基于or-tools的护士排班问题建模求解
基于or-tools的护士排班问题建模求解
- 护士排班问题（Nurse Rostering Problem，NRP）
- ortools官网例题1：A nurse scheduling problem
-
- 代码解析
- 完整代码
- ortools官网例题2：Scheduling with shift requests
-
- 代码解析
- 完整代码
)
护士排班问题（Nurse Rostering Problem，NRP）
护士排班问题（Nurse Rostering Problem，NRP）或护士排程问题（ nurse scheduling problem，NSP）是员工调度问题（Employee Scheduling）的一种。医院需要反复为护理人员制作值班表，通常情况下，护理人员要花费大量的时间来编制值班表，特别是在有许多工作人员提出要求的情况下，而且在处理对当前值班表的临时更改时可能会花费更多的时间。由于人工调度繁琐、耗时，以及其他种种原因，护士排班问题(NRP)或护士排程问题(NSP)引起了人们的广泛关注。
相关文献：
- http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.1030.5363&rep=rep1&type=pdf
- https://arxiv.org/pdf/1804.05002.pdf
ortools官网例题1：A nurse scheduling problem
or-tools官网给出了一个使用CP-SAT求解器解决NRP的算例（https://developers.google.cn/optimization/scheduling/employee_scheduling#java_4）：
医院主管需要满足一个为期 3 天的护士计划，让其在 3 天内满足 4 个护士的条件，但需满足以下条件：
- 每个班次（shift）分为三个 8 小时。
- 每天，每个班次都会分配给一名护士，而每个护士都不例外。
- 在这 3 天时间里，每个护士都至少分配到两次班次。
∑ s = 1 S x n d s = 1 , ∀ d = 1 , 2 , ⋯ , D ; n = 1 , 2 , ⋯ , N ∑ s = 1 S x n d s ≤ 1 , ∀ S × D N ≤ ∑ d = 1 D ∑ n = 1 N x n d s ≤ S × D N + ( S × D ) % N , ∀ s = 1 , 2 , ⋯ , S \sum_{s=1}^Sx_{nds}=1, \quad \forall d=1,2,\cdots,D ; n=1,2,\cdots,N \\ \sum_{s=1}^Sx_{nds} \leq 1,\quad \forall \\ \frac{S \times D}{N} \leq \sum_{d=1}^D\sum_{n=1}^Nx_{nds} \leq \frac{S \times D}{N}+(S \times D)\%N, \quad \forall s=1,2,\cdots,S s=1∑Sxnds=1,∀d=1,2,⋯,D;n=1,2,⋯,Ns=1∑Sxnds≤1,∀NS×D≤d=1∑Dn=1∑Nxnds≤NS×D+(S×D)%N,∀s=1,2,⋯,S
代码解析
1、导入ortools库
from ortools.sat.python import cp_model
2、构造数据
num_nurses = 4 # 护士人数
num_shifts = 3 # 每天有3个班次
num_days = 3 # 3天
all_nurses = range(num_nurses)
all_shifts = range(num_shifts)
all_days = range(num_days)
3、创建模型
model = cp_model.CpModel()
4、创建变量
# 如果将 班次s在d天分配给护士n,则等于 1
shifts = {}
for n in all_nurses:
for d in all_days:
for s in all_shifts:
shifts[(n, d, s)] = model.NewBoolVar(f"shift_n{n}_d{d}_s{s}")
5、约束条件
# 每天每个班次都会分配给一名护士：每天每个班次分配的护士人数之和=1
for (int d : allDays) {
for (int s : allShifts) {
List<Literal> nurses = new ArrayList<>();
for (int n : allNurses) {
nurses.add(shifts[n][d][s]);
}
model.addExactlyOne(nurses);
}
}
# 每个护士每天最多上一个班次
for n in all_nurses:
for d in all_days:
model.AddAtMostOne(shifts[(n, d, s)] for s in all_shifts)
每个护士上的班次尽可能均分，有4个护士，3天*3班次/天=9班次
则每个护士平均分配9 / 4 = 2.25班次，则每个护士至少上2个班次，至多上3个班次。
# Try to distribute the shifts evenly, so that each nurse works
# min_shifts_per_nurse shifts. If this is not possible, because the total
# number of shifts is not divisible by the number of nurses, some nurses will
# be assigned one more shift.
min_shifts_per_nurse = (num_shifts * num_days) // num_nurses
if num_shifts * num_days % num_nurses == 0:
max_shifts_per_nurse = min_shifts_per_nurse
else:
max_shifts_per_nurse = min_shifts_per_nurse + 1
for n in all_nurses:
shifts_worked = []
for d in all_days:
for s in all_shifts:
shifts_worked.append(shifts[(n, d, s)])
model.Add(min_shifts_per_nurse <= sum(shifts_worked))
model.Add(sum(shifts_worked) <= max_shifts_per_nurse)
6、设置模型参数
# 在非优化模型中，可以启用对所有解决方案的搜索
solver = cp_model.CpSolver()
solver.parameters.linearization_level = 0
# Enumerate all solutions.
solver.parameters.enumerate_all_solutions = True
7、调用回调函数
class NursesPartialSolutionPrinter(cp_model.CpSolverSolutionCallback):
"""Print intermediate solutions."""
def __init__(self, shifts, num_nurses, num_days, num_shifts, limit):
cp_model.CpSolverSolutionCallback.__init__(self)
self._shifts = shifts
self._num_nurses = num_nurses
self._num_days = num_days
self._num_shifts = num_shifts
self._solution_count = 0
self._solution_limit = limit
def on_solution_callback(self):
self._solution_count += 1
print(f"Solution {self._solution_count}")
for d in range(self._num_days):
print(f"Day {d}")
for n in range(self._num_nurses):
is_working = False
for s in range(self._num_shifts):
if self.Value(self._shifts[(n, d, s)]):
is_working = True
print(f" Nurse {n} works shift {s}")
if not is_working:
print(f" Nurse {n} does not work")
if self._solution_count >= self._solution_limit:
print(f"Stop search after {self._solution_limit} solutions")
self.StopSearch()
def solution_count(self):
return self._solution_count
# Display the first five solutions.
solution_limit = 5
solution_printer = NursesPartialSolutionPrinter(
shifts, num_nurses, num_days, num_shifts, solution_limit
)
8、调用求解器求解
solver.Solve(model, solution_printer)
完整代码
import numpy as np
from ortools.sat.python import cp_model
import collections
num_nurses = 4 # 护士人数
num_shifts = 3 # 每天有3个班次
num_days = 3 # 3天
all_nurses = range(num_nurses)
all_shifts = range(num_shifts)
all_days = range(num_days)
print(all_nurses)
model = cp_model.CpModel()
# 如果将 班次shift s 在d天分配给护士n,则等于 1
shifts = {}
for nurse in all_nurses:
for day in all_days:
for shift in all_shifts:
shifts[(nurse, day, shift)] = model.NewBoolVar(f"nurse{nurse}_day{day}_shift{shift}")
# 每天每个班次都会分配给一名护士：每天每个班次分配的护士人数之和=1
# Each shift is assigned to a single nurse per day.
for day in all_days:
for shift in all_shifts:
model.AddExactlyOne(shifts[(nurse, day, shift)] for nurse in all_nurses)
# 每个护士每天最多上一个班次
for nurse in all_nurses:
for day in all_days:
model.AddAtMostOne(shifts[(nurse, day, shift)] for shift in all_shifts)
"""
每个护士上的班次尽可能均分，有4个护士，3天*每天3班次=9班次
则每个护士平均9 // 4 = 2
"""
# Try to distribute the shifts evenly, so that each nurse works
# min_shifts_per_nurse shifts. If this is not possible, because the total
# number of shifts is not divisible by the number of nurses, some nurses will
# be assigned one more shift.
min_shifts_per_nurse = (num_shifts * num_days) // num_nurses
if num_shifts * num_days % num_nurses == 0:
max_shifts_per_nurse = min_shifts_per_nurse
else:
max_shifts_per_nurse = min_shifts_per_nurse + 1
for n in all_nurses:
shifts_worked = []
for d in all_days:
for s in all_shifts:
shifts_worked.append(shifts[(n, d, s)])
model.Add(min_shifts_per_nurse <= sum(shifts_worked))
model.Add(sum(shifts_worked) <= max_shifts_per_nurse)
# 在非优化模型中，可以启用对所有解决方案的搜索
solver = cp_model.CpSolver()
solver.parameters.linearization_level = 0
# Enumerate all solutions.
solver.parameters.enumerate_all_solutions = True
class NursesPartialSolutionPrinter(cp_model.CpSolverSolutionCallback):
"""Print intermediate solutions."""
"""
调用回调函数，打印中间结果
"""
def __init__(self, shifts, num_nurses, num_days, num_shifts, limit):
cp_model.CpSolverSolutionCallback.__init__(self)
self._shifts = shifts
self._num_nurses = num_nurses
self._num_days = num_days
self._num_shifts = num_shifts
self._solution_count = 0
self._solution_limit = limit
def on_solution_callback(self):
self._solution_count += 1
so = np.zeros(shape=(num_days, num_shifts), dtype=np.int64)
print(f"Solution {self._solution_count}")
for d in range(self._num_days):
# print(f"Day {d}")
for n in range(self._num_nurses):
is_working = False
for s in range(self._num_shifts):
if self.Value(self._shifts[(n, d, s)]):
is_working = True
# print(f" Nurse {n} works shift {s}")
so[d][s] = n
if not is_working:
# print(f" Nurse {n} does not work")
pass
if self._solution_count >= self._solution_limit:
print(f"Stop search after {self._solution_limit} solutions")
self.StopSearch()
print(f' shift1 shift2 shift3')
for i in range(len(so)):
print(f'day{i + 1}', end='\t')
for j in range(len(so[i])):
print(f'nurse{so[i][j] + 1}', end='\t')
print()
def solution_count(self):
return self._solution_count
# Display the first five solutions.显示前5个解
solution_limit = 5
solution_printer = NursesPartialSolutionPrinter(
shifts, num_nurses, num_days, num_shifts, solution_limit
)
solver.Solve(model, solution_printer)
输出结果为：
ortools官网例题2：Scheduling with shift requests
例题2相比于例题1，增加了特定班次的护士需求，目标函数为最大化护士需求满足的人数（尽可能满足护士需求）。对于大多数调度问题，输出所有解不太可能，因此需要有一个目标函数。例题2和例题1约束条件相同。
代码解析
1、导入库
from ortools.sat.python import cp_model
2、导入数据
shift_requests 是一个5 * 7 * 3的矩阵，表示5个护士7天，每一天3个班次的值班需求。如shift[2][0][1]代表护士护士2在第0天想上班次1。
num_nurses = 5
num_shifts = 3
num_days = 7
all_nurses = range(num_nurses)
all_shifts = range(num_shifts)
all_days = range(num_days)
shift_requests = [
[[0, 0, 1], [0, 0, 0], [0, 0, 0], [0, 0, 0], [0, 0, 1], [0, 1, 0], [0, 0, 1]],
[[0, 0, 0], [0, 0, 0], [0, 1, 0], [0, 1, 0], [1, 0, 0], [0, 0, 0], [0, 0, 1]],
[[0, 1, 0], [0, 1, 0], [0, 0, 0], [1, 0, 0], [0, 0, 0], [0, 1, 0], [0, 0, 0]],
[[0, 0, 1], [0, 0, 0], [1, 0, 0], [0, 1, 0], [0, 0, 0], [1, 0, 0], [0, 0, 0]],
[[0, 0, 0], [0, 0, 1], [0, 1, 0], [0, 0, 0], [1, 0, 0], [0, 1, 0], [0, 0, 0]],
]
3、创建模型
model = cp_model.CpModel()
4、模型变量
shifts = {}
for n in all_nurses:
for d in all_days:
for s in all_shifts:
shifts[(n, d, s)] = model.NewBoolVar(f"shift_n{n}_d{d}_s{s}")
5、约束条件
for d in all_days:
for s in all_shifts:
model.AddExactlyOne(shifts[(n, d, s)] for n in all_nurses)
for n in all_nurses:
for d in all_days:
model.AddAtMostOne(shifts[(n, d, s)] for s in all_shifts)
# Try to distribute the shifts evenly, so that each nurse works
# min_shifts_per_nurse shifts. If this is not possible, because the total
# number of shifts is not divisible by the number of nurses, some nurses will
# be assigned one more shift.
min_shifts_per_nurse = (num_shifts * num_days) // num_nurses
if num_shifts * num_days % num_nurses == 0:
max_shifts_per_nurse = min_shifts_per_nurse
else:
max_shifts_per_nurse = min_shifts_per_nurse + 1
for n in all_nurses:
num_shifts_worked = 0
for d in all_days:
for s in all_shifts:
num_shifts_worked += shifts[(n, d, s)]
model.Add(min_shifts_per_nurse <= num_shifts_worked)
model.Add(num_shifts_worked <= max_shifts_per_nurse)
5、目标函数
# pylint: disable=g-complex-comprehension
model.Maximize(
sum(
shift_requests[n][d][s] * shifts[(n, d, s)]
for n in all_nurses
for d in all_days
for s in all_shifts
)
)
这里python用了嵌套的列表推导式，转换成一般写法，更直观：
expr = 0
for n in all_nurses:
for d in all_days:
for s in all_shifts:
expr += shift_requests[n][d][s] * shifts[(n, d, s)]
model.Maximize(expr)
6、调用求解器
solver = cp_model.CpSolver()
status = solver.Solve(model)
solver.Solve(model)返回的是求解状态（是否得到最优解、可行解等），这里可以从Java语法来看返回值类型，更直观，以上两行代码等价于：
CpSolver solver = new CpSolver();
CpSolverStatus status = solver.solve(model);
7、结果输出
if (status == CpSolverStatus.OPTIMAL || status == CpSolverStatus.FEASIBLE) {
System.out.printf("Solution:%n");
for (int d : allDays) {
System.out.printf("Day %d%n", d);
for (int n : allNurses) {
for (int s : allShifts) {
if (solver.booleanValue(shifts[n][d][s])) {
if (shiftRequests[n][d][s] == 1) {
System.out.printf(" Nurse %d works shift %d (requested).%n", n, s);
} else {
System.out.printf(" Nurse %d works shift %d (not requested).%n", n, s);
}
}
}
}
}
System.out.printf("Number of shift requests met = %f (out of %d)%n", solver.objectiveValue(),
numNurses * minShiftsPerNurse);
} else {
System.out.printf("No optimal solution found !");
}
完整代码
"""Nurse scheduling problem with shift requests."""
from ortools.sat.python import cp_model
def main():
# This program tries to find an optimal assignment of nurses to shifts
# (3 shifts per day, for 7 days), subject to some constraints (see below).
# Each nurse can request to be assigned to specific shifts.
# The optimal assignment maximizes the number of fulfilled shift requests.
num_nurses = 5
num_shifts = 3
num_days = 7
all_nurses = range(num_nurses)
all_shifts = range(num_shifts)
all_days = range(num_days)
shift_requests = [
[[0, 0, 1], [0, 0, 0], [0, 0, 0], [0, 0, 0], [0, 0, 1], [0, 1, 0], [0, 0, 1]],
[[0, 0, 0], [0, 0, 0], [0, 1, 0], [0, 1, 0], [1, 0, 0], [0, 0, 0], [0, 0, 1]],
[[0, 1, 0], [0, 1, 0], [0, 0, 0], [1, 0, 0], [0, 0, 0], [0, 1, 0], [0, 0, 0]],
[[0, 0, 1], [0, 0, 0], [1, 0, 0], [0, 1, 0], [0, 0, 0], [1, 0, 0], [0, 0, 0]],
[[0, 0, 0], [0, 0, 1], [0, 1, 0], [0, 0, 0], [1, 0, 0], [0, 1, 0], [0, 0, 0]],
]
# Creates the model.
model = cp_model.CpModel()
# Creates shift variables.
# shifts[(n, d, s)]: nurse 'n' works shift 's' on day 'd'.
shifts = {}
for n in all_nurses:
for d in all_days:
for s in all_shifts:
shifts[(n, d, s)] = model.NewBoolVar(f"shift_n{n}_d{d}_s{s}")
# Each shift is assigned to exactly one nurse in .
for d in all_days:
for s in all_shifts:
model.AddExactlyOne(shifts[(n, d, s)] for n in all_nurses)
# Each nurse works at most one shift per day.
for n in all_nurses:
for d in all_days:
model.AddAtMostOne(shifts[(n, d, s)] for s in all_shifts)
# Try to distribute the shifts evenly, so that each nurse works
# min_shifts_per_nurse shifts. If this is not possible, because the total
# number of shifts is not divisible by the number of nurses, some nurses will
# be assigned one more shift.
min_shifts_per_nurse = (num_shifts * num_days) // num_nurses
if num_shifts * num_days % num_nurses == 0:
max_shifts_per_nurse = min_shifts_per_nurse
else:
max_shifts_per_nurse = min_shifts_per_nurse + 1
for n in all_nurses:
num_shifts_worked = 0
for d in all_days:
for s in all_shifts:
num_shifts_worked += shifts[(n, d, s)]
model.Add(min_shifts_per_nurse <= num_shifts_worked)
model.Add(num_shifts_worked <= max_shifts_per_nurse)
# pylint: disable=g-complex-comprehension
model.Maximize(
sum(
shift_requests[n][d][s] * shifts[(n, d, s)]
for n in all_nurses
for d in all_days
for s in all_shifts
)
)
# Creates the solver and solve.
solver = cp_model.CpSolver()
status = solver.Solve(model)
if status == cp_model.OPTIMAL:
print("Solution:")
for d in all_days:
print("Day", d)
for n in all_nurses:
for s in all_shifts:
if solver.Value(shifts[(n, d, s)]) == 1:
if shift_requests[n][d][s] == 1:
print("Nurse", n, "works shift", s, "(requested).")
else:
print("Nurse", n, "works shift", s, "(not requested).")
print()
print(
f"Number of shift requests met = {solver.ObjectiveValue()}",
f"(out of {num_nurses * min_shifts_per_nurse})",
)
else:
print("No optimal solution found !")
# Statistics.
print("\nStatistics")
print(f" - conflicts: {solver.NumConflicts()}")
print(f" - branches : {solver.NumBranches()}")
print(f" - wall time: {solver.WallTime()}s")
if __name__ == "__main__":
main()