在matlab 中求解$$\max \sum_{i=1}^n\sum_{j=1}^{10}x_{i,j}$$ $$\text{s.t.}\begin{cases}y_i=10\sum_{j=1}^{10}x_{i,j}, i=1,2,\cdots,n\z_i=\frac{1}{4}\pi h_i^2, h_i=\sum_{j=1}^{10}jx_{i,j}, i=1,2,\cdots,n\d_{i,j}\geq 2.5+r_i+r_j, r_i=\frac{1}{2}\sum_{k=1}^{10}jx_{i,j}, r_j=\frac{1}{2}\sum_{k=1}^{10}jx_{j,k},i,j=1,2,\cdots,n\c_i=10h_i+10, i=1,2,\cdots,n\y_i\leq 500, i=1,2,\cdots,n\z_i\leq 2.8x_{i,1}+5.5x_{i,2}+8.5x_{i,3}+11.9x_{i,4}+14.5x_{i,5}, i=1,2,\cdots,n\x_{i,j}\in{0,1}, i=1,2,\cdots,n, j=1,2,\cdots,10\end{cases}$$
时间: 2024-01-21 22:16:36 浏览: 31
这是一个混合整数规划问题,可以使用MATLAB中的intlinprog函数进行求解。其中,目标函数的系数向量为10的向量,线性不等式约束和等式约束右侧的系数矩阵分别为$n\times 10$和$n\times n$的矩阵,线性等式约束右侧的系数向量为$n$维向量,所有变量的上下界均为0或1。以下是MATLAB代码实现:
```matlab
% 目标函数系数向量
f = -10 * ones(1, 10 * n);
% 不等式约束系数矩阵和右侧系数向量
A = zeros(4 * n, 10 * n);
b = zeros(4 * n, 1);
for i = 1:n
% y_i = 10 * sum_j(x_ij)
A(i, (i-1)*10+1:i*10) = -10 * ones(1, 10);
b(i) = 0;
% delta_ij >= 2.5 + r_i + r_j
for j = 1:n
if j ~= i
r_i = 0.5 * (1:10) * x((i-1)*10+1:i*10)';
r_j = 0.5 * (1:10) * x((j-1)*10+1:j*10)';
A(n+(i-1)*n+j, (i-1)*10+1:i*10) = 0.5 * (1:10);
A(n+(i-1)*n+j, (j-1)*10+1:j*10) = 0.5 * (1:10);
b(n+(i-1)*n+j) = 2.5 + r_i + r_j;
end
end
% y_i <= 500
A(2*n+i, (i-1)*10+1:i*10) = 10 * ones(1, 10);
b(2*n+i) = 500;
% z_i <= 2.8*x_i,1 + 5.5*x_i,2 + 8.5*x_i,3 + 11.9*x_i,4 + 14.5*x_i,5
A(3*n+i, (i-1)*10+1:i*10) = -1/4*pi*(1:10).^2;
A(3*n+i, (i-1)*10+1:i*10) = A(3*n+i, (i-1)*10+1:i*10) + ...
[2.8 5.5 8.5 11.9 14.5 zeros(1,5) zeros(1,5) zeros(1,5) zeros(1,5) zeros(1,5)];
b(3*n+i) = 0;
end
% 等式约束系数矩阵和右侧系数向量
Aeq = zeros(2 * n, 10 * n);
beq = zeros(2 * n, 1);
for i = 1:n
% z_i = 1/4*pi*h_i^2
Aeq(i, (i-1)*10+1:i*10) = 0.25 * pi * (1:10).^2;
beq(i) = 0;
% c_i = 10*h_i + 10
Aeq(n+i, (i-1)*10+1:i*10) = 10 * ones(1, 10);
beq(n+i) = 10;
end
% 变量的上下界
lb = zeros(10 * n, 1);
ub = ones(10 * n, 1);
% 求解混合整数规划问题
[x, fval, exitflag] = intlinprog(f, 1:10*n, A, b, Aeq, beq, lb, ub);
% 输出结果
if exitflag > 0
fprintf('最大值为:%f\n', -fval);
for i = 1:n
fprintf('x_%d = [', i);
for j = 1:10
fprintf('%d ', x((i-1)*10+j));
end
fprintf(']\n');
end
else
fprintf('求解失败!\n');
end
```