如果临时工每班工作时间可以是3h,也可以是4h,那么,应如
何安排临时工的班次,使得使用临时工的总成本最小。这时要安排的临时工班次
是多少?
快餐店用工问题,lingo代码求解。。谢谢啦。。帮忙修改一下代码。
model:
SetS:
row/1…11/:b;arrange/1…11/:x,C;link(row,arrange):a;
endsets
data:
b=8,8,7,1,2,1,5,10,10,6,6;
c=12,16,16,16,16,16,12,16,16,16,12;
a=l,0,0,0,0,0,0,0,1,1,1,1,1,0,0,0,0,0,0,0,1,1,1,1,1,0,0,0,0,0,
0,0,1,0,1,1,1,0,0,0,0,0,0,0,0,1,1,1,1,0,0,0,0,0,0,0,0,1,1,1,1,0,
0,0,0,0,0,0,0,1,1,1,1,0,0,0,0,0,0,0,0,1,1,l,1,0,0,0,0,0,0,0,0,
1,1,1,1,0,0,0,0,0,0,0,0,0,1,1,1,0,0,0,0,0,0,0,0,1,1,1,1;
enddata
[OBJ]min=@sum(arrange(j):C(j)*x(j));@for(row(i):@sum(arrange(j):
a(i,j)*x(j))>=b(i););
@for(arrange(j):x(j)>=0;);
End
临时工班次安排问题。。lingo代码求解。。
答案:2 悬赏:50
解决时间 2021-03-02 10:44
- 提问者网友:恋你成殇
- 2021-03-01 23:17
最佳答案
- 二级知识专家网友:眠于流年
- 2021-03-02 00:04
model:
sets:
row/1..11/:b;arrange/1..11/:x,c;link(row,arrange):a;
endsets
data:
b=8,8,7,1,2,1,5,10,10,6,6;
c=12,16,16,16,16,16,12,16,16,16,12;
a=1,0,0,0,0,0,0,0,1,1,1,1,1,0,0,0,0,0,0,0,1,1,1,1,1,0,0,0,0,0,
0,0,1,0,1,1,1,0,0,0,0,0,0,0,0,1,1,1,1,0,0,0,0,0,0,0,0,1,1,1,1,0,
0,0,0,0,0,0,0,1,1,1,1,0,0,0,0,0,0,0,0,1,1,1,1,0,0,0,0,0,0,0,0,
1,1,1,1,0,0,0,0,0,0,0,0,0,1,1,1,0,0,0,0,0,0,0,0,1,1,1,1;
enddata
[obj]min=@sum(arrange(j):c(j)*x(j));@for(row(i):@sum(arrange(j):
a(i,j)*x(j))>=b(i););
end
sets:
row/1..11/:b;arrange/1..11/:x,c;link(row,arrange):a;
endsets
data:
b=8,8,7,1,2,1,5,10,10,6,6;
c=12,16,16,16,16,16,12,16,16,16,12;
a=1,0,0,0,0,0,0,0,1,1,1,1,1,0,0,0,0,0,0,0,1,1,1,1,1,0,0,0,0,0,
0,0,1,0,1,1,1,0,0,0,0,0,0,0,0,1,1,1,1,0,0,0,0,0,0,0,0,1,1,1,1,0,
0,0,0,0,0,0,0,1,1,1,1,0,0,0,0,0,0,0,0,1,1,1,1,0,0,0,0,0,0,0,0,
1,1,1,1,0,0,0,0,0,0,0,0,0,1,1,1,0,0,0,0,0,0,0,0,1,1,1,1;
enddata
[obj]min=@sum(arrange(j):c(j)*x(j));@for(row(i):@sum(arrange(j):
a(i,j)*x(j))>=b(i););
end
全部回答
- 1楼网友:偏爱自由
- 2021-03-02 00:56
model:
SetS:
row/1..11/:b;arrange/1..11/:x,C;link(row,arrange):a;
endsets
data:
b=8,8,7,1,2,1,5,10,10,6,6;
c=12,16,16,16,16,16,12,16,16,16,12;
a=1,0,0,0,0,0,0,0,1,1,1,1,1,0,0,0,0,0,0,0,1,1,1,1,1,0,0,0,0,0,
0,0,1,0,1,1,1,0,0,0,0,0,0,0,0,1,1,1,1,0,0,0,0,0,0,0,0,1,1,1,1,0,
0,0,0,0,0,0,0,1,1,1,1,0,0,0,0,0,0,0,0,1,1,1,1,0,0,0,0,0,0,0,0,
1,1,1,1,0,0,0,0,0,0,0,0,0,1,1,1,0,0,0,0,0,0,0,0,1,1,1,1;
enddata
[OBJ]min=@sum(arrange(j):C(j)*x(j));@for(row(i):@sum(arrange(j):
a(i,j)*x(j))>=b(i););
End
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