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	5
	Noon–4:00 p.m.	6
	4:00 p.m.–8:00 p.m.	8
	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

Hi

I think this will help you and make you understand the logic behind it.

Here

Let assume, x1= the number of officers beginning work at 8:00 a.m., x2 = the number of officers beginning work at noon, and so on

So We have to minimize = X1+X2+X3+X4+X5+X6

Constrains are

1) X1 >= 5

2) X1+ X2 >= 6

3) X2+X3 >= 8

4) X3+X4 >= 7

5) X4+X5 >= 4

6) X5+X6 >= 6

From 1 and 2 we get

X2 >= 1

Likewise, X3 >= 7

X4 >= 0

X5 >= 4

X6 >= 2

So, if we take all the minimum value of the Xi’s

X1+X2+X3+X4+X5+X6= 19