2004年数学建模电力市场的输电阻塞管理部分代码_2004 数模国赛b

通过出力方案得到段价：

function y = select(x)
dj = [ -505  -505   124   124   124   252   252   252   252   489
  -560  -560   182   203   245   300   320   320   320   495
  -610  -610   152   152   233   233   308   356   356   500
  -500   150   170   200   255   302   325   325   325   800
  -590     0   116   116   188   215   215   310   396   510
  -607  -607   159   173   173   252   305   380   380   520
  -500   120   180   251   260   306   315   335   348   548
  -800  -800   183   183   253   253   303   318   400   800];
drl = [    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];
[m,n]=size(drl);
for i = 1:m
    ch = x(i);
    for j=1:n
        if j==1
            if ch<=drl(i,j)
              y(i) = dj(i,j);
              break;
            elseif ch>drl(i,j)&&ch<drl(i,j+1)
                y(i) = dj(i,j+1);
                break;
            end
        elseif j==n
            if ch>=drl(i,j)
                y(i) = dj(i,j);
                break;
            elseif ch>drl(i,j-1)&&ch<drl(i,j)
                y(i) = dj(i,j);
                break;
            end
        else
            if ch==drl(i,j)
                y(i) = dj(i,j);
                break;
            elseif ch>drl(i,j)
                for k = j+1:n-1
                    if ch<dj(i,k)
                        y(i) = dj(i,k);
                        break;
                    end
                end 
        end
    end
    end
 
end

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通过负荷得到基本出力方案：

clear;clc;
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
];
dj = [-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];
Fh = input('请输入负荷需求量(MW)：');
S = 0;
pri = 0;
A = [dj(:) drl(:)];
AA = sortrows(A,1);
SS = zeros(1,8);
v = [2.2	1	3.2	1.3	1.8	2	1.4     1.8];
g = [120	73	180	80	125	125	81.1	90];
gmax = g+v*15;
gmin = g-v*15;
for i = 1:80
    if sum(SS)<Fh
        pri = pri+AA(i,1);
        [row,~] = (find((AA(i,1)==dj)&(AA(i,2)==drl)));
        for j =1:length(row)
            if (SS(row(j))<gmax(row(j)))&&(SS(row(j)) + AA(i,2))<gmax(row(j))
                SS(row(j)) = SS(row(j)) + AA(i,2);
                
            elseif (SS(row(j))<gmax(row(j)))&&(SS(row(j)) + AA(i,2))>=gmax(row(j))
                SS(row(j)) = gmax(row(j));
                
            end
        end
    else
        disp('清算价为:')
        qs = AA(i-1,1)
        break;
        
    end
end
 remain = sum(SS)-Fh;
 [row1,~] = (find((AA(i-1,1)==dj)&(AA(i-1,2)==drl)));
 SS(row1) = SS(row1) - remain;
 disp('各个机组出力为:')
 SS
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