STATISTICS

STATISTICS.xlsx

QUESTION 2

OPTION 1:		3-MONTH MA		OPTION 2:		3 WMA		OPTION 3		$ 0.6
JAN	650			JAN	650			JAN	650
FEB	725			FEB	725			FEB	725	$ 650.0
MARCH	850			MARCH	850			MARCH	850	$ 695.0
APRIL	825	846.67		APRIL	825	775.00		APRIL	825	$ 788.0
MAY	865	800.00		MAY	865	816.67		MAY	865	$ 825.2
JUNE	915	855.67		JUNE	915	849.17		JUNE	915	$ 849.1
JULY	900	868.33		JULY	900	883.33		JULY	900	$ 888.6
AUG	930	893.33		AUG	930	899.17		AUG	930	$ 895.5
SEPT	950	915.00		SEPT	950	917.50		SEPT	950	$ 916.2
OCT	899	926.67		OCT	899	935.00		OCT	899	$ 936.5
NOV	935	926.33		NOV	935	921.17		NOV	935	$ 914.0
DEC		928.00		DEC		925.50		DEC		$ 926.6
MAD					MAD				MAD
			differences				differences				differences
MAD	846.67		37.78			775.00	105.28		650.00		194.05
	800.00		84.44			816.67	63.61		695.00		149.05
	855.67		28.78			849.17	31.11		788.00		56.05
	868.33		16.11			883.33	3.06		825.20		18.85
	893.33		8.89			899.17	18.89		849.08		5.03
	915.00		30.56			917.50	37.22		888.63		44.58
	926.67		42.23			935.00	54.72		895.45		51.40
	926.33		41.89			921.17	40.89		916.18		72.13
	928.00		43.56			925.50	45.22		936.47		92.42
TOTAL	7,959.99		334.22		TOTAL	7,922.50	400.00		913.99		69.93
mean	884.44		37.14		MEAN	880.28	44.44		926.60		82.54
									9,284.60		836.04
									844.05		76.00
Based on the analysis the best method to use is exponential smoothing since the errors are averaged and therefore minimzed to the least.

Decision Variables: We need to identify the decision variables so that we can formulate our linnera programming model to solve the problem. This is as below:
Let xi represent the number of the employees in a given shift i.
That is i = 1, 2, 3 ,4, 5
Objective function: (WHAT IS THE MAIN OBJECTIVE OF THE RESTAURANT?)
Minimizing the total number of the employees in the restaurant thus minimize cost. This will be represented by the function below:
Min. Z = x1 + x2 + x3 + x4 + x5
Constrained: This is the contarints within whohc we have to operate. That is the time constraints. They are represented as below:
Constraint time shifts (time constraintes as below)	Equation
Shift 1: 10 AM-1 PM	x1 >= 3
Shift 2: 1:00 PM - 4:00 PM	x2 >= 4
Shift 3: 4:00 PM- 7:00 PM	x5 >= 6
Shift 4: 7:00 PM- 10:00 PM	x6 >= 7
Shift 5: 10:00 PM- 1:00 AM	x7 >= 4
Min. number of employees	x1 + x2 + x3 + x4 + x5 <= 15
Non-negativity Constraint	xi >=0
The optimal solution
Optimal Solution: Time shifts with the optimal employees location so that the number of the employees remains 15 to inimize the total operational costs of the restaurant. (Main objecyive)
Time Shift shifts	Number of Employees allocation
Shift 1: 10 AM-1 PM	3
Shift 2: 1:00 PM - 4:00 PM	4
Shift 3: 4:00 PM- 7:00 PM	0
Shift 4: 7:00 PM- 10:00 PM	4
Shift 5: 10:00 PM- 1:00 AM	4
	total = 15
The Total number of the Employee Required is 15
(Note: Problem have multiple optimal solutions)