During each 4-hour period of the day, the Old San Juan Police Department needs a

Question

During each 4-hour period of the day, the Old San Juan Police Department needs at least the following number of policemen. Each policeman has to work two consecutive four-hours shift (a total of eight hours):

Period

Minimum Demand

12:00am-04:00am

18

04:00am-08:00am

22

08:00am-12:00pm

18

12:00pm-04:00pm

16

04:00pm-08:00pm

22

08:00pm-12:00am

16

Formulate (standard form), show, and solve an optimization model that helps you find out the policemen schedule on this situation? Explain.

Period

Minimum Demand

12:00am-04:00am

18

04:00am-08:00am

22

08:00am-12:00pm

18

12:00pm-04:00pm

16

04:00pm-08:00pm

22

08:00pm-12:00am

16

Explanation / Answer

Let us assume following decision variables:

X1 = No. of police added in slot 1

X2 = No. of police added in slot 2

X3 = No. of police added in slot 3

X4 = No. of police added in slot 4

X5 = No. of police added in slot 5

X6 = No. of police added in slot 6

Obj. Function is to minimize total manpower = (X1+X2+X3+X4+X5+X6)

Constraints:

Each time slot police requirements shown below:

X6+X1 >= 18    

X1+X2 >= 22   

X2+X3 >= 18    

X3+X4 >= 16    

X4+X5 >= 22   

X5+X6 >= 16  

X1, X2, X3, X4.X5, X6 >=0

LP solver solution shown below:

Optimum Police Shedule :

X1 = 18

X2 = 8

X3 = 10

X4 = 6

X5 = 16

X6 = 0

12:00am-04:00am 04:00am-08:00am 08:00am-12:00pm 12:00pm-04:00pm 04:00pm-08:00pm 08:00pm-12:00am Decision Variables X1 X2 X3 X4 X5 X6 Police added in Shifts (CC) 18 8 10 6 16 0 Constraints Min. in Shift 1 18 >= 18 Min. in Shift 2 26 >= 22 Min. in Shift 3 18 >= 18 Min. in Shift 4 16 >= 16 Min. in Shift 5 22 >= 22 Min. in Shift 6 16 >= 16 Non. Negative X1, X2,X3,X4,X5,X6 >= 0 Objective Function 58

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