003.00001.00001.00003.00000.33331.00001.00000.33331.00000.33330.20000.33333.00003.00001.0000可得专业选择问题第二层计算结果为最大特征值’=5.3195，经检验一致性比率小于0.1，权向量为：w=-（0.4717，0.6460，0.1951，0.5397，0.1757）同样的方法构造第三层（方案层，见图1）对第二层（准则层，见图一）的每个准则的成对比较矩阵B（i=1,2...5）B1=1.00000.33330.20000.14290.14293.00001.00001.00000.33330.20000.14290.14293.00001.00001.00001.00000.33335.00001.00001.00001.00002.00007.00003.00001.00001.00003.00007.00001.00000.50000.33331.00000.50000.20000.25000.25000.33332.0000B2=5.00004.00004.00003.00001.00001.00000.33333.00000.50002.00000.50003.00001.00001.00001.00001.00000.25000.33331.00001.00001.00001.00000.20002.00001.00001.00001.00001.00000.33330.50001.00001.00001.00001.00000.25002.00004.00005.00003.00004.00001.0000B3=1.00000.25000.20000.20000.16670.20004.00001.00000.33332.00000.25000.25005.00003.00001.00001.00000.33330.50005.00000.50001.00001.00000.50000.33336.00004.00003.00002.00001.00000.50005.00004.00002.00003.00002.00001.0000B4=1.00000.33330.25000.20000.25002.00003.00001.00000.50000.25000.50003.00004.00002.00001.00000.50001.00003.00005.00004.00002.00001.00003.00005.00004.00002.00001.00000.33331.00003.00000.50000.33330.33330.20000.33331.0000B5=1.00002.00005.00006.00002.00002.00000.50001.00003.00005.00002.00002.00000.20000.33331.00002.00000.33330.33330.16670.20000.50001.00000.33330.25000.50000.50003.00003.00001.00001.00000.50000.50003.00004.00001.00001.0000经过检验，上述5个矩阵的一致性比率均小于0.1,可得专业选择问题第三层的计算结果如下：表1:第二层计算结果k12345最大特征值-6.47706.55846.47546.20956.1019W；30.65860.29470.07260.12420.7059W(3)0.18490.31780.21710.24540.5039W3)0.12680.19180.32550.39900.1408W43)0.11630.28320.23880.78150.0943W3)0.18020.20550.56600.37800.3232W3)0.68560.80800.68140.10730.3386三・模型求解计算最下层对目标层组合权向量，按照组合权向量表示的结果进行决策。基础数学权重为：0.6586*0.4717+0.2947*0.6460+0.0726*0.1951+0.1242*0.5397+0.7059*0.1757=0.7063应用数学权重为：0.1849*0.4717+0.3178*0.6460+0.2171*0.1951+0.2454*0.5397+0.5039*0.1757=0.5558计算机科学权重为：0.1268*0.4717+0.1918*0.6460+0.3255*0.1951+0.3990*0.5397+0.1408*0.1757=0.4873经济金融权重为为：0.1163*0.4717+0.2832*0.6460+0.2388*0.1951+0.7815*0.5397+0.0943*0.1757=0.7227统计学权重为：0.1802*0.4717+0.2055*0.6460+0.5660*0.1951+0.3780*0.5397+0.3232*0.1757=0.5890数学教育权重为：0.6856*0.4717+0.8080*0.6460+0.6814*0.1951+0.1073*0.5307+0.3386*0.1757=1.0947结果分析权重由小到大排序为：数学教育，经济金融，基础数学，统计学，应用数学，计算机科学。层次分析法做出的决策结果为数学教育专业。模型优缺点分析层次分析法将定性问题定量化，便于计算和研究，给出可以比较大小的度量结果，利于决策。恰恰也是因为定性问题定量化，带来了一下弊端。第一，只能从原有方案中选优，不能生成新的方案；第二，它的比较、判断知道结果都是粗糙的，不适于精度要求很高的问题；第三，从建立层次结构模型到给出成对比较举证，人的主观因素的作用很大，这就使得决策结果可能难以接受。附录：matlab源代码A=[11/3311/5;31111/3;1/3111/33;11313;531/31/31]1.00000.33333.00001.00000.20003.00001.00001.00001.00000.33330.33331.00001.00000.33333.00001.00001.00003.00001.00003.00005.00003.00000.33330.33331.0000>>[x,y]=eig(A)x=-0.3324-0.2826+0.3076i-0.2826-0.3076i0.38830.4371-0.36360.1935+0.2361i0.1935-0.2361i-0.6709-0.8356-0.3814-0.14110.4497i-0.1411+0.4497i0.1108-0.0793-0.5811-0.33760.2072i-0.3376+0.2072i-0.5874-0.1541-0.52360.59430.59430.20410.2842y=6.870100000-0.6696+3.4113i00000-0.6696-3.4113i00000-0.127300000-0.403>>CI=6.8701-5/4CI=>>CR=CI/1.12CR=5.0179>>CI=(6.8701-5)/4CI=0.4675>>CR=CI/1.12CR=0.4174>>CR=5CR=5>>CI=5CI=5>>fprintf(CR)ErrorusingfprintfNoformatstring.>>A=[11/221/26;21413;1/21/411/21;21212;1/61/311/21]A=1.00000.50002.00000.50006.00002.00001.00004.00001.00003.00000.50000.25001.00000.50001.00002.00001.00002.00001.00002.00000.16670.33331.00000.50001.0000>>(X,Y)=eig(A)(X,Y)=eig(A)IError:Expressionorstatementisincorrect--possiblyunbalanced(,{,or[.>>(x,y)=eig(A)(x,y)=eig(A)IError:Expressionorstatementisincorrect--possiblyunbalanced(,{,or[.>>[x,y]=eig(A)-0.47170.70300.70300.1031+0.0815i0.1031-0.0815iy=5.319500000-0.1609+1.2406i00000-0.1609-1.2406i000000.0012+0.3676i0000-0.6460-0.13830.3144i-0.1383+0.3144i-0.8162-0.8162-0.1951-0.0755+0.0097i-0.0755-0.0097i0.0683-0.2365i0.0683+0.2365i-0.5397-0.34070.4670i-0.3407+0.4670i0.3555+0.3575i0.3555-0.3575i-0.1757-0.0709+0.2072i-0.0709-0.2072i-0.0065+0.0418i-0.0065-0.0418i>>CI=0.3195/4

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