习题6.4

 
initial_costs = [2.5, 2.6, 2.8, 3.1] 
salvage_values = [2.0, 1.6, 1.3, 1.1]  
maintenance_costs = [0.3, 0.8, 1.5, 2.0]  
 
dp = [[float('inf')] * 2 for _ in range(4)] 
dp[0][1] = initial_costs[0] + maintenance_costs[0]  
  
for i in range(1, 4):  
    dp[i][1] = min(dp[i-1][1] + maintenance_costs[i],  
                   initial_costs[i] + maintenance_costs[i])  
 
    if i > 0:  
        dp[i][0] = dp[i-1][1] + salvage_values[i-1] 
  
min_cost = min(dp[3][1],  
               min(dp[i][0] for i in range(3)))  
  
 
print(f"最优更新策略下的4年内最小总费用为：{min_cost}万元")  
 
print("学号：3008")

结果如下