如何更改脚本以使其成本最小化，同时计算所选的体积和重量是可行的？

Question

我使用纸浆创建了一个函数，该函数可容纳一系列物品和卡车。 每个项目都有一个容量，每辆卡车都有一个容量和一个成本。 我使用纸浆求解器来计算所需的最少卡车数量，同时又保持了最低成本。 限制因素是容量，而优化因素是装箱成本。 但是现在，我想添加另一个约束量。 能做到吗？ 您能建议一种使用求解器本身的方法吗？ 我试图在不使用求解器的情况下完成该部分，但这完全没有帮助。 非常感谢。

我尝试创建一个函数，以便在收到消息“卡车不足”时将最小的卡车替换为第二大的卡车。

更新：@kabdulla帮助我得到了答案，但同时指出，它有一个很大的缺陷，即它仅考虑了整个项目的体积，而不是考虑要放入内部的项目的几何形状。 2 X 2 X 2物品应该很容易装入卡车，但是8 X 1 X 1的几何形状不同，约束也不同。 这是我更新的代码（全部感谢@kabdulla），其中包含物品和卡车的详细信息。

from pulp import *
import numpy as np

# Item details
item_name = ["Mudguard1","Mudguard2","Mudguard3","Mudguard4",]
item_id = ["1001474185","1001474401","1001474182","1001474154"]
item_length = [14,22,16,16]
item_breadth = [12,24,14,12]
item_height = [9,26,12,17]
item_quantity = [14,8,10,1]
item_vol = [12.25,63.55,15.55,1.88]
item_mass= [160,528,83,5]
n_items = len(item_id)
set_items = range(n_items)

# Details of trucks
item_name = ["Tata-407-9.5","Tata-407-13","Tata-407-17","Tata-407-16",]
item_id = ["1","2","3","4"]
truck_length = [14,22,16,16]
truck_breadth = [12,24,14,12]
truck_height = [9,26,12,17]
truck_quantity = [14,8,10,1]
truck_vol = [12.25,63.55,15.55,1.88]
truck_mass= [160,528,83,5]
truck_cost = [7,1.5,18,100]


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 allocation 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")

这是我的输出：

Status: Optimal
Total Cost is:  126.5
x_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: 4.0
Item 0 is packed in vehicle 0
Item 1 is packed in vehicle 1
Item 2 is packed in vehicle 2
Item 3 is packed in vehicle 3
The volume of used trucks is 93.22999999999999
Trucks are sufficient

这是一个可玩的colab链接。 https://colab.research.google.com/drive/1uKuS2F4Xd1Lbps8yAwivLbhzZrD_Hv0-

Answer 1

以下代码的版本假定总载重量和每辆卡车的总载重量限制。 注意：这不能保证可以组合在卡车内的物品（体积为8m x 1m x 1m的物品的体积为8立方米，但是不能容纳在内部空间为2m x 2m x 2m = 8立方米的卡车中）例如，但使用下面的音量限制不会违反该限制）。

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