我如何更改我的脚本，以便将成本降至最低，同时计算出所选的体积和重量是否可行？

Question

我使用纸浆创建了一个函数，它接收一系列物品和卡车。每件物品都有容量，每辆卡车都有容量和成本。我使用纸浆求解器来计算所需的最少卡车数量，同时保持最低成本。约束因素是容量，优化因素是仓位成本。但是现在，我想添加另一个体积限制因素。可以做到吗？你能建议一种使用求解器本身来做到这一点的方法吗？我试图在不使用求解器的情况下完成那部分，但它根本没有帮助。非常感谢。我试图创建一个函数，每当我收到消息“卡车不够用”时，它就会用下一个最大的卡车替换最小的卡车更新：@kabdulla 帮助我得到了答案，但也指出它有一个很大的缺陷，它只考虑了项目的整体体积，而不是项目的几何形状以适应内部；一个 2 X 2 X 2 的物品应该很容易装进卡车，但是一个 8 X 1 X 1 的物品有不同的几何形状和不同的限制。这是我更新的代码（全部归功于 @kabdulla），其中包含物品和卡车的详细信息。这是我的输出：Status: OptimalTotal Cost is:  126.5x_soln[[1. 0. 0. 0.] [0. 1. 0. 0.] [0. 0. 1. 0.] [0. 0. 0. 1.]]y_soln[1. 1. 1. 1.]Trucks used: 4.0Item 0 is packed in vehicle 0Item 1 is packed in vehicle 1Item 2 is packed in vehicle 2Item 3 is packed in vehicle 3The volume of used trucks is 93.22999999999999Trucks are sufficient这是一个可以玩的 colab 链接。 https://colab.research.google.com/drive/1uKuS2F4Xd1Lbps8yAwivLbhzZrD_Hv0-

Answer 1

下面的代码版本假设每辆卡车都有总载重量和总载重量限制。注意：这不能保证适合卡车的物品组合（8m x 1m x 1m 的物品体积为 8 立方米，但不适合内部空间为 2m x 2m x 2m = 8 立方的卡车例如米，但使用低于此的体积约束不会违反约束）。

from pulp import *

import numpy as np

# Item masses, volumes

item_mass = [16, 10, 5]

item_vol = [25, 12, 1]

n_items = len(item_vol)

set_items = range(n_items)

# Mass & volume capacities of trucks

truck_mass = [7, 15, 15, 15, 18, 38, 64, 100]

truck_vol = [25, 50, 50, 50, 100, 125, 250, 500]

# Cost of using each truck

truck_cost = [7, 1.5, 1.5, 1.5, 18, 380, 640, 1000]

n_trucks = len(truck_cost)

set_trucks = range(n_trucks)

y = pulp.LpVariable.dicts('truckUsed', set_trucks,

    lowBound=0, upBound=1, cat=LpInteger)

x = pulp.LpVariable.dicts('itemInTruck', (set_items, set_trucks), 

    lowBound=0, upBound=1, cat=LpInteger)

# Model formulation

prob = LpProblem("Truck allocatoin problem", LpMinimize)

# Objective

prob += lpSum([truck_cost[i] * y[i] for i in set_trucks])

# Constraints

for j in set_items:

    # Every item must be taken in one truck

    prob += lpSum([x[j][i] for i in set_trucks]) == 1

for i in set_trucks:

    # Respect the mass constraint of trucks

    prob += lpSum([item_mass[j] * x[j][i] for j in set_items]) <= truck_mass[i]*y[i]

    # Respect the volume constraint of trucks

    prob += lpSum([item_vol[j] * x[j][i] for j in set_items]) <= truck_vol[i]*y[i]

# Ensure y variables have to be set to make use of x variables:

for j in set_items:

    for i in set_trucks:

        x[j][i] <= y[i]

prob.solve()

x_soln = np.array([[x[i][j].varValue for i in set_items] for j in set_trucks])

y_soln = np.array([y[i].varValue for i in set_trucks])

print (("Status:"), LpStatus[prob.status])

print ("Total Cost is: ", value(prob.objective))

print("x_soln"); print(x_soln)

print("y_soln"); print(y_soln)

print("Trucks used: " + str(sum(([y_soln[i] for i in set_trucks]))))

for i in set_items:

    for j in set_trucks:

        if x[i][j].value() == 1:

            print("Item " + str(i) + " is packed in vehicle "+ str(j))

totalitemvol = sum(item_vol)

totaltruckvol = sum([y[i].value() * truck_vol[i] for i in set_trucks])

print("Volume of used trucks is " + str(totaltruckvol))

if(totaltruckvol >= totalitemvol):

  print("Trucks are sufficient")

else:

  print("Items cannot fit")

应该返回以下内容：

('Status:', 'Optimal')

('Total Cost is: ', 19.5)

x_soln

[[0. 0. 0.]

 [0. 0. 0.]

 [0. 0. 0.]

 [0. 1. 1.]

 [1. 0. 0.]

 [0. 0. 0.]

 [0. 0. 0.]

 [0. 0. 0.]]

y_soln

[0. 0. 0. 1. 1. 0. 0. 0.]

Trucks used: 2.0

Item 0 is packed in vehicle 4

Item 1 is packed in vehicle 3

Item 2 is packed in vehicle 3

Volume of used trucks is 150.0

Trucks are sufficient