数学建模 原料
| 原料 |
X1 |
X2 |
X3 |
X4 |
X5 |
X6 |
X7 |
X8 |
| 方案一 |
2 |
5 |
3 |
3 |
2 |
1 |
3 |
1 |
| 剩余总根数 |
23 |
4 |
6 |
8 |
5 |
17 |
1 |
10 |
| 方案二 |
3 |
4 |
3 |
3 |
1 |
2 |
3 |
1 |
| 剩余总根数 |
10 |
15 |
6 |
8 |
16 |
6 |
1 |
10 |
| 方案三 |
3 |
5 |
2 |
2 |
2 |
2 |
3 |
1 |
| 剩余总根数 |
10 |
4 |
19 |
21 |
5 |
6 |
1 |
10 |
成品一中各原料的总根数分别为43,59,39,41,27,28,34,21，利用lingo可求出第一次搭配成品的最大捆数为11捆（单位根），然后用Matlab编程可得出有三种不同方案，则成品一各原料的分配方案如下表所示：
Lingo软件编程：
max=y;
3*x1+3.5*x2+4*x3+4.5*x4+5*x5+5.5*x6+6*x7+6.5*x8=89;
x1+x2+x3+x4+x5+x6+x7+x8=20;
y*x1<=43;
y*x2<=59;
y*x3<=39;
y*x4<=41;
y*x5<=27;
y*x6<=28;
y*x7<=34;
y*x8<=21;
@gin(x1);@bnd(0,x1,43);
@gin(x2);@bnd(0,x2,59);
@gin(x3);@bnd(0,x3,39);
@gin(x4);@bnd(0,x4,41);
@gin(x5);@bnd(0,x5,27);
@gin(x6);@bnd(0,x6,28);
@gin(x7);@bnd(0,x7,34);
@gin(x8);@bnd(0,x8,21);
@gin(y);
Matlab编程：
function y1
c=1;
for x1=0:43/11
for x2=0:59/11
for x3=0:39/11
for x4=0:41/11
for x5=0:27/11
for x6=0:28/11
for x7=0:34/11
for x8=0:21/11
k1=x1*3+x2*3.5+x3*4+x4*4.5+x5*5+x6*5.5+x7*6+x8*6.5;
k2=x1+x2+x3+x4+x5+x6+x7+x8;
if(k1==89)&(k2==20)
fprintf('µÚ%dÖÖ \n %d,%d,%d,%d,%d,%d,%d,%d \n',c,x1,x2,x3,x4,x5,x6,x7,x8)
c=c+1;
end
end
end
end
end
end
end
end
end
由最终讨论需要选择方案二，则成品一第一次搭配后各原料的剩余总根数分别为10,15,6,8,16,6,1,10，利用lingo可求出第二次搭配成品的最大捆数为3捆（单位根），然后用Matlab编程可得出有三种不同方案，则成品一剩余的原材料的分配方案如下表所示：
| 长度 |
3-3.4 |
3.5-3.9 |
4-4.4 |
4.5-4.9 |
5-5.4 |
5.5-5.9 |
6-6.4 |
6.5-6.9 |
| 方案一 |
3 |
4 |
2 |
2 |
5 |
2 |
0 |
2 |
| 剩余总根数 |
1 |
3 |
0 |
2 |
1 |
0 |
1 |
4 |
| 方案二 |
3 |
5 |
2 |
1 |
5 |
1 |
0 |
3 |
| 剩余总根数 |
1 |
0 |
0 |
5 |
1 |
3 |
1 |
1 |
| 方案三 |
3 |
5 |
2 |
2 |
3 |
2 |
0 |
3 |
| 剩余总根数 |
1 |
0 |
0 |
2 |
7 |
0 |
1 |
1 |
Lingo编程：
max=y;
3*x1+3.5*x2+4*x3+4.5*x4+5*x5+5.5*x6+6*x7+6.5*x8=89;
x1+x2+x3+x4+x5+x6+x7+x8=20;
y*x1<=10;
y*x2<=15;
y*x3<=6;
y*x4<=8;
y*x5<=16;
y*x6<=6;
y*x7<=1;
y*x8<=10;
@gin(x1);@bnd(0,x1,10);
@gin(x2);@bnd(0,x2,15);
@gin(x3);@bnd(0,x3,6);
@gin(x4);@bnd(0,x4,8);
@gin(x5);@bnd(0,x5,16);
@gin(x6);@bnd(0,x6,6);
@gin(x7);@bnd(0,x7,1);
@gin(x8);@bnd(0,x8,10);
@gin(y);
Matlab编程：
function y2
c=1;
for x1=0:10/3
for x2=0:15/3
for x3=0:6/3
for x4=0:8/3
for x5=0:16/3
for x6=0:6/3
for x7=0:1/3
for x8=0:10/3
k1=x1*3+x2*3.5+x3*4+x4*4.5+x5*5+x6*5.5+x7*6+x8*6.5;
k2=x1+x2+x3+x4+x5+x6+x7+x8;
if(k1==89)&(k2==20)
fprintf('第%d种\n %d,%d,%d,%d,%d,%d,%d,%d\n',c,x1,x2,x3,x4,x5,x6,x7,x8)
c=c+1;
end
end
end
end
end
end
end
end
end
由以上两次搭配后的剩余原料为12根，小于20，不满足约束条件，停止搭配。因此， 成品一的总捆数为z1=11+3=14。
综上，成品一的搭配方案如下表：
| 3-3.4 |
3.5-3.9 |
4-4.4 |
4.5-4.9 |
5-5.4 |
5.5-5.9 |
6-6.4 |
6.5-6.9 |
| 11捆 |
|||||||
| 3 |
4 |
3 |
3 |
1 |
2 |
3 |
1 |
| 3捆 |
|||||||
| 3 |
5 |
2 |
1 |
5 |
1 |
0 |
3 |
成品三第四次搭配后各原料的剩余总根数分别为4,5,3,11,1,0,3,7,8,10,28,1,7,4,0,16,0,2,0,6,0，0,0,1
利用lingo可求出第五次搭配成品的最大捆数为7捆（单位根），然后用Matlab编程可得出有四种不同方案，如下表所示：
| 原料 |
X23 |
X24 |
X25 |
X26 |
X27 |
X28 |
X29 |
X30 |
| 根数 |
0 |
0 |
0 |
1 |
0 |
0 |
1 |
0 |
| 剩余总根数 |
4 |
5 |
3 |
4 |
1 |
0 |
3 |
0 |
|
|
X31 |
X32 |
X33 |
X34 |
X35 |
X36 |
X37 |
X38 |
| 根数 |
1 |
0 |
2 |
0 |
0 |
0 |
0 |
0 |
| 剩余总根数 |
1 |
10 |
14 |
1 |
7 |
4 |
0 |
16 |
|
|
X39 |
X40 |
X41 |
X42 |
X43 |
X44 |
X45 |
X46 |
| 根数 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
| 剩余总根数 |
0 |
2 |
0 |
5 |
0 |
0 |
0 |
1 |
Lingo编程：
max=y;
14*x1+14.5*x2+15*x3+15.5*x4+16*x5+16.5*x6+17*x7+17.5*x8+18*x9+18.5*x10+19*x11+19.5*x12+20*x13+21*x15+21.5*x16+22*x17+22.5*x18+23*x19+23.5*x20+24*x21+24.5*x22+25*x23+25.5*x24=89;
x1+x2+x3+x4+x5+x6+x7+x8+x9+x11+x12+x13+x14+x15+x16+x17+x18+x19+x20+x21+x22+x23+x24=5;
y*x1<=4;
y*x2<=5;
y*x3<=3;
y*x4<=11;
y*x5<=1;
y*x6<=0;
y*x7<=3;
y*x8<=7;
y*x9<=8;
y*x10<=10;
y*x11<=28;
y*x12<=1;
y*x13<=7;
y*x14<=4;
y*x15<=0;
y*x16<=16;
y*x17<=0;
y*x18<=2;
y*x19<=0;
y*x20<=6;
y*x21<=0;
y*x22<=0;
y*x23<=0;
y*x24<=1;
@gin(x1);@bnd(0,x1,4);
@gin(x2);@bnd(0,x2,5);
@gin(x3);@bnd(0,x3,3);
@gin(x4);@bnd(0,x4,11);
@gin(x5);@bnd(0,x5,1);
@gin(x6);@bnd(0,x6,0);
@gin(x7);@bnd(0,x7,3);
@gin(x8);@bnd(0,x8,7);
@gin(x9);@bnd(0,x9,8);
@gin(x10);@bnd(0,x10,10);
@gin(x11);@bnd(0,x11,28);
@gin(x12);@bnd(0,x12,1);
@gin(x13);@bnd(0,x13,7);
@gin(x14);@bnd(0,x14,4);
@gin(x15);@bnd(0,x15,0);
@gin(x16);@bnd(0,x16,16);
@gin(x17);@bnd(0,x17,0);
@gin(x18);@bnd(0,x18,2);
@gin(x19);@bnd(0,x19,0);
@gin(x20);@bnd(0,x20,6);
@gin(x21);@bnd(0,x21,0);
@gin(x22);@bnd(0,x22,0);
@gin(x23);@bnd(0,x23,0);
@gin(x24);@bnd(0,x24,1);
@gin(x8);
@gin(y);
@bnd(0,y,136);
Matlab编程：
function y34
c=1;
for x1=0:4/7
for x2=0:5/7
for x3=0:3/7
for x4=0:11/7
for x5=0:1/7
for x6=0:0/7
for x7=0:3/7
for x8=0:7/7
for x9=0:8/7
for x10=0:10/7
for x11=0:28/7
for x12=0:1/7
for x13=0:7/7
for x14=0:4/7
for x15=0:0/7
for x16-0:16/7
for x17=0:0/7
for x18=0:2/7
for x19=0:0/7 for x20=0:6/7 for x21=0:0/7 for x22=0:0/7 for x23=0:0/7 for x24=0:1/7 k1=14*x1+14.5*x2+15*x3+15.5*x4+16*x5+16.5*x6+17*x7+17.5*x8+18*x9+18.5*x10+19*x11+19.5*x12+20*x13+20.5*x14+21*x15+21.5*x16+22*x17+22.5*x18+23*x19+23.6*x20+24*x21+24.5*x22+25*x23+25.5*x24; k2=x1+x2+x3+x4+x5+x6+x7+x8+x9+x10+x11+x12+x13+x14+x15+x16+x17+x18+x19+x20+x21+x22+x23+x24; if(k1==89)&(k2==5) fprintf('第%d种 \n %d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,\n',c,x1,x2,x3,x4,x5,x6,x7,x8,x9,x10,x12,x13,x14,x15,x16,x17,x18,x19,x20,x21,x22,x23,x24) c=c+1; end end end
end
end
end
end
end
end
end
end
end
end
end
end
end
end
end
end
end
end
end
end
end
第五次搭配后各原料的剩余总根数分别为4，5,3,4,1,0,3,0,1,10,14,1,0,16,0,2,0,6,0,0,0,1，
利用lingo可求出第五次搭配成品的最大捆数为4捆（单位根），然后用Matlab编程可得出有四种不同方案，如下表所示：
| 原料 |
X23 |
X24 |
X25 |
X26 |
X27 |
X28 |
X29 |
X30 |
| 方案一 |
1 |
0 |
0 |
1 |
0 |
0 |
0 |
0 |
| 方案二 |
1 |
0 |
0 |
1 |
0 |
0 |
0 |
0 |
| 方案三 |
1 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
| 方案四 |
1 |
1 |
0 |
1 |
0 |
0 |
0 |
0 |
|
|
X31 |
X32 |
X33 |
X34 |
X35 |
X36 |
X37 |
X38 |
| 方案一 |
0 |
0 |
1 |
0 |
1 |
1 |
0 |
0 |
| 方案二 |
0 |
0 |
2 |
0 |
0 |
0 |
0 |
1 |
| 方案三 |
0 |
2 |
2 |
0 |
0 |
0 |
0 |
0 |
| 方案四 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
1 |
|
|
X39 |
X40 |
X41 |
X42 |
X43 |
X44 |
X45 |
X46 |
| 方案一 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
| 方案二 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
| 方案三 |
0 |
0 |
0 |
0 |
0 |
00 |
0 |
0 |
| 方案四 |
0 |
0 |
0 |
1 |
0 |
0 |
0 |
0 |
Lingo编程：
max=y;
14*x1+14.5*x2+15*x3+15.5*x4+16*x5+16.5*x6+17*x7+17.5*x8+18*x9+18.5*x10+19*x11+19.5*x12+20*x13+21*x15+21.5*x16+22*x17+22.5*x18+23*x19+23.5*x20+24*x21+24.5*x22+25*x23+25.5*x24=89;
x1+x2+x3+x4+x5+x6+x7+x8+x9+x11+x12+x13+x14+x15+x16+x17+x18+x19+x20+x21+x22+x23+x24=5;
y*x1<=4;
y*x2<=5;
y*x3<=3;
y*x4<=4;
y*x5<=1;
y*x6<=0;
y*x7<=3;
y*x8<=0;
y*x9<=1;
y*x10<=10;
y*x11<=14;
y*x12<=1;
y*x13<=7;
y*x14<=4;
y*x15<=0;
y*x16<=16;
y*x17<=0;
y*x18<=2;
y*x19<=0;
y*x20<=6;
y*x21<=0;
y*x22<=0;
y*x23<=0;
y*x24<=1;
@gin(x1);@bnd(0,x1,4);
@gin(x2);@bnd(0,x2,5);
@gin(x3);@bnd(0,x3,3);
@gin(x4);@bnd(0,x4,4);
@gin(x5);@bnd(0,x5,1);
@gin(x6);@bnd(0,x6,0);
@gin(x7);@bnd(0,x7,3);
@gin(x8);@bnd(0,x8,0);
@gin(x9);@bnd(0,x9,1);
@gin(x10);@bnd(0,x10,10);
@gin(x11);@bnd(0,x11,14);
@gin(x12);@bnd(0,x12,1);
@gin(x13);@bnd(0,x13,7);
@gin(x14);@bnd(0,x14,4);
@gin(x15);@bnd(0,x15,0);
@gin(x16);@bnd(0,x16,16);
@gin(x17);@bnd(0,x17,0);
@gin(x18);@bnd(0,x18,2);
@gin(x19);@bnd(0,x19,0);
@gin(x20);@bnd(0,x20,6);
@gin(x21);@bnd(0,x21,0);
@gin(x22);@bnd(0,x22,0);
@gin(x23);@bnd(0,x23,0);
@gin(x24);@bnd(0,x24,1);
@gin(x8);
@gin(y);
@bnd(0,y,136);
Matlab编程（如上搭配）
选择方案三，第六次搭配后的的剩余量为 0,5,3,4,1,0,3,0,1,2,6,1,7,4,0,16,0,2,0,6,0,0,0,1
利用lingo可求出第七次搭配成品的最大捆数为3捆（单位根），然后用Matlab编程可得出有八种不同方案，则成品一剩余的原材料的分配方案如下表所示：
| 原料 |
方案一 |
方案二 |
方案三 |
方案四 |
方案五 |
方案六 |
方案七 |
方案八 |
| X23 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
| X24 |
0 |
0 |
01 |
1 |
1 |
1 |
1 |
1 |
| X25 |
1 |
1 |
0 |
0 |
1 |
1 |
1 |
1 |
| X26 |
1 |
1 |
1 |
1 |
0 |
0 |
0 |
1 |
| X27 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
| X28 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
| X29 |
0 |
1 |
0 |
1 |
0 |
0 |
1 |
0 |
| X30 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
| X31 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
| X32 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
| X33 |
2 |
0 |
1 |
0 |
1 |
2 |
1 |
0 |
| X34 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
| X35 |
0 |
1 |
2 |
0 |
1 |
0 |
0 |
0 |
| X36 |
1 |
0 |
0 |
1 |
1 |
0 |
0 |
1 |
| X37 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
| X38 |
0 |
1 |
0 |
1 |
0 |
1 |
0 |
0 |
| X39 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
| X40 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
| X41 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
| X42 |
0 |
0 |
0 |
0 |
0 |
0 |
1 |
1 |
| X43 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
| X45 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
| X46 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
Lingo编程：
max=y;
14*x1+14.5*x2+15*x3+15.5*x4+16*x5+16.5*x6+17*x7+17.5*x8+18*x9+18.5*x10+19*x11+19.5*x12+20*x13+21*x15+21.5*x16+22*x17+22.5*x18+23*x19+23.5*x20+24*x21+24.5*x22+25*x23+25.5*x24=89;
x1+x2+x3+x4+x5+x6+x7+x8+x9+x11+x12+x13+x14+x15+x16+x17+x18+x19+x20+x21+x22+x23+x24=5;
y*x1<=0;
y*x2<=5;
y*x3<=3;
y*x4<=4;
y*x5<=1;
y*x6<=0;
y*x7<=3;
y*x8<=0;
y*x9<=1;
y*x10<=2;
y*x11<=6;
y*x12<=1;
y*x13<=7;
y*x14<=4;
y*x15<=0;
y*x16<=16;
y*x17<=0;
y*x18<=2;
y*x19<=0;
y*x20<=6;
y*x21<=0;
y*x22<=0;
y*x23<=0;
y*x24<=1;
@gin(x1);@bnd(0,x1,0);
@gin(x2);@bnd(0,x2,5);
@gin(x3);@bnd(0,x3,3);
@gin(x4);@bnd(0,x4,4);
@gin(x5);@bnd(0,x5,1);
@gin(x6);@bnd(0,x6,0);
@gin(x7);@bnd(0,x7,3);
@gin(x8);@bnd(0,x8,0);
@gin(x9);@bnd(0,x9,1);
@gin(x10);@bnd(0,x10,2);
@gin(x11);@bnd(0,x11,6);
@gin(x12);@bnd(0,x12,1);
@gin(x13);@bnd(0,x13,7);
@gin(x14);@bnd(0,x14,4);
@gin(x15);@bnd(0,x15,0);
@gin(x16);@bnd(0,x16,16);
@gin(x17);@bnd(0,x17,0);
@gin(x18);@bnd(0,x18,2);
@gin(x19);@bnd(0,x19,0);
@gin(x20);@bnd(0,x20,6);
@gin(x21);@bnd(0,x21,0);
@gin(x22);@bnd(0,x22,0);
@gin(x23);@bnd(0,x23,0);
@gin(x24);@bnd(0,x24,1);
@gin(x8);
@gin(y);
@bnd(0,y,136);
Matlab编程（如上搭配）
选择方案二，第七次搭配后的的剩余量为 0,2,3,1,1,0,3,0,1,10,3,1,1,4,0,16,0,2,0,6,0,0,0,1
利用lingo可求出第七次搭配成品的最大捆数为2捆（单位根），然后用Matlab编程可得出有种不同方案，则成品一剩余的原材料的分配方案如下表所示：
| 原料 |
X23 |
X24 |
X25 |
X26 |
X27 |
X28 |
X29 |
X30 |
| 方案一 |
0 |
1 |
1 |
0 |
0 |
0 |
0 |
0 |
| 方案二 |
0 |
1 |
1 |
0 |
0 |
0 |
1 |
0 |
|
|
X31 |
X32 |
X33 |
X34 |
X35 |
X36 |
X37 |
X38 |
| 方案一 |
0 |
0 |
1 |
0 |
0 |
0 |
2 |
0 |
| 方案二 |
0 |
0 |
1 |
0 |
0 |
0 |
0 |
0 |
|
|
X39 |
X40 |
X41 |
X42 |
X43 |
X44 |
X45 |
X46 |
| 方案一 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
| 方案二 |
0 |
0 |
0 |
1 |
0 |
0 |
0 |
0 |
Lingo编程：
max=y;
14*x1+14.5*x2+15*x3+15.5*x4+16*x5+16.5*x6+17*x7+17.5*x8+18*x9+18.5*x10+19*x11+19.5*x12+20*x13+21*x15+21.5*x16+22*x17+22.5*x18+23*x19+23.5*x20+24*x21+24.5*x22+25*x23+25.5*x24=89;
x1+x2+x3+x4+x5+x6+x7+x8+x9+x11+x12+x13+x14+x15+x16+x17+x18+x19+x20+x21+x22+x23+x24=5;
y*x1<=0;
y*x2<=2;
y*x3<=3;
y*x4<=1;
y*x5<=1;
y*x6<=0;
y*x7<=3;
y*x8<=0;
y*x9<=1;
y*x10<=10;
y*x11<=3;
y*x12<=1;
y*x13<=1;
y*x14<=4;
y*x15<=0;
y*x16<=16;
y*x17<=0;
y*x18<=2;
y*x19<=0;
y*x20<=6;
y*x21<=0;
y*x22<=0;
y*x23<=0;
y*x24<=1;
@gin(x1);@bnd(0,x1,0);
@gin(x2);@bnd(0,x2,2);
@gin(x3);@bnd(0,x3,3);
@gin(x4);@bnd(0,x4,1);
@gin(x5);@bnd(0,x5,1);
@gin(x6);@bnd(0,x6,0);
@gin(x7);@bnd(0,x7,3);
@gin(x8);@bnd(0,x8,0);
@gin(x9);@bnd(0,x9,1);
@gin(x10);@bnd(0,x10,10);
@gin(x11);@bnd(0,x11,3);
@gin(x12);@bnd(0,x12,1);
@gin(x13);@bnd(0,x13,1);
@gin(x14);@bnd(0,x14,4);
@gin(x15);@bnd(0,x15,0);
@gin(x16);@bnd(0,x16,16);
@gin(x17);@bnd(0,x17,0);
@gin(x18);@bnd(0,x18,2);
@gin(x19);@bnd(0,x19,0);
@gin(x20);@bnd(0,x20,6);
@gin(x21);@bnd(0,x21,0);
@gin(x22);@bnd(0,x22,0);
@gin(x23);@bnd(0,x23,0);
@gin(x24);@bnd(0,x24,1);
@gin(x8);
@gin(y);
@bnd(0,y,136);
Matlab编程（如上搭配）
选择方案一 第八次搭配后剩余量为0,0,1,1,1,0,3,0,1,0,3,1,1,0,0,16,0,2,0,6 0,0,0,1
利用lingo可求出第九次搭配成品的最大捆数为1捆（单位根），然后用Matlab编程可得出有种25不同方案，经选择其最优方案为如下：
| 原料 |
X23 |
X24 |
X25 |
X26 |
X27 |
X28 |
X29 |
X30 |
| 根数 |
0 |
0 |
1 |
1 |
1 |
0 |
0 |
0 |
| 剩余总根数 |
0 |
0 |
0 |
0 |
0 |
0 |
3 |
0 |
|
|
X31 |
X32 |
X33 |
X34 |
X35 |
X36 |
X37 |
X38 |
| 根数 |
0 |
0 |
1 |
0 |
0 |
0 |
0 |
0 |
| 剩余总根数 |
1 |
0 |
2 |
1 |
1 |
0 |
0 |
16 |
|
|
X39 |
X40 |
X41 |
X42 |
X43 |
X44 |
X45 |
X46 |
| 根数 |
0 |
0 |
0 |
1 |
0 |
0 |
0 |
0 |
| 剩余总根数 |
0 |
2 |
0 |
5 |
0 |
0 |
0 |
1 |
Lingo编程（如上搭配）
Matlab编程（如上搭配）
利用lingo可求出第十次搭配成品的最大捆数为1捆（单位根），经选择其最优方案为如下：
| 原料 |
X23 |
X24 |
X25 |
X26 |
X27 |
X28 |
X29 |
X30 |
| 根数 |
0 |
0 |
0 |
0 |
0 |
0 |
3 |
0 |
| 剩余总根数 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
|
|
X31 |
X32 |
X33 |
X34 |
X35 |
X36 |
X37 |
X38 |
| 根数 |
1 |
0 |
0 |
0 |
1 |
0 |
0 |
0 |
| 剩余总根数 |
0 |
0 |
2 |
1 |
0 |
0 |
0 |
16 |
|
|
X39 |
X40 |
X41 |
X42 |
X43 |
X44 |
X45 |
X46 |
| 根数 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
| 剩余总根数 |
0 |
2 |
0 |
5 |
0 |
0 |
0 |
1 |
Lingo编程（如上搭配）
Matlab编程（如上搭配）