设生产瓦楞纸A
A=[a31a51a32a52a33a53a34a54]T
瓦楞纸尺寸Asize
Asize=25002000180015002500200018001500
生产瓦楞纸价格(五层瓦楞板价格为三层板的1.3倍算出)
Aw=[10.013.06.58.454.86.244.05.2]T
设生产纸箱个数B
B=[b31b51b32b52b33b53b34b54b35b55b36b56b37b57b38b58⋯⋯]T
纸箱尺寸Bsize
Bsize=53053043035029026023021019517514513029023021019017015013011010595858037029027023019018016014013511510590
生产纸箱约束Bst(需求个数)
Bst=[509067958680897510870121881237612979140691166096658672]T
生产纸箱售价Bw
Bw=[2.963.782.212.821.762.251.351.711.021.300.851.080.680.870.550.700.500.640.390.530.340.430.290.37]T
(1) 假设生产的纸箱为需求数
即假设最优情况时,生产纸箱的个数不会溢出(即便生产多了也卖不出去)且能全部售出给下游快递公司
此时B=Bst,wB=fsum(Bst ⋅ Bw)=2462.4,其中定义fsum为求矩阵元素和
Matlab表达式:
Bst=[50,67,86,89,108,121,123,129,140,116,96,86;
90,95,80,75,70,88,76,79,69,60,65,72].';
Bw=[2.96,2.21,1.76,1.35,1.02,0.85,0.68,0.55,0.50,0.39,0.34,0.29;
3.78,2.82,2.25,1.71,1.30,1.08,0.87,0.70,0.64,0.53,0.43,0.37].';
wB=sum((sum(Bst.*Bw))')
此时售价不变,节约成本以获取最大利润
(2) 纸箱结构假设
假设每个纸箱均有6面,其中长宽面为顶底面
且假设纸箱的6个面只有其中一条侧边可以进行拼接,即纸板都是一体的,不能拼接而成
且假设封顶与封底的突出长度均为较短边(宽边)的一半
纸箱的展开图如下
即纸箱需要使用纸板的尺寸BsizeX=(长+宽)×2,BsizeY=高+宽
Matlab表达式:
Bsize=[530,290,370;
530,230,290;
430,210,270;
350,190,230;
290,170,190;
260,150,180;
230,130,160;
210,110,140;
195,105,135;
175,95,115;
145,85,105;
130,80,90];
BsizeArea(:,1)=(Bsize(:,1)+Bsize(:,2)).*2;
BsizeArea(:,2)=(Bsize(:,2)+Bsize(:,3))
即纸箱需要使用纸板的尺寸BsizeArea
BsizeArea=1640152012801080920820720640600540460420660520480420360330290250240210190170
列出每种纸板可能制作的纸箱(每种方案令再无边料可以制作纸箱)(转化为方案问题)
纸板1(2500x2500)
方案 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
---|
1 | 3 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 |
2 | | | | | | | | | | | | |
3 | | | | | | | | | | | | |
纸板2(2000x2000)
纸板3(1800x1800)
纸板4(1500x1500)
目标:利润w
w
subject to
s.t.{