有关matlab优化求解取整的问题我用matlab中的linprog函数求解编的程序如下:>> A=[1,1,1,1,1,0,0,0,0,0,0,0,0,0,0;0,0,0,0,0,1,1,1,1,1,0,0,0,0,0;0,0,0,0,0,0,0,0,0,0,1,1,1,1,1;-1,0,0,0,0,-1,0,0,0,0,-1,0,0,0,0;0,-1,0,0,0,0,-1,0,0,0,0,
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![有关matlab优化求解取整的问题我用matlab中的linprog函数求解编的程序如下:>> A=[1,1,1,1,1,0,0,0,0,0,0,0,0,0,0;0,0,0,0,0,1,1,1,1,1,0,0,0,0,0;0,0,0,0,0,0,0,0,0,0,1,1,1,1,1;-1,0,0,0,0,-1,0,0,0,0,-1,0,0,0,0;0,-1,0,0,0,0,-1,0,0,0,0,](/uploads/image/z/5420227-67-7.jpg?t=%E6%9C%89%E5%85%B3matlab%E4%BC%98%E5%8C%96%E6%B1%82%E8%A7%A3%E5%8F%96%E6%95%B4%E7%9A%84%E9%97%AE%E9%A2%98%E6%88%91%E7%94%A8matlab%E4%B8%AD%E7%9A%84linprog%E5%87%BD%E6%95%B0%E6%B1%82%E8%A7%A3%E7%BC%96%E7%9A%84%E7%A8%8B%E5%BA%8F%E5%A6%82%E4%B8%8B%EF%BC%9A%3E%3E+A%3D%5B1%2C1%2C1%2C1%2C1%2C0%2C0%2C0%2C0%2C0%2C0%2C0%2C0%2C0%2C0%3B0%2C0%2C0%2C0%2C0%2C1%2C1%2C1%2C1%2C1%2C0%2C0%2C0%2C0%2C0%3B0%2C0%2C0%2C0%2C0%2C0%2C0%2C0%2C0%2C0%2C1%2C1%2C1%2C1%2C1%3B-1%2C0%2C0%2C0%2C0%2C-1%2C0%2C0%2C0%2C0%2C-1%2C0%2C0%2C0%2C0%3B0%2C-1%2C0%2C0%2C0%2C0%2C-1%2C0%2C0%2C0%2C0%2C)
有关matlab优化求解取整的问题我用matlab中的linprog函数求解编的程序如下:>> A=[1,1,1,1,1,0,0,0,0,0,0,0,0,0,0;0,0,0,0,0,1,1,1,1,1,0,0,0,0,0;0,0,0,0,0,0,0,0,0,0,1,1,1,1,1;-1,0,0,0,0,-1,0,0,0,0,-1,0,0,0,0;0,-1,0,0,0,0,-1,0,0,0,0,
有关matlab优化求解取整的问题
我用matlab中的linprog函数求解编的程序如下:
>> A=[1,1,1,1,1,0,0,0,0,0,0,0,0,0,0;
0,0,0,0,0,1,1,1,1,1,0,0,0,0,0;
0,0,0,0,0,0,0,0,0,0,1,1,1,1,1;
-1,0,0,0,0,-1,0,0,0,0,-1,0,0,0,0;
0,-1,0,0,0,0,-1,0,0,0,0,-1,0,0,0;
0,0,-1,0,0,0,0,-1,0,0,0,0,-1,0,0;
0,0,0,-1,0,0,0,0,-1,0,0,0,0,-1,0;
0,0,0,0,-1,0,0,0,0,-1,0,0,0,0,-1];
f=[10,8,6,5,4,6,5,4,3,6,3,4,5,5,9];
b=[310;260;280;-180;-80;-200;-160;-220];
lb=[0;0;0;0;0;0;0;0;0;0;0;0;0;0;0];
>> [x,fval]=linprog(f,A,b,[],[],lb)
出来的结果是:
x =
0.0000
0.0000
39.6224
40.3776
220.0000
0.0000
0.0000
140.3776
119.6224
0.0000
180.0000
80.0000
20.0000
0.0000
0.0000
fval =
3.2000e+003
但是我希望的是x的每个值(一共15个变量)都是整数,该怎么改,
有关matlab优化求解取整的问题我用matlab中的linprog函数求解编的程序如下:>> A=[1,1,1,1,1,0,0,0,0,0,0,0,0,0,0;0,0,0,0,0,1,1,1,1,1,0,0,0,0,0;0,0,0,0,0,0,0,0,0,0,1,1,1,1,1;-1,0,0,0,0,-1,0,0,0,0,-1,0,0,0,0;0,-1,0,0,0,0,-1,0,0,0,0,
加入取整函数round(x)