In: Operations Management
Based on the number of students dine in Deer Park Tavern, the manager determines that following number of waiters and waitresses are needed for each day of a week:
Mon | Tue | Wed | Thur | Fri | Sat | Sun |
5 | 7 | 9 | 8 | 10 | 9 | 5 |
The manager hires full time workers (waiters or waitresses) who normally work consecutively for 5 days followed by 2 day off. Additional part time workers can be hired who are required to work two days in a row. Part time workers are paid 25% more daily. Assuming that all workers are equally paid (you may assume $1/day for full time workers) within each category, respectively. In addition, to maintain the quality of service, part time workers on each day should be no more than 40% of full time workers.
Set up an Excel LP model to find a shift schedule for the manager to minimize the operating cost.
Decision variables:
Let Fi be the number of full-time employees in ith shift, where i = 1, 2, 3, 4, 5, 6, 7 for the shifts starting from Monday…to... Sunday respectively.
Let Pi be the number of part-time employees in ith shift, where i = 1, 2, 3, 4, 5, 6, 7 for the shifts starting from Monday…to... Sunday respectively.
The seven shifts will be required to schedule full time employees for 5 consecutive days and next 2 days off. The seven shifts will be required to schedule part-time employees for 2 consecutive days. Following the schedule for each shift where 1 represent working day and 0 represent off-day:
Full-time employee shift |
Part-time employee shift |
|||||||||||||
Day |
F1 |
F2 |
F3 |
F4 |
F5 |
F6 |
F7 |
P1 |
P2 |
P3 |
P4 |
P5 |
P6 |
P7 |
# of Employees -Monday |
1 |
0 |
0 |
1 |
1 |
1 |
1 |
1 |
0 |
0 |
0 |
0 |
1 |
1 |
# of Employees -Tuesday |
1 |
1 |
0 |
0 |
1 |
1 |
1 |
1 |
1 |
0 |
0 |
0 |
0 |
0 |
# of Employees -Wednesday |
1 |
1 |
1 |
0 |
0 |
1 |
1 |
0 |
1 |
1 |
0 |
0 |
0 |
0 |
# of Employees -Thursday |
1 |
1 |
1 |
1 |
0 |
0 |
1 |
0 |
0 |
1 |
1 |
0 |
0 |
0 |
# of Employees -Friday |
1 |
1 |
1 |
1 |
1 |
0 |
0 |
0 |
0 |
0 |
1 |
1 |
0 |
0 |
# of Employees -Saturday |
0 |
1 |
1 |
1 |
1 |
1 |
0 |
0 |
0 |
0 |
0 |
1 |
1 |
0 |
# of Employees -Sunday |
0 |
0 |
1 |
1 |
1 |
1 |
1 |
0 |
0 |
0 |
0 |
0 |
1 |
1 |
The objective of scheduling is minimize the requirement of employees.
The cost of each full time employee is $1 per day and that of part-time employee is $1.25 per day
Objective Function:
Min Z = $1*∑Fi + $1.25∑Pi
Subject To:
Constraint:
Employees required per day constraint:
Day |
Equation |
Monday |
F1 + 0F2 + 0F3 + F4 + F5 + F6 + F7 + P1 + 0P2 + 0P3 + 0P4 + 0P5 + 0P6 + P7 >= 5 |
Tuesday |
F1 + F2 + 0F3 + 0F4 + F5 + F6 + F7 + P1 + P2 + 0P3 + 0P4 + 0P5 + 0P6 + 0P7 >= 7 |
And so on for rest of days
Part-time employees should be not more than 40% o full-time employees on each day
Day |
Equation |
Monday |
P1 + 0P2 + 0P3 + 0P4 + 0P5 + 0P6 + P7 <= 0.4(F1 + 0F2 + 0F3 + F4 + F5 + F6 + F7) -0.4F1 + 0F2 + 0F3 -0.4*F4 -0.4*F5 -0.4*F6 -0.4*F7 + P1 + 0P2 + 0P3 + 0P4 + 0P5 + 0P6 + P7 <= 0 |
Tuesday |
P1 + P2 + 0P3 + 0P4 + 0P5 + 0P6 + 0P7 <= 0.4(F1 + 1F2 + 0F3 + 0F4 + F5 + F6 + F7) -0.4F1 + -0.4F2 + 0F3 + 0F4 -0.4*F5 -0.4*F6 -0.4*F7 + P1 + P2 + 0P3 + 0P4 + 0P5 + 0P6 + 0P7 <= 0 |
And so on for the rest of days
Excel model is as follows:
Full-time employee shift |
Part-time employee shift |
|||||||||||||
Day |
F1 |
F2 |
F3 |
F4 |
F5 |
F6 |
F7 |
P1 |
P2 |
P3 |
P4 |
P5 |
P6 |
P7 |
# of employees |
2 |
4 |
2 |
2 |
0 |
1 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
Total cost = $11