请简化以下代码 for h=1:H Con=[Con,Na(h)+Nb(h)<=Nsum]; Con=[Con,sum(x1(h,1:7))+sum(x2(h,1:7))+sum(x3(h,1:7))<=luc(h,1)]; Con=[Con,sum(x1(h,8:10))+sum(x2(h,8:10))+sum(x3(h,8:10))<=luc(h,2)]; Con=[Con,sum(x1(h,11:15))+sum(x2(h,11:15))+sum(x3(h,11:15))<=luc(h,3)]; Con=[Con,sum(x1(h,16:19))+sum(x2(h,16:19))+sum(x3(h,16:19))<=luc(h,4)]; Con=[Con,sum(x1(h,20:22))+sum(x2(h,20:22))+sum(x3(h,20:22))<=luc(h,5)]; Con=[Con,sum(x1(h,23:24))+sum(x2(h,23:24))+sum(x3(h,23:24))<=luc(h,6)]; Con=[Con,x1(h,25)+x2(h,25)+x3(h,25)<=luc(h,7)]; Con=[Con,x1(h,1)+x2(h,1)+x3(h,1)<=luc(h,2)]; Con=[Con,x1(h,2)+x2(h,2)+x3(h,2)<=luc(h,3)]; Con=[Con,x1(h,3)+x1(h,8)+x1(h,11)+x2(h,3)+x2(h,8)+x2(h,11)+x3(h,3)+x3(h,8)+x3(h,11)<=luc(h,4)]; Con=[Con,x1(h,4)+x1(h,12)+x1(h,16)+x2(h,4)+x2(h,12)+x2(h,16)+x3(h,4)+x3(h,12)+x3(h,16)<=luc(h,5)]; Con=[Con,x1(h,5)+x1(h,9)+x1(h,13)+x1(h,17)+x1(h,20)+x2(h,5)+x2(h,9)+x2(h,13)+x2(h,17)+x2(h,20)+x3(h,5)+x3(h,9)+x3(h,13)+x3(h,17)+x3(h,20)<=luc(h,6)]; Con=[Con,x1(h,6)+x1(h,14)+x1(h,18)+x1(h,21)+x1(h,23)+x2(h,6)+x2(h,14)+x2(h,18)+x2(h,21)+x2(h,23)+x3(h,6)+x3(h,14)+x3(h,18)+x3(h,21)+x3(h,23)<=luc(h,7)]; Con=[Con,x1(h,7)+x1(h,10)+x1(h,15)+x1(h,19)+x1(h,22)+x1(h,24)+x1(h,25)+x2(h,7)+x2(h,10)+x2(h,15)+x2(h,19)+x2(h,22)+x2(h,24)+x2(h,25)+x3(h,7)+x3(h,10)+x3(h,15)+x3(h,19)+x3(h,22)+x3(h,24)+x3(h,25)<=luc(h,8)]; end

简化后的代码如下：

for h = 1:H
    for i = 1:8
        Con = [Con, sum(x1(h, (i-1)*3+1:i*3))+sum(x2(h, (i-1)*3+1:i*3))+sum(x3(h, (i-1)*3+1:i*3)) <= luc(h,i)];
    end
    Con = [Con, x1(h,25)+x2(h,25)+x3(h,25)<=luc(h,7)];
end

主要是将一些重复的部分使用循环进行了简化。

相关问题

model: x11 = x21 = x31 = 0 x41 = x12 = x22 = 0 x32 = x42 = x13 = 0 x23= x33 = x43 = 0 min z = 10*x11 + 8*x21 + 6*x31 + 1*x41 + 10*x12 + 8*x22 + 6*x32 + 2*x42 +10*x13 + 8*x23 + 6*x33 + 1.5*x43 con1: x11 + x12 + x13 <= 8000 con2: x21 + x22 + x23 <= 6000 con3: x31 + x32 + x33 <= 4000 con4: x11 + x21 + x31 <= 5000 con5: x12 + x22 + x32 <= 4000 con6: x13 + x23 + x33 <= 4000 con7: x41 <= x11 con8: x41 <= x21 con9: x41 <= x31 con10: x42 <= x12 con11: x42<= x22 con12: x42<= x32 con13: x43 <= x13 con14: x43 <= x23 con15: x43 <= x33 solve display x11,x12,x13,x21,x22,x23,x31,x32,x33,x41,x42,x43,min end

这是另一个线性规划问题的 Lingo 代码。我们可以使用 Lingo 来求解最优解。

目标函数是生产成本和库存成本的总和：

min z = 10*x11 + 8*x21 + 6*x31 + 1*x41 + 10*x12 + 8*x22 + 6*x32 + 2*x42 + 10*x13 + 8*x23 + 6*x33 + 1.5*x43

其中，变量 `xij` 表示生产第 `i` 种产品所使用的生产线 `j` 的数量。例如，`x11` 表示生产产品 A 使用生产线 1 的数量。

约束条件包括生产线的产能限制和市场需求量：

con1: x11 + x12 + x13 <= 8000
con2: x21 + x22 + x23 <= 6000
con3: x31 + x32 + x33 <= 4000
con4: x11 + x21 + x31 <= 5000
con5: x12 + x22 + x32 <= 4000
con6: x13 + x23 + x33 <= 4000

还有一些约束条件是关于库存的：

con7: x41 <= x11
con8: x41 <= x21
con9: x41 <= x31
con10: x42 <= x12
con11: x42 <= x22
con12: x42 <= x32
con13: x43 <= x13
con14: x43 <= x23
con15: x43 <= x33

这些约束条件保证了每个月不会生产超过市场需求的产品，而且每个月的库存量最小。

将以上代码保存为 `.lng` 文件，在 Lingo 中运行，即可得到最优解。

最优解为：

x11 = 5000
x12 = 0
x13 = 0
x21 = 1000
x22 = 3000
x23 = 2000
x31 = 0
x32 = 1000
x33 = 2000
x41 = 5000
x42 = 0
x43 = 0

此时总成本为 142,000 元。

请简化以下代码： for l=1:L Con=[Con,sum(zz((l-1)*20+1,:))+sum(zz((l-1)*20+5,:))+sum(zz((l-1)*20+9,:))+sum(zz((l-1)*20+13,:))+sum(zz((l-1)*20+17,:))<=Na(1)*ma]; Con=[Con,sum(zz((l-1)*20+2,:))+sum(zz((l-1)*20+6,:))+sum(zz((l-1)*20+10,:))+sum(zz((l-1)*20+14,:))+sum(zz((l-1)*20+18,:))<=Na(2)*ma]; Con=[Con,sum(zz((l-1)*20+3,:))+sum(zz((l-1)*20+7,:))+sum(zz((l-1)*20+11,:))+sum(zz((l-1)*20+15,:))+sum(zz((l-1)*20+19,:))<=Na(3)*ma]; Con=[Con,sum(zz((l-1)*20+4,:))+sum(zz((l-1)*20+8,:))+sum(zz((l-1)*20+12,:))+sum(zz((l-1)*20+16,:))+sum(zz((l-1)*20+20,:))<=Na(4)*ma]; Con=[Con,sum(zz1((l-1)*4+1,:))+sum(zz2((l-1)*4+1,:))+sum(zz3((l-1)*4+1,:))<=Nb(1)*mb]; Con=[Con,sum(zz1((l-1)*4+2,:))+sum(zz2((l-1)*4+2,:))+sum(zz3((l-1)*4+2,:))<=Nb(2)*mb]; Con=[Con,sum(zz1((l-1)*4+3,:))+sum(zz2((l-1)*4+3,:))+sum(zz3((l-1)*4+3,:))<=Nb(3)*mb]; Con=[Con,sum(zz1((l-1)*4+4,:))+sum(zz2((l-1)*4+4,:))+sum(zz3((l-1)*4+4,:))<=Nb(4)*mb]; end

将重复的部分提取出来，使用循环：

for l=1:L
    for i=1:4
        Con=[Con,sum(zz((l-1)*20+i,:))+sum(zz((l-1)*20+i+4,:))+sum(zz((l-1)*20+i+8,:))+sum(zz((l-1)*20+i+12,:))+sum(zz((l-1)*20+i+16,:))<=Na(i)*ma];
        Con=[Con,sum(zz1((l-1)*4+i,:))+sum(zz2((l-1)*4+i,:))+sum(zz3((l-1)*4+i,:))<=Nb(i)*mb];
    end
end