lingo程序问题 求助

Question

model:
!问题三：无网络安全约束的电力市场分配模型;
sets:
jizhu/jz1..jz8/:u,PG0,PGMIN,PGMAX,PG,MCP;
duan/d1..d10/:;
links(jizhu,duan):duanjia,drl,drll,MCPG;
endsets
data:
u=2.2 1 3.2 1.3 1.8 2 1.4 1.8;
PG0=120 73 180 80 125 125 81.1 90;
duanjia=-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;
drl=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;
drll=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

!目标函数;
min=@max(jizhu(i):MCP(i))*982.4;
!PGMAX;
@for(jizhu(k):PGMAX(k)=@smin(PG(k)+15*u(k),@sum(links:drl)));
!PGMIN;
@for(jizhu(k):PGMIN(k)=@smin(PG(k)-15*u(k),0));
!PGMIN<PG<PGMAX;
@for(jizhu(i):PGMIN(i)<PG(i);PG(i)<PGMAX(i));
!p;
@sum(jizhu(i):PG(i))=982.4;
!MCP;
@for(jizhu(i):@for(duan(j)|j#lt#10:b(i)=@if(drll(i,j)#gt#PG(i)#le#drll(i,j+1),j+1)));
!获取PG的段位,即在集合;
@for(jizhu(i):MCP(i)=@min(duanjia(i，b(i)));

Answer 1

错误有下：1，max后面要跟=，正确的为max=2,乘号*不能省略，如目标函数应写为：max=10*x1+9*x2+30*x3+20*x4-15*x5-4*x6-3*x7;3，每行结束时一定要用分号;来结束，如上行，4，约束的标号不能写，如2)15x5+12x6+10x7=0.2*(x1+x2+x3+x4);(x1+x6)/2+(x2+x7)/3=x5;x6=0.8*x3;x7=0.7*x4;结果：Globaloptimalsolutionfound.Objectivevalue:2320.346Infeasibilities:0.000000Totalsolveriterations:3VariableValueReducedCostX10.0000004.400433X2129.87010.000000X386.580090.000000X40.0000000.000000X577.922080.000000X669.264070.000000X70.0000000.6901670

追问

你回答的不是我的问题吧。。。