高志通运筹学实验报告

PAGE运筹学实验报告姓名：高志通学号：20120404302班级：信息与计算科学1203指导老师：武梦梦

目录1、线性规划问题： 22、运输问题： 33、一般整数规划问题： 44、指派问题： 5相关问题说明：实验性质和教学目的本实验是运筹学课内安排的上机操作实验。目的在于了解、熟悉计算机Lingo软件在运筹学模型求解中的作用，激发学习兴趣，提高学习效果，增强自身的动手能力，提高实际应用能力。实验基本要求要求学生：1.实验前认真做好理论准备，仔细阅读实验指导书；2.遵从教师指导，认真完成实验任务，按时按质提交实验报告。主要参考资料1．LINGO软件3.优化建模与LINDO/LINGO软件，清华大学出版社，20054．运筹学编写组主编，运筹学（第四版），清华大学出版社，20125．胡运权主编，运筹学教程（第二版），清华大学出版社，2003—PAGE13—实验1线性规划问题(1)给出原始代码；max4x1+3x2st9x1+8x2<127x1+11x2<249x1+11x2<13end(2)计算结果(包括灵敏度分析，求解结果粘贴)；Globaloptimalsolutionfound.Objectivevalue:5.333333Infeasibilities:0.000000Totalsolveriterations:2VariableValueReducedCostX11.3333330.000000X20.0000000.5555556RowSlackorSurplusDualPrice15.3333331.00000020.0000000.4444444314.666670.00000041.0000000.000000Rangesinwhichthebasisisunchanged:ObjectiveCoefficientRangesCurrentAllowableAllowableVariableCoefficientIncreaseDecreaseX14.000000INFINITY0.6250000X23.0000000.5555556INFINITYRighthandSideRangesRowCurrentAllowableAllowableRHSIncreaseDecrease212.000001.00000012.00000324.00000INFINITY14.66667413.00000INFINITY1.000000(3)回答下列问题(手写)：a)最优解及最优目标函数值是多少；答：最优解：X1=1.333333X2=0.000000；最优目标函数值=5.333333b)资源的对偶价格各为多少，并说明对偶价格的含义；答：y1=0.4444444；y2=0.000000；y3=0.000000；对偶价格的含义：表示每增加一个单位(约束右边的常数)，目标值改变的数量（在最大化问题中目标函数值是增加，在最小化问题中目标函数值是减少）。c)为了使目标函数值增加最多，让你选择一个约束条件，将它的常数项增加一个单位，你将选择哪一个约束条件？这时目标函数值将是多少？答：选择第一个约束条件，若常数项增加一个单位，目标函数值将=5.77777d)对x2的目标函数系数进行灵敏度分析；答：目标函数中x2变量原来的费用系数为3，允许增加（AllowableIncrease）=0.5555556、允许减少（AllowableDecrease）=INFINITY，说明当它在[3-∞，3+0.5555556]=[-∞，3.5555556]范围变化时，最优基保持不变。e)对第2个约束的约束右端项进行灵敏度分析；答：当它在[24-14.66667，24+∞]=[9.33333，∞]范围变化时，最优基保持不变。f)结合本题的结果解释“ReducedCost”的含义。答：ReducedCost指为了使某个变量在解中的数值增加一个单位，目标函数值必须付出的代价。在本题中x1的ReducedCost值为0，当x1增加一个单位时，目标函数值将不变。x2的ReducedCost值为0.5555556，即当x2增加一个单位时，目标函数值将减少0.5555556；实验2运输问题如下是一个最小费用运输问题。产销量及单位运价如下表。(1)给出原始代码；Sets:warehouse/1..6/:WH;customer/1..8/:V;routes(warehouse,customer):c,x;endsetsdata:WH=60,55,51,43,41,52;V=35,37,22,32,41,32,43,38;c=6,2,6,7,4,2,5,9,3,6,5,3,8,9,8,2,7,6,1,5,7,4,3,3,5,2,7,3,9,2,7,1,2,3,9,5,5,2,6,5,5,7,2,2,3,1,4,3;enddata[obj]min=@sum(routes:c*x);@for(warehouse(i):[sup]@sum(customer(j):x(i,j))<=WH(i));@for(customer(j):[dem]@sum(warehouse(i):x(i,j))=V(j));end(2)计算结果(决策变量求解结果粘贴)Globaloptimalsolutionfound.Objectivevalue:634.0000Infeasibilities:0.000000Totalsolveriterations:16VariableValueReducedCostWH(1)60.000000.000000WH(2)55.000000.000000WH(3)51.000000.000000WH(4)43.000000.000000WH(5)41.000000.000000WH(6)52.000000.000000V(1)35.000000.000000V(2)37.000000.000000V(3)22.000000.000000V(4)32.000000.000000V(5)41.000000.000000V(6)32.000000.000000V(7)43.000000.000000V(8)38.000000.000000C(1,1)6.0000000.000000C(1,2)2.0000000.000000C(1,3)6.0000000.000000C(1,4)7.0000000.000000C(1,5)4.0000000.000000C(1,6)2.0000000.000000C(1,7)5.0000000.000000C(1,8)9.0000000.000000C(2,1)3.0000000.000000C(2,2)6.0000000.000000C(2,3)5.0000000.000000C(2,4)3.0000000.000000C(2,5)8.0000000.000000C(2,6)9.0000000.000000C(2,7)8.0000000.000000C(2,8)2.0000000.000000C(3,1)7.0000000.000000C(3,2)6.0000000.000000C(3,3)1.0000000.000000C(3,4)5.0000000.000000C(3,5)7.0000000.000000C(3,6)4.0000000.000000C(3,7)3.0000000.000000C(3,8)3.0000000.000000C(4,1)5.0000000.000000C(4,2)2.0000000.000000C(4,3)7.0000000.000000C(4,4)3.0000000.000000C(4,5)9.0000000.000000C(4,6)2.0000000.000000C(4,7)7.0000000.000000C(4,8)1.0000000.000000C(5,1)2.0000000.000000C(5,2)3.0000000.000000C(5,3)9.0000000.000000C(5,4)5.0000000.000000C(5,5)5.0000000.000000C(5,6)2.0000000.000000C(5,7)6.0000000.000000C(5,8)5.0000000.000000C(6,1)5.0000000.000000C(6,2)7.0000000.000000C(6,3)2.0000000.000000C(6,4)2.0000000.000000C(6,5)3.0000000.000000C(6,6)1.0000000.000000C(6,7)4.0000000.000000C(6,8)3.0000000.000000X(1,1)0.0000004.000000X(1,2)31.000000.000000X(1,3)0.0000003.000000X(1,4)0.0000005.000000X(1,5)29.000000.000000X(1,6)0.0000000.000000X(1,7)0.0000000.000000X(1,8)0.0000008.000000X(2,1)0.0000000.000000X(2,2)0.0000003.000000X(2,3)0.0000001.000000X(2,4)32.000000.000000X(2,5)0.0000003.000000X(2,6)0.0000006.000000X(2,7)0.0000002.000000X(2,8)1.0000000.000000X(3,1)0.0000007.000000X(3,2)0.0000006.000000X(3,3)8.0000000.000000X(3,4)0.0000005.000000X(3,5)0.0000005.000000X(3,6)0.0000004.000000X(3,7)43.000000.000000X(3,8)0.0000004.000000X(4,1)0.0000003.000000X(4,2)6.0000000.000000X(4,3)0.0000004.000000X(4,4)0.0000001.000000X(4,5)0.0000005.000000X(4,6)0.0000000.000000X(4,7)0.0000002.000000X(4,8)37.000000.000000X(5,1)35.000000.000000X(5,2)0.0000001.000000X(5,3)0.0000006.000000X(5,4)0.0000003.000000X(5,5)0.0000001.000000X(5,6)6.0000000.000000X(5,7)0.0000001.000000X(5,8)0.0000004.000000X(6,1)0.0000004.000000X(6,2)0.0000006.000000X(6,3)14.000000.000000X(6,4)0.0000001.000000X(6,5)12.000000.000000X(6,6)26.000000.000000X(6,7)0.0000000.000000X(6,8)0.0000003.000000RowSlackorSurplusDualPriceOBJ634.0000-1.000000SUP(1)0.0000001.000000SUP(2)22.000000.000000SUP(3)0.0000003.000000SUP(4)0.0000001.000000SUP(5)0.0000001.000000SUP(6)0.0000002.000000DEM(1)0.000000-3.000000DEM(2)0.000000-3.000000DEM(3)0.000000-4.000000DEM(4)0.000000-3.000000DEM(5)0.000000-5.000000DEM(6)0.000000-3.000000DEM(7)0.000000-6.000000DEM(8)0.000000-2.000000实验3一般整数规划问题新天地购物广场某一层各时段（每2h为一时段）需要的服务员人数见下表。按规定，服务员连续工作8h（即四个时段）为一班。现要求安排服务员的工作时间，使服务部门服务员总数最少。时段12345678服务员最少数目201618222616106(1)给出原始代码；sets:time/1..8/:required,start;endsetsdata:required=201618222616106;enddatamin=@sum(time:start);@for(time(J):@sum(time(I)|I#le#4:start(@wrap(J+I+2,8)))>=required(J));end(2)计算结果(决策变量求解结果粘贴)Globaloptimalsolutionfound.Objectivevalue:46.00000Infeasibilities:0.000000Totalsolveriterations:3VariableValueReducedCostREQUIRED(1)20.000000.000000REQUIRED(2)16.000000.000000REQUIRED(3)18.000000.000000REQUIRED(4)22.000000.000000REQUIRED(5)26.000000.000000REQUIRED(6)16.000000.000000REQUIRED(7)10.000000.000000REQUIRED(8)6.0000000.000000START(1)26.000000.000000START(2)0.0000000.000000START(3)0.0000000.000000START(4)4.0000000.000000START(5)16.000000.000000START(6)0.0000000.000000START(7)0.0000000.000000START(8)0.0000000.000000RowSlackorSurplusDualPrice146.00000-1.00000020.000000-1.00000030.0000000.00000048.0000000.00000054.0000000.00000060.000000-1.000000714.000000.000000810.000000.000000914.000000.000000答：第1时段安排26人，第2、3时段不增加人，第4时段安排再4人，第5时段再安排16人，总共至少安排服务员46人。实验4指派问题已知如下效率矩阵，求极大化指派问题。B1B2B3B4B5A14871512A279171410A3691287A46714610A56912106(1)给出原始代码；sets:AD/A1..A5/:A;BD/B1..B5/:B;link(AD,BD):c,x;endsetsdata:c=48715127917141069128767146106912106;enddatamax=@sum(link(i,j):c(i,j)*x(i,j));@for(BD(j):@sum(AD(i):x(i,j))=1);@for(AD(i):@sum(BD(j):x(i,j))=1);@for(link:@bin(x));(2)计算结果(决策变量求解结果粘贴)Globaloptimalsolutionfound.Objectivevalue:57.00000Objectivebound:57.00000Infeasibilities:0.000000Extendedsolversteps:0Totalsolveriterations:0VariableValueReducedCostA(A1)0.0000000.000000A(A2)0.0000000.000000A(A3)0.0000000.000000A(A4)0.0000000.000000A(A5)0.0000000.000000B(B1)0.0000000.000000B(B2)0.0000000.000000B(B3)0.0000000.000000B(B4)0.0000000.000000B(B5)0.0000000.000000C(A1,B1)4.0000000.000000C(A1,B2)8.0000000.000000C(A1,B3)7.0000000.000000C(A1,B4)15.000000.000000C(A1,B5)12.000000.000000C(A2,B1)7.0000000.000000C(A2,B2)9.0000000.000000C(A2,B3)17.000000.000000C(A2,B4)14.000000.000000C(A2,B5)10.000000.000000C(A3,B1)6.0000000.000000C(A3,B2)9.0000000.000000C(A3,B3)12.000000.000000C(A3,B4)8.0000000.000000C(A3,B5)7.0000000.000000C(A4,B1)6.0000000.000000C(A4,B2)7.0000000.000000C(A4,B3)14.000000.000000C(A4,B4)6.0000000.000000C(A4,B5)10.000000.000000C(A5,B1)6.0000000.000000C(A5,B2)9.0000000.000000C(A5,B3)12.000000.000000C(A5,B4)10.000000.000000C(A5,B5)6.0000000.000000X(A1,B1)0.000000-4.000000X(A1,B2)0.000000-8.000000X(A1,B3)0.000000-7.000000X(A1,B4)1.000000-15.00000X(A1,B5)0.000000-12.00000X(A2,B1)0.000000-7.000000X(A2,B2)0.000000-9.000000X(A2,B3)1.000000