急向高手请教一线性规划问题
大家好！由于毕业论文需要用到一多元线性规划问题，下面的程序用1STOPT1.5可以运行，但得不到正确的解。恳求dingd主任及其他高手赐教！急。。。。谢谢！菜鸟上。Constant
A(1:30)=;
Constant
B(1:30)=;
Constant
C(1:30)=;
Constant
D(1:30)=;
Constant
E(1:30)=;
Constant
F(1:30)=;
Constant
G(1:30)=;
Constant
H(1:30)=;
Constant
I(1:30)=;
Constant
J(1:30)=;
Parameters x(1:30);
MaxFunction sum(i=1:30)(x*A);
sum(i=1:30)(x*B)<=1503800*0.85;
sum(i=1:30)(x*C)<=77729400*0.84;
sum(i=1:30)(x*E)<=68117000*0.70;
sum(i=1:30)(x*F)<=6174.17*0.75;
sum(i=1:30)(x*G)<=8647.14*0.85;
sum(i=1:30)(x*H)<=2600*0.87;
sum(i=1:30)(x*I)<=6503.75*0.88;
sum(i=1:30)(x*J)<=70000*0.85; 估计你的版本太老了：
算法：单纯形线性规划法
该线性规划的最大(Max)为：207663.949488004
参数最优解为：
x1: 0
x2: 0
x3: 0
x4: 0
x5: 0
x6: 0
x7: 0
x8: 0
x9: 0
x10: 0
x11: 0
x12: 0
x13: 0
x14: 0
x15: 0
x16: 0
x17: 0
x18: 0
x19: 0
x20: 0
x21: 0
x22: 0
x23: 0
x24: 0
x25: 0
x26: 980009.200037771
x27: 0
x28: 0
x29: 0
x30: 0
怎么购买最新版本1STOPT2.5啊？
谢谢dingd主任！软件公司的主页打不开，还有什么方式购买软件呢？dingd主任你应该有2.5版本的吧？请问多少钱呢？急啊。。。另外，假如用lingo做这么多变量的线性规划，编程方便吗？还没接触过lingo.
Lingo实现
Lingo实现:---------------
model:
sets:
Data1/1..30/: X, A, B, C, D, E, F, G, H, II, JJ;
endsets
data:
A=0.3708,0.2587,0.3091,0.3770,0.2557,0.2854,0.2727,0.2795,0.2697,0.2755,0.3209,0.2561,0.1651,0.2685,0.0710,0.2710,0.2510,0.3054,0.2690,0.2431,0.2582,0.2796,0.2763,0.2437,0.2571,0.2119,0.2636,0.2804,0.3216,0.4519;
B=0.7441,0.1908,0.3121,0.3100,0.3904,0.1013,0.0891,0.2759,0.0742,0.7759,0.1940,0.1289,1.3394,1.4505,0.6534,0.4316,0.2150,1.8177,1.7159,1.0255,0.3092,0.1661,0.1968,0.1430,0.0930,0.0563,0.0712,0.5637,0.9781,1.1058;
C=19.644,21.689,25.817,30.491,26.506,5.486,10.230,8.575,0.870,161.38,1.667,1.073,40.857,55.956,50.553,7.318,1.537,14.310,40.926,16.527,5.408,4.413,4.131,6.193,1.529,1.366,9.60,2.009,490.11,207.24;
D=0.5965,0.4954,0.5093,0.5637,0.3518,0.4354,0.1516,0.3130,0.6243,0.4581,0.5151,0.7145,0.8498,0.8726,0.8730,0.8274,0.6177,0.6624,0.8799,0.8311,0.3619,0.6064,0.7667,0.5875,0.4915,0.7456,0.8044,0.5749,0.7126,0.1464;
E=36.971,24.352,21.267,16.212,29.563,19.065,15.835,18.330,3.673,117.60,4.255,4.566,5.832,31.856,27.653,4.856,3.636,8.998,12.269,8.454,6.388,5.040,4.229,3.457,2.092,2.640,2.978,5.802,39.302,134.60;
F=
0.0017,0.0082,0.0058,0.0094,0.0034,0.0004,0.0028,0.0023,0.0001,0.0487,0.0002,0.0001,0.0009,0.0047,0.0059,0.0005,0.0002,0.0007,0.0012,0.0006,0.0003,0.0002,0.0003,0.0005,0.0001,0.0001,0.0007,0.0002,0.0010,0.0033;
G=
0.4950,0.1635,0.2859,0.3909,0.2540,0.0321,0.0670,0.2397,0.0267,1.0853,0.0115,0.0210,0.7801,0.8326,0.5767,0.2149,0.0247,1.7927,0.8215,1.1323,0.0409,0.0536,0.0567,0.0282,0.0224,0.0075,0.0540,0.0236,7.8313,0.0846;
H=
2.908,0.424,0.079,0.194,0.134,0.020,0.000,0.055,0.000,0.192,0.000,0.000,0.442,0.428,0.038,0.000,0.000,1.838,0.694,0.504,0.031,0.113,0.181,0.025,0.007,0.000,0.000,0.000,0.275,0.000;
II=
0.006094,0.004926,0.002922,0.006672,0.001887,0.000262,0.000808,0.005097,0.000218,0.010922,0.000184,0.000286,0.009784,0.006877,0.003541,0.001533,0.000168,0.025378,0.005921,0.00611,0.000507,0.000745,0.000521,0.000762,0.000121,0.000104,0.000293,0.000191,0.044437,0.000788;
JJ=
0.0980,0.0310,0.0486,0.0446,0.0655,0.1032,0.0881,0.0787,0.0722,0.0506,0.0728,0.1040,0.0086,0.0317,0.0217,0.0533,0.0555,0.0844,0.0179,0.0236,0.0551,0.0514,0.0532,0.0298,0.0370,0.0193,0.0473,0.0940,0.0187,0.0990;
enddata
max=@sum(Data1:X*A);
@sum(Data1:X*B)<1503800*0.85;
@sum(Data1:X*C)<77729400*0.84;
@sum(Data1:X*E)<68117000*0.70;
@sum(Data1:X*F)<6174.17*0.75;
@sum(Data1:X*G)<8647.14*0.85;
@sum(Data1:X*H)<2600*0.87;
@sum(Data1:X*II)<6503.75*0.88;
@sum(Data1:X*JJ)<70000*0.85;
-----------------------------------------
求解结果:
------------------------------------------
Global optimal solution found.
Objective value: 207663.9
Total solver iterations: 1
Variable Value Reduced Cost
X( 1) 0.000000 13.61460
X( 2) 0.000000 4.360720
X( 3) 0.000000 7.768528
X( 4) 0.000000 10.66723
X( 5) 0.000000 6.920647
X( 6) 0.000000 0.6215320
X( 7) 0.000000 1.620273
X( 8) 0.000000 6.492824
X( 9) 0.000000 0.4846640
X( 10) 0.000000 30.38784
X( 11) 0.000000 0.4013333E-02
X( 12) 0.000000 0.3372200
X( 13) 0.000000 21.87533
X( 14) 0.000000 23.25523
X( 15) 0.000000 16.22270
X( 16) 0.000000 5.800641
X( 17) 0.000000 0.4468573
X( 18) 0.000000 50.34435
X( 19) 0.000000 22.94111
X( 20) 0.000000 31.74815
X( 21) 0.000000 0.8973613
X( 22) 0.000000 1.234779
X( 23) 0.000000 1.325664
X( 24) 0.000000 0.5530440
X( 25) 0.000000 0.3757747
X( 26) 980009.2 0.000000
X( 27) 0.000000 1.262080
X( 28) 0.000000 0.3863787
X( 29) 0.000000 220.9387
X( 30) 0.000000 1.938332
----------------------------------------------------------------------
用Lingo得到的结果相同, 如果将B到J的数据做成矩阵, 将更简单了.
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