Question

EZ-Windows, Inc., manufactures replacement windows for the home remodeling business. In January, the company produced 14,000 windows and ended the month with 9,500 windows in inventory. EZ-Windows' management team would like to develop a production schedule for the next three months. A smooth production schedule is obviously desirable because it maintains the current workforce and provides a similar month-to-month operation. However, given the sales forecasts, the production capacities, and the storage capabilities as shown, the management team does not think a smooth production schedule with the same production quantity each month is possible. February March April Sales forecast 14,000 15,500 20,000 Production capacity 15,000 14,500 19,000 Storage capacity 6,000 6,000 6,000 The company's cost accounting department estimates that increasing production by one window from one month to the next will increase total costs by $1.00 for each unit increase in the production level. In addition, decreasing production by one unit from one month to the next will increase total costs by $0.65 for each unit decrease in the production level. Ignoring production and inventory carrying costs, formulate and solve a linear programming model that will minimize the cost of changing production levels while still satisfying the monthly sales forecasts. If required, round your answers to two decimal places. If an amount is zero, enter "O". Let: F = number of windows manufactured in February M = number of windows manufactured in March A = number of windows manufactured in April Im increase in production level necessary during month m

Answer 1

Answer :

MIN z = I1 + I2 + I3 + 0.65D1 + 0.65D2 + 0.65D3
subject to
F - s1 = 4500
s1 + M - s2 = 15500
s2 + A - s3 = 20000
F - I1 + D1 = 14000
M - F - I2 + D2 = 0
A - M - I3 + D3  = 0
F <= 15000
M <= 14500
A <= 19000
s1 <= 6000
s2 <= 6000
s3 <= 6000
and I1,I2,I3,D1,D2,D3,F,M,A,s1,s2,s3 >= 0

Here only the solution is provided as the whole process is too long and exceeds the word limit

Iteration-9		Cj	1	1	1	0.65	0.65	0.65	0	0	0	0	0	0	0	0	0	0	0	0
B	CB	XB	I1	I2	I3	D1	D2	D3	F	M	A	s1	s2	s3	S1	S2	S3	S4	S5	S6	MinRatio
F	0	10500	0	0	0	0	0	0	1	0	0	0	0	0	0	0	0	1	0	0
s1	0	6000	0	0	0	0	0	0	0	0	0	1	0	0	0	0	0	1	0	0
I3	1	500	0	0	1	0	0	-1	0	0	0	0	0	-1	0	-2	0	-1	0	0
D1	0.65	3500	-1	0	0	1	0	0	0	0	0	0	0	0	0	0	0	-1	0	0
M	0	14500	0	0	0	0	0	0	0	1	0	0	0	0	0	1	0	0	0	0
A	0	15000	0	0	0	0	0	0	0	0	1	0	0	-1	0	-1	0	-1	0	0
S1	0	4500	0	0	0	0	0	0	0	0	0	0	0	0	1	0	0	-1	0	0
I2	1	4000	0	1	0	0	-1	0	0	0	0	0	0	0	0	1	0	-1	0	0
S3	0	4000	0	0	0	0	0	0	0	0	0	0	0	1	0	1	1	1	0	0
s2	0	5000	0	0	0	0	0	0	0	0	0	0	1	0	0	1	0	1	0	0
S5	0	1000	0	0	0	0	0	0	0	0	0	0	0	0	0	-1	0	-1	1	0
S6	0	6000	0	0	0	0	0	0	0	0	0	0	0	1	0	0	0	0	0	1
z=6775		Zj	-0.65	1	1	0.65	-1	-1	0	0	0	0	0	-1	0	-1	0	-2.65	0	0
		Zj-Cj	-1.65	0	0	0	-1.65	-1.65	0	0	0	0	0	-1	0	-1	0	-2.65	0	0

Since all Zj-Cj≤0

Hence, optimal solution is arrived with value of variables as :
I1=0,I2=4000,I3=500,D1=3500,D2=0,D3=0,F=10500,M=14500,A=15000,s1=6000,s2=5000,s3=0

Min z=6775

so cost = $6775

	February	March	April
production level	10500	14500	15000
increase in production	0	4000	500
decrease in production	3500	0	0
ending inventory	6000	5000	0

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