1、 D5,E1 D5,E6 D3,E4 D3,E1 D2,E3 D2,E1 D1,E2 D1,E1 E6,F E4,F E3,F E2,F E1,F /: c, x;endsetsdata: c = 17.3 -20.2 15.7 -30.2 18.4 -0.2 13.8 -50.2 17.3 -20.2 18.4 -0.2 12.2 -70.2 15.7 -30.2 17.3 -20.2 18.4 -0.2 5 30 50 60 80; enddatan = size(year);max = sum(trans: c * x);for(year(i)| i #ne# 1 #and# i #ne
2、# n: sum(trans(i,j): x(i,j) - sum(trans(j,i): x(j,i) = 0;);sum(trans(i,j)| i #eq# 1 : x(i,j) = 1;sum(trans(j,i)| i #eq# n : x(j,i) = 1;for(trans: bin(x);求解得到 Global optimal solution found. Objective value: 139.0000 Objective bound: Infeasibilities: 0.000000 Extended solver steps: 0 Total solver iter
3、ations: Elapsed runtime seconds: 0.09 Model Class: PILP Total variables: 25 Nonlinear variables: Integer variables: Total constraints: 17 Nonlinear constraints: Total nonzeros: 75 Nonlinear nonzeros: Variable Value Reduced Cost N 16.00000 0.000000 C( A2, B3) 17.30000 0.000000 C( A2, B1) -20.20000 0.
4、000000 C( B3, C4) 15.70000 0.000000 C( B3, C1) -30.20000 0.000000 C( B1, C2) 18.40000 0.000000 C( B1, C1) -0.2000000 0.000000 C( C4, D5) 13.80000 0.000000 C( C4, D1) -50.20000 0.000000 C( C2, D3) 17.30000 0.000000 C( C2, D1) -20.20000 0.000000 C( C1, D2) 18.40000 0.000000 C( C1, D1) -0.2000000 0.000
5、000 C( D5, E1) 12.20000 0.000000 C( D5, E6) -70.20000 0.000000 C( D3, E4) 15.70000 0.000000 C( D3, E1) -30.20000 0.000000 C( D2, E3) 17.30000 0.000000 C( D2, E1) -20.20000 0.000000 C( D1, E2) 18.40000 0.000000 C( D1, E1) -0.2000000 0.000000 C( E6, F) 5.000000 0.000000 C( E4, F) 30.00000 0.000000 C(
6、E3, F) 50.00000 0.000000 C( E2, F) 60.00000 0.000000 C( E1, F) 80.00000 0.000000 X( A2, B3) 1.000000 -17.30000 X( A2, B1) 0.000000 20.20000 X( B3, C4) 1.000000 -15.70000 X( B3, C1) 0.000000 30.20000 X( B1, C2) 0.000000 -18.40000 X( B1, C1) 0.000000 0.2000000 X( C4, D5) 1.000000 -13.80000 X( C4, D1)
7、0.000000 50.20000 X( C2, D3) 0.000000 -17.30000 X( C2, D1) 0.000000 20.20000 X( C1, D2) 0.000000 -18.40000 X( C1, D1) 0.000000 0.2000000 X( D5, E1) 1.000000 -12.20000 X( D5, E6) 0.000000 70.20000 X( D3, E4) 0.000000 -15.70000 X( D3, E1) 0.000000 30.20000 X( D2, E3) 0.000000 -17.30000 X( D2, E1) 0.00
8、0000 20.20000 X( D1, E2) 0.000000 -18.40000 X( D1, E1) 0.000000 0.2000000 X( E6, F) 0.000000 -5.000000 X( E4, F) 0.000000 -30.00000 X( E3, F) 0.000000 -50.00000X( E2, F) 0.000000 -60.00000 X( E1, F) 1.000000 -80.00000 Row Slack or Surplus Dual Price1 0.000000 0.000000 2 139.0000 1.000000 3 0.000000
9、0.000000 4 0.000000 0.000000 5 0.000000 0.000000 6 0.000000 0.000000 7 0.000000 0.000000 8 0.000000 0.000000 9 0.000000 0.000000 10 0.000000 0.000000 11 0.000000 0.000000 12 0.000000 0.000000 13 0.000000 0.000000 14 0.000000 0.000000 15 0.000000 0.000000 16 0.000000 0.000000 17 0.000000 0.000000 18
10、0.000000 0.000000该设备的使用情况为A2B3C4D5E1F。即在使用了5年后对设备进行更新,按此策略执行,总收益最大为139万元。2. 运输问题设Xi是从A煤矿到第i个城市的总运量;Yi是从B煤矿到第i个城市的总运量。按照题意用Lingo建模X1 + Y1 = 290;X2 + Y2 = 250;X3 + Y3 = 270;X1 + X2 + X3 = 400;Y1 + Y2 + Y3 = 450;min = 15*X1 + 18*X2 + 22*X3 + 21*Y1 + 25*Y2 + 16*Y3; 14650.00 3 0.06 LP 6Nonlinear constrai
11、nts: 18 X1 150.0000 0.000000 Y1 140.0000 0.000000 X2 250.0000 0.000000 Y2 0.000000 1.000000 X3 0.000000 12.00000 Y3 310.0000 0.0000001 0.000000 -5.000000 2 0.000000 -8.000000 3 40.00000 0.000000 4 0.000000 -10.00000 5 0.000000 -16.00000 6 14650.00 -1.000000A往甲运送150万吨,B往甲运送140万吨。A往乙运送250万吨,B往乙运送0万吨。A
12、往丙运送0万吨,B往丙运送310万吨。最小总开销为14650万元。3. 生产计划与库存管理(1)生产分为4个季度。每个季度的生产量为;订货量为;成本为,包括第i季度的生产成本,j-i季度的储存成本之和,i-j季度的延期成本之和。设为第i个季度的生产在第j季度交货的交货量。用Lingo建立模型 season /1.4/: a, b; Routes(season, season): a = 13, 15, 15, 13; b = 10, 14, 20, 8; c = 5, 6, 7, 8, 8, 5, 6, 7, 12, 9, 6, 7, 15, 12, 9, 6;min = sum(Routes:for(season(i):sum(season(j): x(i,j) = a(i);for(season(j):sum(season(i): x(i,j) = b(j); Objective valu
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