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. 9
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. 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.

The number of policemen Officers starting each shift is given as follows;

Starting  Time                       Officers Starting

8:00 a.m.                              3

Noon                                    3  

4:00 p.m.                               6

8:00 p.m.                               1

Midnight                               3

4:00 a.m.                               4

Step-by-step explanation:

The given parameters are;

The Time of the Day                       Number of Officers

8:00 a.m. --- Noon                              7

Noon --- 4:00 p.m.                             6  

4:00 p.m. --- 8:00 p.m.                      9

8:00 p.m. --- midnight                         7

Midnight --- 4:00 a.m.                        4

4:00 a.m. --- 8:00 a.m.                        6

From the table, we have;

The total number of hours worked = (7 + 6 + 9 + 7 + 4 + 6) × 4 = 156 hours

The approximate number of policemen = 156/8 = 19.5 ≈ 20

Let a, b, c, d, e, and f be the number of policemen that start their 8 hour shift starting from 4:00 a.m.

Let b be the number of policemen that start their 8 hour shift at 4:00 a.m.

∴ a + b = 7, b + c = 6, c + d = 9, d + e = 7, e + f = 4, f + a = 6

Given that there are 6 variables, by noting that the maximum value of the variables will be about 5, and by trial and error, we have

a - c = 1

d - b = 3

c - e = 2

d - f = 3

a - e = 2

a - f = 1

Whereby a + b + c + d + e + f = 20

Therefore, we have;

b = 3, d = 6, f = 3, a = 4, c = 3, e = 1

Which gives;

Starting  Time                       Officers Starting

8:00 a.m.                              3

Noon                                    3  

4:00 p.m.                               6

8:00 p.m.                               1

Midnight                               3

4:00 a.m.                               4

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