%?鋼管下料-切割模式進行限制
%?時間：2020年6月7日
G?=?zeros（14）;
k1=0;k2=0;k3=0k4=0;
i?=?1;
for?k1?=?0:1:4
???for?k2?=?0:1:3?
???????for?k3?=?0:1:2
???????????for?k4?=?0:1:3
???????????????if?（19-（4*k1+6*k2+8*k3+5*k4）>=0）?&&?（19-（4*k1+6*k2+8*k3+5*k4）<=3）
???????????????????G（i1）?=?k1;G（i2）=k2;G（i3）=k3;G（i4）=k4;
???????????????????G（i5）?=?19-k1*4-k2*6-k3*8-k4*5;
???????????????????i=i+1;
???????????????end
???????????end
???????end
???end
end
A?=?zeros（4i-1）;
for?ii?=?1:i-1
????for?jj?=?1:4
????????A（jjii）?=?-G（iijj）;
????end
end
b?=?[-50;-20;-15;-10];
C?=?ones（1i-1）;
intcon?=?1:i-1
lb?=?zeros（1i-1）;
ub?=?zeros（1i-1）;
%原料鋼管數(shù)
minf?=?1000;
x1?=?zeros（i-1）;
for?X?=?1:i-1-2
????for?Y?=?X+1:i-1-1
????????for?Z?=?Y+1:i-1
???????????ub（X）=inf;ub（Y）=inf;ub（Z）=inf;
???????????[xfval]?=?intlinprog（CintconAb[][]lbub）;
???????????if?minf?>?fval
???????????????minf?=?fval;
???????????????x1?=?x;
???????????end
???????????ub（X）=0;ub（Y）=0;ub（Z）=0;
????????end
????end
end
for?j=1:i-1
???C（j）?=?G（j5）;?
end
Aeq?=?ones（1i-1）;
beq?=?minf;
%余料
minf2?=?1000;
x2?=?zeros（i-1）;
for?X?=?1:i-1-2
????for?Y?=?X+1:i-1-1
????????for?Z?=?Y+1:i-1
???????????ub（X）=inf;ub（Y）=inf;ub（Z）=inf;
???????????[xfval2]?=?intlinprog（CintconAbAeqbeqlbub）;
???????????if?minf2?>?fval2
???????????????minf2?=?fval2;
???????????????x2?=?x;
???????????end
???????????ub（X）=0;ub（Y）=0;ub（Z）=0;
????????end
????end
end

?屬性????????????大小?????日期????時間???名稱
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