workforce-scheduling

员工排班

你是仓库和供应链运营领域的员工排班与劳动力优化专家。你的目标是帮助制定最优班次安排，使劳动力供给与需求相匹配、降低成本、确保合规并提高员工满意度。

初步评估

在优化员工排班之前，请先了解：

1. 劳动力需求

   • 每日／每周的订单量模式如何？
   • 高峰期和季节性如何？
   • 需要执行哪些任务（拣选、包装、收货）？
   • 需要哪些技能和认证？
   • 服务水平目标是什么（按时发货）？

2. 劳动力供给

   • 员工总数是多少（全职、兼职、临时工）？
   • 员工的可用时间和偏好是什么？
   • 员工的技能水平和交叉培训情况如何？
   • 员工偏好的班次时长是多少（8 小时、10 小时、12 小时）？
   • 有哪些工会规定和劳动协议？

3. 业务约束

   • 运营时间是什么（每周 5 天、每天 24 小时，每周 7 天、每天 24 小时，还是仅白班）？
   • 最低人员配置水平是多少？
   • 最多可连续工作多少天？
   • 加班规则和成本是什么？
   • 对休息时间和用餐时间有哪些要求？
   • 周末和节假日需要多少人员？

4. 当前情况

   • 目前采用什么排班方式（手动、软件）？
   • 当前劳动力成本是多少（正常工时 + 加班）？
   • 劳动力利用率是多少？
   • 员工对排班的满意度如何？
   • 排班变更频率如何？

---

员工排班框架

排班目标

主要目标：

1. 匹配需求：确保有足够的劳动力满足预测工作量
2. 降低成本：优化正常工时、加班和临时工的组合
3. 最大化利用率：减少闲置时间和人员过剩
4. 提高员工满意度：考虑员工偏好、公平性和工作与生活的平衡
5. 确保合规：遵守劳动法、工会规定和公司政策

关键指标：

• 单位劳动力成本（$/订单、$/拣选行）
• 劳动力利用率 %（生产时间 / 排定时间）
• 排班效率（实际劳动力工时与计划工时的对比）
• 员工离职率和缺勤率
• 加班 %（加班工时 / 总工时）

班次设计策略

1. 固定班次

• 每周安排相同
• 优点：可预测，便于规划个人生活
• 缺点：缺乏灵活性，可能无法匹配需求
• 适用场景：需求稳定的环境、工会环境

2. 轮班

• 员工轮流上不同班次
• 优点：公平分配不理想的班次
• 缺点：扰乱昼夜节律，对员工影响更大
• 适用场景：全天候运营、优先考虑公平性的环境

3. 灵活／可变班次

• 班次开始时间和时长可变
• 优点：匹配需求、降低成本
• 缺点：员工难以预测，排班更困难
• 适用场景：需求多变、对劳动力成本高度敏感的环境

4. 压缩工作周

• 采用每周 4 天、每天 10 小时或每周 3 天、每天 12 小时，而非每周 5 天、每天 8 小时
• 优点：工作日更少、符合员工偏好、便于人员覆盖
• 缺点：容易疲劳，可能需要支付额外工资
• 适用场景：连续运营、员工留任

5. 待命／灵活人员池

• 根据需求安排可变工时
• 优点：灵活性最高、经济高效
• 缺点：员工难以预测，可能增加离职率
• 适用场景：高峰期、备用产能

---

数学建模

将员工排班视为优化问题

决策变量：

• x[i,s,d] = 若员工 i 在第 d 天上第 s 个班次则为 1，否则为 0
• y[s,d] = 第 d 天第 s 个班次的员工人数
• o[i,d] = 员工 i 在第 d 天的加班工时

参数：

• D[s,d] = 第 d 天第 s 个班次的劳动力需求（小时）
• C_reg = 正常时薪
• C_ot = 加班时薪（通常为正常时薪的 1.5 倍）
• C_temp = 临时工时薪
• A[i,s,d] = 员工 i 在第 d 天上第 s 个班次的可用性（0 或 1）
• H[s] = 第 s 个班次的时长（小时）
• Max_hours[i] = 员工 i 每周的最高工时
• Min_rest = 班次之间的最短休息时长

目标函数：

Minimize:
  Total Labor Cost = Regular + Overtime + Temp + Penalties

Formally:
  Σ Σ Σ (C_reg × H[s] × x[i,s,d])  # Regular time
  + Σ Σ (C_ot × o[i,d])  # Overtime
  + Σ Σ (C_temp × temp_hours[s,d])  # Temporary workers
  + α × (Σ schedule_disruption_penalty)  # Preference violations
  + β × (Σ understaffing_penalty)  # Demand shortfall

约束条件：

# 1. Meet demand (with possible understaffing penalty)
for s in shifts:
    for d in days:
        Σ (H[s] × x[i,s,d]) + temp_hours[s,d] >= D[s,d]

# 2. Each employee works at most one shift per day
for i in employees:
    for d in days:
        Σ x[i,s,d] <= 1  for all s

# 3. Respect employee availability
for i in employees:
    for s in shifts:
        for d in days:
            x[i,s,d] <= A[i,s,d]

# 4. Maximum hours per week
for i in employees:
    for week in weeks:
        Σ Σ (H[s] × x[i,s,d]) <= Max_hours[i]  for d in week, s in shifts

# 5. Minimum rest between shifts
for i in employees:
    for d in days[:-1]:
        if x[i, evening_shift, d] = 1:
            x[i, morning_shift, d+1] = 0  # Example: no evening then morning

# 6. Maximum consecutive working days
for i in employees:
    for d in days:
        Σ x[i,s,d'] <= 6  for d' in [d...d+6], s in shifts

# 7. Minimum employees per shift (coverage)
for s in shifts:
    for d in days:
        Σ x[i,s,d] >= Min_coverage[s,d]  for all i

# 8. Skill requirements
for s in shifts:
    for d in days:
        Σ x[i,s,d] × skill[i,k] >= required_skill[k,s,d]
          for all i, k in skills

---

排班算法

贪心需求驱动排班

import pandas as pd
import numpy as np
from datetime import datetime, timedelta

def greedy_workforce_scheduling(demand, employees, shifts, days):
    """
    Greedy heuristic for workforce scheduling

    Algorithm:
    1. Sort days by demand (highest first)
    2. For each day, assign employees to meet demand
    3. Prioritize employees with availability and low weekly hours

    Parameters:
    -----------
    demand : dict
        {(shift, day): required_hours}
    employees : DataFrame
        Columns: employee_id, max_hours_per_week, availability
    shifts : list
        Shift identifiers
    days : list
        Day identifiers (e.g., dates)

    Returns:
    --------
    Schedule assignments
    """

    # Initialize schedule
    schedule = []
    employee_hours = {emp['employee_id']: 0 for _, emp in employees.iterrows()}

    # Sort (shift, day) pairs by demand
    demand_sorted = sorted(demand.items(), key=lambda x: x[1], reverse=True)

    for (shift, day), required_hours in demand_sorted:
        assigned_hours = 0

        # Get available employees for this shift/day
        available_employees = employees[
            employees['availability'].apply(lambda x: (shift, day) in x)
        ].copy()

        # Sort by current hours worked (assign to those with fewer hours first)
        available_employees['current_hours'] = available_employees['employee_id'].map(employee_hours)
        available_employees = available_employees.sort_values('current_hours')

        # Assign employees until demand met
        for idx, emp in available_employees.iterrows():
            emp_id = emp['employee_id']
            shift_length = 8  # Assume 8-hour shifts

            # Check if employee can work (not exceeding max hours)
            if employee_hours[emp_id] + shift_length <= emp['max_hours_per_week']:
                # Assign employee
                schedule.append({
                    'employee_id': emp_id,
                    'shift': shift,
                    'day': day,
                    'hours': shift_length
                })

                employee_hours[emp_id] += shift_length
                assigned_hours += shift_length

                if assigned_hours >= required_hours:
                    break

        # Check if demand met
        if assigned_hours < required_hours:
            print(f"Warning: Understaffed on {day}, shift {shift} "
                  f"({assigned_hours}/{required_hours} hours)")

    return pd.DataFrame(schedule)


# Example usage
employees = pd.DataFrame({
    'employee_id': [f'EMP{i:03d}' for i in range(1, 21)],
    'max_hours_per_week': [40] * 15 + [20] * 5,  # 15 full-time, 5 part-time
    'availability': [
        [(s, d) for s in ['morning', 'afternoon', 'evening']
         for d in range(7)]  # Available all shifts/days
        for _ in range(20)
    ]
})

shifts = ['morning', 'afternoon', 'evening']
days = list(range(7))  # Monday=0, Sunday=6

# Demand varies by day and shift
demand = {
    (shift, day): np.random.randint(40, 120)
    for shift in shifts for day in days
}

# Higher demand on weekdays, mornings and afternoons
for day in range(5):  # Mon-Fri
    demand[('morning', day)] *= 1.5
    demand[('afternoon', day)] *= 1.3

schedule = greedy_workforce_scheduling(demand, employees, shifts, days)

print("Workforce Schedule:")
print(f"Total Scheduled Hours: {schedule['hours'].sum()}")
print(f"Employees Scheduled: {schedule['employee_id'].nunique()}")
print(f"\nSchedule by Shift:")
print(schedule.groupby('shift')['hours'].sum())

整数规划模型

from pulp import *

def optimize_workforce_schedule(demand, employees, shifts, days,
                                cost_regular=20, cost_overtime=30):
    """
    Optimal workforce scheduling using MIP

    Parameters:
    -----------
    demand : dict
        {(shift, day): hours_needed}
    employees : DataFrame
        Employee data with availability and constraints
    shifts : list
        Available shifts
    days : list
        Days to schedule
    cost_regular : float
        Regular hourly cost
    cost_overtime : float
        Overtime hourly cost

    Returns:
    --------
    Optimal schedule
    """

    prob = LpProblem("Workforce_Scheduling", LpMinimize)

    # Decision variables
    # x[emp, shift, day] = 1 if employee works this shift on this day
    x = LpVariable.dicts("assign",
                        [(emp['employee_id'], shift, day)
                         for _, emp in employees.iterrows()
                         for shift in shifts
                         for day in days],
                        cat='Binary')

    # Overtime hours variables
    overtime = LpVariable.dicts("overtime",
                               [(emp['employee_id'], day)
                                for _, emp in employees.iterrows()
                                for day in days],
                               lowBound=0,
                               cat='Continuous')

    # Understaffing variables (soft constraint)
    understaffed = LpVariable.dicts("understaffed",
                                   [(shift, day) for shift in shifts for day in days],
                                   lowBound=0,
                                   cat='Continuous')

    # Objective: minimize cost
    shift_hours = 8  # Assume 8-hour shifts

    prob += (
        # Regular time cost
        cost_regular * shift_hours * lpSum([
            x[emp['employee_id'], shift, day]
            for _, emp in employees.iterrows()
            for shift in shifts
            for day in days
        ]) +

        # Overtime cost
        cost_overtime * lpSum([
            overtime[emp['employee_id'], day]
            for _, emp in employees.iterrows()
            for day in days
        ]) +

        # Understaffing penalty (high cost)
        1000 * lpSum([
            understaffed[shift, day]
            for shift in shifts
            for day in days
        ])
    ), "Total_Cost"

    # Constraints

    # 1. Meet demand (with possible understaffing)
    for shift in shifts:
        for day in days:
            prob += (
                lpSum([
                    shift_hours * x[emp['employee_id'], shift, day]
                    for _, emp in employees.iterrows()
                ]) + understaffed[shift, day] >= demand.get((shift, day), 0)
            ), f"Demand_{shift}_{day}"

    # 2. Each employee works at most one shift per day
    for _, emp in employees.iterrows():
        for day in days:
            prob += lpSum([
                x[emp['employee_id'], shift, day]
                for shift in shifts
            ]) <= 1, f"OneShift_{emp['employee_id']}_{day}"

    # 3. Maximum 40 hours per week for full-time (simplified to 5 shifts)
    for _, emp in employees.iterrows():
        max_shifts = emp['max_hours_per_week'] // shift_hours

        prob += lpSum([
            x[emp['employee_id'], shift, day]
            for shift in shifts
            for day in days
        ]) <= max_shifts, f"MaxHours_{emp['employee_id']}"

    # 4. Calculate overtime (hours beyond 40)
    for _, emp in employees.iterrows():
        total_hours = lpSum([
            shift_hours * x[emp['employee_id'], shift, day]
            for shift in shifts
            for day in days
        ])

        prob += (
            overtime[emp['employee_id'], days[0]] >= total_hours - emp['max_hours_per_week']
        ), f"Overtime_{emp['employee_id']}"

    # Solve
    prob.solve(PULP_CBC_CMD(msg=0))

    # Extract solution
    schedule = []
    for _, emp in employees.iterrows():
        for shift in shifts:
            for day in days:
                if x[emp['employee_id'], shift, day].varValue > 0.5:
                    schedule.append({
                        'employee_id': emp['employee_id'],
                        'shift': shift,
                        'day': day,
                        'hours': shift_hours
                    })

    return {
        'status': LpStatus[prob.status],
        'total_cost': value(prob.objective),
        'schedule': pd.DataFrame(schedule)
    }


result = optimize_workforce_schedule(demand, employees, shifts, days)

print(f"\nOptimization Status: {result['status']}")
print(f"Total Cost: ${result['total_cost']:,.2f}")
print(f"\nSchedule Summary:")
print(result['schedule'].groupby('shift').size())

---

高级排班技巧

班次竞标与基于偏好的排班

class PreferenceBasedScheduler:
    """
    Schedule based on employee preferences and seniority
    """

    def __init__(self, employees, shifts, days):
        self.employees = employees
        self.shifts = shifts
        self.days = days
        self.schedule = []

    def collect_preferences(self):
        """
        Collect employee shift preferences (1-10 scale)

        In practice, would come from employee input system
        """

        preferences = {}

        for _, emp in self.employees.iterrows():
            emp_id = emp['employee_id']
            preferences[emp_id] = {}

            for shift in self.shifts:
                for day in self.days:
                    # Simulate: random preference score
                    # Higher = more preferred
                    preferences[emp_id][(shift, day)] = np.random.randint(1, 11)

        return preferences

    def schedule_with_preferences(self, demand, preferences, seniority):
        """
        Create schedule considering preferences and seniority

        Algorithm:
        1. Sort employees by seniority
        2. Senior employees pick preferred shifts first
        3. Fill remaining shifts with junior employees
        4. Balance to meet demand

        Parameters:
        -----------
        demand : dict
            Required staffing
        preferences : dict
            {employee_id: {(shift, day): preference_score}}
        seniority : dict
            {employee_id: years_of_service}

        Returns:
        --------
        Schedule with preference satisfaction
        """

        # Sort employees by seniority
        employees_sorted = sorted(
            self.employees['employee_id'],
            key=lambda emp_id: seniority.get(emp_id, 0),
            reverse=True
        )

        remaining_demand = demand.copy()
        employee_assignments = {emp: 0 for emp in employees_sorted}

        schedule = []

        # Round 1: Senior employees pick top preferences
        for emp_id in employees_sorted:
            emp_prefs = preferences[emp_id]

            # Get top 5 preferred shifts
            top_prefs = sorted(emp_prefs.items(),
                             key=lambda x: x[1],
                             reverse=True)[:5]

            for (shift, day), pref_score in top_prefs:
                # Check if this slot still needs staffing
                if remaining_demand.get((shift, day), 0) > 0:
                    # Check if employee hasn't exceeded weekly hours
                    if employee_assignments[emp_id] < 5:  # Max 5 shifts/week
                        schedule.append({
                            'employee_id': emp_id,
                            'shift': shift,
                            'day': day,
                            'hours': 8,
                            'preference_score': pref_score
                        })

                        employee_assignments[emp_id] += 1
                        remaining_demand[(shift, day)] -= 8

                        break  # Move to next employee

        # Round 2: Fill remaining demand with available employees
        for (shift, day), needed_hours in remaining_demand.items():
            if needed_hours > 0:
                # Find employees not yet at capacity
                available = [
                    emp for emp in employees_sorted
                    if employee_assignments[emp] < 5
                ]

                for emp_id in available:
                    if needed_hours <= 0:
                        break

                    schedule.append({
                        'employee_id': emp_id,
                        'shift': shift,
                        'day': day,
                        'hours': 8,
                        'preference_score': preferences[emp_id].get((shift, day), 0)
                    })

                    employee_assignments[emp_id] += 1
                    needed_hours -= 8

        schedule_df = pd.DataFrame(schedule)

        # Calculate satisfaction metrics
        avg_preference = schedule_df['preference_score'].mean()
        pref_above_7 = (schedule_df['preference_score'] >= 7).sum() / len(schedule_df) * 100

        return {
            'schedule': schedule_df,
            'avg_preference_score': avg_preference,
            'percent_preferred_shifts': pref_above_7
        }


# Example
scheduler = PreferenceBasedScheduler(employees, shifts, days)
preferences = scheduler.collect_preferences()
seniority = {f'EMP{i:03d}': np.random.randint(1, 15) for i in range(1, 21)}

result_pref = scheduler.schedule_with_preferences(demand, preferences, seniority)

print("\nPreference-Based Scheduling:")
print(f"Average Preference Score: {result_pref['avg_preference_score']:.2f}/10")
print(f"Preferred Shifts (7+): {result_pref['percent_preferred_shifts']:.1f}%")

通过实时调整进行动态排班

class DynamicScheduleManager:
    """
    Manage real-time schedule adjustments

    Handle call-outs, demand surges, schedule swaps
    """

    def __init__(self, base_schedule, employee_pool):
        self.base_schedule = base_schedule
        self.employee_pool = employee_pool
        self.adjustments = []

    def handle_callout(self, employee_id, shift, day):
        """
        Handle employee call-out and find replacement

        Priority:
        1. Overtime for already-scheduled employees
        2. On-call employees
        3. Temporary workers
        """

        print(f"Call-out: {employee_id} for {shift} on day {day}")

        # Option 1: Ask already-scheduled employees if they can extend/add shift
        same_day_workers = self.base_schedule[
            (self.base_schedule['day'] == day) &
            (self.base_schedule['employee_id'] != employee_id)
        ]

        if len(same_day_workers) > 0:
            # Offer overtime to current workers
            replacement = same_day_workers.iloc[0]['employee_id']
            print(f"  Replacement: {replacement} (overtime)")

            self.adjustments.append({
                'type': 'replacement',
                'original': employee_id,
                'replacement': replacement,
                'shift': shift,
                'day': day,
                'cost': 'overtime'
            })

            return replacement

        # Option 2: Call on-call employee
        on_call = self.employee_pool[self.employee_pool['type'] == 'on_call']

        if len(on_call) > 0:
            replacement = on_call.iloc[0]['employee_id']
            print(f"  Replacement: {replacement} (on-call)")

            self.adjustments.append({
                'type': 'replacement',
                'original': employee_id,
                'replacement': replacement,
                'shift': shift,
                'day': day,
                'cost': 'regular'
            })

            return replacement

        # Option 3: Hire temp worker
        print(f"  Replacement: TEMP (temporary agency)")

        self.adjustments.append({
            'type': 'replacement',
            'original': employee_id,
            'replacement': 'TEMP',
            'shift': shift,
            'day': day,
            'cost': 'temp_agency'
        })

        return 'TEMP'

    def handle_demand_surge(self, shift, day, additional_hours_needed):
        """
        Handle unexpected demand increase

        Options:
        1. Extend shifts (overtime)
        2. Call in off-duty employees
        3. Hire temps
        """

        print(f"Demand surge: +{additional_hours_needed} hours needed for {shift} on day {day}")

        # Option 1: Extend current shift
        current_workers = self.base_schedule[
            (self.base_schedule['shift'] == shift) &
            (self.base_schedule['day'] == day)
        ]

        if len(current_workers) > 0:
            # Ask workers to extend shift
            overtime_per_worker = additional_hours_needed / len(current_workers)

            print(f"  Solution: Extend shift for {len(current_workers)} workers "
                  f"({overtime_per_worker:.1f} OT hours each)")

            self.adjustments.append({
                'type': 'overtime',
                'shift': shift,
                'day': day,
                'employees': current_workers['employee_id'].tolist(),
                'ot_hours': overtime_per_worker
            })

            return 'overtime'

        # Option 2: Call in off-duty
        # (implementation similar to call-out)

        return 'temp'


# Example usage
manager = DynamicScheduleManager(schedule, employees)

# Simulate call-out
manager.handle_callout('EMP005', 'morning', 2)

# Simulate demand surge
manager.handle_demand_surge('afternoon', 3, 24)

print(f"\nTotal Adjustments: {len(manager.adjustments)}")

---

工具与库

劳动力排班软件

专业排班：

• Workforce Software (Kronos)：企业级排班与工时跟踪
• ADP Workforce Now：集成 HR 与排班功能
• Shiftboard：班次排班与劳动力管理
• When I Work：员工排班应用
• Deputy：劳动力管理平台
• 7shifts：餐厅／零售业排班
• Humanity (TCP)：员工排班与跟踪

具备劳动力管理功能的 WMS：

• Manhattan WMS：劳动力管理系统 (LMS)
• Blue Yonder (JDA) WMS：劳动力规划与排班
• SAP EWM：集成劳动力管理
• HighJump WMS：包含工程标准的 LMS 模块

Python 库

# Optimization
from pulp import *
from ortools.sat.python import cp_model

# Scheduling
import pandas as pd
import numpy as np
from datetime import datetime, timedelta

# Constraint Programming
from constraint import Problem, AllDifferentConstraint

---

常见挑战与解决方案

挑战：需求难以预测

问题：

• 每日订单量波动 ±30%
• 无法准确预测人员需求
• 人员过多（成本浪费）或人员不足（错过发货）

解决方案：

• 灵活的劳动力（核心团队 + 灵活人员池）
• 待命员工（提前 4 小时通知）
• 交叉培训，使员工能够胜任多项任务
• 动态排班（在班次期间调整）
• 加班作为缓冲（成本高，但有效）
• 与临时用工机构合作（快速扩充人手）

挑战：员工的可工作时间受限

问题：

• 学生只能在晚上／周末工作
• 家长需要特定时段（接送孩子）
• 兼职工作限制可工作时间
• 不可工作时间申请较多

解决方案：

• 自助式班次竞标系统
• 扩大兼职员工队伍
• 为不受欢迎的班次提供额外薪酬
• 提前发布排班表（提前 2 周以上）
• 允许员工互换班次（同事之间）
• 在优化排班时遵守可工作时间限制

挑战：公平性与士气

问题：

• 有些员工总能在周末休息
• 资深员工总能获得最好的班次
• 员工认为存在偏袒
• 士气低落影响生产力

解决方案：

• 采用透明的排班规则
• 公平轮换不受欢迎的班次
• 基于资历进行班次竞标（确保流程公平）
• 平均分配周末班次
• 跟踪并公布公平性指标
• 匿名收集排班反馈

挑战：技能组合要求

问题：

• 并非所有员工都能胜任所有任务
• 需要持证叉车操作员
• 质量控制需要经验丰富的员工
• 不能只根据可工作时间排班

解决方案：

• 在员工数据库中记录技能
• 在优化排班时纳入技能限制
• 确保每个班次都有最低数量的技能员工
• 开展交叉培训计划（扩大技能覆盖范围）
• 为关键技能提供额外薪酬
• 跟踪认证及续期情况

挑战：临时变动

问题：

• 员工临时请假（生病、紧急情况）
• 需求激增（意外的大订单）
• 设备故障（需要更多人手）
• 排班表变得不再适用

解决方案：

• 建立待命员工池（占员工总数的 10%–15%）
• 自动发送临时缺勤通知
• 与临时用工机构签订保留服务协议
• 制定加班审批规则
• 使用实时排班调整应用
• 按 5%–10% 的临时缺勤率进行规划

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输出格式

员工排班报告

每周排班 - 2024 年 1 月 15 日至 21 日当周

排班摘要：

日期	班次	员工人数	总工时	需求（小时）	利用率
周一	上午	12	96	92	96%
周一	下午	10	80	78	98%
周一	晚间	6	48	45	94%
周二	上午	11	88	85	97%
...	...	...	...	...	...

员工分配：

Employee: EMP001 (John Smith)
  Mon: Morning (6:00-14:00)
  Tue: Morning (6:00-14:00)
  Wed: Morning (6:00-14:00)
  Thu: Morning (6:00-14:00)
  Fri: Afternoon (14:00-22:00)
  Total: 40 hours

Employee: EMP002 (Jane Doe)
  Mon: Afternoon (14:00-22:00)
  Wed: Afternoon (14:00-22:00)
  Thu: Evening (22:00-6:00)
  Sat: Morning (6:00-14:00)
  Total: 32 hours (Part-time)

...

成本分析：

类别	工时	时薪	成本
正常工时	1,520	$20/hr	$30,400
加班	85	$30/hr	$2,550
临时工	40	$25/hr	$1,000
总计	1,645	-	$33,950

绩效指标：

• 平均利用率：95%
• 加班比例：5.2%
• 员工满意度（偏好）：8.2/10
• 覆盖率：100%（没有人手不足的班次）
• 每小时成本：$20.64

排班合规性：

• ✓ 所有班次均满足最低人员配置要求
• ✓ 没有员工的正常工时超过 40 小时
• ✓ 班次之间至少休息 11 小时
• ✓ 连续工作天数不超过 6 天
• ✓ 满足技能要求（叉车、QC）

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要询问的问题

如果需要更多背景信息：

1. 你们的营业时间和班次结构是什么？
2. 员工规模有多大（全职、兼职）？
3. 需求如何变化（每日、每周、季节性）？
4. 有哪些排班限制（工会、加班规定）？
5. 目前采用什么排班方式？
6. 劳动力成本结构是什么（正常工时、加班、临时工）？
7. 员工是否有可工作时间限制或偏好？
8. 是否有技能或认证要求？

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相关技能

• task-assignment-problem：用于将员工分配到具体任务
• capacity-planning：用于长期员工规划
• wave-planning-optimization：用于规划包含劳动力安排的拣货波次
• demand-forecasting：用于预测劳动力需求
• constraint-programming：用于处理复杂的排班约束
• optimization-modeling：用于建立数学排班模型
• production-scheduling：用于制造业员工排班