1、中国石油大学-建模作业 数学规划上机实践作业一1 首先练习一下书中线性规划案例。2 某企业和用户签定了设备交货合同，该企业各季度的生产能力、每台设备的生产本钱和每季度末的交货量见下表，假设生产出的设备当季度不交货，每台设备每季度需要支付保管费0.1万元，试问在遵守合同的条件下，企业应如何安排生产方案，才能使年消消耗用最低？季度工厂生产能力（台）交货量（台）每台设备生产本钱（万元/台）1251512.02352011.03302511.54202012.5变量定义：设第季度生产台，；模型假设：，每个季度都交货交满第四季度交货后无设备剩余；目标函数：约束条件： ， ，。Lingo程序：model:
2、min=12xx1+(x1-15)x0.1+11xx2+(x2+x1-15-20)x0.1+x3x11.5+(x3+x2+x1-15-20-25)x0.1+x4x12.5;x2+x1=35;x3+x2+x1=60;x4+x3+x2+x1=80;x1=15;x2=35;x3=30;x4=20;gin(x1);gin(x2);gin(x3);gin(x4);End程序运行结果：Global optimal solution found. Objective value: 913.5000 Objective bound: 913.5000 Infeasibilities: 0.000000 Ext
3、ended solver steps: 0 Total solver iterations: 0 Variable Value Reduced Cost X1 15.00000 12.30000 X2 35.00000 11.20000 X3 30.00000 11.60000 X4 0.000000 12.50000 Row Slack or Surplus Dual Price 1 913.5000 -1.000000 2 0.000000 0.000000 3 15.00000 0.000000 4 20.00000 0.000000 5 0.000000 0.000000 6 0.00
4、0000 0.000000 7 0.000000 0.000000 8 20.00000 0.000000建模结果：第一季度生产15台，第二季度生产35台，第三季度生产30台，第四季度生产0台，最小值913.5万元。3 五名选手的百米成绩如上所示；如何选拔队员组成4100米混合泳接力队 讨论:丁的蛙泳成绩退步到1152；戊的自由泳成绩进步到575, 组成接力队的方案是否应该调整？变量定义：设表示第个人参加第个工程， 表示第个人没有参加第个工程，为第个人的第个工程的成绩（）。目标函数：。约束条件：每个人最多参加一个工程，所以，每个工程必须有且最多有一个人选择，所以，同时有。Lingo程序Mode
5、l:sets:person/1.5/;position/1.4/;link(person,position):x,a;endsetsdata:a=66.8,75.6,87,58.6, 57.2,66,66.4,53, 78,67.8,84.6,59.4, 70,74.2,69.6,57.2, 67.4,71,83.8,62.4;enddatamin=sum(link:xxa);for(person(i):sum(position(j):x(i,j)=1;);for(position(j):sum(person(i):x(i,j)=1;);for(link:bin(x);end程序运行结果Glo
6、bal optimal solution found. Objective value: 253.2000 Objective bound: 253.2000 Infeasibilities: 0.000000 Extended solver steps: 0 Total solver iterations: 0 Variable Value Reduced Cost X( 1, 1) 0.000000 66.80000 X( 1, 2) 0.000000 75.60000 X( 1, 3) 0.000000 87.00000 X( 1, 4) 1.000000 58.60000 X( 2,
7、1) 1.000000 57.20000 X( 2, 2) 0.000000 66.00000 X( 2, 3) 0.000000 66.40000 X( 2, 4) 0.000000 53.00000 X( 3, 1) 0.000000 78.00000 X( 3, 2) 1.000000 67.80000 X( 3, 3) 0.000000 84.60000 X( 3, 4) 0.000000 59.40000 X( 4, 1) 0.000000 70.00000 X( 4, 2) 0.000000 74.20000 X( 4, 3) 1.000000 69.60000 X( 4, 4)
8、0.000000 57.20000 X( 5, 1) 0.000000 67.40000 X( 5, 2) 0.000000 71.00000 X( 5, 3) 0.000000 83.80000 X( 5, 4) 0.000000 62.40000 A( 1, 1) 66.80000 0.000000 A( 1, 2) 75.60000 0.000000 A( 1, 3) 87.00000 0.000000 A( 1, 4) 58.60000 0.000000 A( 2, 1) 57.20000 0.000000 A( 2, 2) 66.00000 0.000000 A( 2, 3) 66.
9、40000 0.000000 A( 2, 4) 53.00000 0.000000 A( 3, 1) 78.00000 0.000000 A( 3, 2) 67.80000 0.000000 A( 3, 3) 84.60000 0.000000 A( 3, 4) 59.40000 0.000000 A( 4, 1) 70.00000 0.000000 A( 4, 2) 74.20000 0.000000 A( 4, 3) 69.60000 0.000000 A( 4, 4) 57.20000 0.000000 A( 5, 1) 67.40000 0.000000 A( 5, 2) 71.000
10、00 0.000000 A( 5, 3) 83.80000 0.000000 A( 5, 4) 62.40000 0.000000 Row Slack or Surplus Dual Price 1 253.2000 -1.000000 2 0.000000 0.000000 3 0.000000 0.000000 4 0.000000 0.000000 5 0.000000 0.000000 6 1.000000 0.000000 7 0.000000 0.000000 8 0.000000 0.000000 9 0.000000 0.000000 10 0.000000 0.000000模
11、型结论甲乙丙丁4人组队分别参加自由泳、蝶泳、仰泳、蛙泳的比赛，成绩为253.2s=4132 。模型分析当队员丁的蛙泳成绩有较大退步，只有1152；而戊的自由泳成绩进步到575，对lingo程序进行修改，重新建模。程序以及运行结果为Model:sets:person/1.5/;position/1.4/;link(person,position):x,a;endsetsdata:a=66.8,75.6,87,58.6, 57.2,66,66.4,53, 78,67.8,84.6,59.4, 70,74.2,75.2,57.2, 67.4,71,83.8,57.5;enddatamin=sum(l
12、ink:xxa);for(person(i):sum(position(j):x(i,j)=1;);for(position(j):sum(person(i):x(i,j)=1;);for(link:bin(x);end Global optimal solution found. Objective value: 257.7000 Objective bound: 257.7000 Infeasibilities: 0.000000 Extended solver steps: 0 Total solver iterations: 0 Variable Value Reduced Cost
13、X( 1, 1) 0.000000 66.80000 X( 1, 2) 0.000000 75.60000 X( 1, 3) 0.000000 87.00000 X( 1, 4) 0.000000 58.60000 X( 2, 1) 1.000000 57.20000 X( 2, 2) 0.000000 66.00000 X( 2, 3) 0.000000 66.40000 X( 2, 4) 0.000000 53.00000 X( 3, 1) 0.000000 78.00000 X( 3, 2) 1.000000 67.80000 X( 3, 3) 0.000000 84.60000 X(
14、3, 4) 0.000000 59.40000 X( 4, 1) 0.000000 70.00000 X( 4, 2) 0.000000 74.20000 X( 4, 3) 1.000000 75.20000 X( 4, 4) 0.000000 57.20000 X( 5, 1) 0.000000 67.40000 X( 5, 2) 0.000000 71.00000 X( 5, 3) 0.000000 83.80000 X( 5, 4) 1.000000 57.50000 A( 1, 1) 66.80000 0.000000 A( 1, 2) 75.60000 0.000000 A( 1, 3) 87.00000 0.000000 A( 1, 4) 58.60000 0.000000 A( 2, 1) 57.20000 0.000000 A( 2, 2) 66.00000 0.000000 A( 2, 3) 66.40000 0.000000
