Question

The Clark County Sheriff’s Department schedules police officers for 8-hour shifts. The beginning times for the shifts are 8:00 a.m., noon, 4:00 p.m., 8:00 p.m., midnight, and 4:00 a.m. An officer beginning a shift at one of these times works for the next 8 hours. During normal weekday operations, the number of officers needed varies depending on the time of day. The department staffing guidelines require the following minimum number of officers on duty:

	Time of Day	Minimum No. of Officers on Duty
	8:00 a.m.–noon	7
	Noon–4:00 p.m.	6
	4:00 p.m.–8:00 p.m.	7
	8:00 p.m.–midnight	7
	Midnight–4:00 a.m.	4
	4:00 a.m.–8:00 a.m.	6

Determine the number of police officers that should be scheduled to begin the 8-hour shifts at each of the six times to minimize the total number of officers required. (Hint: Let x₁ = the number of officers beginning work at 8:00 a.m.,x₂ = the number of officers beginning work at noon, and so on.) If your answer is zero, enter “0”.

	Starting Time	Officers Starting
	8:00 a.m.
	Noon
	4:00 p.m.
	8:00 p.m.
	Midnight
	4:00 a.m.

Answer 1

Variables:

Let the number of officers starting at each shift be denoted by:

Time of Day	Number of officers
8:00 a.m.–noon	x1
Noon–4:00 p.m.	x2
4:00 p.m.–8:00 p.m.	x3
8:00 p.m.–midnight	x4
Midnight–4:00 a.m.	x5
4:00 a.m.–8:00 a.m.	x6

Minimize:

x1 +x2 +x3 +x4 +x5 + x6

Such that:

x6 + x1 >= 6

x1 + x2 >= 7

x2 + x3 >= 6

x3 + x4 >= 7

x4 + x5 >= 7

x5 + x6 >= 4

x1, x2, x3, x4, x5, x6 >= 0

Solve using LP software. Solution is:

Shift	Number of officers
8:00 a.m.–noon	6
Noon–4:00 p.m.	2
4:00 p.m.–8:00 p.m.	4
8:00 p.m.–midnight	3
Midnight–4:00 a.m.	4
4:00 a.m.–8:00 a.m.	0

Total number of officers = 19