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发表于 2013-1-16 19:08 |只看该作者 |倒序浏览
MODEL:
    title economic pipe;
    sets:
         tubulation/1..9/:L,Q,A,EK;
         caliber/1..5/;
         LINK(tubulation,caliber):C,X,D;
    ENDSETS
    DATA:
    !各段管长;
    L=700,700,700,500,500,500,500,500,500;
    !各段流量;
    Q=0.135,0.060,0.015,0.015,0.045,0.015,0.015,0.015,0.015;
    !节点服务水压标高;
    EK=47,46,44,46,44,43,45,46,43;
    !水源到节点的管段数;
    A=1,2,3,2,2,3,3,3,3;
    !标准管径单价;
    C=42.0 55.2 64.0 83.2 103.2   
      38.7 42.0 55.2 64.0
      25.0 30.8
      25.0 30.8
      30.8 38.7 42.0 55.2
      25.0 30.8
      25.0 30.8
      25.0 30.8
      25.0 30.8;
    !标准管径;
    D=0.25 0.30 0.35 0.40 0.50  
      0.20 0.25 0.30 0.35
      0.1 0.15
      0.1 0.15
      0.15 0.20 0.25 0.30
      0.1 0.15
      0.1 0.15
      0.1 0.15  
      0.1 0.15;
    ENDDATA
    !OBJECTIVE FUNCTION(目标函数);
    [OBJ]MIN=@SUM(LINK(I,J):C(I,J)*X(I,J));
    !pipe range CONSTRAINTE(管长约束);
    @FOR(tubulation(I):[RANGE_CON]@SUM(caliber(J):X(I,J))=L(I););
    !Stress constraint(压力约束);
    @FOR(caliber(J):(@SUM(tubulation(I):(0.135^2*X(1,J)/D(1,J)^5.342))*(1.942*10^-3))<(60-47));
    @FOR(CALIBER(J):(@SUM(TUBULATION(I):(0.135^2*X(1,J)/D(1,J)^5.342+0.060^2*X(2,J)/D(2,J)^5.342))*(1.942*10^-3))<(60-46));
    @for(caliber(j):(@SUM(TUBULATION(I):(0.135^2*X(1,J)/D(1,J)^5.342+0.060^2*X(2,J)/D(2,J)^5.342+0.015^2*X(3,J)/D(3,J)^5.342))*(1.942*10^-3))<(60-44));
    @for(caliber(j):(@SUM(TUBULATION(I):(0.135^2*X(1,J)/D(1,J)^5.342+0.015^2*X(4,J)/D(4,J)^5.342))*(1.942*10^-3))<(60-46));
    @FOR(CALIBER(J):(@SUM(TUBULATION(I):(0.135^2*X(1,J)/D(1,J)^5.342+0.045^2*X(5,J)/D(5,J)^5.342))*(1.942*10^-3))<(60-44));
    @for(caliber(j):(@SUM(TUBULATION(I):(0.135^2*X(1,J)/D(1,J)^5.342+0.045^2*X(5,J)/D(5,J)^5.342+0.015^2*X(6,J)/D(6,J)^5.342))*(1.942*10^-3))<(60-43));
    @for(caliber(j):(@SUM(TUBULATION(I):(0.135^2*X(1,J)/D(1,J)^5.342+0.045^2*X(5,J)/D(5,J)^5.342+0.015^2*X(7,J)/D(7,J)^5.342))*(1.942*10^-3))<(60-45));
    @for(caliber(j):(@SUM(TUBULATION(I):(0.135^2*X(1,J)/D(1,J)^5.342+0.060^2*X(2,J)/D(2,J)^5.342+0.015^2*X(8,J)/D(8,J)^5.342))*(1.942*10^-3))<(60-46));
    @for(caliber(j):(@SUM(TUBULATION(I):(0.135^2*X(1,J)/D(1,J)^5.342+0.060^2*X(2,J)/D(2,J)^5.342+0.015^2*X(9,J)/D(9,J)^5.342))*(1.942*10^-3))<(60-43));
    !速度要求;
    !非负假定;
    @FOR(LINK(I,J):X(I,J)>0;);
这个程序是有些问题的,求大神帮我解答!!!
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2#
发表于 2014-6-19 16:54 |只看该作者
    C=42.0 55.2 64.0 83.2 103.2   
      38.7 42.0 55.2 64.0
      25.0 30.8
      25.0 30.8
      30.8 38.7 42.0 55.2
      25.0 30.8
      25.0 30.8
      25.0 30.8
      25.0 30.8;

Required elements number is 9*5,unknown elements should be expressed like :1,,3,,,,,,3,,3;
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