python - 使用 PuLP 优化器实现具有额外弹性约束的装箱问题

Question

我正在使用 python 作为编程语言并实现将相似长度分组在一起的约束以满足线性规划。引用下图代码

import pulp
from itertools import product
import pandas as pd
import numpy as np

# DataFrame of item, weight, and length
df_updated = pd.DataFrame([['item1', 10, 'A'], ['item2', 20, 'B'],  ['item3', 20, 'C'], 
        ['item4', 20, 'B'], ['item5',10, 'A'], ['item6',10, 'B']], 
        columns = ['itemname', 'QuantityToGroup', 'Length'])

# Max weightage per bin
max_weight = 40

# Max bin to use
min_bins = int(np.ceil(round((df_updated['QuantityToGroup'].sum() / max_weight))))
max_bins = 3

problem = pulp.LpProblem("Grouping_lengths", pulp.LpMinimize)

# Variable to check, if we are using the bin or not
bin_used = pulp.LpVariable.dicts('is_bin_used', range(max_bins), lowBound=0, upBound=1, cat='Binary')

# Possible combinations to put the item in the bin
possible_item_in_bin = [(item_index, bin_num) for item_index, bin_num in product(df_updated.index, range(max_bins))]
item_in_bin = pulp.LpVariable.dicts('is_item_in_bin', possible_item_in_bin, lowBound=0, upBound=1, cat = 'Binary')

# Only one item in each bin
for item_index in df_updated.index:
    problem += pulp.lpSum([item_in_bin[item_index, bin_index] for bin_index in range(max_bins)]) == 1, f"Ensure that item {item_index} is only in one bin"

# Sum of quantity grouped in each bin must be less than max weight
for bin_index in range(max_bins):
    problem += pulp.lpSum(
            [item_in_bin[item_index, bin_index] * df_updated.loc[item_index, 'QuantityToGroup'] for item_index in df_updated.index]
        ) <= max_weight * bin_used[bin_index], f"Sum of items in bin {bin_index} should not exceed max weight {max_weight}"

# Length Constraints
lengths = list(df_updated.Length.unique())
for length in lengths:
    items_n = df_updated.index[df_updated['Length'] == length].tolist()
    if len(items_n) > 1:
        for bin in range(max_bins - 1):
            first_index = items_n[0]
            for item in items_n[1:]:
                constr = pulp.LpConstraint(item_in_bin[first_index, bin] - item_in_bin[item, bin], sense = 0, rhs = 0, name = f"place item {item} in bin {bin} if length number {length} is chosen for this bin")
                problem += constr

# Objective function to minimize bins used
problem += pulp.lpSum(bin_used[bin_index] for bin_index in range(max_bins)), "Objective: Minimize Bins Used"

problem.solve(pulp.PULP_CBC_CMD(msg = False))


for val in problem.variables():
    if val.varValue == 1:
       print(val.name, val.varValue)

对于给定的输入代码，无法对长度为 B 的项目进行分组，因为长度 B 的总权重为(项目 2 -> 20，项目 4 -> 20，项目 6 -> 10)50，大于最大值权重 40。代码按预期工作。

但我必须使长度约束具有弹性，这意味着违反约束是可以的，但如果违反约束则应增加惩罚。我探索了Elastic Constraints我认为这正是我的要求。

但是我在实现它们时遇到了一个问题，即保持问题的线性。我必须以不同的方式制定我的约束吗？感谢您的帮助。

代码的可能预期输出，确保尊重最小化浪费的目标并遵循约束。如果不遵守约束，则增加处罚。

# item 1 (A - 10), item 5 (A - 10), item3 (C - 20) on 1st bin. 
# item 2 (B) and item 4 (B) on 2nd bin.
# item 6 (B - 10) on 3rd bin

我还尝试了其他方法来制定长度约束部分，如下所示:

# Length Variable
lengths = list(df_updated.length.unique())

# Possible combinations to put the lengths in the bin
possible_length_in_bin = [(length, bin_num) for length, bin_num in product(range(len(lengths)), range(max_bins))]

# Constraint to group similar lengths together on same bin
length_in_bin = pulp.LpVariable.dicts('LengthInBin', possible_length_in_bin, cat = 'Binary')
for item, length, bins_index in product(df_updated.index, range(len(lengths)), range(max_bins)):
    problem += pulp.lpSum(item_in_bin[(item, bins_index)] == length_in_bin[(length, bins_index)]), (f"Only place item {item} in bin {bins_index} if length number {length} is chosen for this bin")

该部分的其余部分与上面相同。但是，该解决方案仍然没有返回预期的结果。

Answer 1

这是我认为可以回复邮件的解决方案。它仍然是线性的。您需要引入几个变量来计算特定 bin 中不同长度的数量。

其中一些变量需要“大 M”类型约束才能将二进制变量链接到求和。

然后有了这个变量，您可以为“重载”具有多种长度类型的 bin 添加一个小(或大？)惩罚。

再看一遍，可以删除 tot_bins 变量，只需在任何地方用 lpSum(bin_used[b] for b in bins) 替换，但很清楚如所写。

我用 black 重新格式化了代码，我不确定我是否喜欢它，但至少它是一致的。 :)

代码

import pulp
from itertools import product
import pandas as pd
import numpy as np

# DataFrame of item, weight, and length
df_updated = pd.DataFrame(
    [
        ["item1", 10, "A"],
        ["item2", 20, "B"],
        ["item3", 20, "C"],
        ["item4", 20, "B"],
        ["item5", 10, "A"],
        ["item6", 10, "B"],
    ],
    columns=["itemname", "QuantityToGroup", "Length"],
)
lengths = list(df_updated.Length.unique())

# Max weightage per bin
max_weight = 40

# big M for number of items
big_M = len(df_updated)

# Max bin to use
min_bins = int(np.ceil(round((df_updated["QuantityToGroup"].sum() / max_weight))))
max_bins = 3
bins = list(range(max_bins))

problem = pulp.LpProblem("Grouping_lengths", pulp.LpMinimize)

# Variable to check, if we are using the bin or not
bin_used = pulp.LpVariable.dicts(
    "is_bin_used", bins, cat="Binary"
)

# Indicator that items of dimension d are located in bin b:
loaded = pulp.LpVariable.dicts(
    "loaded", [(d, b) for d in lengths for b in bins], cat="Binary"
)

# the total count of bins used
tot_bins = pulp.LpVariable("bins_used")

# the total count of overloads in a bin.  overload = (count of dimensions in bin) - 1
overload = pulp.LpVariable.dicts("bin_overloads", bins, lowBound=0)

# Possible combinations to put the item in the bin
possible_item_in_bin = [
    (item_index, bin_num) for item_index, bin_num in product(df_updated.index, bins)
]
item_in_bin = pulp.LpVariable.dicts(
    "is_item_in_bin", possible_item_in_bin, cat="Binary"
)

# Force each item to be loaded...
for item_index in df_updated.index:
    problem += (
        pulp.lpSum([item_in_bin[item_index, bin_index] for bin_index in bins]) == 1,
        f"Ensure that item {item_index} is only in one bin",
    )

# Sum of quantity grouped in each bin must be less than max weight
for bin_index in bins:
    problem += (
        pulp.lpSum(
            [
                item_in_bin[item_index, bin_index]
                * df_updated.loc[item_index, "QuantityToGroup"]
                for item_index in df_updated.index
            ]
        )
        <= max_weight * bin_used[bin_index],
        f"Sum of items in bin {bin_index} should not exceed max weight {max_weight}",
    )

# count the number of dimensions (lengths) in each bin
for b in bins:
    for d in lengths:
        problem += loaded[d, b] * big_M >= pulp.lpSum(
            item_in_bin[idx, b] for idx in df_updated.index[df_updated.Length == d]
        )

# attach the "bin used" variable to either the "loaded" var or "item in bin" var...
for b in bins:
    problem += bin_used[b] * big_M >= pulp.lpSum(
        item_in_bin[idx, b] for idx in df_updated.index
    )

# count total bins used
problem += tot_bins >= pulp.lpSum(bin_used[b] for b in bins)

# count the overloads by bin
for b in bins:
    problem += overload[b] >= pulp.lpSum(loaded[d, b] for d in lengths) - 1


# Objective function to minimize bins used, with some small penalty for total overloads
usage_wt = 1.0
overload_wt = 0.2

problem += (
    usage_wt * tot_bins + overload_wt * pulp.lpSum(overload[b] for b in bins),
    "Objective: Minimize Bins Used, penalize overloads",
)

problem.solve(pulp.PULP_CBC_CMD(msg=False))
status = pulp.LpStatus[problem.status]
assert(status=='Optimal')  # <--- always ensure this before looking at result if you don't print solve status


print(f"total bins used: {tot_bins.varValue}")
print("bin overloads:")
for b in bins:
    if overload[b].varValue > 0:
        print(f"    bin {b} has {overload[b].varValue} overloads")

for idx, b in possible_item_in_bin:
    if item_in_bin[idx, b].varValue == 1:
        print(
            f"load {df_updated.itemname.iloc[idx]}/{df_updated.Length.iloc[idx]} in bin {b}"
        )

结果:

total bins used: 3.0
bin overloads:
    bin 0 has 1.0 overloads
load item1/A in bin 0
load item2/B in bin 2
load item3/C in bin 1
load item4/B in bin 2
load item5/A in bin 0
load item6/B in bin 0