未在 cplex 中顯示任何解決方案

Question

在我對 pevch[i][j] 和 pevdis[i][j] 的方程進行積分后，CPLEX 未顯示任何解

詮釋噸= 24; 整數 n=10； 整數j=0； 范圍數=1..n； 范圍tavail=1..t； float soc[number][tavail]=...; // 0..4 float pl [tavail] = [10000000,70000,9000000,60000,60000,12000000,120000,60000,0000,4000000,150000000000,000000,000000,00000000,00000000,000000000000,0000000000000000000000000000000000000000000000000000000000,000000009000099999999999999系別因此里6000000,4000000,15000000,9000000,12000000,6000000,8000000]; //EV 的 soc 在 0..11 //generation float pg[tavail]=[10000000,9500000,8500000,11000000,600000,7500000,10000000,9500000,8500000,11000000,600000,750000000050000,1000000,100 11000000,600000,7500000,10000000,9500000,8500000,11000000,600000,7500000]; //target load at 0..11 float pt[tavail]=[10000000,10000000,10000000,10000000,10000000,10000000,10000000,10000000,10000000,10000000,10000000,10000000,10000000,10000000,10000000,10000000,10000000, 10000000,10000000,10000000,10000000,10000000,10000000,10000000]; //總線電壓為 0..11 float v[tavail]=[240,232,229,233,230,235,228,234,227,229,231,230,226,232,233,230,236,233,231,232,232,233,233,230]; //目標總線電壓為 0..11 時的總線電壓 float vt[tavail]=[230,230,230,230,230,230,230,230,230,230,230,230,230,230,230,230,230,230,230,230,230,230,230,230]; //決策變量充電功率 ev dvar float pevch[tavail][number] in 0..100000; //ev 的決策變量放電功率 dvar float pevdis[tavail][number] in 0..100000; //目標 function 最小化 sum(i in tavail)((pt[i]-pl[i])+sum(j in number)-pevch[i][j]+sum(j in number)pevdis[i][ j]); 受制於

forall(i in tavail,j in number) if(pt[i]-pl[i]<0 && soc[j][i]>0.7&& v[i]<vt[i]) pevch[i][j ]==(0.9-soc[j][i])*100000/(24-i); 否則 pevdis[i][j]==0;

forall(i in tavail,j in number) if(pt[i]-pl[i]>0 && soc[j][i]<=0.7 && v[i]>vt[i]) pevch[i][ j]==(0.7-soc[j][i])*100000/(24-i); 否則 pevch[i][j]==0; }

Answer 1

如果你命名你的約束，cplex 將提供松弛，這將幫助你調試你的 model。

你可以改變

forall(i in tavail,j in number) if(pt[i]-pl[i]<0 && soc[j][i]>0.7&& v[i]<vt[i]) pevch[i][j]==(0.9-soc[j][i])*100000/(24-i); else pevdis[i][j]==0;

    forall(i in tavail,j in number) if(pt[i]-pl[i]>0 && soc[j][i]<=0.7 && v[i]>vt[i]) pevch[i][j]==(0.7-soc[j][i])*100000/(24-i); else pevch[i][j]==0; }

進入

forall(i in tavail,j in number) ct1:if(pt[i]-pl[i]<0 && soc[j][i]>0.7&& v[i]<vt[i]) pevch[i][j]==(0.9-soc[j][i])*100000/(24-i); else pevdis[i][j]==0;

    forall(i in tavail,j in number) ct2:if(pt[i]-pl[i]>0 && soc[j][i]<=0.7 && v[i]>vt[i]) pevch[i][j]==(0.7-soc[j][i])*100000/(24-i); else pevch[i][j]==0; }