Ｌｉｎｇｏ进阶篇

dianslm · 发表于 2005-9-14 06:12:34

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>2004电力市场输电阻塞管理
<

>要准确计费，必须给出第i台机组出力和对应其当前段价的函数关系式，这是个难点，没几个队能正确写出，好在当年没有最佳答案，大家就可各显神通，错了评委也难知道．
<

>想用LINGO解，必用其自定义函数@user(），我们可定义其为阶跃函数，从而解决本问题．
 
MODEL: SETS: i6/1..6/:y,c; i8/1..8/:x,x0,u,v,T,TT,q,vv,uu,xx; i12/1..12/:z6; i10/1..10/:; cc(i6,i8):A; dd(i8,i10):e,h,ee; ENDSETS p=0.5; !j0=303; !r=0; j0=356; r=0.397; <a href="http://www.shumo.com/bbs/mailtmin=0.25*@sum(i8(i):p*j0*(-x(i))*@user(-x(i))+T(i)*@user(x(i" target="_blank" >min=0.25*@sum(i8(i):p*j0*(-x(i))*@user(-x(i))+T(i)*@user(x(i</A>)));
@for(i8(i):T(i)=x(i)*(TT(i)-j0));
@for(i8(i):TT(i)=h(i,1)*@user(x0(i)+x(i))+(h(i,2)-h(i,1))*@user(x0(i)+x(i)-ee(i,1))+(h(i,3)-h(i,2))*@user(x0(i)+x(i)-ee(i,2))+(h(i,4)-h(i,3))*@user(x0(i)+x(i)-ee(i,3))+(h(i,5)-h(i,4))*@user(x0(i)+x(i)-ee(i,4))+(h(i,6)-h(i,5))*@user(x0(i)+x(i)-ee(i,5))+(h(i,7)-h(i,6))*@user(x0(i)+x(i)-ee(i,6))+(h(i,8)-h(i,7))*@user(x0(i)+x(i)-ee(i,7))+(h(i,9)-h(i,8))*@user(x0(i)+x(i)-ee(i,8))+(h(i,10)-h(i,9))*@user(x0(i)+x(i)-ee(i,9)));
@sum(i8(i):x(i))=0; @sum(i8(j):A(1,j)*(x(j)+x0(j)))+c(1)<z6(1)*(1+z6(7)*r); @sum(i8(j):A(2,j)*(x(j)+x0(j)))+c(2)<z6(2)*(1+z6(8)*r); <a href="http://www.shumo.com/bbs/mailt-@sum(i8(j):A(3,j)*(x(j)+x0(j)))+c(3)<z6(3)*(1+z6(9)*r" target="_blank" >-@sum(i8(j):A(3,j)*(x(j)+x0(j)))+c(3)<z6(3)*(1+z6(9)*r</A>); @sum(i8(j):A(4,j)*(x(j)+x0(j)))+c(4)<z6(4)*(1+z6(10)*r); @sum(i8(j):A(5,j)*(x(j)+x0(j)))+c(5)<z6(5)*(1+z6(11)*r); @sum(i8(j):A(6,j)*(x(j)+x0(j)))+c(6)<z6(6)*(1+z6(12)*r);
@for(i6(i):y(i)=@sum(i8(j):A(i,j)*(x0(j)+x(j)))+c(i)); @for(i8(i):xx(i)=x0(i)+x(i)); @FOR(i8(i)

BND(vv,xx,uu)); @FOR(i8(i)

BND(-v,x,u)); @for(i8(i)

free(x(i))); @for(i8(i):@free(xx(i))); @for(i6(i):@free(y(i))); @for(i8(i):@free(q(i))); @for(i8(i):@free(TT(i))); @for(i8(i):@free(T(i)));
DATA: A=0.0828, 0.0483, 0.0530, 0.1199, -0.0254, 0.1220, 0.1216, -0.0012, -0.0546, 0.1279, 0, 0.0333, 0.0868, -0.1124, -0.0189, 0.0987, -0.0695, 0.0616, -0.1566, -0.0099, 0.1245, 0.0021, -0.0025, -0.2014, -0.0345, -0.1024, 0.2052, -0.0208, -0.0118, 0.0060, 0.1449, 0.0765, 0.0005, 0.2433, -0.0646, -0.0411, -0.0652, 0.0703, -0.0043, -0.0089, 0.2378, -0.0602, -0.0779, 0.0930, 0.0469, 0.0001, 0.1659, 0.0007;
 c=110.2965 131.2289 -108.8732 77.4817 132.9745 120.6633;
 !v=63.0000 21.0000 48.0000 39.0000 27.0000 45.0000 34.9000 50.9000; !u=3.0000 ,9.0000 ,48.0000 , 0 , 27.0000 , 15.0000 , 7.1000 , 3.1000; u=3.0000 7.0000 9.8000 0 17.0000 5.0000 0.0000 0; v= 63.0000 23.0000 86.2000 39.0000 37.0000 55.0000 42.0000 54.00;
z6=165, 150, 160, 155, 132, 162, 0.13, 0.18, 0.09, 0.11, 0.15, 0.14 ; !x0=150.0000 79.0000 180.0000 99.5000 125.0000 140.0000 95.0000 113.9000; x0= 150.0000 81.0000 218.2000 99.5000 135.0000 150.0000 102.100 117.0000;
uu=190 89 280 116 155 180 125 160; vv=0 0 0 0 0 0 0 0; !x1=153 86.87 228 90.1124 152 95.3222 60.1 117;
e=70 0 50 0 0 30 0 0 0 40 30 0 20 8 15 6 2 0 0 8 110 0 40 0 30 0 20 40 0 40 55 5 10 10 10 10 15 0 0 1 75 5 15 0 15 15 0 10 10 10 95 0 10 20 0 15 10 20 0 10 50 15 5 15 10 10 5 10 3 2 70 0 20 0 20 0 20 10 15 5;
h=-505 0 124 168 210 252 312 330 363 489 -560 0 182 203 245 300 320 360 410 495 -610 0 152 189 233 258 308 356 415 500 -500 150 170 200 255 302 325 380 435 800 -590 0 116 146 188 215 250 310 396 510 -607 0 159 173 205 252 305 380 405 520 -500 120 180 251 260 306 315 335 348 548 -800 153 183 233 253 283 303 318 400 800;
 ee=70 70 120 120 120 150 150 150 150 190 30 30 50 58 73 79 81 81 81 89 110 110 150 150 180 180 200 240 240 280 55 60 70 80 90 100 115 115 115 116 75 80 95 95 110 125 125 135 145 155 95 95 105 125 125 140 150 170 170 180 50 65 70 85 95 105 110 120 123 125 70 70 90 90 110 110 130 140 155 160;
 ENDDATA END </TD></TR>
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shug_d · 发表于 2005-9-15 01:06:40

<

>顶!!!
<

>敢问Lingo达到这个高度需要几个月？[em01][em01][em01]

dianslm · 发表于 2005-9-14 06:22:46

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0cm 0cm 0pt; mso-layout-grid-align: none"> 结果为：（按１５分钟计）
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0cm 0cm 0pt; mso-layout-grid-align: none">　　　W=982.4MW时 阻塞费用 F=5805元 J0=303
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0cm 0cm 0pt 24.05pt; TEXT-INDENT: 24pt"><v:shapetype><v:stroke joinstyle="miter"></v:stroke><v:formulas><v:f eqn="if lineDrawn pixelLineWidth 0"></v:f><v:f eqn="sum @0 1 0"></v:f><v:f eqn="sum 0 0 @1"></v:f><v:f eqn="prod @2 1 2"></v:f><v:f eqn="prod @3 21600 pixelWidth"></v:f><v:f eqn="prod @3 21600 pixelHeight"></v:f><v:f eqn="sum @0 0 1"></v:f><v:f eqn="prod @6 1 2"></v:f><v:f eqn="prod @7 21600 pixelWidth"></v:f><v:f eqn="sum @8 21600 0"></v:f><v:f eqn="prod @7 21600 pixelHeight"></v:f><v:f eqn="sum @10 21600 0"></v:f></v:formulas><v:path connecttype="rect" gradientshapeok="t" extrusionok="f"></v:path><lock aspectratio="t" v:ext="edit"></lock></v:shapetype><v:shape><v:imagedata></v:imagedata></v:shape>Ｘ增量=[0.492 9 48 -9.42 27 -43.27 -34.9 3.1] 
<v:shape><v:imagedata></v:imagedata></v:shape>Ｘ０=[150.0 79.0 180.0 99.5 125.0 140.0 95.0 113.9 ]
 由 x =<v:shape> <v:imagedata></v:imagedata></v:shape>+<v:shape> <v:imagedata></v:imagedata></v:shape> 得分配方案即8个电机组出力如下: 
 X = [150.49 88 228 90 152 96.73 60.1 117]
 由 Y=A*x+c; 及分配预案,得到6条线路的潮流值如下:
 Y= [165 150 -155.1 124.63 131.62 158.96] 
 
W =1052.8 Mw时用二分法搜索出最小的R =0.397，对应如下结果：
 阻塞费用 F=2512元 J0=356
 <v:shape><v:imagedata></v:imagedata></v:shape>=[1.425 7 9.8 0 17 -35.22 0 0 ] 
<v:shape><v:imagedata></v:imagedata></v:shape>=[150.0 81.0 218.2 99.5 135 150 102 117 ]
 由x =<v:shape> <v:imagedata></v:imagedata></v:shape>+<v:shape> <v:imagedata></v:imagedata></v:shape>得8个电机组出力为: 
 x=[151.43 88 228 99.5 152 114 102.1 117 ]
6条线路的潮流值为: Y= [173.52 147.44 -155.32 130.6 132.32 167.03 ]

dianslm · 发表于 2005-9-14 06:25:30

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>1997（A）产品参数设计的 Lingo 程序 
<

>MODEL: SETS: NODES/1..7/:x,d,c,t,b,a; ENDSETS
<

>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 2005-9-14 06:28:34

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>1997（A）产品参数设计的 Lingo 程序没用１０８个循环，若要用也可，但难度大增，我这就不帖出了，帖出一般人也看不懂！

dianslm · 发表于 2005-9-14 06:31:41

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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="http://www.shumo.com/bbs/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</TD></TR></TABLE>

dzhdiv · 发表于 2005-9-14 06:33:47

<

>太难了吧

dianslm · 发表于 2005-9-14 06:36:13

想学就简单！

dianslm · 发表于 2005-9-14 06:59:27

<

>2004电力市场输电阻塞管理-----最简明的模型

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>min R （R为裕量的统一升缩因子，在0~1中变化）
<

>min F （F堵塞费用）
sum（dx(i)）=0 dx（i）为x（i）的增量
0<=x（i）<=x（i）的上界（即第i台机组最大发电量）
|y（i）|<y（i）的上限*（1+R*裕度T（i）） （6个不等式） 
vlb<dx<ulb ulb（i）----爬坡对dx（i）的上限限制
------------------------------------------------------------------------------------------------------------ 上为求解最精巧的模型，其中：
sum（x（i））=1052.8 
x（i）=x0（i）+dx（i） 
 x=（x（1） x（2）...x（8）） 
<a href="http://www.shumo.com/bbs/mailt用Lingo8.0解，须会使用@USE()函数，即定义阶跃函数，此外，还须懂一些特定的小计巧才行" target="_blank" >用Lingo8.0解，须会使用@USER()函数，即定义阶跃函数，此外，还须懂一些特定的小计巧才行</A>。

dianslm · 发表于 2005-9-14 07:13:22

<

>论创新：
<

>--------人人都爱说创新，其实，真正懂得辩证法的人应该知道，创新是一把双刃剑。人类上月球，有的人构思了用气球载人登月球；大学导数符号有的学生不按国际规范来书写；自动化生产线上，有的工人不按操作规范来做；家长规定的纪律孩子不遵守；高速公路上来个“创意”----违犯交通规则行车，....。其时，创新是有条件的，创新在大多数情况下要合理，合情。一个好的“创新”须要一段时间的“不创新”来发扬光大！我们的社会需要大量不创新的人，也需要一定数量创新的人，当然也需要少数意想天开的人！
<

>我们以前太强调整齐划一了,这可能会伤害到我们的创新,反过来,当我们太强调....!
---------------------------------------------------------------------------------------------------------
何时该创新，何时不该创新？------这是有较高水平和素质的人才能掌握的！

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