MATLAB练习题及求解.doc

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1、建立模型并求解练习1：如图,P,Q到河岸的距离分别为10,8,试选一点R,使得R到河岸的距离及RP,RQ的和最小.PQR6解：设点R的坐标为（x，y）则有：=PR+R(y)+RQ=+y+在lingo中输入：RP=@sqrt((x)^2+(10-y)^2); RY=y; RQ=@sqrt((x-6)^2+(y-8)^2); min=RP+RY+RQ; 则可得到结果：Localoptimalsolutionfound.Objectivevalue:14.19615Infeasibilities:0.3330669E-14

2、Extendedsolversteps:5Totalsolveriterations:200VariableValueReducedCostRP5.4641020.000000X4.732051-0.2154924E-07Y7.2679490.000000RY7.2679490.000000RQ1.4641020.000000RowSlackorSurplusDualPrice10.000000-1.00000020.000000-1.00000030.000000-1.000000414.19615-1.00000

3、0所以R（4.732,7.268）。练习2：要把7种规格的包装箱装到两辆铁路平板车上去，箱子宽、高相同，而厚度和重量不同，下表给出了它们的厚度、重量和数量。c1c2c3c4c5c6c7厚度（厘米）重量(千克)数量48.72000852.03000761.31000972.0500648.74000652.02000464.010008每辆平板车有10.2米长的地方用于装箱（像面包片那样），载重40吨。由于货运限制，对c5，c6，c7三种包装箱的装载有如下特殊约束：它们所占的空间（厚度）不得超过302.7厘米。试把包装

4、箱装到平板车上，使浪费的空间最小。解：程序为：min=2040-x11*48.7-x21*48.7-(x12+x22)*52-(x13+x23)*61.3-(x14+x24)*72-(x15+x25)*48.7-(x16+x26)*52-(x17+x27)*64;x11+x21<=8;x12+x22<=7;x13+x23<=9;x14+x24<=6;x15+x25<=6;x16+x26<=4;x17+x27<=8;x11*48.7+x12*52+x13*61.3+x14*72+x15*48.7+x16*52+x17*

5、64<=1020;x21*48.7+x22*52+x23*61.3+x24*72+x25*48.7+x26*52+x27*64<=1020;x15*47.8+x16*52+x17*64<=302.7;x25*47.8+x26*52+x27*64<=302.7;x11*2+x12*3+x13*1+x14*0.5+x15*4+x16*2+x17*1<=40;x21*2+x22*3+x23*1+x24*0.5+x25*4+x26*2+x27*1<=40; @gin(x11);@gin(x12);@gin(x13);@gin

6、(x14);@gin(x15);@gin(x16);@gin(x17);@gin(x21);@gin(x22);@gin(x23);@gin(x24);@gin(x25);@gin(x26);@gin(x27);运行结果：Globaloptimalsolutionfound.Objectivevalue:0.000000Objectivebound:0.000000Infeasibilities:0.1154632E-12Extendedsolversteps:140191Totalsolveriterations:

7、262446VariableValueReducedCostX110.000000-48.70000X216.000000-48.70000X125.000000-52.00000X222.000000-52.00000X132.000000-61.30000X236.000000-61.30000X145.000000-72.00000X240.000000-72.00000X152.000000-48.70000X250.000000-48.70000X161.000000-52.00000X260.000000

8、-52.00000X172.000000-64.00000X274.000000-64.00000RowSlackorSurplusDualPrice10.000000-1.00000022.0000000.00000030.0000000.00000041.0000000.00000051.0000000.00000064.0000000.0