九组交通规划课程设计

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=1,j=3350253023005819000185.85793315

2、415.576638982.890371758i=1,j=4250253016504174500205.52146091815.244505152.995732274i=1,j=5130253010702707100304.8675344514.811388513.401197382i=2,j=1500319024207719800126.21460809815.859299012.48490665i=2,j=2175031903080982520047.46737106716.100461071.386294361i=2,j=3600319023007337000106.3969296551

3、5.80844062.302585093i=2,j=4140319016505263500254.94164242315.476306763.218875825i=2,j=5200319010703413300205.29831736715.043190122.995732274i=3,j=1420216024205227200156.04025471115.469386322.708050201i=3,j=2400216030806652800105.99146454715.710548382.302585093i=3,j=390021602300496800086.80239476315.

4、41852792.079441542i=3,j=4360216016503564000155.88610403115.086394072.708050201i=3,j=580216010702311200354.38202663514.653277433.555348061i=4,j=1200164024203968800205.29831736715.193974342.995732274i=4,j=2190164030805051200255.24702407215.43513643.218875825i=4,j=3370164023003772000155.91350300615.143

5、115922.708050201i=4,j=4800164016502706000106.68461172814.810982092.302585093i=4,j=580164010701754800304.38202663514.377865453.401197382i=5,j=1100100024202420000354.60517018614.69927813.555348061i=5,j=2140100030803080000254.94164242314.940440153.218875825i=5,j=380100023002300000354.38202663514.648419

6、683.555348061i=5,j=4100100016501650000354.60517018614.316285853.555348061i=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)Qij612260045.291046580.1604246521444.63013779240

7、048.036104620.0729457513369.637626581900044.732482340.0373899746121.858039417450041.250335920.0331687346363.763211270710037.11342620.0209176629078.902089771980047.926517840.0592885634136.37943982520050.831309910.2067558421258.024006733700047.335450930.0729457513320.488157526350043.650679540.02573606

8、98678.890638341330039.273044390.0331687346058.722904522720043.577033720.0460027974847.176602665280046.218206660.0729457513242.115684496800043.039608010.0940127342342.594069356400039.689241360.0460027974414.728256231120035.708890540.01755474910408.71555396880040.744861720.0331687346285.7828505120043.

9、214378730.0257360698592.142687377200040.242364550.0460027974476.253457270600037.109745960.0729457512603.175201175480033.388087320.0209176628167.588035242000036.111950740.01755474910526.20279308000038.300670320.0257360697615.169627230000035.666590210.01755474910396.38551165000032.890167290.0175547499

10、587.091355107000029.591681350.0592885632553.960261 OD12345合计11444.6303369.6386121.8586363.7639078.90226378.79124136.3791258.0243320.4888678.8906058.72323452.50434847.1773242.1162342.5944414.72810408.71625255.33146285.7838592.1434476.2532603.1758167.58830124.942510526.2037615.17010396.3869587.0912553

11、.96040678.81合计27240.17224077.09126657.57931647.64736267.889145890.378重新计算和 550026378.791480027240.1720.2085008370.176210341480023452.504500024077.0910.2046689770.207666283350025255.331400026657.5790.1385846020.150051136300030124.942300031647.6470.0995852540.094793777240040678.81240036267.8890.058998

12、7760.06617424 通过无约束重力模型计算得到的OD表不满足出行分布的约束条件，因此还要用其他方法继续进行迭代，这里采用平均增长系数法进行迭代运算。 用平均增长系数法第一次迭代计算OD表 O D12345合计增长系数1277.883701.1661097.502965.0481246.8744288.4731.28252787.731259.364588.9221299.502820.4823756.0011.2283762.933561.293338.078515.1511065.6383243.0931.07924866.7961319.975558.718253.001676.9

13、283675.4180.816251237.9291015.351086.682737.212159.8434237.0160.5664合计3933.2723857.1483669.9023769.9143969.76519200.001增长系数1.22041.29631.08990.79580.6046 用平均增长系数法第二次迭代计算OD表 55004288.47348003933.2721.2825077831.22035801248003756.00150003857.1481.2779549311.29629456835003243.09340003669.9021.079216661

14、1.08994736130003675.41830003769.9140.8162336910.79577412124004237.01624003969.7650.5664363790.604569792O D12345合计增长系数1347.752904.0841301.8871002.8211176.4744733.0181.1622983.999333.834697.2551347.406772.2894134.7831.16093877.211666.679366.673482.952897.1533290.6681.06364882.6551394.242532.509203.924

15、08.8913422.2170.876651105.962945.662899.981502.11993.5893547.3130.6766合计4197.5794244.5013798.3053539.2183348.39619127.999增长系数1.14351.1781.05310.84760.7167 用平均增长系数法第三次迭代计算OD表 55004733.01848004197.5791.1620492461.14351629848004134.78350004244.5011.1608831711.17799477535003290.66840003798.3051.06361383

16、21.05310131830003422.21730003539.2180.8766247140.84764487524003547.31324003348.3960.6765684340.7167611O D12345合计增长系数1400.8831057.7981441.9381007.6821105.1865013.4871.09721133.763390.398771.856775.582725.0423796.6411.26433968.059747.217388.071461.523798.6343363.5041.04064891.5441432.318513.798175.807

17、325.7613339.2280.898451006.472876.895778.335382.66865.23109.570.7718合计4400.7214504.6263893.9982803.2623019.82318622.43增长系数1.09071.111.02721.07010.7947 用平均增长系数法第四次迭代计算OD表 55005013.48748004400.7211.0970408421.0907303648003796.64150004504.6261.2642754481.10997006235003363.50440003893.9981.0405814891.02

18、722189430003339.22830002803.2620.898411251.0701818124003109.5724003019.8230.7718108930.794748566O D 12345合计增长系数1438.521167.2861531.5281091.9361045.395274.661.042721335.009463.45884.353905.282746.4394334.5331.107431031.618803.464401.227487.023732.8783456.211.01274886.7041438.32494.693173.046275.78332

19、68.5460.91785937.298825.062700.125352.43651.072865.9910.8374合计4629.1494697.5824011.9263009.7232851.5619199.94增长系数1.03691.06440.9970.99680.8416 第四次迭代之后，满足设定收敛条件=10%，停止迭代，第四次迭代计算后的OD表就为最终的预测OD表。a.第一次分配60%（1） 确定最短路线。OD点对最短路线节点号OD点对最短路线节点号141-2-4515-6-4-2-1151-2-4-6-5545-6-4171-2-4-6-7575-6-7414-2-1717-

20、6-4-2-1454-6-5747-6-4474-6-7757-6-5（2）分配od量。（3） 路权修正。第二次分配30%（1） 确定最短路线。OD点对最短路线节点号OD点对最短路线节点号141-3-4515-3-1151-3-5545-3-4171-3-5-7575-7414-3-1717-5-3-1454-3-5747-5-3-4474-3-5-7757-5（2） 分配od量。（3） 路权修正。第三次分配10%（1） 确定最短路线。OD点对最短路线节点号OD点对最短路线节点号141-3-4515-3-1151-3-5545-3-4171-3-5-7575-7414-2-1717-5-3-1

21、454-3-5747-6-4474-6-7757-5（2） 分配od量。将三次分配加和。b. 第一次分配60% 1-4 交通量为330×60%=198（1） 计算各交通节点i至出行终点4的最短路权节点号1234567521.503.534.5（2） 令i=出行起点r，即从出行起点r开始进行分配，本例中，i=r=。（3） 判别有效路段(4) 计算边权（5） 计算节点点权(6) 计算各有效路段的OD量分配率(7) 计算有效路段交通分配量（8） 取i=2或3只有一条有效路段所以1-5 交通量为258（1）计算各交通节点i至出行终点5的最短路权节点号12345678.55.543.500.5

22、2（2） 令i=出行起点r，即从出行起点r开始进行分配，本例中，i=r=。（3） 判断有效路段。（4） 计算边权（5） 计算节点点权.（6） 计算各有效路段的OD量分配率（7）计算有效路段交通分配量I=2时 ，有效路径只有一个Q（2，4）=Q(1,2)=141I=4时，有效路径只有一个Q(4,6)=Q（2，4）=141I=6时,有效路径只有一条Q（6，5）=Q(4,6)=141i=3时L（3-4，5）=5 L（3-5，5）=4Q(3,4)=38 Q(3,5)=79I=4时Q（4，6）=Q(6,5)=Q（3，4）=38节点号12345679.56.564.521.50节点号12345670355

23、8.589.5同理得出：1-4 1-5 1-7 4-1 4-5Q(1,2)=139 Q(1,2)=141 Q(1,2)=186 Q(4,2)=192 Q(4,6)=60Q(1,3)=59 Q(1,3)=117 Q(1,3)=114 Q(2,1)=192 Q(6,5)=60Q(2,4)=139 Q(2,4)=141 Q(2,4)=186Q(3,4)=59 Q(4,6)=141 Q(4,6)=186 Q(6,5)=141 Q(6,7)=186 Q(3,4)=348 Q(3,4)=57 Q(3,5)=79 Q(3,5)=57 Q(4,6)=38 Q(4,6)=57 Q(6,5)=38 Q(6,7)=

24、57 Q(5,6)=45 Q(5,7)=12 Q(6,7)=454-7 5-1 5-4 5-7 7-1Q(4,6)=120 Q(5,3)=112 Q(5,3)=13 Q(5,6)=284 Q(7,5)=104Q(6,7)=120 Q(5,6)=134 Q(5,6)=59 Q(5,7)=76 Q(7,6)=196 Q(3,1)=112 Q(3,4)=13 Q(6,7)=284 Q(5,3)=47 Q(6,4)=134 Q(6,4)=59 Q(5,6)=57 Q(4,2)=134 Q(3,1)=47 Q(2,1)=134 Q(6,4)=57 Q(4,2)=57 Q(2,1)=57 Q(6,4)=196 Q(2,1)=1967-4 7-5Q(7,5)=35 Q(7,6)=237Q(7,6)=115 Q(7,5)=63Q(5,3)=16 Q(6,5)=237Q(5,6)=19Q(3,4)=16Q(6,4)=19Q(6,4)=115计算路权同理算出40%最后与60%所得出的数据加和得到（3） 逐次平均法 Q=0时，路权为最短路径为OD1-2-41-2-4-6-51-2-4-6-