lingo 一例通!!!!!

爱情无声 · 发表于 2005-7-21 19:27:30

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>一般来说LINGO多用于解决大规模数学规划。
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>用时要注意以下几点：
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>（1） 每条语句后必须使用分号“；”结束。问题模型必须由MODEL命令开始，END结束。
（2） 用MODEL命令来作为输入问题模型的开始，格式为MODEL：statement （语句）。
（3） 目标函数必须由“min =”或“max =”开头。
则上面的例子的输入就为
modul:
min=(x1-1)^2+(x2-2)^2;
x2-1=1;
x1+x2<=2;
end
我们即可得到最优值0.5。当X1=0.5，X2=1.5，及灵敏度分析。我们还可得作图分析。
对于大规模规划求解请参见LINGO的HELP文件。

爱情无声 · 发表于 2005-7-21 20:02:54

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>呵呵 ,很简单吧!!

ringer · 发表于 2005-7-21 22:16:43

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>你说的很简单哦，是这样吗？
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>这样的解法能获奖吗？

爱情无声 · 发表于 2005-7-21 22:50:04

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>意在解题而不在于深与浅！！

limw · 发表于 2005-7-22 00:47:40

呵呵，这个问题就简单，那其它的呢？

爱情无声 · 发表于 2005-7-22 02:00:33

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>规划问题用他解
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>其他的用matlab

dianslm · 发表于 2005-7-22 23:14:47

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>1997（A）产品参数设计的 Lingo 程序 

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> MODEL: SETS: NODES/1..7/:x,d,c,t,b,a; ENDSETS
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>min=ff; ff=1000*(@SUM(NODES:C)+1000*100*(f-1.5)^2+1000*100/9*@sum(NODES

d*x*t)^2)); f=174.42*(x(1)/x(5))*((x(3)/(x(2)-x(1)))^0.85)*g/h; g=(1-2.62*q*(x(4)/x(2))^1.16)^0.5; q=(1-0.36*(x(2)/x(4))^0.56)^1.5; h=(x(6)*x(7))^0.5; d(1)=f*(1/x(1)+0.85/(x(2)-x(1))); d(3)=0.85*f/x(3); d(5)=f/x(5); d(6)=f/(2*x(6)); d(7)=f/(2*x(7)); d(4)=f*1.31/x(2)^1.16*(z41+z42)/f4; z40=1-0.36*x(2)^0.56/x(4)^0.56; f4=1-2.62*z40^1.5*x(4)^1.16/x(2)^1.16; z41=1.16*x(4)^0.16*z40^1.5; z42=1.5*0.36*0.56*z40^0.5*x(2)^0.56/x(4)^0.4; d(2)=f*0.85/(x(2)-x(1))+1.31*f*x(4)^1.16*(z21+z22)/f4; z21=-1.16*z40^1.5/x(2)^2.16; z22=-1.5*0.36*0.56*z40^0.5/x(4)^0.56/x(2)^1.6;
@FOR(NODES( I): @BND(a,x,b)); @free(f); @free(g); @free(q); @free(h); @free(v); @free(z40); @free(f4); @free(z41); @free(z42); @free(z21);@free(z22); @free(ff);
DATA: a=0.075,0.225,0.075,0.075,1.125,12,0.5625; b=0.125,0.375,0.125,0.125,1.875,20,0.935; C=25,50,50,50,50,25,25; t=0.05,0.05,0.05,0.1,0.1,0.05,0.05; ENDDATA END
---------------------------------
Rows= 19 Vars= 26 No. integer vars= 0 Nonlinear rows= 18 Nonlinear vars= 24 Nonlinear constraints= 18 Nonzeros= 90 Constraint nonz= 84 Density=0.175
 ** WARNING ** Problem is poorly scaled. The units of the rows and variables should be changed so the coefficients cover a much smaller range.
 Optimal solution found at step: 56 Objective value: 748736.8=74.87万元/1000件产品
 Variable Value Reduced Cost FF 748736.8 1.000000 F 1.496842 0.0000000E+00 G 0.9350423 0.0000000E+00 H 2.598093 0.0000000E+00 Q 0.1798687 0.0000000E+00 Z41 0.1486261 0.0000000E+00 Z42 0.2301376 0.0000000E+00 F4 0.8743041 0.0000000E+00 Z40 0.3186425 0.0000000E+00 Z21 -1.735826 0.0000000E+00 Z22 -2.687810 0.0000000E+00 V 1.234568 0.0000000E+00 X( 1) 0.7500000E-01 0.0000000E+00 X( 2) 0.3750000 0.0000000E+00 X( 3) 0.1249971 0.0000000E+00 X( 4) 0.1200215 0.0000000E+00 X( 5) 1.494412 0.0000000E+00 X( 6) 12.00000 0.0000000E+00 X( 7) 0.5625072 0.0000000E+00

dianslm · 发表于 2005-7-22 23:20:23

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>1998（A）大基金投资组合的 Lingo 程序 
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> MODEL: SETS: QUARTERS/A0,A1,A2,A3,A4,A5,A6,A7,A8,A9,A10,A11,A12,A13,A14, A15/

,Q,X,R; ENDSETS MIN=0.3884*Y-0.6116*Z; <a href="mailtZ=5+@SUM(QUARTERS

(R-P)/(1+P)-0.05)*X" target="_blank" >Z=5+@SUM(QUARTERS

(R-P)/(1+P)-0.05)*X</A>); @FOR(QUARTERS:Y>X/(1+P)*Q); @SUM(QUARTERS:X)=100; DATA: R=0.05,0.096,0.185,0.494,0.239,0.081,0.14,0.407,0.312,0.336,0.368,0.118, 0.09,0.35,0.094,0.15;

=0,0.021,0.032,0.06,0.015,0.076,0.034,0.056,0.031,0.027,0.029,0.051, 0.057,0.027,0.045,0.076; Q=0,0.42,0.54,0.60,0.42,0.012,0.39,0.68,0.334,0.533,0.4,0.31,0.055, 0.46,0.053,0.23; ENDDATA END
-------------------------------------
Rows= 19 Vars= 18 No. integer vars= 0 ( all are linear) Nonzeros= 69 Constraint nonz= 63( 33 are +- 1) Density=0.191 Smallest and largest elements in absolute value= 0.311005E-02 100.000 No. < : 0 No. =: 2 No. > : 16, Obj=MIN, GUBs <= 4 Single cols= 0
 Optimal solution found at step: 12 Objective value: -16.64271
 Variable Value Reduced Cost Y 7.611293 0.0000000E+00 Z 32.04536 0.0000000E+00
 X( A0) 0.0000000E+00 0.1358472 X( A1) 0.0000000E+00 0.1215006 X( A2) 0.0000000E+00 0.7575390E-01 X( A3) 13.44662 0.0000000E+00 X( A4) 0.0000000E+00 0.3145336E-01 X( A5) 0.0000000E+00 0.1635851 X( A6) 0.0000000E+00 0.1037293 X( A7) 11.81989 0.0000000E+00 X( A8) 23.49474 0.0000000E+00 X( A9) 14.66566 0.0000000E+00 X( A10) 19.58005 0.0000000E+00 X( A11) 0.0000000E+00 0.1274384 X( A12) 0.0000000E+00 0.1473327 X( A13) 16.99304 0.0000000E+00 X( A14) 0.0000000E+00 0.1377493 X( A15) 0.0000000E+00 0.1243654

dianslm · 发表于 2005-7-22 23:24:44

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>1999（A）自动化车床的 Lingo 程序 .......
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>2001（B）公交车时刻表的 Lingo 程序 .......
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>............................
2004（B）电力系统管理的 Lingo 程序 .......

行者无疆 · 发表于 2005-7-23 18:51:07

可不简单啊！

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