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import numpy as np
import cvxpy as cpx=cp.Variable(6,pos=True)
obj=cp.Minimize(x[5])
a1=np.array([0.025, 0.015, 0.055, 0.026])
a2=np.array([0.05, 0.27, 0.19, 0.185, 0.185])
a3=np.array([1, 1.01, 1.02, 1.045, 1.065])
k=0.05; kk=[]; qq=[]
while k<0.27:con=[cp.multiply(a1,x[1:5])-x[5]<=0,a2@x[:-1]>=k, a3@x[:-1]==1]prob=cp.Problem(obj,con)prob.solve(solver='GLPK_MI')kk.append(k); qq.append(prob.value)k=k+0.005plt.rc('text',usetex=False); plt.rc('font',size=16); plt.rc('font',family='SimHei')
plt.plot(kk,qq,'k')
plt.plot(kk,qq,'b.')
plt.xlabel("收益 k"); plt.ylabel("风险 Q",rotation=0)
plt.show()print("学号：3028")print("学号：3028")print("学号：2023310143028")
runfile('C:/Users/ddrrg/.spyder-py3/PYTHON2/习题4.4.py', wdir='C:/Users/ddrrg/.spyder-py3/PYTHON2')
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MAX_B = 24
MAX_DEBUG = 5 products = [ {"name": "Ⅰ", "A_hours": 1, "B_hours": 6, "debug_hours": 1, "profit": 2}, # 假设产品Ⅰ至少使用1小时设备A {"name": "Ⅱ", "A_hours": 5, "B_hours": 2, "debug_hours": 1, "profit": 1}
] max_profit = 0
best_plan = {} for i in range(MAX_A // products[0]["A_hours"] + 1): for j in range(MAX_B // products[1]["B_hours"] + 1): # 计算调试时间是否足够 if (i + j) * max(products[0]["debug_hours"], products[1]["debug_hours"]) > MAX_DEBUG: continue total_A_hours = i * products[0]["A_hours"] + j * products[1]["A_hours"] total_B_hours = i * products[0]["B_hours"] + j * products[1]["B_hours"] if total_A_hours > MAX_A or total_B_hours > MAX_B: continue total_profit = i * products[0]["profit"] + j * products[1]["profit"] if total_profit > max_profit: max_profit = total_profit best_plan = {"Ⅰ": i, "Ⅱ": j} print(f"最优生产计划：产品Ⅰ生产{best_plan['Ⅰ']}件，产品Ⅱ生产{best_plan['Ⅱ']}件")
print(f"最大利润为：{max_profit}元")print("学号：3028")print("学号：2023310143028")
最优生产计划：产品Ⅰ生产3件，产品Ⅱ生产2件
最大利润为：8元
学号：3028
