第一次写程序，求改错
本帖最后由 华清池主 于 2015-3-14 18:40 编辑论文需要，没学过，匆匆忙忙用其他的改改了，不知道错误在哪里，求大神指正
model:
SETS:
et1/e11..e112/:e1;
et2/e21..e112/:e2;
yt/yt1..yt12/:t;
yt1/y11..y112/:y1;
yt2/y21..y212/:y2;
L:/l1,l2/:;
link（L,et1）:l,e;
link(l,y):l,y;
ENDSETS
DATA:
e1=-0.827014218,-0.447129909,1,1,0.613069647,1,-0.457264957,-0.348615091,-1,-0.628324057,-0.15125,0.224107143;
e2=0.172985782 0.552870091 0.04 0.265277778 -0.386930353 0.018867925 -1 0.651384909 -0.270500532 0.371675943 0.84875 -0.775892857;
t=0.172985782 0.552870091 1 1 0.613069647 1 0 0.651384909 0 0.371675943 0.84875 0.224107143;
y1=1 1 0 0 0 0 0.457264957 1 1 1 1 0;
y2=0 0 0.96 0.734722222 1 0.981132075 1 0 0.270500532 0 0 1;
ENDDATA
MAX=1-(2*@SUM(et1(t): @ABS(l1*e1(t)+l2*e2(t))))/(@SUM(yt(t): (yt(t)+l1*y1(t)+l2*y2(t)))+@SUM(et1(t): @ABS(l1*e1(t)+l2*e2(t))));
@for（@SUM（L(i):l(i)）=1）;
END
你的表述中，数学公式用得十分不标准，所以我不能完全明白你的意思，只能靠猜测，如此得到如下的LINGO代码：MODEL:
SETS:
P1/1,2/:L;
P2/1..12/:T;
LINK(P1,P2):E,Y;
ENDSETS
DATA:
E=-0.827014218,-0.447129909,1,1,0.613069647,1,-0.457264957,-0.348615091,-1,-0.628324057,-0.15125,0.224107143
0.172985782,0.552870091,0.04,0.265277778,-0.386930353,0.018867925,-1,0.651384909,-0.270500532,0.371675943,0.84875,-0.775892857;
T=0.172985782,0.552870091,1,1,0.613069647,1,0,0.651384909,0,0.371675943,0.84875,0.224107143;
Y=1,1,0,0,0,0,0.457264957,1,1,1,1,0
0,0,0.96,0.734722222,1,0.981132075,1,0,0.270500532,0,0,1;
ENDDATA
MAX=1-((2*@SUM(P2(J):@ABS(@SUM(P1(I):L(I)*E(I,J)))))/(@SUM(LINK(I,J):Y(I,J))+@SUM(LINK(I,J):L(I)*Y(I,J))+@SUM(P2(J):@ABS(@SUM(P1(I):L(I)*E(I,J))))));
@SUM(P1(I):L(I))=1;
END由此得到的局部最优解的结果是：
Linearization components added:
Constraints: 96
Variables: 96
Integers: 24
Local optimal solution found.
Objective value: 0.6298245
Objective bound: 0.6298245
Infeasibilities: 0.000000
Extended solver steps: 26
Total solver iterations: 2085
Model Class: MINLP
Total variables: 98
Nonlinear variables: 26
Integer variables: 24
Total constraints: 98
Nonlinear constraints: 1
Total nonzeros: 292
Nonlinear nonzeros: 26
Variable Value Reduced Cost
L( 1) 0.3869304 0.000000
L( 2) 0.6130696 0.000000
T( 1) 0.1729858 0.000000
T( 2) 0.5528701 0.000000
T( 3) 1.000000 0.000000
T( 4) 1.000000 0.000000
T( 5) 0.6130696 0.000000
T( 6) 1.000000 0.000000
T( 7) 0.000000 0.000000
T( 8) 0.6513849 0.000000
T( 9) 0.000000 0.000000
T( 10) 0.3716759 0.000000
T( 11) 0.8487500 0.000000
T( 12) 0.2241071 0.000000
E( 1, 1) -0.8270142 0.000000
E( 1, 2) -0.4471299 0.000000
E( 1, 3) 1.000000 0.000000
E( 1, 4) 1.000000 0.000000
E( 1, 5) 0.6130696 0.000000
E( 1, 6) 1.000000 0.000000
E( 1, 7) -0.4572650 0.000000
E( 1, 8) -0.3486151 0.000000
E( 1, 9) -1.000000 0.000000
E( 1, 10) -0.6283241 0.000000
E( 1, 11) -0.1512500 0.000000
E( 1, 12) 0.2241071 0.000000
E( 2, 1) 0.1729858 0.000000
E( 2, 2) 0.5528701 0.000000
E( 2, 3) 0.4000000E-01 0.000000
E( 2, 4) 0.2652778 0.000000
E( 2, 5) -0.3869304 0.000000
E( 2, 6) 0.1886793E-01 0.000000
E( 2, 7) -1.000000 0.000000
E( 2, 8) 0.6513849 0.000000
E( 2, 9) -0.2705005 0.000000
E( 2, 10) 0.3716759 0.000000
E( 2, 11) 0.8487500 0.000000
E( 2, 12) -0.7758929 0.000000
Y( 1, 1) 1.000000 0.000000
Y( 1, 2) 1.000000 0.000000
Y( 1, 3) 0.000000 0.000000
Y( 1, 4) 0.000000 0.000000
Y( 1, 5) 0.000000 0.000000
Y( 1, 6) 0.000000 0.000000
Y( 1, 7) 0.4572650 0.000000
Y( 1, 8) 1.000000 0.000000
Y( 1, 9) 1.000000 0.000000
Y( 1, 10) 1.000000 0.000000
Y( 1, 11) 1.000000 0.000000
Y( 1, 12) 0.000000 0.000000
Y( 2, 1) 0.000000 0.000000
Y( 2, 2) 0.000000 0.000000
Y( 2, 3) 0.9600000 0.000000
Y( 2, 4) 0.7347222 0.000000
Y( 2, 5) 1.000000 0.000000
Y( 2, 6) 0.9811321 0.000000
Y( 2, 7) 1.000000 0.000000
Y( 2, 8) 0.000000 0.000000
Y( 2, 9) 0.2705005 0.000000
Y( 2, 10) 0.000000 0.000000
Y( 2, 11) 0.000000 0.000000
Y( 2, 12) 1.000000 0.000000
Row Slack or Surplus Dual Price
1 0.6298245 1.000000
2 0.000000 -0.1970252
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