Question

A university service center hires part-time student workers and opens 10:00am-4:00pm daily. The number of student workers required hour-by-hour is given below.

Time	Required # of student workers
10:00 - 11:00	8
11:00 - noon	10
noon - 1:00	5
1:00 - 2:00	8
2:00 - 3:00	10
3:00 - 4:00	6

Student workers are scheduled for 4-hour shifts and work for 4 consecutive hours once they report on duty. The goal is to find the staff scheduling plan with the minimum total number of student workers hired.

Hint: There are only 3 possible shifts to assign student workers to, i.e., starting at 10:00am, 11:00am, noon.

Formulate this problem as a linear programming model. DO NOT SOLVE.

Clearly type your Decision Variables, Objective Function, and Constraints.

Answer 1

Students can join the shift at 10:00 am, 11:00 am and 12:00 noon only as they have to work for 4 hour shifts and each student can only be part of one shift. Hence we have an objective function with three variables

• X₁ = No. of students starting at 10:00 am.
• X₂ = No. of students starting at 11:00 am.
• X₃ = No. of students starting at 12:00 noon.

The objective function is therefore to Minimize Y = X₁ + X₂ + X₃.

The Constraints are the number of Students available every hour. This will be principally equal to the number of students available from the three shits as they join and leave which can be easily computed. There will be a constraint equation corresponding to every one hour block as specified in the question

• 10:00 - 11:00 : Minimum of 8 students needed. Hence. X₁ >= 8
• 11:00 - 12:00 : Minimum 10 students needed. Hence X₁ + X₂ >= 10. Note that all students from previous shift would be continuing to work
• 12:00 - 1:00 : Minimum of 5 students needed. Hence X₁ + X₂ + X₃ >= 5
• 1:00 - 2:00 : Minimum of 8 students needed, however no more students joined. Hence X₁ + X₂ + X₃ >= 8.
• 2:00 - 3:00 : Minimum of 10 students needed, now the first shift has finished. Hence X₂ + X₃ >= 10
• 3:00 - 4:00 : Minimum of 6 students needed, now the second shift has also finished. Hence X₃ >= 6

Note that the third constraint is redundant as the the fourth constraint subsumes it. Hence we have 5 constraints along with the fact that all X_i are positive integers.