線性總和分配（SciPy）和平衡成本

Question

我很難使用scipy.optimize.linear_sum_assignment將任務（成本）平均分配給工人，其中每個工人都可以分配多個任務。 成本矩陣表示每個工人的每個任務的工作量。

我們希望最小化所有工人的總成本，同時平均分配每個工人的成本。

在這個例子中，我們有 3 個名為a 、 b和c的工人。 每個工人總共可以分配 4 個任務，因此在成本矩陣中，我們有代理a_1 、 a_2等等。

linear_sum_assignment確實為我們提供了總成本最小化的分配。 簡單來說，我們的示例使用了一個成本矩陣，這樣任何分配都會給我們相同的總成本。

然而，成本並沒有平均分配給 3 名工人。 在我們的示例中，3 名工人的192分別為65和163 。

是否可以盡可能地降低成本，同時將每個工人的成本更平均地分配給 3 名工人？

from scipy.optimize import linear_sum_assignment
import numpy as np

worker_capacities = [
    "a_1", "a_2", "a_3", "a_4",
    "b_1", "b_2", "b_3", "b_4",
    "c_1", "c_2", "c_3", "c_4",
]
n_tasks = len(worker_capacities)
c = np.array([
 [27, 42, 65, 33, 67, 45, 60, 76,  6,  6, 43, 26],
 [27, 42, 65, 33, 67, 45, 60, 76,  6,  6, 43, 26],
 [27, 42, 65, 33, 67, 45, 60, 76,  6,  6, 43, 26],
 [27, 42, 65, 33, 67, 45, 60, 76,  6,  6, 43, 26],
 [27, 42, 65, 33, 67, 45, 60, 76,  6,  6, 43, 26],
 [27, 42, 65, 33, 67, 45, 60, 76,  6,  6, 43, 26],
 [27, 42, 65, 33, 67, 45, 60, 76,  6,  6, 43, 26],
 [27, 42, 65, 33, 67, 45, 60, 76,  6,  6, 43, 26],
 [27, 42, 65, 33, 67, 45, 60, 76,  6,  6, 43, 26],
 [27, 42, 65, 33, 67, 45, 60, 76,  6,  6, 43, 26],
 [27, 42, 65, 33, 67, 45, 60, 76,  6,  6, 43, 26],
])

_, assignments = linear_sum_assignment(c)
print("Costs for worker a:", sum(c[i][j] for i, j in enumerate(assignments[0:4])))
print("Costs for worker b:", sum(c[i+4][j] for i, j in enumerate(assignments[4:8])))
print("Costs for worker c:", sum(c[i+8][j] for i, j in enumerate(assignments[8:12])))

給出輸出

Costs for worker a: 65
Costs for worker b: 163
Costs for worker c: 192

Answer 1

linear_sum_assignment方法不支持約束或自定義目標，所以我認為這是不可能的。

但是，您可以將您的問題表述為混合整數線性規划問題 (MILP)，並通過PuLP ¹解決它。 為了平均分配每個工人的總成本，您可以最小化每個工人的最大和最小總成本之間的差異。 這是一個可能的公式：

Sets:

- workers = ["a", "b", "c"]
- tasks   = [1, 2, ..., 12]

Variables:

- x[w,t] = 1 iff worker w is assigned to task t, 0 otherwise
- min_val
- max_val

Model:

min  max_val - min_val

s.t. 

# each worker is assigned to exactly n_tasks_per_worker tasks
sum(x[w,t] for t in tasks) == n_tasks_per_worker for all w in workers

# each task can only be assigned once
sum(x[w,t] for w in workers) == 1 for all t in tasks

# evenly distribute the tasks
sum(x[w,t] for t in tasks) <= max_val for all w in workers
sum(x[w,t] for t in tasks) >= min_val for all w in workers

代碼很簡單：

import pulp
import numpy as np

workers = ["a", "b", "c"]
n_workers = len(workers)
n_tasks_per_worker = 4
n_tasks = n_workers * n_tasks_per_worker

c = np.array([[27, 42, 65, 33, 67, 45, 60, 76,  6,  6, 43, 26],
              [27, 42, 65, 33, 67, 45, 60, 76,  6,  6, 43, 26],
              [27, 42, 65, 33, 67, 45, 60, 76,  6,  6, 43, 26]])

# create the model
mdl = pulp.LpProblem("even_assignment")

# decision variables
x = {}
for w in workers:
    for t in range(n_tasks):
        x[w, t] = pulp.LpVariable(f"x[{w}, {t}]", cat="Binary")

max_val = pulp.LpVariable("max_val", cat="Continuous")
min_val = pulp.LpVariable("min_val", cat="Continuous")

# objective: minimize the difference between the maximum and the minimum
#            costs per worker
mdl.setObjective(max_val - min_val)

# constraint: each worker gets assigned exactly n_tasks_per_worker
for w in workers:
    mdl.addConstraint(sum(x[w, task] for task in range(n_tasks)) == n_tasks_per_worker)

# constraint: each task can only be assigned once
for task in range(n_tasks):
    mdl.addConstraint(sum(x[w, task] for w in workers) == 1)

# constraint: evenly distribute the tasks
for i_w, w in enumerate(workers):
    assignment_cost = sum(x[w,task]*c[i_w,task] for task in range(n_tasks))
    mdl.addConstraint(assignment_cost <= max_val)
    mdl.addConstraint(assignment_cost >= min_val)

# solve the problem
mdl.solve()

# Output
for i_w, w in enumerate(workers):
    worker_cost = sum(x[w, t].varValue*c[i_w, t] for t in range(n_tasks))
    print(f"costs for worker {w}: {worker_cost:.2f}")

這給了我

costs for worker a: 165.00
costs for worker b: 167.00
costs for worker c: 164.00

---

[1] 確切地說，PuLP 不是求解器，它只是一個建模框架，可以將 MILP 傳遞給 CBC、SCIP、HiGHS 等 MILP 求解器。