【工程】Pulp-案例混合问题（四）

一、问题描述

Ben这个人，想要尽可能便宜地生产他们的猫粮产品，同时确保他们满足罐装食品的营养分析要求。因此，他们想要改变每种食材的用量（主要成分是鸡肉、牛肉、羊肉、大米、小麦和凝胶），同时还能满足他们的营养标准。
鸡肉、牛肉和羊肉的成本分别为0.013美元、0.008美元和0.010美元，而大米、小麦和凝胶的成本分别为0.002美元、0.005美元和0.001美元（所有费用都是每克）。
食材与营养之间的关系如下，每克食材的贡献（克）：

image.png

假设只想用两种食材来制作猫粮：鸡肉和牛肉。

决策变量：
$x_{1}$：鸡肉
$x_{2}$：牛肉

目标函数：
$$min(0.013*x_{1}+0.008x_{2})$$

约束条件：
$$\left{
\begin{aligned}
x_{1}+x_{2} & = 100\
0.1x_{1}+0.2x_{2} & \geqslant 8\
0.08x_{1}+0.1x_{2} & \geqslant 6\
0.001x_{1}+0.005x_{2} & \leqslant 2\
0.002x_{1}+0.005x_{2} & \leqslant 0.4\
\end{aligned}
\right.$$

使用python程序来解决：

# Import PuLP modeler functions
from pulp import *

# Create the 'prob' variable to contain the problem data
prob = LpProblem("The Whiskas Problem",LpMinimize)

# The 2 variables Beef and Chicken are created with a lower limit of zero
x1=LpVariable("ChickenPercent",0,None,LpInteger)
x2=LpVariable("BeefPercent",0)

# The objective function is added to 'prob' first
prob += 0.013*x1 + 0.008*x2, "Total Cost of Ingredients per can"

# The five constraints are entered
prob += x1 + x2 == 100, "PercentagesSum"
prob += 0.100*x1 + 0.200*x2 >= 8.0, "ProteinRequirement"
prob += 0.080*x1 + 0.100*x2 >= 6.0, "FatRequirement"
prob += 0.001*x1 + 0.005*x2 <= 2.0, "FibreRequirement"
prob += 0.002*x1 + 0.005*x2 <= 0.4, "SaltRequirement"

# The problem data is written to an .lp file
prob.writeLP("WhiskasModel.lp")

# The problem is solved using PuLP's choice of Solver
prob.solve()

# The status of the solution is printed to the screen
print "Status:", LpStatus[prob.status]

# Each of the variables is printed with it's resolved optimum value
for v in prob.variables():
    print v.name, "=", v.varValue
    
# The optimised objective function value is printed to the screen
print "Total Cost of Ingredients per can = ", value(prob.objective)