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1、xyo线性规划的简单应用 Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Evaluation only. Created with Aspose.Slid
2、es for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.使z=2x+y取得最大值的可行解为 ，且最大值为 ；复习引入1.已知二元一次不等式组x-y0 x+y-10 y-1（1）画出不等式组所表示的平面区域；满足 的解(x,y)都叫做可行解；z=2x+y 叫做 ；（2）设z=2x+y，则式中变量x,y满足的二元
3、一次不等式组叫做x,y的 ；y=-1x-y=0x+y=12x+y=0返回(-1,-1)(2,-1)使z=2x+y取得最小值的可行解 ，且最小值为 ；这两个最值都叫做问题的 。线性约束条件线性目标函数线性约束条件(2,-1)(-1,-1)3-3最优解xy011Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2
4、004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.例题分析 例1:某工厂
5、生产甲、乙两种产品.已知生产甲种产品1t需消 耗A种矿石10t、B种矿石5t、煤4t；生产乙种产品1吨需消 耗A种矿石4t、B种矿石4t、煤9t.每1t甲种产品的利润是600 元,每1t乙种产品的利润是1000元.工厂在生产这两种产品的 计划中要求消耗A种矿石不超过300t、消耗B种矿石不超过 200t、消耗煤不超过360t.甲、乙两种产品应各生产多少(精 确到0.1t),能使利润总额达到最大?甲产产品（1t）乙产产品（1t）资资源限额额（t） A种矿矿石（t） B种矿矿石（t） 煤（t） 利润润（元） 产品消耗量 资源列表：5104 6004 4 9 1000300200 360设生产甲、乙
6、两种产品.分别为x t、yt,利润总额为z元Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Evaluation only. Created with As
7、pose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.例题分析甲产产品（1t）乙产产品（1t）资资源限额额（t） A种矿矿石（t） B种矿矿石（t） 煤（t） 利润润（元） 产品消耗量资源列表：510460044 9 1000300200 360把题中限制条件进行转化：约束条件10x+4y
8、300 5x+4y200 4x+9y360 x0 y 0z=600x+1000y. 目标函数：设生产甲、乙两种产品.分别为x t、yt,利润总额为z元xtytEvaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose
9、 Pty Ltd.Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.例题分析解:设生产甲、乙两种产品.分别为x t、yt,利润总额为z=600x+1000y. 元, 那么10x+4y300
10、5x+4y200 4x+9y360 x0 y 0 z=600x+1000y. 作出以上不等式组所表示的可行域作出一组平行直线 600x+1000y=t，解得交点M的坐标为(12.4,34.4)5x+4y=2004x+9y=360由10x+4y=3005x+4y=2004x+9y=360600x+1000y=0M答:应生产甲产品约12.4吨，乙产品34.4吨，能使利润总额达到最大 。(12.4,34.4)经过可行域上的点M时,目标函数 在y轴上截距最大. 90300 xy10 2010754050 40此时z=600x+1000y取得最大值.Evaluation only.Evaluation
11、only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.C
12、reated with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.例题分析例2 要将两种大小不同规格的钢板截成A、B、C三种规格， 每张钢板可同时截得三种规格的小钢板的块数如下表所示 ： 解：设需截第一种钢板x张，第一种钢板y张，则 规格类型 钢板类型第一种钢板第二种钢板A规格B规格C规格2121312x+y15,x+2y18, x+3y27, x0 y0 作出可行域（如图）目标函数为 z=x+y
13、今需要A,B,C三种规格的成品分别为15，18，27块，问 各截这两种钢板多少张可得所需三种规格成品，且使所 用钢板张数最少。X张y张Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.Eva
14、luation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.例题分析x0y2x+y=15x+3y=27x+2y=18x+y =02x+y15,x+2y18, x+3y27, x0, xN y0 yN直线x+
15、y=12经过的整点是B(3,9)和C(4,8)，它们是最优解. 作出一组平行直线z=x+y ，目标函数z= x+yB(3,9) C(4,8) A(18/5,39/5)当直线经过点A时z=x+y=11.4,x+y=12解得交点B,C的坐标B(3,9)和C(4,8)调整优值法2 4 6181282724681015但它不是最优整数解.作直线x+y=12答（略）Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Evaluation only. Created with As
