CC BY 4.0 （除特别声明或转载文章外）
如果这篇博客帮助到你，可以请我喝一杯咖啡~
Formulate the following Diet Problem. Transform the formulation into LP standard form. Use some LP solver to solve the problem. Try to find the dual of the problem (optional). Suppose that we have three foods and two nutritional requirements to satisfy:
Food Cost / Serving Vitamin A Calories Maximum Servings Corn (C) $0.18 107 72 10 2% milk (M) $0.23 500 121 10 Wheat bread (W) $0.05 0 65 10
\ Minimum Amount Nutrient in Diet Maximum Amount in Diet Calories (Cal) 2000 2250 Vitamin A (VA) 5000 50000
Formulate the problem
\[\begin{aligned} \min\, && 0.18x_1+0.23x_2+0.05x_3 \\ \textit{s.t.}\, & 2000\le & 72x_1+121x_2+65x_3 &\le 2250 \\ & 5000\le & 107x_1+500x_2 &\le 50000 \\ & 0\le & x_1&\le 10 \\ & 0\le & x_2&\le 10 \\ & 0\le & x_3&\le 10\\ \end{aligned}\]Transform the formulation into LP standard form
\[\mathbf{A}=\begin{bmatrix} 107 & 500 & 0 \\ -107& -500&0 \\ 72 & 121 & 65 \\ -72 & -121 & -65 \\ 1 & 0 & 0 \\ -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & -1 \end{bmatrix}\\ \mathbf{b}=\left[50000,-5000,2250,-2000,10, 0, 10, 0, 10, 0\right]^T\\ \mathbf{c}=\left[0.18,0.23,0.05\right]\]Use some LP solver to solve the problem
此处使用了 scipy.optimize.linprog，方便又快捷。
from scipy.optimize import linprog
if __name__ == "__main__":
A_ub = [[107, 500, 0],
[-107, -500, 0],
[72, 121, 65],
[-72, -121, -65],
[1, 0, 0],
[-1, 0, 0],
[0, 1, 0],
[0, -1, 0],
[0, 0, 1],
[0, 0, -1]]
b_ub = [50000, -5000, 2250, -2000, 10, 0, 10, 0, 10, 0]
c = [0.18, 0.23, 0.05]
print(linprog(c, A_ub, b_ub))
运行上面这段代码，得到结果，与参考答案一致。
con: array([], dtype=float64)
fun: 3.1500000382242486
message: 'Optimization terminated successfully.'
nit: 9
slack: array([4.47919444e+04, 2.08055583e+02, 2.49999993e+02, 7.18327624e-06,
8.05555529e+00, 1.94444471e+00, 2.22118501e-09, 1.00000000e+01,
1.78711305e-07, 9.99999982e+00])
status: 0
success: True
x: array([ 1.94444471, 10. , 9.99999982])