I am trying to create an optimal shift schedule where employees are assigned to shift times. The output should aim to spend the least amount of money. The tricky part is I need to account for specific constraints. These being:
1) At any given time period, you must meet the minimum staffing requirements
2) A person has a minimum and maximum amount of hours they can do
3) An employee can only be scheduled to work within their available hours
4) A person can only work one shift per day
The staff_availability df contains the employees to choose from ['Person'], the available min - max hours they can work ['MinHours']-['MaxHours'], how much they get paid ['HourlyWage'], and availability, expressed as hours ['Availability_Hr'] and 15min segments ['Availability_15min_Seg'].
Note: Available employees don't have to be assigned shifts if not required. They're just available to do so.
The staffing_requirements df contains the time of day ['Time'] and the staff required ['People'] during those periods.
The script returns a df 'availability_per_member' that displays how many employees are available at each point in time. So 1 indicates available to be scheduled and 0 indicates not available. It then aims to allocate shift times, while accounting for the constraints using pulp.
I am getting an output but the shift times aren't applied to employees consecutively.
import pandas as pd
import matplotlib.pyplot as plt
import matplotlib.dates as dates
import pulp
staffing_requirements = pd.DataFrame({
'Time' : ['0/1/1900 8:00:00','0/1/1900 9:59:00','0/1/1900 10:00:00','0/1/1900 12:29:00','0/1/1900 12:30:00','0/1/1900 13:00:00','0/1/1900 13:02:00','0/1/1900 13:15:00','0/1/1900 13:20:00','0/1/1900 18:10:00','0/1/1900 18:15:00','0/1/1900 18:20:00','0/1/1900 18:25:00','0/1/1900 18:45:00','0/1/1900 18:50:00','0/1/1900 19:05:00','0/1/1900 19:07:00','0/1/1900 21:57:00','0/1/1900 22:00:00','0/1/1900 22:30:00','0/1/1900 22:35:00','1/1/1900 3:00:00','1/1/1900 3:05:00','1/1/1900 3:20:00','1/1/1900 3:25:00'],
'People' : [1,1,2,2,3,3,2,2,3,3,4,4,3,3,2,2,3,3,4,4,3,3,2,2,1],
})
staff_availability = pd.DataFrame({
'Person' : ['C1','C2','C3','C4','C5','C6','C7','C8','C9','C10','C11'],
'MinHours' : [3,3,3,3,3,3,3,3,3,3,3],
'MaxHours' : [10,10,10,10,10,10,10,10,10,10,10],
'HourlyWage' : [26,26,26,26,26,26,26,26,26,26,26],
'Availability_Hr' : ['8-18','8-18','8-18','9-18','9-18','9-18','12-1','12-1','17-3','17-3','17-3'],
'Availability_15min_Seg' : ['1-41','1-41','1-41','5-41','5-41','5-41','17-69','17-79','37-79','37-79','37-79'],
})
staffing_requirements['Time'] = ['/'.join([str(int(x.split('/')[0])+1)] + x.split('/')[1:]) for x in staffing_requirements['Time']]
staffing_requirements['Time'] = pd.to_datetime(staffing_requirements['Time'], format='%d/%m/%Y %H:%M:%S')
formatter = dates.DateFormatter('%Y-%m-%d %H:%M:%S')
# 15 Min
staffing_requirements = staffing_requirements.groupby(pd.Grouper(freq='15T',key='Time'))['People'].max().ffill()
staffing_requirements = staffing_requirements.reset_index(level=['Time'])
staffing_requirements.index = range(1, len(staffing_requirements) + 1)
staff_availability.set_index('Person')
staff_costs = staff_availability.set_index('Person')[['MinHours', 'MaxHours', 'HourlyWage']]
availability = staff_availability.set_index('Person')[['Availability_15min_Seg']]
availability[['first_15min', 'last_15min']] = availability['Availability_15min_Seg'].str.split('-', expand=True).astype(int)
availability_per_member = [pd.DataFrame(1, columns=[idx], index=range(row['first_15min'], row['last_15min']+1))
for idx, row in availability.iterrows()]
availability_per_member = pd.concat(availability_per_member, axis='columns').fillna(0).astype(int).stack()
availability_per_member.index.names = ['Timeslot', 'Person']
availability_per_member = (availability_per_member.to_frame()
.join(staff_costs[['HourlyWage']])
.rename(columns={0: 'Available'}))
''' Generate shift times based off availability '''
prob = pulp.LpProblem('CreateStaffing', pulp.LpMinimize) # Minimize costs
timeslots = staffing_requirements.index
persons = availability_per_member.index.levels[1]
# A member is either staffed or is not at a certain timeslot
staffed = pulp.LpVariable.dicts("staffed",
((timeslot, staffmember) for timeslot, staffmember
in availability_per_member.index),
lowBound=0,
cat='Binary')
# Objective = cost (= sum of hourly wages)
prob += pulp.lpSum(
[staffed[timeslot, staffmember] * availability_per_member.loc[(timeslot, staffmember), 'HourlyWage']
for timeslot, staffmember in availability_per_member.index]
)
# Staff the right number of people
for timeslot in timeslots:
prob += (sum([staffed[(timeslot, person)] for person in persons])
== staffing_requirements.loc[timeslot, 'People'])
# Do not staff unavailable persons
for timeslot in timeslots:
for person in persons:
if availability_per_member.loc[(timeslot, person), 'Available'] == 0:
prob += staffed[timeslot, person] == 0
# Do not underemploy people
for person in persons:
prob += (sum([staffed[(timeslot, person)] for timeslot in timeslots])
>= staff_costs.loc[person, 'MinHours']*4) # timeslot is 15 minutes => 4 timeslots = hour
# Do not overemploy people
for person in persons:
prob += (sum([staffed[(timeslot, person)] for timeslot in timeslots])
<= staff_costs.loc[person, 'MaxHours']*4) # timeslot is 15 minutes => 4 timeslots = hour
prob.solve()
print(pulp.LpStatus[prob.status])
output = []
for timeslot, staffmember in staffed:
var_output = {
'Timeslot': timeslot,
'Staffmember': staffmember,
'Staffed': staffed[(timeslot, staffmember)].varValue,
}
output.append(var_output)
output_df = pd.DataFrame.from_records(output)#.sort_values(['timeslot', 'staffmember'])
output_df.set_index(['Timeslot', 'Staffmember'], inplace=True)
if pulp.LpStatus[prob.status] == 'Optimal':
print(output_df)
Below is an output for the first two hours (8 15 min time slots). The issue is the shifts aren't consecutive. The employees scheduled for the first 8 time slots are mainly different. I'd have 5 people starting within the first 2 hours. Employees should only work one shift per day.
Timeslot C
0 1 C2
1 2 C2
2 3 C1
3 4 C3
4 5 C6
5 6 C1
6 7 C5
7 8 C2
Here's an answer to your revised question, i.e., how to add a constraint that requires each employee to work consecutive time periods.
I suggest that you add the following constraint (written here algebraically):
x[t+1,p] <= x[t,p] + (1 - (1/T) * sum_{s=1}^{t-1} x[s,p]) for all p, for all t < T
where x is your staffed variable (written here as x for compactness), t is the time index, T is the number of time periods, and p is the employee index.
The logic of the constraint is: If x[t,p] = 0 (the employee is not working in period t) and x[s,p] = 1 for any s < t (the employee was working in any prior period), then x[t+1,p] must = 0 (the employee cannot be working in period t+1. Thus, once the employee stops working, they can't re-start. Note that if x[t,p] = 1 or x[s,p] = 0 for every s < t, then x[t+1,p] can equal 1.
Here's my implementation of this constraint in pulp:
# If an employee works and then stops, they can't start again
num_slots = max(timeslots)
for timeslot in timeslots:
if timeslot < num_slots:
for person in persons:
prob += staffed[timeslot+1, person] <= staffed[timeslot, person] + \
(1 - (1./num_slots) *
sum([staffed[(s, person)] for s in timeslots if s < timeslot]))
I ran the model and got:
Optimal
Staffed
Timeslot Staffmember
1 C2 1.0
2 C2 1.0
3 C2 1.0
4 C2 1.0
5 C2 1.0
6 C2 1.0
7 C2 1.0
8 C2 1.0
9 C2 1.0
C6 1.0
10 C2 1.0
C6 1.0
11 C2 1.0
C6 1.0
12 C2 1.0
C6 1.0
13 C3 1.0
C6 1.0
14 C3 1.0
C6 1.0
etc. So, employees are working in consecutive time periods.
Note that the new constraints slow down the model a bit. It still solves in <30 seconds or so. But if you are solving much larger instances, you might have to re-think the constraints.