优化方法课程设计优化方法课程设计一用一用 lindo 求解求解 LP程序程序max 2x1+x2st 3x1+5x2=0;-x1-x22+2=0;运行结果：运行结果：Local optimal solution found at iteration: 56Objective value: 10.00000Variable Value Reduced CostX1 1.000000 0.000000X2 1.000000 0.000000Row Slack or Surplus Dual Price1 10.00000 -1.0000002 0.000000 -0.40000013 0.000000 1.200000三用三用 lingo 求解非线性规划求解非线性规划 (1)非线性规划模型非线性规划模型min=xijtij. 411ixij 411jxij(2)程序程序model:sets:r/1.4/;d/1.4/;link(r,d):t,x;endsetsdata:t=56 74 61 6363 69 65 7157 77 63 6755 76 62 62;enddatamin=sum(link(i,j):t(i,j)*x(i,j);for(d(j):sum(r(i):x(i,j)=1);for(r(i):sum(d(j):x(i,j)=1);end（3）运行结果）运行结果Global optimal solution found at iteration: 9Objective value: 249.0000Variable Value Reduced CostT( 1, 1) 56.00000 0.000000T( 1, 2) 74.00000 0.000000T( 1, 3) 61.00000 0.000000T( 1, 4) 63.00000 0.000000T( 2, 1) 63.00000 0.000000T( 2, 2) 69.00000 0.000000T( 2, 3) 65.00000 0.000000T( 2, 4) 71.00000 0.000000T( 3, 1) 57.00000 0.000000T( 3, 2) 77.00000 0.000000T( 3, 3) 63.00000 0.000000T( 3, 4) 67.00000 0.000000T( 4, 1) 55.00000 0.000000T( 4, 2) 76.00000 0.000000T( 4, 3) 62.00000 0.000000T( 4, 4) 62.00000 0.000000X( 1, 1) 0.000000 0.000000X( 1, 2) 0.000000 0.000000X( 1, 3) 1.000000 0.000000X( 1, 4) 0.000000 0.000000X( 2, 1) 0.000000 12.00000X( 2, 2) 1.000000 0.000000X( 2, 3) 0.000000 9.000000X( 2, 4) 0.000000 13.00000X( 3, 1) 1.000000 0.000000X( 3, 2) 0.000000 2.000000X( 3, 3) 0.000000 1.000000X( 3, 4) 0.000000 3.000000X( 4, 1) 0.000000 0.000000X( 4, 2) 0.000000 3.000000X( 4, 3) 0.000000 2.000000X( 4, 4) 1.000000 0.000000Row Slack or Surplus Dual Price1 249.0000 -1.0000002 0.000000 -55.000003 0.000000 -73.000004 0.000000 -60.000005 0.000000 -62.000006 0.000000 -1.0000007 0.000000 4.0000008 0.000000 -2.0000009 0.000000 0.000000z=(0,0,1,0,0,1,0,0,1,0,0,1,0,0,0,0,0,0,1) )0(T则则A为蝶泳，为蝶泳，B为蛙泳，为蛙泳，C为自由泳，为自由泳，D为仰泳。为仰泳。