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1、《交通规划》课程设计任务书九组组员：杜锋121378王凯121391薛万冲111368一采用下面的无约束重力模型两边取对数，得令y=，=，，，，则次方程为二元线性回归方程、、为待标定系数通过已知表格获取25个样本数据，如表1所示。表格1样本点QijOiDjOiDjCijyx1x2i=1,j=1120025302420612260057.09007683615.62749741.609437912i=1,j=2600253030807792400106.39692965515.868659462.302585093i=
2、1,j=3350253023005819000185.85793315415.576638982.890371758i=1,j=4250253016504174500205.52146091815.244505152.995732274i=1,j=5130253010702707100304.8675344514.811388513.401197382i=2,j=1500319024207719800126.21460809815.859299012.48490665i=2,j=2175031903080982520
3、047.46737106716.100461071.386294361i=2,j=3600319023007337000106.39692965515.80844062.302585093i=2,j=4140319016505263500254.94164242315.476306763.218875825i=2,j=5200319010703413300205.29831736715.043190122.995732274i=3,j=1420216024205227200156.04025471115.469386
4、322.708050201i=3,j=2400216030806652800105.99146454715.710548382.302585093i=3,j=390021602300496800086.80239476315.41852792.079441542i=3,j=4360216016503564000155.88610403115.086394072.708050201i=3,j=580216010702311200354.38202663514.653277433.555348061i=4,j=12001
5、64024203968800205.29831736715.193974342.995732274i=4,j=2190164030805051200255.24702407215.43513643.218875825i=4,j=3370164023003772000155.91350300615.143115922.708050201i=4,j=4800164016502706000106.68461172814.810982092.302585093i=4,j=580164010701754800304.38202
6、663514.377865453.401197382i=5,j=1100100024202420000354.60517018614.69927813.555348061i=5,j=2140100030803080000254.94164242314.940440153.218875825i=5,j=380100023002300000354.38202663514.648419683.555348061i=5,j=4100100016501650000354.60517018614.316285853.555348
7、061i=5,j=5580100010701070000126.36302810413.883169212.48490665采用最小二乘法对这25个样本数据进行标定，得出=1.633，=0.244，=1.137，则获得的二元线性回归方程为。通过=，，，可得，，，即标定的重力模型为利用已标定的重力模型求解分布交通量Oi*Dj(Oi*Dj)^0.244Cij^(-1.137)Qij612260045.291046580.1604246521444.63013779240048.036104620.072945751336
8、9.637626581900044.732482340.0373899746121.858039417450041.250335920.0331687346363.763211270710037.11342620.0209176629078.902089771980047.926517840.0592885634136.379439825200
