python数学建模第五章-优快云博客

5.4

import numpy as np  
from scipy.optimize import minimize  
  
def objective(x):  
    return -np.sum(np.sqrt(x) * np.arange(1, 101))  
  
def constraint1(x):  
    return x[1] - 10  
  
def constraint2(x):  
    return 20 - (x[1] + 2*x[2])  
  
def constraint3(x):  
    return 30 - (x[1] + 2*x[2] + 3*x[3])  
  
def constraint4(x):  
    return 40 - (x[1] + 2*x[2] + 3*x[3] + 4*x[4])  
  
def constraint5(x):  
    return 1000 - np.dot(x, np.arange(1, 101))  
  
constraints = [  
    {'type': 'ineq', 'fun': constraint1},  
    {'type': 'ineq', 'fun': constraint2},  
    {'type': 'ineq', 'fun': constraint3},  
    {'type': 'ineq', 'fun': constraint4},  
    {'type': 'ineq', 'fun': constraint5}  
]  
  
bounds = [(0, None)] * 100  
x0 = np.ones(100) * 0.1 
  
result = minimize(objective, x0, method='SLSQP', constraints=constraints, bounds=bounds)  
  
print('Optimal solution:', result.x)  
print('Objective function value at optimal solution:', -result.fun)

5.5

import numpy as np  
from scipy.optimize import minimize  
  
def objective(x):  
    return 2*x[0] + 3*x[0]**2 + 3*x[1] + x[1]**2 + x[2]  
  
def constraint1(x):  
    return 10 - (x[0] + 2*x[0]**2 + x[1] + 2*x[1]**2 + x[2])  
  
def constraint2(x):  
    return 50 - (x[0] + x[0]**2 + x[1] + x[1]**2 - x[2])  
  
def constraint3(x):  
    return 40 - (2*x[0] + x[0]**2 + 2*x[1] + x[2])  
  
 
def constraint4(x):  
    return x[0]**2 + x[2] - 2  
  
def constraint5(x):  
    return 1 - (x[0] + 2*x[1])  
  
constraints = [  
    {'type': 'ineq', 'fun': constraint1},  
    {'type': 'ineq', 'fun': constraint2},  
    {'type': 'ineq', 'fun': constraint3},  
    # {'type': 'eq', 'fun': constraint4}, 
    {'type': 'ineq', 'fun': constraint5}  
]  
  
 
bounds = [(0, None)] * 3  
x0 = np.array([0.1, 0.1, 0.1])  
  
result = minimize(objective, x0, method='SLSQP', constraints=constraints, bounds=bounds)  
  
print('Optimal solution:', result.x)  
print('Objective function value at optimal solution:', result.fun)

5.7

import numpy as np  
  
demands = [40, 60, 80]  
max_production = 100  
total_demand = sum(demands)  
  
dp = np.full((4, total_demand + 1), float('inf'))  
dp[0][0] = 0  
  
prev_production = np.full((4, total_demand + 1), -1) 
  
for i in range(1, 4):  
    prev_demand = sum(demands[:i-1])  
    for j in range(total_demand + 1):  
        if j < prev_demand + demands[i-1]:  
            
            continue  
        for x in range(max(0, j - prev_demand - demands[i-1] + 1), min(max_production + 1, j - prev_demand + 1)):  
            production_cost = 50 * x + 0.2 * x**2  
            storage_cost = 4 * (j - prev_demand - x)  
            total_cost = dp[i-1][j-x] + production_cost + storage_cost  
            if total_cost < dp[i][j]:  
                dp[i][j] = total_cost  
                prev_production[i][j] = x  
   
min_cost = float('inf')  
final_state = -1  
for j in range(total_demand, total_demand + 1):  
    if dp[3][j] < min_cost:  
        min_cost = dp[3][j]  
        final_state = j  
  
production_plan = [0] * 3  
current_state = final_state  
for i in range(3, 0, -1):  
    production_plan[i-1] = prev_production[i][current_state]  
    current_state -= prev_production[i][current_state]  
 
print(f"最小总费用为: {min_cost} 元")  
print("生产计划为:")  
for i, plan in enumerate(production_plan, 1):  
    print(f"第{i}季度生产: {plan} 台")