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谁帮我运行一下,告诉我哪里有错,谢谢

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lidianqin        

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  • TA的每日心情
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    2013-1-27 16:01
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    [LV.1]初来乍到

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    1#
    发表于 2011-10-19 22:33 |只看该作者 |倒序浏览
    |招呼Ta 关注Ta
    model:
    sets:
    gj/g1..g15/:c1_sj,c2_sj,c3_sj,c4_sj,c5_sj,c6_sj,c7_sj,ST1,r1,d1,c1_c1;
    sx/s1..s15/:c1_t,c2_t,c3_t,c4_t,c5_t,c6_t,c7_t,c1_c,c2_c,c3_c,c4_c,c5_c,c6_c,c7_c,c1_d,c2_d,c3_d,c4_d,c5_d,c6_d,c7_d,ST,r,d,y;
    links(sx,gj):note;
    links1(sx,sx):x;
    endsets
    data:
    c1_sj=130 130 130 130 130 130 20 20 20 20 20 20 120 125 125;
    c2_sj=230 230 230 230 230 230 30 30 30 30 30 30 225 225 225;
    c3_sj=250 250 250 250 250 250 29 29 29 29 29 29 257 257 257;
    c4_sj=160 160 160 160 160 160 38 38 38 38 38 38 142 160 160;
    c5_sj=200 200 200 200 200 200 25 25 25 25 25 25 150 215 215;
    c6_sj=0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
    c7_sj=300 300 300 300 300 300 0 0 0 0 0 0 300 0 0;
    ST1=20 20 20 20 20 20 10 10 10 10 10 10 15 16 16;
    r1=1 1 1 1 1 1 0 0 0 0 0 0 1 0 0;
    enddata
    [OBJ]min=@sum(gj(j)|j#LT#7smax(@smax(d1(j),d1(j+6))+200-1920,0))+@sum(sx(i)smax((d(i)-1920),0));!目标总延迟时间最小;
    @for(sx(i):ST(i)=@sum(gj(j):note(i,j)*ST1(j)));!转换准备时间;
    @for(sx(i):r(i)=@sum(gj(j):note(i,j)*r1(j)));!转换种类;
    @for(sx(i):c1_t(i)=@sum(gj(j):note(i,j)*c1_sj(j)));!转换加工时间;
    @for(sx(i):c2_t(i)=@sum(gj(j):note(i,j)*c2_sj(j)));
    @for(sx(i):c3_t(i)=@sum(gj(j):note(i,j)*c3_sj(j)));
    @for(sx(i):c4_t(i)=@sum(gj(j):note(i,j)*c4_sj(j)));
    @for(sx(i):c5_t(i)=@sum(gj(j):note(i,j)*c5_sj(j)));
    @for(sx(i):c6_t(i)=@sum(gj(j):note(i,j)*c6_sj(j)));
    @for(sx(i):c7_t(i)=@sum(gj(j):note(i,j)*c7_sj(j)));
    @for(sx(i)sum(gj(j):note(i,j))=1;);!每个顺序只有一个工件;
    @for(gj(j)sum(sx(i):note(i,j))=1;);!每个工件只占一个顺序位置;
    @for(sx(i)|i#LT#3:c1_c(i)=ST(i)+c1_t(i));!A/B上第一个加工的工件在第一个工位上的完工时间;
    @for(sx(i)|i#LT#3:c2_c(i)=c1_c(i)+c2_t(i));
    @for(sx(i)|i#LT#3:c3_c(i)=c2_c(i)+c3_t(i));
    @for(sx(i)|i#LT#3:c4_c(i)=c3_c(i)+c4_t(i));
    @for(sx(i)|i#LT#3:c5_c(i)=c4_c(i)+c5_t(i));
    @for(sx(i)|i#LT#3:c6_c(i)=c5_c(i)+c6_t(i));
    @for(sx(i)|i#LT#3:c7_c(i)=c6_c(i)+c7_t(i));!A/B上第一个加工的工件在其余工位上的完工时间;
    @for(sx(i)|i#LT#3:c1_d(i)=c1_c(i));
    @for(sx(i)|i#LT#3:c2_d(i)=c2_c(i));
    @for(sx(i)|i#LT#3:c3_d(i)=c3_c(i));
    @for(sx(i)|i#LT#3:c4_d(i)=c4_c(i));
    @for(sx(i)|i#LT#3:c5_d(i)=c5_c(i));
    @for(sx(i)|i#LT#3:c6_d(i)=c6_c(i));
    @for(sx(i)|i#LT#3:c7_d(i)=c7_c(i));!A/B上第一个加工的工件离开每个工位的时间;

    @for(sx(i)|i#GT#2:c1_c(i)=@sum(sx(k)|k#LT#i:x(k,i)*(c1_d(k)+@if((c1_t(i)-c1_t(k))#NE#0,1,0)*ST(i)+c1_t(i))));
    @for(sx(i)|i#GT#2:c2_c(i)=@sum(sx(k)|k#LT#i:x(k,i)*(@smax(c2_d(k),c1_c(i))+c2_t(i))));
    @for(sx(i)|i#GT#2:c3_c(i)=@sum(sx(k)|k#LT#i:x(k,i)*(@smax(c3_d(k),c2_c(i))+c3_t(i))));
    @for(sx(i)|i#GT#2:c4_c(i)=@sum(sx(k)|k#LT#i:x(k,i)*(@smax(c4_d(k),c3_c(i))+c4_t(i))));
    @for(sx(i)|i#GT#2:c5_c(i)=@sum(sx(k)|k#LT#i:x(k,i)*(@smax(c5_d(k),c4_c(i))+c5_t(i))));
    @for(sx(i)|i#GT#2:c6_c(i)=@sum(sx(k)|k#LT#i:x(k,i)*(@smax(c6_d(k),c5_c(i))+c6_t(i))));
    @for(sx(i)|i#GT#2:c7_c(i)=r(i)*(c6_c(i)+c7_t(i)));
    @for(sx(i)|i#GT#2:c1_d(i)=c2_c(i)-c2_t(i));
    @for(sx(i)|i#GT#2:c2_d(i)=c3_c(i)-c3_t(i));
    @for(sx(i)|i#GT#2:c3_d(i)=c4_c(i)-c4_t(i));
    @for(sx(i)|i#GT#2:c4_d(i)=c5_c(i)-c5_t(i));
    @for(sx(i)|i#GT#2:c5_d(i)=c6_c(i)-c6_t(i));
    @for(sx(i)|i#GT#2:c6_d(i)=r(i)*(c7_c(i)-c7_t(i))+(1-r(i))*(y(i)*1.2*c6_c(i)+(1-y(i))*c6_c(i)));
    @for(sx(i)|i#GT#2:c7_d(i)=c7_c(i));

    @for(sx(i)|i#GT#2sum(sx(k)|k#LT#i:x(k,i))=1);!每个工件只有一个紧前工件;
    @for(sx(k)|k#LT#15sum(sx(i)|i#GT#2#and#i#GT#k:x(k,i))<=1);!每个工件最多有一个紧后工序;
    @for(sx(i)|i#LT#15#and#i#GT#1:y(i)=x(i,i+1)*y(i+1)+(1-x(i,i+1))*(1-y(i+1)));
    @for(sx(i):d(i)=r(i)*c7_d(i)+(1-r(i))*c6_d(i));
    @for(gj(j):d1(j)=@sum(sx(i):d(i)*note(i,j)));
    @for(gj(j):c1_c1(j)=@sum(sx(i):c1_c(i)*note(i,j)));
    c1_c1(3)>280;
    c1_c1(9)>170;
    y(1)=1;
    y(2)=0;
    @for(links(i,j)bin(note));
    @for(sx(i)bin(y));
    @for(links1(k,i)bin(x));
    end
    zan
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    hejia88        

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